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IL Ranker Series Chemistry for JEE Grade 12 Module 2
ISBN 978-81-983474-9-7 [SECOND
EDITION]
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Preface
The Joint Entrance Examination (JEE) is a critical threshold that students aspiring to study at the prestigious IITs need to cross. Cracking this examination requires rigorous preparation, and good study materials are as indispensable in this journey as hard work and determination.
The quality of study materials can be determined by how well they cater to the needs of the aspirants and how comprehensive they are in their scope and reach. IL Ranker Series for JEE was first published in 2024 and was quickly picked up by JEE aspirants. We actively reached out to the users and sought critical feedback. User feedback started pouring in and the relevant and valuable ones were incorporated into the books. Hence, the revised edition of IL Ranker Series was shaped and honed by the expert suggestions and recommendations of practising teachers and subject matter experts.
The second edition of Ranker includes substantial changes and improvements. New questions have been added where required, and all the JEE Main questions have been categorised into Levels 1, 2, and 3. While Level 1 questions are more aligned to NCERT, Levels 2 and 3 include more challenging, multi-concept questions of the same rigour as in JEE main exam. Relevant topic-level changes have also been introduced to make the content clearer and more accessible. In mathematics, a new chapter has been added to help students get a comprehensive understanding of the concepts.
The second edition of IL Ranker Series stands out because it is built to propel the JEE aspirants towards achieving their dream of passing the joint entrance examination. A variety of features in the books and an array of practice questions cater to their learning needs and prepare them thoroughly for the test.
The comprehensive coverage of the NCERT syllabus provides a strong foundation and helps in holistic preparation for JEE. The series includes diverse question types for JEE Main and JEE Advanced exams. Expert tips, theory-based questions, and a mix of difficulty levels aim to improve students’ conceptual understanding, cross-topic synthesis, and retention of key concepts.
The second edition of IL Ranker Series is more than just a set of books. It is a commitment to helping students learn, grow, and succeed. We have designed it to ignite your passion for engineering and to foster a lifelong love of learning in you. With every page you turn, you will move one step closer to materialising your dream of cracking JEE with a good rank.
Key Features of the Book
Chapter Outline
1.1 Types of Solutions
1.2 Methods of Concentration
1.3 Solubility
This outlines topics or learning outcomes students can gain from studying the chapter. It sets a framework for study and a roadmap for learning.
Specific problems are presented along with their solutions, explaining the application of principles covered in the textbook. Solved Examples
1. What is the molality of a solution of H2SO4 having 9.8% by mass of the acid?
Sol. 9.8% by mass of H2SO4 contains 9.8 g of H2SO4 per 100 g of solution.
Therefore, if mass of solution = 100 g, mass of solute, H2SO4 = 9.8 g,
Try yourself:
1. In a solution of H 2 SO 4 and water, mole fraction of H2SO4 is 0.9. How many grams of H2SO4 is present per 100 g of the solution?
Ans: 98
Try Yourself enables the student to practice the concept learned immediately.
This comprehensive set of questions enables students to assess their learning. It helps them to identify areas for improvement and consolidate their mastery of the topic through active recall and practical application.
CHAPTER REVIEW
Types of Solutions
■ A solution is a homogeneous mixture of two or more non–reacting components. Formation of solution is a physical process.
TEST YOURSELF
1. The mole fraction of a solvent in aqueous solution of a solute is 0.6. The molality of the aqueous solution is (1) 83.25 (2) 13.88 (3) 37 (4) 73
It offers a concise overview of the chapter’s key points, acting as a quick revision tool before tests.
Organised as per the topics covered in the chapter and divided into three levels, this series of questions enables rigorous practice and application of learning.
These questions deepen the understanding of concepts and strengthen the interpretation of theoretical learning. These complex questions combining fun and critical thinking are aimed at fostering higher-order thinking skills and encouraging analytical reasoning.
Exercises
JEE MAIN LEVEL
LEVEL 1, 2, and 3
Single Option Correct MCQs
Numerical Value Questions
THEORY-BASED QUESTIONS
Single Option Correct MCQs
Statement Type Questions
Assertion and Reason Questions
JEE ADVANCED LEVEL
BRAIN TEASERS
FLASHBACK
CHAPTER TEST
This comprehensive test is modelled after the JEE examination format to evaluate students’ proficiency across all topics covered, replicating the structure and rigour of the JEE examination. By taking this chapter test, students undergo a final evaluation, identifying their strengths and areas of improvement.
Level 1 questions test the fundamentals and help fortify the basics of concepts. Level 2 questions are higher in complexity and require deeper understanding of concepts. Level 3 questions perk up the rigour further with more complex and multi-concept questions.
This section contains special question types that focus on in-depth knowledge of concepts, analytical reasoning, and problem-solving skills needed to succeed in JEE Advanced.
Handpicked previous JEE questions familiarise students with the various question types, styles, and recent trends in JEE examinations, enhancing students’ overall preparedness for JEE.
THE p -BLOCK ELEMENTS CHAPTER 4
Chapter Outline
4.1 Group 15 Elements
4.2 Group 16 Elements
4.3 Group 17 Elements
4.4 Group 18 Elements
Nitrogen (N), phosphorus (P), arsenic (As), antimony (Sb), and bismuth (Bi) are the five elements of group VA. They are located in the p-block of the long form of the periodic table. These elements are assigned group 15 by the I.U.P.A.C. These are called elements of nitrogen family or sometimes called elements of nitrogen-phosphorus family. These elements are named pnictogens. Their position in the long form of the periodic table is indicated in Table. 4.1.
4.1 GROUP 15 ELEMENTS
Each of the group VA elements has five electrons in the valence shell. Two electrons are filled in the s–sub shell and the remaining three in the p–sub shell. The general electronic
Table 4.1 Position of group VA elements
configuration of these elements is ns 2 np 3 and is given in the box method in Fig. 4.1 . Configurations are summarised in Table 4.2.
Electronic configuration in the ground state
Fig. 4.1 Configuration of group VA electrons
Table 4.2 Atomic numbers and electronic configuration of group VA elements
In accordance with the Hund’s rule of maximum spin multiplicity, the three electrons are distributed equally in the p-sub shell. Thus, these elements have half-filled orbitals.
CHAPTER 4: The p-BLOCK ELEMENTS 2
Due to the orbital symmetry, elements with exactly half filled orbitals are stable. Hence, the group VA elements are not so reactive and fairly stable.
Only two electrons are present in the penultimate shell of nitrogen, eight electrons in phosphorus and eighteen each in arsenic, antimony, and bismuth. This shows why nitrogen differs in some of its properties from phosphorus and these two elements differ in their properties from the remaining elements of the group.
4.1.1 Occurrence
Nitrogen occurs even in free state, in air as dinitrogen, to the extent of 78% by volume and 75% by weight. In combined state, it exists as nitrate. Nitrogen is tasteless and nitrogen atom has two stable isolopes N 14 and N15.
Chile salt petre: NaNO3
Indian salt petre: KNO3
Norwegian salt petre: Ca(NO3)2
Most abundant element of group VA in the earth’s crust is phosphorus. Its relative abundance in the earth’s crust is 11.2 ppm. It is the eleventh most abundant element. The important source of phosphorus is phosphorite rocks which is mainly calcium phosphate. The same chemical form is also present in bones and teeth. Phosphoproteins are present in egg and milk.
Phosphorite: Ca 3(PO4)2
Apatite family: () 942 6 CaPO.CaX
(X = F, Cl, OH)
Fluorapatite: 3Ca3(PO4)2.CaF2
Chlorapatite: 3Ca3(PO4)2.CaCl2
Hydroxyapatite: 3Ca3 (PO4)2.Ca(OH)2
Arsenic, antimony, and bismuth are available mainly as sulphide minerals.
4.1.2 Physical Properties
The similarities and gradation in the general physical properties of the group VA elements are summarised in Table 4.3
Gas at room temperature, @ White phosphorous, ** Grey a -form at 38 atm, @@ Sublimes
Table 4.3 Gradation in the physical constants of group VA elements
Atomic radius: From nitrogen to bismuth, the atomic number and the atomic weight gradually increase. The differentiating electron enters in the next higher energy shell, suggesting an increase in the atomic radius. However, the increase in the radius is less predominant beyond phosphorus. This is due to the shielding effect by inner d-orbitals.
Physical state: Nitrogen is a gas at room temperature. Other elements of the group VA are solids. The melting and boiling points gradually increase from nitrogen to antimony. The smaller values of bismuth are attributed to a different crystal structure.
Metallic nature: The values of Pauling’s electronegativity gradually decreases from nitrogen to bismuth. The metallic nature increases gradually with an increase in the atomic size. Nitrogen and phosphorus are nonmetals. Arsenic and antimony are metalloids. A metalloid is the element which exhibits both metallic and non-metallic properties. Bismuth is clearly a metal of the group VA.
Ionisation potential: The ionisation potential values decrease gradually on descending the group as the atomic number increases. These values are relatively high compared to the corresponding elements of adjacent groups.
High ionisation value is attributed to the stable half-filled electronic configuration in the p sub-shells of these elements. The difference in the ionisation potential values after phosphorus is less predominant, which is attributed to shielding effect by inner d-orbitals.
Atomicity: Nitrogen is a diatomic molecule with a triple bond between two bonded atoms. The triple bond contains one sigma and two pibonds, with a bond energy of 946 kJ mol–1. The bond is very strong, difficult to dissociate and is hence responsible for the chemical inertness of nitrogen.
Atomicity of phosphorus, arsenic, and antimony is 4 each. They form tetra-atomic
molecules, with four atoms occupying each corner of a regular tetrahedron, as shown in Fig.4.2
In the tetrahedral structure of phosphorus molecule, each phosphorus atom forms three covalent bonds with a P–P bond order equal to one and a total of six sigma bonds per molecule. The P–P bond length is 221 pm.
Catenation: The first three elements of the group VA exhibit catenation, but this property is much less than that of carbon of IVA group and sulphur of VIA group. Catenation ability is relatively more for phosphorous than for nitrogen.
Three nitrogen atoms are present in azide (N 3 –) and four nitrogen atoms in tetrazine (H2N–N=N–NH2). Compounds having a chain of eight nitrogen atoms have been reported in ditetrazines. Such tendency, however, decreases down to phosphorus and antimony. This decrease is attributed to decrease in M–M bond energies.
Allotropy: The existence of an element in two or more forms of same physical state is called allotropy. Except nitrogen, all the elements show allotropy. Nitrogen gas has no allotropes (solid nitrogen has allotropes).
P4
Fig. 4.2 Structure of phosphorus molecules
Phosphorus has three important allotropes, white (or yellow), red and black. Other more doubtful allotropes like violet phosphorus and scarlet phosphorus are also reported.
White phosphorus is a colourless waxy translucent solid. It is poisonous and glows in dark due to chemiluminescence. It is insoluble in water but soluble in non-polar organic solvents such as carbon disulphide. It undergoes disproportionation on boiling with aqueous caustic soda, liberating phosphine.
White phosphorus is less stable, because of angular strain in its molecule at 60°. It catches fire readily in air to give dense white fumes of phosphorus pentoxide. For this reason of high reactivity in solid phase, it is stored under water.
P 4 + 5O2→ P4O10
It dissolved in boiling NaOH solution in an inert atmosphere giving PH 3
P 4 + 3NaOH + 3H2O → PH3 + 3NaH2PO2
Red phosphorus is obtained by heating white phosphorus to 573 K in inert atmosphere for about 3 days. Red phosphorus has iron grey lustre. It is non-poisonous and odourless. It is insoluble in water as well as in organic solvents. Red phosphorus is less reactive and does not glow in dark. Lower reactivity of red phosphorus is due to its polymeric chain structure.
Black phosphorus has two forms a - black and b -black. a -Black phosphorus is formed when red phosphorus is heated in a sealed tube at 803 K for several days. It is stable, not oxidised in air, but can be sublimed in air. It has opaque monoclinic or rhombohedral structure. b -Black phosphorus is prepared under high pressure, by heating white phosphorus at 473 K. It is also stable in air and does not burn upto 673 K.
Bond ing and valency: Each element of the group VA has five valence electrons and requires three more for attaining stable octet configuration. Thus, the usual valency of these elements is 3. The maximum valency, however, is the group number 5.
Nitrogen cannot be pentavalent in its compounds due to lack of valence d-orbitals. The excitation of electron from 2s to 3s orbital is difficult in nitrogen, as the energy gap is high between L and M shells. In ammonium salts, the maximum covalency of nitrogen is four with three covalent bonds and one dative bond.
Due to its small size and high electronegativity, nitrogen can form ionic compounds with metals. By gaining 3 electrons, it forms metal nitrides, like Li3N and AlN. Phosphorus also has the tendency of forming metal phosphides.
The heavier elements of group VA, have vacant valence d-orbitals. These orbitals can be used for covalency and expand their valency up to even + 6, as in 6PF .
Oxidation states: The general oxidation numbers of the group VA elements are–3, +3 and +5. The negative oxidation state (– 3), is more predominant for nitrogen, due to its higher electronegativity and non-metallic nature.
Nitrogen exhibits maximum number of types of oxidation states in its compounds. These oxidation states are listed in Table 4.4. All oxidation states of nitrogen from + 1 to + 4 tend to disproportionate in acid solution.
3HNO2 → HNO3 + 2NO + 3H2O
In case of phosphorus, all intermediate oxidation states disproportionate into – 3 and + 5, both in acid as well as in alkali. Compounds of As, Sb, and Bi do not show disproportionation.
Table 4.4
Oxidation states of nitrogen
S.No. Name of
1. Ammonia NH3 – 3
2. Hydrazine N2H4 – 2
3. Hydroxylamine NH2OH – 1
4. Hydrazoic acid N3H – 1/3
5. Ammonium nitrite NH4NO2 0
6. Nitrous oxide N2O + 1
7. Nitric oxide NO + 2
8. Nitrous acid HNO2 + 3
9. Nitrogen dioxide NO2 + 4
10. Nitric acid HNO3 + 5
The importance of –3 oxidation state decreases from nitrogen to phosphorus and then almost vanishes for bismuth. The +3 state becomes increasingly stable for arsenic, antimony and bismuth. The tendency of exhibiting +5 oxidation state by the group VA elements decreases from nitrogen to bismuth. This is because of the non-involvement of s–electron pair in bonding, called inert pair effect. Due to the inert pair effect, +3 oxidation state is more stable for bismuth than +5 state, but bismuth can exhibit +5 oxidation state in BiF 5
All the elements of group VA with metals to form their binary compounds containing –3 oxidation state such as Ca 3N2; Ca3P2; Na3As2; Zn3Sb2 and Mg3Bi2.
4.1.3 Chemical Properties
Hydrides: The binary compounds formed by an element with hydrogen are called hydrides. All the elements of nitrogen family form the hydrides of the type MH 3 , where M is the group VA element. The hydrides along with their characteristics are listed in Table 4.5.
The M–H bond length increases and bond energy decreases from NH3 to BiH3. Ammonia is the most stable. The stability gradually decreases from NH3 to BiH3. In the same order of reactivity, the reduction ability increases down the group.
The central atom of these hydrides has a lone pair of electrons. According to Lewis theory of acids and bases, these hydrides act as bases. Ammonium salts are known, but phosphonium salts are less common. Ammonia is a weak base. Basic nature of these hydrides decreases from NH 3 to SbH3. BiH3
These hydrides form dative bonds, as they act as electron pair donors. The ability to donate electron pairs and form complex compounds decreases from NH 3 to BiH3
H3P : + AlCl3 → [H3PAlCl3]
CoCl3 + 6NH3 →[Co(NH3)6]Cl3
Table 4.5 Hydrides of the group VA and their bond characteristics
4: The p-BLOCK ELEMENTS 6
These hydrides are all good reducing agents, except ammonia. The reduction ability increases with increase in the instability of the hydride.
Ammonia is the most soluble gas in water, as ammonia forms intermolecular hydrogen bonds with water. Moreover, both ammonia and water are polar molecules. With a decrease in the electronegativity of the central atom in MH3, the polarity of M-H bond decreases and solubility in water also decreases.
Ammonia is prepared in the laboratory by the action of caustic soda on an ammonium salt. Metal phosphides, upon hydrolysis, give phosphine. Similarly, metal arsenide or antimonide upon hydrolysis gives arsine or stibine.
Mg3As2 + 6 HCl → 2AsH3 + 3MgCl2
All these hydrides are colourless, highly volatile, foul smelling and poisonous gases. Their ease of formation and the ease of substitution with groups like –Cl, –CH3, etc., decreases from ammonia to bismuthine.
The bond angle in NH 3 is nearer to tetrahedral angle and in the other hydrides of the group VA it is nearer to 90°. This is due to the use of sp 3 hybridised orbitals by the central atom in NH 3 and unhybridised p–orbitals by the central atoms of the other hydrides. The decrease in the bond angle from ammonia to phosphine is due to an increase in the size of the central atom and a decrease in the electronegativity of the central atom.
Ammonia can be liquefied easily. Its volatility is less. Melting and boiling points of ammonia are higher than the expected values. This is due to intermolecular hydrogen bonding in ammonia.
The order of volatility is:
PH 3 > AsH3 > NH3 > SbH3
The order of basic nature is:
NH3 > PH 3 > AsH3 > SbH3
The order of stability is :
NH3 > PH 3 > AsH3 > SbH3
Besides these MH3 type hydrides, nitrogen forms other hydrides like hydrazine (N 2H4) and hydrazoic acid (N 3 H). Phosphorus is also known to form diphosphine (P 2H4) and tetraphosphine (P4H6). Diphosphine is neutral and has no action on litmus.
4.1.4 Dinitrogen Preparation
Dinitrogen (N2) is commercially prepared by the fractional distillation of liquid air. Liquid nitrogen distils at 77 K. In the laboratory, nitrogen is obtained by treating aqueous ammonium chloride solution with potassium or sodium nitrite.
NH4Cl + KNO2 → N2 + 2H2O + KCl
The possible impurities, NO and HNO 3 are removed by passing the gas through aqueous sulphuric acid containing potassium dichromate.
Nitrogen can be obtained by heating ammonium dichromate. Pure nitrogen can be obtained by the thermal decomposition of azide.
()4272232 2 NHCrONCrOHO ∆ →++
32 2NaN2Na3N; ∆ →+ ()32 2 BaNBa3N ∆ →+
Dinitrogen is a colourless non-toxic gas. It is very less soluble in water (23 cc dm –3 at 273 K and 1 bar).
Dinitrogen is chemically inert at room temperature because of high bond enthalpy. At high temperature, nitrogen directly combines to give ionic nitrides with some metals and covalent nitrides with non-metals.
3MgNMgN +→
2 2BN2BN +→
At about 2000°C (also during lightnings), nitrogen combines with oxygen to form nitric oxide.
N2 + O2
2NO
Phosphorus is prepared by the reduction of phosphate with coke in the presence of silica.
2Ca3(PO4)2+ 10C + 6SiO2 → P 4 + 6CaSiO3 + 10CO
Uses: Dinitrogen is used in the manufacture of ammonia, calcium, cyanamide, etc. It is used to create inert atmosphere in iron and steel industry. It is used as inert diluent for reactive chemicals. Liquid nitrogen is used as a refrigerant, to preserve biological materials, food items and in cryosurgery.
4.1.5 Oxides of Nitrogen
The binary compounds formed by an element with oxygen are called oxides. Nitrogen forms many oxides, because p p –p p bonds between nitrogen and oxygen are relatively stronger. These oxides, their common preparation methods and their nature are listed in Table 4.6
Colourless
monoxide
(I) oxide]
Nitrogen monoxide [Nitrogen (II) oxide] NO +2 2NaNO2 + 2FeSO4 + 3H2SO4 → Fe2(SO4)3 + 2NaHSO4 + 2H2O + 2NO
Dinitrogen trioxide Nitrogen (III) oxide
Colourless gas, neutral
Blue acidic Nitrogen dioxide
Dinitrogen tetroxide [Nitrogen (IV) oxide]
Dinitrogen pentoxide [Nitrogen (V) oxide]
Brown gas, acidic
Colourless
Colourless Solid acidic
Table 4.6 Oxides of nitrogen
CHAPTER 4: The p-BLOCK ELEMENTS 8
Non-metallic oxides are usually acidic. However, nitrous oxide and nitric oxide are neutral oxides. With an increase in the oxidation number of nitrogen, the acidic nature of the oxide increases. Nitrogen pentoxide is most acidic among oxides of nitrogen.
Nitrous oxide is called laughing gas. It is a linear molecule. NO is colourless and is oxidised in the atmosphere to NO 2 . Nitric oxide is absorbed in freshly prepared ferrous sulphate to form a brown coloured complex, [Fe(H2O)5NO]SO4, called brown ring.
The familiar brown ring test for nitrates depends on the ability of Fe2+ to reduce nitrates to nitric oxide and nitric oxide react with ferrous to form brown complex.
N 2 O 4 is a dimer of NO 2 . NO 2 has one unpaired electron. It is brownish yellow and paramagnetic. But due to absence of unpaired electrons, N2O4 is diamagnetic and colourless.
The group VA elements form two important types of oxides. All these trioxides and pentaoxides are generally covalent. Trioxides have the general formula M2O3 and pentoxides with the general formula M2O5. The oxides of phosphorus, arsenic and antimony exist as dimers Table 4.7 and acidic. Sb 4O 6 is amphoteric and Bi2O3 is basic.
Acidic nature decreases from the oxide of nitrogen to that of bismuth. Pentoxides are more acidic than the corresponding trioxides.
Table 4.7 Oxides of the group VA elements
Nitrogen N2O3 N2O5
Phosphorus P4O6 P4O10
Arsenic As4O6 As4O10
Antimony Sb4O6 Sb4O10
Bismuth Bi2O3
Tri and pentoxides of nitrogen and phosphorus are important. Phosphorus
pentoxide is a powerful dehydrating agent. The structures of the oxides of phosphorus are given in Fig.4.3. and those of nitrogen in Fig.4.4.
Fig. 4.3 Structures of oxides of phosphorus
While N 2 O 3 and P 4 O 6 have only covalent bonds, N 2 O 5 and P 4 O 10 have covalent and co-ordinate covalent bonds. The number of oxygen atoms present around N atom in N2O3 is 2 and in N 2O 5 is 3 and around P atom in P4O6 is 3 and in P4O10 is 4.
Trioxides on reaction with water give the corresponding –ous acids. Similarly pentoxides on reaction with water give –ic acids.
N2O3 + H2O → HNO2 Nitrous acid
N2O5 + H2O → HNO3 Nitric acid
4.1.6 Halides of Nitrogen
The binary compounds formed by an element with halogen are called halides. The elements of nitrogen family form two types of halides. Trihalides with the formula MX 3 have the oxidation number of the group VA element (M)+3. Pentahalides with the formula MX5 have the oxidation number of the group VA element +5.
Nitrogen does not form pentahalides, due to the lack of valence d–orbitals. The important and most stable halide of nitrogen is nitrogen trifluoride, NF3. Nitrogen trichloride is yellow liquid and explosive.
Nitrogen halides are normally obtained by the action of excess of halogen on ammonia.
NH3 + 3Cl2 → NCl3 + 3HCl
NH3 + 3F2 → NF3 + 3HF
4.1.7
Halides of Phosphorus
Phosphorus trichloride is obtained by passing dry chlorine over heated white phosphorus. Phosphorus pentachloride is obtained with excess chlorine. Phosphorus trifluoride is formed when fluorine reacts with phosphorus trichloride.
P 4 + 6Cl2 → 4 PCl3
P 4 + 10Cl2 → 4 PCl5
2PCl3 + 3F2 → 2PF 3 + 3Cl2
Chlorides of phosphorus are also obtained from phosphorus by the action of thionyl
chloride or sulphurylchloride
P 4 + 8SOCl2 → 4PCl3 + 4SO2 + 2S2Cl2
PCl 3 is colourless oily liquid and PCl5 is yellow powder. PCl3 is more stable than PCl5. On warming PCl5 sublimes, but decomposes on strong heating and liberates chlorine.
PCl5 exists as ionic solid, [PCl4]+ [PCl6]–. In liquid or vapour states, PCl 5 has trigonal bipyramidal structure. [PCl4]+ is tetrahedral and [PCl6]– is octahedral.
Both PCl3 and PCl5 are used in the synthesis of ethyl chloride and acetyl chloride.
All the halides of the group VA are covalent, except BiF3. These halides undergo hydrolysis, except NF3. Hydrolysis of NCl3 gives ammonia and hypochlorous acid. Aqueous NCl 3 is a bleaching agent.
Fig.4.4 Structures of oxides of nitrogen
NCl3 + 3H2O →NH3 + 3HOCl
Trihalides of phosphorus and arsenic undergo hydrolysis to give the corresponding –ous acids. Pentahalides of phosphorus undergo hydrolysis to give corresponding –ic acids.
P 4 + 10SO2Cl2 → 4PCl5 + 10SO2
PCl5 → PCl3 + Cl2
PCl5 + C2H5OH → C2H5Cl + POCl3 + HCl
PCl5+CH3COOH → CH3COCl+POCl3+HCl
PCl3 + 3C2H5OH → 3C2H5Cl + H3PO3
PCl3 + 3CH3COOH → 3CH3COCl + H3PO3
Finely divided metals on heating with phosphorus pentachloride give corresponding chlorides.
Sn + 2PCl5 → SnCl4 + 2PCl3
PCl3 + 3H2O →3HCl + H3PO3
Phosphorus acid
PCl5 + H2O → POCl3 + 2HCl
POCl3 + 3H2O → 3HCl + H3PO4
Phosphoric acid
2Ag + PCl5 → AgCl + PCl3
Trihalides use the sp3 hybridised orbitals of the central atom. They have ammonia like trigonal monopyramidal structure. Pentahalides use the sp3d hybridised orbitals of the central atom. They have trigonal bipyramidal structure as shown in Fig.4.5
structure with nitrogen atom at the apex, as shown in Fig.4.6.
Structure of PCl3 and PCl5
4.1.8 Ammonia
Ammonia Structure :The central atom in ammonia undergoes sp 3 hybridisation. Ammonia has 3 bond pairs and one lone pair of electrons. It has trigonal pyramidal
Fig.4.6. Structure of ammonia
Due to the tetrahedral hybridisation, the bond angle in NH3 is expected to be 109°28’. The observed bond angle in NH 3 is 107.8°. The decrease in the bond angle is attributed to the presence of a lone-pair on nitrogen atom and greater repulsions between lone-pair and bond pairs.
Preparation: Ammonia can be prepared in the laboratory by the action of dry slaked lime or caustic soda on an ammonium salt.
2NH4Cl + Ca(OH)2→CaCl2 + 2H2O + 2NH3
NH4Cl + NaOH→NaCl + H2O + NH3
Ammonia is present in small quantities in soil and in air, where it is formed by the decay of nitrogenous organic matter
H2NCONH2 + 2H2O →(NH4)2CO3 2NH3 + H2O + CO2
In the Haber’s synthesis, ammonia is prepared on a large scale from its elements hydrogen and nitrogen in the volume ratio, 3 : 1. The flow chart block diagram of Haber’s method is shown in Fig.4.7.
The reaction between nitrogen and hydrogen to give ammonia is reversible.
N2 + 3H2 2NH3 + 22 kcal.
The favourable conditions for the Haber’s synthesis are:
1. High pressure, 200 – 400 atm.
2. Moderate (optimum working) temperature of 450 – 500°C.
H2NCONH2 + 2H2O→(NH4)2CO3
2NH3 + H2O + CO2
Fig.4.5.
Recirculating pump Reaction bed with Fe + Mo
Fig. 4.7 Flow chart block diagram of ammonia synthesis
The raw materials, nitrogen and hydrogen, are mixed in 1 : 3 volume ratio. The mixture is compressed to over 200 atmospheres pressure and then introduced into the reaction bed containing the catalyst. The mixture is allowed to react at 450 – 500°C. The reaction mixture, with about 10% ammonia, is cooled to get the liquid ammonia. The liquid is tapped out. The unreacted gases are recirculated to the reaction bed.
Properties: Ammonia is a colourless gas with pungent characteristic smell. It freezes at 198 K and boils at 240 K. It is highly soluble in water
NH3 + H2O → NH4OH
Ammonia is a Lewis base. Due to its basic nature, it cannot be dried over sulphuric acid, phosphorus pentoxide or anhydrous calcium chloride. Quick lime is used to dry ammonia. Ammonia forms salts with acids.
NH3 + HCl → NH4Cl
Ammonia precipitates metal hydroxides from their salt solutions
ZnSO4+2NH4OH → Zn(OH)2+(NH4)2SO4
2FeCl3+3NH4OH → Fe2O3. xH2O + 3NH4Cl
Ammonia acts as a reductant, when it reacts with cupric oxide and as acid when heated with an active metal like sodium.
Ammonia has a lone pair of electrons on the central atom and acts as Lewis base. It is a ligand and forms complex compounds with metal ions, a blue complex with Cu2+ and colourless complex with Ag +
1. Liquid ammonia is a good solvent for ionic substances as well as for covalent substances.
2. Ammonia is used in the manufacture of fertilisers like urea, ammonium sulphate, calcium ammonium nitrate, etc.
3. Liquid ammonia is used as a refrigerant.
4. Ammonia is used in the manufacture of nitric acid by Ostwald’s method.
5. Ammonia is used in the qualitative analysis, as a reagent for the detection of ions like Cu2+ cation, Ag+ cation, etc.
4.1.9 Nitric Acid
Nitrogen forms oxyacids like hyponitrous acid (H 2N 2O 2), nitrous acid (HNO 2), and nitric acid (HNO3).
Nitric acid is the most important oxyacid of nitrogen. It is prepared in the laboratory by heating potassium or sodium nitrate with concentrated sulphuric acid.
NaNO3 + H2SO4→NaHSO4 + HNO3
Nitric acid is manufactured by Ostwald’s process.
In Ostwald’s process ammonia is mixed with air in 1 : 7 ratio, when passed over hot platinum gauge catalyst, ammonia is oxidised to nitric oxide.
4NH3+ 5O2 → 4NO + 6H2O.
Nitric oxide thus formed combines with oxygen to give nitrogen dioxide, which is dissolved in water under pressure to give nitric acid.
2NO + O2→2NO2;
3NO2 + H2O→2HNO3 + NO
Nitric oxide thus formed is recycled and concentrated by distillation. The concentration is improved to 98% by dehydration with concentrated sulphuric acid. Pure nitric acid crystals are obtained by cooling in a freezing mixture.
Laboratory grade nitric acid is 68% by mass with specific gravity 1.5. Its melting point is 231 K and boiling point is 355 K.
Nitric acid is a strong acid and acts as a strong oxidant. Iodine is oxidised to iodic acid, sulphur to sulphuric acid, phosphorus to phosphoric acid and carbon to carbon dioxide.
I2 + 10 HNO3→2HIO3 + 10NO2 + 4H2O
S8 + 48HNO3 → 8H2SO4 + 48NO2 + 16H2O
P 4 + 20HNO3 →4H3PO4 + 20NO2 + 4H2O
C + 4HNO3→CO2 + 4 NO2 + 2H2O
It attacks metals, except noble metals such as Au and Pt. The products of the oxidation
reaction depend upon concentration of acid, temperature and electro-positivity of metal. Be, Al, Cr, Fe and Ni are rendered passive with conc. nitric acid.
3Cu+8HNO3(dil)→3Cu(NO3)2+2NO+4H2O
Cu+4HNO3(conc)→Cu(NO3)2+2NO2+2H2O
4Zn+10HNO3(dil)→4Zn(NO3)2+N2O+5H2O
Zn+4 HNO3(conc)→ Zn(NO3)2+2NO2+2H2O
The familiar brown ring test for nitrates depends on the ability of ferrous iron to reduce nitrates to nitric oxide. This brown ring test is usually carried out by adding dilute solution of ferrous sulphate to an aqueous solution containing nitrate ion. A brown ring is formed at the interface when concentrated sulphuric acid is carefully added along the sides of the test tube.
3Fe2+ + NO3– + 4H+ → NO + 2H2O + 3Fe3+ NO + [Fe(H2O)6]2+→[Fe(H2O)5NO]2++ H2O brown ring
Some metals like Al, Cr, etc., are passive to concentrated nitric acid due to the formation of protective oxide film.
The main use of nitric acid is to manufacture ammonium nitrate fertilisers. Other nitrates are used for pyrotechnics and explosives like ammonal, nitroglycerin, and trinitrotoluene.
In laboratory, a 1 : 1 mixture of concentrated nitric acid and concentrated sulphuric acid is used as nitration mixture. It is used in the preparation of artificial silk (cellulose nitrate), perfumes, dyes and medicines.
Other major uses of nitric acid are in the pickling of metal surfaces, etching of metals and an oxidiser in rocket fuel.
The structure of nitric acid is given in Fig.4.8. N – O bond length 1.22 A0 N N 1300 104.6 pm ONO bond angle 1300 NOH bond angle 1020 1160 127pm H O O O O O H 102° O O – H bond length 0.96 A0
Fig.4.8. Structures of nitric acid
4.1.10
Preparation of Phosphine
P hosphine is prepared by the reaction of calcium phosphide with water or dilute hydrochloric acid. In laboratory, phosphine is obtained by the action of concentrated caustic soda on hot white phosphorus, in carbon dioxide atmosphere.
Ca3P2 + 6H2O →2PH3 + 3Ca(OH)2
Ca3P2 + 6HCl →2PH3 + 3CaCl2
P 4 + 3NaOH + 3H2O →PH3 + 3NaH2PO2
The impurities present in phosphine are generally P2H4 and vapours of P4. Phosphine is absorbed in HI to form phosphonium iodide, which on treating with KOH gives off pure phosphine.
HI + PH 3 →PH 4 I
PH4I + KOH →KI + H2O + PH3
Phosphine is colourless and has the rotten fish odour. It is slightly soluble in water and aqueous solution decomposes in light giving red phosphorus.
Phosphine is a weak base. It forms phosphonium ions in acids, but it has no action on red litmus. Phosphine explodes when
treated with oxidants like chlorine, bromine and conc. nitric acid.
PH 3 + HBr →PH 4 Br
When phosphine is absorbed in solutions of metal salts, the corresponding phosphides are formed.
2PH 3 + 3CuSO4 → Cu3P2 + 3H2SO4
2PH 3 + HgCl2 Hg3P2 + 6HCl
A mixture of CaC 2 and Ca 3 P 2 is used in Holme’s signals and also in smoke screens.
4.1.11 Oxyacids
Oxyacid is a ternary compound of hydrogen, oxygen and non-metalic element. Elements of group VA form two important types of oxyacids. The oxyacid with lower oxidation state is usually called –ous acid and the one with higher oxidation state is called –ic acid.
Oxyacids of Phosphorus
Phosphorus forms a number of oxyacids. Four -ous acids and five –ic acids are known. Some common oxyacids of phosphorus are listed in Table 4.8.
Hypophosphorus acid H3PO2 +1
Orthophosphorus acid H3PO3 +3
Pyrophosphorus acid H 4P2O5 +3
One P–OH
Two P–H
One P = O with P4 + alkali + acid
Two P–OH
One P–H
One P = O P2O3 + H2O
Two P–OH
Two P–H
Two P = O PCl3 + H3PO3
Table 4.8 Oxyacids of phosphorus (A: – ous acids)
Table 4.9 Oxyacids of phosphorus (B: – ic acids)
Hypophosphoric acid H 4P2O6 + 4
Orthophosphoric acid H3PO4 + 5
Pyrophosphoric acid H 4P2O7 + 5
Metaphosphoric acid (HPO3)3 Trimer + 5
Peroxyphosphoric acid H3PO5 + 5
Cyclotrimetaphosphoric acid, (HPO3)3
Fig.4.9 Oxyacids of phosphorus
H ypophosphorus acid and meta–phosphoric acids are monobasic, while hypophosphoric acid and pyrophosphoric acid are tetrabasic acids. Since hypophosphorus acid is monobasic, it cannot form acidic salt. Phosphorus acid is prepared by dissolving phosphorus trioxide or phosphorus trichloride in water. The structure of phosphorus acid is shown in Fig.4.9.
Orthophosphorus acid on heating undergoes disproportionation to give phosphine and orthophosphoric acid.
Four P–OH
Two P = O
One P = P red P4 + alkali
Three P–OH
One P = O P4O10 + H2O
Two P–OH
Two P = O
Two P –O–P heat phosphoric acid
Three P–OH
Three P = O
Three P–O–P
Three P–OH
One P = O
One O – O
Phosphorus acid + Br2, heat in a sealed tube
Phosphoric acid + H2O2
Polymetaphosphoric acid, (HPO3)n
4 H3PO3→3H3PO4+ PH 3
Orthopho sphoric acid is commonly called phosphoric acid. It is prepared on a large scale from bone ash or from phosphorite by treating with sulphuric acid. Phosphoric acid is a weak and tribasic acid. It forms three series of salts such as NaH 2 PO 4 , Na 2 HPO 4 and Na 3PO 4 known as primary phosphate, secondary phosphate and tertiary phosphate. Na2HPO4 and NaH2PO4 are acidic salts, but Na3PO4 is a normal salt.
All oxyacid s contain P = O and P–OH bonds. But those acids in which oxidation state is less than +5, have P–H or P–P bonds but both bonds are not present together.
The number of –OH groups determines the basicity of the acid and the number of P–H bonds denotes the reduction ability of the oxyacid. Thus, hypophosphorus acid is a stronger reductant and gives metallic silver with silver nitrate.
H3PO2 + 2H2O + 4AgNO3 → 4Ag + 4HNO3 + H3PO4
Aqueous hypophosphorus acid reduces benzene diazonium salt to benzene. Phosphorus acid also reduces aqueous metallic salts to precipitate metals (like Cu, Hg).
In general, phosphorus undergoes sp 3 hybridisation and has tetrahedral geometry in its oxyacid. Phosphorus usually forms five bonds.
4.1.12 Anamolous Behaviour of Nitrogen
The following are the important points in which nitrogen differs from the rest of the group VA elements.
■ Nitrogen is a gas, while other elements of the group VA are solids at room temperature.
■ Nitrogen is a diatomic molecule, while phosphorus, arsenic and antimony are tetra-atomic molecules. Nitrogen molecule alone has the ability to form bonds, while in other molecules the bond order is one.
■ The hydride of nitrogen, NH3 is stable and has intermolecular hydrogen bonding. Hydrides of other elements are reactive.
■ Nitrogen does not form pentahalide, while P, As and Sb form pentahalides.
■ Nitrogen forms many oxides, but for other elements only tri and pentoxides are important.
■ Nitrogen predominantly exhibits –3 oxidation state, while for other elements +3 and +5 states are important.
■ Nit rogen is regarded as inert gas, due to high triple bond energy, while other elements are reactive.
■ Nitrogen cannot form as the heavier elements can, e.g., R3P = O or R3P = CH2 (R is alkyl group). Phosphorus and arsenic can form bond also with transition elements when their compounds like P(C 2H5)3 and As(C6H5)3 act as ligands.
The abnormal behaviour of nitrogen and the difference in the properties listed above are mainly due to the following factors.
i. Small size of nitrogen atom and absence of valence d-orbitals in nitrogen atom.
ii. High value of electronegativity of nitrogen.
iii. Tendency of forming multiple bonds by nitrogen atom.
iv. Having only two electrons in the penultimate, nitrongen.
TEST YOURSELF
1. Which of the following VA group hydrides is thermally most stable?
(1) SbH3 (2) AsH3
(3) PH 3 (4) NH3
2. Bond angles of NH3, PH3, AsH3, and SbH3 are in the order
(1) PH 3 > AsH3 > SbH3 > NH3
(2) SbH3 > AsH3 > PH 3 > NH3
(3) SbH3 > AsH3 > NH3 > PH 3
(4) NH3 > PH 3 > AsH3 > SbH3
3. The most stable trihalide of nitrogen is (1) NF3 (2) NCl3 (3) NBr3 (4) Nl3
4. In the reaction 546 2PClPClPCl+− + ,the change in hybridisation is from (1) sp3d to sp3 and sp3d2
(2) sp3d to sp2 and sp3
(3) sp3d to sp3d2 and sp3d3
(4) sp3d2 to sp3 and sp3d
5. Which is a mild reducing agent among the following group 15 hydrides?
(1) BiH 3 (2) PH 3 (3) NH3 (4) AsH3
6. N 2 is a stable molecule but N 4 molecule is unknown. P 4 is much more stable than molecular P2. Which is the best explanation for this difference?
(1) N2 has valence electrons only in bonding and non-bonding orbitals, while P 2 has some valence electrons in antibonding orbtials.
(2) The greater electronegativity of N compared to P stabilises the molecules with lower molar masses
(3) The greater size of P compared to N results in decreased overlap in pi bonds.
(4) The preference of P to adopt smaller bond angles than N favours formation of tetrahedral P4 molecules.
7. Bones glow in the dark because of slow combustion of white phosphorus in air. This is known as
(1) fluorescence
(2) phosphorescence
(3) chemiluminescence
(4) thermoluminiscence
8. Which of the following molecules has the negative standard enthalp y of formation (ΔHf)?
Table 4.10 Position of group VI A elements
A = NH3, B = PH3, C = AsH3, D = SbH3
(1) A (2) B
(3) C (4) D
9. p p –d p bond(s) is/are present in
(1) NH3 (2) NO3
(3) PCl3 (4) 3 4 PO
Answer Key
(1) 4 (2) 4 (3) 1 (4) 1
(5) 3 (6) 3 (7) 3 (8) 1 (9) 4
4.2 GROUP 16 ELEMENTS
Oxy g en (O), sulphur (S), selenium (Se), tellurium (Te), polonium (Po) and livermorium (Lv) are the five elements of group VIA. They are located in the p-block of the long form of the periodic table. These elements are assigned group 16 by the I.U.P.A.C.
These are called elements of oxygen family or sometimes elements of oxygen sulphur family. These elements are also called chalcogens because of their ability to form minerals. Most of the minerals of copper contains either oxygen or sulphur.
The position of group VIA elements in the long form of the periodic table is indicated by shaded portion in Table 4.10
E ach of the group VIA elements has six electrons in the valence shell. Two electrons are filled in the s sub-shell and the remaining four in the p-sub-shell. The general electronic configuration of these elements is ns 2 np 4 . Configuration of these elements is given in the box method in Fig.4.10.
Fig.4.10. Electronic configuration of group VIA elements
The elctronic configuration of each element of group 16 is given in table Table.4.11
Table 4.11 Atomic numbers and electronic Configuration of the group VIA elements
In accordance with the Hund’s rule, the fourth electron will be paired up in the p-sub shell. Therefore, the number of unpaired electrons in an atom of each of these elements is 2.
Only two electrons are present in the penultimate shell of oxygen, eight electrons in sulphur, eighteen electrons each in selenium, tellurium, and polonium. This shows why oxygen differs in some of its properties from sulphur and these two elements differ from the remaining elements of the group.
4.2.1 Occurrence
Oxygen and sulphur occur widely in nature as oxides, sulphides, and sulphates. The important minerals are:
Sulphides Sulphates
Galena : PbS Gypsum : Ca SO4.2H2O
Zinc blende: ZnS Epsom : MgSO 4.7H2O
Copper pyrites: CuFeS2 Barytes : BaSO4
Oxygen is also available in free state as dioxygen. Oxygen comprises about 21% of the earth’s atmosphere by volume. Oxygen is the most abundant element of the earth’s crust. It makes up 46 percent of mass of earth’s crust.
4.2.2. Physical Properties
The similarities and gradation in the general physical properties of the group VIA elements are summarised in Table 4.12 The large difference between the melting and boiling points of oxygen and sulphur is due to difference in atomicity (oxygen is diatomic, while sulphur is octaatomic).
Atomic Radius
From oxygen to polonium, the atomic radius and the atomic weight gradually increase. The differentiating electron enters into the next higher energy shell, suggesting an increase in the atomic radius. However, the increase in the radius is less predominant beyond sulphur. This is due to the shielding effect by inner d–orbitals.
Physical State
Oxygen is a gas at room temperature. Other elements of the group VIA are solids.
The melting and boiling points gradually increase from oxygen to tellurium. The smaller values of polonium are attributed to a different crystal structure.
Metallic Nature
The electronegativity values on Pauling’s scale gradually decrease from oxygen to polonium. The metallic nature increases gradually with an increase in the atomic size. Oxygen and sulphur are non–metals. Selenium and tellurium are metalloids. Polonium, though radioactive and least abundant, is clearly a metal.
Table 4.12 Gradation in the physical constants of the group VIA elemetnts
Ionisation Potential
The ionisation potential values decrease gra dually on descending the group as the atomic number increases.
Ionisation potential values of the group VIA elements are relatively low compared to the corresponding elements of adjacent groups. Higher ionisation potential value of the group VA elements is attributed to the stable half-filled p-orbitals in group VA. The higher values of ionisation potential of the group VIIA elements are attributed to the larger nuclear charge of these elements. The decrease in ionisation potential values after sulphur is less predominant, which is attributed to shielding effect of inner d–orbitals.
Electron Affinity
Electron gain enthalpy of oxygen is less negative than that of sulphur because oxygen is compact in nature.
Atomicity
Th e oxygen is a diatomic molecule with a double bond between two bonded atoms. The double bond contains one sigma and one pi–bond. Double bond of oxygen molecule based on orbital overlap is shown in Fig.4.11.
Oxygen is a paramagnetic molecule due to the presence of two unpaired electrons. The magnetic behaviour of oxygen is explained by molecular orbital theory. Atomicity of sulphur, selenium, and tellurium is eight each. They form crown shaped molecules. The structure of these molecules is referred to as puckered ring. In the sulphur molecule the S–S bond length is 2.04 Å and SSS bond angle is 107°, as shown in Fig.4.12 and chair molecule of sulphurshown in Fig.4.13.
Fig.4.11 Orbital overlaps in O2 molecule
Fig.4.12 Puckered ring S8 molecule
Fig.4.13. S6 Chair molecule
Allotropy
All the group VIA elements exhibit allotropy. T he only gaseous element that exhibits allotropy is oxygen. Its allotrope is triatomic oxygen, called ozone.
Sulphur exhibits two types of allotropes, crystalline and amorphous. Yellow rhombic sulphur and monoclinic sulphur are the crystalline forms of sulphur. At room temperature, the most stable form of sulphur is rhombic. It is the most common form of sulphur and stable up to 95.6°C. Above this temperature monoclinic sulphur is stable.
Rhombic sulphur crystals are formed on evaporating sulphur in carbondisulphide solution. It is yellow and melts at 386 K with specific gravity 2.06. It is insoluble in water and dissolves to some extent in benzene, ether and alcohol. It is readily soluble in carbon disulphide. Monoclinic sulphur is insoluble in carbon tetrachloride.
The melting point of monoclinic sulphur is 393 K and specific gravity is 1.98. It is soluble in CS2. This form of sulphur is prepared by melting rhombic sulphur in a dish and cooling, till crust is formed. Two holes are made in the crust and the remaining liquid poured out. On removing the crust, colourless needle shaped crystals of b - sulphur are formed. It is stable above 369 K and transforms into a -sulphur below it. Conversely, a- sulphur is stable below 369 K and transforms into a - sulphur above this temperature. At 369 K both the forms are stable. This temperature is called transition temperature. Monoclinic sulphur has two forms b– and g–. All a–, b–, g–are crystalline for ms and differ in their densities.
Several other modifications of sulphur containing 6 to 20 sulphur atoms per ring have been synthesised. Among these, cyclo-S6 is important and the ring adopts chair form. At temper atures greater than 1000 K, S 2 is dominant and is paramagnetic like O 2.
Bonding and Valency
Each element of the group VIA has six valence electrons and requires two more for attaining stable octet configuration. The usual valency of chalcogen is thus two. The maximum valency of these elements, however, is the group number 6.
Oxygen cannot exhibit tetravalency and hexavalency like the other elements of this group. This is due to lack of valence d–orbitals in oxygen. In hydronium ion, the maximum covalency of oxygen is three. Due to its high electronegativity and small size, oxygen can form ionic compounds with metals. By gaining 2 electrons, it forms metal oxides. Sulphur also has the tendency of forming metal sulphides.
Oxidation Numbers
The general oxidation numbers of the group VIA elements are – 2, + 2, + 4 and + 6. The negative oxidation state (– 2) is predominant for oxygen. This is due to its non-metallic nature and higher electronegativity value.
Oxygen exhibits positive oxidation numbers in fluorides, + 1 in O2F2 and +2 in OF2. Oxygen exhibits – 1 state in peroxides. Oxygen also exhibits fractional oxidation state, – 1/2 in superoxides.
The importance of – 2 oxidation state decreases from oxygen to sulphur and then almost vanishes for tellurium and polonium. The tendency of exhibiting + 6 oxidation state by the group VIA elements decreases from oxygen to polonium. This is due to inert pair effect.
The common oxidation states of sulphur in its compounds are –2, +2, +4 and +6. These are all even numbers. Two unpaired electrons can form two bonds in the ground state of sulphur. Electronic configuration of sulphur in different states is shown in Fig 4.14
Configuration in the ground state
Configuration in the first excited state
Configuration in the second excited state
Fig.4.14 Electronic configuration of sulphur in different states
With four unpaired electrons, sulphur forms four bonds in the first excited state. With six unpaired electrons, six bonds are formed in the second excited state.
In the ground state, sulphur exhibits –2 oxidation state, e.g., H2S, ZnS, etc. In the first excited state, sulphur may undergo sp2 or sp3d hybridisation. The oxidation state exhibited is +4, e.g., SO 2, SF 4, SCl 4, etc. In the second excited state, sulphur may undergo sp2 or sp3d2 hybridisation. The oxidation state exhibited is +6, e.g., SO3, SF6, etc.
S and Se are more stable in their +4 or +6 states than +2 state. Te is stable in its +4 and +6 states, though + 2 state is also considerably stable.
4.2.3 Chemical Properties
Hydrides
The binary compounds of hydrogen with chalcogens are called hydrides. Their general formula is MH 2 , also represented as H 2 M.
General characteristics of hydrides of group VIA are given in Table 4.13.
Water is formed by the combination of hydrogen and oxygen in the presence of a catalyst. It is the neutralisation product of a protonic acid with a base. MH2 type hydrides of chalcogens are usually prepared by the action of acid on a binary compound with metal.
The shape of MH2 type hydride is angular. The hybridisation of oxygen in water molecule is sp 3. It has tetrahedral geometry with two lone pairs on the central atom. Water is a bent molecule with ‘V’ shape. The bond angle in water is 104.5°. The decrease in the bond angle from the tetrahedral angle is due to greater repulsions between two lone pairs.
In the hydrides of other elements also, the central atom has two bond pairs and two lone pairs. The bond angles are almost equal to 90°. This is due to the use of unhybridised pure p – orbitals for bonding.
All these hydrides, except water, are gases. They are highly volatile substances, foul smelling and poisonous. Hydrogen sulphide is most volatile. Abnormal high boiling point of water and its low volatility are attributed to intermolecular hydrogen bonding. Only van der Waals forces of attractions are present in other hydrides.
Table 4.13 Characteristic of hydrides of group VIA elements
Movi ng down from H 2 O to H 2 Te, the M–H bond length increases and bond energy decreases as a result of which thermal stability decreases. Water is most stable and its reactivity is very less.
The order of volatility is H2S > H2Se > H2Te > H2O
The order of stability is H2O > H2S > H2Se > H2Te
The order of boiling points is H2O > H2Te > H2Se > H2S
The order of K a values is H2O > H2S > H2Se > H2Te
Water is practically a neutral liquid. These hydrides dissolve in water and ionise to produce proton.
MH2 + H2O H3O + + MH–
The dissociation constants are low, but the value increases from H 2O to H2Te. These hydrides are weak dibasic acids and form two types of salts, e.g., NaHS and Na 2S. The conjugate base of weak acid water is hydroxyl ion, OH– and is a strong base. H2Te is relatively more acidic and TeH– is relatively a weaker base.
Oxygen has two binary compounds with
Types of oxides
Basic oxides
Acidic oxide
Amphoteric oxide
Neutral oxide
hydrogen, H2O and H2O2. Sulphur has many binary compounds with hydrogen H2S, H2S2 and poly-sulphides like H 2S3, H2S4, H2S5, etc. H 2S 5 is called hydrogen pentasulphide and also as hydrogen polysulphide. As the number of sulphur atoms increases, the stability of hydride decreases.
Oxides
Binary compounds of oxygen with another element are called oxides. Oxygen reacts with most of the elements to form oxides. Ozone may be treated as dioxide of oxygen. In many cases, one element forms two or more oxides.
Oxides can be simple, where rules of valency and stoichiometry are not violated e.g., MgO, MnO2, Al2O3, CO2, etc. If the content of oxygen is high, it is called a superoxide (e.g., KO2, CsO2, etc.) and less is called a suboxide (e.g., C3O2, S2O, etc.). Magnetic oxide (Fe3O4), red lead (Pb3O4), etc., are called mixed oxides.
Simple oxides can be classified, based on their acid-base behaviour into 4 types. These four types of oxides are l isted in Table 4.14.
In general, metallic oxides are basic. Some of the metallic oxides and oxides of some metalloids are amphoteric. Alumina reacts with acids as well as alkalies, exhibiting dual nature.
Behaviour of oxides
Neutralises hydrochloric acid
Neutralises caustic soda solution
Neutralises hydrochloric acid as well as caustic soda solution
Neutralises neither hydrochloric acid nor caustic soda solution
Examples of oxides
Na 2O, K 2O, MgO, CaO, CrO, CuO, Ti2O, Fe2O3
B 2 O 3 , CO 2 , NO 2 , P 4 O 6 , P 4 O 10 , SO2, SO3, Cl2O, Cl2O7, Cr2O6
In general, non–metallic oxides are acidic. Some of the non–metallic oxides are neutral. Neutral oxides do not react with acids as well as bases. If an element is forming two or more binary oxides, that oxide with more oxygen content is generally more acidic. As a general rule, only non-metal oxides are acidic but oxides of some metals in high oxidation state also have acidic character
Example: Mn2O7, CrO3, V2O5
Oxygen reacts with other elements of group VIA to form dioxides and trioxides. These oxides are listed in Table 4.15. The acidic nature of the oxides decreases down the group. The stability and solubility also decreases. SO 2 is reducing while TeO2 is an oxidising agent
Table 4.15 Oxides of group VIA elements
Sulphur SO2 SO3 S2O, S6O S8O
Selenium SeO2 SeO3Tellurium TeO2
4.2.4 Dioxygen Preparation
Oxygen is commercially prepared by the fractional distillation of liquid air. It is prepared in the laboratory by thermal decomposition of potassium chlorate in the presence of manganese dioxide as catalyst, decomposition of KMnO4 and nitrates of alkali metals, except lithium. 42422 2KMnOKMnOMnOO
Oxygen is also available by the decomposition of higher oxides of some metals, oxides of relatively lower electropositive metals and peroxides.
2Pb3O4→ 6PbO + O2
2HgO→ 2Hg + O2
2PbO2→ 2PbO + O2
2Ag2O → 4Ag + O2
2H2O2 → 2H2O + O2
On large scale, oxygen is obtained at anode by the electrolysis of acidulated water. Molecular oxygen is also called dioxygen and it is unique in being paramagnetic with two unpaired electrons.
Dioxygen is colourless, odourless gas, liquefies at 90 K and freezes at 55K. Its solubility in water is less, about 3.08 cc in 100 cc of water at 293 K, which is just sufficient, as dissolved oxygen, for vital marine life. Oxygen has three stable isotopes: 16O, 17O, and 18O. Molecular oxygen is paramagnetic.
Oxygen directly reacts with nearly all non–metals and metals, except metals like Au, Pt, etc., and noble gases. Combination of oxygen is often exothermic. However, to over come the double bond dissociation, the reactions are initiated by external heating.
Oxygen is important in normal respiration, oxyacetylene welding, manufacture of steel, combustion process, and constituent in rocket fuels.
Sulphur is the sixteenth most abundant element in the earth’s crust. Abundance of sulphur is 340 ppm.
Organic materials such as wool, hair, proteins, egg, garlic, onion, and mustard contain sulphur. In vapour state, Sulpher partially exists as S 2 and exhibits
paramagnetism. Though native sulphur is also available, sulphur is usually recovered by the distillation of iron pyrites
3FeS2 → Fe3S4 + 2S
H2S and SO2 are present in volcanic gases. These gases react to give sulphur.
2H2S + SO2 → 3S + 2H2O
Selenium and tellurium are very less abundant in the earth’s crust. Polonium is radioactive and is a decay product of minerals of thorium and uranium. It was first isolated by Curie from the mineral, pitchblende.
4.2.5 Ozone
Ozone is trioxygen with atomicity 3. It is an allotrope of oxygen. Ozone is present in the stratosphere and prevents the harmful ultraviolet radiations coming from the sun reaching the earth.
Structure: Ozone is an angular molecule. The ‘V’ shaped molecule can be represented with two canonical structures.
The actual structure of ozone is a resonance hybrid of these two canonical forms. It has a pair of delocalised pi electrons. The experimental O–O bond length in O 3 is 1.28 Å, which is shorter than a O–O single bond (1.48 Å), but longer than O = O double bond (1.2 Å) length. The bond angle in ozone is 117°.
The canonical structures and the resonance hybrid structure of ozone is shown in Fig.4.15 The bond order in ozone is 1.5.
Ozone of purity greater than 10% can be obtained by using a battery of ozonizers. Pure ozone can be condensed in a vessel surrounded by liquid oxygen.
Ozone is thermodynamically unstable with respect to oxygen since its decomposition into oxygen results in the liberation of heat (ΔH is negative) and an increase in entropy (ΔS is positive). These two effects reinforce each other, resulting in large negative Gibbs energy change (ΔG) for its conversion into oxygen. It is not really surprising, therefore, high concentrations of ozone can be dangerously explosive.
Ozone liberates atoms of nascent oxygen (O 3 →O 2 + O). Due to this ease, it acts as a powerful oxidising agent. For example, it oxidises lead sulphide to lead sulphate and iodide ions to iodine.
PbS(s) + 4O3(g) → PbSO4(s) + 4O2(g)
2I–(aq) + H2O(l) +O3(s) → 2OH–(aq) + I2(s)+O2(g)
Metals like Ag and Hg are oxidised to their respective oxides.
2Ag + O3 →Ag2O + O2
2Hg + O3 →Hg2O + O2
Due to the dissolution of Hg, mercury loses its meniscus and starts sticking to the sides of the glass (container). This is called ‘tailing of mercury’. However, the meniscus can be regained by shaking it with water, which dissolves Hg2O.
Fig.4.15 Resonance structures of ozone
Ozone is produced from oxygen by silent electric discharge.
3O2(g) 2O3(g); ΔH° = 142 kJ mol–1
Ozonised oxygen is collected from the outlet with about 5 to 10% ozone.
When ozone is bubbled through the solution of an alkene or alkyne in an inert solvent like CH 2 Cl 2 , CCl 4 , etc., at 195 K, ozonides are formed.
When ozone reacts with an excess of potassium iodide solution buffered with a borate buffer (pH 9.2), iodine is liberated which can be titrated against a standard solution of sodium thiosulphate. This is a quantitative method for estimating O3 gas.
4: The p-BLOCK ELEMENTS
Experiments have shown that nitrogen oxides (particularly nitric oxide) combine very rapidly with ozone and thus the possibility that nitrogen oxides emitted from the exhaust systems of supersonic jet aeroplanes might be slowly depleting the concentration of the ozone layer in the upper atmosphere.
NO(g) + O3(g) → NO2(g) + O2(g)
Ozone is used as a germicide, disinfectant and for sterilising water. It is used as a bleaching agent and oxidising agent in the manufacture of potassium permanganate.
4.2.6 Sulphur Dioxide
Sulphur dioxide is obtained by the burning of sulphur or by treating a sulphide with mineral acid or by roasting a sulphide mineral.
S + O2→ SO2
Na2SO3 + H2SO4 → Na2SO4 + H2O + SO2
4FeS2 + 11H2O → 2Fe2O3 + 8SO2
Sulphur dioxide is easily liquefiable gas and it liquefies at room temperature under a pressure of two atmospheres and boils at 263 K. It is used as non-aqueous solvent.
Sulphur dioxide, when passed through water, forms a solution of sulphurous acid
SO2 + H2O H2SO3
In its reaction with water and alkalies, the behaviour of sulphur dioxide is very similar to that of carbon dioxide.
Sulphur dioxide is a reducing agent in neutral and acid media. In the presence of moisture it liberates nascent hydrogen and bleaches vegetable colouring matter. Schiff’s reagent is prepared based on the bleaching action of sulphur dioxide.
SO2 + 2H2O → 2(H) + H2SO4
Sulphur dioxide reacts readily with sodium hydroxide forming sodium sulphite, which then reacts with more sulphur dioxide to form sodium hydrogen sulphite.
2NaOH +SO2 → Na2SO3 + H2O
Na2SO3 + H2O + SO2 → 2NaHSO3
Sulphur dioxide reduces ferric salts to ferrous salts and acidified permanganate to colourless manganous salts in solutions
2Fe+3 + SO2 + 2H2O → 2Fe+2 + 4H+ + SO42–
2KMnO4 + 2H2O + 5SO2 →K2SO4 + 2MnSO4 + 2H2SO4
Sulphur dioxide forms addition compounds. In the presence of sun light or charcoal it adds with chlorine to give sulphuryl chloride.
SO2 + Cl2 →SO2Cl2
All dioxides are solids, except sulphur dioxide. All trioxides are solids. Reducing property of dioxides decrease from SO 2 to TeO2
Sulphur dioxide is used in refining petroleum and sugar, in bleaching wool and silk and as an antichlor, disinfectant and preservative. Sulphuric acid, sodium hydrogen sulphite and calcium hydrogen sulphite (industrial chemicals) are manufactured from sulphur dioxide. Liquid SO2 is used as a solvent to dissolve a number of organic and inorganic chemicals.
Structure of gaseous SO2 : Sulphur forms SO 2 in the first excited state and SO3 in the second excited state. Sulphur undergoes sp 2 hybridisation in both oxides with planar trigonal geometry. These structures are given in Fig.4.16
S = O bond length 1.43A0
Angular
OSO bond angle 119.50 S O O
Fig.4.16. Structures of sulphur dioxide
A molecule of SO2 has two double bonds and a lone pair at vertex. It is angular and the bond angle is 119.5°.
4.2.7 Halides
Oxygen forms only one type of halides known as fluorides. A large number halides of sulphur SX6 SX4, SX2 and S2X2 are known, where X is a halogen. The stability of the halides decreases in the order F – > Cl – > Br – > I – . Amongst hexahalides, hexafluorides are the only stable halides. All hexafluorides are gaseous in nature. They have octahedral structure. Sulphur hexafluoride, SF 6 is exceptionally stable for steric reasons.
Amongst tetrafluorides, SF4 is a gas, SeF4 a liquid and TeF4 a solid. These fluorides have sp 3 d hybridisation and thus, have trigonal bipyramidal structures in which one of the equatorial positions is occupied by a lone pair of electrons. This geometry is also regarded as see-saw geometry.
All elements except selenium form dichlorides and dibromides. These dihalides are formed by sp 3 hybridisation and thus, have tetrahedral structure. The well known monohalides are dimeric in nature. Examples are S2F2, S2Cl2, S2Br2, Se2Cl2 and Se2Br2
These dimeric halides undergo disproportionation as given below:
2Se2Cl2 → SeCl4 + 3Se
4.2.8 Anomalous Behaviour of Oxygen
Oxygen differs from rest of the chalcogens. The important points of difference between oxygen and other elements of group VIA are:
■ Oxygen is a gas, other elements are solids at room temperature.
■ Oxygen is simple diatomic, other elements have puckered ring octa atomic structures.
■ Oxygen is a non–metallic element and forms oxides of different nature.
■ The valency of oxygen is only 2. Other elements exhibit even hexavalency.
■ In H3O+ andO3, valency of oxygen is 3.
■ Though all molecules of chalcogens have even number of electrons, oxygen is paramagnetic.
■ The hydride of oxygen is a liquid. While hydrides of other elements are foul odoured gases.
■ Water is neutral. Hydrides of other elements act as weak acids in aqueous solutions.
■ Oxygen forms only fluorides, while other elements form numerous halides.
The anomalous behaviour of oxygen is attributed to the small atomic size, high electronegativity and lack of vacant valence d–orbitals.
4.2.9 Oxyacids of Sulphur
The element known to have maximum number of oxyacids is sulphur. Oxyacids of sulphur are divided into four types.
(a) Sulphurous acid series, (b) Sulphuric acid series, (c) Thionic acid series and (d) Peroxy acid series.
Some of the oxyacids have more than one structure. But generally it is observed that the central sulphur atom undergoes sp3 hybridisation with tetrahedral geometry around it. Each oxyacid of sulphur has two –OH groups and basicity is 2. The structural information of the oxyacids of sulphur is listed in Table 4.16.
Table 4.16 Oxyacids of sulphur and their structural aspects. Oxyacid
a) Sulphurous acid series H2SO3 + 4
Thiosulphurous acid H2S2O2 – 2, + 4
Dithionous acid H2S2O4 + 3, + 3
Pyrosulphurous acid H2S2O5 + 3, + 5
b) Sulphuric acid series
Sulphuric acid H2SO4 + 6
Thiosulphuric acid H2S2O3 – 2, + 6
Pyrosulphuric acid H2S2O7 + 6, + 6
c) Thionic acid series
Dithionic acid H2S2O6 + 5, + 5
Polythionic acid (x = n + 2) H2SxO6 + 5, + 5
d) Peroxy acid series
Peroxymono–sulphuric acid H2SO5 + 6
Peroxydi–sulphuric acid H2S2O8 + 6, + 6
Tautomeric structures are possible
p-d bond between sulphur atoms
Unstable and is known as hyposulphurous form
Average oxidation state of sulphur is + 4
Has 2 p-d bonds
Hydrated thiosulphate is familiar as hypo.
It is called oleum or fuming sulphuric acid
Has 2 p-d bonds on each sulphur atom
Middle sulphur atoms have zero oxidation state
Called Caro’s acid. Fairly stable
Called Marshall’s acid. Persulphates are well known
4.2.10 Sulphuric Acid
Sulphuric acid is one of the most important industrial chemicals world wide. It is called ‘oil of vitriol’ or ‘king of chemicals’.
Sulphuric acid was prepared earlier by lead chamber process, which is an example of homogeneous catalysis. It is now manufactured by contact process in the present times. Contact process is an example of heterogeneous catalysis and 96% to 98% pure H2SO4 can be prepared. Three main aspects of contact process are:
Sulphur dioxide required for the process is obtained by roasting of iron pyrities.
4FeS2 + 11O2→ 2Fe2O3 + 8SO2
Sulphur dioxide is oxidised catalytically with atmospheric air to sulphur trioxide. The catalyst commonly used is vanadium pentoxide.
2SO2 + O2 → 2SO3
Sulphur trioxide formed is absorbed in conc. sulphuric acid to get oleum, which is diluted with water to obtain sulphuric acid.
SO3 + H2SO4→ H2S2O7
H2S2O7 + H2O → 2H2SO4
Catalysts are often poisoned in contact
process. Hence, gases used must be pure. In modern plants, excess pure oxygen is used. Various parts of the plant used in the manufacture of sulphuric acid is as shown in Fig.4.17
The SO2 to SO3 is reversible, exothermic and leads to a decrease in volume in forward direction. Hence low temperature and high pressure are favourable. In practice, the plant is operated at 2 bar pressure and a temperature of 720 K.
Sulphuric acid is dense, colourless oily liquid with a specific gravity 1.84 at room temperature. It freezes at 283 K and boils at 610 K. It is soluble in water, but the dissolution is largely exothermic. Hence care is to be taken in preparing an aqueous solution the concentrated acid must be added slowly into water with constant stirring. Sulphuric acid is less volatile and hence, it is used to prepare volatile acids from their corresponding salts.
2KF + H2SO4→ 2HF + K2SO4
2NaNO3 + H2SO4→ 2HNO3 + Na2SO4
Sulphuric acid is a strong diprotic acid (Ka1 > 10 and Ka2 is 1.2 × 10–2). It forms two types of salts upon neutralisation.
Fig.4.17 Manufacture of H2SO4 by contact process
H2SO4 + NaOH → NaHSO4 + H2O
H2SO4 + 2NaOH → Na2SO4 + 2H2O
Concentrated sulphuric acid is a strong dehydrating agent. It removes water from organic compounds and charrs carbohydrates. HSO24 122211 2 CHO 12C + 11HO →
Hot concentrated sulphuric acid is a moderately strong oxidant. It is less powerful than HNO3, but more powerful than H3PO4 in its oxidation ability. It oxidises metals and non–metals and is reduced to sulphur dioxide.
Cu + 2H2SO4 → CuSO4 + 2H2O + SO2
C + 2H2SO4 → CO2 + 2H2O + 2SO2
3S + 2H2SO4 → 2H2O + 3SO2
Bulk of sulphuric acid is used in manufacturing fertilisers. It is also used in petroleum refining, detergent industry, manufacture of paints and pigments. It is used in electroplating process, galvanisation process, manufacture of nitrocellulose, storage batteries and also as a laboratory reagent.
TEST YOURSELF
1. Oxidation state of oxygen in OF2 and CO is, respectively,
(1) –2; +2 (2) +2; +2
(3) –2; –2 (4) +2; –2
2. Bond dissociation energy of E-H bond of the H2E hydrides of group 16 elements (given below) follows the order
A) S B) O C) Se D) Te
(1) A > B > C > D (2) D > C > B > A
(3) B > A > C > D (4) A > B > D > C
3. Which of the following gas dissolves with H2SO4 to give oleum?
(1) SO3
(2) SO2
(3) H2S
(4) O2
4. Which is the correct thermal stability order for H2E (E = O, S, Se, Te, and PO)?
(1) H2Se < H2Te < H2Po < H2O < H2S
(2) H2S < H2O < H2Se < H2Te < H2Po
(3) H2O < H2S < H2Se < H2Te < H2Po
(4) H2Po < H2Te < H2Se < H2S < H2O
5. Oxygen is more electronegative than sulphur, yet H2S is acidic while H2O is neutral. This is because
(1) water is a highly associated compound
(2) H–S bond is weaker then H–O bond
(3) H2S is gas while H2O is a liquid
(4) the molecular weight of H 2S is more
6. Which of the following statements regarding sulphur is incorrect?
(1) S2 molecule is paramagnetic.
(2) Melting point of S8 (monoclinic) > S8 (rhombic)
(3) Both rhombic and monoclinic sulphur are soluble in CS2
(4) The oxidation state of sulphur is never less than +4 in its compounds.
7. Which among KI, FeSO 4 , K 2 MnO 4 , and KMnO4 can’t be oxidised by O 3?
(1) KI (2) FeSO4
(3) K2MnO4 (4) KMnO4
8. The number of sigma and pi bonds in peroxodisulphuric acid are, respectively, (1) 9 and 4 (2) 11 and 4 (3) 4 and 8 (4) 4 and 9
9. Shape of SF4 molecule is (1) tetrahedral (2) square planar
(3) see – saw
(4) trigonal pyramidal
10. The numer of P p –d p bonds present in SO3 molecule is (1) 1 (2) 2
(3) 3 (4) 4
11. The correct order of O-O bond length in O2, H2O2, and O3 is
(1) H2O2 > O3 > O2 (2) O2 > H2O2 > O3
(3) O3 > H2O2 > O2 (4) O2 > O3 > H2O2
12. Which of the following is not the source of sulphur?
(1) Gypsum (2) Epsomite
(3) Galena (4) Pitchblend
13. Among the following, which bond has the highest bond energy?
(1) O – O (2) S – S
(3) Se – Se (4) Te – Te
14. Oxygen is always divalent while sulphur can form 2, 4 and 6 bonds because
(1) oxygen is more electronegative than sulphur
(2) sulphur has vacant d-orbitals while oxygen does not
(3) sulphur has large atomic radius than oxygen
(4) sulphur is more electronegative than oxygen
Answer Key
(1) 4 (2) 3 (3) 1 (4) 4
(5) 4 (6) 4 (7) 4 (8) 2
(9) 3 (10) 2 (11) 1 (12) 4 (13) 2 (14) 2
4.3 GROUP 17 ELEMENTS
Fluorine (F), chlorine (Cl), bromine (Br), iodi ne (I) and astatine (At) are the five elements of group VIIA. They are located in the p-block of the long form of the periodic table. These elements are assigned group 17
by the IUPAC. These are called halogens as their salts are formed in sea water. Halogen is a Greek word meaning sea salt. 'At' is a radioactive element.
The position of these elements in the long form of the periodic table is indicated by the shaded portion in Table 4.17
Each of the group VIIA elements has seven electrons in the valence shell. Two electrons are filled in the s- sub-shell and the remaining five in the p-sub-shell. The general outer electronic configuration of these elements is ns2np5. Configuration of these elements is summarised in Table 4.18.
Table 4.18 Atomic numbers and electronic configuration of the group VIIA elements
The similarities in the properties of elements result very largely from their similar electronic configuration. Halogens show similarities among themselves. So much similarity is not shown by other groups of elements.
IA 0
Table 4.17 Position of group VII elements
Only two electrons are present in the penultimate shell of fluorine, eight electrons in chlorine and eighteen electrons each in bromine, iodine and astatine. This shows why fluorine differs in some of its properties from chlorine and these two elements differ in their properties from the remaining elements of the group.
4.3.1 Occurrence
Halogens are not available in free state. They are naturally available in combined state, due to their high reactivity. The common natural form of the halogen is halide. Fluorine occurs in soil, river water, plants, bones, and teeth of animals. Sea is a potential source of halides. Fluorine is less available in sea water, only to the extent of few ppm. Most available halide in sea water is chloride. Sea water contains about 2.5% (w/v) sodium chloride. Bromine is available as MgBr2 and iodine as NaIO3. Solid deposits of sodium chloride on rocks is called rock salt. Chile salt (NaNO3) petre contains upto 0.2% sodium iodate.
Chlorine and bromine are extracted from brine solution, which is concentrated sea water. The ash o f se a weeds contain about 0.5% of iodine.
Important fluoride minerals:
Fluorspar : CaF2
Cryolite : Na3AlF6
Fluorapatite : 3Ca3(PO4)2.CaF2
Important chloride minerals
Carnallite : KCl.MgCl2.6H2O
Sylvine : KCl
Rock salt : NaCl
The similarities and gradation in the general physical properties of four available halogens are summarised in Table 4.19.
4.3.2 Physical Properties
Atomic radius: From fluorine to iodine, t he atomic number and the atomic weight gradually increase. The differentiating electron enters in the next higher energy shell, suggesting an increase in the atomic radius. However, the increase in the radius is less predominant beyond chlorine. This is due to the screening effect by the presence of inner d-electrons.
Table 4.19 Gradation in the physical constants of halogens
The radius of the halide ions is much larger than the covalent radius of the corresponding halogens. Larger radius of the anion is attributed to a less effective nuclear charge of anion than that of the parent atom. The radius of the halide ions increase gradually from fluoride to iodide.
Physical State
The tendency to form condensed molecules increases with increase in atomic number and size. Fluorine and chlorine are gases. Bromine is the only non-metallic liquid at room temperature. Iodine is a volatile solid and can be sublimed.
Halogens exist as diatomic covalent molecules.
The van der Waals forces are present between their molecules. The halogens are volatile on account of weak intermolecular forces. With an increase in the size and number of electrons, these forces increase from fluorine to iodine and the molecules are more condensed. Thus bromine is a liquid and iodine is a solid.
Electronegativity
The halogens have high values of electronegativity. Fluorine is the most electronegative element and the best nonmetal. The electronegativity values decrease from fluorine to iodine. As a result of the decrease of electronegativity, the non-metallic nature decreases down the group. All halogens are non-metals. However, iodine shows some metallic character in few cases.
Electron Affinity
Halogens have maximum negative electron gain enthalpy in the corresponding periods. This is due to the fact that the atoms of these elements have only one electron less than stable noble gas configurations. Electron gain enthalpy of the elements of the group becomes less negative down the group. However, the
negative electron gain enthalpy of fluorine is less than that of chlorine. It is due to small size of fluorine atom. As a result, there are strong interelectronic repulsions in the relatively small 2p orbitals of fluorine and, thus, the incoming electron does not experience much attraction.
Ionisation Potential
Ionisation potential values of halogens are very high among representative elements. Therefore, they have little tendency to form cation, X + . The ionisation potential values decrease gradually from fluorine to iodine, suggesting the tendency to from cation increases to iodine. Molten iodine monochloride (ICl) and molten iodinecyanide (ICN) conduct electricity, showing the existence of I+ cation.
Colour
Halogens are all coloured. The reason for the colours exhibited by halogens is absorption of energy in the visible region.
Halogens absorb the visible radiation and consequently their electrons are excited. The energy for excitation depends upon the size of halogen molecule. Fluorine is smallest in size among halogens. It requires high energy for excitation. It absorbs high energy violetindigo radiation and hence appears yellow. On the other hand, iodine is larger in size. It requires relatively less energy for excitation. It absorbs, low energy orange-red radiation and hence appears violet. Chlorine appears greenish yellow and bromine red.
Bonding and Valency
Each of the halogen element has seven valence electrons and requires one more for attaining stable octet configuration. Thus the usual valency of halogens is 1. The maximum valency, however, is the group number 7.
Fluorine is always monovalent in its compounds. It cannot expand its valency greater than 1. This is due to lack of valence d-orbitals in fluorine.
Halogens are covalent and the atomicity of their molecules is 2. One electron pair is shared between atoms of a halogen molecule to form a sigma bond. Bond dissociation energy is highest for chlorine among halogens. Bond energy decreases from chlorine to iodine gradually. Bond dissociation energy of fluorine is less than that of chlorine or bromine. The decreasing order of bond energies of halogens is: Cl2 > Br2 > F2 > I2. The exceptionally lower bond energy of fluorine is explained as follows.
Fluorine atom is very small. The internuclear distance in fluorine is only 1.44 A°. Fluorine has notoriously weak bond. This was explained successfully by Coulson. The non-bonding electron pairs of adjacent fluorine atoms in F2 molecule are very close to each other. These lone pairs of electrons are under strong repulsion forces. Hence the bond strength in fluorine is reduced.
Earlier, Mulliken explained that fluorine has no d-orbitals and hence multiple bonds are not possible in fluorine molecule. But Coulson’s explanation is simpler and widely accepted.
The decrease in the bond energy from chlorine to iodine is attributed to the higher internuclear distances.
Oxidation Numbers
The general oxidation numbers of the group VIIA elements are – 1, + 1, + 3, + 5 and + 7. states of + 4 and + 6 occur in oxides and oxyacids of chlorine and bromine. The element that never exhibits positive oxidation numbers is fluorine. In all its compounds, fluorine has the universal oxidation number, –1.
The oxidation numbers of chlorine are usually odd numbers. Chlorine has one
unpaired electron in its ground state as shown in Fig.4.18
Configuratin in ground state
Configuratin in first excited state
Configuratin in second excited state
Configuratin in third excited state
Fig.4.18 Electronic configurations of chlorine
In its ground state, halogen forms only one bond. In its first excited state, halogen usually undergoes sp 3 d hybridisation. Out of five hybridised orbitals two are occupied by lone pairs and the remaining three are useful for bonding. In its second excited state, halogen usually undergoes sp 3d 2 hybridisation. Out of six hybridised orbitals one is occupied by lone pair and the remaining five are useful for bonding. In its third excited state, halogen usually undergoes sp3d3 hybridisation. All the seven hybridised orbitals are used for bonding.
4.3.3 Chemical Properties
Fluorine is the most reactive of all elements in the periodic table. It reacts with all the other elements except the lighter noble gases helium and neon. It reacts with xenon under mild conditions to form xenon fluorides. Reactions of fluorine with many elements are vigorous and explosive. In the massive form a few metals such as copper, nickel, iron and aluminium acquire a protective fluoride layer and become passive to fluorine.
Halogen atoms have seven electrons in the valence state. They acquire one more electron to form stable octet. The ability of attracting the electron is high for halogens, hence they are very reactive, their reactivity compared to other non-metals is high due to low dissociation enthalpies of halogens. The reactivity of halogens gradually decrease from fluorine to iodine.
With Hydrogen
The trend in the reactivity of halogens is evidenced in the reaction with hydrogen. Fluorine reacts with hydrogen even in dark. Chlorine reacts with hydrogen in the presence of diffused light. Bromine reacts with hydrogen upon heating and the reaction is complete in the presence of a catalyst, like platinum. Iodine reacts with hydrogen upon heating and the reaction is never complete since a reversible equilibrium is established. Properties of hydrogen halides are compared in Table 4.20
Table.4.20 Properties of Hydrogen halides
Hydrogen halides are gases. Hydrogen fluoride can be liquified easily. Boiling point of hydrogen fluoride is higher than that of other hydrogen halides. This is because of intermolecular hydrogen bonding in hydrogen fluoride. Most volatile hydrogen halide is hydrogen chloride.
Hydrogen fluoride is the most stable hydrogen halide. Stability of hydrogen halides is in the order: HF > HCl > HBr > HI. In aqueous solutions, HF is a weak acid.
With Water
Halogens are all covalent. They are all soluble in water, but the extent to which they react with water and the reaction mechanism that is followed, vary among halogens. Fluorine is so strong as an oxidant that it oxidises water to oxygen. The reaction is strongly exothermic and spontaneous. The free energy change is large and negative.
F2 + 3H2O → 2H3O+ + 2F– + O2;
∆ Go = –795 kJ mol–1
Similar reaction with other halogens is possible, but the reaction is slow because of high activation energies. An alternative disproportionation reaction occurs with other halogens.
X2 + H2O → HX + HOX
Bromine is the slightly soluble in water. The reaction of iodine with water is nonspontaneous, since the free energy change for the reaction is positive. Iodine is more soluble in 10% aqueous potassium iodide solution. This is due to the formation of a complex tri-iodide ion (I 3–). Ions like I 3–, IBr2–, ICl4 –, etc., are called polyhalides.
With Oxygen
Halogens form many oxides with oxygen but most of them are unstable. Fluorine forms two compounds OF2 and O2F2. However, only OF2 is thermally stable at 298 K. These oxides are essentially oxygen fluorides because of the higher electronegativity of fluorine than oxygen. Both are strong fluorinating agents. O 2 F 2 oxidises plutonium to PuF 6 and the reaction is used in removing plutonium as PuF6 from spent nuclear fuel.
Chlorine, bromine and iodine form oxide in which the oxidation states of these halogens range from +1 to +7. A combination of kinetic and thermodynamic factors lead to the generally decreasing order of stability of oxides formed by halogens, I > Cl > Br. The higher oxides of halogens tend to be more stable than the lower ones.
Chlorine oxides, Cl 2 O, ClO 2 , Cl 2 O 6 and Cl2O7 are highly reactive oxidising agents and tend to explode. ClO2 is used as a bleaching agent for paper pulp and textiles and in water treatment.
The bromine oxides, Br2O, BrO2, BrO3 are the least stable halogen oxides (middle row
anomaly and exist only at low temperatures, They are very powerful oxidising agents.
The ioidine oxides, I 2 O 4 , I 2 O 5 , I 2 O 7 are insoluble solids and decompose on heating. I2O5 is a very good oxidising agent and is used in the estimation of carbon monoxide.
With Metals
Halogens react with metals to form metal halides. The ionic character of the halides decreases in the order: MF > MCl > MBr > MI. If a metal exhibits more than one oxidation state, the halides in higher oxidation state will be more coavalent. SnCl 4, PbCl 4, SbCl 5 and UF6 are more covalent than SnCl2, PbCl4, SbCl3 and UF4.
With Halogens
Halogens react with each other forming binary compounds, called inter-halogen compounds
Table 4.21. Inter halogens are four types as listed in Table 4.22 . Interhalogens are diamagnetic. Atomicity of an interhalogen compound is an even number and the highest atomicity is eight.
Interhalogen compounds can be prepared by the direct combination or by the action of halogen on lower interhalogen compounds.
The interhalogen compound formed depends upon specific condition. 437K
ClF2ClF;I3Cl2ICl +→+→
I2 + Cl2 → 2ICl;
Br2 + 5F2→ 2BrF 5
Physical properties of interhalogen compounds are intermediate between those of constituent halogens, except that the melting and boiling points are a little higher than expected.
Interhalogen compounds have heteroatomic bonds. These bonds are polar and hence interhalogens are more reactive
than halogens. However, the reactivity of interhalogen is less than that of fluorine.
Hydrolysis of interhalogens give halide ion derived from the smaller halogen and oxyanion derived from larger halogen.
XX' + H2O → HX' + HOX
ClF5 + 3H2O → 5HF + HClO3
The structure of interhalogen compounds are shown in Fig.4.19.
Fig.4.19 Structure of interhalogen compounds
Oxidation Ability
Electron affinity of halogen is high, hence, they are good oxidants. Fluorine is the best oxidising agent. The oxidation ability of halogens decreases from fluorine to iodine. Halogen atoms gain electrons and form halide ions. When halogens of lower molecular size react with halides of higher ionic size, the corresponding halides are formed.
Fluorine can displace all halides from their salts.
F2 + 2KCl → 2KF + Cl2;
F2 + 2KBr →2 KF + Br2 and
F2 + 2KI → 2KF + I 2
Chlorine cannot oxidise fluoride. It displaces bromide and iodide. Bromine can displace only iodide. Iodine cannot displace halides.
Cl2 + 2KBr → 2KCl + Br2;
Cl2 + 2KI → 2KCl + I2
Br2 + 2KI → 2KBr + I2
Table 4.21 Some properties and structures of interhalogen compounds
XX' ClF Colourless gas
BrF pale brown gas
IF(a) detected spectroscopically
BrCl(b) gas
ICl ruby red solid (α-form)
Linear
Linear
Linear
Linear
Linear brown red solid (β-form)
IBr black solid
Linear
XX' 3 ClF3 colourless gas Bent T-shaped
BrF 3 yellow green liquid
IF 3 yellow powder
ICl3(c) orange solid
Bent T-shaped
Bent T-shaped
Bent T-shaped
XX' 5 IF 5 colourless gas but solid below 77K Square Pyramidal
BrF 5 colourless liquid Square pyramidal
ClF5 colourless liquid Square pyramidal
XX' 7 IF 7 colourless gas Pentagonal bipyramidal
a Very unstable; b The pure solid is known at room temperature; c Dimerises and forms Cl-bridged dimer (I2Cl6)
Table 4.22 Four types of interhalogen compounds
XX' monohalide Ground state ns2np2np2np1nd0
XX' 3 trihalide First excited state
XX' 5 Pentahalide Second excited state
XX' 7 Heptahalide Third excited state
ns2np2np1np1nd1 sp3d hybridisation 3 bond pairs and 2 lone pairs 'T' shaped
ns2np1np1np1nd1nd1 sp3d2 hybridisation 5 bond pairs and 1 lone pair square pyramidal
ns1np1np1np1nd1nd1nd1 sp3d3 hybridisation 7 bond pairs and no lone pairs Pentagonal bi-pyramidal
Fluorine, chlorine and bromine oxidise aqueous ferrous iron to ferric ion.
2Fe2+(aq) + X2→ 2Fe3+(aq) + 2X–
(Here X2 = F2, Cl2 or Br2, but not I2)
Iodine, however, is such a weak oxidising agent that it cannot remove electrons from ferrous ion.
The standard reduction potentials of halogens are positive and decrease from fluorine to iodine. Thus, the halogen that has highest reduction potential acts as strongest oxidant. With a decrease in the reduction potential, their oxidising ability decreases.
Oxidation ability can also be explained with the net enthalpy change of the reaction. The enthalpy change in the reaction for the conversion of halogen to hydrated halide can be estimated by the application of Hess law using Born - Haber cycle.
X2(s)→ X2(l) ; ΔH1 = enthalpy of fusion
X2(l)→ X2(g) ;
ΔH2 = enthalpy of vapourisation
X2(g)→ 2X(g) ;
ΔH 3 = enthalpy of dissociation
X(g) + e–→ X–(g) ;
ΔH 4 = electron gain enthalpy
X–(g)+ aq → X–(aq);
ΔH 5 = enthalpy of hydration
Overall enthalpy change,
Since enthalpy change is more negative for fluorine, it is the best oxidant. This is because of low bond dissociation energy of fluorine and high hydration energy of fluoride ion.
4.3.4 Anamolous Behaviour of Fluorine
Halogens are very reactive and act as good oxidants. Fluorine is the most reactive halogen and is called superhalogen. The following are
the important points in which fluorine differs from the rest of the group VIIA elements.
1. Fluorine does not exhibit positive oxidation number in its compounds. It shows only –1 oxidation state, while other halogens exhibit positive oxidation states, ranging from +1 to +7.
2. Fluorine is always univalent in its compounds, while the other halogens are even heptavalent.
3. Hydrogen fluoride is a liquid at 20°C, while the other hydrogen halides are gases at room temperature. Hydrofluoric acid is a weak acid, but aqueous HCl, HBr and HI are strong acids.
4. Silver fluoride is water soluble, but other silver halides are insoluble in water. Calcium fluoride is insoluble in water, while other calcium halides are water soluble.
5. Fluorine can displace all halides from their salts, while other halogens cannot displace all halides.
6. The only halogen that oxidises water to oxygen is fluorine. Its reaction with water is spontaneous. Other halogens form hydrohalic acid and hypohalous acid with water.
7. Fluorine can directly combine with carbon to form fluorocarbons, while other halogens cannot combine with carbon even in drastic conditions.
8. Fluorine can react even with inert gases like xenon. Other halogens do not react.
9. Fluorine forms two fluorides of oxygen OF2 and O2F2. However, only OF2 is thermally stable at 298 K. These oxids are essentially oxygen fluorides because of the higher electronegativity of fluorine than oxygen. Both are strong fluorinating agents. O 2F2 oxidises plutonium to PuF 6 and the reaction is used in removing plutonium as PuF6 from spent nuclear fuel.
Halogens react with metals to form metal halides. For example, bromine reacts with magne-sium to give magnesium bromide.
Mg(s) + Br2(l) → MgBr2(s)
The ionic character of the halides of the halides decreases in the order MF > MCl > MBr > MI where M is a monovalent metal. If a metal exhibits more than one oxidation state, the halides in higher oxidation state will be more covalent than the one in lower oxidation state. SnCl4, PbCl4, SbCl5 and UF6 are more covalent than SnCl 2, PbCl 2, SbCl 3 and UF4 respectively.
The anamalous behaviour of fluorine and the difference in the properties listed above are mainly due to the following factors.
1. Small size of fluorine atom.
2. Absence of d-orbitals in the valence shell of fluorine.
3. Highest electronegativity of fluorine.
4. Low bond dissociation enthalpy of fluorine.
5. Fluorine has only two electrons in its penultimate shell.
4.3.5 Chemistry of Chlorine
Isolation
Chlorine is naturally available as chloride. Scheele was first to prepare chlorine, by the oxidation of chlorides. Davy established its elementary nature and suggested the name chlorine on account of its colour. Chlorine is prepared in the laboratory by the oxidation of hydrochloric acid. The oxidising agents used are manganese dioxide (MnO 2 ), potassium permanganate (KMnO4), potassium dichromate (K2Cr2O7) and bleaching powder (CaOCl2).
Bulk isolation of chlorine in commercial scale also involves the oxidation of chloride. In Weldon’s process, pyrolusite mineral (MnO2) is heated with concentrated hydrochloric acid, to obtain chlorine. In Deacon’s process, chlorine is manufactured by the atmospheric oxidation of hydrochloric acid in the presence of cupric chloride catalyst at 400°C. 2 CuCl,400C 2 22 4HClO 2HO2Cl ° +→+
Electrolysis of brine solution in Nelson’s cell or Castner–Kellner’s cell is a modern method of preparing chlorine on a large scale. Down’s process is also used for the commercial production of chlorine.
Properties
Chlorine is pungent odoured gas with greenish yellow colour. It is about 2.5 times heavier than air. Chlorine has pungent odour. It is pollutant, poisonous and effects mucous membrane. On freezing chlorine gives yellow liquid and finally pale yellow solid.
Chlorine is also a very reactive element .
Reaction with Hydrogen
Chlorine reacts with hydrogen in diffused light to form hydrogen chloride. H2 + Cl2 Hϑ → 2HCl
Reaction with Metals
Chlorine reacts with all metals, except noble metals. It forms metal chlorides.
2Na + Cl2→ 2NaCl
2Al + 3Cl2→ 2AlCl3
Iron can form two halides with chlorine, but the metal on reaction directly with chlorine forms a compound with its highest stable oxidation state. Hence union of iron with
chlorine gives ferric chloride. Similarly with copper, cupric chloride is formed.
2Fe + 3Cl2→ 2FeCl3
Cu + Cl2→ CuCl2
Reaction with Non–metals
Chlorine reacts with non–metals except nitrogen and oxygen at room temperature to form non-metal chlorides.
P 4 + 6Cl2→ 4PCl3
2B + 3Cl2→ 2BCl3
S8 + 4Cl2→ 4S2Cl2
Reaction with Water
Chlorine reacts with water forming yellow colour solution of hydrochloric and hypochlorous acids.
H2O + Cl2→ HCl + HOCl
Hypochlorous acid is unstable and decomposes to give nascent oxygen. Thus moist chlorine acts as oxidant and bleaching agent.
HOCl → HCl + (O)
Reaction with Halides
Ch lorine does not oxidise fluoride, but liberates bromine and iodine from their respective aqueous halid e salts.
2KBr + Cl2 → 2KCl + Br2
2KI + Cl2 → 2KCl + I2
Reaction with Alkali
Ch lorine reacts with dilute alkali at room temperature forming chloride and hypochlorite. With hot, concentrated alkali chlorine forms chlorid e and chlorate.
Cl2 + 2NaOH (cold, dil) →
NaCl + NaOCl + H2O
3Cl2 + 6NaOH (hot, conc) →
5NaCl + NaClO3 + 3H2O
With dry slaked lime, chlorine gives bleaching powder.
2Cl2 + 2Ca(OH)2→
Ca(OCl)2 + CaCl2 + 2H2O
The composition of bleaching power is Ca(OCl)2.CaCl2.Ca(OH)2.2H2O
Reaction with Ammonia
Chlorine reacts with excess ammonia to form nitrogen and ammonium chloride. Excess chlorine giv es nitrogen trichloride and ammonium chloride.
8NH3 + 3Cl2 → N2 + 6NH4Cl ;
4NH3 + 3Cl2 → NCl3 + 3NH4Cl ;
NH3 + 3Cl2 → NCl3 + 3HCl
Reaction with Hydrocarbons
Chlorine gives substitution products with saturated hydrocarbons and addition products with unsaturated hydrocarbons.
CH4 + Cl2→ CH3Cl + HCl
C2H4 + Cl2 → C2H4Cl2
Oxidation Ability
Chlorine is a good oxidising agent. It oxidises bromides to bromine, iodides to iodine, ferrous salts to ferric salts.
2FeCl2 + Cl2→ 2FeCl3 ;
2FeSO4 + H2SO4 + Cl2→ Fe2(SO4)3 + 2HCl
Moist chlorine oxidises vegetable colouring matter and bleaches it. Chlorine oxidises hydrogen sulphide to sulphur and sulphur dioxide to sulphuric acid. It oxidises sulphite and thiosulphate to sulphate.
H2S + Cl2→ 2HCl + S
SO2 + Cl2 + 2H2O → H2SO4 + 2HCl
I2 + 6H2O + 5Cl2→ 2HIO3 + 10HCl
Na2SO3 + H2O + Cl2→ Na2SO4 + 2HCl
C10H16 + 8Cl2→ 16HCl + 10C
Uses
Chlorine is a good oxidising agent. It is used in the sterilisation of water, because it kills bacteria.
It is used in manufacturing bleaching powder and as a bleaching agent for wood pulp, cotton, rayon or linen.
It is used in the extraction of noble metals like gold and platinum.
It is used in the preparation of solvents like carbon tetrachloride, chloroform, ethylene dichloride and acetylene tetrachloride; refrigerants like freons; insecticides like DDT, BHC, plastics like PVC and neoprene rubber.
It is used in the preparation of poisonous gases of the war-fare like,
■ phosgene, also called carbonyl chloride (COCl2),
■ tear gas, also called nitrochloroform or chloropicrin (CCl3NO2),
■ mustard gas (ClC2H4–S–C2H4Cl),
4.3.6 Hydrogen Chloride
Hydrogen chloride was first prepared by Glauber, by heating common salt with concentrated sulphuric acid. The same method is used in laboratory to prepare it now-a-days.
NaCl + H2SO4 → NaHSO4 + HCl
NaHSO4 + NaCl → Na2SO4 + HCl
Hydrogen chloride gas can be dried by passing the gas through concentrated sulphuric acid.
Hydrogen chloride is a colourless and pungent odoured gas. It is easily liquefied to a colourless liquid and freezes to a white crystalline solid at 159 K. It is extremely soluble in water and forms acid. Hydrochloric acid is a strong acid and the pKa for its ionisation in water is –7.
HCl(g) + H2O(l)→H3O+(aq) + Cl–(aq)
Hydrochloric acid reacts with ammonia and gives white fumes of ammonium chloride.
NH3 + HCl →NH4Cl
Aquaregia is a mixture of three parts of concentrated hydrochloric acid and one part of concentrated nitric acid. This mixture is used to dissolve noble metals like gold and platinum.
Au + NO–3 + 4H+ + 4Cl– →
AuCl4– + NO + 2H2O
3Pt + 4NO–3 + 16H+ + 18Cl– →4NO –3
3PtCl62– + 4NO + 8H2O
Hydrochloric acid decomposes salts of weaker oxyacids to liberate acidic gaseous oxides.
Na2SO3 + 2HCl → 2NaCl + H2O + SO2
CaCO3 + 2HCl → CaCl2 + H2O + CO2
Na2CO3 + 2HCl → 2NaCl + H2O + CO2
NaHCO3 + HCl → NaCl + H2O + CO2
2NaNO 3 + 2HCl → 2NaCl + NO + NO 2 + H2O
Hydrogen chloride is used in the manufacture of chlorine, ammonium chloride and glucose from corn starch. Hydrochloric acid is also used as a laboratory reagent, in medicine, in extracting glue from bones and purifying bone black.
4.3.7 Oxyacids of Halogens
Fluorine does not form Oxyacids (except HOF). Chlorine has four oxyacids. The Oxyacids of halogens are listed in Table 4.23 The oxyacids of chlorine are more acidic than those of bromine and Iodine. With a decreases in the electronegativity of halogen, the acidic strength of oxyacids decreses.
Table 4.23 Different oxoacids of halogens
Fluorine HOF (Hypofluorous acid)
Chlorine HOCl (Hypochlorous acid)
Bromine HOBr (hypobromous acid)
Iodine HOI (Hypoiodus acid)
Structures
HOClO (Chlorous acid)
HOBrO (Bromous acid)
Chlorine undergoes sp3 hybridisation in all these oxyacids. The anion of the oxyacid and the corresponding structure shown in Fig 4.20.
Except in hypochlorite, all these anions are stabilized by resonance structures. Perchlorate ion has four canonnical structures and is most stable. Perchoric acid ionises readily and acts as strongest acid.
Fig.4.20 Structure of anions of chlorine oxyacids
Table 4.24 Oxyanions of chlorine
HOClO2 (Chloric acid)
HOBrO2 (Bromic acid)
HOIO2 (Iodic acid)
TEST YOURSELF
HOClO3 (Perchloric acid)
HOBrO3 (Perbromic acid)
HOIO3 (Periodic acid)
1. Among halogens, the one that can oxidise water to oxygen most readily is (1) chlorine (2) bromine (3) fluorine (4) iodine
2. Which of the following has maximum negative electron gain enthalpy?
(1) F (2) Cl (3) Br (4) I
3. The hydrogen halides with highest melting point and highest boiling point are respectively (1) HF, HI (2) HF, HF (3) HI, HI (4) HI, HF
4. Fluorine is a strong oxidising agent in aqueous solution. This is best explained by (1) low bond dissociation enthalpy of fluorine, high electron gain enthalpy of fluorine.
(2) low bond dissociation enthalpy of fl uorine, high hydration enthalpy of fluoride ion
(3) high hydration enthalpy of fluoride ion, high electron gain enthalpy of fluorine
(4) high hydration enthalpy of fluoride ion, high ionisation enthalpy of fluorine
5. Best reducing agent among the following is (1) HI (2) HBr
(3) HCl (4) HF
6. The possible oxidation numbers shown by Cl atom in the 2nd and 3rd excited states are (1) +1, +5 (2) +3, +5
(3) +5, +7 (4) +3, +7
7. Which of the following statements is not true for halogens?
(1) All form oxoacid of the type HXO 2
(2) All are oxidising agents.
(3) All but fluorine show positive oxidation states.
(4) Chlorine has the highest electron-gain enthalpy.
8. Which of the following is correct for bond dissociation enthalpy in halogen molecule?
(1) F2 > Cl2 > Br2 > I2
(2) F2 > Cl2 < Br2 < I2
(3) Cl2 > Br2 > F2 > I2
(4) I2 > Cl2 > Br2 > F2
9. Which one of the following reaction does not occur?
(1) 22 F2Cl2FCl +→+
(2) 22 Cl2Br2ClBr +→+
(3) 22 Br2I2BrI +→+
(4) 22 Cl2F2ClF +→+
10. Which one of the following represents the reaction between fluorine and cold dilute NaOH?
(1) 22 2F4NaOH4NaF2HO +→+
(2) 23 3FNaOH5NaFNaFO +→+
(3) 23 3FNaOH5NaFNaFO +→+
(4) 2 22 2F2NaOH2NaFOFHO +→++
11. Concentrated H 2 SO 4 cannot be used to prepare HBr or HI from KBr or KI because it
(1) reacts too slowly with KBr or KI
(2) reduces HBr or HI
(3) oxidises KBr or HI
(4) oxidises KBr to KBrO3 or KI to KIO3
12. Which anion can undergo both oxidation and reduction?
(1) Cr2O72– (2) NO3–
(3) OCl– (4) S2–
13. Acid used for making permanent marking on glass surface is
(1) HNO3 (2) HF
(3) HIO3 (4) H2SO4
Answer Key
(1) 3 (2) 2 (3) 4 (4) 2 (5) 1 (6) 3 (7) 1 (8) 3
(9) 4 (10) 4 (11) 3 (12) 3 (13) 2
4.4
GROUP 18 ELEMENTS
4.4.1 Introduction to Compounds of Noble Gases
N oble gases have stable electronic configuration. Hence, these gases are not expected to form chemical compounds with other elements.
226 4222 88 286 RaHeRn →+
The electron affinity values of these elements are zero. The atoms of these elements thus have no tendency to accept electrons. Hence, these elements do not form anions.
The ionisation potentials of these elements are very high. The atoms of these elements, thus have little tendency to lose electrons. Hence, these elements do not form cations.
The first and reasonably stable compound of any noble gas was prepared by Bartlett in 1962. It was based on the fact that first ionisation enthalpy of molecular oxygen (1175 kJ mol –1) was almost equal with that of xenon (1170 kJ mol –1 ). Bartlett made xenon react with platinum hexafluoride and obtained red solid compound of composition xenon hexafluoroplatinate (V), Xe[PtF 6 ]. The excitement caused by this discovery led to active search for the other noble gas compounds.
Among noble gases, xenon is known to be reactive. It is known to form compounds with most electronegative elements fluorine, oxygen and more electronegative groups like OSeF 6, OTeF5. The important compounds of xenon are fluorides, oxides and oxyfluorides. Xenon cannot form compounds with nitrogen and chlorine because of their low electronegativity
4.4.2 Occurrence
The inert gases are called noble gases as they show some chemical activity under certain specified conditions. The inert gases are also called rare gases as their natural availability is very less. They are also called aerogens, since most of them are available in air.
Radon is a radioactive element and short lived. The other five gases of zero group always occur in free state.
The main sources of noble gases are air, natural gas, spring water and radioactive minerals. Air is potential source of the first five noble gases. They all form about 1% by volume of air. Argon is the principle noble gas constituent of the air.
Helium was believed to be the decay product of radioactive metals like thorium, uranium and radium. Minerals like pitchblende and monazite contain considerable amounts of occluded helium.
New noble gas element Z = 118 called oganesson was prepared by the collision of caliofornium with calcium atoms.
24948 2941 9820 1180 CfCaOg3n +→+
Og has atomic mass 294 and electronic configuration (Rn) 5f146d107s27p6. Only very small amount of Og has been produced. Its half life is 0.7 millisecond.
The zero group of the long form of the periodic table consists of six elements, collectively known as inert gases. The elements are: helium, neon, argon, krypton, xenon and radon. The elements of zero group are the end members of each series of the long form of the periodic table.
First series ends with helium, second with neon, third with argon and so on. The zero group is the only group in the long form of the periodic table, in which all members are gases.
The elements of the zero group are chemically inactive under ordinary conditions and hence, are called inert gases or inactive gases.
The general outermost electronic configuration of inert gases is ns 2 np6, except for helium. Helium has 1s2 configuration. The electronic configuration of the inert gases is shown in the Table 4.25
Table 4.25 Electronic configuration of noble gases
4.4.3
Physical Properties
B eside s similarity in the electronic configuration, the following points justify the inclusion of all six gases in the same group of the periodic table.
A gradation in the properties is also observed with the increase of atomic numbers. All these elements are colourless, odourless and tasteless gases. These gases cannot be oxidised. They do not undergo combustion in air and also do not help in burning. Some physical properties of these gases are listed in Table 4.26.
All the inert gases possess low melting and boiling points, due to weak intermolecular attractive forces. The boiling points increase down the group. The ionisation potentials are high, suggesting that the elements are very stable and it is difficult to remove electrons. The electron affinities of the elements are taken as zero.
These gases are sparingly soluble in water and solubility slightly increases down the group. All, except helium, are absorbed by coconut charcoal at suitable temperatures.
All these gases are monoatomic and give characteristic line spectra. The atomic radius of these elements is large, as the practically measured radius is van der Waals radius.
Table 4.26 Physical properties of noble gases
4.4.4 Compounds of Xenon
Xenon Difluoride
Xenon difluoride is obtained when a 2:1 mixture of xenon and fluorine are heated in a nickel tube at 673K and 1 bar
Xe+F2→ XeF2
The central atom xenon has two bond pairs and three lone pars in XeF 2. The molecule is linear and is non-polar. Xenon difluoride acts as a strong oxidant. XeF 2 on hydrolysis gives Xe, HF and O2
2XeF2 + 2H2O → 2Xe + 4HF + O2
Xenon Tetrafluoride
Xenon tetrafluoride is obtained when a 1 : 5 mixture of xenon and fluorine are heated in a nickel tube at 873K, at a pressure of 7 bar
Xe + 2F2 → XeF4
It undergoes sublimation. Xenon tetrafluoride molecule has four bond pairs and two lone pairs on the central atom. The shape of the molecule is square planar and is non-polar. Xenon tetrafluoride, on treating with water, forms xenon trioxide. It is a disproportional reaction. Xenon trioxide is a solid at room temperature, but explosive.
6XeF4 + 12H2O →
4Xe + 24HF + 3O2 + 2XeO3
Xenon Hexafluoride
Xenon hexafluoride is obtained when a 1 : 20 mixture of xenon and fluorine are heated in a nickel tube at 573 K, at a pressure of 60-70 bar. It is also prepared by the reaction of XeF4 with O2F2 at 143 K.
Xe + 3F2→ XeF6
XeF4 + O2F2→ XeF6 +O2
Xenon hexafluoride has six bond pairs and a lone pair on the central atom. The shape of the molecule is distorted octahedral and it is expected to be a polar molecule.
Xen on hexafluoride, upon partial hydrolysis, gives xenon oxyfluorides and on complete hydrolysis, gives xenon trioxide.
XeF6 + H2O → XeOF4 + 2HF
XeF6 + 2H2O → XeO2F2 + 4HF
XeF6 + 3H2O → XeO3 + 6HF
XeF2, XeF4 and XeF6 are colourless and are crystalline solids.
All the three types of xenon fluorides are covalent molecules. The hybridisation of xenon in these molecules and the shapes of molecules are summarise d in Table 4.27
The structures of the xenon fluoride molecules are shown in the Fig.4.21.
Bonding in XeF 2 , XeF 4 and XeF 6 are explained by using the first, second and third excited state electronic configuration of xenon, respectively.
Similar to xenon, binary compounds of krypton with fluorine are known. Radon difluoride is also reported. However, radon is almost not available in the atmosphere.
Xenontrioxide is a colourless explosive solid. It has pyramidal structure shown in Fig.4.23 . Xenonmonoxy tetrafluoride is a colourless volatile liquid. It has square pyramidal structure as shown in Fig.4.22.
Fig. 4.23. Structure of XeO3
XeF2, XeF4 and XeF6 are colourless crystalline solids and sublime readily at 298 K.
Fig.4.21 Structures of xenon fluorides
Fig. 4.22. Structure of XeOF4
Table 4.27 Hybridisation and shape of xenon fluoride molecules
The structures of the three xenon fluorides can be deduced from VSEPR. XeF2 and XeF4 have linear and square planar structures respectively. XeF6 has seven electron pairs (6 bonding pairs and one lone pair) and would, thus, have a distorted octahedral structure as found experimentally in the gas phase. Xenon fluorides react with fluoride ion acceptors to form cationic species and fluoride ion donors to form fluoroanions.
XeF2 + PF 5→ [XeF]+[PF6]– ;
XeF4 + SbF5→ [XeF3]+[SbF6]– ;
XeF6 + MF → M+ [XeF7]–
(M = Na, K, Rb or Cs)
Uses: Helium is a non-inflammable and light gas. Hence, it is used in filling balloons for meteorological observations. It is also used in gas- cooled nuclear reactors. Liquid helium (b.p. 4.2K) finds use as cryogenic agent for carrying out various experiments at low temperatures. It is used to produce and sustain powerful superconducting magnets which form an essential part of modern NMR spectrometers and Magnetic Resonance Imaging (MRI) systems for clinical diagnosis. It is used as a diluent for oxygen in modern diving apparatus because of its very low solubility in blood.
Neon is used in discharge tubes and fluorescent bulbs for advertisement display purposes. Neon bulbs are used in botanical gardens and in green houses.
Argon is used mainly to provide an inert atmosphere at high temperature.
Metallurgical processes (arc welding of metals or alloys) and for filling electric bulbs.
Argon is a lso used in the laboratory for
handling substances that are air-sensitive. There are no significant uses of xenon and krypton. They are used in light bulbs designed for special purposes.
TEST YOURSELF
1. Why is helium preferred to nitrogen gas by the deep sea divers?
(1) Nitrogen has no tendency to dissolve in blood at high pressure, deep in the sea, but He has that tendency.
(2) He’ has no tendency to dissolve in blood at high pressure, deep in the sea, but N2 has the tendency to dissolve in blood
(3) Both N2 and He have no tendency to dissolve in blood.
(4) Both N2 and He have tendency to dissolve in blood.
2. The noble gas that behaves abnormally in liquid state is
(1) Xe
(2) Ne
(3) He
(4) Ar
3. Which of the following is T-shaped?
(1) XeOF2
(2) XeO3
(3) XeOF4
(4) XeF4
4. The increasing d-character in hybridisation of Xe in XeF2, XeF4, and XeF6 is
(1) XeF2< XeF4< XeF6
(2) XeF4 <XeF2< XeF6
(3) XeF6 < XeF4 < XeF2
(4) XeF2 < XeF6 < XeF4
Answer Key
(1) 2 (2) 3 (3) 1 (4) 1
CHAPTER REVIEW
Group 15
■ Nitrogen (N), Phosphorus (P), Arsenic (As), Antimony (sb) and Bismuth (Bi) are the elements of group VA.
■ Elements of group VA are called pnicogens. The order of percentage abundance in the earth crust, P > N > As > Sb > Bi.
■ The valency shell configuration of VA group elements is ns2np3.
■ Nitrogen is diatomic. Phosphorous, arsenic and antimony are tetra-atomic. Bismuth is monoatomic.
■ Nitrogen can form two p p –p p bonds, due to small size and greater overlaping character, while other element can’t.
■ In P 4 molecule, all the four atoms lie at the corners of a tetrahedron. The P-P-P bond angle is 600, oxidation state of P is zero and covalency of P is 3.
■ Most abundant gas in the earth’s atmosphere is nitrogen.
■ About 75% by mass and 78% by volume in air is nitrogen.
■ In combined state ‘N’ is available as KNO3 (Indian salt petre) and NaNO 3 (Chile salt petre)
■ Phosphorus is the eleventh most abundant element in the earth’s crust and the most abundant in the VA group.
■ Sources of phosphrus minerals in the earth’s crust are : Phosphorite Ca 3(PO4)2, Fluorapatite 3Ca 3 (PO 4 ) 2 .CaF 2 and Chlorapatite 3Ca3(PO4)2. CaCl2.
■ Except Bi, all elements of VA group exhibit allotropy.
■ Nitrogen has two allotropes in solid state. It has no allotropes in gaseous state.
■ a -Nitrogen, b -Nitrogen
■ Allotropes of phosphorus are yellow or white phosphorus, red phosphorus and black phosphorus.
■ White phosphorus molecule has a regular tetrahedral structure. It has six P-P bonds.
■ White phosphorus is more reactive, due to high bond angle strain.
■ Metallic character of the VA group elements increases as the atomic number increases.
■ N and P are non metals, As and sb are metalloids. Bi is a true metal.
■ Electronegativity decreases from N to Bi. As and Bi have same electronegativity values.
■ The common oxidation states of these elements are –3, +3 and +5.
■ Nitrogen is gas and other elements of group VA are solids.
■ Dinitrogen is inert, due to high bond dissocia-tion energy, 945.4 kJ/mol. 1σ and 2 p bonds present in N2 molecule.
■ Catenation power decreases from N to As with decreasing bond energies from N to As.
■ Catenation of nitrogen is observed in azide. Tetrazenes having organic substitutents give chains with 8 atoms of nitrogen.
■ Group VA elements form MH 3 type hydrides. Shape of MH3 is pyramidal with one lone pair on the central atom.
■ From NH 3 to BiH 3 ease of formation of hydrides, stability, ease of replacing hydrogen atom by Cl or methyl group decrease.
■ Basic nature, bond angle, water solubility and ionic character decrease from NH 3 to BiH3.
■ NH 3 and PH 3 are volatile and colourless gases. The boi ling points of VA group hydrides is in the order : SbH 3 > NH 3 > AsH3 > PH 3
■ The volatility of hydrides of group VA elements is in the reverse order of their boiling points.
■ As the atomic size of the central atom increases the lone pair is spread over a large surface area, as the result electron density decreses. Hence basic nature decreases.
■ Order of basic nature of group VA hydrides is, NH3 > PH 3 > AsH3
■ Ammonia is the only VA hydride which has hydrogen bonds in liquid state.
■ Ammonia is more readily formed and more stable than phosphine.
■ Ammonia forms co-ordinate bonds readly. Phosphine acts as an electron pair donor and can form complexes.
■ The central atom in NH 3 molecule undergoes sp 3 hybridization, a lone pair of electrons present at one of the vertices.
■ The bond angles in the hydrides of group VA decrease from NH 3 to BiH 3 due to increase in size of central atom.
■ Pure ‘p’ orbitals are involved in the formation of hydrides of group VA, except in NH3
■ Group VA elements form M2O3, M2O4 and M2O5 type oxides.
■ Nitrogen alone forms many oxides. This is due to pp-pp multiple bonding between N and O atoms.
■ Pentoxides are more acidic than trioxides. The oxide of an element with higher oxidation state is more acidic.
■ Among VA oxides, N2O5 is the most acidic oxide
■ Acidic nature of oxides decreases with increase in atomic number. The oxides of As and Sb are amphoteric.
■ N2O3 is more acidic than Bi 2O3. The basic nature increases from N2O3 to Bi2O3 because of increase in the size of central atom, which influences the metallic properties.
■ Trioxides dissolve in water to formous acids.
232 NOHO2HNO2 +→
Pentoxides dissolve in water to form –ic acids.
+→ 2523 NOHO2HNO
■ The –ic acids are generally more acidic than corrosponding -ous acids.
■ The oxidising nature of oxides decreases from N2O3 to Bi2O3
■ The oxides of nitrogen and phosphorus are chemically similar although their structures are different.
■ The VA oxide which acts as dehydrating agent is P2O5
■ Trioxides and pentoxides of P, As and Sb are dimeric. P 4 O 6 , As 4 O 6 and Sb 4 O 6 are trioxides. P 4 O 10 , As 4 O 10 and Sb 4 O 10 are pentoxides.
■ Nitrous oxide N 2 O is colourless and odourless is called laughing gas.
■ N2O is prepared by heating a mixture of NH4Cl and NaNO3. It is stable, relatively unreactive and neutral.
■ Nitric oxide (NO) is formed as an intermediate in the manufacture of nitric acid by the catalytic oxidation of ammonia.
■ NO is a colourless gas, paramagnetic and is very reactive.
■ Nitric oxide readily reacts with O2 to form reddish brown NO2 gas.
■ No is absorbed by cold FeSO 4 solution to form brown coloured FeSO 4NO.
■ Nitrogen dioxide (NO2) is reddish brown gas and paramagnetic.
■ NO 2 is obtained by heating lead nitrate. It is an odd electron molecule and is very reactive.
■ N2O4 is a mixed anhydride of HNO2 and HNO3.
■ N2O4 has no unpaired electron and hence it is colourless and diamagnetic.
■ Nitrogen pentoxide (N2O5) is obtained by dehydration of HNO 3 by P 4O 10. It is the anhydride of nitric acid
■ N2O molecule is linear. NO is linear with odd electron bonding
■ NO2 is trigonal planar molecule. It is stable due to resonating structures.
■ In N2O5 each nitrogen atom is surrounded by three oxygen atoms. N 2 O 5 has both covalent and dative bonds.
■ P4O6 and P4O10 are dimers. Oxygen atoms act as bridges in both the oxides. In both the oxides, number of bridge oxygen atoms is six.
■ Number of oxygen atoms surrounded by ‘p’ atom in P4O6 is three and in P4O10 is four.
■ Group VA elements form trihalides MX 3 and pentahalides MX5
■ M is sp 3 hybridised in MX 3 and is sp 3 d hybridised in MX5. Shape of MX3 is trigonal pyramidal and MX5 is trigonal bipyramidal.
■ Hydrolysis of MX 3 gives –ous acids and MX 5 gives –ic acids.
3233 PCl3HOHPO3HCl +→+
52 3 PCl3HO2HClPOCl +→+
■ NCl5 is not formed due to the absence of d-orbitals in the valency shell of ‘N’.
■ Hydrolysis of NCl3 gives NH3 and HOCl.
■ Hydolysis of PCl 3 gives HCl and H 3PO 3 PCl5 on hydrolysis gives HCl and H3 PO4.
■ Phosphorus forms two series of oxoacids. In all oxoacids phosphorous atom is tetrahedrally surrounded by other atoms. In these acids at least one P–OH bond is present.
■ The ionisable ‘H’ atoms are responsible for basicity of the oxy acids of phosphorus.
■ Phosphorous series of acids have P–H bonds and these H atoms are responsible reducing properties.
■ Metaphosphorous acid (HPO 2) normally exists as cyclic compound.
■ Phosphoric series of acids not contain P-H bonds in their structure. The oxidation state of ‘P’ in these acids is +5.
■ Heating an ammonium salt with an alkali gives ammonia.
■ NH3 gas is identified by its smell (or) with a glass rod dipped in HCl solution.
■ On large scale NH3 is prepared from coal, by Haber’s process and by cyanamide process.
■ Haber synthesised ammonia from elements. It is a reversible exothermic process. The reaction proceeds with a decrease in volume.
■ According to lechatlier principle, the favourable conditions for Haber’s synthesis of ammonia are low temperature and high pressure.
■ For Habers process, optimum conditions are, Temperature 725–775 K, Pressure 200–300 atm. Finely divided iron as catalyst and molybdenum as promoter.
■ Ammonia is used as refrigerant. It is used to produce fertilisers and for the manufacture of HNO3.
■ Liquid NH3 is a good solvent for both ionic and covalent compounds.
■ A mmonia is used in the manufacture of Na 2CO 3 by solvay process and in the prepara-tion of rayon and artificial silks.
■ On a large scale, nitric acid is manufactured by Ostwald’s process, which is based on oxidation of NH3 by atmospheric oxygen. Pt or Rh is used as catalyst.
■ HNO 3 obtained by ostwald’s process is about 61%. If is further concentrated by distillation until 68% is obtained distillation by mixing with conc. H2SO4 to get 98% and cooling in freezing mixture to get 100% acid.
■ Nitric acid is a colourless liquid. It exists as a planar molecule in gaseous state.
■ In aqueous solutions, nitric acid behaves as a strong acid, giving hydronium and nitrate ions.
■ A mixture of conc HNO3 and conc H2SO4 in 1 : 1 volume ratio is called nitration mixture. It is used to convert benzene to nitrobenzene.
■ HNO3 is used in manufacturing of fertilizers like {CaO.Ca(NO 3) 2}, manufacturing of explosives, perfumes, dyes and drugs.
Group 16
■ The elements Oxygen (O), Sulphur (S), Selenium (Se), Tellurium (Te) and Polonium (Po) belong to VI A group.
■ Group VIA elements are called chalcogens as they are mineral forming elements.
■ Pyrolusite (MnO2), Haematite (Fe2O3) are oxide minerals.
■ Gypsum (CaSO4.2H2O), Barytes (BaSO4) and Epsom salt (MgSO4.7H2O) are sulphate minerals
■ Oxygen is the most abundant element in earth’s crust. Sulphur is the 16 th most abundant element in earth’s crust.
■ The order of abundance of VI A group elements is, O > S > Se > Te > Po
■ Oxygen ocurs as oxides, carbonates, sulphates, nitrates and borates.
■ Sulphur mainly occurs as sulphides and sulphates.
■ Oxygen and sulphur are available even in the native state.
■ All the group VIA elements have six electrons (ns2np4) in their outermost energy level.
■ Atoms of group VIA elements tend to attain nearest inert gas electronic configuration by gaining or sharing of two electrons.
■ Oxygen is diatomic. Sulphur, Selenium and Tellurium are octaatomic. Polonium is monoatomic.
■ Tendency to form octaatomic rings, among the elements of group VIA is more in S and also in Se.
■ Due to its small size oxygen is capable of forming bonds. Hence it exists as a diatomic gas. The remaining elements of group VIA are not capable of forming bonds due to their large size and do not exist as diatomic molecules.
■ Atomic and ionic radii increase gradually with increase in atomic number.
■ Oxygen is a gas, but other elements are solids at room temperature.
■ The most common oxidation state of VI A group elements is – 2.
■ Oxygen exhibits –2, –1, –1/2, + 1 and +2 oxidation states.
■ In oxides, the oxidation state of ‘O’ is –2, in peroxides, – 1 and in super oxides, –1/2, in O2F2, +1 and in OF2, +2.
■ Oxygen does not exhibit higher oxidation states due to lack of ‘d’ orbitals in its valency shell.
■ Other elements of group VIA exhibit – 2, + 2 in ground state, + 4 in first excited state and + 6 in second excited state.
■ Tendency to form – 2 state decreases down the group VIA due to decrease in electronegativity.
■ Oxygen has two allotropes, both are non metallic gases. oxygen is stable diatomic gas and paramagnetic. Ozone is unstable triatomic gas and diamagnetic.
■ Sulphur exists in several non-metallic allotropic forms. The different forms arise partly from the extent to which S has polymerized and partly from the crystal structures adopted.
■ Rhombic or octahedral or α-sulphur is the common crystalline form of sulphur. Pale yellow in colour. It consists of S8 structural units packed together into octahedral shape. It is the stable variety at ordinary temperature.
■ Monoclinic or prismatic or ß-sulphur is stable above 95.6oC. Thus 368.5K is the transition temperature of S (R) S(M)
■ a,b and g-forms are all in puckered ring structures and differ in density. The atoms in the ring lie in two parallel planes with angle 1070 and bond length equal to 2.04A°.
■ The S 8 units break around 160°C. On further heating above it’s melting point S8 units dissociates in to S 6, S4, S2 units.
■ Elements of VI A form H2M type covalent hydrides. These hydrides are produced by the action of water or acid on metal chalcogenides.
■ H2O and H2S are exothermic compounds, while H 2 Se and H 2 Te are endothermic compounds.
■ Non-poisonous hydride of VI A group which exists in liquid state is H2O. other gases are poisonous and bad smelling.
■ H 2S is gas but H 2O is liquid, because of absence of molecular association in H 2S
■ The tendency of formation of hydrides decreases from o to Po, due to decrease in electronegativity.
■ Thermal stability and bond energy values of the hydrides decrease from H2O to H2Po. Reduction ability increases from H 2O to H2Po.
■ Hydrides are weak acids. Acidic nature increases from H2O to H2Te due to increase in their Ka values.
■ Order of boiling points : H 2O > H 2Te > H2Se > H2S. Volatility is in reverse order.
■ Water has higher boiling point than other VIA hydrides due to inter molecular hydrogen bonding.
■ In H 2 O molecule, sp 3 hybrid orbitals are involved in bonding. While in other hydrides, pure p-orbitals are involved in bonding.
■ In H2O the bond angle is 104°28’. In other hydrides of group VIA the bond angle is around 90°.
■ All elements of group VIA generally form dioxides and trioxides. MO 2 type oxides are: SO2, TeO2 and PoO2. MO3 type oxides are: SO3, SeO3 and TeO3.
■ Except SO2 all other dioxides of group VIA elements are solids.
■ Sulphur dioxide is produced by burning sulphur or roasting of metal sulphides in air.
■ Solubility of dioxides decreases from SO 2 to PoO2. SO2 is highly soluble in water.
■ Dioxides dissolve in water and form –ous acids. The strength of the oxy acids is in the order : H2So3 > H2SeO3 > H2TeO3.
■ SO2 can act as reducing agent in acid and neutral media.
■ SO2 reduces acidified yellow K2Cr2O7 into green Cr2(SO4)3
■ SO2 bleaches by reduction in presence of moisture. It’s bleaching action is temporary. During the bleaching action, SO2 is oxidised to H2SO4.
■ SO 2 forms addition compounds with halogens. e.g., SO2Cl2 (Sulphuryl chloride)
■ SO2 is the anhydride of sulphurous acid. It is angular with bond angle 119º 30’. It has one and one bond. Its bond length is 1.43A°.
■ Liquid SO2 is used as solvent for organic and inorganic chemicals.
■ VIA group elements form M2X2, MX2, MX4 and MX6 type halides. Oxidation states of VI A group elements in these halides are + 1, + 2, + 4 and + 6 respectively.
■ Compounds of oxygen with fluorine are called oxygen fluorides, because fluorine is more electronegative than oxygen.
■ Group VIA elements form MX 6 type of hexahalides. Among hexahalides, only hexa fluorides are stable. They have octahedral structures.
■ SF 4 is highly reactive gas. It is thermally stable. It is fluorinating agent. Scl4 is formed by direct reaction of sulphur with chlorine. Scl4 is an unstable liquid.
■ Many oxyacids of sulphur are known as anions and salts. Oxo anions have strong bonds which prevent polymerization.
■ H2SO5 is called Caro’s acid and H2S2O8 is called Marshall’s acid.
■ In all oxyacids of sulphur the basicity is 2. sp3 hybridisation takes place in all acids. All pi bonds are bonds.
■ Ozone can be prepared by silent electric discharge on pure dry oxygen gas.
■ Formation of ozone from oxygen is reversible and endothermic process.
■ Ozone is heavier than air, poisonous and slightly soluble in water. It is highly soluble in turpentine oil or glacial acetic acid
■ Ozone is thermodynamically unstable. The decomposition is exothermic.
■ Ozone is a powerful oxidizing agent, next to F2. Ozone oxidises black lead sulphide to white lead sulphate, HCl to Cl2 and iodide to I2
■ Nitric oxide can cause depletion of ozone in the upper atmosphere
■ Ozone acts as bleaching agent. Ozone bleaches by oxidation. Ozone is dry bleach for oils, ivory etc.
■ Ozone is used as an insecticide, bactericide and in the manufacture of potassium permanganate.
■ Ozone is used in purifying drinking water and cleaning atmosphere.
■ Sulphuric acid (H2SO4) is also called as ‘oil of vitriol’ or ‘king of chemicals’.
■ Sulphuric acid is commercially prepared by Contact process. Three main steps are : formation of SO2 by burning of ‘S’ (or) iron pyrites in oxygen, catalytic oxidation of SO2 into SO3 and absorption of SO3 into H2SO4
■ Pure SO 3 obtained is absorbed in 98% conc. H 2 SO 4 to form oleum, also called pyrosulphuric acid (H 2S2O7).
■ Oleum is diluted with water to obtain sulphuric acid of desired concentration.
■ Because of its low volatility, sulphuric acid is used to prepare volatile acids from their corresponding salts
■ Sulphuric acid is a strong dehydrating agent.
■ Sodium salt of thiosulphuric acid is called hypo. It is a pentahydrate.
Group
17
■ Fluorine, chlorine, bromine, and iodine are collectively known as halogens.
■ Halogens are very reactive non metals. They do not occur in free state.
■ Halogens occur in the combined state as halides. Iodine is also available as NaIO 3
■ Synthetically made element of group VIIA is astatine. It is radioactive short lived element.
■ The general outer electronic configuration of the halogens is ns 2np5
■ Fluorine is the most reactive element. It is regarded as super halogen
■ Halogens are diatomic molecules. van der Waals forces of attraction between molecules increases from fluorine to iodine.
■ Gaseous elements in VII A group are fluorine and chlorine. Bromine is a liquid and iodine is a subliming solid.
■ The melting points and the boiling points increase from fluorine to iodine due to increase in van der Waals forces. But volatile nature decreases down the group.
■ Atomic volume and density increase from fluorine to iodine. The size of atoms and the ions also increase gradually.
■ Ionisation energies decrease from fluorine to iodine.
■ Fluorine has abnormally high ionisation energy due to its small size and electrons in it are held strongly by the nucleus
■ The decreasing order of bond energies of halogens is Cl2 > Br2 > F2 > I2
■ Low bond energy of fluorine is responsible for its high reactivity.
■ Fluorine has the highest electronegativity (4.0) on Pauling scale. Electronegativity decreases down the group.
■ Order of electron affinity of halogens is, Cl > F > Br > I.
■ Small size and greater repulsion of electrons are responsible for low electron affinity of fluorine than that of chlorine.
■ Covalent bond is formed between two halogen atoms. The compounds formed between halogens and other non metals are covalent.
■ Fluorine is always univalent. Fluorine exhibits -1 oxidation state, since it is the most electro-negative element.
■ Most common covalency of Cl, Br or I is 1, but highest valency is 7.
■ Common oxidation state of halogens is –1. Except fluorine other halogens can also exhibit + 1, + 3, + 5 and + 7.
■ Fluorine cannot exhibit variable valencies due to the absence of ‘d’ orbitals.
■ Cl, Br and I exhibit higher oxidation states in different excited states and undergo various types of hybridisations.
■ Halogens are strong oxidising agents due to high electron affinity values. The oxidising capacity decreases from fluorine to iodine.
■ Lighter halogen displaces a heavier halogens from their halides.
■ Halogens absorb one of the seven colours of the visible light. This absorbed radiation promotes an electron to higher state.
■ Fluorine is yellow gas, Br 2 is red liquid, Cl 2 is greenish yellow coloured gas and iodine is violet.
■ The reactivity of halogens with H2 decreases from F 2 to I 2. F 2 reacts violently even in dark with hydrogen. The reaction of Cl2 with H2 is slow in dark but fast in sunlight.
■ The order of stability or boiling points of hydrogen halides is, HF > HCl > HBr > HI.
■ Order of acidic strength and reducing ability of hydrogen halides is, HI > HBr > HCl > HF.
■ The anomalous behaviour of fluorine is due to its small size, highest electronegativity, low F-F bond dissociation enthalpy and non availability of d orbitals in valence shell.
■ Fluorine is monovalent, unlike the other halogens.
■ Sea water contains about 2.5% (w/v) by weight of sodium chloride.
■ Important minerals of chlorine are : Rock salt, Carnalite KCl . MgCl 2. 6H 2O, Horn silver AgCl and Sylvine KCl.
■ Chlorine is commercially produced as a byproduct in the electrolysis of aqueous NaCl in Nelson cell method and from fused NaCl in Down’s process.
■ Except with nitrogen, oxygen and noble gases chlorine reacts with non- metals directly at ordinary temperature.
■ The bleaching action and disinfectant property of chlorine water is due to oxidation.
■ With cold dilute alkali chlorine gives chloride and hypochlorite.
■ With hot concentrated alkali chlorine gives chloride and chlorates.
■ With dry slaked lime chlorine gives bleaching powder.
■ Chlorine oxidises ferrous salts to ferric salts, hydrogen sulphide to sulphur, sulphites and thiosulphates to sulphates.
■ Chlorine acts as a bleaching agent in the presence of moisture. Chlorine bleaches vegetable colours like litmus and indigo.
■ Bleaching action of chlorine is due to production of nascent oxygen.
■ Chlorine is used for the purification of drinking water.
■ Chlorine is used in the preparation of poisonous gases like phosgene (COCl 2), mustard gas S(CH 2 -CH 2 -Cl) 2 , teargas CCl 3.NO 2 and D.D.T. (dichlorodiphenyl trichloroethane).
■ Fluorine has no known oxyacids (except HOF).
■ Four types of oxyacids of other halogens are known.
■ Acidic nature and thermal stability of oxyacids increases with increasing oxidation state of chlorine : HClO < HClO2 < HClO3 < HClO4
■ Perchloric acid is the strongest oxyacid and perchlorate is the weakest base known.
■ Order of bond length of Cl–O in oxyanions is, ClO4– < ClO3– < ClO2– < ClO–.
■ In the case of metals which can form more than one type of chloride, always the chloride in the higher oxidation state is formed.
■ Order of bond energy of Cl–O in oxyanions of chlorine is, ClO– < ClO2– < ClO3– < ClO4–
■ There are four types of interhalogen compounds. Their composition are: XX', XX' 3 and xx'7
■ XX' 3 is T shaped, XX'5is square pyramidal and has pentagonal bipyramidal structure.
Group 18
■ Helium, neon, argon, krypton, xenon and radon are noble gas elements.
■ Noble gases constitute the elements of zero group in the long form of the periodic table.
■ The general outermost electronic configuration of noble gases is ns 2np6.
■ Noble gases are called inert gases and are monoatomic.
■ Noble gases are chemically inert because of ‘octet’ configuration.
■ The main source of occurence of noble gases is air.
■ Argon is the principal noble gas constituent in the air.
■ Helium is totally inert because of its small size, high ionisation potential and zero electron affinity.
■ The first and reasonable stable compound of any noble gas, xenonplatinum hexafluoride was prepared by Bartlett.
■ The compounds of xenon are usually three types: fluorides, oxides and oxyfluorides.
■ Xenon has three fluorides, XeF2, XeF4 and XeF6
■ The structure of xenon difluoride is linear, xenon tetrafluoride is square planar and xenon hexafluoride is distorted octahedral.
■ Xenon has two oxides, XeO 3 and XeO4.
■ Xenon trioxide has pyramidal shape and xenontetraoxide has tetrahedral shape (XeO4).
■ Xenonmonoxy tetrafluoride (XeOF4) has square pyramidal structure.
■ Helium is preferred to nitrogen by deep sea divers and asthma patients. The respiration mixture is 20% O2 and 80% He by volume.
■ He is used to provide inert atmosphere, in gas thermometers.
■ Liquid helium is used as cryogenic agent for carrying experiments at low temperatures.
■ Neon is used in advertisement discharge lamps and in green houses
Argon is used to provide inert atomosphere in metallurgy and for filling electric bulbs.
■ Phosphorus has following main allotropes: white P4 tetrahedral, soluble in CS2, alcohol, ether and benzene, Red (P 4) n polymeric chain; Black phosphorus having layer structure with P-P-P angle 99 0
■ White phosphorus: Phosphorite produces phosphorus on strong heating with silica and coke.
■ SO2 is angular in shape. Sulphur in SO 2 is in sp2 hybridisation.
■ Bond angle of SO2 is 119.5° and bond length is 143 Pm or 1.43A°.
■ SO3 has planar triangular structure. Bond angle in SO3 is 120° and bond length is 143 pm or 1.43A°. In SO3 sulphur atom is in sp2 hybridisation.
■ Out of three bonds in double bonds in SO3 one is formed by overlap but the other two are formed by the overlapping of bonds between oxygen and sulphur atoms.
■ Solid SO3 exists in three forms and forms. form SO3 the most stable form and is made of cross-linked chains.
■ In lab reaction of F 2 and H 2 O at low temperature, is used to obtain ozonised oxygen.
■ Ozone oxidises many compounds but reduces only peroxides.
Group 17
■ Oxides of Chlorine : (For JEE) Chlorine forms four oxides
■ The oxidation state of chlorine in Cl 2O, ClO2, Cl2O6, and Cl2O7 are +1, +4, +6 and +7 respectively.
■ Stability and acidic character of oxides increases with increase of oxidation state of chlorine.
■ Stability and acidic character order of oxides of chlorine is Cl 2O < ClO2 < Cl2O6 < Cl2O7.
■ The decreasing order of oxidising power is Cl2O > ClO2 > Cl2O6 > Cl2O7.
■ In Cl2O molecule, oxygen atom undergoes SP3 hybridisation.
■ The shape of Cl2O molecule is angular and bond angle is 111° due to steric crowding of larger chlorine atoms. But in OF 2 the bond angle is 103°.
■ ClO2 is paramagnetic due to presence of odd electron bond.
■ In ClO2 molecule, chlorine atom undergoes sp3 hybridisation. The shape of ClO 2 molecule is angular and its bond angle is 118°.
■ Pure liquid Cl 2O 6 is diamagnetic, but in vapour state it is paramagnetic, because of existence of ClO3 monomer
■ Cl2O6 is dissolved in water to form chloric acid and perchloric acid. It is a mixed anhydride of chloric and perchloric acid.
Cl2O6 + H2O → HClO3 + HClO4
■ In solid state, Cl2O6 is considered to have ClO2+ and ClO4 – ions.
■ Oxides of Bromine, Br2O, BrO2 and BrO3 are least stable halogen oxides, and they exist only at low Temperature.
■ Perchloric acid is heated with phosphorous pent-oxide to form chlorine heptoxide equation quation
■ Cl 2 O 7 is dissolved in water to form perchloric acid. So it is called anhydride of perchloric acid.
Cl2O7 + H2O2 → HClO4
■ I2O4 exists in the form of [IO] + [IO3]– and I4O9 exists in the form of I 3 + (IO3)3(–).
■ The order of acidic strength of hypohalous acids
HOF > HOCl > HOBr > HOI
■ In hypohalous acids (HOX) thermal stability of ‘O-X’ bond increases, hence oxidizing power decreases.
■ Order of oxidizing power of hypohalites is: OCl–>O– Br>OI–
■ The stability and acidic character order of oxy acids of chlorine is HOCl < HClO2 < HClO3 < HClO4
■ Oxidising power oder of oxyacids of chlorine is HOCl > HClO 2 > HClO 3 > HClO4
■ Basic nature of oxy anions decreases in the order ClO– > ClO2– > ClO3– > ClO4–
■ Increasing order of bond length of Cl–O bond in oxy anions of chlorine
ClO4– < ClO3– < ClO2 – < ClO–
■ Increasing order of bond energy of Cl–O bond in oxyanions of chlorine
ClO– < ClO2– < ClO3– < ClO4–
■ Per halates are strong oxidising agents, the order of oxidizing power is ClO 4 – > BrO4 – > IO4–
■ Bleaching powder is also called calcium chlorohypochlorite because it is considered as a mixed salt of hydrochloric acid and hypochlorous acid. It is manufactured by the action of chlorine on dry slaked lime at 400C .
■ In cold water bleaching powder ionises into Ca2+, Cl– and ClO– ions. In hot water
bleaching powder forms Ca 2+ , Cl – and ClO3–
■ A good quality bleaching powder contains 35 - 38% available chlorine. (Theoritically 56% of Cl2 if CaOCl2 is taken and if CaOCl2. H2O is taken it is 49%)
■ Pseudohalogens: They are certain uninegative ions made up of two electronegative atoms which have properties similar to the halide ions. These are called pseudo halides.
■ Like halides, the pseudohalides have the corresponding pseudohalogens which behave like halogens in their chemistry.
■ Some examples are pseudohalides pseudohalides pseudohalogens cyanide CN– Cyanogen (CN)2
Thiocyanate SCN– Thicyanogen (SCN)2
Cyanate OCN– Oxycyanogen (OCN)2
Selenocyanate SeCN – Selenocyanogen (SeCN)2
Tellurocyanate TeCN – Tellurocyanogen (TeCN)2
■ Halogens or interhalogens combine with halide ions to form polyhalide ions.
■ The most common example of polyhalide ion formation is furnished by the increase in solubility of iodine in water in the presence of KI which is due to the formation of tri iodide ion : I– + I2 → I 3 –
■ Contain two different types of halogens.
■ Thermal stability of ‘AX’ type of interhalogen compounds is in the order of IF > BrF > ClF > ICl > IBr > BrCl
Group 18
■ Helium was discovered in the chromosphere of sun during the total solar eclipse in 1868. Because it was first discovered near the sun, it was named helium (Helios = Sun).
■ The spectrum of helium is similar to that of sodium, which contains D 1 and D 2 in the yellow region with an D 3 line in the spectrum of He.
■ Rayleigh discovered that nitrogen prepared from air will have more density than the nitrogen prepared by chemical methods.
■ The adsorption capacity of different noble gases on charcoal depends on two points:
i) At low temperature, the adsorption capacity of noble gases increases with increase in the atomic weight.
ii) Adsorption capacity also depends on temperature. Adsorption capacity is inversely proportional to temperature.
■ Helium condenses as helium-I at 4.2 K and Helium-II at 2.2 K. Helium II behaves abnormally like a gas. Generally, liquids flow downwards but helium-II flows upwards against the gravitational force, like a gas. The viscosity of helium-II is less than that of helium-I
■ Clathrate compounds are those formed by the trapping of inert gas atoms in the cavities of crystal lattices of certain organic and inorganic compounds.
■ Helium and neon do not form clathrate com-pounds; because of their small size, they escape.
■ Clathrate compounds are nonstoichiometric compounds.
Exercises
JEE MAIN LEVEL
Level-I
GROUP-15 Elements
Single Option Correct MCQs
1. Which of the following does not show allotropy?
(1) P
(2) As
(3) N
(4) Sb
2. In the 15th group, which of the following elements has the same electronegativity
(1) N and P
(2) P and As
(3) P and Sb
(4) Sb and Bi
3. Which among the following is ionic?
(1) PF 3
(2) BiF 3
(3) NF3
(4) SbF3
4. Moscovium belongs to which of the following group and period in the periodic table?
(1) 16, 7
(2) 15, 7
(3) 15, 5
(4) 16, 6
5. Which one of the following has the lowest boiling point?
(1) NH3
(2) PH 3
(3) AsH3
(4) SbH3
6. Which of the following fluorides does not exist?
(1) SbF5
(2) AsF5
(3) PF 5
(4) NF5
7. The VA group element which exhibits wide range of oxidation states is _____.
(1) P
(2) N
(3) As
(4) Bi
8. Which of the following is Amphoteric oxide?
(1) N2O3
(2) P2O3
(3) As2O3
(4) Bi2O3
9. Which of the following halide is stable?
(1) NF3
(2) NCl3
(3) NBr3
(4) Nl3
10. VA group hydrides are lewis bases due to presence of (1) unpaired electrons.
(2) high electron affinity values.
(3) low electronegativity.
(4) lone pair of electrons.
11. The element which has 18 electrons in its penultimate and anti-penultimate shells is ___.
(1) Sb
(2) As
(3) P
(4) Bi
12. Acidic and paramagnetic oxide of nitrogen is ____.
(1) NO
(2) N2O3
(3) NO2
(4) N2O5
13. Which of the following is a neutral oxide?
(1) N2O
(2) Cl2O7
(3) SO2
(4) Al2O3
14. Fluorapatite is a mineral of ____.
(1) Phosphorus
(2) Arsenic
(3) Nitrogen
(4) Antimony
15. Aquaregia is a mixture of
(1) Conc. HNO3 + Conc. H2SO4 in 1:1 ratio.
(2) Conc. HNO3 + Conc. H2SO4 in 1:3 ratio.
(3) Conc. HNO3 + Conc. HCl in 1:3 ratio.
(4) Conc. HNO3 + Conc. HCl in 1:1 ratio.
16. Non-combustible hydride is _____.
(1) PH 3
(2) AsH3
(3) SbH3
(4) NH3
17. The maximum covalency of nitrogen is ____.
(1) 2
(2) 3
(3) 4
(4) 5
18. Which of the following exists as a diatomic molecule?
(1) N
(2) P
(3) As
(4) Bi
19. Acidic hydride of nitrogen is ____.
(1) NH3
(2) N3H
(3) N2H4
(4) N2H2
20. The element which forms oxides in all the oxidation states from +1 to +5 is _____.
(1) N
(2) P
(3) As
(4) Sb
21. Phosphonic acid is _____.
(1) H3PO3
(2) H3PO2
(3) H3PO4
(4) HPO3
22. Formula of Red Phosphorous is _____.
(1) P 4
(2) (P4)2
(3) (P4)3
(4) (P4)n
23. HNO3 can not be used
(1) for pickling of steel.
(2) etching of metals.
(3) reductant in rocket fuel.
(4) preparation of Nitroglycerin,TNT, and TNB.
24. Nitrolim is _____.
(1) CaCN2+coke
(2) Ca(CN) 2+graphite
(3) CaCN2+graphite
(4) Ca(CN) 2 + coke
25. Ammonia is used as refrigerant because of (1) low latent heat of vapourization.
(2) high latent heat of vapourization.
(3) pleasant smell.
(4) hydrogen bonding capacity.
26. In elementary state, only single covalent bonds are present in _____.
(1) N
(2) O
(3) C
(4) P
27. Thermodynamically most stable form of phosphorous is _____.
(1) White
(2) α-black
(3) β- black
(4) Red
GROUP-16 Elements
Single Option Correct MCQs
28. a k p value is highest for _____.
(1) H2O
(2) H2S
(3) H2Se
(4) H2Te
29. What is the molecular formula of monoclinic sulphur?
(1) S2
(2) S6
(3) S8
(4) S
30. The 16th group element with highest melting point is _____.
(1) Se
(2) Te
(3) Po
(4) S
31. Negative electron gain enthalpy is least for _____.
(1) O
(2) S
(3) Se
(4) Te
32. The hydride of VIA group which does not possess reducing property is ____.
(1) H2Te
(2) H2Se
(3) H2S
(4) H2O
33. In which of the following compounds, oxygen exhibits +2 oxidation state?
(1) H2O
(2) H2O2
(3) OF2
(4) KO2
34. Which one has the lowest atomicity in its elementary state?
(1) Oxygen
(2) Sulfur
(3) Tellurium
(4) Selenium
35. The substance which does not liberate oxygen on treatment with ozone is ____.
(1) PbS
(2) HCl
(3) SO2
(4) Hg
36. Among Group 16 elements, which one does not show –2 oxidation state?
(1) Te
(2) Po
(3) O
(4) Se
37. Sulphur and oxygen exist as
(1) diatomic and polyatomic.
(2) diatomic and diatomic.
(3) polyatomic and diatomic.
(4) polyatomic and polyatomic.
38. Which of the following is solid?
(1) SF4
(2) SeF4
(3) SF6
(4) TeF4
39. Among the following the weakest conjugate base is _____.
(1) SH–
(2) OH–
(3) TeH–
(4) SeH–
40. Which one of the following is not a gas?
(1) SO2
(2) O3
(3) SeF4
(4) SF4
41. The oxidation state of sulphur in H2S is ____.
(1) +2
(2) –2
(3) 1 2
(4) 1 3
42. Anomalous behavior of oxygen is due to its (1) small size and low electronegativity. (2) large size and high electronegativity. (3) large size and low electronegativity.
(4) small size and high electronegativity.
43. Among the following, which is the most volatile compound?
(1) H2S
(2) H2O
(3) H2Se
(4) H2Te
44. Which does not form dichlorides and dibromides is ____.
(1) Se
(2) Te
(3) S
(4) O
45. The most stable allotropic form of sulphur is ____.
(1) rhombic
(2) monoclinic
(3) plastic
(4) milk of sulphur
46. According to VBT, the bond angle of H 2O is ____.
(1) 104° (2) 90°
(3) 107° (4) 109°
GROUP-17 Elements
Single Option Correct MCQs
47. Bond dissociation enthalpy is least in ____.
(1) F2
(2) Cl2
(3) Br2
(4) I2
48. Fluoroapatite is _____.
(1) CaF2
(2) Na3AlF6
(3) 3Ca3(PO4) 2. CaF2
(4) Ca3(PO4)2.3CaF2
49. Which of the following is used in the estimation of carbon monoxide?
(1) I2O5
(2) I2O7
(3) Cl2O7
(4) BrO3
50. Pure chlorine is obtained by
(1) heating PtCl4
(2) heating MnO2 with HCl.
(3) treating bleaching power with HCl.
(4) heating mixture of NaCl and MnO2 with conc. H2SO4.
51. Which of the following having highest electron gain enthalpy?
(1) Br (2) Cl
(3) I (4) F
52. Which of the following is not disproportinate?
(1) ClO
(2) 4 ClO
(3) ClO3
(4) 1 ClO2
53. Which of the following is incorrect match?
(1) F2 = Yellow (2) Cl2 = colourless
(3) Br2 = Red (4) I2 = Violet
54. Among the following, the weakest acid is ____.
(1) HF
(2) HCl
(3) HBr
(4) Hl
55. The bond dissociation energy is highest for ____.
(1) Cl2
(2) I2
(3) Br2
(4) F2
56. 0.2% of sodium iodate is present in ____.
(1) KNO3
(2) Na3AlF6
(3) NaNO3
(4) NaCl
57. Best reducing agent is ______.
(1) HI
(2) HBr
(3) HCl
(4) HF
58. The highest electronegative element is ____.
(1) carbon
(2) oxygen
(3) fluorine
(4) nitrogen
59. Which of the following attacks on glass?
(1) HCl
(2) Hl
(3) HBr
(4) HF
60. Silica can react with ______.
(1) HCl
(2) O2
(3) HF
(4) Cl2
61. Paramagnetic oxide of chlorine is ______.
(1) ClO3
(2) Cl2O6
(3) Cl2O
(4) Cl2O7
62. Which acid can combine with its own salt?
(1) HF
(2) HBr
(3) HCl
(4) Hl
63. Chemical formula of Fluorspar is ______.
(1) AgF (2) CaF2
(3) Na3AlF6
(4) NaF
64. Fluorine reacts with water to give (1) HF and O2.
(2) HF and OF2 (3) HF and O3 (4) HF, O2 and O3
65. ClO3 is the mixed anhydride of (1) HClO2 and HClO3 (2) HClO3 and HClO4 (3) HClO2 and HClO4. (4) HClO2 and HClO.
66. Hydrochloric acid at 25 oC is ______. (1) Ionic and liquid (2) Covalent and gas
(3) Ionic and gas (4) Covalent and solid
67. What is the oxidation state of chlorine in hypochlorous acid?
(1) +7
(2) +5
(3) +3
(4) +1
68. Which of the following reaction does not takes place.
(1) 22 F2Br2FBr +→+
(2) 22 Cl2I2ClI +→+
(3) 22 Br2F2BrF +→+
(4) 22 Cl2Br2ClBr +→+
69. The halogen that undergoes sublimation is ______.
(1) F2
(2) Cl2
(3) Br2
(4) I2
70. Superhalogen is ______.
(1) F2
(2) Cl2
(3) Br2
(4) I2
71. The following is not a mineral of chlorine?
(1) Carnalite
(2) Horn silver
(3) Sylvine
(4) Cryolite
72. T-shaped interhalogen is ______.
(1) ICl
(2) ClF3
(3) BrF 5
(4) IF 7
73. When Cl2 reacts with Fe (or) Cu it forms
(1) Fe2O3
(2) Cu2O.
(3) FeCl3 and CuCl2.
(4) FeCl2 and CuCl.
GROUP-18 Elements
Single Option Correct MCQs
74. 143K 46 XeFXeF, → which is the most suitable reagent for the above conversion?
(1) F2
(2) O2F2
(3) OF2
(4) ClF3
75. IP1 of the following is least in ______.
(1) He
(2) Xe
(3) Ne
(4) Kr
76. What is the approximate mole fraction of inert gases in air?
(1) 0. 2
(2) 0. 02
(3) 0. 01
(4) 0. 005
77. The first compound of a noble gas known is ______.
(1) Xe. 6H2O
(2) Xe. [PtF6]
(3) XeO3
(4) XeF6
78. The 18th group element that does not occur in the atmosphere is ______.
(1) He
(2) Kr
(3) Ne
(4) Rn
79. Components of gaseous mixture useful for sea divers are (1) O2 and He.
(2) O2 and H2
(3) O2 and N2.
(4) O2 and CO2
80. Which noble gas has least tendency to form compounds?
(1) Ne
(2) Ar
(3) Kr
(4) He
81. Which one has the highest boiling point?
(1) Ne
(2) Kr
(3) Xe
(4) He
82. Welding of Mg can be done in an atmosphere of ______.
(1) Xe (2) Ar
(3) He (4) O2
83. What is the atomic number (Z) of the noble gas that reacts with fluorine?
(1) 10 (2) 54
(3) 18 (4) 2
84. The noble gas with highest value of positive electron gain enthalpy is ______.
(1) He
(2) Ne
(3) Rn
(4) Xe
85. The most polarisable noble gas is ______.
(1) Kr
(2) Ne
(3) Xe
(4) Ar
86. 67 XeFMFMXeF +→+ . Here ‘M’ is ______.
(1) Alkali metals
(2) Alkaline earth metals
(3) Transition metals
(4) Inner transition metals
87. Which of the following is more volatile?
(1) He
(2) Xe
(3) Kr
(4) Ne
88. Which has the highest first ionization potential value?
(1) He
(2) Nr
(3) Kr
(4) Xe
89. Which of the following molecule has only one lone pair of electrons on the central atom?
(1) XeF2
(2) XeF4
(3) XeO3
(4) XeO4
90. The percentage by volume of argon in atmosphere is nearly ______.
(1) 1%
(2) 2%
(3) 10%
(4) 0. 2%
91. Which noble gas is generally used in safety devices for protecting electrical instruments?
(1) He
(2) Ne
(3) Ar
(4) Kr
92. Argon is used in arc welding because of its (1) low reactivity with metal.
(2) ability to lower the melting point of metal.
(3) flammability.
(4) high calorific value.
93. Which of the following is the life saving mixture for an asthma patient?
(1) Mixture of helium and oxygen
(2) Mixture of helium and nitrogen
(3) Mixture of xenon and oxygen
(4) Mixture of xenon and nitrogen
94. The compound which will not exist is ______.
(1) XeF2 (2) XeF4
(3) XeF6 (4) XeF8
95. The oxidation state of ‘Xe’ in Xe[PtF 6] is ______.
(1) +1 (2) +2
(3) +4 (4) +6
96. Most reactive inert gas ____.
(1) He (2) Ne
(3) Xe (4) Kr
97. Noble gases exists only in monoatomic state. This is due to
(1) non availability of unpaired electrons.
(2) high ionization energy.
(3) large size.
(4) zero electron affinity.
98. Noble gases are only sparingly soluble in water due to
(1) dipole - dipole interactions.
(2) induced dipole- induced dipole interactions.
(3) dipole-induced dipole interactions.
(4) hydrogen bonding.
99. Helium is used in gas balloons instead of hydrogen, because (1) it is monoatomic.
(2) it is lighter.
(3) it is not radioactive.
(4) it is non - combustible.
100. The electronic configuration of 1s2 2s22p53s1 shows
(1) ground state of fluorine atom.
(2) excited state of fluorine atom.
(3) excited state of neon atom.
(4) excited state of ion O2 –.
Level-II
GROUP-15 Elements
Single Option Correct MCQs
1. Which of the following statements is incorrect?
(1) N-N bond length in N2H4 is greater than N-N bond length in N2F4.
(2) In PCl5 the P-Cl axial bond length is greater than P–Cl equatorial bond length.
(3) In SF6, all S-F bonds have equal bond length.
(4) In PCl3F2, the P-F bond length is greater than P-Cl bond length.
2. One of the acids listed below is formed only from P2O3; the rest are formed from P2O5. The acid formed from P2O3 is
(1) HPO3
(2) H 4P2O7
(3) H3PO4
(4) H3PO3
3. Which of the following compounds doesn't give nitrogen on heating?
(1) NaN3
(2) (NH4)2SO4
(3) NH4NO2
(4) (NH4)2Cr2O7
4. Which one of the following is the most acidic oxide?
(1) N2O3 (2) N2O5
(3) P4O6 (4) P4O10
5. Which of the following has maximum complex forming ability with a given metal ion?
(1) PH 3 (2) BiH 3
(3) NH3 (4) SbH3
6. The reaction of white phosphorus with aqueous NaOH gives phosphine along with another phosphorus containing compound. The reaction type and the oxidation states of phosphorus in phosphine and the other product are, respectively,
(1) Redox reaction, -3, and -5
(2) Redox reaction, +3, and +5
(3) Disproportionation reaction,–3, and +1
(4) Disproportionation reaction,–3, and +3
7. Which one of the following fluorides does not exist?
(1) NF5
(2) PF 5
(3) AsF5
(4) SbF5
GROUP-16 Elements
Single Option Correct MCQs
8. The acidic character of dioxides of members of oxygen family decreases in the order
The type of hybridisation in the central atom of ‘tx’ is
(1) sp3 (2) sp3d
(3) sp3d2 (4) sp2
10. Which does not give oxygen on heating?
(1) Zn(ClO3)2 (2) K2Cr2O7
(3) (NH4)2Cr2O7 (4) KClO3
11. Identify the correct statements among the following.
A) S8 ring has a crown shape.
B) The ∠ S–S–S bond angles in S8 and S6 rings are the same.
C) S2 (vapour) is paramagnetic like oxygen gas.
D) Rhombic and monoclinic sulphur have S8 molecules.
(1) A only
(2) A and B only
(3) A, C, and D
(4) A, B, C, and D
GROUP-17 Elements
Single Option Correct MCQs
12. The order of electron gain enthalpy among the following is (in magnitude)
(1) Cl > F > Br > I
(2) F > Cl > Br > I
(3) Br > Cl > F > I
(4) Cl > Br > F > I
13. Iodine reacts with concentrated HNO3 to yield 'Y' along with other products. The oxidation state of iodine in 'Y' is (1) 5 (2) 7
(3) 3 (4) 1
14. Which of the following halide ions is the strongest conjugate base?
(1) F–
(2) Cl–
(3) Br–(4) l–
GROUP-18 Elements
Single Option Correct MCQs
15. Inert gases have positive electron gain enthalpy. Their correct order is
(1) He<Ne<Kr<Xe
(2) He<Xe<Kr<Ne
(3) He<Kr<Xe<Ne (4) Xe<Kr<Ne<He
16. Which of the following is a redox reaction? (1) XeF6+H2O→ (2) XeF6+SbF5→ (3) XeF4+H2O→ (4) XeF6+CsF→
Multiple concept Questions
Single Option Correct MCQs
17. 422 432 NHNOAHO; NHNOBHO ∆ ∆ →+ →+
Then, (1) A=B=N2
(2) A=B=N2O
(3) A=N2O; B=N2O (4) A=N2; B=N2O
18. ()()() Hot/conc 22 2g aq 2 ClBaOH XBaClHO+→++
X+H2SO4→Y+BaSO4 heated to dryness 22 Y ZHOO→++
Y and Z are respectively, (1) HClO4, ClO2 (2) HClO3, Cl2O6 (3) HClO3, Cl2O6 (4) HClO4, Cl2O7
19. Which of the following are peroxoacids of sulphur?
(1) H2SO5 and H2S2O8 (2) H2SO5 and H2S2O7 (3) H2S2O7 and H2SO4 (4) H2S2O7 and H2S2O6
20. C o mpound A undergoes hydrolysis to pr oduce a colourless gas with rotten fish smell. The gas gives a vortex ring. The gas is
(1) PH 3 (2) P2O3 (3) P2O5 (4) P2S3
Numerical Value Questions
21. HF is a weak acid but on addition of SbF5, it becomes a very strong acid. The number of 90° angles in the anionic part of the product is ___.
22. The total number of diprotic acids among the following is H3PO4,H2SO4,H3PO3,H2S2O7,H3BO3, H3PO2,H2CrO4,H2SO3,H2CO3
23. How many of the following oxides have higher bond angle than F 2O?
H2O,Cl2O,ClO2,Br2O
Level-III
1. Chlorine on reaction with iodate ion in basic medium , according to the given equation: IO3 + x OH + Cl2 → (A) + yH2O + zCl . x, y, z are stoichiometric coefficients in balanced equations] Compound (A) on
heating in acidic medium undergoes, ()() 00 2 100C200C 4HO C AB→→
Identify the compound (C)?
(1) I2O3
(2) I2
(3) I2O7
(4) I2O5
2. ()()() i1HHi2 2 ggg SSS; ∆+∆+→→
The total enthalpy change for the process is 3200kJ/Mole.
()() gg SeS; +−+→
The enthalpy change for the process is –1000 kJ/Mole. Then ∆iH2
(1) Equal to 2∆iH1
(2) Equal to –2∆iH1
(3) Greater than 2∆iH1
(4) Lesser than 2∆iH1
3. In the scheme given below, X and Y, respectively, are () aq.NaOH Metal halide Whiteparticipate(P)Filtrate Q →+
P () 2 PbOexcess heat → X (a coloured species in solution)
Q () 2 24 MnO(OH) Conc. HSO Warm Y gives blue coloration with KIstarch paper →
(1) 2 42 CrOandBr
(2) 2 42 MnOandCl
(3) 42 MnOandCl
(4) MnSO4 and HOCl
4. 2 HO 22 S2ClXYHCl;YZHO ∆ +→→+→+ oxidation state of S in Z is (1) +4
(2) +2
(3) +6
(4) +1
5. A greenish yellow gas reacts with an alkali metal hydroxide to form a halate which can be used in fireworks and safety matches. The gas and halate respectively are
(1) Br2 and KBrO3
(2) Cl2 and KClO3
(3) I2 and NaIO3
(4) Cl2 and NaClO3
6. A metal X on heating in nitrogen gives Y. Y on treatment with H2O gives a colourless gas which when passed through sodium hypochlorite liberate N2 gas. The compound Y is
(1) Mg(NO3)2
(2) Mg3N2
(3) NH3
(4) MgO
7. If two litre of air is passed repeatedly over heated copper and heated Mg till no further reduction in volume takes place, the volume finally obtained will be approximately
(1) 200mL
(2) 20mL
(3) zero
(4) 10mL
8. In the following reaction sequence in aqueous solution, the species X, Y and Z, respectively, are S2O32– Ag+ Clear solution White precipitate black precipitate Ag+ with time X Y Z
(1) () 3 23 2232 2 AgSO,AgSO,AgS
(2) () 5 23 232 3 AgSO,AgSO,AgS
(3) () 3 3 223 2 AgSO,AgSO,Ag
(4) () 3 324 3 AgSO,AgSO,Ag
9. Consider the following reaction: 425 HClOPOAB +→+
Product A is an acidic oxide. Select correct statement(s):
(1) B is oxy acid of phosphorous
(2) B undergoes cyclisation for stability
(3) In product A, at least one element is present in its lowest oxidation state
(4) Product A has six equivalent bonds
10. Oxyacid(s) of phosphorus has the following properties. Complete neutralisation of the acid with sodium hydroxide solution gives an aqueous solution of sodium ions and oxyacid anions in the ratio 2:1. When a solution of the acid is warmed with silver nitrate solution metallic silver is deposited. What is/are the structure(s) of the oxyacid(s)?
11. The inter halogen compound formed when Uranium (solid) is treated with ClF3 (liquid) is X then the incorrect statement regarding X is
(1) X is a solid at room temperature
(2) X is a colourless gas at room temperature
(3) X upon hydrolysis produces an oxoacid in which the oxidation state of halogen is +3
(4) X consist of total 11 lone pairs
12. A certain element X was belong to 6 th group found to form three compounds with Flourine having the formula XF2,XF4
(1) HO — P — OH
(2)
(3)
(4)
and XF6. One of its oxide has the formula XO3 and X reacts with sodium to form the compound Na2X.
(1) Among 24 XF,XF XF6 and XO3 XF2 and XF 4 are polar.
(2) Some hybrid orbitals of XF4 have zero s-character
(3) XF6 has lesser bond angle than XO 3
(4) All the four XF2, XF4, XF6 and XO3 have non planar electron domain geometry
13. Amount (in grams) of product obtained when 6.4 grams of sulphur dioxide reacts with 7.1 grams of chlorine in presence of charcoal catalyst is ______ (AW: S = 32, O = 16, Cl = 35.5)
14. Nitric oxide reacts with ozone in 1:1 stoichiometric ratio in upper atmosphere to form reddish brown gas mixture. How many number of moles of unpaired electrons in the gas mixture, after the reaction if one mole of nitric oxide is used up for completing the reaction. Consider all species electronically in ground state.
15. In how many of the following reactions XeOF4 is formed as one other product?
i. 1:1ratio
62 XeFHO+→
ii. 2:1ratio
63 XeFXeO+→
iii. 2:1ratio
62 XeFSiO+→
iv. 42 XeFHO+→
v. 22 XeFHO+→
vi. 3 XeONaOH+→
16. Under ambient conditions, the total number of gases released as products in the final step of the reaction scheme shown below is
XeF6 P + other product products slow disproportionation in OH –/H2O OH –/H2O Q Complete hydrolysis
17. The total number of compounds having atleast one bridging oxo group among the molecules given below is ______
23254647425 O,NO,PO,PO,HPO, N 10 2
33223225227 HPO,HSO,HSO,HSO
18. Xenon trioxide reacts with NaOH in two steps forming a compound B
3 XeONaOHA +→
22 2A2NaOHB2HOXeO +→+++
The number of pπ−dπ bonds present in B is
19. At 143 K the reaction of XeF 4 with O 2 F 2 produces a xenon compound Y and another product Z. The total number of lone pairs of electrons in Y + number of unpaired electrons in Z is equal to _____
20. The concentration of ozone rises above about X ppm, breathing becomes uncomfortable resulting in headache and nausea. What is value of X 10
21. 3 XeONaOHA +→
2A2NaOHBXeO2HO +→+++
22
The bond order in, B is
22. KClO 3 exhibit different kinds of decomposition at low and high temperatures respectively. Let us assume that at a particular temperature it exhibits both kinds of decomposition by the participation of KClO 3 in equimolar proportion. Find out the no. of moles of KCl produced by decomposition of eight moles of KClO 3 by the above said process.
23. Among the following the no of correct equations are
A. 7626 2CsXeFXeFCsXeF →+
B. 4 622 2HXeO2OHXeOXeO2HO ++ +→+++
C. 34 XeONaOHNaHXeO +→+
D. 6 32 H XeO XeOO (rapid decomposition) water + + →+
E. 44 XeFPtPtFXe +→+
F. 422 XeF4NOXe4NOF +→+
G. 42262 XeFOFXeFO +→+
H. 2 22 1 XeF2NaOHXeO2NaFHO (slow) 2 +→+++
I. 22 XeF2HO2Xe4HFO2 (fast) +→++
J. 24 873k Xe2F XeF 7bar/Ni tube +→
24. In an attempt to establish the formula of an oxide of nitrogen, a known volume of the pure gas was mixed with hydrogen and passed over a catalyst at a suitable temperature. Conversion of the oxide to ammonia and water was to have taken place.
()2g H,catalyst xy 32 NO xNHyHO→+
2400 cm3 of the nitrogen oxide, measured at room temperature and pressure produced 7.20g of water. The ammonia produced was
THEORY-BASED QUESTIONS
Statement Type Questions
Each question has two statements, statement I and statement II. In light of the given statements. Mark the correct answer as
(1) if both statement I and statement II are correct,
(2) if both statement I and statement II are incorrect,
(3) if statement I is correct but statement II is incorrect.
(4) if statement I is incorrect but statement II is correct.
1. S-I : Both H 3PO 3 and H 3PO 4 have same number of hydrogen atoms but H 3PO 4 is a tribasic acid and H 3PO 3 is a dibasic acid.
neutralized by 200 cm3 of 1.0 mol dm3 HCl. What is the oxidation number of the nitrogen in the nitrogen oxide?
25. The number of correct statements about H3PO4, H3PO2 and H3PO3
i) All are weaker acids than HCl
ii) In all the three acids, the central atom is sp2 hybridized
iii) The order of Ka1 of H3PO2 > H3PO3 > H3PO4
iv) The basicities of H 3 PO 4 , H 3 PO 3 and H3PO2 are 3, 2 and 1 respectively.
26. Gaseous xenon fluoride can be prepared by shining light on a mixture of Xe and F 2 gases. Assume that xenon gas was added to a 0.25 L flask until its pressure was 0.12 atm at 0.0 °C. Fluorine gas was then added until the total pressure becomes 0.36 atm at 0.0 °C. After the reaction was complete, the xenon had been consumed completely and the total pressure of the gases remaining in the flask was still 0.24 atm at 0°C. The formula of compound prepared from Xe and F2 is XeF2n then n is_____
S-II : 1 mole of H 3 PO 3 is neutralised by 2 moles of NaOH, while 1 mole of H 3PO 4 is neutralised by 3 moles of NaOH.
2. S-I : SbCl5 is more covalent than SbCl 3.
S-II : The higher oxides of halogens also tend to be more stable than the lower ones.
3. S-I : Elements of group 16 generally show lower value of first ionisation enthalpy compared to corresponding periods of group 15.
S-II : Pure ozone is a pale blue gas, dark blue solid, and violet-black liquid.
Assertion and Reason Questions
In each of the following questions, a statement of Assertion (A) is given, followed by a corresponding statement of Reason (R). Mark the correct answer as
(1) if both (A) and (R) are true and (R) is the correct explanation of (A),
(2) if both (A) and (R) are true but (R) is not the correct explanation of (A),
(3) if (A) is true but (R) is false,
(4) if both (A)and (R) are false.
JEE ADVANCED LEVEL
Multiple Option Correct MCQs
1. Whic h of the following molecules has a dative bonding (dπ–pπ)?
(1) P4O10
(2) (SiH3)3 N
(3) P4O6
(4) N2O5
2. Which halide of sulphur undergoes disproportionation in water?
(1) SCl4
(2) S2F2
(3) S2Cl2
(4) SF6
3. Regarding S8 molecule choose the correct statements.
(1) Sulphur atom is involved in sp 3 hybridisation
(2) Sulphur exhibits enantiotropic allotropy.
(3) Four sulphur atoms are in one plane and the other four sulphur atoms are in another plane.
(4) The valency and the covalency of sulphur are 1.
4. (A) : NCl3, on hydrolysis, gives NH 3.
(R) : PCl3, on hydrolysis, gives PH 3
5. (A) : Flourine forms one oxoacid.
(R) : Flourine has smallest size amongst all halogens and is highly electronegative.
6. (A) : Xenon tetrafluoride molecule is denoted as AB4 E4, where E is a lone pair.
(R) : All AB 4 type molecules have no dipole moment, because of symmetrical tetrahedral structure.
4. A mixture of conc. HCl and HNO3 made in 3 : 1 ratio contains
(1) NOCl (2) NCl3
(3) Cl2 (4) ClO2
5. Out of the following, which is incorrectly indicated order?
(1) Cl2>Br2>I2>F2 : Bond dissociating energy
(2) HF<HCl<HBr<HI : Boiling point
(3) HF>HCl>HBr>HI : Bond dissociation energy
(4) I2>Br2>Cl2>F2 : Oxidising property
6. An element (E)+NaOH→ disproportionation products onacidification (E) → is again produced.
(1) Cl2 and Br2 (2) F2 and Br2
(3) P and F2 (4) Br2 and I2
7. Based on the compounds of group 15 elements, choose the correct statement(s).
(1) Bi2O5 is more basic than N2O5
(2) NF3 is more covalent than BiF 3.
(3) PH 3 boils at lower temperature than NH3
(4) The N-N single bond is stronger than the p–p single bond.
Numerical Value Questions
8. Reaction of Br 2 with Na 2 CO 3 in aqueous solution gives sodium bromide and sodium bromate with evolution of CO2. The number of sodium bromide molecules involved in the balanced chemical equation is ______.
9. The action of concentrated H2SO4 on urea, NH2CONH2, results in the production of a white crystalline solid X. This is a monobasic acid, which, on treating with nitrite ion and dil. HCl, liberates one mole of N2, and on addition of aq. BaCl2, the resulting solution yields one mole of BaSO4 per mole of X. The total number of atoms in X is _____.
10. When ClO 2 is dissolved in alkali, two products are formed. The difference in the oxidation state of Cl in products is _____
11. In 1896, a chemist reported that by combining potassium hexafluoromanganite (IV) K2MnF6 with SbF5 at 150 °C, elemental fluorine is generated. To get one mole of fluorine gas, the total number of reactant molecules that should be taken is ______.
12. XeF4 reacts with SbF5 to form [XeFm]n+[SbFy]z–. Then, m+n+y+z = ______
Integer Value Questions
13. How many of the following reactions would give HCl as one of the products?
(a) CH4+Cl2→
(b) FeSO4+H2SO4+Cl2→
(c) I2+H2O+Cl2→
(d) H2O+SO2+Cl2→
(e) H2O+Cl2→
(f) SO3+Cl2→
(g) NaCl(aq) Electrolysis →
(h) Cl2O7+H2O→
(i) Cl2+NaOH(conc)→
14. Among the following properties, how many increase down the group for halogens?
(i) Chemical reactivity of halogens
(ii) Reducing character of halogens
(iii) Melting point of halogens
(iv) Covalent character of metal halides
(v) Dissociation enthalpy of H-X bond
(vi) pka of aqueous HX solutions
(vii) Boiling point of halogens
15. How many of the following ammonium salts will evolve N2 gas on heating?
16. Number of compounds that can be reduced by PH3 is (either anhydrous or in aqueous solution) ______.
CuSO4, AgNO3, AlCl3, CuCl2, N2O,Cl2
17. How many of the following reactions yield H3PO4 as one of the products?
i) Ca3(PO4)2+H2SO4→
ii) P4O6+H2O→
iii) PCl5+H2O→
iv) P4S10+H2O→
v) H3PO2+ N2Cl–+
Passage–based Questions
(Q:18–19)
Upon heating KClO3 in the presence of catalytic amount of MnO 2 a gas 'W' is formed. Excess amount of 'W' reacts with white phosphruos to give 'X'. The reaction of 'X' with pure HNO 3 gives 'Y' and 'Z'.
18. 'W' consists 'A' number of oxygen atoms. 'X' consists 'B' number of oxygen atoms.
Then, () () 3A×B B-A is equal to P . Find the value of 10P.
19. ‘y’ molecule central atom oxidation number equal to 'M'.
‘Z’ molecule central atom oxidation number equal to 'N'.
Then, M×N M+N is equal to R. Find the value of 2R.
(Q:20–21)
An aqueous solution of [A] gives pale yellow precipitate with aq. AgNO3 solution. Compound [C] in concentrated form is used as a drying agent for Cl2 gas. The equivalent weight of Br2 is X. (atomic weight of Br = 80)
Hot con. NaOH A + B [C] Br2 75% yield
20. Calcu late the value of X . 5 (Round off to nearest integer)
21. Calculate the weight of compound ‘B’ produced when 0.1 mole of Br2 is taken in the reaction.
(Round off to nearest integer)
(Q:22–23)
Halogens react with each other to form a number of compounds called interhalogen compounds. Their general formula is AXn, where A is less electronegative halogen while X is a more electronegative halogen, and n is its number. The interhalogen compounds are essentially
polar. The reactions of interhalogens are similar to those of halogens.
22. Among halogens and interhalogen compounds, the total number of electrons in the most reactive species is N. Calculate the value of N/9.
23. The number of orbitals involved in the hybridisation of central atom of IF 5 is X. The number of lone pairs of electrons in the central atom of ClF 3 molecule is Y. The atomic weight of the element showing maximum oxidation state in interhalogen compounds is Z. Calculate the value of Y+Z 10
(Q:24–25)
White–P4, on reaction with SO2Cl2, forms a compound (A) and an acidic gas (B). ‘A’ can react with NH4Cl to form cyclophosphazenes (PNCl 2) n( n =3,4,5..etc.)'A', on reaction with NH4Cl, one such cyclic compound ‘C’, with n=3. Then, answer the following questions.
24. Number of pπ−dπ bonds in cyclic compound C is ____.
25. Total number of lone pairs in acidic gas ‘B’ is ________.
(Q:26–27)
Sulphur containing the compound ‘P’ on boiling with sulphur, gives an additional compound ‘Q’. ‘Q’, on reaction with PbCl 2, forms a lead compound. This lead compound on further boiling, gives one more lead compound ‘R’. Same compound ‘Q’, on reaction with cupric salt (CuCl2), forms, at the end a complex salt ‘S’.
26. The oxidation number of sulphur in lead compound ‘R’ is
(1) +4
(2) +6
(3) 2.5
(4) –2
27. The number of copper atoms in the complex salt ‘S’ is
(1) 2 (2) 5 (3) 6 (4) 1
Matrix Matching Questions
28. Match Column-I (reaction) with Column-II (products) and choose the correct option.
Column - I
Column - II
(A) 4NH3 + 5O2 (p) PbO+NO2+H2O
(B) P4+3NaOH+3H2O (q) N2O+ H2O
(C) NH4NO3 (r) 3NaH2PO2+PH3
(D) Pb(NO3)2 (s) 4NO + 6H2O
Choose the correct option.
(A) (B) (C) (D)
(1) s r q p
(2) q p s r
(3) r s p q
(4) p q s r
29. Match Column-I (xenon compound) with Column-II (shape of molecule).
Column - I
Column - II
(A) XeF4 (p) Distorted octahedral
(B) XeF6 (q) Tetrahedral
(C) XeO3 (r) Square planar
(D) XeO4 (s) Pyramidal
Choose the correct option.
(A) (B) (C) (D)
(1) r q p s
(2) r p s q
(3) p s q r
(4) s q r p
30. Match Column-I (oxide)with Column-II (type of bond)
Column-I (Oxide) Column-II (Type of Bond)
(A) N2O4 (i) 1N=O bond
(B) NO2 (ii) 1N–O–N bond
(C) N2O5 (iii) 1N–N bond
(D) N2O (iv) 1N=N/N ≡ N bond
Choose the correct answer from the options given below.
(A) (B) (C) (D)
(1) ii i iii iv
(2) iii i ii iv
(3) ii iv iii i
(4) iii i iv ii
31. Match Column-I (Reaction) with Column-II (catalyst).
Choose the correct answers from the options given below:
(A) (B) (C) (D)
(1) ii iii i iv
(2) iii ii i iv
(3) iii iv ii i
(4) iii ii iv i
BRAIN TEASERS
1. S S+ ONa ∆ ONa O +2HCl 2NaCl H2O + SO2 + S +
(S =Radio isotope of sulphur)
Choose the correct option.
(1) S appears only in precipitate.
(2) S appears in SO2.
(3) S appears in both SO 2 and S due to exchange.
(4) Non-radioactive S appears only in precipitate.
2. All the halogens are coloured. The colours arise due to
(1) weak van der Waals forces between the halogen molecules
(2) strong oxidising power of the halogens
(3) absorption of light in the visible region due to excitation of electron from HOMO to LUMO
(4) emission of light due to transfer of an electron from a higher state to the ground state
3. In the scheme given below, X and Y, respectively, are
Metal halide aq.NaOH → white precipitate (P)+filtrate(Q)
24 2 aq.HSO PbO(excess) PX → (a coloured species in solution)
2 24 MnO(OH) Conc.HSO Warm QY → (gives blue colouration with KI–starch paper)
(1) 242 CrOandBr
(2) 242 MnOandCl
(3)42 MnOandCl
(4) MnSO4 and HOCl
4. A greenish yellow gas reacts with an alkali metal hydroxide to form a halate that can be
used in fireworks and safety matches. The gas and halate, respecively, are
(1) Br2 KBrO3
(2) Cl2, KClO3
(3) l2, NalO3
(4) Cl2, NaClO3
5. How many of the following reactions evolve a paramagnetic gas?
(a) Cu+dil HNO3→
(b) Zn+dil HNO3→
(c) Pb+ dil HNO3→
(d) Zn+NaOH→
(e) (NH4)2SO4+NaOH
(f) KO2+H2O→
6. Which is the correct combination for gases released from Kipp’s apparatus and reagents taken?
(1) FeS+dil.H2SO4→H2S
(2) Cu+dil.HNO3→NO
(3) Mg3N2+H2O→NH3
(4) MnO2+HCl→Cl2
7. A potassium salt (F) gives a brown coloured gas with dil.HCl in presence of air and forms a dark brown coloured ring (G) with ferrous sulphate and acetic acid at the junction of liquids. The salt (F) decolourises the acidified KMnO4 solution. Choose the correct option(s) for (F) and (G).
(1) Addition of potassium iodide solution to the salt solution (F), followed by freshly prepared starch solution, and acidification with acetic acid produces blue colour.
(2) (G) is paramagnetic complex Fe(H2O)5(NO2)]2+
(3) Addition of acidified Co(NO3)2 to the salt solution (F) produces a yellow coloured low spin complex.
(4) Anion of F can be removed using thiourea and NH2SO3H in acidic medium during de tection of NO3 ion using Brown ring test.
FLASHBACK (Previous JEE Questions)
JEE Main
1. The number of bridged oxygen atoms present in compound B formed from the following reactions is (2022)
(1) 0 (2) 1
(3) 2 (4) 3
2. Consider the following reaction :
A+alkali→B(Major Product) (2022)
If B is an oxoacid of phosphorus with no P–H bond, then A is
(1) White P4 (2) Red P4 (3) P2O3 (4) H2PO3
3. The conversion of hydroxyapatite occurs due to presence of F– ions in water. The correct formula of hydroxyapatite is (2021)
(1) [3Ca3(PO4)2Ca(OH)2]
(2) [3Ca(OH)2CaF2]
(3) [Ca3(PO4)2CaF2]
(4) [3Ca3(PO4)2CaF2]
4. On heating, compound (A) gives a gas (B), which is a constituent of air. This gas,
CHAPTER TEST – JEE MAIN
Section-A
1. Which of the following are correct stability orders
a) Pb2+> Pb4+,Tl+ >Tl3+
b) Bi3+ < Sb3+; Sn2+<Sn4+
c) Pb2+> Pb4+, Bi2+ >Bi5+
d) Sn2+<Pb2+, Sn4+>Pb4+
e) Sn2+<Pb2+, Sn4+<Pb4+
f) Tl3+<In3+, Sn2+>Sn4+
(1) c and f (2) a, c, and d
(3) c and e (4) b and d
when treated with H 2 in the presence of a catalyst gives another gas (C), which is basic in nature. (A) should not be (2020)
(1) (NH4)2Cr2O7 (2) NaN3
(3) Pb(NO3)2 (4) NH4NO2
JEE Advanced
5. Dissolving 1.24 g of white phosphorus in boiling NaOH solution in an inert atmosphere gives a gas Q. The amount of CuSO 4 (in g) required to completely consume the gas Q is _______. (2022)
[Given: Atomic mass of H = 1, O = 16, Na = 23, P = 31, S = 32, Cu = 63](Round off to nearest integer)
6. With reference to aqua regia, choose the correct option(s). (2019)
(1) Reaction of gold with aqua regia produces NO2 in the absence of air.
(2) Reaction of gold with aqua regia produces an anion having Au in +3 oxidation state.
(3) Aqua regia is prepared by mixing conc. HCl and conc.HNO3 in 3 : 1(v/v) ratio.
(4) The yellow colour of aqua regia is due to the presence of NOCl and Cl 2
2. Given below are two statements.
Statement I : Chlorine can easily combine with oxygen to form oxides; and the product has a tendency to explode.
Statement II : Chemical reactivity of an element can be determined by its reaction with oxygen and halogens.
In light of the above statements, choose the correct answer from the options given below
(1) Both statement I and statement II are incorrect.
(2) Both statement I and statement II are correct.
(3) Statement I is incorrect but statement II is correct.
(4) Statement I is correct but statement II is incorrect.
3. Which of the following reactions is not spontaneous?
(1) 2F2(g)+2H2O(l)→4H+ (aq) +4F– (aq) +O2(g)
(2) Cl2(g)+H2O(l)→HCl (aq) +HOCl (aq)
(3) Br2(g)+H2O(l)→HBr (aq) +HOBr (aq)
(4) 2I2(s)+2H2O(l)→4I– (aq) +4H+(aq)+O2(g)
4. For electron gain enthalpies of the elements denoted as ∆ eg H, the incorrect option is
(1) ∆ eg H(Se)< ∆ eg H(S)
(2) ∆ eg H(I)< ∆ eg H(At)
(3) ∆ eg H (Te)< ∆ eg H(Po)
(4) ∆ eg H(Cl)< ∆ eg H(F)
5. 2S(fused)+Cl2→A
A+H2O → HCl+B+C
A, B, C in the above equation, respectively, are:
(1) S2 Cl2, SO2, S
(2) SCl2, SO2, SO3
(3) S2Cl2, SO3, SO2–4
(4) S2Cl2, SO3, H2SO4
6. Concentrated hydrochloric acid, when kept in open air, sometimes produces a cloud of white fumes. The explanation for it is that
(1) concentrated hydrochloric acid emits strongly smelling HCl gas all the time
(2) oxygen in air reacts with the emitted HCl gas to form a cloud of chlorine gas
(3) strong affinity of HCl gas for moisture in air results in forming of droplets of liquid solution, which appears like a cloudy smoke
(4) due to strong affinity for water, concentrated hydrochloric acid pulls moisture of air towards itself forming water droplets that convert into fumes
7. Which of the following statements is correct for the iodine molecule?
(1) In liquid state, it conducts electricity, though very slightly, due to selfionisation:
3I2 I+ 3 + I–3
(2) Its conductivity increases with the rise in temperature.
(3) It behaves like an intrinsic semiconductor.
(4) 1, 2, and 3
8. When Cu2+ ion is treated with KI, a white precipitate, X appears in solution. The solution is titrated with sodium thiosulphate. The compound Y is formed. X and Y, respectively, are
(1) X=Cu2I2, Y=Na2S4O6
(2) X=CuI2, Y=Na2S4O6
(3) X=CuI2, Y=Na2S4O3
(4) X=Cu2I2, Y=Na2S4O5
9. In gaseous state, PCl5 has two types of bonds. They are axial and equatorial bonds. Their bond lengths are, respectively,
(1) 202 pm, 240 pm
(2) 240 pm, 202 pm
(3) 202 pm, 202 pm
(4) 150 pm, 160pm
10. O–O bond length in H2O2 is X than the O–O bond length in F2O2.The O–H bond length in H2O2 is Y than that of the O–F bond in F2O2. Choose the correct option for X and Y from those given below.
(1) X-longer, Y-shorter
(2) X-shorter, Y-longer
(3) X-longer, Y-longer
(4) X-shorter, Y-shorter
11. When SO2 gas is passed into chlorine water, the product(s) formed is / are
(1) SCl2 and S2Cl2
(2) H2SO4 and HCl
(3) SO2Cl2
(4) SOCl2 and HCl
12. For OF2 molecule, consider the following:
A. Number of lone pairs on oxygen is 2.
B. FOF angle is less than 104.5°.
C. Oxidation state of O is –2.
D. Molecule is bent ‘v’ shaped.
E. Molecular geometry is linear. Choose the correct statements.
(1) A, B, D only
(2) A, C, D only
(3) C, D, E only
(4) B, E, A only
13. A peroxy monosulphate anion HSO –5 has
(1) five sulphur–oxygen bonds and no oxygen–oxygen bond
(2) four sulphur–oxygen bonds and one oxygen–oxygen bond
(3) three sulphur–oxygen bonds and two oxygen–oxygen bonds
(4) five sulphur–oxygen bonds and two oxygen–oxygen bonds
14. The correct group of halide ions that can be oxidised by oxygen in acidic medium is (1) I– only
(2) Br– and I–(3) Br– only (4) Cl–, Br– and I– only
15. The number of P–O–P bonds in H 4 P 2O 4, (HPO3)3, and P4O10 are, respectively, (1) 1, 2, 4 (2) 0, 3, 6 (3) 1, 3, 6 (4) 0, 3, 4
16. A pale yellow solid (A) having crown shape is heated with concentrated H 2SO4. It gives suffocating smell of gas B, which is passed through moistened starch iodide paper and turns it blue. The gas B and solid A are respectively,
(1) SO3 and S8
(2) SO2 and S8
(3) SO3 and S2N2
(4) SO3 and SO2
17. 'A', obtained by Ostwald’s method, involving air oxidation of NH 3 , upon further air oxidation, produces 'B'. 'B' on hydration, forms an oxoacid of nitrogen along with evolution of 'A'. The oxoacid also produces 'A' and gives positive brown ring test.
Identify A and B, respectively.
(1) NO2, N2O4
(2) N2O3, NO2
(3) NO, NO2
(4) NO2, N2O5
18. The following are some statements related to VA group hydrides.
I) Reducing property increases from NH 3 to BiH3.
II) Tendency to donate lone pair increases from NH3 to BiH3.
III) Ease of formation of hydrides decreases from NH3 to BiH3.
IV) Stability of hydrides decreases from NH3 to BiH3
Correct statements are
(1) I, II, III, IV
(2) I, III, IV
(3) I, II, IV
(4) II, III, IV
19. CIF5 at room tempertature is a (1) colourless liquid with trigonal bipyramidal geometry
(2) colourless gas with trigonal bipyramidal geometry
(3) colourless liquid with square pyramidal geometry
(4) colourless gas with square pyramidal geometry
Section-B
20. In spite of having lower dissociation energies, bromine and iodine are weaker oxidising agents than chlorine due to their
(1) smaller electron affinities and greater hydration energies
(2) smaller electron affinities and smaller hydration energies
(3) greater electron affinities and greater hydration energies
(4) greater electron affinities and smaller hydration energies
21. Which of the following statements are correct?
a) Ozone is thermodynamically more stable.
b) Tailing of mercury is due to the formation of mercuric oxide layer.
c) S2-molecule is paramagnetic.
d) Bond angle in P4 molecule is 60°.
e) White phosphorus is more reactive than red phosphorus.
f) Oxidation state of Fe in [Fe(H2O)5NO]SO4 is +2.
22. How many of the following statements are correct regarding FNNF molecule?
i) It exists as two distinct geometric isomers.
CHAPTER TEST – JEE ADVANCED
2023 P1 Model
Section-A
[Multiple Option Correct MCQs]
1. Which of the following reaction does not occur?
(1) F2+2Cl–→2F–+Cl2 (2) Cl2+2F–→2Cl–+F2
(3) Br2+2l–→2Br–+2 (4) I2+2Cl–→2l–+Cl2
2. Which of the following statements are correct about the compound of Xe?
(1) Hybridisation of Xe in XeO 4 is sp3.
(2) With SbF5, XeF2 acts as a Lewis acid but XeF4 acts as a Lewis base.
ii) It is planar.
iii) Its nitrogen–nitrogen bond is longer than that in N2F4
iv) It is more stable than its structural isomer with both fluorine atoms bonded to the same nitrogen.
23. When Cl 2 gas reacts with one mole of turpentine oil (C10H16), the number of moles of HCl formed is_____.
24. A 62 g quantity of white phosphorus was burned in an excess of oxygen and the product was dissolved in water to form an acid. Calculate the number of moles of acid obtained.
25. Consider the following molecule.
Ca lculate the value of p÷q. Here, p and q are total number of dπ−pπ bonds and total number of sp3 hybridised atoms, respectively, in given molecule.
(3) With SbF5, both XeF2 and XeF4 act as Lewis acids.
(4) With SbF5, both XeF2 and XeF4 act as Lewis acids.
3. Although electron gain enthalpy of fluorine is less negative compared to chlorine, fluorine is a stronger oxidising agent than chlorine because of
(1) low enthalpy of dissociation of F–F bond
(2) high hydration energy of F – ion
(3) high electron affinity of fluorine than chlorine
(4) Cl2 has pale green colour and F2 has pale yellow colour
4. When a solution of a mixture having two inorganic salts was treated with freshly prepared ferrous sulphate in acidic medium, a dark brown ring was formed, whereas, on treatment with neutral FeCl3, it gave deep red colour, which disappeared on boiling, and a brown red ppt was formed. The mixture contains
(1) 2 24 3 a CO nd NO (2) 2 2- 234 and O C SO
(3) 33 a C d HCO n O NO
(4)3 2 3 and S CH O COO
5. One mole of P4 reacts with 8 moles of SOCl2 to give 4 moles of A, x mole of SO2, and 2 moles of B. A, B, and x, respectively, are (1) PCl3,S2Cl2, and 2 (2) POCl3,S2Cl2, and 2 (3) PCl3,S2Cl2, and 4 (4) POCl3,S2Cl2, and 4
6. Reaction of thionyl chloride with white phosphorus forms a compound [A], which, on hydrolysis, gives [B], a dibasic acid. [A] and [B] are, respectively, (1) P4O6 and H3PO3 (2) PCl5 and H3PO4 (3) PCl3 and H3PO3 (4) PCl3 and H3PO4
7. Which of the phosphorus oxoacids can create silver mirror from AgNO 3 solution? (1) H 4P2O7 (2) (HPO3)n (3) H 4P2O6 (4) H 4P2O5
Section-B
[Numerical Value MCQs]
8. Reaction of conc.H2SO4 with pure KMnO4 yields a green, highly explosive product. Number of unpaired electrons present in that product is ___.
9. NaClO 3 , on reaction with SO 2 in the presence of H2SO4, gives a yellow coloured gas (A), which, upon ozonolysis, produces a compound (B), along with dioxygen. The oxidation state of central atom in A is x, and the average oxidation state of chlorine in compound ‘B’ is y. Find |x–y|.
11. Thermal decomposition of AgNO3 produces two paramagnetic gases. The total number of electrons present in the antibonding molecular orbitals of the gas that has the higher number of unpaired electrons is _______.
12. How many of the following have bond angles more than tetrahedral bond angle?
+ 222 NO,NO,NO
13. Sum of the number of halogen atoms present in tear gas, phosgene, and mustard gas is.
Section-C
[Matrices Matching Questions]
14. Match List-I (the reactions in the given stoichiometry of the reactants) with List-II (products).
List-I
List-II
(A) P2O3+3H2O→ (i) P(O)(OCH3) Cl2
(B) P 4 + 3NaOH + 3H2O → (ii) H3PO3
(C) PCl5 + CH3COOH → (iii) PH 3
(D) H3PO2 + 2H2O + 4AgNO3 → (iv) POCl3 (v) H3PO4
Choose the correct answer from the options given below.
(A) (B) (C) (D)
(1) ii iii i v
(2) iii v iv ii
(3) v ii i iii
(4) ii iii iv v
15. Match List-I with List-II and choose the correct answer.
List-I List-II
(A) Acidic hydride of N (p) Nitrogen dioxide
(B) Basic hydride of N (q) Ammonia
(C) Brown colour gas (r) Hydrazoic acid
(D) Colourless gas (s) Nitric oxide
(A) (B) (C) (D)
(1) r q p s
(2) q r s p
(3) s p r q
(4) p q r s
16. Match List-I (species) with List-II (number of lone pair on central atom).
List-I (Species) List-II (Number of lone pairs of electrons on the central atom)
(A) XeF2 (i) 0
(B) XeO2F2 (ii) 1
(C) XeO3F2 (iii) 2
(D) XeF4 (iv) 3
ANSWER KEY
JEE Mains
Level-I
Choose the correct answer from the options given below.
(A) (B) (C) (D)
(1) iv i ii iii
(2) iii iv ii i
(3) iii ii iv i
(4) iv ii i iii
17. Match the List-I (acid) with List-II (acid description).
List-I List-II
(A) H2S2O8 (p) It contains one peroxy linkage
(B) H2SO4 (q) It is dibasic acid
(C) H3PO3 (r) It contains two dπ–pπ bonds
(D) HClO3 (s) Oxidation number of central atom is +5
Choose the correct answer from the options given below.
(A) (B) (C) (D)
(1) p,q,s p,q,s q,s r
(2) p,q p,q,r r q
(3) p p,q,s r,s p,q,s
(4) p,q q q r,s
Theory-based Questions
Brain Teasers
Flash Back
Chapter Test-JEE Main
Chapter Test-JEE Advanced
THE d - AND f - BLOCK ELEMENTS CHAPTER 5
Chapter Outline
5.1 Electronic Configuration
5.2 General Properties of the Transition Elements
5.3 Some Important Compounds of Transition Elements
5.4 The Lanthanoids
5.5 The Actinoids
In the mo dern periodic table, elements are classified into four blocks on the basis of electronic configuration. Elements in which differentiating electron enters d-orbitals in penultimate shell are called d-block elements.
Elements are placed in between s- and pblock elements in four series. They are 3d, 4d, 5d and 6d series. First three series of d-block
are completely filled with ten elements in each series and 6d series is incomplete.
3d-series elements are from scandium (Z = 21) to zinc (Z = 30), 4d-series elements are from yttrium.
(Z = 39) to cadmium (Z = 48) and 5d-series elements are from lanthanum (Z = 57), hafnium (Z = 72) to mercury (Z = 80). d-Block consists of ten groups: IIIB, IVB, VB, VIB, VIIB, VIII, IB and IIB or there are placed from 3 rd group to 12 th group in modern periodic table. Names and atomic numbers of 3d, 4d and 5d series elements are given in the Table 5.1 .
In a broader sense, d-block elements are called transition elements. This is because, d-block elements represent transition in properties from most electropositive s-block elements to least electropositive p-block elements. In transition elements, ultimate and penultimate shells are incompletely filled with electrons.
Table 5.1 Atomic number (Z), names and symbols of d-block elements
5: The d- and f- Block Elements
Elements that have partially filled d-sub sh ells either in their elemental form or in any of their chemically significant oxidation state, are called transition elements. The electronic configurations of Zn, Cd, and Hg are represented by the general formula (n–1)d10ns2. The d-orbitals in these elements are completely filled in the atomic state as well as in their common oxidation states. Therefore, they are not regarded as transition elements. All transition elements are d-block elements but all d-block elements are not transition elements.
5.1 ELECTRONIC CONFIGURATION
General electronic configuration of d-block elements is (n–1)d 1-10 ns 1 or 2. Here n values are 4 for 3d series, 5 for 4d, 6 for 5d and 7 for 6d series. Half filled and completely filled subshells give extra stability to the atom because of greater exchange energy. Thus in some transition elements, electrons move between ns and (n–1)d sub shells and exhibit anamalous electronic configuration. These configurations are given in the Table 5.2.
When compared with 4d and 5d series elements, 3d series elements are more widely distributed in the nature.
Another interesting point is that in 3d series elements with even atomic numbers
are largely available than the elements with odd atomic numbers. Iron is the fourth most abundant among all elements in the earth’s crust.
The important mineral of iron is haematite (Fe 2 O 3 ). Pyrolusite (MnO 2 ) and magnite (Mn 2 O 3 .H 2 O) are the main minerals of manganese. Chromite (FeO.Cr 2 O 3 ) is the important mineral of chromium.
There are greater horizontal similarities in the properties of the transition elements in contrast to the main group elements. However group similarities usually exist. We shall first study the general characteristics and their trends in the horizontal rows (particularly 3d). General and characteristic properties of transition elements are:
■ Metallic nature and electropositivity
■ High melting and boiling points
■ Extremely hard
■ Good conductors of heat and electricity
■ Malleable and ductile
■ Variable oxidation states
■ Paramagnetic and ferromagnetic nature
■ Formation of hydrated coloured ions and salts
■ Alloy forming ability
Table 5.2 Anomalous electronic configurations of some elements
Table 5.3 Lattice structures of transition metals
■ Complex compound forming ability
■ Interstital compound forming ability and
■ Catalytic property.
with the exceptions of Zn, Cd, Hg and Mn, transition elements have one or more typical metallic structures at normal temperatures as shown in Table 5.3. Nearly all of them display typical metallic properties such as high tensile strength, ductility, malleability high thermal and electrical conductivity and metallic lustre.
TEST YOURSELF
1. In which of the following sets, all have d 10 configuration in their ground state?
(1) Fe, Cr, Co, Ni (2) Sc, Ti, Fe, Zn (3) Cd, Au, Hg, Ni (4) Cu, Zn, Cd, Ag
2. Which of the following groups of transition metals are called coinage metals? (1) Cu, Ag, Au (2) Ru, Rh, Pb (3) Fe, Co, Ni (4) Os, Ir, Pt
3. Liquid metal among d-block elements is (1) Hg (2) Zn (3) Nb (4) Cd
4. Which of the following is a transition element?
(1) Zn (2) Pb (3) Cu (4) Sn
Answer Key (1) 4 (2) 1 (3) 1 (4) 3
5.2 GENERAL PROPERTIES OF THE TRANSITION ELEMENTS
Some important general characteristics of 3d series of transition elements are given in Table 5.4
Table 5.4 Some general characteristic properties of 3d- series elements
Electric configuration
5.2.1 Atomic and Ionic Radii
Atomic radii of the d-block elements are intermediate between those of s- and p-block elements.
Trend in Series
The atomic radius decreases in a period in the beginning, because with increase in the atomic number, the nuclear charge goes on increasing progressively. However, we know that the electrons enter the penultimate shell, the number of d-electrons increases and screening effect also increases. This neutralises the effect of increased nuclear charge due to increase in atomic number. Consequently, atomic radius remains almost unchanged after the middle of series.
At the end of any series, there are increased electron – electron repulsions between the added electrons in the same orbitals. These repulsions exceed the attractive forces due to increased nuclear charge. Therefore, electron cloud expands and the atomic radius increases at the end of any series. Slight irregularities in size have been ascribed to crystal field effects.
Trend in Groups
In IIIB group from Sc to La, atomic radius increases. In the remaining groups of d-block, atomic radius increases from 3d series element to 4d series element. But atomic radius of 4d series element and 5d series element of same group are nearly equal due to the effect of lanthanoid contraction.
For example, atomic radius of IVB group elements Ti, Zr and Hf are 1.47 A 0, 1.60 A 0 and 1.58 A 0 respectively. Atomic radius of VB group elements V, Nb and Ta are 1.34 A0,
1.46 A0 and 1.46 A0 respectively. Atomic radii of d-block elements are represented in Fig. 5.1.
Ionic Radii
The ionic radii follow the same trend as the atomic radii. Since the metals exhibit different oxidation states, the radii of ions also differ. The ionic radii of transition metals decreases, with increase in oxidation state. For the same oxidation state, in a series form left to right, ionic radius decreases due to increase in nuclear charge.
5.2.2 Ionisation Energies
Ionisation energies of d-block elements are more than that of s-block elements and less than that of p-block elements. In a given series with an increase in the atomic number, atomic size decreases and nuclear charge increases. As a result, nuclear attraction over the valence electrons increases. Hence, ionisation energy increases in the series. In any series, ionisation energy is highest for the last element. First three successive ionisation energy values of 3d series elements and first ionisation energy of 4d and 5d series elements are given in the Table 5.5
Fig. 5.1 Trends in atomic radii of d-block elements
Table 5.5 Ionization energy values of d-block elements
Table 5.6 Electronic configurations and the oxidation states of 3d-element
The irregular trend in the first ionisation enthalpy of the 3d metals, though of little chemical significance, can be accounted for by considering that the removal of one electron alters the relative energies of ‘4s’ and ‘3d’ orbitals.
In IIIB-group ionisation potential decreases from scandium to lanthanum, but in the remaining groups trend is not the same. From 3d to 4d series element ionisation potential decreases in some groups and increases to little extent in some other groups.
In a group from 4d to 5d series element, there is no change in atomic radii but nuclear charge increases by 32 units. Thus from 4d to 5d series ionization energy is increased by greater extent.
Chromium and copper have unusually high second ionisation energy when compared with their neighbouring elements and with their first ionization energy. This is because Cr+ and Cu+ have d5 and d10 stable electronic configurations.
Second ionisation energy of zinc is consideringly low because removal of second electron from zinc leads to the stable d 10 c onfiguration. Removal of electron from the d5 configuration is difficult. Thus, third ionisation energy of manganese is higher than that of the neighbours. The higher third ionisation energy values of Ni, Cu and Zn also indicate why it is difficult to obtain oxidation state greater than +2 for these elem ents.
Other General Properties
Transition elements have high density due to their small size. In a period, densities increases. as the atomic radii decreases. The decrease in metallic radius coupled with increase in atomic mass results in a general increase in the density of these elements. Density increases down the groups in transition elements. Densities of second transition series are higher and third transition series are still higher. Iridium has the highest density (22.7 g cm –3) among all the elements. Osmium also has a very high density (22.6 g cm–3).
All the transition elements are metals. They have high melting and boiling points, high tensile strength, ductility, malleability, hardness, high thermal and electrical conductivities and lustre. With the exceptions of Zn, Cd, Hg and Mn, d-block elements have one or more metallic structures at normal temperatures.
High melting points, boiling points and heat of atomisation values of transition elements are due to the involvement of (n–1)d electrons in addition to ns electrons in the interatomic metallic bonding. As we move along a transition series, the metallic bond strength increases upto the middle due to increase in number of unpaired d-electrons upto d5 configuration and then decreases due to decrease in number of unpaired d-electrons. Therefore, in any transition series, melting points of metals rise to a maximum at d 5 (except for anomalous values of Mn and Tc) and then falls regularly as the atomic number increases.
In any series of d-block, last element (IIB group) has the least melting and boiling points because there is no unpaired d-electrons for metallic bonding.
IIB group elements (Zn, Cd and Hg) are called volatile metals of d-block due to their less boiling points. Melting points of 3d, 4d and 5d transition metals are given in the Fig. 5.2. In a group, generally melting point increases from 3d to 5d series element. Same trend is observed in heat of atomisation values. Since the alkali and alkaline earth metals have only one and two valence electrons respectively available for metallic bond, their melting and boiling points are relatively low compared to transition elements. Among all metals tungston has the highest melting point. Mercury is a liquid at room temperature.
5.2
I n general, the resultant bonding will be stronger elements with greater number of valence electrons. Enthalpy of atomisation is an important factor in determining the metal electrode potential. Metals with high enthalpy of atomisation tend to be noble in their reaction. Metals of 4d and 5d series have greater enthalpies of atomisation than the corresponding 3d metals Fig. 5.3. This is an important factor in accounting for the occurrence of much more frequent
metal–metal bonding in compounds of the heavy transition metals.
Fig. 5.3 Enthalpy of atomisation of transition elements
All transition elements are metals and electropositive. They are all good conductors of electricity. Silver is the best conductor at room temperature.
5.2.3
Variable Oxidation States
Transition elements exhibit variable oxidation states due to the involvement of (n–1) d electrons along with ns electrons in bonding. The elements that give the greatest number of oxidation states occur in the middle of the series. The lesser number of oxidation states at the extreme ends is due to the presence of a few electrons to lose or to share or the presence of many d electrons for higher valence (Cu,Zn) variable oxidation states of 3d series elements are given in the Table 5.6
For transition elements, energy difference between ns electrons and (n–1) d electrons is very small number. Hence, all ns electrons and variable number of (n–1) d electrons can be involved in bonding and thus, exhibit more than one oxidation state.
For example manganese ([Ar] 4s 2 3d 5 ) exhibit, +2 oxidation state by using two 4s electrons, +4 oxidation state by using two 4s and two of five 3d electrons and +7 oxidation state by using two 4s- and all the five 3d electrons in bonding.
d-Block elements exhibit the common oxidation state +2 by lossing two electrons from ns- subshell. Chromium and copper
Fig.
Trends in melting points of transition elements
exhibit +1 oxidation state by losing the one 4s electron.
In any d- series of elements, for the first five elements the minimum oxidation state is numerically equal to the number of ns electrons and the maximum oxidation state is numerically equal to the sum of ns and (n–1) d electrons. For the second five elements, the minimum oxidation state is numerically equal to the number of ns electrons and the number of unpaired (n–1) d electrons.
In 3d series Mn exhibit the highest oxidation state +7 in the compound KMnO4. In 4d and 5d series Ru and Os respectively exhibit the highest oxidation state +8 in the compounds RuO4 and OsO4. Some transition elements exhibit the lower oxidation state zero in their complex compounds with carbon monoxide. For example, oxidation state of nickel in [Ni(CO)4] is zero.
In the +2 and +3 oxidation states, the bonds formed by transition metals are mostly ionic. In the compounds of higher oxidation states, the bonds are essentially covalent. For example, in MnO4–, Mn2O7 (oxidation state of manganese is +7) all the bonds are covalent. In the groups of p-block, the lower oxidation states are favoured by the heavier members due to inert pair effect. The opposite is true in the groups of d-block. For example, in group 6, Mo (VI) and W (VI) are found to be more
stable than Cr (VI). Thus Cr (VI) in the form of dichromate ion in acidic medium is a strong oxidising agent, whereas MoO 3 and WO3 are not.
Stability of Higher Oxidation State
d-Block elements exhibit their highest oxidation state in fluorides and oxides. The oxides of 3d metals are given in Table 5.7. The ability of oxygen to form multiple bonds with transition metals explain why transition metals exhibit the highest oxidation state in their oxides.
The highest oxidation number in the oxides coincides with the group number upto 7 th group. Beyond group 7, no higher oxides of element above M 2 O 3 are known although ferrates (VI) (FeO42–) are formed in alkaline medium but they readily decompose to give Fe2O3 and O2.
The ability of fluorine to stabilise the highest oxidation state is due to either higher lattice energy as in the case of ionic compounds like CoF3 or higher bond energy as in the case of covalent compounds like VF 5 and CrF6.
Another feature of fluorides is their instability in the lower oxidation state. e.g., VX, CuX (X = Cl, Br or I). On the other hand, all Cu(II) halides are known except the iodide because Cu2+ oxidises I– to I2. Halides of 3d- metals are given in Table 5.8.
Table 5.7 Oxides of 3d-metals
M any copper (I) compounds are unstable in aqueous solution and undergo disproportionation to Cu 2+ and Cu. The stability of aqueous Cu2+ rather than aqueous Cu+ is due to the higher hydration energy of Cu2+ than Cu+, which compensates the second ionisation energy of
Trends in Stability of Higher Oxidation States
Besides the oxides, oxocations stabilise V V as VO 2 + , V IV and VO 2+ , and Ti IV as TiO 2+ The ability of oxygen to stabilise these high oxidation states exceeds that of fluorine. Thus, the highest manganese fluoride is MnF4, whereas the highest oxide is Mn2O7. The ability of oxygen to form multiple bonds to metals explains its superiority. In the covalent oxide Mn2O7, each Mn is tetrahedrally surrounded by O’s including a Mn-O-Mn bridge. The tetrahedral [MO4]n– ions are known for VV, Cr VI, Mn V, Mn VI, and Mn VII (M = transition metal)
5.2.4 Electrode Potentials and Chemical Activity
Transition metals vary widely in their chemical reactivity. Many of them are sufficiently electro-positive to dissolve in mineral acids. A few are unaffected.
The metals of the first series (exception of copper) are relatively more reactive and are oxidised by 1 M H+.
The Electrode potential values for M2+/M ( Table 5.9 ) indicate a decreasing tendency to form divalent ions due to the increase in the sum of the first and second ionisation enthalpies. It is interesting to note that the Electrode potentials values for Mn, Ni, and Zn are more negative than expected from the general trend. Whereas the stabilities of half-filled d subshell (d 5 ) in Mn 2+ and completely filled d-subshell (d10) in zinc are related to their E° values; for nickel, E° value is related to the highest negative enthalpy of hydration.
Variation of observed 2 M/M E + ο values with atomic number for 3d- series elements given in the Fig. 5.4 . If the net enthalpy change in the conversion of M(s) to M2+(aq) is more positive, then is more (positive or less negative).
Fig. 5.4 Trends in standard electrode potentials of 3d–elements
Table 5.8 Halides of 3d-metals
5.2.5 Trends in M3+/M2+ Standard Electrode Potentials
On examination of the E°(M 3+/M 2+) values, the lo w value of ‘Sc’ reflects the stability of Sc3+ which has a noble gas configuration. The highest value of ‘Zn’ is due to the removal of an electron from the stable d10 configuration of Zn2+. The comparatively high value for ‘Mn’ shows that Mn 2+ (d 5 ) is particularly stable, whereas co mparatively low value for ‘Fe’ indicates the extra stability of Fe3+(d 5). The comparatively low value for V is related to the stability of V2+ (half-filled t2g level).
An examination of the redox couple M3+ / M2+, given in Table 5.4, shows that Mn3+ and Co3+ ions are the strongest oxidising agents in aqueous solutions. The ions Ti2+, V2+ and Cr2+ are strong reducing agents and will liberate hydrogen from a dilute acid.
()()()() 23 aqaq 2Cr2H2CrHaq2g ++++→+
5.2.6 Magnetic Properties
When a magnetic field is applied on substances, mainly two types of behaviours are observed. They are diamagnetism and paramagnetism.
Diamagnetic substances are repelled by the applied field. If permitted, diamagnetic substances move from stronger part of the field to a weaker part of the field. Magnetic lines of forces do not pass easily through the
diamagnetic substance when compared with vaccum. Substances having all the paired electrons behave as diamagnetic substances.
Examples:
Sc3+, Ti4+, Cu+, ZnO, KCl, K4[Fe(CN)6], O3, O22–, etc.
Paramagnetic substances are attracted by the applied field. If permitted, they move from a weaker part of the field to a stronger part of the field. Magnetic lines of force passes easily through the paramagnetic substances when compared with vacuum. Substances having at least one unpaired electron behave as paramagnetic substances.
Examples:
Ti3+, V3+, Cu2+, Cr2O3, O2, O2–, O2+, NO, NO2, ClO2, K3[Fe(CN)6], etc.
Ferromagnetism is a special case of paramagnetism. Ferromagnetic substances are strongly attracted by the applied field and magnetised permanently. In ferromagnetic substances the magnetic moments of the individual atoms are aligned in the same direction.
Ferromagnetism in the crystalline substances disappears in the solution form of the substance. Ferromagnetism is exhibited by some transition metals and their compounds.
Examples: Fe, Co, Ni, CrO2, etc.
Table 5.9 Thermo chemical data (kJ mol–1) and electrode potentials of 3d elements
The magnetic moment is expressed in Bohr magnetons (BM). One Bohr magneton is equal to 9.273 × 10–24 JT–1 and is obtained from the relation.
Bohr magneton = eh/4mc, where e is charge of electron (in esu), h is Planck’s constant (in erg s), m is mass of electron (in g), and c is light velocity (in cm/s).
The magnetic moment of any substance arises from the spin and orbital motion of unpaired electrons. In many compounds of the 3d-series elements, the angular momentum due to orbital motion of unpaired electrons is small and can be ignored. Therefore, the spin only magnetic moment, () S 2 µ=+ nn BM, Here, n is number of unpaired electrons in the substance.
Magnetic moment (μ) is zero, if number of unpaired electrons (n) is zero. Such a substance is diamagnetic.
Many of the transition metal ions are paramagnetic. Paramagnetism arises from the presence of unpaired electrons, each such electron having a magnetic moment associated with its spin angular momentum and orbital angular momentum.
The spin only magnetic moment values of metal ions of 3d- series are given in Table 5.10 For some ions like Co2+, the observed magnetic moment is higher than the calculated magnetic
moment. This suggest, that there is also an orbital contribution for these ions. In 4d- and 5d- series elements, the orbital contribution is significant.
5.2.7 Coloured Salts and Solutions
Most of the compounds of s- and p-block are colourless but hydrated ions and compounds of many transition metals are coloured.
Examples: MnCl2 is pink, CuCl2 is blue, NiSO4 is green, CuSO4.5H2O is blue coloured, etc., The colour of the hydrated transition metal ions and compounds may be attributed to the presence of unpaired d-electrons. In the gaseous state or isolated state of metal ion, all the five d-orbitals in a subshell are degenerate. But under the influence of anions in the compound or under the influence of water molecules in the hydrated metal ion, five degenerate orbitals are splitted into two sets of d-orbitals having different energies. This phenomenon is called crystal field splitting and it is shown in Fig. 5.5.
Fig. 5.5 Splitting of the five degenerate d orbitals
Table 5.10 The spin only magnetic moment values of metal ions of 3d- series
Th e excitation of electrons from a set of d-orbitals having less energy to another set of d-orbitals having high energy within the same d subshell is called d–d transition. The amount of energy required for d–d transition is equal to the energy of any one of the colours in white light. Therefore, when white light falls on a hydrated transition metal-ion having at least one unpaired d-electron, it absorbs a characteristic colour from visible region for d–d transition and transmit the complementary colour.
For example, [Ti(H 2O) 6] 3+ has a lone 3d electron. It absorbs green and yellow light (λ= 500–570 nm) from visible region and transmits a mixture of red and violet light (λ = 400–450 nm). Hence hydrated Ti 3+ ion has purple colour. Similarly, hydrated cupric ion (Cu2+) absorbs orange-red light from visible region and transmits its complementary colour greenish-blue. Thus, cupric compounds have greenish blue colour. Colours of some hydrated ions of 3d- series metals are given in the Table 5.1 1
Same metal exhibits different colours in the different oxidation states. For example, Fe 2+ is green, while Fe 3+ is pinkish-yellow. Similarly, colours of Cr 2+ and Cr 3+ ions are blue and green respectively. Colours of Mn2+, Mn3+, and Mn6+ ions are pink, blue and green respectively.
Hydrated transition metal ions like Sc 3+ , Ti4+ and V5+ with empty d-subshell and metal ions like Cu + , Zn 2+ with completely filled d-subshell are colourless. This is due to the absence of unpaired d-electrons.
Even though Cr 6+ and Mn 7+ have no unpaired d-electrons, some of their oxo anions like Cr 2 O 7 2– (orange), CrO 4 2– (yellow) and MnO4– (purple) are coloured. These colours are due to a different process, known as charge transfer phenomenon.
5.2.8 Formation of Complex Compounds
The transition metals form a large number of complex compounds due to the comparatively small sizes of the metal ions, their high ionic charges and the ability of d-orbitals for bond formation. A few examples for complexes are [Fe(CN)6]3–, [Fe(CN)6]4–, [Cu(NH3)4]2+ etc.
Table 5.11 Colours of some hydrated of 3d- series metal ions
Purple
5.2.9
Catalytic Activity
Tra nsition metals and their compounds act as catalysts. Vanadium (V) oxide in contact process for manufacture of H 2SO 4, finely divided iron in Haber’s process for manufacture of ammonia and nickel in catalytic hydrogenation are some of the examples. The catalytic activity of transition elements is due to their ability to exhibit variable oxidation states and to form complex compounds. At a
5: The d- and f- Block Elements
solid surface, catalysts involve in the formation of bonds between reactant molecules and atoms of the surface of the catalyst. This has the effect of increasing the concentration of the reactants at the catalyst surface and also weakening of the bonds in the reacting molecules. Transition metals become more effective catalysts, because the metal ions can change their oxidation states. In some cases catalyst forms unstable intermediates with the reactants.
In contact process for the manufacture of sulphuric acid, vanadium pentoxide or a vanadate is used as catalyst to oxidise sulphur dioxide to sulphur trioxide.
3 254 VO(or)VO
2SOO 2SO +→
2(g)2(g) 3(g)
This reaction takes place in two steps. In the first step SO2 is oxidised to SO3 and V5+ is reduced to V4+. In the second step V4+ oxidised to V5+
Ferric ion acts as catalyst for the reaction between iodide and perdisulphate ions. Fe3 22 28 24 2ISOI2SO + +→+
This reaction takes place in the following two steps.
32 2Fe2I2FeI2 +−+ +→+
2232 28 4 2FeSO2Fe2SO
5.2.10
Interstitial Compounds
I n terstitial compounds are those that are formed when small atoms like hydrogen, carbon or nitrogen are trapped in the holes or interstitices of the metal lattice. They are usually non stoichiometric and are neither ionic nor covalent. For example, steel and cast iron are interestitial compounds of iron and carbon. Some more examples are TiC, MoC, Mo2C, VH0.56, TiH1.7 and Mn4N.
Hydrogen always occupies the tetrahedral holes, whereas carbon and nitrogen occupy the larger octahedral holes in the lattice. In
these compounds the metal lattice is not altered but the lattice may expand a little. Thus, the density of the interstitial compounds is less than that of the metals. Component elements do not combine in definite ratios and hence these compounds are also called non-stoichiometric compounds. The formulae of non-stoichiometric compounds do not correspond to any normal oxidation states of metal. The oxides and sulphides of transition metals that show variable oxidation states are generally non-stoichiometric compounds.
The interstitial compound of iron and oxygen is represented as Fe0.93O. In the lattice of ferrous oxide, some iron (II) are replaced by iron (III). Some examples for interstitial compounds are WO2.88 – 2.92, Fe0.89 – 0.96S, etc.
The principle physical and chemical properties of interstitial compounds are as follows:
i. They retain metallic conductivity.
ii. They have high melting point than those of pure metals.
iii. They are very hard, some borides approach diamond in hardness.
iv. They are chemically inert.
5.2.11 Alloys
An alloy is an intimate mixture of metals prepared by mixing the components. Alloys are homogeneous solid solutions in which the atoms of one metal are distributed randamly along the atoms of the other metals or nonmetals or metalloids. These alloys are formed by atoms with metallic radii within about 15 percent of each other. Alloys are reading formed by transition metals because of similar radii and other characteristics. These alloys are hard and have often high melting points.
The best known ferrous alloys are variety of steels. The main constituent element in steel is iron. The alloying element may be manganese, chromium, vanadium, tungsten, molybdenum, etc.
Copper is an important constituent element in non-ferrous alloys like brass with zinc and bronze with tin.
TEST YOURSELF
1. The 3d element with least density is (1) Cr (2) Sc (3) Mn (4) Cu
2. Greatest tendency to convert from +2 → +3 in aqueous solution is for (1) Cr (2) Co (3) Mn (4) Fe
3. Identify the incorrect statement from the following.
(1) Melting point of W> Mo > Cr (2) Density of Cu > Mn (3) IP2 of Cr > Mn
(4) Stability of +6 oxidation state in Cr > Mo > W
4. The set having ions that are coloured and paramagnetic is (1) Cu2+, Cr3+, Sc+ (2) Cu2+, Zn2+, Mn4+ (3) Sc3+, V5+, Ti4+ (4) Ni2+, Mn7+, Hg2+
5. The nature of oxides V 2 O 3 and CrO is indexed as ‘X’ and ‘Y’ type, respectively. The correct set of X and Y is (1) X=basic, Y=amphoteric (2) X= amphoteric, Y= basic (3) X=acidic, Y=acidic (4) X=basic, Y=basic
6. At 298 K, a 0.1 M CH 3COOH solution is 1.34% ionised. The ionisation constant K a for acetic acid will be
(1) 1.82 × 10–5 (2) 18.2 × 10–5 (3) 0.182 × 10–5 (4) none of these
7. Four successive members of the first row transition elements are listed below with atomic number. Which one of them is expected to have the highest M/M32 E ++ ο value?
8. Among the following, choose the incorrect statement about transition metals.
(1) Cr, Mo and W have high melting points. (2) With increase in number of unpaired electrons melting point increases.
(3) Mn3+ is more stable than Mn2+
(4) They show variable oxidation states.
9. The order of melting points of Cr, Mo, and W is
(1) Cr > Mo > W
(2) MO > Cr > W
(3) W > Mo > Cr
(4) W > Cr > Mo
10. Which of the following transition elements does not exhibit variable oxidation states?
(1) Mn (2) Ni (3) Ti (4) Sc
11. The number f-electrons in +3 oxidation state of gadolinium (Z= 64) is x and in +2 oxidation state of ytterbium (Z=70) is y. The sum of x and y is (1) 13 (2) 20
(3) 18 (4) 21
12. Highest IP 2 (second ionisation enthalpy) among 3d-series elements is possible for (1) Mn (2) Zn (3) Cu (4) Co
13. The metal not present in the alloy german silver is
(1) Cu (2) Zn (3) Ni (4) Sn
Answer Key
(1) 2 (2) 1 (3) 4 (4) 1
(5) 4 (6) 1 (7) 4 (8) 3
(9) 3 (10) 4 (11) 4 (12) 3
(13) 4
5.3 SOME IMPORTANT COMPOUNDS OF TRANSITION ELEMENTS
Transition elements form ionic compounds and also covalent compounds some of them are given below
5.3.1 Oxides
The transition metals generally react with oxygen at high temperature to form oxides. Since the transition metals exhibit variable oxidation states, they form a large variety of oxides. All the 3d-series elements except scandium form MO type oxides, which are ionic as well as basic. The highest oxidation number in the oxides, coincides with the group number upto group 7 and is attained in Sc2O3 to Mn2O3. Beyond group 7, no higher oxides of metal, above M2O3 are known.
As the oxidation number of a metal increases, ionic character decreases. In the case of manganese, Mn2O7 is a covalent green oil. In the case of chromium and vanadium, CrO3 and V2O5 have low melting points. Transition metal oxides with lower oxidation state of metals are ionic oxides and are generally basic. As the oxidation number of a metal increases, acidic character increases. In case of manganese MnO and Mn 2 O 3 are basic, Mn3O4 and MnO2 are amphoteric, and Mn2O7 is acidic. In case of chromium CrO is basic, Cr2O3 and CrO2 are amphoteric and CrO 3 is acidic. In case of vanadium, VO, V2O3, and V2O4 are basic but V2O5 is amphoteric. V2O5 reacts with alkali as well as acids to given VO43–and VO2+ respectively.
5.3.2 Oxoanions
Oxonium ions formed by transition metals are 22 4274 CrO,CrO,MnO etc., some of the corresponding salts of the ons are as follows.
Potassium Dichromate
Preparation: Potassium dichromate is manufactured from chromite ore, FeO.CrO3 Chromite ore is fused with sodium carbonate in free access of air.
The yellow solution of sodium chromate is filtered and acidified with sulphuric acid to give a solution from which orange sodium dichromate, Na 2Cr2O7.2H2O can be crystallised.
2Na2CrO4 + 2H+→ Na2Cr2O7 + 2Na+ + H2O
Sodium dichromate is more soluble in water than potassium dichromate. The latter is therefore, prepared by treating the solution of sodium dichromate with potassium chloride.
Na2Cr2O7 + 2KCl → K2Cr2O7 + 2NaCl
Properties: K2Cr2O7 is orange red crystalline solid having melting point 670 K. It is moderately soluble in cold water but readily soluble in hot water.
Action of heat: When heated to a white heat, it decomposes to give potassium chromate, chromic oxide and oxygen gas.
4K2Cr2O7 ∆
4K2CrO4 + 2Cr2O3 + 3O2
Action of alkalies and acids: When an alkali is added to an orange-red solution of potassium dichromate, a yellow solution of potassium chromate results. This on acidification again gives back orange-red. Dichromate in acidic solutions is called chromic acid. Dichromate ion, Cr 2O 7 2– and chromate ion, CrO 4 2– exist in equilibrium which are inter convertible by changing the pH of the solution. + +− +−
2 2 2OH(HO) 22 274 2H(HO) CrO2CrO
It indicates that the acidic solution contains Cr2O72– ions, while alkaline solution contains CrO42– ions.
Chrom yl chloride test: When potassium dichromate is heated with concentrated sulphuric acid and metal chloride, salt, deep red vapours of chromyl chloride are evolved.
K2Cr2O7+4NaCl+3H2SO4→2CrO2Cl2+
K2SO4 + 2Na2SO4 + 3H2O
Oxidising properties: Potassium dichromate acts as powerful oxidising agent in acidic medium. Both Na 2 Cr 2 O 7 and K 2 Cr 2 O 7 are oxidising agents but K 2 Cr 2 O 7 is preferred because it is not hygroscopic and can be used as primary standard in volumetric analysis. In acidic solution, oxidising action of dichromate ion can be represented as follows.
Cr2O72– + 14H+ + 6e–→ 2Cr3+ + 7H2O; E° = +1.33 V
Acidified potassium dichromate will oxidise iodides to iodine, sulphides to sulphur, tin (II) to tin (IV) and iron (II) salts to iron (III) salts. The half reactions are noted below.
6I–→ 3I2 + 6e–
3H2S → 6H+ + 3S + 6e–
3Sn2+ → 3Sn4+ + 6e–.
6Fe2+ → 6Fe3+ + 6e–
The full ionic equation may be obtained by adding half reactions of oxidising agent and reducing agent.
Cr2O72– + 6Fe2+ + 14H+→ 2Cr3+ + 6Fe3+ + 7H2O
Structure: Structures of chromate ion, CrO42–and dichromate ion, Cr 2 O 7 2– are given in Fig.5.6 . Both in chromate and dichromate ions, chromium undergo sp 3 hybridisation. The chromate ion is tetrahedral whereas the dichromate ion consists of two tetrahedral sharing one corner with Cr–O–Cr bond angle of 126°. In chromate ion chromium form pd
ππ bonds with two oxygens.
Fig. 5.6 Structure of chromate and dischromatic ions
In chromate ion each chromium forms bonds with three oxygens.
Uses: It is a very important chemical used in leather industry. It is used as an oxidant for preparation of many azo compounds.
Potassium Permanganate
Potassium permanganate is the salt of permanganic acid, HMnO 4 which is an unstable acid and exists only in solution
Preparation: KMnO 4 is prepared from the pyrolusite ore (MnO 2) in two steps. In the first step, dark green K 2 MnO 4 is prepared by fusing the finely powdered pyrolusite ore with an alkali metal hydroxide and oxidising agent like KNO3. In the second step K2MnO4 is acidified, then it disproportionates to give potassium permanganate.
Commercially potassium permanganate is prepared by the alkaline oxidative fusion of MnO2 followed by the electrolytic oxidation of manganate ion.
In the laboratory it is prepared by oxidation of manganous ion salt with peroxodisulphate.
2Mn2+ + 5S2O82– + 8H2O
2MnO4– + 10SO42– + 16H+
Properties: Potassium permanganate forms dark purple (almost black) lustrous crystals. It gives deep pink colour in solution.
Its solubility in water at 20°C is only about 6.4 g per 100 g of water, but solubility increases with increase in the temperature.
At 2400C, potassium permanganate decomposes and gives off oxygen gas.
2KMnO4 → K2MnO4 + MnO2 + O2
Permanganate ion is diamagnetic. The ion possess weak temperature dependent paramagnetism.
Oxidising properties: Potassium permanganate is a powerful oxidising agent in neutral, alkaline and acidic solutions. Permanganate ion is reduced to Mn2+ ion in acidic solutions, reduced to manganate ion in strongly alkaline solutions and reduced to MnO 2 in neutral or feebly alkaline solutions.
MnO4– + 8H++ 5e– → Mn2+ + 4H2O ;
E° = +1.52 V
MnO4– + e– → MnO42– ; E° = +0.56 V
MnO4– + 4H+ + 3e– → MnO2 + 2H2O;
E° = +1.69 V
Oxidising properties in acidic solutions:
To acidify the KMnO4 sulphuric acid is used but not hydrochloric acid because it oxidises hydrochloric acid to chlorine.
Permanganate oxidises iodide ions to iodine violet,
2MnO4– + 10I– + 16H+ → 2Mn2+ + 5I2 + 6H2O
It oxidises ferrous ion (green) to ferric ion (yellow).
MnO4– + 5Fe2+ + 8H+→ Mn2+ + 5Fe3+ +4H2O
It oxidises oxalate ions to carbon dioxide gas,
2MnO4– + 5C2O42– + 16H+ →
2Mn2+ + 10CO2 + 8H2O
It oxidises nitrite ions to nitrate ions.
2MnO4– + 5NO2– + 6H+ →
2Mn2+ + 5NO3– + 3H2O
It oxidises sulphite or sulphurous acid to sulphate or sulphuric acid.
2MnO4– + 5SO32– + 6H+ →
2Mn2+ + 5SO42– + 3H2O
It oxidises hydrogen sulphide to sulphur.
2MnO4– + 5S2– + 16H+ → 2Mn2+ + 5S + 8H2O
Oxidising properties in neutral or fairy alkaline solutions: Permanganate oxidises iodide ion to iodate ion.
2MnO4– + I– + H2O →2MnO2 + IO3– +2OH–
Permanganate oxidises thiosulphate ion to sulphate ion.
8MnO4– + 3S2O32– + H2O →
8MnO2 + 6SO42– + 2OH–
It oxidises manganous ion to manganese dioxide in the presence of zinc sulphate or zinc oxide catalyst.
2MnO4– + 3Mn2+ + 2H2O → 5MnO2 + 4H+
Structures of manganate and permanganate ions: Permanganate ion is isostructural with perchlorate ion. Both in permanganate and manganate ions, manganese undergo sp 3 hybridisation. Structures of permanganate and manganate ions both are tetrahedral and are given in Fig. 5.7
In permanganate ion, three bonds are formed by overlapping of d-orbitals of manganese and p-orbital of oxygen. In manganate ion, two bonds are formed by overlapping of d-orbitals of manganese and p-orbitals of oxygen. Permanganate ion is diamagnetic but manganate ion is paramagnetic with one unpaired d-electron.
Uses: Permanganate is used in qualitative and quantitative analysis. Due to oxidising property, it is used for bleaching wool, cotton, silk and other textile fibres. It is also used for decolourisation of oils.
Cold dilute alkaline potassium permanganate is used as oxidant in organic chemistry under the name Baeyer’s reagent.
(1) S2O72- (2) sulphur (3) SO32– (4) SO42–
4. Permanganometric titrations in presence of HCl, are not satisfactory because (1) HCl is a weak acid (2) HCl is a volatile acid (3) KMnO4 will oxidise HCl into Cl 2 (4) KMnO4 will show disproportination in presence of HCl
5. Which order is correct for oxidising power? (1) 2 272 CrOVO −+ < (2) 24 VOMnO +− > (3) 2 227 VOCrO +− < (4) 2 274 CrOMnO >
6. Ferrous ion changes to ‘X’ ion on reacting with acidified KMnO 4 . The number of d-electrons in ‘X’ and its magnetic moment are (1) 6 and 6.93 BM (2)5 and 5.92 BM (3) 5 and 4.9 BM (4) 4 and 5.92 BM
Answer Key
(1) 3 (2) 2 (3) 4 (4) 3 (5) 3 (6) 2
Fig. 5.7 Structures of permanganate and manganate ions
TEST YOURSELF
1. () Ä 23 2 pink X+KCO+airY ,Y+ClZ→→
Which of the following is correct?
(1) X = black, MnO2; Y = blue, K2CrO4; Z=KMnO4
(2) X = green, Cr2O3; Y = yellow, K2CrO4; Z=K2Cr2O7
(3) X = black, MnO2; Y = green, K2MnO4; Z = KMnO4
(4) X = black, Bi2O3; Y = colourless, KBiO2; Z = KBiO3
2. Permanganate ion is isostructural with (1) Chlorate ion (2) Perchlorate ion (3) Chlorite ion (4) Nitrate ion
3. In neutral or alkaline solution, MnO 4 –oxidises thiosulphate ion to
5.4 THE LANTHANOIDS
The fourteen elements from cerium (Z = 58) to lutetium (Z = 71) that come after lanthanum are called lanthanoids. These elements are also called rare earth elements. The fourteen elements, from thorium (Z = 90) to lawrencium (Z = 103) that come after actinium are called actinoids. lanthanoids belong to IIIB group and sixth period, actinoids belong to IIIB group and seventh period in the modern periodic table.
In the f-block elements differentiating electron enters f-subshell of the prepenultimate shell. These elements are placed separately at the bottom of periodic table to avoid excessively large width of the table. Since lanthanum closely resembles lanthanoids, it is usually included in any discussion of the lanthanoids, with general symbol ‘Ln. Monazite sand was the oldest source of lanthanoids.
5.4.1 Electronic Configuration
Lanthanum has the electronic configuration [Xe]5d16s2. It might be expected that the 14 elements from cerium to lutetium would be formed by adding 1, 2, 3, ....., 14 electrons into the 4f level. However, it is energetically favourable to move the single 5d electron into the 4f level in most of the elements, but not in the case of Ce, Gd and Lu. The names, symbols, electronic configurations and radii are given in the Table 5.12
General electronic configuration of lanthanoid atoms is [Xe]6s 2 5d 14 f n or [Xe]6s25d04f(n+1), otherwise [Pd]4fn5s25p65d16s2 or [Pd]4f(n+1)5s25p65d06s2. General electronic configuration of trivalent lanthanoid ion (Ln3+) is [Xe]4fn, where n increases gradually from 1 to 14. It may be noted that atoms of these elements have electronic configuration with 6s2 common but with valuable occupancy of 4f level.
5.4.2 Oxidation States
Lanthanoids exhibit the common oxidation state +3. Even though some lanthanoids exhibit +2 and +4 oxidation states, they are always less stable than +3 state. The higher oxidation states occur in the fluorides and oxides but the lower oxidation states occur in the other halides, particularly bromides and iodides.
The +2 and +4 oxidation states occur, particularly when they lead to either a noble gas configuration (Ce4+) or a half filled f sub shell (Eu 2+ and Tb 4+) or a completely filled f- subshell (Yb 2+ ). In addition, +2 and +4 states exist for the elements that are close to these stable electronic configurations. Thus Sm 2+ and Tm 2+ occur with f 6 and f 13 configurations and Pr4+ and Nd4+ occur with f1 and f2 configurations.
Table 5.12 Electronic configuration of lanthanides and some of their cations
The +3 state is only the most common and the most stable. Thus Ce4+ act as a strong oxidant reverting to the common state +3. The E° value for Ce4+, Ce3+ is +1.74 V, suggests that Ce4+ can oxidise water but the reaction rate is very slow. Similarly Tb4+ acts as oxidant. Eu2+ and Yb2+ act as reductants by reverting back to the common state +3.
The E° value for Ce++/Ce3+ is +1.74 V which suggests that it can oxidise water. However, the reaction rate is very slow and hence Ce(IV) is a good analytical reagent. Pr, Nd, Tb and Dy also exhibit +4 state but only in oxides, MO2. Eu2+ is formed by losing the two s electrons and its f7 configuration accounts for the formation of this ion. However, Eu2+ is a strong reducing agent changing to the common +3 state. Similarly, Yb2+ which has f14 configuration is a reductant. Tb4+ has half-filled f-orbitals and is an oxidant.
5.4.3 Atomic and Ionic Radii
Atomic and ionic radii decrease gradually with increase in atomic number. The decrease in atomic (metallic) radii is not quite regular as it is in Ln3+ ions as shown in Fig. 5.8
The contraction in size is similar to that observed in transition series. It is attributed to the same cause, the imperfect shielding of one electron by another in the same sub-shell. However the sheilding of one 4f- electron by another is less than one d-electron by another, with the increase in nuclear charge along the series. Thus there is fairly regular decrease in the size with increasing atomic number. The cumulative effect of the contraction of the lanthanoid series is known as lanthanoid contraction.
As a result of lanthanoid contraction, second and third transition series elements have nearly same atomic radii. Nearly same size of pairs of elements, Zr and Hf, Nb and Ta, Mo and W is because of lanthanoid contraction. Due to this reason, in a group second and third transition series elements resembles each other much more than the elements of first and second transition series. As a result of lanthanoid contraction ionisation energy of third transition series elements is more than that of second transition series elements. Because of very small change in atomic radii of lanthanoids, their chemical properties are quite similar. Thus it is very difficult to separate these elements from one another using simple chemical separation methods.
5.4.4 Ionisation Energies
The first ionisation enthalpy values of the lanthanoids are around 600 kJ mole –1 , the second ionisation values are about 1200 kJ mole –1 comparable with those of calcium. A detailed discussion of the variation of the third ionisation enthalpies indicater that the variation of the third ionisation enthalpies indicater that the exchange enthalpy considerations (as in 3d orbitals of the first transition series) appear to import a certain degree of stability to empty, half-filled and
Fig. 5.8 Ionic radius of lanthanoids
5: The d- and f- Block Elements
comp letely filled orbitals of f-level. This is indicated from the abnormally low value of the third ionisation enthalpy of lanthanum, Gadolinium and lutetium.
Colour and Magnetic Properties
Many trivalent lanthanoid ions are coloured both in the solid state and in aqueous solution. The colour of the ions is due to the presence of unpaired f- electrons. However absorption bands are narrow, probably because of the excitation within f- level. La 3+ and Lu 3+ are colourless due to the absence of unpaired felectrons. The colour of Ln 3+ ion depends on the number of unpaired f- electrons. Ln 3+ ion with ‘n’ 4f- electrons and Ln3+ ion with ‘(14–n)’ 4f-electrons have similar colour. For example Sm3+ (4f5) and Dy3+ (4f9) are yellow in colour. Colours of Ln3+ ions are listed in Table 5.13.
Table 5.13 Colour of some lanthanoid ions
Lanthanum La3+ 0 colourless
Cerium Ce3+ 1 colourless
Praseodyium Pr3+ 2 green
Neodyminum Nd3+ 3 lilac
Promethium Pm3+ 4 pink
Samarium Sm3+ 5 yellow
Europium Eu3+ 6 pale pink
Gadolinium Gd3+ 7 colourless
Lutetium Lu3+ 14 colourless
Ytterbium Yb3+ 13 colourless
Thulium Tm3+ 12 pale green
Erbium Er3+ 11 pink
Holmium Ho3+ 10 pale yellow
Dysprosium Dy3+ 9 yellow
Terbium Tb3+ 8 pale pink
Gadolinium Gd3+ 7 colourless
Ln2+ and Ln4+ ions do not have colours similar to their isoelectronic L n3+ ions.
The lanthanoid ions other than the f0 type (La3+ and Ce4+) and f14 type (Yb2+ and Lu3+) are all paramagnetic. The paramagnetic behaviour is explained based on unpaired electrons in f-subshells.
All the lanthanoids are silvery white soft metals and tarnish rapidly in air. The hardness increases with increasing atomic number, samarium being steel hard. They have typical metallic structures and are good conductors of heat and electricity. Density and melting point increases smoothly across the series except for Eu and Yb.
5.4.5 Chemical Properties
Properties of lanthanoids are alike. This is due to electrons being filled into the low lying f orbitals and similar outer electronic configuration. In lanthanoids 4f electrons are very effectively shielded by 5s and 5p electrons. Consequently the 4f electrons do not take part in bonding. This is one main difference between d block and f block elements.
In their chemical behaviour, in general, the earlier members of the series are quite reactive similar to calcium, but with increasing atomic number, they behave more like aluminium.
3 o M/M E + of lanthanoids are in the range of –2.2 to –2.4 V.
Lanthanoids react slowly with cold water and quickly with hot water to form hydroxides with liberation of hydrogen gas. These hydroxides are ionic and basic. They are less basic than Ca(OH) 2 but more basic than Al(OH)3. The basic nature of the hydroxides decreases from Ce(OH) 3 to Lu(OH) 3 due to decrease in ionic radius. Lanthanoids on heating in oxygen gas form Ln 2 O 3 oxides. These oxides are ionic and basic in nature. Basic strength of oxides decreases from Ce to Lu.
Carbides like LnC2 and Ln4(C2)3 are formed when the metals are heated with carbon in inert atmosphere. They liberate hydrogen from dilute acids and burn in halogens to form halides. Lanthanods form nitrides LnN on heating with nitrogen gas and form sulphides Ln 2 S 3 on heating with sulphur. Chemical reactions of lanthanoids are given in Fig. 5.9 .
and traces of carbon, silicon, calcium and aluminium. A good deal of mischmetal is used in magnesium-based alloy to produce bullets, shell and lighter flint. Mixed oxides of lanthanoids are employed as catalysts in petroleum cracking. Some individual lanthanoid oxides are used as phosphors in television screens and similar fluorescing surfaces.
TEST YOURSELF
1. Wh i ch one of the following acts like a reducing agent in aqueous solution?
(1) Eu2+ (2) Gd3+
(3) Yb3+ (4) Ce4+
Fig.5.9 Chemical reactions of lanthandes
The lanthanoid ions Ln3+ have a high charge, which favours the formation of complexes. But lanthanoids can not form complex compounds as readily as d-block metals because of their larger sizes. However, complexes with chelating ligands are well known.
Separation of Lanthanoids
The method is the most highly worked. Basic nature of Lanthanoids decreases with the increasing atomic number. Addition of a small amount of a base can precipitate the metal. The basic Lu is precipitated first. Ion exchange method is used to separate first the bigger hydrate ion followed by smaller hydrate ion. Solvent extraction method mainly depends on the solubility of lanthanoids salts. Some of the lanthanoids salts are soluble in water while a few of the lanthanoids salts are soluble in organic solvents.
Uses: The best single use of the lanthanoids is for the production of steels which are used in making plates and pipes. A well known alloy of lanthanoids is mischmetal which consists of a lanthanoids metal ( 95%), iron ( 5%)
2. Magnetic moment of Gd3+ ion (Z = 64) is (1) 3.62 BM (2) 9.72 BM (3) 7.9 BM (4) 10.60 BM
3. Among the following pairs, the one in which both transition metal ions are colourless is (1) Sc3+, Zn2+ (2) Ti4+, Cu2+ (3) V2+, Ti3+ (4) Zn2+, Mn2+
4. Percentage of iron in the alloy mischmetal is (1) 2% (2) 40% (3) 15% (4) 5%
5. Select the smallest ion among Ce 3+, La 3+ , Pm3+, and Yb3+ (1) La3+ (2) Pm3+ (3) Ce3+ (4) Yb3+
6. Which of the following statements is not correct?
(1) La(OH)3 is less basic than Lu(OH) 3.
(2) In lanthanoid series, ionic radius of Ln+3 ions decreases.
(3) La is actually an element of transition series rather than lanthanoid series.
(4) Atomic radii of Zr and Hf are same before lanthanoid contraction.
7. Zr (Z = 40) and Hf (Z = 72) have similar atomic and ionic radii because of (1) belonging to same group
(2) diagonal relationship
(3) lanthanoid contraction
(4) having similar chemical properties
8. Cerium is an important member of lanthanoids. Which of the following statements about cerium is incorrect?
(1) Cerium exhibits +3 and +4 oxidation states.
(2) The +3 oxidation state is more stable than +4 state for cerium.
(3) The +4 oxidation state of cerium is not known in solution.
(4) Ce (iv) is a strong oxidising agent.
9. Which one of the following acts like a reducing agent in aqueous solution?
(1) Eu2+ (2) Gd3+ (3) Yb3+ (4) Ce4+
Answer Key
(1) 1 (2) 3 (3) 1 (4) 4
(5) 4 (6) 1 (7) 3 (8) 3
(9) 1
5.5 THE ACTINOIDS
Fourteen elements from thorium (Z = 90) to lawrencium (Z = 103) are actinoids. The names, symbols and electronic configurations of actinium and actinoids are given in the Table 5.14 . In lanthanoid promethium is the only radioactive element but all actinoids are radioactive elements. In actinide series first three elements (Th, Pa, and U) are naturally occurring elements but the remaining are manmade elements. Natural radioactive elements have long half-lives but artificial radioactive elements have lower half lives.
The heaviest naturally available element is uranium. Elements with Z > 92 are all synthetic.
Elements after uranium are called transuranic elements. Most of the transuranic elements were discovered by Seaborg. Some of the actinoids are prepared only in nanogram quantities only. This factor render their study more difficult.
Table 5.14 Electronic configuration of lanthanides and some of their cations
5.5.1
Electronic Configuration
All t he actinoids are believed to have the electronic configuration of 7s 2 and variable occupancy of the 5f- and 6d-subshells. The fourteen electrons are formally added to 5f, though not in thorium (Z = 90), but from protactinium onwards. The irregularities in the electronic configurations of the actinoids, like those in the lanthanoids are related to the stabilities of the f 0, f 7 and f 14. Thus, the configurations of americium and curium are [Rn]5f7 7s2 and [Rn] 5f7 6d1 7s2. The 5f-orbitals extended into space beyond the 6s- and 6porbitals and participate in bonding. This is indirect contrast to the lanthanoids where the 4f- orbitals are burred deep inside the atom and thus unable to take part in bonding. The participation of 5f orbitals explains the higher oxidation states shown by the earlier actinide elements.
5.5.2
Atomic and Ionic Radii
Similar to lanthanoids, actinoids also exhibit actinide contraction. But actinide contraction is greater from element to element than lanthanoid contraction. This is because the 5f electrons themselves provide poor shielding from element to element in the series. Actinide contraction is less important because the chemistry of element succeeding the actinoids are much less known at the present.
5.5.3 Oxidation States
Like the lanthanoids, common oxidation state of actinide is +3. However, this is not always the most stable oxidation state in actinoids. Oxidation states of actinoids are given in the Table 5.15
In contrast to lanthanoids, the first half of the actinoids exhibit a greater range of higher oxidation states. This may be due to the fact
that 5f, 6d and 7s electrons have comparable energies and participate in bonding. The maximum oxidation state increases from +4 in Th to +5, +6 and +7 in Pa, U, and Np, respectively.
5.5.4 Chemical Reactivity
Actinoids are silvery white metals in appearance but display a variety of structures. The structural variability is obtained due to irregularities in metallic radii which are far greater than in lanthanoids.
Actinoids are highly reactive metals, especially when finely divided. Hydrochloric acid attacks all metals but most of the metals are slightly affected by nitric acid due to the formation of protective oxide layers. Actinoids do not react with alkalies.
5.5.5 Paramagnetism
Actinide ions are generally paramagnetic due to the presence or unpaired electrons. The magnetic properties of actinoids are more complex than those of the lanthanoids. Although the variation in the magnetic susceptibility of the actinoids with the number of unpaired 5f electrons is roughly parallel to the corresponding results for the lanthanoids, the latter have higher values.
5.5.6 Ionisation Potentials
Ionisation energies of early actinoids are lesser than that of the early lanthanoids. This is quite reasonable since it is to be expected that when 5f orbitals are begining to be occupied, they will penetrate less into the inner core of electrons. The 5f- electrons, will therefore, be more effectively shielded from the nuclear charge than the 4f- electrons of the corresponding lanthanoids.
Table 5.15 Oxidate states of actinoids (stable state is given in bold)
Comparison with Lanthanoids
Actinoids are relative more reactive than lanthanoids, because the outer electrons are less firmly held and are available for bonding. The lanthanoid and actinoids contractions, have extended effects on the sizes, and therefore the properties of the elements succeeding them in their respective periods.
The lanthanoid contraction is more important because the chemistry of elements succeeding the actinoids are much less known at the present time.
Application of d- and f-block Elements
Iron and steels are the most important construction materials. Their production is based on the reduction of iron oxides, the removal of impurities and the addition of carbon and alloying metals such as Cr, Mn, and Ni.
Some compounds are manufactured for special purposes such as TiO for the pigment industry and MnO2 for use in dry battery cells. The battery industry also requires Zn and Ni/Cd.
The elements of Group 11 are still worthy of being called the coinage metals, although Ag and Au are restricted to collection items and the contemporary UK ‘copper’ coins are copper-coated steel. The ‘silver’ UK coins are a Cu/Ni alloy.
Many of the metals and/or their compounds are essential catalysts in the chemical industry. V 2 O 5 catalyses the oxidation of SO 2 in the manufacture of sulphuric acid. TiCl 4 with Al(CH3)3 forms the basis of the Ziegler catalysts used to manufacture polyethylene (polythene).
Iron catalysts are used in the Haber process for the production of ammonia from N 2/H 2 mixtures. Nickel catalysts enable the hydrogenation of facts to proceed. In the Wacker process the oxidation of ethyne to ethanal is catalysed by PdCl2, nickel complexes
are useful in the polymerisation of alkynes and other organic compounds such as benzene. The photographic industry relies on the special light-sensitive properties of AgBr.
TEST YOURSELF
1. The lanthanoid that does not show +4 oxidation state is
(1) Dy (2) Ce
(3) Eu (4) Tb
2. The reason for greater range of oxidation states in actinoids is attributed to
(1) actinoid contraction
(2) 5f, 6d, and 7s levels having comparable energies
(3) 4f and 5d levels being close in energies
(4) the redioactive nature of actinoids
3. The actinoids showing +7 oxidation state are
(1) U, Np (2) Pu, Am (3) Np, Pu (4) Am, Cm
4. Which of the following actinoids has / have one electron in 6d orbital?
I. U(Z = 92) II. Np(Z = 93)
III. Pu(Z = 94) IV. Am (Z = 95)
(1) I and III only (2) I and IV only (3) I and II only (4) Only III
5. Which of the following does not contain 5f electrons in its electronic configuration?
(1) Thorium (Z=90)
(2) Americium (Z=95)
(3) Curium (Z=96)
(4) Nobelium (Z=102)
6. Which of the following is not an actinoid?
(1) Td (2) Cf
(3) U (4) Cu
Answer Key
(1) 3 (2) 2 (3) 3 (4) 3
(5) 1 (6) 1
CHAPTER REVIEW
d-Block Elements
■ Elements in which differentiating electron enters into (n-1) d orbitals are called d-block elements.
■ the d-block elements are present in group 3 to 12 of the long form periodic table and lie in between s-block and p-block.
■ Elements of d-block represent a transition from electropositive to electronegative character.
■ Transition elements are defined as the elements having partially filled d-orbitals in elemental form or in their stable oxidation state.
■ All d-block elements except Zn, Cd, and Hg (group 12) are transition elements.
■ There are four transition series, namely 3d, 4d, 5d, which are completely filled and 6d series is incomplete.
■ Elements of 3d Series are : Sc, Ti, V, Cr, Mn, Fe, Co, Ni, Cu and Zn.
■ Elements of 4d Series are : y, Zr, Nb, Mo, Tc, Ru, Rh, Pd, Ag and Cd
■ Elements of 5d Series are : La, Hf, Ta, w, Re, Os, Ir, Pt, Au and Hg.
Electronic Configuration
■ The general electronic configuration of d-block elements is (n–1)d 1–10 ns0-2
■ Only the element Pd has pseudo inert gas configuration. Pd [Z = 46] = 4d 10
■ The atomic and ionic radii of 3d elements are smaller than 4d and 5d elements.
■ The radii of 4d and 5d elements do not differ significantly due to lanthanoid contraction.
■ In first transition series as the electrons are added to 3d orbitals, the atom shows slow contraction in size.
■ These slight irregularities in size have been ascribed to crystal field effects.
■ The atomic and ionic radii of terminal elements of each d-series increases due to repulsions between paired electrons which reduces attractive forces.
■ All transition elements are metals. They are less metallic compared to s-block metals.
■ Transition metals possess metallic lustre. they are hard and brittle. They easily form alloys.
■ Transition metals are less reactive due to high ionisation energy and high heat of sublimation.
■ Pt and Au are almost unreactive. They are called noble metals.
■ The transition metal ions with the configuration d5 or d10 will be more stable.
■ Ferric ion with d5 configuration is more stable than ferrous ion.
General Properties of the Transition Elements
■ Transition elements possess high melting and boiling points. This is due to involvement of electrons from both ns and (n-1)d subshell in metallic bonding.
■ In any series highest melting point is for VI B group elements, due to the presence
5: The d- and f- Block Elements
of maximum number of unpaired electrons that gives strong metallic bonding.
■ Zn, Cd and Hg have very low melting points due to non involvement of d10 configuration in metallic bonding.
■ As number of valence electrons increases, strength of metallic bond, enthalpy of atomisation increases.
■ In a period decrease in atomic size is less compared to decrease in atomic size of representative elements. This is due to added (n-1)d electrons providing screening effect.
■ Conductivities of transition elements are very high. Silver metal is the best conductor.
■ Densities of transition metals are very high due to their low atomic volumes.
■ In 3d series highest density is observed in ‘Cu’ and least density in ‘Sc’.
■ VIII group elements possess maximum densities. Very high densities are exhibited by Ir and Os.
■ Ionisation potential of ‘3d’ transition metals increases from Sc to Fe, then decreases to Ni and increases in Cu and Zn.
■ Transition elements have high ionisation potentials and low heat of hydration.
■ Transition elements are less reactive than s-block elements. The least reactive metals are Pt, Ir, and Au.
■ The increase in I.P. while passing from 4d to 5d is larger than 3d to 4d. The increase in I.P. from ‘Zr’ to ‘Hf’ is 86 kJ/mole, whereas from ‘Ti’ to ‘Zr’ is 18 kJ/mole.
■ 5d’ transition elements possess higher ionisation energies than 3d and 4d elements. This is due to the greater effective nuclear charge ineffective shielding of the nucleus by 4f-electrons.
■ Decreasing order of IP1 values :
Sc > Y > La, ; Hg > Zn > Cd, ; Hf > Zr > Ti
■ IP 2 values are much higher for Cr and Cu as they have stable configurations in unipositive state, with considerable loss of exchange energy.
■ IP 3 values relatively high for Zn as it is getting stable configuration after IP 2
■ d-Block elements form ionic bonds in lower oxidation states and covalent bonds in higher oxidation states.
■ Transition metals possess good catalytic properties due to the free valencies and ability to show variable oxidation states.
Some Important Compounds of Transition Elements
■ The characteristic properties of transition elements are: variable oxidation states, formation of coloured ions, paramagnetism, formation of alloys, complex compounds, etc.
■ Transition metals exhibit variable oxidation states as they use outer ns electrons as well as inner (n–1) d-electrons in bonding.
■ The oxidation states of transition metals in their compounds will be in between 0 and +8.
■ The common oxidation state +2, due to loss of electrons either from ns or (n-1)d levels.
■ In 3d series lowest oxidation state is +1, shown by Cr and Cu. Highest oxidation state of 3d element is +7 and is shown by Mn.
■ In 4d series highest oxidation state is +8 (unstable) shown by Ru in RuO 4.
■ In 5d series highest is +8 (stable) shown by Os in OsO4.
■ Substances are classified as paramagnetic, diamagnetic and ferromagnetic on the basis of their magnetic behaviour in an external magnetic field.
■ If the magnetic lines of forces are drawn into the substance the field (B) in the substance is greater than the applied field (H) such a substance is called paramagnetic substance.
■ Paramagnetic substances are weakly attracted in to a magnetic field.
■ Examples paramagnetic substances are : O2, NO, NO2, ClO2, Fe +2, Cu +2, K 3[Fe(CN)6], Cr3+.
■ Paramagnetic nature is due to the presence of unpaired electrons.
■ If the magnetic lines of forces are repelled by the substance then the field (B) is less than the applied field (H). Such a substance is called diamagnetic substance
■ Diamagnetic substances are weakly repelled in a magnetic field. They do not contain unpaired electrons.
■ Examples of diamagnetic substances are O 3, H 2O 2, Zn 2+, alkaline earth metal vapours.
■ In ferromagnetic substances all magnetic moments are alligned in the same direction. Ferromagnetism is a special case of paramagnetism.
■ In ferromagnetic substances the field strength, (B) >>> (H).
■ Fe, Co and Ni are examples of ferromagnetic d-block elements.
■ Ferromagnetism disappear in the solution form of the substance.
■ Transition metal ions are generally paramag-netic since they contain unpaired electrons.
■ Magnetic property is measured in Bohr magnetons (BM). One BM is 9.273×10 –24 JT–1 (S.I.Unit).
■ If ‘n’ is the number of unpaired electrons, the magnetic moment is (2) BM µ=+ nn
■ Au+ is diamagnetic due to 5d10configuration and is colourless.
■ Au +3 is paramagnetic due to 5d 8 configuration and is green in colour.
■ The solution of solids of transition metal ions absorb light in visible region and exhibit colours.
■ The d-d transition causes colour, if the ion contains unpaired d-electrons.
■ [Ti (H2O)6]Cl3 absorbs light of wave length 500 to 575 nm and transmit red and violet. Hence it appears purple.
■ Fe 2+ is green, Fe 3+ is yellow, Cr 2+ is blue, Cr3+ is green, Mn2+ is pink, Mn6+ is green.
■ Hydrated Cu + ,Zn +2 , Sc 3+ , Ti 4+ , Mn +7 are colourless as d–d transition is not possible.
■ But in oxo ions 2 CrO27 , 2 4 CrO , 4 MnO the colour is due to charge transfer phenomenon.
■ An intimate mixture having physical properties similar to that of the metal, formed by a metal with other metals or metalloid or sometimes a non-metal, is called an alloy.
■ Alloys are generally prepared to modify malleability, ductility, toughness, resistance to corrossion etc, so as to suit industry.
■ Alloys are classified into ferrous type and non ferrous type depending upon the presence or absence of iron in the alloy.
■ Bronze contains Cu–75% to 90%, Sn–10 to 25%. It is used in utensils, coins and statues.
■ Brass contains Cu 60% to 80%, Zn 20% to 40%. It is used in machinery parts.
■ Transition metals form interstetial or nonstoichiometric compounds with nonmetals such as carbon, hydrogen, boron and nitrogen. The non metal (guest) atoms are accommodated in the voids of crystals.
■ The non metal atoms occupy holes without altering metal lattice. But the lattice
CHAPTER 5: The d- and f- Block Elements
expands a little. Due to this the density of interstetial compounds is less than that of metal.
■ Metals like Ti, V, Zr, Nb, Hf, and Ta form interstetial compounds. Hydrogen always occupy tetrahedral voids and C and N occupy larger octahedral voids.
■ Interstitial compounds will not possess any chemical bond. Since no bond is present between guest and host elements they are called no bond compounds.
■ Catalytic activity of transition metals depends on their ability to exist in different states, due to variable oxidation states and presence of vacant ‘d’ orbitals.
■ Potassium dichormate is prepared from chromite ore fusing with sodium carbonate in air.
■ Potassium permanganate is prepared from pyrolusite ore.
■ Chromate, manganate and permanganate are all tetrahedral.
The Lanthanoids
■ The elements from 57La to 71Lu are known as lanthanoids. They are characterized by the filling of 4f orbitals in their atoms.
■ The lanthanoids occur as orthophosphates in monazite sand. It is a mixture containing mostly ‘La’ phosphate and trivalent phosphates of Ce, Pr and Nd.
■ The general electronic configuration of the lanthano is [Xe]4f0–14 5d0–1 6s2 (or) [Pd]
4f 0–14 5d0–1 6s2
■ Lanthanoids are silvery white in colour. They are malleable, ductile, have low tensile strength and are good conductors of heat.
■ Lanthanoids have relatively high density and posses high melting points.
■ Many of the lanthanoid ions are coloured due to the electronic transition between different 4f levels.
■ The majority of the lanthanoid ions exhibit paramagnetism due to the presence of unpaired electrons.
■ The oxides and hydroxides of the lanthanoids are basic in character.
■ Lanthanoids are called rare earths because of their rare abundance in the earth’s crust.
■ Lanthanoids have high electropositive character due to low ionisation energies.
■ Lanthanoids mostly exhibit +3 in their compounds. They exhibit +2 and +4 states where they are able to get half filled or completely filled orbitals.
■ Ce +4 has noble gas configuration. Still it comes back to +3 by accepting electron, and hence, acts as oxidant.
■ Eu+2 has F7 configuration. It returns back to Eu+3 by losing electrons. Hence Eu +2 is reductant.
■ Lanthanoids react slowly with cold water and quickly with hot water liberating H 2 gas.
■ Lanthanoids form basic oxides M 2O3 and hydroxides M(OH)3
■ The basic nature of hydroxides decreases from Ce(OH)3 to Lu(OH)3.
■ In air lanthanoids get oxidised to form M 2 O 3 . Their oxides are ionic in nature. These oxides are basic in nature and the basic nature decreases with increase in atomic weights.
■ Lanthanoids do not form coordination compounds as readily as d-block elements due to their larger sizes. However, complexes with chelating ligands are well known for lanthanoids.
■ The magnetic moment and colours show periodicity in lanthanoids.
■ The colour of lanthanoid (III) ion with 4fn configuration is similar to the lanthanoid (III) ion with 4f(14-n) configuration.
■ 4d series elements are larger than 3d series. The size of 4d and 5d are almost same, due to lanthanoid contraction.
■ Hf & Zr, Nb & Ta, Mo & W pairs shows same size and same chemical properties.
The Actinoids
■ The elements from 90 Tb to 103Lw are known as actinoids. They are mostly synthetic.
■ Actinoids are highly reactive metals in their finally divided state.
■ Their differentiating electron enters 5f subshell.
■ Their common oxidation state is +3
■ They generally show paramagnetism.
■ Alloys of iron are the most important construction materials.
■ MnO2 is used in dry cells.
■ Oxides of Lanthanoids are used as catalysts in petroleum cracking
■ Lanthanoids oxides are used in television screens.
Exercises
JEE MAIN LEVEL
Level - I
Position in the periodic table
Single Option Correct MCQs
1. d - block elements are placed in the modern periodic table.
(1) After Mg in 3rd period
(2) After Ca in 3rd period
(3) After Mg in 4th period
(4) After Ca in 4th period
2. The correct statement about the fourth transition series is
(1) It corresponds to the filling of the 4d sublevel
(2) It corresponds to the filling of the 3d sublevel
(3) It corresponds to the filling of the 5d sublevel
(4) It corresponds to the filling of the 6d sublevel
Electronic configuration
3. Which of the following d-block metal has 4f14- electronic configuration in their inner shell?
(1) Zn (2) Cd
(3) Hg (4) Ag
4. The electronic configuration of Cr 3+is
(1) [Ar]3d54s1
(2) [Ar]3d24s1
(3) [Ar]3d34s0
(4) [Ar]3d44s2
5. Which of the elements possess only one electron in 5d-orbital.
(1) Tm, Pm
(2) Pr, Lu
(3) La, Pm
(4) La, Lu
6. The outer electronic configuration of Ag is 4d 10 5s1, it belongs to (1) 5th period, group 4 (2) 4th period, group 5
(3) 5th period, group 11
(4) 6th period, group 9
7. Element with atomic number 30 belongs to (1) s-block
(2) d-block
(3) p-block
(4) f-block
Numerical Value Questions
8. Number of unpaired electrons in Cr-atom in its ground state electronic configuration.
9. Number of electrons that vanadium (Z = 23) has in p-orbitals is equal to _____.
10. The sum of ns and (n – 1)d electrons in Tc are:
11. Number of electrons present in the outer most shell of Cu atom is ____.
General trends in properties
Single Option Correct MCQs
12. The 3d-series element having the highest atomic radii is
(1) Sc (2) Zn
(3) N (4) V
13. Atomic radius is least for (1) Sc (2) Cr
(3) Fe (4) Ni
14. The ionic radius of V+2 is 84 pm, then the ionic radius of Cr+2 is
(1) 80 pm (2) 76 pm
(3) 88 pm (4) 72 pm
15. The ionic radius of V+3 is 74 pm, then the ionic radius of Cr+3 is
(1) 69 pm
(2) 76 pm
(3) 99 pm
(4) 33 pm
16. The highest melting point of 3d-series element is for
(1) W
(2) Cu
(3) Mn
(4) Cr
General trends in properties
Single Option Correct MCQs
17. Maximum oxidation state exhibited by Os is
(1) +7 (2) +6
(3) +8 (4) +5
18. Which of the following is not a coinage metal?
(1) Cu
(2) Ag
(3) Au
(4) Zn
19. Gold and silver are called noble metals, because:
(1) They do not normally react
(2) Even acids cannot dissolve them
(3) They are used in jewellery
(4) They are very chemical reactive
20. Which is not correct for transition metals?
(1) Variable oxidation states
(2) Complex formation
(3) Partially filled ��-orbitals
(4) All the ions are colorless
21. The element which does not show variable oxidation state
(1) Sc
(2) V
(3) Fe
(4) Hg
22. The electronic configuration of Ti +3 is
(1) (Ar)4503d1
(2) (Ar)4503d2
(3) (Ar)4523d1
(4) (Ar)4503d3
23. Transition elements make complex compounds due to
(1) Presence of empty d-Orbital
(2) large size
(3) variable valency
(4) presence of high nuclear charge
24. Incorrect statement regarding interstitial compounds is
(1) They have high melting points, higher than those of pure metals
(2) They are very hard
(3) They retain metallic conductivity
(4) They are chemically reactive
25. In the 3d – series, the element with highest melting point is
(1) Mn
(2) Fe
(3) Cr
(4) Cu
26. Which of the following element has high density?
(1) Sc
(2) Zn
(3) Cu
(4) Co
27. In 3d series the element with least density is
(1) Zn (2) Sc
(3) Cu (4) Mn
28. Enthalpy of atomization in 3d series is least for
(1) Zn (2) Cr
(3) Fe (4) V
29. Highest oxidation state shown by Mn when reacts with flourine is
(1) +7
(2) +6
(3) +2
(4) +4
30. Colourless ion in aqueous solution is (1) Cu2+
(2) Co2+
(3) Zn2+
(4) Ni2+
31. d-block elements show variable oxidation states due to
(1) participation of ns electrons
(2) participation of ns and (n−1)d electrons
(3) participation of ( n−1)d electrons
(4) participation of np electrons
32. The oxidation state of ‘Ni’ in Ni(CO) 4 is (1) +II
(2) +III
(3) +VIII
(4) Zero
33. Liquid metal among d-block elements is (1) Hg
(2) Zn
(3) Nb
(4) Cd
34. Highly volatile metal among the following is
(1) V
(2) Zn
(3) Cr
(4) Mn
35. The 3d element with positive E°(M +2/M) valueis
(1) Sc
(2) Ti
(3) Cr
(4) Cu
36. Which one of the following oxide is not basic?
(1) V2O3
(2) V2O4
(3) CrO
(4) Mn2O7
37. The highest oxidation state is exhibited by the transition metal with electronic configuration
(1) (n–1)d5ns1
(2) (n–1)d5ns2
(3) (n–1)d8ns2
(4) (n–1)d6ns2
38. Which of the following would be attracted towards magnetic field
(1) Zn
(2) Mn
(3) Mg
(4) Cd
39. The common oxidation state exhibited by transition elements is
(1) +II
(2) +IV
(3) +VI
(4) +VII
40. Number of unpaired electrons in Ni2+(Z=28) is
(1) 4
(2) 2
(3) 6
(4) 8
41. The ions from among the following which are colourless are:
(i) Ti4+ (ii) Cu+ (iii) Co3+ (iv) Fe2+
(1) (i) and (ii) only
(2) (i), (ii) and (iii)
(3) (iii) and (iv) (4) (ii) and (iii).
42. All the following cannot exist e xcept.
(1) FeI3
(2) CuF
(3) MnF 7
(4) MnO3F
43. Interstitial hydride is given by (1) Cr
(2) Mo
(3) W
(4) Mn
44. The major compound of German silver are:
(1) Cu, Zn and Ni
(2) Ge, Cu and Ag
(3) Zn, Ni and Ag
(4) Cu, Zn and Ag
Numerical Value Questions
45. How many of the 3d-series elements have atomic radius greater than Cr.
46. How many 4d-series elements have an atomic radius more than Mo?
47. The common oxidation state of transition elements is
48. The oxidation number of the metal in Zeisel’s salt is
49. How many 3d-series elements have fewer densities than Zn?
50. How many 3d- series elements have the standard M 2+ (aq)/M(s) potential with positive sign?
51. Among the 3d series elements, the maximum oxidation state exhibited is
52. Out of the following how many of them are coloured
53. The percentage of silver in the alloy of German silver is ____.
54. Oxide of vanadium(X) is used as a catalyst in the manufacture of H2SO4. Atomicity of (X) is Y and the oxidation state of vanadium in (X) is (Z). Find Z+Y value
55. V +x , Cr +y have the same outer electronic configuration as 3d3. Find the value of xy?
56. The number of 3d electrons in Mn +3 and Mn +2 are respectively ‘x’ and ‘y’. Find the product of xy?
57. The calculated magnetic momentum of Ti+x and Cu+y is the same and equal to 1. 73 BM. Find the sum of 3d electrons present in both the ions?
58. The oxidation states of transition metal atoms in K2Cr2O7, KMnO4 and K2FeO4 are x, y and z respectively. The sum of x, y and z is ______
Some important compounds
Single Option Correct MCQs
59. Potassium manganate (K2MnO4) is formed when
(1) Chlorine is passed into aqueous KMnO4 solution
(2) Manganese dioxide is fused with potassium hydroxide in air
(3) Formaldehyde reacts with potassium permanganate in presence of strong alkali
(4) Potassium permanganate reacts with concentrated sulphuric acid
60. The equilibrium 22 27 4 CrO2CrO is shifted to the right in
(1) an acidic medium
(2) a basic medium
(3) a weakly acidic medium
(4) a neutral medium
61. The yellow colour of chromates changes to orange on acidification due to the formation of ___?
(1) Cr+3 (2) Cr2O3
(3) 2 CrO27 (4) 2 4 CrO
62. Guignet’s green is known as
(1) Cr2O3.2H2O
(2) FeO3.2H2O
(3) Cu2O3
(4) FeCO3.Cr2O3
63. Permanganate ion is isostructural with (1) Chlorate ion
(2) Perchlorate ion
(3) Chlorite ion
(4) Nitrate ion
64. Structure of 2 CrO27 has (1) Two tetrahedra sharing one corner
(2) Two tetrahedra sharing one edge
(3) Two tetrahedra sharing one face
(4) Irregular octahedral
65. Number of Cr – O sigma bonds in dichromate ion 2 CrO27 is
(1) 6
(2) 7
(3) 8
(4) 4
66. The colour of KMnO4 is due to:
(1) L → M charge transfer transition
(2) σ−σ∗ transition
(3) M→ L charge transfer transition
(4) d - d transition
67. The starting material for the manufacture of KMnO4 is
(1) pyrolusite
(2) manganite
(3) magnatite
(4) haematite
68. Green colour of manganate ion is due to (1) Charge transfer
(2) d-d transition
(3) Half filled d-sub shells
(4) Increasing number of unpaired electrons
69. In potassium manganate, the oxidation state of manganese is
(1) +5 (2) +6 (3) +7 (4) +8
Numerical Value Questions
70. The number of oxygen atoms that are directly attached to one chromium in dichromate ion are
71. The total change in the oxidation state of manganese involved in the reaction of KMnO4 and potassium iodide in the acidic medium is
72. When K 2Cr 2O 7 is heated to a white heat, liberating oxygen gas and forming chromium compounds (X) and (Y). Find the product of oxidation states per Cr atoms in (X) and (Y)?
73. What is the change in oxidation state of “Mn”when solid KMnO 4 is heated in a current of H2?
Lanthanoids
Single Option Correct MCQs
74. Name a member of the lanthanoid series which is well known to exhibit +4 oxidation state
(1) Lanthanum (2) Cerium
(3) Lutetium (4) Terbium
75. The elements in which electrons are progressively filled in 4f orbital are called (1) actinoids
(2) transition elements
(3) lanthanoids
(4) halogens
76. The general electronic configuration of lanthanoids is
(1) [xe]4f0-14, 5d0-10, 6s2
(2) [xe]4f1-14, 5d0-1, 6s2
(3) [xe]4f1-14, 5d0-1, 6s1-2
(4) [xe]4f0-14, 5d0-1, 6s1-2
77. Zr and Hf have almost equal atomic and ionic radii due to
(1) Actinoid contraction
(2) Lanthanoid contraction
(3) Belonging to the d block
(4) diagonal relation ship
78. Which of the following oxidation state is common for all Lanthanoids?
(1) +2
(2) +3
(3) +4
(4) +5
79. Which one of the following acts like a reducing agent in aqueous solution ?
(1) Eu2+
(2) Gd3+
(3) Yb3+
(4) Ce4+
80. Which one of the following is steel hard with high melting point
(1) Eu
(2) Yb
(3) Ce
(4) Sm
81. Mischmetal is an alloy consisting mainly of:
(1) Lanthanoid metals
(2) Actinoid and transition metals
(3) actinoid metals
(4) Lanthanoid and actinoid metals
82. The lanthanide ion that would show colour is
(1) Gd3+
(2) Sm3+
(3) La3+
(4) Lu3+
83. Maximum number of unpaired electrons are present in:
(1) Gd+3
(2) Yb+2
(3) Tb+2
(4) Pm+3
84. The only lanthanoid which is radioactive is (1) cerium
(2) promethium
(3) praseodyrnium
(4) lanthanum
85. When Lanthanoid element is heated with sulphur then compound obtained is
(1) Ln2S2
(2) Ln2S3 (3) Ln2S
(4) Ln3S4
Numerical Value Questions
86. Number of colorless lanthanide ions among the following is _____ Eu3+, Lu3+, Nd3+, La3+, Sm3+
87. The number of 4f electrons in ground state electronic configuration of Gd (Z=64) is____
88. (X) is the only synthetic radio active element of lanthanoid. Its electronic configuration is [Xe}4fx5dy6sz. Find the value of (xyz).
Actinoids
Single Option Correct MCQs
89. The highest possible oxidation states of uranium and plutonium respectively are (1) 6 and 7 (2) 6 and 4 (3) 7 and 6 (4) 4 and 6
90. Common oxidation state of Actinoids is (1) +3
(2) +2
(3) +1
(4) +7
91. Highest oxidation state exhibited by actinoid is ____
(1) Th (2) U
(3) Np (4) Lr
5: The d- and f- Block Elements
92. Nobel gas core in the electronic configuration of lanthanides and actinides respectively are
(1) [Xe],[Rn]
(2) [Kr],[Xe]
(3) [Rn],[Xe]
(4) [Rn],[Rn]
93. The element with the configuration [Rn]7s26d2 is
(1) d - Block element
(2) s - Block element
(3) f - Block element
(4) p - Block element
94. Actinide that is naturally available with highest isotopic number (1) U (2) Pu (3) Cu (4) Fm
95. Number of ‘f’ electrons in thorium are (1) 14 (2) 16 (3) zero (4) 15
Numerical Value Questions
96. Number of radioactive elements in rare earth elements is ‘x’ Number of radioactive elements amongst a ctin ides is ‘y’. Then 2xy 2 + is _______
97. The number of ‘f ’ electrons in the ground state electronic configuration of Np(Z=93) is
98. How many of the following statements are correct?
I) Lanthanoids are silvery white metals and tarnish in air
II) Actinoids are silvery in appearance
III) The chemistry of the actinoids is much more complicated than the lanthanoids
IV) Both La3+ and Lu3+ are yellow coloured in aqueous solutions
V) Among actinoids, the elements in the first half of the series frequently exhibit higher oxidation states.
VI) The magnetic properties of the actinoids are more complex than those of the lanthanoids
99. Number of lanthanoid ions (Ln3+) having half filled f7 configuration = x
Number of actinoid ions (An3+) having half filled f7 configuration = y
Then (x + y) is _______
100.
Lanthanoid ion/ Actinoid ion
Number of unpaired electrons
Pr+3 X
Er+2 Y
Pu+7 Z
Then (X+Y+Z) is
Level - II
Electronic Configuration
Single Option Correct MCQs
1. All the following have anomalous electronic configuration except (1) Cr (2) Cu (3) Ni (4) Pd
2. Outer electronic configuration of palladium is
(1) 4d5 5s1 (2) 4d9 5s2
(3) 4d10 5s1 (4) 4d10 5s0
3. Which set of elements among the following are called non-transitional elements?
(1) Cu, Ag, and Au
(2) Fe,Co, and Ni
(3) Zn, Cd, and Hg
(4) Re, Os, And Ir
General Properties
Single Option Correct MCQs
4. Element having minimum enthalpy of atomisation is
(1) Cr (2) Zn
(3) V (4) Co
5. The 3d series element with positive SRP (M2+/M) is
(1) Zn (2) Co
(3) Cu (4) Mn
6. Choose the wrong statement among the following about transition element.
(1) The highest oxidation is achieved with oxygen and fluorine, and lower oxidation states are achieved with iodine and carbon monoxide.
(2) Highest oxidation of d-block elements is stabilised mainly by oxygen rather than fluorine.
(3) Highest oxidation state is observed in the middle elements.
(4) Stability of lower oxidation increases down the group in d-block elements.
7. Select the set of correct statements is:
i. Manganese exhibits +7 oxidation state in its oxide.
ii. Ruthenium and osmium exhibit +8 oxidation in their oxides.
iii. Scandium shows +4 oxidation state, which is oxidising in nature.
iv. Cr shows oxidising nature in +6 oxidation state.
(1) (i) and (iii)
(2) (ii), (iii) and (iv)
(3) (i), (ii), and (iv)
(4) (ii) and (iii)
8. Magnetic moment 2.84 BM is given by (At. no. Ni =28, Ti = 22, Cr = 24, Co = 27)
(1) Ti3+ (2) Cr2+
(3) Co2+ (4) Ni2+
9. Which of the following 3d-metal ions will give the lowest enthalpy of hydration (ΔhydH) when dissolved in water?
(1) Cr2+ (2) Mn2+
(3) Fe2+ (4) Co2+
10. Among the following, which is the strongest oxidising agent?
(1) Mn3+ (2) Fe3+
(3) Ti3+ (4) Cr3+
11. The correc t order of basicity of oxides of vanadium is
(1) V2O5 > V2O4 > V2O3
(2) V2O3 > V2O4 > V2O5
(3) V2O4 > V2O3 > V2O5
(4) V2O3 > V2O5 > V2O4
12. Regarding interstitial compounds select the incorrect statement.
(1) They have melting point higher than those of pure metals.
(2) They are very hard.
(3) They retain metallic conductivity.
(4) They are chemically reactive.
Numerical Value Questions
13. How many of the following alloys match with given major component metals in them?
German silver = Ni + Zn + Cu
Magnalium= Mg+Al
Bronze= Zn+Sn
Misch metal = Lanthanoid+Fe
Brass = Cu+Zn
Coinage alloy = Ni + Cu
14. How many of the following metal ions have similar value of spin only magnetic moment in gaseous state?
V3+, Cr3+, Fe2+, Ni3+
(Given: Atomic number: V=23;Cr=24;Fe=26;Ni=28)
15. The number of hydrogen bonded water molecules associated with CuSO4.5H2O is ____.
16. How many of the following statements are correct?
i. Interstitial compounds contain nonmetal atoms trapped inside the metal crystal whereas alloys are homogeneous blends of metals.
ii. Steel and bronze are alloys of transition and non-transition metals.
iii. Some boride containing interstitial compounds are very hard and approach diamond in hardness.
iv. Interstitial compounds are chemically more reactive than parent metal.
17. Out of the following, how many of them are coloured in aqueous solution?
21. In neutral or alkaline solution, MnO 4 –oxidises thiosulphate to (1) 2-SO27 (2) 2-SO28 (3) 2 3SO (4) 24 SO
22. ()() 23 X+KCO+Air Y → air
Which of the following is correct?
(1) X = black, MnO2, Y = blue, K2CrO4, Z = KMnO4
(2) X = green Cr2O3, Y = yellow, K2CrO4, Z = K2Cr2O7
(3) X= black, MnO2, Y = green, K2MnO4, Z= KMnO4
(4) X= black Bi2O3, Y = colourless, KBiO2, Z = BiO3
23. KMnO 4 oxidises I – in acidic and neutral/ faintly alkaline solution, respectively, to
(1) I2 and I2
(2) IO3– and IO3–
(3) IO3– and I2
(4) I2 and IO3–
24. Hydrochloric acid cannot be used to make the medium acidic in the oxidation reactions of KMnO4. Why?
(1) Both HCl and KMnO4 act as oxidising agents.
(2) KMnO4 oxidises HCl into Cl2, which is also an oxidising agent.
(3) KMnO4 is a weaker oxidising agent than HCl.
(4) KMnO 4 acts as a reducing agent in presence of HCl.
25. When H2O2 is added to ice-cold solution of acidified potassium dichromate containing ether, and the contents are shaken and allowed to stand,
(1) a blue colour is obtained in ether due to formation of a Cr2(SO4)3
(2) a blue colour is obtained in ether due to formation of CrO5
(3) CrO3 is formed, which dissolves in ether to give blue colour
(4) chromyl chloride is formed
26. For dichro mate dianion, which of the following properties is incorrect?
(1) Two 24 CrO tetrahedrons are joined through oxygen.
(2) Terminal Cr−O bonds are shorter than bridge Cr−O bond lengths.
(3) The hybridisation of chromium in 2-CrO27 is sp3.
(4) O−Cr−O bond angle is less than that of Cr−O−Cr bond angle.
27. Identify the incorrect statement from the following
(1) Solubility of Na2Cr2O7 is more than that of K2Cr2O7.
(2) K 2 Cr 2 O 7 is the primary standard in volumetric analysis.
(3) Na2CrO4 is orange coloured.
(4) All Cr−O bond lengths in 24 CrO are equal.
Numerical Value questions
28. On heating x grams of potassium dichloronate, 9.6 g of O 2 is liberated. What is the weight of x taken? (At. wt. K = 39, Cr = 52, O =16)
29. The number of electrons involved in oxidation of iodide ion by permanganate in alkaline solution per iodide ion is _____.
30. 24 MnO is quite strongly oxidising and is stable only in very strong alkali. In acidic solution it disproportionates as a 24 MnO
+H+ → ____MnO2+H2O Find the value of ‘a’ in the above balanced reaction: Lanthanides
Single Option Correct MCQs
31. Which of the following elements have half-filled f-orbitals in their ground state?
(Given: Atomic number of Sm = 62; Eu = 63; Tb = 65; Gd = 64; Pm = 61)
A) Sm B) Eu
C) Tb D) Gd
E) Pm
Choose the correct answer from the options given below.
(1) A and B only (2) C and D only
(3) B and D only (4) A and E only
32. Gadolinium belongs to 4f series. Its atomic number is 64. Which of the following is the correct electronic configuration of gadolinium?
(1) [Xe]4f95s1 (2) [Xe]4f75d1s2
(3) [Xe]4f45d46s2 (4) [Xe]4f86d2
33. Identify the wrong order for the property indicated against it.
(1) NCl3 > NH3 >NF3: Bond angle
(2) Lu(OH)3 > Eu(OH)3 > La(OH)3: Basic strength
(3) Ba(OH)2 > Ca(OH)2 > Be(OH)2: Solubility in water at same temperature (4) HOCl > HClO > HClO3: Oxidisation power
34. Among Ce(4f1 5d1 6s2), Nd(4f4 6s2), Eu(4f76s2), and Dy 4f106s2, the elements having highest and lowest third ionisation energies, respectively, are (1) Nd and Ce (2) Eu and Dy (3) Eu and Ce (4) Dy and Nd
35. Misch metal is a pyrophoric alloy. It is an alloy of lanthanoid elements. Percentage of lanthanoid elements in the alloy misch metal is (1) 95% (2) 80% (3) 60% (4) 75%
Numerical Value Questions
36. The number of unpaired electrons present Gd+2 ion (At. no. of Gd=64) is _____.
37. Number of electrons present in 4f−orbital of Ho +3 ion is (given atomic number of Ho = 67) _______.
38. The total number of unpaired electrons in Gd (Z =64) is _______.
Actinides
Single Option Correct MCQs
39. Which of the following electronic configurations of the corresponding element is wrongly matched?
(1) Thulium - [Xe]4f136s2
(2) Americium - [Rn]5f77s2
(3) Uranium - [Rn]5f47s2
(4) Terbium - [Xe]4f96s2
40. The highest oxidation state exhibited by an actinoid element in its compound is
(1) +6 (2) +5
(3) +7 (4) +4
41. Choose the correct statements among the following.
A) Actinoids show variable oxidation states.
B) Highest oxidation state shown by actinoids in their compounds is +7.
C) All actinoids are radioactive elements.
D) Actinoids react with acids and alkalies.
(1) A, B, C, and D only
(2) C and D only
(3) A and B only
(4) A, B, and C only
Numerical Value Questions
42.
Lanthanoid ion/ Actinoid ion
Number of unpaired electrons
Pr+3 x
Er+2 y
Pu+7 z
Then (x+y+z) is:
43. Numb er of lanthanoid ions (Ln 3+ ) having half filled f 7 configuration = x.
Number of actinoid ions (An 3+ ) having half filled f 7 configuration = y.
Then, (x+y) is ___
44. How many of the following statements are correct?
i. Lanthanoids are silvery white metals and tarnish in air.
ii. Actinoids are silvery in appearance.
iii. The chemistry of the actinoids is much more complicated than that of the lanthanoids.
iv. Both La3+ and Lu3+ are yellow coloured in aqueous solutions.
v. Among actinoids, the elements in the first half of the series frequently exhibit higher oxidation states.
vi. The magnetic properties of the actinoids are more complex than those of the lanthanoids.
Multi-concept Questions
Single Correct MCQs
45. The spin only magnetic moment value of the most basic oxide of vanadium among V2O3, V2O4, and V2O5 is _____BM.
(1) 2 (2) 3 (3) 4 (4) 1
46. Potassium permanganate, on heating at 513 K gives a product that is (1) paramagnetic and colourless (2) diamagnetic and colourless (3) diamagnetic and green (4) paramagnetic and green
47. The maximum number of possible oxidation states of actinides are shown by (1) nerkelium (BK) and californium (cf)
(2) nobelium (No) and lawrencium (Lr)
(3) actinium (Ac) and thorium (Th) (4) neptunium (Np) and plutonium (Pu)
48. Given below are two statements.
Statement I : CeO 2 can be used for oxidation of aldehydes and ketones.
Statement II : Aqueous solution of EuSO4 is a strong reducing agent.
Choose the correct answer from the options given below
(1) Both statement I and II are correct
(2) Both statement I and II are incorect
(3) Statement I is true but statement II is incorrect
(4) Statement I is false but statement II is correct
49. Number of electrons transferred in each case when KMnO4 acts as an oxidising agent to give MnO 2,Mn 2+, Mn(OH) 3, and MnO 4 2− , are, respectively, (1) 3, 5, 4 and 1 (2) 4, 3, 1 and 5 (3) 1, 3, 4 and 5 (4) 5, 4, 3 and 1
51. (NH4)2Cr2O7, on heating, liberates a gas. The same gas will be obtained by (1) heating NH4NO3 (2) heating NH4NO2 (3) treating H2O2 with NaNO2 (4) treating Mg3N2 with H2O
52. Thermal decomposition of a Mn compound (X) at 513 K results in compound (Y), MnO2 and a gaseous product. MnO2 reacts with N aCl and concentrated H 2 SO 4 to give a pungent gas. Z, X, Y, and Z, respectively, are (1) K3MnO4, K2MnO4 and Cl2 (2) K2MnO4, KMnO4 and SO2 (3) KMnO4, K2MnO4 and Cl2 (4) K2MnO4, KMnO4 and Cl2
53. Select the wrong statement regarding transition metals among the following (1) 4s electrons penetrate towards the nucleus more than 3d electrons.
(2) Atomic radii of transition metals increase rapidly with increase in atomic number because of poor shielding of nuclear attraction by (n-1)d electrons.
(3) Second and third transition series elements have nearly the same size
(4) Their densities are higher, and densities of the 5d series elements are higher than those of 4d series elements.
54. Ce3+, La3+, Pm3+, and Yb3+ have ionic radii in the increasing order as (1) La3+ > Ce3+ > Pm3+> Yb3+
(2) Yb3+ < La3+ < Ce3+ < Pm3+
(3) La3+ < Ce3+ < Pm3+ < Yb3+
(4) Yb3+ < PM3+ < La3+ < Ce3+
55. In the dichromate dianion, (1) 4 Cr–O bonds are equivalent (2) 6 Cr–O bonds are equivalent (3) All Cr–O bonds are equivalent (4) All Cr–O bonds are non-equivalent
56. 24 CrO (yellow) change to 2-CrO7 (orange) in pH = X and vice versa in pH = Y. Hence, X and Y are
(1) 6, 8 (2) 6, 5 (3) 8, 6 (4) 7, 7
Numerical Value Questions
57. Among Co 3+, Ti 2+,V 2+, and Cr 2+ions, one, if used as a reagent, cannot liberate H 2 from dilute mineral acid solution. Its spinonly magnetic moment in gaseous state is _______BM. (Nearest integer)
58. In alkaline medium KMnO4 can be converted into MnO2. The decrease in oxidation state is ‘n’. So equivalent mass of
Molar mass
KMnO . = n What is the value of ‘n’ ?
4
59. How many of the following are paramagnetic La3+, Ce4+, Yb2+, Lu3+, O2, S2,Th3+,Pd
60. The sum of ns and (n-1)d electrons in Tc are
61. Ferrous ion changes to ‘X’ ion, on reacting with acidified hydrogen peroxide. The magnetic moment of the ion is______
Level - III
1. The mass of potassium dichromate crystals required to oxidize 750 cm3 of 0.6 M Mohr’s salt solution is (Given, molar mass: potassium dichromate=294, Mohr’s salt=392)
(1) 0.49 g
(2) 0.45 g
(3) 22.05 g
(4) 2.2 g
2. Consider the following four statements and pick the correct list of statements form the options
(I) Both Chromium (II) ion and Manganese (III) ion have the same ground state electronic configuration in isolated gaseous ions
(II) Chromium (II) is a reducing agent and Manganese (III) ion is an oxidizing agent in water
(III) Chromium (II) is an oxidizing agent and Manganese (III) ion is a reducing agent in water
(IV) Both Chromium (0) and Manganese(0) are paramagnetic in nature
(1) I, II, and IV
(2) I, III, and IV
(3) I and II
(4) I and IV
3.
CrClABC . In this sequence, the compound (C) formed is ____.
(1) Na2CrO4
(2) Na2Cr2O7
(3) Cr(OH)3
(4) PbCrO4
4. Identify the correct statement(s) in the light of quantitative estimation of aqueous ferrous ions by using acidified permanganate or acidified dichromate solution, under conditions of concentration of all species as 1 M.
(1) MnO4− can be used in aqueous HCl
(2) Cr2O72− can be used in aqueous HCl
(3) MnO4 can be used in aqueous H 2SO4
(4) Cr2O72− can be used in aqueous H 2SO4
5. Which are the true statements regarding ionization energy of d–block elements?
(1) First ionization energy of Mn is greater than Cr
(2) First ionization energy of Pd is greater than Rh and Ag
(3) First ionization energies in general decreases from 3d to 4d transition series but it increases from 4d to 5d series
(4) First ionization energy of Mo is greater than that of W
6. Consider the following reactions(unbalanced)
Zn + hot conc.H2SO4 → G + R + X
Zn + conc.NaOH → T + Q ↑
G + H2S + NH4OH → Z + X + Y
Choose the correct options.
(1) Z is black in colour
(2) The oxidation state Zn in T is +1
(3) R is a V – shaped molecule
(4) Bond order of Q is 1 in its ground state
7. Among CrO, Cr 2 O 3 and CrO 3 , the sum of spin-only magnetic moment values of basic and amphoteric oxides is _____10 −2 BM(nearest integer). (Given atomic number of Cr is 24).
ii) Zn
8. White fumes (C) Moisture Air MCl4 (Colourless liquid at room temperature m=transition element)
iii) H2O (B) Purple colour compound
Number of unpaired electrons in the compound ‘B’ is ‘x’ and oxidation state of transition element in the compound ‘C’ is ‘y’. Then the value of x + y is_____
9. An acidified manganate solution undergoes disproportionation reaction. The spin-only magnetic moment value of the product having, manganese in higher oxidation state is ______B.M. (Nearest integer)
10. Find out the total number of reaction(s) (with their given condition(s)) in which Mn2+ converts into MnO4 ?
i) ++ → H 2+Mn+BiO3
ii) ∆ → 2+ 323 Mn+KNO+NaCO
iii) → fused 22 MnO+KOH+O
iv) ++ → H 2+ Mn+PbO2
v) → + air 2+ Mn+NaOH
vi) → 2+2282 Mn+SO+HO
11. Fusion of chromite ore with sodium carbonate in free excess of air gives two compounds X and Y(contains transition metals) along with CO2.Sum of the spin –
THEORY-BASED QUESTIONS
Statement Based Questions
Each question has two statements: statement I (S-I) and statement II (S-II). Mark the correct answer as(1)Both statement I and statement II are correct.
(1) if both statement I and statement II are correct,
(2) if both statement I and statement II are incorrect
only magnetic moments (nearest integer) of the transition metal ions in X and Y is____ (Given that: Among X and Y , if any compound contains two or more than two transition metals, then consider average oxidation state for calculating spin – only magnetic moments)
12. Chromite ore is processed through the following sequence:
In this sequence, determine the number of species (from A to H) that show green colour?
13. Pyrolusite on heating with KOH in the presence of air gives a dark green compound (A). The solution of (A) on treatment with carbon dioxide gives a purple coloured compound (B) along with other compound. The solution is concentrated and the dark purple crystals are separated. Compound (B) with conc. H 2 SO 4 gives(C) which decomposes to (D) and O2. The oxidation state of transition metal in compound D is___
(3) if statement I is correct but statement II is incorrect, (4) if statement I is incorrect but statement II is correct.
1. S-I : The first ionisation energy of Zn is high.
S-II : Zn +2 is diamagnetic in aqueous solution.
2. S-I : The first ionisation energy of Zn is greater than that of Cu
S-II : Atomic size of Zn is greater than that of Cu.
3. S-I : Iron (III) separately catalyses, acidified K2Cr2O7 and neutral KMnO4 have the ability to oxidise I– to I2.
S-II : Manganate ion is paramagnetic in nature and involves pπ-pπ bonding.
4. S-I : O2, Cu2+, and Fe3+ are weakly attracted by magnetic field and are magnetised in the same direction as magnetic field.
S-II : NaCl and H2O are weakly magnetised in opposite direction to magnetic field.
5. S-I : The Eo value for Ce 4+/Ce3+ is +1.74V.
S-II : Ce is more stable in Ce4+ state than Ce3+ state.
Assertion and Reason Questions
In each of the following questions, a statement of Assertion (A) is given followed by a corresponding statement of Reason (R).
(1) if both (A) and (R) are true and (R) is the correct explanation of (A),
(2) if both (A) and (R) are true but (R) is not the correct explanation of (A),
JEE ADVANCED LEVEL
Multi Option Correct MCQs
1. Though the outer electronic configuration is same for the elements in 1B or 11th group, Cu, Ag, and Au, gold and silver are less reactive than copper because
(1) Cu has less ionisation energy than silver and gold
(2) the higher effective nuclear charge due to poor shielding effect of inner d-electrons in Ag and d and f-electrons in Au make them more difficult to oxidise
(3) if (A) is true but (R) is false,
(4) if both (A) and (R) are false.
6. (A) : In CuSO 4 .5H 2 O, Cu−O bonds are present
(R) : In CuSO4.5H2O, ligands co-ordinating with Cu(II) ion are O-and S-bonded ligands.
7. (A) : To a solution of potassium chromate, if a strong acid is added, it changes its colour from yellow to orange.
(R) : The colour change is due to the oxidation of potassium chromate.
8. (A) : KMnO 4 is purple in colour due to charge transfer spectra.
(R) : There is no electron present in d-orbitals of manganese in MnO 4
9. (A) : Melting point of Fe is more than that of Mn.
(R) : Mn has higher number of unpaired electrons than Fe in atomic state.
10. (A) : Cu + disproportionates in aqueous medium.
(R) : Cu+2 has much more negative ∆hyd H° than Cu+
11. (A) : Transition elements form alloys easily.
(R) : Transition elements have high melting points.
(3) the valence electron orbital energies for most common Lewis bases match the orbitals of Cu and its cations more closely and would interact with them more favourably to form products
(4) heat of atomisation of Ag and Au is more than that of Cu
2. Which of the following statements are true?
(1) The chemistry of molybdenum is more similar to that of tungsten than that of chromium.
(2) Higher oxidation states are more prevalent in the second and third rows of transition elements.
(3) Transition metals have high atomic volumes and are, therefore not very dense.
(4) Enthalpies of atomisation reach maximum in the middle of a row, with some exceptions.
3. VF5 is known but other vanadium (V) halides are not known. Why?
(1) Fluorine can oxidise the vanadium to its higher oxidation state but not other halogens.
(2) There is steric hindrance around vanadium in VX5 as the size of other halogen atoms is large.
(3) Vanadium (V) can act as oxidising agent while Br and I act as reducing agents.
(4) There is no reaction between vanadium and the lower halogens Cl2, Br2, and I2
4. Cu2+(3d9) is more stable than Cu+(3d10) in aqueous medium. This is due to
(1) greater charge and smaller size of Cu 2+ ion than Cu+ ion
(2) more extensive hydration of Cu 2+ than Cu+ ion.
(3) free movement of one 3d electron over all the bonding orbitals in Cu 2+ ion
(4) in aqueous solution, Cu2+ is at stable electronic configuration
5. When chromite ore is heated with Na 2CO3 powder in the presence of air,
(1) one gaseous product is formed
(2) one product is water insoluble, which is dark brown in colour
(3) one product is water soluble, which is red in colour
(4) one product is water soluble, which is yellow in colour
6. K2MnO4 is unstable in acidic solution and the green solution obtained is changed into purple colouration. Choose the correct statements regarding the above changes.
(1) It produces KMnO4
(2) It is disproportionation reaction
(3) It produces black precipitate of MnO 2
(4) It produces red precipitate of hydrated MnO2
7. Choose the correct statement(s) about Cr2+ and Mn3+. [Atomic numbers of Cr = 24 and Mn = 25]
(1) Cr2+ is a reducing agent.
(2) Mn3+ is an oxidizing agent.
(3) Both Cr2+ and Mn3+exhibit d4 electronic configuration.
(4) When Cr2+ is used as a reducing agent, the chromium ion attains d 5 electronic configuration.
8. If a mixture of NaCl, conc.H 2 SO 4 , and K 2 Cr 2 O 7 is heated in a dry test tube, a red vapour (P) is formed. This vapour (P) dissolves in aqueous NaOH to form a yellow solution, which, upon treatment with AgNO3, forms a red solid (Q). P and Q are, respectively,
(1) CrO2Cl2 (2) Ag2Cr2O7
(3) Ag2CrO4 (4) Na[CrOCl5]
9. Which of the following statement(s) is/are correct?
(1) Cr < Mn : Third ionisation enthalpy
(2) Ni < Cu < Zn : Metallic radius
(3) Cu < Zn : Density
(4) Cu > Zn : Enthalpy of atomisation
Numerical Value Questions
10. What volume (in mL) of 0.4 M acidified KMnO 4 solution can completely oxidise 0.9612 moles of Fe2+ into Fe3+ions (nearest integer)?
11. Potassium permanganate oxidises hypo in faintly alkaline or neutral medium. What is the mole ratio of KMnO4 to Na2S2O3 in the balanced chemical reaction?
12. When permanganate is added to MnSO 4, a black precipitate is formed in neutral or faintly alkaline medium. What is the weight of black precipitate formed when 15.8 g of KMnO 4 is added to MnSO 4 ?
(At. wt. Mn = 55, K = 39, O = 16)
13. Heating of dry crystals of KMnO4 gives a black residue. Addition of little water to this residue, followed by filtration, gives green filtrate and black residue. The difference in oxidation states of Mn in green filtrate and black residue is _____.
Integer Value Questions
14. How many of the following statements is/ are correct regarding A, B, C, and D?
i) A is black precipitate.
ii) B is white precipitate.
iii) Four atoms are in one plane in compound ‘C’.
iv) Shape of anionic part of D is square planar.
v) Oxidation states of Cu in B and D are same
vi) Oxidation state of Cu in B is same as that in CuSO4 quickly decomposes excess KCN ↑ CuSO4 KCN B + C
A
15. The oxidation number of Mn in the product of oxidative fusion of MnO 2 is ____.
16. Consider the following reagents. Acidified K 2 Cr 2 O 7 , Alkaline KMnO 4 CuSO 4 , H 2 O 2 , Cl2, O3, FeCl3, HCl, and Na2S2O3. The total number of reagents that can oxidise queous iodide to iodine is _____.
17. During preparation of K2Cr2O7, chromite ore (FeCr2O4) is fused with sodium carbonate in the presence of oxygen gas to produce a yellow coloured substance and other byproducts. Number of electrons transferred per molecule of chromite ore during this process is _____.
Passage – based Questions
(Q: 18-19)
The magnetic moment of d-block elements can be calculated based on unpaired electrons in Bohr magnetons (BM).
18. The oxidation state of vanadium in V 2O x with a magnetic moment of 2.89 BM is__. (1) 2 (2) 3
(3) 5 (4) 1
19. Find number of compounds which are diamagnetic ZnSO 4 , MnSO 4 , FeSO 4 , Fe2(SO4)3, CoCl3, K2Cr2O7, and KMnO4 (1) 1 (2) 0 (3) 3 (4) 4
(Q:20-21)
When a metal rod ‘X’ dipped into an aqueous colorless concentrated solution of compound ‘Y’ the solution turns light blue. Addition of aqueous NaCl to the blue solution gives a white precipitate ‘Z’ addition of aqueous NH3 dissolves ‘Z’ and gives intense blue color solution.
20. The metal ‘X’ atomic number is
21. The final solution contains two complexes sum of their coordination number is
Matrix Matching Questions
22. Match List-I (process) with List-II (catalyst). List-I (processes/ reactions) List-II (catalyst)
(A) 2SO2(g)+O2(g)→ 2SO3(g) (I) Fe(s)
(B) 4NH 3(g) +5O 2(g) → 4NO(g) +6H2O(g) (II) Pt(s)–Rh(s)
1 Potassium dichromate is preferred in place of sodium dichromate in quantitative analysis for use as primary standard because
(1) potassium dichromate is less soluble in water than sodium dichromate
(2) potassium dichromate is weak oxidising agent than sodium dichromate
(3) sodium dichromate is hygroscopic while potassium dichromate is not
(4) molecular mass of potassium dichromate is more than sodium dichromate
(C) Elements with maximum unpaired electrons (r) Tc
(D) Radioactive transition element (s) Ru
(A) (B) (C) (D)
(1) q,s q p r
(2) q s p r
(3) s p q r
(4) p,r s q s
28. Match column-I (metal) with column-II (electronic configuration).
Column–I Column–II
(A) Pr (p) 4f12
(B) Gd+2 (q) 4f36s2
(C) Eu (r) 4f76s2
(D) Tm+3 (s) 4f75d1 (t) 4f14
Choose the correct option.
(A) (B) (C) (D)
(1) s t p q
(2) q r s t
(3) q s r p
(4) p q r s
2. Which of following statements is wrong?
(1) In any group of d-block elements, third element (5d series) has more 1 st ionisation energy, which is less than the other two elements above it.
(2) Irregularities in 2 0 / MM E + of 3d series elements is due to irregularities in sublimation energies and ionisation energies
(3) Melting points of 5d series elements are more than the corresponding 3d and 4d elements in their groups.
(4) Highest oxidation state of an element is more stable for the 5d series element than the other two elements above it in that group.
3. Hot concentrated H2SO4 reacts with KMnO4 and forms dangerously explosive, green oily compound (X). (X) is anhydride of HMnO4. Oxidation state of Mn in (X) is (1) +4 (2) +6 (3) +3 (4) +7
4. Some properties of Au, Ag and Cu are given below:
Cu 128 1085 +2
144 961 +1 Au 144 1064 +3
To make Au stronger and harder, it is often alloyed with other metals such as Cu and Ag. Consider two alloys, one of Au & Cu and the other Au & Ag, each with same mole fraction of Au. If Au/Cu alloy is harder than the Au/
FLASH BACK (Previous JEE Questions)
JEE Main
1. Strong reducing and oxidising agents among the following, respectively, are (2023) (1) Eu2+ and Ce4+ (2) Ce4+ and Tb4+ (3) Ce3+ and Ce4+ (4) Ce4+ and Eu2+
2. In chromyl chloride, the number of d-electrons present on chromium is same as in (Given: at. no. of Ti : 22, V : 23, Cr : 24, Mn: 25, Fe : 26) (2023) (1) Fe (III) (2) Mn (VII) (3) Ti (III) (4) V (IV)
3. Prolonged heating is avoided during the preparation of ferrous ammonium sulphate to (2023) (1) prevent oxidation (2) prevent hydrolysis
Ag alloy then which of the following is the best explanation based on the information in the table above?
(1) Cu has two common oxidation states, but Ag has only one
(2) Cu has a higher m.pt than Au, but Ag has a lower melting point than Au
(3) Cu atoms are smaller than Ag atoms
(4) Cu atoms are less polarizable than Au or Ag atoms thus Cu has weaker inter particle forces
5. The disproportionation of MnO42– in acidic medium resulted in the formation of two manganese compounds A and B. If the oxidation state of Mn in B is smaller than that of A, then the spin-only magnetic moment μ value of B in BM is _______________. (Nearest integer)
6. The color of light absorbed by an aqueous solution of CuSO4 is (1) Orange – red (2) Blue-green (3) Yellow (4) Violet
(3) prevent breaking (4) prevent reduction
4. Which of the following statements are correct? (2023)
A) The M3+/M2+ reduction potential for iron is greater than manganese.
B) The higher oxidation states of first row d-block elements get stabilised by oxide ion.
C) Aqueous solution of Cr+2 can liberate hydrogen from dilute acid.
D) Magnetic moment of V+2 is observed between 4.4–5.2 BM.
(1) A, B and D only
(2) A, B only
(3) B, C only
(4) C, D only
5. Given below are two statements. One is labelled Assertion A and the other is labelled Reason (R). (2023)
Assertion (A) : 5f electrons can participate in bonding to a far greater extent than 4f electrons.
Reason (R) : 5f orbitals are not as buried as 4f orbitals.
In light of the above statements, choose the correct answer from the options given below. (2023)
(1) if both (A) and (R) are true and (R) is the correct explanation of (A),
(2) if both (A) and (R) are true but (R) is not the correct explanation of (A),
(3) if (A) is true but (R) is false,
(4) if both (A) and (R) are false.
6. The mismatched combinations are
A. Chlorophyll – Co
B. Water hardness – EDTA
C. Photography - [Ag(CN)2]–
D. Wilkinson catalyst - [(Ph 3P)3RhCl]
E. Chelating ligand – D-Penicillamine
Choose the correct answer from the options given below. (2023)
(1) A and E Only
(2) D and E Only
(3) A, C, and E Only
(4) A and C Only
7. The pair of lanthanides in which both elements have high third ionisation energy is (2023)
(1) Dy, Gd (2) Eu, Gd (3) Eu, Yb (4) Eu, Yb
8. K2Cr2O7 paper acidified with dilute H2SO4 turns green when exposed to (2023)
(1) hydrogen sulphide
(2) sulphur dioxide
(3) carbon dioxide
(4) sulphur trioxide
9. Which one amongst the following are good oxidising agents? (2023)
A. Sm2
C. Ce4+
B. Ce2+
D. Tb4+
Choose the most appropriate answer from the options given below: (2023)
(1) C and D only (2) A and B only (3) C only (4) D only
10. The observed magnetic moment of the complex [Mn(NCS) 6 ] x – is 6.06 BM. The numerical value of x is ______. (2023)
11. For a metal ion, the calculated magnetic moment is 4.90 BM. This metal ion has _______ number of unpaired electrons. (2023)
12. The ratio of spin-only magnetic moment value μ eff [Cr(CN) 6 ] 3− /μ eff [Cr(H 2 O) 6 ] 3+ is __________. (2023)
13. KMnO4 is titrated with ferrous ammonium sulphate hexahydrate in presence of dilute H2SO4. Number of water molecules produced for 2 molecules of KMnO 4 is _______ (2023)
14. H2S (5 moles) reacts completely with acidified aqueous potassium permanganate solution. In this reaction, the number of moles of water produced is x , and the number of moles of electrons involved is y. The value of (x + y) is ____. (2023)
15. Fusion of MnO2 with KOH in presence of O2 produces a salt (W). Alkaline solution of W, upon electrolytic oxidation, yields another salt (X). The manganese containing ions in W and X are Y and Z, respectively. Correct statement(s) is (are) (2019)
(1) In aqueous acidic solution, Y undergoes disproportion reaction to give Z and MnO2
(2) Both Y and Z are coloured and have tetrahedral shape.
(3) Y is diamagnetic in nature while Z is paramagnetic.
(4) In both Y and Z, π-bonding occurs between p-orbitals of oxygen and dorbitals of manganese.
16. Consider the following reactions (unbalanced).
Zn + hot conc.H2SO4 → G + R + X
Zn + conc.NaOH → T + Q
CHAPTER TEST – JEE MAIN
Section A
1. Zr a nd Hf have almost equal atomic and ionic radii because of (1) diagonal relationship (2) lanthanoid contraction (3) actinoid contraction (4) belonging to the same group
2. Highest oxidation state shown by Mn, when it reacts with flourine, is (1) +7 (2) +6 (3) +2 (4) +4
3. Given below are two statements. One is labeled Assertion (A) and the other is labeled Reason (R).
Assertion (A) : Permanganate titrations are not performed in presence of hydrochloric acid.
Reason (R) : Chlorine is formed as a consequence of oxidation of hydrochloric acid.
In light of the above statements, choose the correct answer from the options given below.
(1) if both (A) and (R) are true and (R) is the correct explanation of (A), (2) if both (A) and (R) are true but (R) is not the correct explanation of (A), (3) if (A) is true but (R) is false, (4) if both (A) and (R) are false.
4. The total number of Mn = O bonds in Mn2O7 is
(1) 4 (2) 5 (3) 6 (4) 3
G + H2S + NH4OH → Z (a precipitate) + X + Y
Choose the correct options. (2019) (1) Z is black in colour.
(2) The oxidation state Zn in T is +1.
(3) R is a V– shaped molecule.
(4) Bond order of Q is 1 in its ground state.
5. In 3d series, the metal having the highest M2+/M standard electrode potential is (1) Cr (2) Fe (3) Cu (4) Zn
6. An inorganic salt solution gives a red precipitate with silver nitrate. The precipitate dissolves in dilute nitric acid. The solution contains (1) bromide (2) iodide (3) phosphate (4) chromate
7. Nd2+ = _______
(1) 4f26s2 (2) 4f46s2 (3) 4f3 (4) 4f4
8. The magnetic moment of a transition metal compound has been calculated to be 3.87 BM. The metal ion is (1) TI2+ (2) Mn2+ (3) V2+ (4) Cr2+
9. The basic character of the transition metal monoxides follows the order (Atomic no. of Ti = 22, V = 23, Cr = 24, Fe = 26) (1) TiO > VO > CrO > FeO
(2) VO > CrO > TiO > FeO
(3) CrO > VO > FeO > TiO
(4) TiO > FeO > VO > CrO
10. Identify the isoelectronic pair of ions from the following.
(1) Pr3+, Nd3+
(2) TB3+, Dy3+
(3) Eu3+, Gd3+
(4) Pr3+, Ce3+
11. Ti3+ is purple but Ti4+ is colourless. This is because
(1) d1 configuration of Ti3+
(2) d0 configuration of Ti3+
(3) d3 configuration of Ti3+
(4) d10 configuration of Ti3+
12. The pair of species having same value of spin only magnetic moment is (1) Mn3+, Ni2+ (2) Fe2+, Ti2+ (3) Cr3+, Co2+ (4) Zn2+, Ti3+
13. Which of the following is the correct order of second ionisation energy?
(1) V > Cr > Mn (2) V < Cr < Mn
(3) V < Cr > Mn (4) V > Cr < Mn
14. The most common oxidation state of lanthanoid elements is +3. Which of the following is likely to deviate easily from +3 oxidation state?
(1) Ce (At. no. 58) (2) La (At. no. 57) (3) Lu (At. no. 71) (4) Gd (At. no. 64)
15. The f-orbitals are half and completely filled, respectively, in lanthanide ions (Given: Atomic number of Eu, 63; Sm, 62; Tm, 69; Tb, 65; Yb, 70; Dy, 66)
(1) Eu2+ and Tm2+ (2) Sm2+ and Tm2+ (3) Tb2+ and Yb2+ (4) Dy2+ and Yb2+
16. Given below are two statements: One is labeled Assertion (A) and the other is labeled Reason (R).
Assertion (A) : Ionisation of transition metals involves loss of ns electrons before (n−1)d orbitals.
Reason (R) : Filling of ns orbitals take place before the filling of (n−1) dorbital.
In light of the above statements choose the correct answer from the options given below. (1) if both (A) and (R) are true and (R) is the correct explanation of (A),
(2) if both (A) and (R) are true but (R) is not the correct explanation of (A), (3) if (A) is true but (R) is false, (4) if both (A) and (R) are false.
17. Which of the following sulphates of lanthanoid has strong reducing nature?
(1) EuSO4 (2) Eu2(SO4)3
(3) Ce(SO4)2 (4) Ce2(SO4)3
18. In each of the following questions, a statement of Assertion (A) is given, followed by a corresponding statement of Reason (R). Mark the correct answer as
Assertion (A) : The most common oxidation state of lanthanides is +3.
Reason (R) : All Lanthanides have three electrons in their outermost f sub shell.
In the light of the above statements choose the correct answer from the options given below.
(1) if both (A) and (R) are true and (R) is the correct explanation of (A), (2) if both (A) and (R) are true but (R) is not the correct explanation of (A), (3) if (A) is true but (R) is false, (4) if both (A) and (R) are false.
19. Which of the following lanthanoid ions is diamagnetic?
(1) Eu2+ (2) Yb2+
(3) Ce2+ (4) Sm2+
20. Magnetic moment of Cr2+(Z = 24), Mn2+ (Z = 25), Fe2+(Z = 26) are x, y, z. They are in order
(1) x < y < z (2) x > y > z
(3) z < x = y (4) x = z <
21. Find the number of species from the following which has magnetic moment value of 1.73 BM. Fe2+, Ci2+, Ni2+, NO2, Mn+2, Zn+2, Co+2
22. The spin only magnetic moment of the complex present in Fehling’s reagent is ______ B.M. (Nearest integer )
23. H2S (3 mol) reacts completely with acidified aqueous potassium dichromate solution. In this reaction, the number of mole of water produced is ‘y’ and the number of moles of electrons involved is ‘x’. The value of (x + y) is _____.
24. Manganese (VI) has ability to disproportionate in acidic solution. The
CHAPTER TEST – JEE ADVANCED
2022 P2 Model
Section A
[Single Correct MCQs]
1. Which of the following arrangements does not represent the correct order of the property stated against it?
2. The transition element having least enthalpy of atomisation is
(1) Zn (2) V
(3) Fe (4) Cu
3. Three elements chromium, manganese and iron have heat of atomisations 400 kJ, 420 kJ, and 280 kJ per 1 mol of metal but not respectively. Arrange the values in the order for chromium, manganese, and iron in kJ/ mol.
(1) 400, 420, 280 (2) 280, 400, 420
(3) 400, 280, 420 (4) 420, 400, 280
4. Select the correct order of decreasing second ionization enthalpy.
Ti(22), V(23), Cr(24), Mn(25)
(1) V > Mn > Cr > Ti
(2) Mn > Cr > Ti > V
(3) Ti > V > Cr > Mn
(4) Cr > Mn > V > Ti
difference in oxidation states of two ions it forms in acidic solution is______.
25. Acidified potassium permanganate solution oxidises oxalic acid. The spin-only magnetic moment of the manganese product formed from the above reaction is______ BM. (nearest integer)
Section B
[Multiple Option Correct MCQs]
5. Which of the following statement(s) is/are correct?
(1) Permanganate ion, pertechnate ion, and perrhenate ion are all dark purple coloured species.
(2) In both haemoglobin and oxyhaemoglobin, iron is in +2 oxidation state.
(3) Both [Ni(CO) 4 ] and [NiCl 4 ] 2− are tetrahedral complexes.
(4) Chromium forms highly oxidising compounds in +6 oxidation state, whereas the equivalent compounds of molybdenum and tungsten are not appreciably oxidising.
6. Choose the correct statement(s) among the following.
(1) Zinc is not a transition element because it does not have incompletely filled d-orbitals in both elemental state as well as in its oxidised state.
(2) Silver can exhibit +2 oxidation state.
(3) Zinc has the lowest enthalpy of atomisation in its period due to unavailability of any unpaired electrons.
(4) Cr(VI) is more stable than Mo( VI).
7. The lanthanides that show +2 oxidation state in addition to +3 state are (1) Ce (2) Eu (3) Yb (4) Ho
Section C
[Integer Value Questions]
8. The number of transition metal atoms which are smaller than manganese is ______. Ti, V, Cr, Fe, Co, Ni, Cu, Zn, Tc, Ra
9. Number of Cr-O linkages present in dichromate ion is ______.
10. How many of the following lanthanoid ions show color in aqueous solution?
La(III), Eu(III), Gd(III), L u(III)
11. The number of pairs of elements having the same number of unpaired electrons in 3d-series is _______.
12. In the chemical reaction between stoichiometric quantities of KMnO4 and KI in weakly basic solution, what is the number of moles of I2 released for 4 moles of KMnO4 consumed?
13. An acidified solution of potassium chromate was layered with an equal volume of amyl alcohol. When it was shaken after the addition of 1 mL of 3% H2O2, a blue alcohol layer was obtained. The blue colour is due to the formation of a chromium (VI) compound ‘X’. What is the number of oxygen atoms bonded to chromium through only single bond in a molecule of X?
(Q: 14-15)
Section D [Passage-Based Questions]
14. What is the difference in the oxidation number of Cr between (Z) and (X)?
15. What is the bond angle in product (P) of two tetrahedrals sharing one corner with Cr-O-Cr at__
(Q: 16-17)
100 mL solution containing x M SO32−and y M S2O32− exactly requires 80 mL of 0.05 M CrO42− in alkaline medium for complete oxidation. CrO42− reduces to CrO21−, and the only sulphur containing product formed is SO 4 2− . After the completion of reaction, the solution is treated with excess BaCl2, and all the SO4 2− is precipitated as BaSO4. The weight of BaSO4, formed is found to be 0.9336 g. (Molecular weight of BaSO4 is 233.4 g/mol)
6.2 Definition of Some Important Terms Pertaining to Coordination Compounds
6.3 Nomenclature of Coordination Compounds
6.4 Isomerism in Coordination Compounds
6.5 Bonding in Coordination Compound
6.6 Stability of Coordination Compound
6.7 Bonding in Metal Carbonyl
6.8 Importance and Applications of Coordination Compounds
Coordination compounds are a special class of compounds in which the central metal atom is surrounded by ions or molecules beyond its normal valency. These are also referred to as complex compounds or simply complexes. These compounds are widely present in minerals, plants, and animals and perform many important functions. Chlorophyll, hemoglobin, and vitamin B12 are coordination
compounds of magnesium, iron, and cobalt, respectively. The transition elements form a large number of complex compounds. This is due to the comparatively small size of the metal ions, their high ionic charges, and the availability of d-orbitals for bond formation.
6.1 WERNER'S THEORY OF COORDINATION COMPOUNDS
Alfred Werner was the first to formulate his ideas about the structures of coordination compounds. He prepared and characterised a large number of coordination compounds and studied their physical and chemical behaviour by simple experimental techniques.
Werner proposed the concept of primary and secondary valence for a metal ion. In a series of compounds of cobalt (III) chloride with ammonia, it was found that some of the chloride ions could be precipitated as silver chloride on adding excess silver nitrate solution but some chlorides remained unprecipitated.
One mole CoCl 3 .6NH 3 (yellow) gave three mole AgCl, one mole CoCl 3 .5NH 3 (purple) gave two mole AgCl, and one mole CoCl 3 .4NH 3 (green) gave one mole AgCl. These observations, together with the results of conductivity measurements in solution, can be explained if six groups in all, either chloride ions or ammonia molecules or both, remain
Table 6.1 Formulation of cobalt (III) chlorideammonia complexes
bonded to the cobalt ion during the reaction. Werner proposed the term secondary valence for the number of groups bound directly to the metal ion. Two compounds in Table 6.1 have identical empirical formula, CoCl 3.4NH3,but distinct properties. Such compounds are termed isomers.
6.1.1 Postulates in Werner's Theory
■ In coordination compounds, metals show two types of valences, primary and secondary valence.
■ The primary valences are normally ionisable (non-directional) and are satisfied by negative ions. Primary valence is numerically equal to the oxidation number of central metal ion and represented by discontinuous lines ( ).
■ The secondary valences are non-ionisable. (directional). These are satisfied by neutral molecules or negative ions. The secondary valence is equal to the coordination number and represented by a continuous line ( ) in the structure.
■ The groups bounded by the secondary linkages to the metal have characteristic spatial arrangements corresponding to different coordination numbers.
■ Werner further postulated that octahedral, tetrahedral and square planar geometrical shapes are more common in coordination compounds of transition metals. For example, [Co(NH3)6]3+ and [CoCl2(NH3)4]+ are octahedral entities, while [Ni(CO)4] is tetrahedral and [PtCl 4]2– is square planar. Werner’s structures of some complex compounds are given in Fig. 6.1, 6.2, 6.3, 6.4
Examples:
In CoCl 3 .6NH 3 complex, primary valence of cobalt is satisfied by three Cl – ions and the secondary valence is satisfied by six NH 3 molecules. This complex can be written as [Co(NH3)6]Cl3. One mole of this complex can
give three moles chloride ions and one mole complex ions [Co(NH 3) 6] 3+ when dissolved in water.
6.1 Structure of CoCl3.6NH3
In CoCl3.5NH3 complex, primary valence of cobalt is satisfied by three Cl – ions and the secondary valence is satisfied by five NH 3 and one Cl – ion. One chloride ion satisfies both primary and secondary valences. This complex can be written as [Co(NH 3)5Cl]Cl2. Species satisfying both primary and secondary valence are also not ionisable. One mole of this complex gives two moles of chloride ions and one mole of complex ions [Co(NH3)5Cl]2+ when dissolved in water.
Fig. 6.2 Structure of CoCl3.5NH3
I n CoCl 3 .4NH 3 complex, primary valence is satisfied by three Cl– ions. The secondary valence is satisfied by four NH 3 molecules and two Cl – ions. Two chloride ions satisfy both primary and secondary valences and this complex can be written as [Co(NH 3)4Cl2]Cl. One mole of this complex gives one mole of chloride ions and one mole of complex ions, [Co(NH3)4Cl2]+ when dissolved in water.
Fig.
Fig. 6.3 Structure of CoCl3.4NH3
In CoCl3.3NH3 complex, three Cl– ions satisfy primary valence. Three NH 3 molecules and three Cl– ions satisfy secondary valence and the complex can be written as [Co(NH 3)3Cl3]. It cannot give any ions when dissolved in water, and it is a neutral complex.
2 The secondary valence of Pt4+ is six. Calculate the number of moles of AgCl participated, when excess of AgNO3 solution is added to 2 L of 0.1M PtCl4.4NH3 solution.
Sol. Secondary valence of Pt 4+ is six and it will be satisfied by four NH 3 and two Cl – ions. The remaining two Cl – ions satisfy only primary valence. PtCl4.4NH3 can be written as [Pt(NH3)4Cl2]Cl2
One mole of this complex gives 2 moles of chloride ions in the aqueous solution.
Try yourself:
1. What is the secondary valency of cobalt in [Co(NH3)5Cl]Cl2 is?
Answer: [Co(NH 3 )5Cl]Cl 2 is an octahedral complex, so secondary valency is 6.
6.1.2 Difference Between a Double Salt and a Complex
Fig. 6.4 Structure of CoCl3.3NH3
All the cobaltamine complexes prepared by Werner are hexacoordinated. All these complexes have octahedral geometry.
1 When excess of silver nitrate solution is added to the aqueous solution containing 0.1 mole of CoCl3.xNH3, if 28.7g of silver chloride is precipitated, what is the value of x?
Sol. Number of moles of AgCl precipitated = 28.7 0.2mole 143.5 = .
0.1 mole of CoCl3.xNH3 gives 0.2 mole AgCl. One mole of CoCl 3 x NH 3 gives 2 moles of AgCl. Thus, in the complex CoCl3.xNH3, two chloride ions satisfy only primary valence. One chloride ion satisfies both secondary valence and primary valence. Since the secondary valence of Co3+ is six and one Cl – is in coordination sphere, complex possesses five NH3. Thus, the value of ‘x’ is five.
When two or more salts combine either physically or chemically, molecular compounds or addition compounds are formed. Double salts are those molecular compounds which exist only in crystal lattice and lose their identity when dissolved in water.
Examples:
Carnallite (KCl MgCl2.6H2O), potash alum(K2SO4 Al2(SO4)3.24H2O), Mohr’s salt (FeSO4(NH4)2SO4 6H2O) etc.
These are formed by physical union of stable compounds. Shape and size of the crystals of double salts are different from those of its component salts. However, their solutions will show same physical and chemical properties as a mixture of solutions of their components. Thus, for example, aqueous solution of potash alum gives the test for K + , Al3+, and 2 4 SO ions.
Complex compounds are those molecular compounds which retain their identity even when dissolved in water. Properties of complexes are completely different from those of the constituents. Complexes are
also called coordination compounds because they contain coordinate covalent bonds, e.g., cupraammonium sulphate, potassium ferrocyanide etc. These are formed by chemical union of stable compounds. The physical properties, such as colour, conductivity etc., of these compounds are distinctly different from their constituents. Complex compounds are new chemical species different from their constituents. For example, potassium ferrocyanide aqueous solution can give the test for ferrocyanide ion, [Fe(CN)6]4–, but not Fe2+ and CN– ions.
3. Which ions can be tested in a solution of Mohr’s salt?
Sol. FeSO4.(NH4)2SO4.6H2O is the composition of Mohr’s salt. It is a double salt. It undergoes complete ionisation when dissolved in water. Thus, its aqueous solution gives the test for Fe2+, NH4+ and ions.
Try yourself:
2. K2SO4 Al2(SO4)3.24H2O is
Answer: potash alum 4SO2K O2.24H3)4(SO2Al is an example of double salt.
TEST YOURSELF
1. What is the amount of white precipitate formed by treating 0.1 moles of PtCl2.2NH3 with excess of AgNO3?
(1) 143.5 g (2) 287 g
(3) Zero g (4) 28.7 g
2. Which of the following has highest molar conductivity in solution?
(1) K4[Fe(CN)6]
(2) K3[Fe(CN)6]
(3) [Pt(NH3)4Cl2]Cl2
(4) [Cu(H2O)]SO4.H2O
3. A six coordinate complex of formula CrCl3.6H2O has green colour. One litre of 0.1 M solution of the complex, when treated with excess of AgNO3, gave 28.7 g of white precipitate. The formula of the complex would be
(1) [Cr(H2O)6]Cl3
(2) [CrCl(H2O)5]Cl2H2O
(3) [CrCl(H2O)4]Cl. 2H2O
(4) [Cr(H2O)3Cl.3]
4. On treatment of 100 mL of 0.1 M solution of CoCl3.6H2O with excess of AgNO3, 1.2 × 1022 ions are precipitated. The complex is (1) [Co(H2O)4Cl2]Cl.H2O
(2) [Co(H2O)3Cl3]3H2O
(3) [Co(H2O)6]Cl3
(4) [Co(H2O)5Cl]Cl2.H2O
Answer Key
(1) 3 (2) 1 (3) 2 (4) 4
6.2 DEFINITION OF SOME IMPORTANT TERMS PERTAINING TO COORDINATION COMPOUNDS
Some of the important terms related to coordination compounds explained below. Coordination entity: A coordination entity constitutes a central metal atom or ion bonded to a fixed number of ions or molecules. For example, [CoCl 3 (NH 3 ) 3 ] is a coordination entity in which the cobalt ion is surrounded by three ammonia molecules and three chloride ions. Other examples are [Ni(CO)4], [PtCl2(NH3)2], [Fe(CN)6]4–, [Co(NH3)6]3+.
Central metal ion: The cation to which some neutral molecules or anions are attached by coordinate covalent bond is called central metal ion or centre of coordination. For example, in [Cu(NH 3 ) 4 ]SO 4 , K 4 [Fe(CN) 6 ] and K 3 [Fe(CN 6 )], central metal ions are, respectively, Cu 2+ , Fe 2+ and Fe 3+ . Some complex compounds possess neutral central metal atoms. For example, in [Ni(CO) 4] and [Fe(CO)5], centre of coordination is Ni and Fe, respectively. These central atoms or ions are also reffered to as Lewis acids.
Ligands: Any atom, ion, or molecule which is capable of donating one or more electron
pairs to the central metal atom or ion is called a ligand or a coordinating group. Further, in a ligand, the atom which donates the electron pair is called a donor atom. Thus, a ligand is a donor of electrons and a metal ion is an acceptor of electrons. However, the ligand may possess acceptor orbitals in addition to donor orbitals. These ligands are also referred to Lewis bases.
Number of ligating groups (electron donating groups) in the ligand is called denticity. Based on denticity, ligands can be classified as mono, bi, and polydentate ligands when they donate one, two, and more than two electron pairs, respectively. Some examples of ligands are given in Table 6.2.
A bidentate ligand which has two donor atoms that are same is called a symmetrical bidentate ligand.
A bidentate ligand which has two donor atoms that are different is called an unsymmetrical bidentate ligand. Bi and polydentate ligands are form the ring structures around the central metal atom is called chelate ligands. Complex
with chelate ligands is called chelate complex and it is more stable.
Ligands that have variable denticity are called flexidentate ligands, for example, EDTA–4 (denticity = 4, 5, 6) and SO42– ( dentitcity = 1, 2).
A monodentate ligand containing more than one ligating group is called ambidentate ligand.
e.g., CN–, 2 NO and SCN–.
2 NO ion can coordinate through nitrogen or oxygen to a central metal ion. CN – ion can coordinate through carbon or nitrogen. SCN– ion can coordinate through the sulphur or nitrogen atom.
Based on the charge, ligands can be classified as neutral ligands (NH 3 , H 2 O, H2NCH2CH2NH2), negative ligands (Cl–, 2 NO, CN –, 2 24CO , H 2 NCH 2 COO – ) and pos itive ligands (NO+). Ethylene can also act as a ligand by donating π electrons.
Coordination number: Coordination number of central metal ion or atom in a complex compound is the number of coordinate
bidentate
Table 6.2 Examples for mono, bi and polydentate ligands
covalent bonds formed by it or the maximum number of monodentate ligands that can be bound to it. For example, in [Fe(CN)6] 4 , [Ni(CO) 4 ] and [Ag(NH 3 ) 2 ] + , coordination numbers of Fe2+, Ni, and Ag+ are, respectively, 6, 4, and 2. Similarly, in [Co(en) 3 ] 3+ and [Fe(C 2 O 4 ) 3 ] 3– , coordination number of both Co 3+ and Fe 3+ is 6, because of the use of bidentate ligands. Since, in the complex formation, the attraction between a metal ion and the dipolar molecule or anions around it is involved, the coordination number will tend to be as large as possible.
Coordination sphere: The central metal ion and the ligands attached to it are enclosed in square bracket, and this is termed as the coordination sphere. For example, in the complex K4[Fe(CN)6], the coordination sphere is [Fe(CN)6]4–.
Complex ion: An electrically charged or neutral species formed by the combination of a central metal atom or ion with more than one ligand is called complex ion. The charge carried by a complex ion is the algebraic sum of the charges carried by the central metal ion and the ligands coordinated to it. Charge of complex ion is balanced by some ionisable groups, which are written outside the bracket, called counter ions. For example, in K4[Fe(CN)6], [Fe(CN)6]4– is a complex ion and the counter ions are four K + ions.
Coordination polyhedron: The arrangement of coordinated ligands around the metal ion must be such that it minimises the electrostatic repulsion between the ligands. The spatial arrangement of ligands that are directly attached to the central metal ion defines a coordination polyhedron about the central atom. Shape of polyhedron can be known from coordination number. Coordination
polyhedron for different coordination numbers is given in the Table 6.3.
Table 6.3 Coordination numbers and shapes of complex species
Coordination corresponding
Two
Three
Four
Shape of the corresponding complex species
Linear
Trigonal planar
Four Tatrahedral Square planar
Five
Five Square pyramidal Trigonal bipyramidal
Six Octahedral
Seven Pentagonal bipyramidal
Oxidation number of central atom: The oxidation number of the central atom in a complex is defined as the charge it would carry if all the ligands are removed along with the electron pairs that are shared with the central atom. The oxidation number is represented by a Roman numeral in parenthesis, following the name of the coordination entity. For example, oxidation number of copper in the complex [Cu(CN)4]3– is + 1 and it is written as Cu (I).
Homoleptic and heteroleptic complexes: Complexes in which the central metal is bound to only one kind of donor group are called homoleptic complexes, e.g., [Co(NH 3 ) 6 ] 3+ , [Fe(CN)6]4–, etc. Complexes in which the central metal is bound to more than one kind of donor group are known as heteroleptic complexes. e.g., [Co(NH3)4Cl2]+.
Mononuclear and polynuclear complexes:
Complexes having only one central atom are called mononuclear complexes, e.g., K 4 [Fe(CN) 6 ], [(Co(NH 3 ) 6 ]Cl 3 , etc. Complexes having more than one central atom are called polynuclear complexes, e.g., [Co2(NH3)6(OH)3]Cl3,[ (CO)3Fe(CO)3Fe(CO)3].
Cationic, anionic and neutral complexes: In the complex compound, if the complex ion is
a cation, it is called a cationic complex, and if th e complex ion is an anion, it is called an anionic complex. Examples of cationic complexes are [Co(NH3)6]Cl3, [Cu(NH3)4]SO4, etc. and examples of anionic complexes are K4[Fe(CN6)], Na3[Ag(S2O3)2], etc. Complexes in which only the coordination sphere is present without counter ions are called neutral complexes, e.g., [Ni(CO)4], [Co(NH3)3Cl3], etc.
4. Identify the ligands and central metal ion in [Co(NH3)2(en)2]3+. Calculate the oxidation number and coordination number of metal ion.
Sol. In [Co(NH 3) 2(en) 2] 3+, ligands are ammonia and ethylene diamine. Oxidation number of cobalt = +3. Central metal ion is Co3+ and it is surrounded by two monodentate ligands NH 3 and two bidentate ligands ethylene diamine. Thus, coordination number of Co 3+ is 6.
Try yourself:
3. Which type of complex is cryolite?
3–]6[AlF is an anion.
Answer: Cryolite, Na 3 [AlF 6 ], is an anionic complex. This is because the complex ion
TEST YOURSELF
1. In Tollen’s reagent, the oxidation number and coordination number of central metal ion respectively, are
(1) +1, 2 (2) + 2, 2 (3) + 1, 1 (4) 0, 2
2. Which of the following represents chelating ligand?
(1) Cl– (2) 2 CO24
(3) OH– (4) H2O
3. Which statement regarding ligands is correct?
(1) Positive ions are good ligands. (2) All ligands are negative ions.
(3) The atom bonding to the metal is positively polarised.
(4) All ligands are Lewis bases.
4. Which of the following is a heteroleptic complex?
A) COCl3.3NH3
B) COCl3.4NH3
C) COCl3.5NH3
D) COCl3.6NH3
(1) A and B (2) B and C
(3) A, B, and C (4) A, B, C, and D
5. Number of EDTA molecules required to make an octahedral complex with a Ca +2 ion is
(1) 2 (2) 6 (3) 1 (4) 3
Answer Key
(1) 1 (2) 2 (3) 4 (4) 3
(5) 3
6.3 NOMENCLATURE OF COORDINATION COMPOUNDS
Nomenclature is important in coordination Chemistry because of the need to have an unambiguous method of describing formulas and writing systematic names, particularly when dealing with isomers. The formulas and names adopted for coordination entities are based on the recommendations of the International Union of Pure and Applied Chemistry (IUPAC).
6.3.1 Formula of Mononuclear Coordination Entities
The following rules are applied while writing the formula of mononuclear complexes as per IUPAC recommendations of 2004.
■ The central atom is listed first.
■ The ligands are then listed in alphabetical order. The placement of a ligand in the list does not depend on its charge. For example, [CoCl2(NH3)4]Cl is wrong, and [Co(NH3)4Cl2]Cl is correct.
■ Polydentate ligands are also listed alphabetically. In case of abbreviated ligand, the first letter of the abbreviation is used to determine the position of the ligand in the alphabetical order. For example, [CoBrCl(NH 3 ) 2 (en)]Cl is wrong, and [Co(NH3)2BrCl(en)]Cl is correct.
■ The formula for the entire coordination entity, whether charged or not, is enclosed in square brackets. When ligands are polyatomic, the formula of liquid is enclosed in parenthesis. Ligand abbreviations are also enclosed in parenthesis.
■ There should be no space between the ligands and the metal with in a coordination sphere.
■ When the formula of a charged coordination entity is to be written, without the counter ion, the charge is indicated outside the square brackets as a right superscript with the number before the sign. For example, [Co(CN)6]3–, [Cr(H2O)6]3+ etc.
■ The charge of the cation(s) is balanced by the charge of the anion(s).
5. List out the incorrect ones among the following, according to IUPAC, and write the correct formulae.
a) [Zn(OH)4]K2
b) [CoCl(NH3)4(H2O)]Cl2
c) [Ag(CN)2] [Ag(NH3)2]
d) [Pt(NH3)2Cl(NO2)]
Sol. Formulae (a), (b), and (c) are wrong and only (d) is the correct formula.
To correct (a) and (c), formula of cation is to be written first. After correction, the correct formula for (a) and (c) are K 2[Zn(OH)4] and [Ag(NH3)2] [Ag(CN)2]. To correct (b), ligands are to be in alphabetical order, irrespective of charge. After correction, the correct formula of (b) is [Co(NH3)4(H2O)Cl]Cl2
Try yourself:
4. Write the IUPAC name of [CoCl 2(en)2]Cl coordination compounds.
Ans : Dichloridobis (ethane–1,2–diammine) cobalt (III) chloride
6.3.2. Naming of Mononuclear Coordination Compounds
IUPAC has recommended the following rules for naming the mononuclear coordination compounds.
■ In ionic complexes, the cation, whether it is complex or not, must be named first as a single word and then the anion is named. Name of neutral complexes are to be written as a single word.
■ In the name of complex ion, the ligands are named in alphabetical order before the name of the central metal atom or ion.
■ Oxidation state of the metal in cation, anion or neutral coordination entity is indicated by Roman numerical in parenthesis.
■ Names of the anionic ligands end in ‘O’ but the names of positive ligands end in ‘ium’. The 2004 IUPAC draft recommends that anionic ligands, particularly halides, will end with ‘-ido’, so that chloro would become chlorido. Neutral ligands have their names either unchanged or have characteristic names. Names of anionic ligands are given in Table 6.4. The names of cationic and neutral ligands are given in Table 6.5.
■ Prefixes di, tri, tetra, penta, etc. are used to indicate the number of the individual ligands in the coordination entity. When the names of the ligands include a numerical prefix, then the terms bis, tris, and tetrakis are used for two, three, and four ligands, respectively, and the name of ligand is placed in parenthesis. For example, the complex [NiCl 2 (PPh 3 ) 2 ] is named dichlorobis (triphenylphosphine) nickel (II).
If the complex ion is a cation, the metal is named the same as the element. For example, Fe in a complex cation is called iron and Pt is called platinum. If the complex ion is an anion, the name of the metal ends with the suffix
‘-ate’. For example, Co in a complex anion is called cobaltate. For some metals, the Latin names are used in the complex anions, e.g., Ferrate, cuprate, hydrargyrate, and argentate are used for Fe, Cu, Hg, and Ag, respectively.
■ There is no need to indicate the number of counter ions with a number prefix in the name of ionic complexes.
■ The neutral complex molecule is named similar to that of the cationic complex.
6 Write the formula of the following coordination compounds:
a. Amminebromidochloridonitrito–O–platinum(II) ion
b. Dichloridobis (ethane–1,2–diammine) latinum (IV) nitrate
c. Diaquatetrahydroxoaluminate (III) ion
d. Mercury (I) tetrathiocyanatocobaltate (III).
Sol. a) [Pt(NH3) BrCl(ONO)]– , b) [PtCl2(en)2] (NO3)2, c) [Al(H2O)2(OH)4]– and d) Hg[Co(SCN)4].
Try yourself:
5. What is the formula of Hexaamminechromium (III) hexacyanocobaltate (III) coordination compounds? ]6)3[Cr(NHAns: ]6[Co(CN)
6.3.3 Examples of Naming
The Following examples illustrate the nomenclature for coordination compounds.
■ [Cr(NH3)4(H2O)2]Cl3 is named tetraammine diaquachromium (III) chloride. In this compound, cation is complex ion and Cl –ions are counter ions. Ammine ligands are named before the aqua ligands, according to alphabetical order.
Four ammine ligands and two aqua ligands are present. Thus, the number prefix tetra is used for ammine and di is used for aqua. After writing the names of ligands, the name of the metal, chronium is written without ending with –ate, since it is a cationic complex. Oxidation state of chromium is indicated in Roman number in parenthesis. The name of the anion, chloride, is written as a separate word. There is no need to indicate the number of counter ions.
■ [Co(en)3]2(SO4)3 is named tris (ethane–1,2–diammine) cobalt (III) sulphate. Sulphate is the counter anion in this molecule. Since it takes three sulphates to bond with two complex cations, the charge on each complex cation must be +3. Further, ethane–1,2–diammine is a neutral ligand, so the oxidation number of cobalt in the complex ion must be +3. Since the number prefix is already present in the name of ligand, we use tris instead of tri for three ligands.
7. What is the IUPAC name of [Co(NH3)4Br2]2[ZnCl4]?
Sol. Tetra ammine dibromocobalt (III) tetrachlorozincate (II)
Try yourself:
6. Write the formula of tetracarbonylnickel (0).
Ans: ]4[Ni(CO)
TEST YOURSELF
1. The IUPAC name of [Co(NH3)4Br2]2[ZnCl4] is
(1) dibromo tetraammine cobalt (III) tetrachlorozinc (II)
(2) tetrammine dibromo cobalt (III) tetrachlorozinc (II)
(3) tetra ammine dibromocobalt (III) tetrachlorozincate (II)
(4) tetrachlorozinc (II) tetra ammine dibromo cobaltate (III)
3. Tris (ethane-1, 2-diammine) cobalt (III) sulphate is denoted by the formula
(1) [Co3(en)]SO4 (2) [Co(en)3]SO4
(3) [Co(en)3]3(SO4)2 (4) [Co(en)3]2(SO4)3
4. The IUPAC name of the complex [CoNO2(NH3)5]Cl2 is
(1) pentaamminenitrito-N-cobalt (III) chloride
(2) nitrito-N-pentaamminecobalt (III) chloride
(3) nitrito-N-pentaamminecobalt (II) chloride
(4) pentaamminenitrito-N-cobalt (II) chloride
Answer Key
(1) 3 (2) 1 (3) 4 (4) 1
6.4 ISOMERISMS IN COORDINATION COMPOUNDS
Two or more compounds that have the same c hemical formula but a different arrangement of atoms are called isomers and the phenomenon is known as isomerism. Isomerism in complexes may be broadly divided into two types, structural isomerism and stereoisomerism.
6.4.1 Structural Isomerism
Structural isomerism arises due to different kinds of bonds between the metal and the ligands. This can be subdivided into ionisation, solvate, linkage, and coordination isomerism.
Ionisation Isomerism
The compounds that have same molecular formula but give different ions in solution are called ionisation isomers. This type of isomerism is possible when the counter ion in a complex salt is itself a potential ligand and can displace a ligand which can then become the counter ion. Generally, anionic complexes cannot exhibit ionisation isomerism.
Ionisation isomers will be produced when ligands inside and counter ions outside the coordination sphere are interchanged. For example, [Co(NH 3 ) 5 Br]SO 4 and [Co(NH 3) 5(SO 4)]Br together are ionisation isomers. Aqueous solution of first compound gives white precipitate with BaCl 2 solution, showing the presence of SO 4 2– ion, and it gives no precipitate with aqueous solution of AgNO 3, showing the absence of Br– ion. Aqueous solution of second compound gives a pale yellow precipitate with AgNO3 solution, showing the presence of Br – ion, and it gives no precipitate with BaCl2 solution, showing the absence of SO 4 2– ion. Other examples are [Pt (NH 3 ) 4 Cl 2 ]Br 2 and [Pt(NH 3 ) 4 Br 2 ] Cl 2 ;[Co(NH 3 ) 4 Cl 2 ]NO 2 and [Co(NH 3 ) 4 Cl(NO2)]Cl2
Solvate Isomerism
The compounds that have same molecular formula but differ in the number of solvent molecules present as ligands or as molecules
of solvation are called solvent isomers. It is similar to ionisation isomerism, in which water molecules may occur inside and outside the coordination sphere. It is also called hydrate isomerism.
CrCl3.6H2O molecular formula represents three possible isomers. They are [Cr(H2O)6]Cl3 (violet), [Cr(H2O)5Cl]Cl2.H2O (grey-green), and [Cr(H 2 O) 4 Cl 2 ] Cl.2H 2 O (dark-green). [Cr(H 2 O) 3 Cl 3 ].3H 2 O may also be possible. With excess of aqueous AgNO3 solution, one mole of violet coloured CrCl 3 .6H 2 O gives three moles of AgCl precipitate, one mole of light green CrCl 3.6H 2O gives two moles of AgCl precipitate and one mole of dark green CrCl 3 .6H 2 O gives one mole of AgCl precipitate. Other examples of hydrate isomers are [Co(H 2 O)Cl(en) 2 ]Cl 2 and [CoCl 2 (en) 2 ] Cl.H2O.
Linkage Isomerism
Linkage isomerism arises in complexes having at least one ambidentate ligand. Compounds that have the same molecular formula but differ they in the mode of attachment of a ligand to the metal ion are called linkage isomers. Jorgensen discovered linkage isomer in the complex [Co(NH3)5(NO2)]Cl2, as shown in Fig. 6.5. Complex in which the nitrite ligand is bound through oxygen (–ONO) is red and the complex in which the nitrite ligand is bound through nitrogen (–NO2) is yellow.
Yellow
Fig 6.5 Linkage isomers of [Co(NH3]5 (NO2)]Cl2
Coordination Isomerism
In i onic complexes, if the cation and anion are complex entities, then coordination of isomerism may arise. Mutual exchange of ligands completely or partially between the cation and anion results in this isomerism. For example, [Co(NH3)6][Cr(CN)6] and [Cr(NH3)6] [Co(CN)6] are coordination isomers. In the first complex, NH3 ligands are bound to Co3+ and CN – ligands to Cr 3+, but in the second complex, NH3 ligands are bound to Cr3+ and CN– ligands are bound to Co3+. [Cu(NH3)4] [PtCl 4] and [Cu(NH 3) 3Cl] [Pt(NH 3)Cl 3] are coordination isomers formed by partial exchange of ligands. [Cu(NH3)4][PtCl4] and [Pt(NH3)4][CuCl4] are coordination isomers formed by complete exchange of ligands.
6.4.2 Stereoisomerism
Stereoisomers have the same chemical formula and chemical bonds but they have different spatial arrangement. Stereoisomerism is mainly of two types: geometrical isomerism and optical isomerism.
Geometrical Isomerism
This type of isomerism arises in heteroleptic complexes due to different possible geometric arrangements of the ligands. In the cis isomer similar ligands are adjacent to each other, but in the trans isomer, similar ligands are opposite to each other. Geometrical isomerism of compounds with coordination numbers 4 and 6 is most important.
The complexes having coordination number four adopt tetrahedral or square planar geometry. Tetrahedral complexes do not exhibit geometrical isomerism because
the relative positions of the ligands attached to the central metal are the same, with respect to each other.
However, square planar complexes show geometrical isomerism. Square planar complexes of the type MA 2 X 2 and MA 2 XY (A, X and Y are monodentate ligands) can exhibit geometrical isomerism.
Square planar complex of the type MABXY shows 3 isomers, 2 cis and 1 trans. Geometrical isomers of [Pt(NH 3 ) 2 Cl 2 ] and [Pt (NH3)2BrNO2] are shown, respectively, in Fig.6.6andFig.6.7.
Fig.6.6
Geometrical isomers of [Pt(NH3)2Cl2]
Fig. 6.7
Geometrical isomers of [Pt(NH3)2BrNO2]
Square planar complexes containing unsymmetrical bidentate ligands, such as [M(AB)2], also show geometrical isomerism. Geometrical isomers of [Pt(gly) 2], where gly denotes H2NCH2COO–, are given in Fig. 6.8
Cis - diglycinatoplatinum (II)
trans-diglycinatoplatinum (II)
Fig.6.8 Geometrical isomers of [Pt(gly)2]
The octahedral complexes of the type [MA4X2],[M(AA)2X2] etc. exhibit geometrical isomerism. Here, A, X are monodentate ligands and (AA) is symmetrical bidentate ligand. Geometrical isomers of [Co(NH3)4Cl2]+ are given in Fig. 6.9 and of [CoCl 2(en) 2] + are given in Fig. 6.10.
-isomer
Fig. 6.10 Geometrical isomers of [Co(en)2Cl2]+
Another type of geometrical isomerism occurs in octahedral coordination entities of the type [Ma3b3] like [Co(NH3)3(NO2)3]. If three donor atoms of the same ligands occupy adjacent positions at the corners of an octahedral face, we have the facial (fac) isomer. When the positions are around the meridian of the octahedron, we get the meridional (mer) isomer Fig. 6.11
Fig. 6.9 Geometrical isomeers of [Co(NH3)4Cl2)+
Fig 6.11 Facial and meridional isomers [Co(NH3)3(NO2)3]
Optical Isomerism
A Coordination compound which can rotate the plane of polarised light is said to be optically active.
The coordination compounds that have the same formula but differ in their ability to rotate directions of the plane of polarised light are said to exhibit optical isomerism and the molecules are optical isomers. The optical isomers are a pair of molecules that are nonsuper impossible mirror images of each other.
This is due to the absence of elements of symmetry in the complex.
Optical isomerism is expected in tetrahedral complexes of the type Mabcd These isomers are not ionisable due to their labile nature. Bisbenzoylacetonato-beryllium(II) is an example .
Fig. 6.14 Enantiomers of [Co(en)2NH3Cl]+2
Octahedral complexes containing hexadentate ligands, such as ethylenediamine tetra acetate, also show optical isomerism. For example, [Co(edta)]– ion is optically active. A 1:1 mixture of d- and l- isomers is called racemic mixture.
Complexes of the type [M(AA) 3 ] n+ also exhibit optical isomerism. Though three isomers are predicted for such type of complex, only two are optically active. The third one is optically inactive and is called meso form. Isomers of [Co(en)3]3+ are shown in Fig. 6.15.
Optical isomerism is common in octahedral complexes. Examples of optical isomers are [Co(en) 3 ] 3+ , cis [CoCl 2 (en) 2 ] + and [Co(NH 3 ) 2 Cl 2 (en)] + . Optical isomers of [Pt(py) 2 (NH 3 ) 2 Cl 2 ] 2+ , [Pt(en) 2 Cl 2 ] 2+ , [CO(en)2NH3Cl]2+ are shown respectively in Fig. 6.12, Fig.6.13 and Fig.6.14.
Fig. 6.12 Optical isomers of cis-[Pt(py)2(NH3)2Cl2]2+
Fig. 6.15 Isomers of the complex ion, [Co(en)3]3+
The complexes with coordination number 6 can exist in octahedral geometry. These are exhibit different number of stereo isomers. The following cases may arise in complexes with octahedral geometry are show in Table 6.6.
Table 6.6 Number of stereoisomers in octahedral complexes
Fig. 6.13 Enantiomers of
[Pt(en)2Cl2]2+
Compound No. of stereoisomers
of geometrical isomers
Ma3bcd 5 4 2 Ma2bcde 15 9 12 Mabcdef 30 15 30
Ma2b2c2 6 5 2
Ma2b2cd 8 6 4
Ma3b2c 3 3 0
M(AA)(BC) de 10 5 10
Sol. [Cr(gly)3] exhibits geometrical isomerism and optical isomerism. It act as cis and trans isomers. Cis or trans isomer has d-form and l-form, with a total of 4 isomers.
Try yourself:
7. CoCl3.xNH3 exhibits geometrical isomerism. What is the value of x?
Ans: The value of x may be 3 or 4. CoCl .3NH3 3 exhibits facial - meridional isomerism.
TEST YOURSELF
1. [Co(NH3)5Br]SO4 and[Co(NH3)5SO4]Br are the examples of
(1) ionisation isomerism
Formula Possible numbers of stereoisomers Possible numbers of geometrical isomers Possible numbers of enantiomer pair
4 3 1
[M(AB)a2b2] 6 4 2
[M(AB)a2b3] 12 7 5
[M(AB)abcd] 24 12 12
M(ABCBA)d 7 4 6
(uppercase letters represent chelating ligands and lowercase letters represent monodenadate ligands. The optical isomers include both the isomers, to get total number of isomers = half of the optical isomers+ geometrical isomers)
8. What type of isomerism is exhibited by the complex [Cr(gly)3]?
(2) linkage isomerism
(3) coordination isomerism
(4) geometrical isomerism
2. The compounds [Cr(H2O)6] Cl3, [Cr(H2O)5Cl] Cl2. H2O, and [Cr(H2O)4 Cl2] Cl2. H2O exhibit (1) linkage isomerism
(2) geometrical isomerism
(3) ionisation isomerism
(4) hydrate isomerism
3. Which of the following pairs is an example of coordination isomerism?
(1) [Cr(H2O)6]Cl3 and [Cr(H2O)5Cl]Cl2. H2O
(2) [Co(NH3)5NO2]Cl2 and [Co(NH3)5ONO] Cl2
(3) [Co(NH3)6][Cr(CN)6] and [Cr(NH3)6] [Co(CN)6]
(4) [Co(NH3)5Br]SO4 and [Co(NH3)5SO4] Br
4. Number of stereo isomers for [Pt(Br)(Cl) (NH 3)(P y)] and [Pt(en) 2Cl 2] 2+ are ‘ X ’ and ‘Y’, respectively. X Y =
(1) 0.666 (2) 1
(3) 1.33 (4) 0.5
5. Which one of the following has an optical isomerism? (en= ethylenediamine)
(1) [Zn(en)(NH3)2]2+ (2) [Co(en)3]3+
(3) [Co(H2O)4(en)]3+ (4) [Zn(en)2]2+
6. Which of the following complex compounds shows optical isomerism?
Werner’s theory failed to answer some basic questions:
■ Why do only certain elements possess the reasonable property of forming coordination compounds?
■ Why do the bonds in coordination compound have directional properties?
■ Why do coordination compounds have characteristic magnetic and optical properties?
Many approaches have been introduced to explain the stability and nature of bonding in
coordination compounds. They are Sidgwick theory, valence bond theory (VBT), crystal field theory (CFT), ligand field theory (LFT) and molecular orbital theory (MOT). We shall focus our attention on elementary treatment of the application of VBT and CFT to coordination compounds.
6.5.1 Sidgwick Theory
According to Sidgwick, primary valence of metal is regarded as formed by electron transfer. The secondary valence corresponds to the coordinate covalent bonds in the complex.
The ligands possess pairs of electrons which are donated to fill the vacant orbitals of the metal and form coordinate covalent bonds.
Sidgwick also suggested that the metal ion will continue accepting electron pairs from ligands until the total number of electrons on the metal in the complex is equal to that of the next inert gas. The total number of electrons at the central metal in a complex after coordination is known as effective atomic number (EAN) of the metal in that complex.
The EAN of a metal in a complex is obtained by subtracting the number of electrons lost by the metal in its ion formation from the atomic number (Z) and then adding the number of electrons gained through coordination.
Table 6.7 EAN values of metals in some complex compounds
The number of electrons lost by the metal is equal to its oxidation state in the complex and the number of electrons gained by the metal is equal to twice of its coordination number. EAN = Number of electrons in metal ion + (2 × coordination number)
In many complexes, EAN of a metal in a complex corresponds to atomic number of the next heavier inert gas. However, as seen from Table 6.7, there are some exceptions to the Sidgwick rule of effective atomic number.
9. If [Fe(CO) x ] follows the Sidgwick rule of stability, what is the value of ‘ x’?
Sol. The complex compound [Fe(CO) x] following the Sidgwick rule of stability, EAN of iron is to be equal to the atomic number of nearest inert gas, 36(Krypton). 36 = 26 – 0 + 2 × x x = 5 Hence the value of x is 5.
Try yourself:
8. What is the EAN value of [Ni(CO) 4]?
Ans: The EAN value of ]is4[Ni(CO) 36
9.5.2 Valence Bond Theory
Valence bond theory is the oldest theory of complex compounds proposed by Linus Pauling. Pauling explained formation of complex between a metal and ligand and structures of complex compounds.
Postulates of Valence Bond Theory
■ The central metal first loses the requisite number of electrons to form the ion and necessary number of vacant orbital s are created.
■ The number of electrons lost is numerically equal to the oxidation state of the metal.
■ The vacant orbitals of the metal ion undergo suitable hybridisation so that the desired shape is obtained by the complex. The type of hybridisation and the shape of complex species are given in the Table 6.8
■ In octahedral complexes, if (n–1)d orbitals are used for hybridisation along with ns and np orbitals, then it is d 2sp3 hybridisation. Such complexes are called inner orbital or low spin or spin paired or strong field or covalent complexes.
■ In octahedral complexes, if nd orbitals are used for hybridisation along with ns and np orbitals, then it is sp 3d2 hybridisation. Such complexes are called outer orbital or high spin or spin free or weak field or ionic complexes.
■ In the formation of some complexes, the electrons in the metal orbital may undergo pairing against Hund’s rule to get the required number of empty (n–1)d orbitals.
■ Completely filled orbitals of ligand overlap the vacant metal orbitals to form a strong coordinate covalent bond to the extent possible.
■ The metal acquires the next inert gas configuration or comes nearest to the inert gas configuration.
■ The electrons in an incompletely filled orbital gives rise to the resultant magnetic moment.
Table 6.8 Hybridisation and shapes of the complexes of 3d–metals
Examples
It is usually possible to predict the geometry of a complex from the knowledge of its magnetic behaviour on the basis of valence bond theory. In the diamagnetic complex, [Co(NH 3) 6] 3+ , cobalt is in +3 oxidation state and it has the electronic configuration 3d 6
Since Co(NH 3 ) 6 3+ is diamagnetic, four unpaired electrons in free Co 3+ pair up to generate two empty 3d orbitals. Co3+ undergoes d2sp3 hybridisation by using two empty 3d–orbitals, one empty 4s–orbital, and three empty 4p-orbitals. Six pairs of electrons, one from each NH 3 molecule, occupy the six empty hybrid orbitals. Thus, [Co(NH3)6]3+ is inner orbital complex and it has octahedral structure. The hybridisation scheme in Co 3+ of the complex [Co(NH3)6]3+ is as shown in Fig. 6.16.
of Co3+ ion
Hybridised orbitals of Co3+
Inner orbital [Co(NH3)6]3+ complex
4p 3d
3d d2sp3 hybrid orbitals
3d 6 pairs from NH3 ligands
Fig 6.16 Scheme of hybridisation of Co3+ in [Co(NH3)6]3+
The paramagnetic octahedral complex [CoF6]3–uses the outer 4d orbitals in hybridisation (sp3d2). It is thus called outer orbital complex. The hybridisation scheme in Co 3+ of this complex is as shown in Fig. 6.17.
Outer orbital [COF6]33d sp3d2 Hybrid orbitals
6 pairs from F– ligands
Fig 6.17 Scheme of hybridisation of Co3+ in [CoF6]3–
In complexes with the coordination number four, square planar structure dsp2 hybridisation is possible if one (n–1)d empty orbital is available along with one empty ns orbital and two empty np orbitals. If empty (n–1)d orbitals
are not available, then tetrahedral structure with sp3 hybridisation is possible.
In paramagnetic complex, [NiCl 4 ] 2–, nickel is in +2 oxidation state and the ion has the electronic configuration 3d8. Since the complex is paramagnetic, two half filled 3d orbitals of Ni2+ remain as they are. Now, Ni2+ undergoes sp3 hybridisation and the complex, [NiCl4]2–, has tetrahedral structure. Similarly, [Ni(CO)4] has tetrahedral geometry but it is diamagnetic, since nickel is in zero oxidation state and contains no unpaired electrons. The hybridisation scheme of in Ni2+ of the complex [NiCl4]2– is given in Fig. 6.18
Hybridised orbitals of Ni2+
Orbitals of Ni2+ ion 3d 3d 4s 4p
sp3 Hybrid orbitals
4 pairs from Cl– ligands
High spin[NiCl4]2–complex
Fig 6.18 Scheme of hybridisation of Ni2+ in [NiCl4]2–
In diamagnetic complex, [Ni(CN) 4]2–, nickel is in +2 oxidation state and the ion has the electronic configuration 3d 8. Since the complex is diamagnetic, electrons in two half filled 3d-orbitals of Ni2+ pair up to generate one empty 3d–orbital. Ni 2+ undergoes dsp 2 hybridisation and the complex has square planar structure. The hybridisation scheme of Ni2+ is as shown in Fig. 6.19
Orbitals of Ni2+ ion
Hybridised orbitals of Ni2+
Low spin[NiCN4]2–complex 3d dsp2 Hybrid orbitals
4 pairs from CN– ligands
Fig 6.19. Scheme of hybridisation of Ni2+ in [Ni(CN)4]2–
Hybridisation and structures of some more complexes are given in Table 6.9. It is important to note that the hybrid orbitals do not actually exist. In fact, hybridisation is a mathematical
Table 6.9 Hybridisation of metal ion and structure of some complexes
[Mn(CN)6]4-
[Cr(NH3)6]3+
[Cr(H2O)6]3+
[Fe(CN)6]3-
[Fe(CO)5]
manipulation of the wave equation for the atomic orbitals involved.
10. Why does the hexaquamanganese(II)ion contain five unpaired electrons, while the hexacyno manganese (II) ion contains only one unpaired electron?
Sol. Mn2+
3d5 4s 4p 4d ↑ ↑↑↑↑↑
H2O is a weak field ligand and CN – is a strong field ligand. Thus, in [Mn(H2O)6]2+, pairing of unpaired electrons is not possible and it possesses five unpaired electrons.
In [Mn(CN)6]4–, pairing of four unpaired electrons takes place, and it possesses one unpaired electron.
Try yourself:
9. What are inner orbital or low spin or spin paired or strong field or covalent complexes?
Ans: In octahedral complexes, if (n–1)d orbitals are used for hybridisation along with ns and np orbitals, then it is d3sp2 hybridisation. Such complexes are called inner orbital or low spin or spin paired or strong field or covalent complexes.
9.5.3 Magnetic Properties of Coordination Compounds
The magnetic moment of coordination compounds can be measured by the magnetic susceptibility experiments. The results can be used to obtain information about the structures adopted by metal complexes. For metal ions with up to three electrons in the d-orbitals, like (d 1 ) Ti 3+ , (d 2 ) V 3+ , and (d 3 ) Cr3+, two vacant 3d-orbitals are available for octahedral hybridisation, along with 4s and 4p orbitals. The magnetic behaviour of these free ions and their coordination entities are similar. When more than three 3d electrons are present, the required pair of vacant 3d orbitals for octahedral hybridisation is not directly available. Thus, for d4 (Cr2+, Mn3+), d5(Mn2+, Fe3+) and d6 (Fe2+, Co3+) cases, a vacant pair of d- orbitals results only by pairing 3d electrons,
which leaves two, one, and zero unpaired electrons, respectively.
In many cases, the magnetic data of complexes agree with partial spin pairing for metal ions with d 4 configuration and maximum spin pairing for metal ions with d 5 and d 6 configuration. [Mn(CN) 6 ] 3– has magnetic moment of two unpaired electrons, while [MnCl6]3– has a paramagnetic moment of four unpaired electrons. This apparent anomaly is explained by the valence bond theory in terms of formation of inner orbital and outer orbital coordination entities. [Mn(CN)6]3–, [Fe(CN)6]3–, and [Co(C2O4)3]3–are inner orbital complexes involving d 2sp 3 hybridisation. The former two complexes are paramagnetic and the latter is diamagnetic. On the other hand, [MnCl6]3–, [FeF6]3–, and [CoF6]3– are outer orbital complexes involving sp3d2 hybridisation. These are paramagnetic corresponding to four, five, and four unpaired electrons.
9.5.4. Limitations
While the VB theory explains the formation, structures and magnetic behaviour of coordination compounds to a larger extent, it suffers from the following shortcomings. Applying VB theory, [Cu(NH3)4]2+ complex is expected to be tetrahedral with sp 3 hybridisation of metal ion. Experimentally, the complex was observed to be square planar.
VB theory involves a number of assumptions, and it does not give quantitative interpretation of magnetic data and colour exhibited. It does not give a quantitative interpretation of the thermodynamic or kinetic stabilities of coordination compounds. It does not explain field strength of the liquids.
11. How many unpaired electrons present in the square planar [Pt(CN)4]2– ion ?
sol. In [Pt(CN)4]2–, Pt2+ has
4d85so5po
It is a squar planar complex in which CN– is a strong ligand, hence, electrons are paired against Hunds rule.
4d85so5po d
In [Pt(CN)6]2–, no unpaired electrons are present and it is diamagnetic.
Try yourself:
10. The spin only magnetic moment of [MnBr42–is 5.9BM. Predict the geometry of the complex ion.
Ans: Since the coordination number of Mn 2+ ion in the complex ion is 4, the geometry may be either tetrahedral (sp )3 or square planar (dsp).2
9.5.5 Crystal Field Theory
The crystal field theory considers metal ligand bond as ionic, which arises purely from electrostatic interaction between the metal ion and ligand.
In crystal field theory, ligands are treated as point charges in case of anionic ligands or dipoles in case of neutral ligands. The five d orbitals in an isolated gaseous metal ion are degenerate. This degeneracy is maintained if a spherically symmetrical field of negative charges surrounds the ion. However, when this negative field is due to ligands in a complex, it becomes asymmetrical and the degeneracy of the d orbitals is lifted. It results in splitting of the d–orbitals. The pattern of splitting depends upon the nature of the crystal field.
Crystal Field Splitting in Octahedral Complexes
In an octahedral coordination entity with six ligands surrounding the metal ion, there will be repulsion between the electrons in metal d–orbitals and the electrons (or negative charges)
of ligand. Such a repulsion is more when the metal d–orbital is directed towards the ligand, than when it is away from the ligand.
In 22 xy d and 2 z d orbitals, lobes are orientated towards axes, but in d xy , d yz, and d xz orbitals, lobes are oriented in between the axes. Since the ligands approach the metal ion along the axes, 22 xy d and 2 z d orbitals experience more repulsion and the energy of orbitals is raised. The energy of dxy, dyz, and dxz orbitals is lowered, relative to average energy of d–orbitals in the spherical crystal field. The set of three lower energy orbitals is called t 2g and the set of two higher energy orbitals is called e g. This splitting of degenerate levels due to the presence of ligands in a definite geometry is termed as crystal field splitting and the energy separation is denoted by ∆ 0, as shown in Fig. 6.20.
The crystal field splitting energy, ∆ 0 , depends upon the field produced by the ligand and charge on the metal ion. In general, ligands can be arranged in a series in the order of increasing field strength.
This series is called spectrochemical series. It is an experimentally determined series based on the absorption of light by complexes with different ligands.
In d1, d2, and d3 coordination entities, the d–electrons occupy t 2g orbitals singly in accordance with the Hund’s rule. For d4 ions, two ways of electron distributions are possible. In one way, the fourth electron enters t2g level and pairs with an existing electron. In the second way, the fourth electron enters eg level without pairing in t2g level. The way of electron distribution depends on the magnitude of crystal field splitting energy, ∆ 0 , and the pairing energy, P. Pairing energy is the energy required for electron pairing in a single orbital.
Energy
d orbitals in free ion
Average energy of the d orbitals in spherical crystal field Splitting of d-orbitals in octahedral crystal field
Fig. 6.20 Splitting of d-orbitals in octahedral complex
The energy of e g - orbitals is increased by (3/5)∆0 and the energy of t2g orbitals is lowered by (2/5)∆ 0 . Each electron entering the t 2g orbital stabilises the complex by 0.4 ∆ 0 units and each electron entering e g destabilises the complex by 0.6∆0 units.
If ∆ 0 < P, the fourth electron enters one of the eg- orbitals, giving the configuration t2g3 e g 1. Ligands for which ∆ 0 < P are known as weak field ligands, and they form high spin complexes or outer orbital complexes.
If ∆ 0 > P, it be comes more energetically favourable for the fourth electron to occupy a t 2g orbital with configuration 4o 2gg te.
Ligands which produce this effect are known as strong field ligands and they form low spin complexes or inner orbital complexes.
Filling up of d–orbitals and crystal field stabilisation energies are given in Table 6.10. Crystal field stabilisation energy calculations show that d 4 to d 7 coordination entities are more stable for strong field as compared to weak field cases.
Table 6.10 Crystal field stabilisation energies in the presence of strong and weak ligand
Crystal
Field Splitting in Tetrahedral Coordination Entities
In the formation of tetrahedral coordination entity, the d–orbital splitting is inverted and is smaller as compared to octahedral field splitting. Splitting of d–orbitals in a tetrahedral crystal field is given in Fig. 6.21
Energy
d orbitals in free ion
Average energy of the d orbitals in spherical crystal field
Splitting of d-orbitals in octahedral crystal field
Fig.6.21 Splitting of d-orbitals in a tetrahedral crystal field
For the same metal, same ligands and metal ligand distance, the relation between tetrahedral splitting energy,∆ t and octahedral splitting energy,∆0, is ∆t = (4/9)∆0. Consequently, the orbital splitting energies are not sufficiently large for forcing pairing and, therefore, low spin configurations are rarely observed.
Jahn Teller Distortion
Any non-Linear molecular system possessing degenerate electronic state will be unstable, and will undergo some kind of distortion to get stability. This is called John Teller (J.T) distortion. This distortion lowers its symmetry and degeneracy as shown in Table 6.11
■ t2g°, t2g3, t2g6 ⇒ t2g(symmterical)
■ t 2g1, t2g2, t2g4, t2g5 ⇒ t2g(unsymmterical)
■ eg°, eg4(symmetrical)
■ e g1, eg3( unsymmetrical)
■ e g 2 – have high spin complex is called symmterical.
■ e g 2 – have low spin complex is called unsymmterical.
■ t 2g (sym)+eg (sym) ⇒ No distortion
■ t2g(sym) +eg (unsym) ⇒ strong distortion
■ t 2g (unsym) +eg (unsym) ⇒ strong distortion
Note: following complexes exhibit J.T distortion
■ d 4 → High spin → strong distortion (Cr+2, Mn+3)
■ d 6 → High spin → slight distortion ( Fe+2, Co+3)
■ d7→ Low spin → strong distortion ( Co+2)
■ d 9 → High spin/ Low spin → strong distortion (Cu+2)
Table 6.11. Jahn teller distortion
dn–
Configura-
d1 t 2g 1 (unsym)
eg° (sym)
d2 t 2g 2 (unsym)
Slight t 2g 1 (unsym) eg° (sym) Slight
eg° (sym) Slight t 2g 2 (unsym) eg° (sym) Slight
d3 t 2g 3 (sym)
eg° (sym) No t 2g 3 (sym) eg° (sym) No
d4 t 2g 3 (sym)
e g 1 (unsym) strong t 2g 4 (unsym)
e g 0 (sym) Slight
d5 t 2g 3 (sym)
e g 2 (sym) No t 2g 5 (unsym)
eg° (sym) Slight
d6 t 2g 4 (unsym)
e g 2 (sym) Slight t 2g 6 (sym)
e g ° (unsym) No
from t 2g level to e g level. Consequently, the complex appears red-violet in colour, which is the complementary colour of absorbed light.
The relationship between the absorbed wavelength and the colour observed is given in Table 6.12, by taking some examples of complex compounds. The crystal field theory attributes the colour of the coordination compounds to d–d transition of the electron.
It is important to note that in the absence of ligand, crystal field splitting does not occur, and hence, the substance is colourless. For example, anhydrous Ti 3+ is colourless. Similarly, anhydrous CuSO 4 is white, but CuSO4.5H2O is blue in colour. The influence of the ligand on the colour of a complex may be illustrated by considering the complex [Ni(H2O)6]2+
If the bidentate ligand, ethane-1,2diamine(en) is progressively added in the molar ratio of metal and ligand, 1:1, 1:2 and 1:3, the following series of reactions occur. Reactions can be observed with the colour changes.
[Ni(H2O)6]+2(aq) + en(aq) →
d7 t 2g 5 (unsym)
e g 2 (sym)
d8 t 2g 6 (sym)
Slight t 2g 6 (sym) e g 1 (unsym) Slight
e g 2 (sym) No t 2g 6 (sym)
e g2(sym) No
d9 t 2g 6 (sym)
e g 3 (unsym) Strong t 2g 6 (sym)
d10 t 2g 6 (sym)
e g 3 (unsym)
e g 4 (sym) No t 2g 6 (sym)
No
e g 4 (sym)
6.5.6. Colour in Coordination Compounds
Crystal field theory can also explain the colour of complexes in the same way of colour of hydrated transition metal ions which was already discussed in the previous unit. For example, in [Ti(H 2 O) 6 ] 3+ , Ti 3+ has one d–electron in t 2g level. If light corresponding to the energy of yellow-green region is absorbed by the complex, it would excite the electron
[Ni (HO)(en)] 2HO ++
2 24(aq) 2 (pale blue)
[Ni(HO)en]en++→
2 24(aq)(aq) (pale blue)
[Ni (HO)(en)] 2HO ++
2 222(aq) 2 (purple)
[Ni(HO)(en)]en++→
2 222(aq)(aq) (purple)
2 3(aq) 2 (violet)
[Ni(en)] 2HO ++
Table 6.12 Relationship between absorbed wave length and coloured observed
Wavelength of light absorbed Colour of the absorbed light Colour of transmitted light
490–500 nm blue green red
500–560 nm green purple
560–580 nm yellow green violet
580–595 nm yellow blue
595–605 nm orange green blue
605–750 nm red blue green
6.5.7
Limitations
The crystal field theory successfully explains the formation, structures, colour and magnetic properties of coordination compounds to a larger extent. However, from the assumption that the ligands are point charges, it follows that anionic ligands should exert the greatest splitting effect than neutral ligands; but actually, anionic ligands, like halides, exert very less splitting effect. Further, it does not take into account the covalent character of bonding between the ligand and the central metal ion. All these weaknesses of CFT were later explained by ligand field theory and molecular orbital theory, which are beyond the scope of this book.
12. How does the gemstone ruby exhibit red colour and emerald exhibit green colour?
Sol. Ruby is aluminium oxide (Al2O3) containing about 0.5–1% Cr 3+ ion (d 3). This chromium (III) is an octahedral complex incorporated into the alumina lattice. The d–d transitions in Cr3+ give the red colour to ruby. Emerald is the beryl mineral (3BeO.Al2O3.6SiO2) having some Cr3+ ions at octahedral sites. The absorption bands seen in ruby shift to longer wavelength, namely yellow-red and blue, causing emerald to reflect green colour.
Try yourself
11. Anhydrous copper sulphate is colourless, but hydrated copper sulphate is blue. Explain.
It absorbs energy in visible region of light. Hence it exhibits colour.
The splitting of d-orbitals is reasonably increased.
Ans: In anhydrous copper sulphate, the splitting is very less. It absorbs energy in infrared region. It is colourless. In hydrated copper sulphate, water molecules are present in the coordination sphere.
TEST YOURSELF
1. According to CFT, the energy of t2g orbitals in an octahedral complex
(1) decreases by 0 2 5 ∆
(2) increases by 0 2 5 ∆
(3) increase by 0 3 5 ∆
(4) decreases by 0 3 5 ∆
2. The spin only magnetic moment of [Mn(Br) 4 ] 2− is 5.9 BM. Then, possible hybridisation of Mn in the complex is (1) sp3d (2) dsp3 (3) d2sp3 (4) sp3
3. The crystal field stabilisation energy (CFSE) for [CoCl6] 4− is 18000 cm −1. The CFSE for [CoCl4]2− will be (1) 6000 cm–1 (2) 16000 cm–1 (3) 18000 cm–1 (4) 8100 cm–1
4. [Ma 6 ] +2 ,[Mb 6 ] +2 ,[Mc 6 ] +2 , and [Md 6 ] +2 octahedral complexes absorb wavelengths in the region of red, green, blue, and yellow, respectively. What is the strongest ligand among a, b, c, and d. (1) a (2) b (3) c (4) d
5. Arrange the following cobalt complexes in the order of decreasing crystal field stabilisation energy (CFSE) value of complexes? 323 6 26 36 ABC [CoF][Co(HO)][Co(NH)] +++ and 3 3 D [Co(en)] +
(1) D > C> A > B (2) A > D >C > B (3) D > C >B > A (4) A > B > C > D
6. Choose the incorrect statement with respect to valence bond theory. (1) The electrons in the metal orbitals may undergo regrouping even against Hund’s rule.
(2) The electrons in an incompletely filled orbital give rise to the resultant magnetic moment.
(3) Ligand orbitals overlap the vacant metal orbitals to form a strong coordinate covalent bond to the extent possible.
(4) The number of electrons lost is equal to the coordination number of the metal.
7. Among [Ni(CO)4], [Ni(CN)4]−2, and [NiCl4]–2 species the hybridisation states of the Ni atom are, respectively,
(1) sp3,dsp2 and dsp2
(2) sp3, dsp2, and sp3
(3) sp3, sp3, and dsp2
(4) dsp2, sp3, and sp3
8. Which of the following systems has maximum number of the unpaired electrons in an inner orbital octahedral complex?
(1) d4 (2) d9 (3) d7 (4) d5
9. Which of the following complex species does not involve inner orbital hybridisation?
(1) [Co(NH3)6]3+ (2) [Cr(NH3)6]3+
(3) [CoF6]3- (4) [Fe(CN)6]3-
10. Geometrical shapes of the complexes formed by the reaction of Ni2+ with Cl–, CN–, and H2O, respectively, are
(1) octahedral, tetrahedral, and square planar (2) tetrahedral, square planar, and octahedral
(3) square planar, tetrahedral, and octahedral (4) octahedral, square planar, and octahedral
Answer Key
(1) 1 (2) 4 (3) 4 (4) 3
(5) 1 (6) 4 (7) 1 (8) 1
(9) 3 (10) 2
6.6 STABILITY OF COMPLEX COMPOUNDS
The equilibrium constant for the formation of a complex compound is the stability constant.
M + 4L ML4; = 4 c4 [ML] K [M][L]
Here, eq uilibrium constant, K C , can be taken as stability constant. Larger the stability constant, the higher is the proportion of ML 4 that exists in the solution. Thus, greater the stability constant, greater is the stability of the complex compound. The complex, ML4, can be considered as it is formed from metal M and ligand L in four steps, with K1, K2, etc., which are referred to as stepwise stability constants.
M + L ML ; K1 = [ML] / [M][L]
ML + L ML2 ; K2 = [ML2] / [ML][L]
ML2+ L ML 3 ; K3 = [ML3] / [ML2][L]
ML 3 + L ML 4 ; K4 = [ML4] / [ML3][L]
Generally, successive stability constant decreases. The overall stability constant is the product of stepwise stability constants.
K c = K1 × K2 × K 3 × K 4
The instability constant or dissociation constant of coordination compound is defined as the reciprocal of the stability constant. Lesser the instability constant, greater is the stability of the complex compound.
6.6.1 Facors Affecting the Stability of Complexese
Some of the factors effecting on the coordination compounds explained in the following manner.
Shape of complex( Hybridisation)
In general, the complexes, have more crystal field stabilisation energy, those are more stable. Stablility order (∆)
In general, metal ion has higher oxidation state. Its effective nuclear charge increases. It leads to strong attraction between central
metal atoms and ligands, so the ligands are closer to the metal ion and cause repulsions of d-orbital electrons of metal leads more splitting and have high ‘∆’, of the complexes becomes more stability.
For examples: [Fe(III)(CN)6]3–>[Fe(II)(CN)6]4–
Size (Principal Quantum Number) of Metal Ion
Size(principal quantum Number) of metal ion is increases the crystal field stabilisation energy increases, the stability of complexes increases
For example:
[Co(NH3)6]3+ < [Rh(NH3)6]3+ < [Ir(NH3)6]3+
Nature of the ligand
According to spectro-chemical series, the crystal field splitting energy, ∆0, depends upon the field produced by the ligand and charge on the metal ion. In general, ligands can be arranged in a series in the order of increasing field strength.
I – < Br – < SCN – < Cl – < N 3– < F – < OH – < C2O42– < H2O < NCS– < EDTA4– < Py NH3 < en < NO2 – < CN– < CO.
For example:
[CrCl6]3+ < [Cr(H2O)6]3+ < [Cr(CN)6]3–
Irving -William Order
Stabilities of high spin complexes of the ions between Mn 2+ and Zn 2+ with a given ligand vary in the order;
Mn2+ < Fe+2<Co2+ < Ni2+ < Cu2+ > Zn 2+
Radii of these Ions are in the order:
Mn2+(0.91Å) < Fe2+(0.83Å)
< Co2+(0.82Å) < Ni2+(0.78Å)
< Cu2+ (0.69Å) > Zn2+(0.74Å)
Chelating Effect
Complexes containing chelate rings are usually more stable than similar complexes
containing no rings with same donor atoms due to increase in entropy of the reactions.
22 336 (log18.1)(log7.99)(aq)(aq)
>
Ni(en)Ni(NH)
Examp le: When ethylene diamine(en) is allowed to react with hydrated complex ion, [M(H2O)6]2+ , it replaces two H2O molecules from it and, in the system, the number of particles increases. Hence, entropy increases.
∆S = +. ∆G = ∆H –T∆S, ∆G = –RT ln k
Thus, the complex [M(H2O)4(en)] 2+ is more stable than [M(H 2 O) 6 ] 2+ or [ M(NH 3 ) 6 ] 2+
[M(H2O)6]2++ en [M(H2O)4(en)]2+ +2H2O
Rings Size:
The larger the number of the chelate rings in a complex, the greater is its stability.
■ The stability of the complexes also depends on the number of the atoms present in the ring.
■ Three-membered rings including the metal, are unstable.
■ Four-membered rings occur in carbonate, nitrate and sulphate.
■ Five-membered chelates are more stable than the six-membered chelates when the atoms in the rings are joined by single bonds only.
■ Six-membered chelates are more stable than five-membered chelates of hetrocyclic ligands or of ligands involving conjugation in the chelate ring
[Cu(trien)]2+ > [Cu(en)2]2+ > [Cu (NH3)4]2+
Trien : triethylene tetraamine
■ Large bulky ligands form less stable complexes due to steric effects.
■ After formation of complex hydrogen bonds is possible in complex that is more stable than other complex.
[Ni(dmg)2]> [Ni(en)2]+2
13. Ag + + NH 3 [Ag(NH 3 )] + ; K 1 = 3.5 × 10 –3 ; [Ag(NH 3 )] + +NH 3 [Ag(NH 3 ) 2 )] + ; K 2 = 1.7 × 10 –3 . Calculate the formation constant of [Ag(NH 3 ) 2 ] + . What is the instability constant?
Sol. Overall formation constant is the product of successive formation constants, K c = K1 × K2
Instability constant is the inverse of the stability constant, c 1 K = × 6 1 5.9510 = 1.7 × 107
Try yourself:
12. What is instability constant?
Ans: Instability constant is the inverse of the stability constant
TEST YOURSELF
1. Most stable ion is
(1) [Fe(H2O)6]+3 (2) [Fe(CN)6]-3
(3) [FeCl6]-3 (4) [Fe(C2O4)3]-2
2. Which complex compound is most stable?
(1) [Co(NH3)6]2(SO4)3
(2) [Co(NH3)4(H2O)Br](NO3)2
(3) [Co(NH3)3(NO3)3]
(4) [CoCl(en)2]NO3
3. Octahedral complexes of copper (II) undergoes structural distortion (Jahn-Teller). Which one of the given copper (II) complexes will show the maximum structural distortion? (en–ethylenediamine; H2N−CH2−CH2−NH2
(1) [Cu(H2O)6]SO4
(2) [Cu(en)(H2O)4]SO4
(3) cis-[Cu(en)2(Cl2]
(4) trans-[Cu(en)2(Cl2]
4. The stability constant of the complexes formed by a metal ion (M2+) with NH3, CN–,
H2O, and en are of the order 1011, 1027, 1015, and 108, respectively. Then, (1) en is the strongest ligand
(2) CN– is the strongest ligand
(3) the strength of the ligands has no relationship with given values (4) all ligands are equally strong Answer Key
6.7 BONDING IN METAL CARBONYLS
Compounds having one or more metal carbon bonds are called organometallic compounds. For example, CH3MgBr is an organometallic compound whereas C 2 H 5 ONa is not an organo-metallic compound since the metal atom is attached to oxygen.
Organometallic compounds can be broadly classified into three types : σbonded organometallics, π bonded organometallics, and metal carbonyls.
σ bonded Organometallics
These compounds contain metal carbon covalent sigma bond. Some examples of this type of compound are Grignard reagent, diethyl zinc, trimethyl aluminium, tetramethyl tin, etc.
π − bonded Organometallics
These are organometallic compounds which involve the use of πbonds present in the organic compounds. Zeise’s salt, K[PtCl 3(C2H4)] and dibenzene chromium are organometallic compounds of this type.
6.7.1 Metal Carbonyls
T hese are the homoleptic complexes of metal and neutral ligands, carbon monoxide. Metal carbonyls have well defined structures.
Tetracarbonyl nickel(0) is tetrahedral, pentacarbonyliron(0) is trigonal bipyramidal, while hexacarbonyl chromium(0) is octahedral. Decarbonyldimanganese(0) is made up of two square pyramidal Mn(CO) 5 units joined by a Mn –Mn bond. Octacarbonyldicobalt(0) is made up of two trigonal pyramidal Co(CO) 4 units joined by a Co–Co bond. Structures of some metal carbonyls are given in Fig. 6.22
atomic orbitals of carbon. The higher energy spz, orbital of oxygen is closer to its 2s orbital and has major contribution from it.
Tetrahedral Trigonal bi pyramidal
[Mn2(CO)10]
A c cording to molecular orbital theory, the energy difference between 2s and 2p–orbital for carbon is small. Therefore, the 2s and 2p z orbital on carbon is closer in energy to its pure 2pz mixed atomic orbitals. The higher energy spz orbital on carbon is closer energy sp z orbital of carbon is closer to its pure 2s orbital, and therefore, has major contribution from it.
Similarly, 2s and 2pz orbitals of oxygen also mix with one another to form two sp z mixed orbitals, but not to the sam e extent to which the orbital of carbon can mix because the energy difference between 2s and 2pz orbitals on oxygen is higher than that between the same
Oxygen is more electronegative than carbon, so the energies of orbitals of oxygen are lower than that of similar orbitals on carbon. The lower energy mixed spz orbital of carbon is comparable in energy with the higher energy mixed spz orbital of oxygen. These two orbitals of same symmetry and comparable energies combine linearly to form s –bonding and antibonding molecular orbitals. The lower energy spz orbital on oxygen and higher energy spx orbital on carbon are far apart A(i.e., large energy difference between them). Though these orbital have same symmetry, that do not combine and, therefore, these orbitals remain almost non-bonding rather than combining to give MOs. They are localised on oxygen and carbon, respectively.
The higher energy sp z orbital of carbon is non-bonding but it has same antibonding character. The non-bonding sp z orbital on oxygen is close in energy to 2s orbital on oxygen and, therefore, has mostly s-character of 2s orbital on oxygen.
On the other hand, non-bonding spzorbital on carbon. These non-bonding orbitals are occupied by lone pairs on carbon and oxygen and are directed away from the CO bond. The degenerate 2p x, 2p y orbitals on oxygen have p -symmetry. The orbitals on carbon combine with same orbitals on oxygen to give two dengerative p -bonding ( p ) and two dengenerative p -antibonding( p *) orbitals.
Fig. 6.22 Structure of some metal carbonyls
Fig. 6.23 Molecular orbital diagram of CO
The lower energy atomic orbitals of oxygen contribute more to the bonding molecula r orbital and the higher energy atomic orbital of carbon contribute more to antibonding molecular orbital. Therefore, in CO, the bonding molecular orbitals will have the character of orbitals of oxygen and the antibonding molecular orbitals have the character of orbitals of less electronegative carbon. This is due to the conversation of orbitals.
Synergetic Effect
The HOMO of CO ligand donates its lone pair of electrons to the empty orbital of suitable symmetry on the metal (e.g., and 2 spdzz hybrid orbital) to form a M–CO σ –bond ( Fig.6.9 )
Carbon monoxide has two LUMO π* orbitals which are also localised on carbon. These orbitals have correct symmetry to overlap with non-bonding metal d-orbitals that have π symmetry, such as the t2g(dxy,dxz or dzx) orbitals in octahedral complex. A metal atom having electrons in a d-orbital of suitable symmetry can donate electron density to the LUMO π*CO ( Fig.6.9 ). The π- interaction leads to the delocalisation of electrons from filled d-orbitals of suitable symmetry on the metal atom into the π* orbitals on the CO ligands. Because the electron density is flowing from the metal d-orbital on to the π *orbitals on ligands, the donation is known as π-back donation or back bonding and CO ligand is said to be a π acceptor (or π-acid). The s donor and π-acceptor interaction is shown in Fig.6.24
The π bond between metal and carbonyl carbon reduces the bond order of C–O bond from triple bond in CO to a double bond. This is shown by increase in C–O bond length from 1.128 Ao in CO to about 1.15 Ao in many carbonyls.
Two important physical evidences supporting the synergic bonding in nonclassical complex are bond lengths and vibrational spectra. As the strength of bond between two atoms increases, stretching frequency of the bond increases. IR Stretching frequency ∝ bond strength ∝ bond order and inversely proportional to bond length.
Factors Affecting the Magnitude of Stretching Frequency
Charge on central metal: A higher electron density on the metal will cause an increase in the extent of back bonding. This leads to an increase in electron density in the antibonding p-orbital on CO lengthening and weakening of C–O bond and decrease of the C–O stretching frequency.
Number of CO Ligands: As number of C–O liagands attached to a metal increases , the formal negative charge on metal increases. Therefore, the extent of back donation increases and hence C–O stretching frequency decreases.
Example:
Stretching frequency order
Fig.6.24 Synergic bonding in a carbonyl complex
The metal to ligand bonding creates a synergic effect which strengthens the bond between carbon monoxide and the metal. The total bonding is, thus, M = C = O.
[Ni(CO)4]> [Fe(CO)5]>[Cr(CO)6]
Presence of other ligands on Metal: The ligands which reduce the electron density on the metal will reduce the back p -bonding in metal carbonyls. As a result, M–C bond will
be longer and weaker and C–O bond will be shorter and stronger. Therefore, M–C bond stretching frequency decreases and the CO stretching frequency increases. The infrared C–O stretching frequency measurements provide the following order of p -acceptor strength of various ligands NO + > CS > CO>PF3> PCl3>P(OR)3 >PR3.
Effective atomic number (EAN): Effective atomic number (EAN) of metal carbonyls Ha[Mx (CO)y]can be calculated as
2 EAN= a xzzyxx x ++− +
x = Number of metal atoms
y = Number of ligands
z = atomic of number of central atom
a = Number of ionisable atoms in outer sphere
v = Number of valency electrons in metal and ligands
Number of metal–metal or d- bonds in metal carbonyls is (18 x–v)/2
Organometallic compounds have a good number of applications. In Mond process, for refining of nickel, impure nickel is first converted into volatile compound, [Ni(CO)4], and then decomposed to get pure nickel. A mixture of TiCl 4 and triakyl aluminium is called Ziegler–Natta catalyst and it is used for polymerisation of alkenes
14. The EAN value of CO in Ha[Co (CO)4] is 36 What is the value of 'x'?
Sol. 2 EAN= a xzzyxx x ++− + 2 1(27)2(4)(1)1 36 1 a ++− =+
36 = 35+ a a = 1
15. The bond length of C–O bond in carbon monoxide is 1.128 A°. The C–O bond length in Fe(CO)5 is.
Sol. C-O bond length of metal carbonyl > C–O bond length in CO. In metal carbonyls, there is synergic bonding between metal and carbon from CO. Due to synergic bond formation between metal and CO becomes stronger. So, bond order between CO decreases and bond length between C–O decreases. The bond order of CO decreases. So, bond length of CO is less than 1.158 Å.
Try yourself:
13. Which of the following ligands are not π acceptors?
bonding with metal in metal carbonyls but water is π donor.
Ans: NO, ,+NO 3PF are π acceptors due to synergic
TEST YOURSELF
1. In [Ni(CO)4], metal act as (1) π−donar σ−acceptor (2) σ−donar π−acceptor (3) π−donar π−acceptor (4) σ−donar σ−acceptor
2. Which of the following complexes has highest C−O bond energy?
3. The number of bridging carbonyl groups in [Co2(CO)3] and [Mn2(CO)10], respectively, are
(1) 2 and 0 (2) 2 and 4 (3) 0 and 2 (4) 2 and 2
4. Which of the following species show(s) synergic bonding?
I. [Mo(CO)6], II. [Mo(CO)6]–
III. [Ni(CN)4]4–, IV. [(PPh3)3Mo(CO)3]
(1) Only III and IV (2) Only II and III
(3) Only I and II (4) I, II, III, and IV
5. The number of bridged ligands in the complex [Mn2(CO)10] is
(1) 1 (2) 3
(3) 6 (4) zero
Answer Key
(1) 2 (2) 2 (3) 1 (4) 4
(5) 3
6.8 IMPORTANCE AND APPLICATIONS OF COORDINATION COMPOUNDS
The coordination compounds are widely present in the mineral, plant and animal worlds. They play many important functions in the area of analytical chemistry, metallurgy, biological systems, industry and medicine. These are described below.
Qualitative and Quantitative Analysis
Cupric salt gives pale blue precipitate with aqueous ammonia solution. It dissolves in excess of ammonia solution due to the formation of a deep blue water soluble complex.
This is the identification test for cupric ion.
(aq) 3422 (deep blue) (aq) 2OH [Cu(NH)(HO)] +
Mercuric chloride (HgCl2) gives a scarlet red precipitate of mercuric iodide with small amount of KI. On adding excess of KI, the precipitate formed again dissolves, forming a complex, K 2 [HgI 4 ], which is a colourless substance. Solution of this complex taken in excess of caustic potash solution is known as Nessler’s reagent. The reagent is used to detect ammonia, and hence, ammonium salts.
HgCl2+2KI HgI2 + 2KCl
HgI2+2KI K2[HgI4] (Nessler’s reagent)
The most common complexes are simple hydrates. All the transition metals form hydrated ions. These simple ions have characteristic colours in aqueous solutions. For example, [Fe(H2O)6]2+ is pale green and [Cu(H2O)4]2+ is blue.
In the Lassaigne’s test for nitrogen, Prussian blue coloured compound formed ferric ferro-cyanide, Fe4[Fe(CN)6]3, which is a complex compound. Silver chloride white precipitate is soluble in excess of ammonia due to the formation of the complex.
AgCl + 2NH3 [Ag(NH3)2]Cl
Hardness of water is estimated by simple titration with edta. The Ca 2+ and Mg 2+ ions form stable complexes with edta. The selective estim-ation of these ions can be done due to difference in the stability constants of calcium and magnesium complexes.
Photography
In black and white photography, the developed film is fixed by washing with hypo solution which dissolves the undecomposed AgBr to form a complex, Na3[Ag(S2O3)2]
AgBr +2Na2S2O3 Na3[Ag(S2O3)2] + NaBr
Metallurgy
Silver is found in the free state and as silver sulphide. Native gold is also known. Both silver and gold are extracted from the mineral with aqueous sodium or potassium cyanide as a complex [Ag(CN)2]– and [Au(CN)2]–. The silver and gold metals present in solutions are displaced from the complex in the solution by adding zinc.
2Ag 2 S+8NaCN+4O 2 → 4Na[Ag(CN) 2 ] +2 + Na2SO4
2Na [Ag(CN)2]+ Zn → Na2 [Zn(CN)4]+2+Ag
Articles can be electroplated with silver and gold much more smoothly and evenly from solutions of the complexes [Ag(CN) 2] – and [Au(CN) 2] –, than from a solution of simple metal ions.
Purification of metals can be achieved through formation and subsequent decomposition of their coordination compounds. Impure nickel is converted to Ni(CO)4, which is decomposed to yield pure nickel in Mond’s process.
Medicine
There is growing interest in the use of chelate therapy in medicinal chemistry. An example is the treatment of problems caused by the presence of metals in toxic proportions in animal systems. Thus, excess of copper and iron are removed by using the chelating ligands, D–penicillamine and desferrioxime B, via the formation of coordination compounds.
Ethylenediamine tetraacetate is used in the treatment of lead poisoning. A platinum complex, cis [PtCl2(NH3)2], known as cis platin, is used in cancer chemotherapy.
Catalysis and Bio-systems
Coordination compounds act as catalysis in different reactions. Welkinson catalyst, [(Ph 3P) 3RhCl], is used in hydrogenation of alkenes.
The green pigment responsible for photosyntheses, chlorophyll, is a coordination compound of Mg 2+. It is called magnesium phorphyrine.
Haemoglobin, the red pigment of blood which acts as the oxygen carrier, is a coordination compound of Fe 2+ .
Vitamin B 12 , cyano-cobalamine, is a coordination compound of Co 3+ . Some enzymes, like carboxypeptidase A and carbonic anhydrase, are also coordinated metal ions.
16. How is the theory of complex compounds used in photography?
Sol. The developed film in photography will be fixed using sodium thiosulphate (hypo) solution. Hypo dissolves the undecomposed AgBr to give a soluble complex compound.
2Na2S2O3+ AgBr Na3[Ag(S2O3)2] + NaBr
Try yourself:
14. What is Wilkenson's catalyst?
Ans: Wilkenson's catalyst is a complex compound with chemical composition RhCl].3P)3[(Ph This catalyst is used in hydrogenation of alkenes
TEST YOURSELF
1. Hardness of water is estimated by simple titration using (1) formate (2) acetate (3) EDTA (4) glyoxile
2. Ammonium cations can be detected using the complex (1) [Cu(NH3)4]2+ (2) [HgI4]2–(3) [Ag(CN)2]– (4) [HgI2]
3. Nickel is purified using the concept of complex compounds. The complex related is (1) Ni(CO)6 (2) [Ni(NH3)4]2+ (3) Ni(CO)4 (4) [Ni(NH3)6]2+
4. The deep blue complex produced by adding excess of ammonia to CuSO 4 solution is (1) [Cu(NH3)2]2+ (2) [Cu(NH3)4]2+ (3) [Cu(NH3)6]2+ (4) [Cu(NH3)4]+
5. Ziegler–Natta catalyst is (1) solution of SnCl4 + trialkylaluminium (2) solution of TiCl4 + trialkylaluminium (3) solution of TiCl4 + trialkylchromium (4) solution of SnCl4 + Tollen’s reagent
6. The number of moles of KI required to prepare one mole of K2[HgI4] is (1) 4 (2) 3 (3) 2 (4) 1
Answer Key
(1) 3 (2) 2 (3) 3 (4) 2 (5) 2 (6) 1
CHAPTER REVIEW
Werner’s Theory of Coordination Compounds
■ Addition compounds are formed when two or more independent stable compounds combine either physically or chemically.
■ A double salt gives test for all the constituent an ions present in it. Carnallite, alum, etc. are examples of double salts.
■ A complex species is either a molecule or ion whose physical properties, such as colour and conductivity, are distinctly different from those of the substances from which it is formed.
■ A complex compound retains its identily and does not exhibit the properties of constituent elements.
■ Werner proposed that transition metals possess two types of valencies: primary (1°) valency and secondary (2°) valency.
■ Primary valency indicates oxidation number. It is ionisable valency and is satisfied by only negative ions.
■ Secondary valency is non-ionisable valency and determines the shape of complex. It is satisfied by positive ions, negative ions and neutral molecules.
■ Coordination number denotes 2 ° valency.
■ Shape of complex can be assumed based on the coordination number.
■ Werner prepared cobalt amine complexes CoCl3. 6NH3, CoCl3.4NH3 and CoCl3.3NH3.
■ One mole of CoCl 3.6NH 3 produces four mol es of ions in solution with excess AgNO3 solution, and three moles of AgCl are precipitated. CoCl 3 .3NH 3 does not ionise in solution and gives no precipitate.
■ Werner’s theory does not correlate electronic configuration of the central metal with the formation of the complex compounds.
■ Werner’s theory fails to explain colour and magnetic and optical properties of complexes.
■ In the formation of a complex ion, the central metal ion acts as a Lewis acid whereas the ligand acts as Lewis base.
■ The formation of coordination complex is a Lewis neutralisation.
Defination of Some Important Terms Pertaining to Coordination Compounds
■ Ligand is an ion or molecule that surrounds the metal ion in a condition complex. Ligand is an electron pair donor.
■ Based on the charge, ligands are of 3 types: neutral ligands, like NH3, H2O, CO, anionic ligands, like CN–, Cl–, NO2–, and cationic ligands like NO+
■ Based on the number of electron pairs donated, ligands are of three types: monodeutate ligands, like NH 3, H2O, Cl–, CN–, bidentate ligands like –OOC–COO–, H 2 NCH 2 COO – and polydentate ligands like ethylenediamine tetracetate (EDTA)4-.
■ Number of coordinate covalent bonds that a metal ion forms with the ligands in a complex compound is called coordination number.
■ Some ligands contain two or more atoms with unshared pairs of electrons. If they coordinate to the same central metal or ion to form a ring, the phenomenon is known as chelation.
■ Transition metal ions form complex compounds due to small size, high effective nuclear charge of the metal ion and presence of vacant d-orbitals.
■ The central metal ion, together with the ligands, is known as complex ion or coordination sphere.
■ Coordination entities are mainly of two types; mononuclear compounds and polynuclear compounds.
■ The coordination sphere of a complex compound carrying negative charge is an anionic complex, e.g., K 4 [Fe(CN) 6 ], K3[Fe(CN)6], K2[Ni(CN)4].
■ The coordination sphere of a complex compound carrying positive charge is cationic complex, e.g., [Ag(NH 3 ) 2 ]Cl, [Cu(NH3)4]SO4.
■ The coordination sphere of a complex compound which does not carry any charge is neutral complex, e.g., [Ni(CO)4], [Co(NH3)3Cl3].
Nomenclature of Coordination Compounds
■ All mononuclear complex entities contain a single central metal atom, which is symbolised first in IUPAC notation of writing formula.
■ If more than one ligand of each type is present, alphabetical order according to the first letter is to be followed in listing them in IUPAC.
■ Formulae of polyatomic ligands are enclosed in parenthesis. Complete coordination entity is enclosed in square brackets.
■ Different species present in the formula are written continuously without leaving space.
■ Charge of anionic or cationic complex is indicated outside the square brackets as a right superscript.
■ Neutral coordination entity is written as one word for naming the complexes as per IUPAC. Cationic part is named first, and then an anionic part. Number of cations or anions is not indicated.
■ Within a coordination entity, ligands are named in alphabetical order and then name of the central metal atom (or ion) follows.
■ The oxidation state of the metal is represented in Roman numerical after the name in parenthesis.
■ The names of cationic ligands and neutral ligands do not change, except aqua for H2O, ammine for NH3, carbonyl for CO, and nitrosyl for NO. The names of anionic ligands end with – ‘O’.
■ Prefixes like di–, tri–, tetra– etc, are used to name simple ligands. bis–, tris–, tetrakis–, etc. are used for ligands whose names have di–, tri– etc, already included. In anionic entities, the name of central metal ion terminates with –ate.
■ The atom of the ligand which is bonded to the metal is indicated by its chemical symbol.
■ If the complex compounds possesses water or solvent of crystallisation, they follows the name of the compound and their numbers are represented by arabic numericals before their names.
Isomerism in Coordination Compounds
■ Two or more complexes with identical composition but different properties are called isomers.
■ Structural isomerism arises due to different kinds of bonds. Ionisation, hydrate, ligand and coordination isomerism are examples of structural isomerism.
■ Ionisation isomerism: Isomers differ in type of ions obtained on dissociation in aqueous solutions. [PtCl2(NH3)4]Br2 gives Br– ions and [PtBr2(NH3)4]Cl2 gives Cl– ions.
■ Hydrate isomerism is a special type of ionisation isomerism, which arises due to the presence of different numbers of water molecules in and outside the coordination sphere.
■ Ligand isomerism: This occurs due to different isomers of the ligand, e.g. [Co(H2NCH2CH2CH2NH2)3]3+ and [Co(CH3CH(NH2) CH2NH2)3]3+.
■ Stereoisomerism is due to difference in spacial orientation.
■ Geometrical isomerism is also known as cistrans isomerism. It is shown by complexes in which the coordination entities have the general formula, [Ma2b2] and [Ma2b4] or [Ma2(bb)2] where M is the central metal atom/ion land a and b are (ligands).
■ In cis-isomer, the same ligands are on the same side, whereas in trans-isomers, they are on the opposite side.
■ When there are three ligands of each type [Ma 3b 3], a different type of geometrical isomerism occurs in octahedral coordination entities. It is known as facial (fac) meridional (mer) isomerism.
■ Cis-trans isomerism is observed only in some square planar and octahedral complexes.
■ Tetrahedral complexes do not show geometrical isomerism.
■ Square planar complexes Ma 4, Ma 3b, do not show geometrical isomerism.
■ Octahedral complexes Ma6, Mab5, do not show geometrical isomerism.
■ Optical isomerism is due to the difference in the rotation of plane of polarised light in a polarimeter.
■ Dextro (d- or +) isomer rotates the light to right of the original plane and laevo (l– or –) isomer rotates to left.
■ Optical isomerism is generally shown by octahedral complexes containing polydentate ligands. They must have the property of chirality.
■ A 1:1 equilibrium mixture of d– and l–isomers gives a recemic mixture, with a net optical rotation of zero.
Bonding in Coordination Compound
■ Pauling’s valence bond theory provided useful understanding of the valence and structure of compounds.
■ Complexes with unpaired electrons are paramagnetic and complexes with no unpaired electons are diamagnetic
■ The hybridisation of central metal atom depends upon the co-ordination number.
■ Tetra coordinated complexes with sp 3 hybridisation are tetahedral and those with dsp2 are square planar.
■ Hexacoordinated complexes are octahedral. The hybridisation of central atom is sp 3d2 or d2sp3.
■ With dsp2 and d2sp3 hybridisation, inner complexes are formed. Inner complex is also called spin paired, low spin, strong field, or covalent complex.
■ With sp 3 and sp 3d 2 hybridisation, outer complexes are formed. Outer complex is also called sping free, high spin, weak field, low field, or ionic complexes.
■ The orbitals in the valence shell of central atom, such as (n-1)d, ns, np, nd orbitals, undergo hybridisation to form identical hybrid orbitals.
■ These empty hybrid orbitals of metal overlap with completely filled orbitals of ligands to form dative bonds with ligands.
■ [Co(NH 3 ) 6 ] 3+ has d 2 sp 3 hybridisation. Oxidation state of Co is +3. It is diamagnetic, inner orbital complex and has octahedral shape.
■ [NiCl4]2– has sp3 hybridisation. Oxidation state of Ni is +2. It is tetrahedral, paramagnetic, and high spin complex.
■ [Ni(CN) 4 ] 2– has dsp 2 hybridisation. Oxidation state of Ni is +2. It is square planar, diamagnetic, and low spin complex.
■ Valency bond theory can explain the structure and magnetic behaviour of coordination compounds.
■ Effective atomic number (EAN) concept was introduced by Sidgwick to explain the stability of complexes.
■ EAN is the sum of number of electrons of the metal ion and the number of electrons gained from ligands.
■ When EAN of central metal cation is equal to the atomic number of the nearest inert gas, the complex is expected to possess greater stability.
Stability of Coordination Compound
■ The equilibrium constant for the formation of a complex compound is the stability constant.
■ Stability of the complex depends on shape of complex, size and charge of the central atom, nature of the ligand, and ring size of the ligand
Bonding in Metal Carbonyls
■ The metal–carbon bond in metal carbonyls possesses both s and p character.
■ The ligand to metal is s bond and metal to ligand is p bond. This unique synergic bonding provides stability to metal carbonyls.
Importance and Applications of Coordination Compounds
■ Optical isomers are called enantiomorphs or enantiomers. They are non-superimposable m irror images and almost identical in physical and chemical properties.
■ Complex formation is an important feature in qualitative analysis. Cu2+ is identified by deep blue complex, [Cu(NH 3)4(H2O)2]2+.
■ Nessler’s reagent is used for detection of NH3 or NH4+ salts.
■ Nessler’s reagent is a solution of K 2[HgI4] complex in excess of KOH. Complex K2[HgI4] is colourless.
■ Depending upon the characteristic colour of hydrate complexes, respective metal ions can be detected. [Fe(H2O)6]2+ is pale green, [Cu( H2O)4]2+ is blue and [Ti(H2O)6]3+ is purple.
■ In photography, the unreacted AgX on the film is dissolved in dilute hypo solution due to the formation of the complex, Na3[Ag(S2O3)2].
■ Metal complexes release metal slowly and, thus, give a uniform coating in electroplating of the metal.
■ In electroplating of silver, K[Ag(CN) 2] is used because it provides Ag + slowly and, thus, gives a uniform coating.
■ Ag and Au are extracted by the use of complex formation. The metal Ag or Au is displaced from the complex in the solution by using Zn.
■ Chlorophyll is the green pigment present in the leaves of plants. Chlorophyll has a tetrapyrrole ring structure with Mg 2+ present in the centre.
■ Haemoglobin is the red pigment in the red-blood cells (erythrocytes). Fe2+ has 6 coordinate linkages.
■ The problem of ‘CO’ pollution is the formation of carboxy haemoglobin complex. The bonding of Fe2+ with CO is very strong and it is difficult to dissociate. CO prevents transport of O 2 and, thus, CO is toxic.
Exercises
JEE MAIN LEVEL
Level - I
Some important terms
Single Option Correct MCQs
1. Which of the following ligands is not chelating?
(1) EDTA
(2) en (3) oxolate
(4) pyridine
2. The donor atoms in EDTA are (1) two N and two O
(2) two N and four O
(3) four N and two O
(4) three N and three O
3. Homoleptic complex among the following (1) K[Co(C2O4)2Cl2]
(2) [Ni(CO)4]
(3) [Co(NH3)3(NO2)3]
(4) [Co(NH3)5Cl]
4. The formula of ‘nitrosyl’ group
(1) NO
(2) NO+
(3) NO–
(4) ONO
5. Coordination number of Ni in [Ni(C2O4)3]4− is
(1) 3 (2) 6 (3) 4 (4) 5
6. According to Lewis theory, the ligands are (1) acidic in nature
(2) basic in nature
(3) neither acidic nor basic
(4) some are acidic and others are basic
7. The co ordination number of a metal in coordination compounds is
(1) same as primary valency
(2) sum of primary and secondary valencies
(3) same as secondary valency
(4) none of these
8. Ligands in a complex salt are:
(1) ions or molecules linked by coordinate bonds to a central metal atom or ion.
(2) molecules linked by coordinate bonds to a central metal atom or ion.
(3) cations linked by coordinate bonds to a central metal atom or ion.
(4) anions linked by coordinate bonds to a central metal atom or ion.
9. The value of x on the [Ni(CN) 4]x is:
(1) +2 (2) 0
(3) -2 (4) +4
10. ‘Metal-Isothiocyanato’ is indicated by its chemical symbol as
(1) M-NCS
(2) M-SCN
(3) M-CNS
(4) M-CSN
11. In which of the following complexes the metal is in zero oxidation state?
(1) [Cu(NH3)4]Cl2
(2) [Cu(NH3)4]Cl2
(3) [Mn2(CO)10]
(4) [Ag(NH3)2]Cl
12 .Which of the following is a π- acceptor ligand?
(1) NH3
(2) CN–
(3) H2O
(4) NO2
13. Which of the following represents chelating ligand?
(1) Cl (2) 2 CO24 (3) OH– (4) H2O
Numerical Value Questions
14. The total number of coordination sites in ethylenediamine tetraacetate (EDTA) 4– is _________.
15. Number of ambidentate ligands among the following is ________. NO2–,SCN–,C2O42–,NH3,CN–,SO42–,H2O
16. The oxidation state of ‘Fe’ in [Fe(CN)6]3− is _____.
17. The denticity of the ligand N(CH2CH2NH2)3 is ____.
18. How many EDTA molecules are required to make an octahedral complex with a Co2+ ion?
Formula for writing and Nomenclature of coordination compounds
Single Option Correct MCQs
19. The IUPAC name of K3[Co(C2O4)3] is:
(1) Potassium trioxalatocobaltate (III) (2) Potassium trioxalatocobaltate (II)
(3) Potassium tris (oxalate) cobaltate (III) (4) Potassium trioxalatocobalt (III)
20. The correct IUPAC name of K 2MnO4 is (1) Potassium tetraoxopermanganate (VI)
49. According to CFT, the bond between metal and ligand is _______.
(1) covalent
(2) ionic
(3) dative
(4) metallic
50. The tetrahedral crystal field splitting is only ______ of the octahedral splitting.
(1) 1 2 (2) 2 9
(3) 4 9 (4) 5 9
51. What is magnetic moment of [FeF 6]3– ?
(1) 6.92 BM (2) 5.92 BM
(3) 7.62 BM (4) 3.14 BM
52. Among the ligands NH 3, en,CN and CO the correct order of their increasing field strength, is:
(1) 3 NHenCNCO <<<
(2) 3 CNNHCOen <<<
(3) 3 enCNNHCO <<<
(4) 3 CONHenCN <<<
53. Which one of the following is a paramagnetic, inner orbital octahedral complex.
(1) ()6 4 FeCN
(2) () 3 3 6 CoNH +
(3) ()6 3 FeCN
(4) 3 CoF6
54. In an octahedral complex with Δ 0 < P the correct electronic configuration of d 6 according to CFT is
(1) 42 etg2g
(2) 33 te2gg
(3) 60 te2gg
(4) 42 te2gg
55. The Crystal field stabilisation energy for high spin d4 octahedral complex is
(1) −0.6Δ0 (2) −1.8Δ0 (3) −1.6Δ0 (4) −1.2Δ0
56. Which of the following systems in an octahedral complex has maximum unpaired electrons?
(1) 9 d(high spin) (2) 6 d(low spin)
(3) 4 d(low spin) (4) 7 d(high spin)
57. According to CFT for the same metal, same ligands and metal-ligand distances the correct relationship is
(1) 0t 4 9 ∆=∆
(2) t0 4 ∆=∆
(3) t0 4 9 ∆=∆
(4) t0 9 4 ∆=∆
58. d2sp3 hybridization is not present in
(1) () 3 3 6 CoNH +
(2) () 3 3 6 CrNH +
(3) () -4 6 FeCN
(4) 3– CoF6
59. Crystal field stabilization energy for the complex, K3[FeF6] is
(1) Zero
(2) o -2 ∆
(3) o -2.4 ∆
(4) o -1.2 ∆
60. The spin only magnetic momentum of complex ion [NiCl4]2 is (1) 1.732 BM
(2) 2.82 BM
(3) 0
(4) 3.87 BM
Numerical Value Questions
61. Spin only magnetic moment of 4 MnBr6 is _____ BM. (round off to close integer)
62. The d-electronic configuration of [CoCl4]2−in tetrahedral crystal field is eg m t 2g n. Sum of m and n is ____.
63. Magnetic moment [Ag(CN)2] is zero, the no. of unpaired electrons is ______.
64. One mole of the metal complex with the formula [Co(en)2Cl3] gives one mole of silver chloride on treatment with an excess silver nitrate. The secondary valency of Co in the complex is ________________ .
65. The total number of unpaired electrons present in the complex () 3 3 KCrox is ____.
66. Number of t 2g Electrons present in the complex ion () 3 24 3 CoCO is “X” and in the complex ion () 2 2 6 MnHO + is “Y”. Find the product of XY?
Stability of Coordination Compounds in solution
Single Option Correct MCQs
67. Most stable ion is (1) [Fe(H2O)6]+3 (2) [Fe(CN)6]-3 (3) [FeCl6]-3 (4) [Fe(C2O4)3]-2
68. A metal ‘M’ forms four tetrahedral homoleptic complexes. The complex with maximum stability is the one with the overall instability constant value of (1) 2 × 10–5 (2) 40 × 10–6 (3) 0.5 × 10–3 (4) 0.03 × 10–4
Numerical Value Questions
69. The product of stability ane instability constant of a given complex is “Y”. Find the value of the (99+Y)?
Organometallic compounds
Single Option Correct MCQs
70. In Fe(CO)5, the Fe – C bond possesses (1) π -character only
(2) Both σ - and π -character (3) Ionic character (4) σ -character only
71. Hybridization of Nickel in [Ni(CO) 4] is (1) sp3 (2) dsp2 (3) sp2d (4) dsp3
72. No of bridged “CO” groups in Fe 2(CO)9: (1) 2 (2) 3 (3) 4 (4) 1
Numerical Value Questions
73. Number of bridging CO ligands in [Mn2(CO)10] is _____________
74. How many electrons are required to make V(CO)6 an 18-electron species?
75. What is the value of 'n' in (CO)n-Co-Co(CO)n?
Importance and applications of coordination compounds
Single Option Correct MCQs
76. Which of the following vitamin contains cobalt?
77. Which is used in cancer – chemotherapy? (1) cis-platin
(2) Zeise’s salt
(3) Both 1 and 2
(4) None
78. Ziegler – Natta catalyst is (1) ()KPtClCH324
(2) () 3 3 PhPRhCl
(3) ()254 3 AlCHTiCl +
(4) () 55 2 FeCH
79. The polydentate ligand used in the treatment of lead poisoning is _______.
(1) EDTA
(2) Glycinate
(3) Oxalate
(4) Ethylene diammine
80. A reagent used for identifying nickel ion is:
(1) potassium ferrocyanide
(2) dimethylglyoxime
(3) phenolphthalein
(4) EDTA
Numerical Value Questions
81. When Ni+2 reacts with DMG, it forms a cherry red complex. Number of π-bonds, dative bonds present are 'a' and 'b' respectively. Find the sum of “a” and “b”?
Importance and applications of organometallics
Single Option Correct MCQs
82. Which of the following metal carbonyl exists in the liquid state?
(1) Iron carbonyl
(2) Chromium carbonyl
(3) Vanadium carbonyl
(4) Manganese carbonyl
83. The σ-bond formed between metal and carbon of Carbon monoxide in metal carbonyl is due to overlapping of
(1) Empty π 2p orbital of the CO with an empty p- orbital of the metal
(2) Empty * 2pπ orbital of the CO with a filled d - orbital of the metal
(3) Filled π 2p orbital of the CO with an empty d- orbital of the metal
(4) Filled * 2pπ orbital of the CO with an empty d- orbital of the metal
Numerical Value Questions
84. The number of CO ligands that can be bonded to a Fe atom is ________, if the complex follows EAN rule.
Level
- II
Introduction-Werner's Theory
Single Option Correct MCQs
1. Select the correct statement for double salt.
(1) Double salts are stable in solid state but lose their identity in aqueous solution.
(2) In double salt, the properties of constituent ions are not changed in their aqueous solution.
(3) Double salts are stable in solid state and do not lose their identity in aqueous solution.
(4) Both A and B statements are true.
2. Consider the following observations made with aqueous solutions.
Formula Mole of AgCl precipitated per mole of the compounds with excess AgNO3
(I) PdCl2.4H2O 2
(II) NiCl2.6H2O 2
(III) PtCl4.2HCl 0
(IV) CoCl3.4NH3 1
(V) PtCl2.2NH3 0
The secondary valencies of I and IV is
(1) 4 and 5, respectively
(2) 4 and 6, respectively
(3) 6 and 4, respectively
(4) 2 and 4, respectively
3. The molecular formulas of various hexacordinates are given below.
(a) CrCl3.6NH3
(b) CrCl3.5NH3
(c) CrCl3.4NH3
If the numbers of NH3 ligands attached to the central metal ion are 6, 5, and 4 respectively. Then, the number of Cl– ions showing dual character in (a), (b), and (c), respectively, are
(1) 3, 3, 3 (2) 0, 1, 2
(3) 3, 2, 1 (4) 6, 5, 4
4. If two Cl– ions are present in a complex X and are precipitated by AgNO3 to form new complex PtCl4.4NH3, then the formula of the complex X is _______.
(1) [Pt(NH3)4]Cl4
(2) [Pt(NH3)4Cl2]Cl2
(3) [Pt(NH3)4Cl3]Cl
(4) [Pt(NH3)3Cl3]Cl NH3
5. Primary and secondary valency of central metal ion in iron(III) hexacyanoferrate(II) [Fe4[Fe(CN)6]3] are ___ and ___, respectively (1) 3, 2 (2) 2, 3 (3) 2, 6 (4) 3, 6
6. The species that satisfies primary and secondary valency of central metal ion in K4[Fe(CN)6] are ___ and ___, respectively.
(1) 2CN– and 6CN– (2) 3CN– and 6CN–(3) 4K+ and 6CN– (4) 2K+ and 6CN–
7. Effective atomic number of the complex [Co(NH3)6]Cl3 is ______.
(1) 24 (2) 27
(3) 35 (4) 36
8. Which of the following does not obey EAN rule?
(1) [Co(C2O4)3]3–
(2) [V(CO)6]–
(3) [Ti( s – C6H5)2 ( p – C5H5)2]
(4) [Fe( p – C5H5)2]
9. If Co2(CO)x follows Sidwick’s rule of EAN, then the correct formula is ______.
(1) Co2(CO)4 (2) Co2(CO)3
(3) Co2(CO)8 (4) Co2(CO)10
Numerical Value Questions
10. When one mole of [Co(NH3)3Cl3] is added to excess of AgNO3 solution, the weight of AgCl precipitate is ____ g.
11. The secondary valency of chromium in [Cr(en)3]Cl3 is _____.
Definition of Some Important Terms
Single Option Correct MCQs
12. The donar atoms of hexadentate ligand EDTA are
(1) four nitrogen atoms and two oxygen atoms
(2) two nitrogen atoms and four oxygen atoms
(3) six nitrogen atoms
(4) three nitrogen atoms and three oxygen atoms
13. Which of the following is homoleptic complex?
(1) [Co(NH3)6]Cl3
(2) [Co (NH3)4Cl2]Cl
(3) [Co (NH3)3Cl3]
(4) [Co (NH3)5Cl]Cl2
14. The co-ordination number and the oxidation state of the element ‘M’ in the complex [M(en)2(C2O4)]NO2 (where (en) is ethan-1, 2-diamine) are, respectively,
(1) 6 and 3 (2) 4 and 3
(3) 6 and 2 (4) 4 and 2
15. Which of the following pairs of ligands are chelating ligands.
(1) 1,2 – ethanediammine, glycinate , EDTA
(2) cyanido, nitrito, thiocyanato
(3) methylamine, nitrosyl, aqua (4) 1–propanamine, chlorido, sulphato
Numerical Value Questions
16. Ag+ salt solution excessKCNsolution
complex ion 'X' + other products
Cu2+ salt solution excessKCNsolution
complex ion 'Y' + other products
excessKCNsolution
Ni2+ salt solution
complex ion 'Z' + other products
Then, the sum of the coordination nulmbers of metal ions present in X, Y, and Z is ___.
17. Maximum number of dative bonds that can be formed by EDTA are ____.
18. The oxidation number of the metal in Zeise’s salt, K[PtCl3(C2H4)] is ____.
Nomenclature of Coordination Compounds
Single Option Correct MCQs
19. The formula of the complex tris (ethylenediamine) cobalt (III) sulphate is
(1) [Cr(en)3]Br3 (2) [Co(en)3SO4]
(3) [Co(en)3]2SO4 (4) [Co(en)3]2(SO4)3
20. Complex X of compositionCr(HO)6Cln has a spin only magnetic moment of 3.83 BM. It reacts with AgNO3 and shows geometrical isomerism. The IUPAC nomenclature of X is
35. Wrong match among the following is (1) [Mn(CN)6]3−…..diamagnetic (2) [Mn(Cl)6]3−….. paramagnetic (3) [Ni(CN)4]2−….. diamagnetic (4) [Ni(Cl)4]2−….. paramagnetic
36. The complex ion [Cu(NH3)4]+2 is (1) tetrahedral and paramagnetic (2) tetrahedral and diamagnetic
(3) square planar and paramagnetic
(4) square planar and diamagnetic
37. Incorrect statement among the following is (1) [CoF 6 ] 3– is paramagnetic with four unpaired electrons
(2) [Co(C2O4)3]3– is diamagnetic complex
(3) [Fe(CN) 6] 3– is paramagnetic complex with one unpaired electron (4) [Mn(CN)6]3– is diamagnetic complex
38. The colour of the coordination compounds depends on the crystal field splitting. What will be correct order of absorption of wavelength of light in the visible region, for the complexes [Co(NH3)6]3+, [Co(CN)6]3–, and [Co(H2O)6]3+?
(1) [Co(CN)6]3– > [Co(NH3)6]3+ > [Co(H2O)6]3+
(2) [Co(NH3)6]3+ > [Co(H2O)6]3+ > [Co(CN)6]3–
(3) [Co(H2O)6]3+ > [Co(NH3)6]3+ > [Co(CN)6]3–
(4) [Co(NH3)6]3+ > [Co(CN)6]3– > [Co(H2O)6]3+
39. Identify the correct increasing order of crystal field splitting strength of the given ligands.
(1) NH3 > Cl– > CN– > F– > CO > H2O
(2) F– > Cl– >NH3 > CN– > H2O > CO
(3) Cl– > F– > H2O > NH3 > CN– > CO
(4) CO– > CN– > NH3 > H2O > F– > Cl–
40. Identify the correct decreasing order of wavelength of light absorbed by the following complexes.
(1) III > I > IV > II
(2) II > IV > I > III
(3) III > I > II > IV
(4) IV > II > III > I
41. I....[Co(NH 3 ) 6 ] +3 , II.....[Co(CN) 6 ] −3 , III.... [Co(NO2)6]−3. The correct decreasing order of wavelength of light absorbed is
(1) II > I > III (2) III > I > II
(3) I > III > II (4) III > II > I
42. What is the correct electronic configuration of the central atom in [Ni(NH3)6]2+, based on crystal field theory?
(1) e4 g t4 2g (2) t62ge4g (3) e4 g t4 2g (4) t62ge2g
43. The electronic spectrum of [Ti(H 2 O) 6 ] 3+ shows a single broad peak with a maximum at 20,300 cm−1. The crystal field stabilisation energy (CFSE) of the complex ion, in kJ mol−1, is (1 kJ mol−1 = 83.7 cm−1)
(1) 83.7 (2) 242.5 (3) 145.5 (4) 97
44. Which complex is most stable?(K d is dissociation constant)
(1) [Cu(CN)2]– Kd = 1 10–16
(2) [Fe(CN)6]4– Kd = 1 10–37
(3) [Fe(CN)6]3– Kd = 1 10–44
(4) [Ag(CN)2]– Kd = 1 10–20
45. Which of the following complexes formed by Cu2+ ions is most stable?
(1) Cu2+ + 4NH3 → [Cu(NH3)4]2+, log k = 11.6
(2) Cu2+ + 4CN– → [Cu(CN)4]2–, log k = 27.3
(3) Cu2+ + 2en → [Cu(en)2]2+, log k = 15.4
(4) Cu2+ + 4H2O → [Cu(H2O)4]2+, log k = 8.9
46. From the stability constants (hypothetical values) given below, predict which is the strongest ligand.
(1)
Then, the formation constant of [Ag(NH3)2]+ from Ag+ and NH3 is _____.
(1) 1.7 ×10–3
(2) 5.92 ×10–6
(3) 1.8 × 103 (4) 1.7 ×107
Numerical Value Questions
50. Spin only magnetic moment of [MnBr 6]4− is____BM. [Round off to the closest integer]
51. Based on VBT number of unpaired electrons present in [MnCl4]–2 is p and oxidation state of central metal atom in the same complex is q Find sum of p and q?
52. Considering that Δ o > P, the magnetic moment (in BM) of [Ru(H2O)6]2+ would be ____.
53. In potassium ferrocyanide, there are ____. pairs of elections in the g 2 t set of orbital
54. The d-electronic configuration of [Co(H2O)6]2+ in tetrahedral crystal field is e g mt 2g n. Sum of m and n is ____.
55. The ratio of crystal field stabilisation energies of [V(H2O)6]+2/[Mn(H2O)6]+3 is _____.
Bonding in Metal Carbonyls
Single Option Correct MCQs
56. The longest CO bond length will be with (1) [Mn(CO)6]+ (2) [V(CO)6]–(3) Cr(CO)6 (4) [Ti(CO)6]2–
57. Metal ‘M’ forms a carbonyl compound, in which it is present in its lower valance state. Which of the following bonding is possible for the metal carbonyl?
(1)
58. Among the following metal carbonyls, the C−O bond order is lowest in
(1) [Mn(CO)6]+ (2) [V(CO)6]–
(3) [Cr(CO)6] (4) [Fe(CO)5]
59. Which of the following species can act as a reducing agent?
(1) [Co(CO)4]+ (2) [Mn(CO)6]
(3) [Mn(CO)5] (4) [Cr(CO)6]
60. The number of bridging carbonyl groups in [Co2(CO)3] and [Mn2(CO)10], respectively, are
(1) 2 and 0 (2) 2 and 4
(3) 0 and 2 (4) 2 and 2
Numerical Value Questions
61. In the complex Fe(CO) x, the value of x is ______.
62. Number of bridging coligands in [Mn2(CO)10] is ____.
63. If NO reacts with [Cr(CO)6], how many NO are required to replace maximum number of CO groups?
Importance and Applications of Coordination Compounds
Single Option Correct MCQs
64. Which of the following statements is incorrect regarding the importance of coordination compounds in biological systems?
(1) Vitamin B12 used to prevent anaemia is a complex compound of zinc.
(2) Haemoglobin is the red pigment of blood and contains Fe2+
(3) Chlorophyll is the green pigment of plants and contains Mg2+
(4) Vitamin B-12 contains Co 3+ .
65. Choose the correct statements among the following.
A) Na2EDTA is used to remove Ca +2, and Mg+2 ions from hard water.
B) The oxidation number of Ni in the complex formed in the refining of Ni by Mond’s process is +2.
C) The oxidation state of Rh in Wilkinson catalyst is +1.
D) D-pencillamine and desferrioxime-B are used to remove excess of lead.
E) cis-platin is used for the treatment of cancer.
(1) B, C, and D only (2) A, C, and E only
(3) A, B, C, D, and E (4) C, D, and E only
66. The complex solutions of gold and silver are used in electroplating process. The coordination number of Au and Ag in these complexes, respectively, are (1) 4, 4 (2) 4, 2 (3)2, 2 (4) 6, 4
Numerical Value Questions
67. How many π bonds are present in ferrocene?
68. Sum of the oxidation numbers of the metals present in the catalyst that is used in the manufacturing of high density polythene is _________.
69. The oxidation state and coordination number of metal in Wilkinson catalyst are +x and y, respectively. (x + y) is __________.
Multiple Concept Questions
Single Option Correct MCQs
70. Nickel and palladium both form complexes of the general formula [M(PR 3 ) 2 Cl 2 ].
The ligand PR 3 is a phosphine such as triphenyl phosphine P(C 6 H 5 ) 3 . It is a Lewis base. Among these two compounds one is paramagnetic but another one is diamagnetic. From this, it can be concluded that
(1) both complexes are tetrahedral and do not exhibit geometrical isomerism
(2) both complexes are square planar and can exhibit geometrical isomerism.
(3) nickel complex is tetrahedral and does not exhibit geometrical isomerism, while palladium complex is square planar and can exhibit geometrical isomerism
(4) nickel complex is square planar and can exhibit geometrical isomerism, while palladium complex is tetrahedral and cannot exhibit geometrical isomerism
71. Which of the following statements are correct?
I) In octahedral complexes, 222 zxy d nd d a orbitals have higher energy than dxy, dyz, and d zx orbitals.
II) In tetrahedral complexes, d xy , dyz, and d zx orbitals have higher energy than 222 zxy dandd orbitals.
III) The colours of complexes are due to electronic transitions from ore sets of d-orbitals to another set of orbitals.
IV) Δtetrahedraltetaahedraloctahedral 9 4 ∆=∆
(1) I, II, and III (2) I and IV
(3) III and IV (4) II, III, and IV
72. According to IUPAC nomenclature, sodium nitroprusside Na2[Fe(CN)5NO] is named (1) sodium pentacyanonitrosyl ferrate(II)
I) Two isomers are produced if the reactant complex ion is a cis-isomer.
II) Two isomers are produced if the reactant complex ion is a trans-isomer.
III) Only one isomer is produced if the reactant complex ion is a trans-isomer.
IV) Only one isomer is produced if the reactant complex ion is a cis - isomer.
The correct statements are
(1) (II) and (IV) (2) (I) and (II) (3) (I) and (III) (4) (III) and (IV)
76. Stable complex based on EAN rule are i. K4[Fe(CN)6] ii. [Co(NH3)5Cl]Cl2
iii. [Ni(CO)4] iv. K2[Ni(CN)4]
(1) i only
(2) i and ii only
(3) i, ii, and iii only
(4) i, ii, and iv only
77. Octahedral complexes of copper (II) undergo structural distortion (Jahn–Teller). Which one of the given copper (II) complexes will show the maximum structural distortion? (en–ethylenediamine; H2N−CH2−CH2−NH2
(1) [Cu(H2O)6]SO4
(2) [Cu(en)(H2O)4]SO4
(3) cis–[Cu(en)2Cl2]
(4) trans–[Cu(en)2Cl2]
78. The octahedral complex of a metal ion M 3+ with four monodentate ligands L 1, L 2, L 3, and L 4 absorb wavelengths in the regions of red, green-yellow, and blue, respectively. The increasing order of ligand strength of the four ligands is
(1) L 4 < L 3 < L2 < L1
(2) L1 < L 3 < L2 < L 4
(3) L 3 < L2 < L 4 < L1
(4) L1 < L2 < L 4 < L 3
79. Total number of cis N−Mn−Cl bond angles (that is, Mn−N and Mn−Cl bonds in cis position) present in a molecule of cis[Mn(en)2Cl2] complex is x, and total number of trans N−Mn−Cl in the same complex is y. Then x–y is _____.
(1) 2 (2) 3
(3) 4 (4) 1
80. Consider the following reaction and statements: [Co(NH 3 ) 4 Br 2 ] + + Br − → [Co(NH3)3Br3] + NH3
I. Two isomers are produced if the reactant complex ion is a cis-isomer.
II. Two isomers are produced if the reactant complex ion is a trans-isomer.
III. Only one isomer is produced if the reactant complex ion is a trans-isomer.
IV. Only one isomer is produced if the reactant complex ion is a cis-isomer.
The correct statements are (1) (III) and (IV) (2) (II) and (IV) (3) (I) and (II) (4) (I) and (III)
Numerical Value Questions
81. In Mond’s process for refining of nickel, it is separated as a volatile complex on heating with carbon monoxide. If oxidation number, co‐ordination number, and the number of unpaired electrons with nickel in the complex are P, Q, and R, respectively, the value of (P+Q+R) is ______.
82. Tetraamminecopper(II) sulphate solution, on acidification with acetic acid and then adding potassium hexacyanoferrate(II) solution, gives a chocolate brown precipitate ‘X’. If oxidation state of metal present in the coordination sphere of ‘X’ is + a and oxidation state of metal present in the outside of the coordination sphere of ‘X’ is +b, then (a+b) is ___.
83. Empirical formula of a complex is CrH12O6Cl3. Consider that all the hydrogens and oxygen atoms are in the form of water molecules. When it is treated with conc. H 2 SO 4 , it loses 13.5% of original mass. Then, find the number of water molecules in the coordination sphere of the complex. (At. wt. Cr = 52; H = 1; Cl = 35.5; O = 16)
84. Na 2 S + Na 2 [Fe(CN) 5 NO] → X (violet complex), if i) number of ligands present in X = a ii) number of d-orbitals of central metal involved in the formation of X = b iii) number of linkage isomers of X = c [if ambident behaviour of –CN is not considered] then, find a/b.
Level - III
1. Reaction of [Ni(CN) 4 ] 2− with metallic potassium in liquid ammonia at −33°C yields complex X. The geometry and magnetic behavior of X, respectively, are:
(1) Square planar and diamagnetic (2) Tetrahedral and diamagnetic (3) Octahedral and paramagnetic (4) Square pyramidal and paramagnetic
2. For a complex (d 6−configuration) having Δ0=25000 cm−1 and P=15000 cm−1, the crystal field stabilization energy is
(1) 30000 cm−1
(2) –60000 cm−1
(3) –30000 cm−1
(4) 60000 cm−1
3. Three arrangements are shown for the complex [Co(en)(NH 3) 2Cl 2] +. Pick up the wrong statement.
(1) P and Q are geometrical isomers
(2) Q and R are optical isomers
(3) P and R are optical isomers
(4) Q and R are geometrical isomers
4. Go through the following complexes,
(1) trans – [Co(NH3)4Cl2]+
(2) cis – [Co(NH3)2(en)2]3+
(3) trans – [Co(NH3)2(en)2]3+
(4) [NiCl42–]
(5) [TiF62–]
(6) [CoF63–]
Choose the incorrect option(s):
(1) (1) and (2) are optically active whereas, (3) is optically inactive
(2) (2) is optically active whereas, (1) and (3) are optically inactive
(3) (4) and (5) are coloured, and (6) is diamagnetic
(4) (4) is coloured, and (5), (6) are paramagnetic
5.
()() +1moleenaq+1moleenaq RS→→
When – en is ethane –1,2 –diamine
Correct statement(s) about these complexes is/are
(1) ‘R’ can exhibit geometrical isomerism and both have same colour.
(2) In P,Q,R and S hybridization of nickel is same.
(3) P,Q,R and S are high spin complexes. (4) For the complex ‘S’ totally two stereo isomers possible.
In both the complexes Co has t 2g 6 e g 0 config uration. Then the correct options among the following
(1) Complex (I) is paramagnetic (2) Complex (II) is Diamagnetic (3) (X) is oxidizing agent
(4) (X) is reducing agent
7. Which of the following compound has/have effective atomic number equal to the atomic number of a noble gas?
(1) K[Co(CO)4]
(2) K2[Fe(CO)4]
(3) [Co(NH3)6]Cl2
(4) [CoCl3(H2O)3]
8. The total numbers of stereoisomers obtained if NH 3 is replaced by bidentate en ligand in complex [Co(CN)2(NH3)2Cl2] (en=NH2CH2CH2NH2)
(1) 2
(2) 5
(3) 6
(4) 4
9. The enthalpy of hydration of the Fe2+ ion is 11.4 k cal/mol which is higher than expected if there were no crystal field stabilization energy. Assuming the aqua complex to be high- spin, estimate the magnitude of Δ0 for [Fe(H2O)6]2+?
10. The wave number of the radiation absorbed by [CoCl 6] 4– is 18000 cm –1, then the wave number of the radiation absorbed by [CoCl4]2– is k × 103 cm–1. Then the value of k is ______.
11. Find the difference between the number of optically active isomers and the number of optically inactive isomers of the octahedral complex, [Ma2bcde]n± where a, b, c, d and e are monodentate ligands and do not contain chiral centre.
12. How many of the following statements are correct
a) K2[NiF6] is a diamagnetic complex
b) [Mn(H 2O) 4] +2 is square planar but [MnCl4]−2 is tetrahedral
c) Spinonly magnetic moment of [Fe(NH3)6]+2 is 24BM
d) In [PtCl6]−2 complex – Cl is a strong field ligand
e) CFSE of [Co(H2O)6]+3 is −2.4 Δ0+2E where ‘E’ is pairing energy
f) All octahedral complexes having strong field ligands are low spin complexes
g) [Co(NO 2 ) 6 ] −4 is an inner orbital complex
h) [Ni(CN)4]−2 is an inner orbital complex but [NiCl4]−2 is outer orbital complex
THEORY-BASED QUESTIONS
Statement Type Questions
Each question has two statements: statement I (S-I) and statement II (S-II). Mark the correct answer as
(1) if both statement I and statement II are correct,
(2) if both statement I and statement II are incorrect
(3) if statement I is correct but statement II is incorrect,
(4) if statement I is incorrect but statement II is correct.
i) In a tetrahedral complex, lower energy d-orbital set is e g and higher energy d-orbital set is t2g.
13. The formation constant is x × 10 5 for the reaction of a tripositive metal ion with a thiocyanate ion to form a mono complex 3+- 2+ M+SCNMSCN . If the total metal concentration in the solution is 2.0 × 10−3M, the total SCN concentration is 1.50 × 10−3M, and the free SCN concentration is 1.0 × 10−5M, then find the value of x (to the nearest integer).
14. If an octahedral homoleptic complex [Fe IIIL x] nI does not exhibit geometrical isomerism but it exhibits optical isomerism, then find out the total number of correct statement(s) (denticity of L ≤ 2).
(i) Number of ligands are 4
(ii) Number of stereoisomers possible are 2
(iii) Complex does not follow Sidgwick rule
(iv) Minimum and maximum number of electron is present in t2g can be 3 and 5 respectively.
(v) The complex will always be paramagnetic in nature.
(vi) Complex will always be a high spin complex
1. S-I : According to VBT, ns, np-orbitals and (n-1)d or nd-orbitals of central metal ion can involve in hybridisation.
S-II : According to VBT, rearrangement of electrons against Hund's rule cannot occur in central metal ion.
2. S-I : Ammonium ion acts as a ligand
S-II : Hybridisation of nitrogen in ammonium ion is sp3 .
S-II : Coordination isomerism is shown by anionic complexes only.
5. S-I : C2 molecule contains 2π bonds.
S-II : [Co(Cl)(en) 2 (NH 2 C 6 H 4 −Me)]Cl 2 compound shows geometrical isomerism and optical isomerism.
6. S-I : [MnCl6]3–, [FeF6]3–, and [CoF6]3− are paramagnetic complexes, having four, five, and four unpaired electrons, respectively.
S-II : [MnCl 6] 3–,[FeF 6] 3– and [CoF 6] 3− are high spin octahedral complexes in which central metal ions undergo sp3d2 hybridisation.
7. S-I : [TiF6]3– is inner orbital complex and paramagnetic complex with one unpaired electron.
S-II : [Co(NH3)6]+3 is outer orbital complex and paramagnetic complex.
8. S-I : Hybridisation of central metal ion in [CrF6]−3 is sp3d2.
S-II : [Cr(CN) 6] −3 is an example of inner orbital octahedral complex.
9. S-I : [Fe(NH 3 ) 6 ] +2 is less stable than [Co(NH3)6]+2.
S-II : The greater the radius of the cation, the greater is the stability of its complex ion.
10. S-I : The metal-carbon bond in metal carbonyls possess 'σ' character only.
S-II : In metal carbonyls, the M–C σ bond is formed by the donation of a pair of electrons from a filled d orbital of metal into the vacant antibonding π* orbital of carbon monoxide.
Assertion and Reason Questions
Assertion and Reasoning
In each of the following questions, a statement of Assertion (A) is given, followed by a corresponding statement of Reason (R). Mark the correct answer as
(1) if both (A) and (R) are true and (R) is the correct explanation of (A),
(2) if both (A) and (R) are true but (R) is not the correct explanation of (A),
(3) if (A) is true but (R) is false,
(4) if both (A) and (R) are false.
11. (A) : In N 2H 4 any one nitrogen atom can coordinate with central metal or both can coordinate with central metal, i.e., it can act as a chelating ligand.
(R) : N2H4 is an ambident neutral ligand.
12. (A) : Tetrahedral complexes having two different types of unidentate ligands (without consideration of stereo centre in ligands) coordinated with central metal ion will show geometrical isomerism.
(R) : Geometrical isomerism arises in homoleptic complexes due to different possible geometric arrangement of the ligands.
(R) : Ambidentate ligand can ligate through two different donor atoms.
14. (A) : The spin only magnetic moment value for [Fe(CN)6]3− is 1.74 BM, whereas for [Fe(H2O)6]3+ is 5.92 BM.
(R) : In both complexes, Fe is present in +3 oxidation state.
15. (A) : Geometrical isomerism is not possible in tetrahedral complex having two different types of unidentate ligands coordinated with central metal ion.
(R) : In tetrahedral geometry the relative position of unidentate ligands attached to the central atom or ion remain the same with respect to each other.
16. (A) : [CrCl2(ox)2]−3 does not shows optical isomerism.
(R) : Optical isomerism is common in octahedral complexes involving bidentate ligands.
17. (A) : (Ph3P)3RhCl is Wilkinson’s catalyst.
(R) : (Ph 3 P) 3 RhCl can show optical isomerism but does not show geometrical isomerism.
18. (A) : Aqueous solution of Mohr’s salt gives test for NH4+, Fe+2, and SO42– ions.
(R) : Mohr’s salt is a double salt.
JEE ADVANCED LEVEL
Multiple Option Correct MCQs
1. CrCl 3 .6H 2 O dissolves i n water to form a complex. If a solution is prepared by dissolving 0.1 mol of CrCl3.6H2O in water and treated with excess AgNO 3 solution, 0.2 mol of AgCl is precipitated. Select the correct option for complex. (Mass number of Cr=52,O=16, Cl = 35.5)
(1) It will exist in hydrate isomerism.
(2) It loses almost 6.75% by mass of its original mass on treatment with conc. H2SO4
(3) It is a paramagnetic complex in which central metal is in d2sp3 hybridisation.
(4) It will exist in optical isomeric form.
2. Which of the following statement(s) is/are true about the complex ion [Pt (en)2 Cl2]2+? [en = H2N CH2 CH2 NH2]
(1) It has two geometrical isomers, cis and trans
(2) Both the cis and trans isomers display optical activity.
(3) Only the cis -isomer has nonsuperimposable mirror image.
(4) The oxidation state of Pt in the complex is +4.
3. Consider the complex [Pt(NH3)(PH3)Cl.Br]. Choose the correct statements regarding this complex.
(1) It has 3 geometrical isomers, of which 2 are cis and one is trans
(2) The isomers in which Cl and Br are adjacent are cis isomers.
19. (A) : In the complex [Co(NH 3 ) 3 C1 3 ], chloride ions cannot be tested.
(R) : In this complex, chloride ions satisfy secondary valencies of cobalt.
20. (A) : [Ni(CO)4] is diamagnetic in nature.
(R) : Ni atom undergoes dsp2 hybridisation.
(3) All the 3 isomers have same magnetic moment.
(4) All the 3 isomers have same colour since the position of ligands change in the same plane.
4. The complex(es), that can exhibit the type of isomerism shown by [Pt(NH3)2Br2] is(are) [en = H2NCH2CH2NH2]
(1) [Pt(en)(SCN)2]
(2) [Zn(NH3)2Cl2]
(3) [Pt(NH3)2Cl4]
(4) [Cr(en)2(H2O)(SO4)]+
5. Choose the correct statements among the following.
(1) Most five-coordinate compounds adopt either the trigonal bipyramidal (TBP) or square pyramidal(sp structures).
(2) When the central atom is a main group element, the axial bonds are longer than the equatorial bonds in TBP structure but reverse is the case for the sp structure.
(3) When the central atom is a transition element in TBP (e.g.,[CuCl 5 ] 3− ) the axial Cu−Cl bonds are shorter than equatorial ones but, on the other hand, in sp (e.g.,[Ni(CN)5]3−), the axial Ni−C bonds are longer than the equatorial ones.
(4) In BrF5, the bromine atom is below the basal plane of square pyramid but in [Ni(CN)5]3−, the Ni atom is above the basal plane.
6. Until the discovery of Alfred Werner, people thought that all optically active compounds should contain carbon. The compound prepared by Werner
Co Co OH (NH3)4 Cl6 OH
is optically active and proved that the earlier theory is wrong. Choose the correct statements about this compound.
(1) All the cobalt atoms are in the same oxidation state.
(2) It cannot exhibit geometrical isomerism.
(3) In this complex, there are four octahedral centres sharing the 3 edges of central cobalt octahedron by the 3 other octahedrons of cobalt ions.
(4) It is a diamagnetic complex.
Numerical Value Questions
7. Among the following, what is the total number of complexes that are optically active?
(i) [Cr(ox)3]3−
(ii) cis−[Pt(Cl2)(en)]
(iii) cis−[Rh(Cl2)(NH3)4]+
(iv) [Ru(dipy)3]3+
( v) cis−[Co(NO2)3(dien)]
(vi) trans−[Co(NO2)3(dien)]
(vii) cis−[Co(NO2)3(NH3)3]
8. The number of stereoisomers possible for [Co(ox)2(Br)(NH3)]2− is ____. [ox = oxalate]
9. An aqueous solution of nitrate salt is mixed with equal volume of freshly prepared FeSO4 solution. When concentrated H2SO4 is added dropwise along the walls of tube, without disturbing, the ‘X’ appears at the interface (magnitude), with cationic ligand. The oxidation state of iron in ‘X’ compound is___.
Integer Value Questions
10. Find the total number of stereo isomers possible for the complex [Mg(EDTA)] 2–
11. How many of the following compounds are paramagnetic? KO 2 , BaO 2 , Na 2 O 2 , [Cu(NH 3) 4] 2+, [Ni(NH 3) 6] 2+, [Cr(NH 3) 6] 3+ , O2, O3, [PtCl2(NH3)2]
12. The difference in the number of unpaired electrons in Co+2 ion in its high spin and low spin octahedral complexes is _____.
13. The number of unpaired electrons in the complex ion [CoF6]−3 is ______.
Passage -based Questions
(Q 14-15)
The homoleptic carbonyls (compounds containing carbonyl ligands only) are formed by most of the transition metals. The metalcarbon bond in metal carbonyl posses both σ and π character.
14. In the cobalt carbonyl complex, [Co2(CO)8], the number of Co–Co bonds is ' p ' and terminal CO ligands is 'q'. Find the value of p + q.
15. Decacarbonyl dimanganese(0) is made up of ‘ x’ number of square pyramidal units. Find the value of ‘x’.
(Q 16-17)
The splitting of the degenerate d-orbitals due to the presence of ligands in a definite geometry is termed crystal field splitting, and the energy separation is denoted by Δ o for octahedral complexes.
16. Magnitude of crystal field splitting energy (CFSE) for the complex [Cr(H 2 O) 6 ]+ 2 is _____, when P = 20925 cm –1 and Δ o = 10462.5 cm–1 (given 1 kJ/mol = 83.7 cm –1).
17. Magnitude Crystal field splitting energy (CFSE) for the complex [Cr(NH 3 ) 6 ] +2 is ______ (in kJ/mol) when P = 125 and Δo = 250 kJ/mol.
Matrix Matching Questions
18. Match the Column-I (ligand) with ColumnII (donor atoms).
Column-I
Column-II
(A) EDTA (p) 3(N,N,N)
(B) acac (q) 6(N,N,O,O,O,O)
(C) glycenate (r) 2(N, O)
(D) en (s) 4(N,N,N,N)
(t) 2(O,O)
(u) 2(N, N)
Choose the correct option.
(A) (B) (C) (D)
(1) q t p r
(2) q r p r
(3) q t r u
(4) s t q t
19. Match Column-I (complex) with Column- II (charge on coordination sphere).
Column-I (Complex)
Column-II (Charge on coordination Sphere)
(A) CoCl3.6NH3 (p) +1
(B) CoCl3.5NH3 (q) +2
(C) CoCl3.4NH3 (r) +3
(D) CoCl3.3NH3 (s) +4 (t) 0
Choose the correct match in terms of the charge on the complex.
(A) (B) (C) (D)
(1) r q t p
(2) r p q t
(3) r q p t
(4) s r q p
20. Match Column-I (mole of AgCl precipitated) with Column-II (secondary valency).
Column-I (Mole of AgCl per mole of compound) Column-II (Secondary valencies)
(A) PtCl2.2NH3 = zero (p) 2
(B) CoCl3.4NH4 -=1 (q) 3
(C) PtCl2.4NH3 = 2 (r) 6
(D) CrCl3.6NH3 = 3 (s) 4
Choose the correct option.
(A) (B) (C) (D)
(1) p q r s
(2) q p s r
(3) s r p r
(4) s r s r
21. Match the column-I (ligand ) with columnII (denticity).
Column-I (Ligand) Column-II (Common denticity)
(A) SO42– (p) Tetradente
(B) C2O42– (q) Hexadentate
(C) EDTA4 (r) Tridentate
(D) Diethylene triamine(dien) (s) Monodentate
(t) Bidentate
Choose the correct option.
(A) (B) (C) (D)
(1) s p q r
(2) t r p t
(3) s t q r
(4) t p q r
22. Match the complex species given in Column-I with the possible isomerism given in Column-II and assign the correct code.
Column-I (Complex species)
Column-II (Isomerism)
(A) [Co(NH3)4Cl2]+ (p) Optical
(B) cis-[Co(en2)4 Cl2]+ (q) Ionisation
(C) [Co(NH3)5 (NO2)]Cl2 (r) Coordination
(D) [Co(NH3)6] [Cr(CN6)] (s) Geometrical
(t) Linkage
(A) (B) (C) (D)
(1) p r s q
(2) r s q p
(3) q p s r
(4) s p q r
23. Match column-I(complex) with column- II (type of isomerism).
Column-I (Complex compounds)
Column-II (Type of isomerism shown)
(A) [Co(H2O)3F3] (p) Geometrical isomerisim
(B) [Co(en)3]Cl3 (q) Optical isomerism
(C) [Co(en)2(NO2)2] Cl (r) Linkage isomerism
(D) K3[Cr(CN)6] (s) Ionisation isomerism
Choose the correct option.
(A) (B) (C) (D)
(1) p q r s
(2) q s s r
(3) p q s r
(4) r q s p
24. Match the column-I(complex) with column-II (type of isomerism).
Column-I Column-II
(A) [Co(NH3)5(NO2)] Cl2 and [Co(NH3)5 (ONO)]Cl2 (p) Ionisation isomerism
(B) [Co(NH3)6 (CO(CN)6]Cl2 and [Co(NH3)5 (Cr(CN)6]Cl2 (q) Hydrate isomerism or solvate isomerism
(C) [Co(NH3)5 (SO4)]Br and [Co(NH3)5 Br–)]SO4 (r) Coordination isomerism
(D) [Cr(H2O)6]Cl3 and [Cr(H2O)5Cl]Cl2. H2O (s) Linkage isomerism
Choose the correct option.
(A) (B) (C) (D)
(1) s r p q
(2) r q s p
(3) p r q s
(4) s q p r
25. Match column-I (complex) with column- II (charge on coordination sphere).
Column-I Column-II
(A) [Co(NH3)5NO2]Cl2 and [Co(NH3)5 ONO] Cl2 (p) Ionisation isomerism
(B) [Cr(NH3)6] [Co(CN)6] and [Cr(CN)6] [Co(NH3)6] (q) Coordination isomerism
(C) [Co(NH3)5(SO4)] Br and [Co(NH3) Br]SO4 (r) Linkage isomerism
(D) [Cr(H2O)6]Cl3 and [Cr(H2O)5Cl] Cl2.H2O (s) Solvate isomerism
Choose the correct option.
(A) (B) (C) (D)
(1) r p q s
(2) q r p s
(3) r q p s
(4) s r q p
26. Match column-I (formula of compound) with column-II (coloiur and type of electrolyte).
Column-I
(Formula)
Column-I (Colour and type of electrolyte)
(A) CoCl3.6NH3 (p) Purple 1:2 electrolyte
(B) CoCl3.5NH3 (q) Green 1:1 electrolyte
(C) CoCl3.4NH3 (r) Yellow 1:3 electrolyte
(D) CoCl3.3NH3 (s) Colourless non electrolyte
Choose the correct option.
(A) (B) (C) (D)
(1) p r q s
(2) q s p r
(3) r p q s
(4) s q r p
BRAIN TEASERS
1. EAN of Na[PtCl3(η2 – C2H4)] is (1) 86 (2) 78 (3) 84 (4) 34
2. The coordination number of metal M in a complex having molecular formula MCl3. (H 2 O) 4 is 6 and there is no molecule of water of crystallisation in it. The volume of 0.1M AgNO3 solution needed to precipitate the free chloride ions in 200 mL of 0.01 M complex is (1) 40 mL (2) 20 mL (3) 60 mL (4) 80 mL
3. A solution containing 2.675 g of CoCl 3. 6 NH3 (molar mass = 267.5 g mol–1) is passed through a cation exchanger. The chloride ions obtained in solution were treated with excess of AgNO 3 to give 4.78 g of AgCl (molar mass = 143.5 g mol–1). The formula of the complex is (At. Mass of Ag = 108 u)
27. Match column-I (complex) with column- II (mole of AgCl precipitated).
Column - I (Complex)
Column - II (Mole of AgCl produced by AgNO3 per mole of complex)
(A) [Co(NH 3 ) 6 ]Cl 3 Yellow (p) 1
(B) [CoCl(NH3)5]Cl2 purple (q) 2
(C) [CoCl2(NH3)4]Cl violet (r) 3 (s) 4 (t) 0
Choose the correct option (A) (B) (C)
(1) r p q
(2) r q p
(3) s q p
(4) r r r
(1) [Co(NH3)6]Cl3
(2) [CoCl2(NH3)4]Cl
(3) [CoCl3(NH3)3]
(4) [CoCl(NH3)5]Cl2
4. The magnetic moment of a certain complex (A) of Co was found to be 4.89 B.M and the EAN as 36. Co also forms complex (B) with magnetic moment 3.87 BM and EAN 37, and complex (C) with EAN as 36 but diamagnetic. Which of the following statements is true regarding the above observation?
(1) The oxidation states of Co in (A), (B) and (C) are +3, +2 , and +3, respectively.
(2) Complexes (A) and (B) have sp 3 d 2 hybridisation state while (C) has dsp 3 hybridisation state.
(3) The spin multiples of Co in (A), (B), and (C) are 4, 3, and 1, respectively.
(4) The oxidation states of Co in (A), (B), and (C) are +6, +8, and +1, respectively.
5. Following fill in the blanks with the given options.
A) For the same metal, the same ligands and metal ligand distance, ∆ o = ......∆ t
B) Anhydrous CuSO 4 is ……… and CuSO4.5H2O is …………. in colour.
C) In tetrahedral crystal field, …………. configurations are rarely observed.
D) Ruby is Al2O3 containing about 0.5-1% Cr3+ ions, which are randomly distributed in position normally occupied by
FLASHBACK ( Previous JEE Questions )
JEE Main
1. The IUPAC name of K3[Co(C2O4)3] is (2023)
(1) potassium trioxalate cobalt (III)
(2) potassium tris(oxalato) cobalt (III)
(3) potassium trioxalate cobaltate (III)
(4) potassium tris(oxalato) cobaltate (III)
2. Given below are two statements. One is labelled Assertion (A) and the other is labelled Reason (R).
Assertion A : In the complex Ni(CO) 4 and Fe(CO) 5 , the metals have zero oxidation state.
Reason R : Low oxidation states are found when a complex has ligands capable of π- Donor character in addition to the σ- bonding.
In light of the above statements, choose the correct answer from the options given below. (2023)
(1) A is correct but R is not correct
(2) Both A and R are correct but R is not the correct explanation of A
9. Which of the following complexes has a possibility to exist as meridional isomer? (2023)
(1) [Pt(NH3)2Cl2] (2) [Co(en)3]
(3) [Co(en)2Cl2] (4) [Co(NH3)3(NO2)3]
10. Number of ambidentate ligands in a representative metal complex M(en)(SCN)4 is _______. [en = ethylenediamine] (2023)
11. In potassium ferrocyanide, there are _____ pairs of electrons is the t 2g set of orbitals. (2023)
12. The homoleptic and octahedral complex of Co2+ and H2O has ____unpaired electrons in the t 2g set of orbitals. (2023)
13. The d-electronic configuration of [CoCl4]2– in tetrahedral crystal field is em t 2g n Sum of 'm' and 'number of unpaired electrons' is___. (2023)
14. The number of paramagnetic species from the following is ______. (2023) [Ni(CN)4]2–, [Ni(CO)4], [NiCl4]2–[Fe(CN)6]4–, [Cu(NH3)4]2+ [Fe(CN)6]3–, and [Fe(H3O)6]2+
JEE Advanced
15. Match the electronic configurations in List-I with appropriate metal complex ions in ListII and choose the correct option. [Atomic Number: Fe = 26, Mn = 25, Co = 27] (2023)
List-I
List-II
(A) t62ge0g (p) [Fe(H2O)6]2+
(B) t33ge2g (q) [Mn(H2O)6]2+
(C) e2 g t3 2g (r) [Co(NH3)6]3+
(D) t42ge2g (s) [FeCl4]–(t) [CoCl4]–2
(A) (B) (C) (D)
(1) p s q r
(2) p q s t
(3) r q t p
(4) r q s p
16. The complex(es), which can exhibit the type of isomerism shown by [Pt(NH3)2Br2], is(are) [en = H2NCH2CH2NH2] (2023)
(1) [Pt(en)(SCN)2]
(2) [Zn(NH3)2Cl2]
(3) [Pt(NH3)2Cl4]
(4) [Cr(en)2(H2O)(SO4)]+
17. List-I contains metal species and List-II contains their properties. (2022)
List-I List-II
(A) [Cr(CN)6]4– (p) t 2g orbitals contain 4 electrons
CHAPTER TEST-JEE MAIN
Section-A
1. Which of the following complexes are heteroleptic?
(i) [Cr(NH3)]3+ (ii) [Fe(NH3)4Cl2]+
(iii) [Mn(CN6)]4– (iv) [Co(NH3)4Cl2]
(1) (i), (iv) (2) (ii) and (iv) (3) (i) and (ii) (4) (i) and (iv)
2. IUPAC name of a complex is tetraamminedichlorido cobalt(III) chloride. The complex is represented as (1) [Co(NH3)4Cl2]Cl3 (2) [Co(NH3)4Cl2]Cl (3) [Co(NH2)4Cl2]Cl (4) [Co(NH2)4Cl2]Cl2
3. What is the EAN of central metal in Ni(gly)2? (At. no. of Ni = 28) (1) 30 (2) 34 (3) 36 (4) 32
4. Among [Ni(CO)4], [Ni(CN)4]2– and [NiCl4]2–species, the hybridisation state of atoms are respectively, Ni (1) sp3, dsp2, dsp2 (2) sp3, dsp2, dsp3 (3) sp3, sp3, dsp2 (4) dsp2, sp3, sp3
(B) [RuCl6]2– (q) (spin-only) = 4.9 BM
(C) [Cr(H2O)6]2+ (r) low spin complex ion
(D) [Fe (H2O)6]2+ (s) metal ion in 4+ oxidation state (t) d4 species
[Given: Atomic number of Cr=24, Ru=44, Fe=26] Match each metal species in LIST-I with their properties in LIST-II, and choose the correct option.
(A) (B) (C) (D)
(1) r,t p,s q,t p,q
(2) r,s p,t p,q q,t
(3) p,r r,s r,t p,t
(4) q,t s,t p,t q,r
5. Arrange the following cyano complexes in increasing order of magnetic moment. (1) [Fe(CN)6]4– < [Fe(CN)6]3– < [Mn(CN)6]3–< [Cr(CN)6]3–(2) [Fe(CN)6]3– < [Cr(CN)6]3– < [Mn(CN)6]3–< [Fe(CN)6]4–(3) [Mn(CN)6]3– < [Fe(CN)6]3– < [Fe(CN)6]4–< [Cr(CN)6]3–(4) [Fe(CN)6]3– < [Cr(CN)6]3– < [Fe(CN)6]4– < [Mn(CN)6]3–
6. The set that does not have ambidentate ligands is (1) NO2–, C2O4–2, EDTA–4 (2) EDTA–4, NCS–, C2O4–2, (3) C2O4–2, ethylene dimine H2O (4) NO2–, C2O4–2, NO2–, NCS–
7. The donor sites of (EDTA) 4− are (1) O atoms only (2) N atoms only
(3) two N atoms and four O atoms (4) three N atoms and three O atoms
8. The formula of the complex tris (ethylenediamine) cobalt (III) sulphate is
(1) [Co(en)2]SO4]
(2) [Co(en)3]SO4]
(3) [Co(en)3]3(SO4)2
(4) [Co(en)3]2(SO4)3
9. Among (a, b, c, and d), the complexes that can display geometrical isomerism are
(a) [Pt(NH3)3Cl]+
(b) [Pt(NH3)Cl5]–
(c) [Pt(NH3)2Cl(NO)2]
(d) [Pt(NH3)4ClBr]2+
(1) (b) and (c)
(2) (d) and (a)
(3) (c) and (d)
(4) (a) and (b)
10. 1 mol of CoCl3 reacts with 5 mol of ammonia to give a purple coloured complex. The number of moles of AgCl precipitated, when 1 mol of complex is treated with excess of AgNO3, is
(1) 1 mol
(2) 2 mol
(3) 3 mol
(4) zero mol
11. How many moles of AgCl would be obtained when 100 mL of 0.1M CoCl3(NH3)5 is treated with excess of AgNO3?
(1) 0.01
(2) 0.02
(3) 0.03
(4) 0.04
12. The geometry and magnetic behaviour of the complex [Ni(CO)4] are (1) square planar geometry and diamagnetic (2) tetrahedral geometry and diamagnetic (3) tetrahedral geometry and paramagnetic (4) square planar geometry and paramagnetic
13. Co +3 forms four complexes with four different ligands, that are [Co(NH 3 ) 6 ] +3 , [CoCl 6] –3, [Co(H 2O) 6] +3, and [Co(CN 6] –3
The order of CFSE (Δ0) in these complexes is in the order
14. Select the incorrect statement(s) among the following.
a) If Δ 0 > P, it becomes more kinetically favourable for the fourth electron to occupy t2g orbital with t 2g 4e g 0 configuration for d4 ions.
b) Nature of bonding in coordination compound cannot be explain by MOT.
c) Solvated isomerism is a special class of coordination isomerism.
(1) b and c only
(2) b only
(3) a and c only
(4) a, b, and c
15. In [Ni(CO)4], metal act as
(1) π− donor σ−acceptor
(2) σ− donor π−acceptor
(3) π− donor π−acceptor
(4) σ− donor σ−acceptor
16. Number of bridged carbonyl groups in [Mn 2 (CO) 10 ] and [Co 2 (CO) 8 ] are, respectively,
(1) 2, 2 (2) 0, 0 (3) 0, 2 (4) 2, 0
17. The stability constants of the complexes formed by a metal ion M2+ with NH3, CN , and H2O are of the order of 1015, 1027, and 1011, respectively. Then,
(1) NH3 is the strongest ligand
(2) CN is the strongest ligand
(3) these values cannot predict the strength of the ligand
(4) all the ligands are equally strong
18. The colour exhibited by one of the iron ions in aqueous solutions is pale green. The primary valency and secondary valency respectively in the green complex are
(1) 2, 4 (2) 2, 6 (3) 3, 4 (4) 3, 6
19. What is the order of spin only magnetic moment of the following system?
I) Mn2+ in presence of weak field ligand in octahedral field
II) Ni2+ in presence of strong field ligand in octahedral field
III) Cr3+ in presence of ligand in octahedral field
IV) Sc3+ in presence of weak field ligand in octahedral field
(1) I > II > III > IV (2) I > III > II > IV
(3) I > III > IV > II (4) I > IV > III> II
Section-B
20. Which of the following statements is false?
(1) [NiCl 4 ] 2– ion is paramagnetic and
CHAPTER TEST-JEE ADVANCED
2022 P1 Model
Section-A
[Numerical Value Questions]
1. How many of the following complex ions have dsp2 hybridisation?
3. How many of the following complexes are expected to have d2sp3 hybridisation?
tetrahedral but [PtCl4]2– is diamagnetic and square planar.
(2) [CoF6]3– ion is paramagnetic but [NiF6]2–ion is diamagnetic and low spin.
(3) Δ o order [CrCl 6 ] 3– < [Cr(NH 3 ) 6 ] 3+ < [Cr(CN)6]3–
(4)Dimethylglyoximato nickel (II) is diamagnetic and tetrahedral complex.
21. When cupric sulphate solution is added with ‘ x’, it produced a deep blue-coloured soluble complex ‘y’. The total value of primary and secondary valencies of copper in ‘y’ is ____.
22. The number of oxygen atoms involved in bonding in the coordination sphere of [Mg(EDTA)]2−is ______.
23 How many geometrical isomers are possible for the complex compound [MA 2B2C2]?
24. What is the possible number of geometrical isomers exhibited by [CrCl 2Br2(NH3)2]–?
25. The spin only magnetic moment of [CoF6]−3 complex is _____BM.
4. If [Cu(H2O)4]2+ absorbs a light of wavelength 600 nm for d-d transition, then the value of octahedral crystal field splitting energy for [Cu(H2O)6]2+ will be____×10−21 J. [Nearest integer](Given: h = 6.63 × 10−34 Js and C = 3.08 × 108 ms−1)
5. [Fe(CN) 6 ] 3− should be an inner orbital complex. Ignoring the pairing energy, the value of crystal field stabilisation energy for this complex is ( – ) __________ Δ 0. (Nearest integer)
6. CSFE value of [CoCl3(H2O)3] is –x 0 + yP. If 0 < P the (–x + y + 1) = ______
7. With respect to the complex ion [Mn(OH2)6]2+, how many of the following statements are true?
I. It is diamagnetic.
II. It is a low spin complex.
III. The metal ion is a d5 ion.
IV. The ligands are weak field ligands.
V. It is octahedral.
8. In Nessler’s reagent, the oxidation number of central metal is _____.
Section-B
[Multiple Correct Option MCQs]
9. Which of the following statement(s) is/are correct?
(1) Both KMnO4 and K2Cr2O7 are coloured due to charge transfer spectra.
(2) EAN of Co in Co2(CO)8 is 35.
(3) [Co(en)3]3+ is optically active.
(4) Both cis and trans isomers of [CoCl2(NH3)4]+ are optically inactive.
10. The pair(s) of coordination complex/ions exhibiting the same kind of isomerism is/ are
(1) [Cr(NH3)5Cl]Cl2 and [Cr(NH3)4Cl2]Cl
(2) [Cr(NH3)4Cl2]+ and [Pt(NH3)2(H2O)Cl]+
(3) [CoBr2Cl2]–2 and [Pt3Br2Cl2]–2
(4) [Pt(NH3)3NO3]Cl and [Pt(NH3)3Cl]Br
11. Which of the following statement(s) is/are correct?
(1) [MnCl6]−3 has four unpaired electrons.
(2) [Co(C2O4)3]–3 is inner orbital complex having d 2 sp 3 hybridisation and diamagnetic.
(3) [CoF6]−3 is outer orbital complex and diamagnetic.
(4) [FeF6]−3 is an outer orbital complex and paramagnetic.
12. Which of the following statements is/are correct?
(1) Greater the stability constant of a complex ion, greater is its stability.
(2) Greater the oxidation state of the central metal ion, greater is the stability of complex
(3) CO stabilises the complex because of its synergic bonding.
(4) Chelate complexes have low stability constants.
13. Choose the correct statements among the following.
(1) The complexes [NiCl]−2 and [Ni(CN)4]−2 differ in magnetic properties.
(2) The complexes [NiCl]−2 and [Ni(CN)4]−2 differ in their geometry.
(3) The complexes [NiCl]−2 and [Ni(CN)4]−2 differ in primary valences of nickel.
(4) The complexes [NiCl]−2 and [Ni(CN)4]−2 differ in primary valences of nickel.
In both the complexes Co has 60 2 gg te configuration. Then the correct options among the following.
(1) Complex (I) is paramagnetic.
(2) Complex (II) is diamagnetic.
(3) (X) is oxidising agent.
(4) (X) is reducing agent.
Section-C
[Matrix Matching Questions]
15. Identify the metals (M) in the complexes given in Column-I and match the complexes with their characteristic property/properties given in Column-II.
Column-I Column-II
(A) EAN for [M(H2O)6]3+ is 36 (p) Paramagnetic complex
(B) EAN for [M(H2O)6]2+ is 33 (q) Diamagnetic complex
(C) EAN for [M(NH3)6]3+ is 35 (r) The value of spin only magnetic moment is equal to corresponding fluoro complex
(D) EAN for [M(NH3)4]2+ is 36 (s) The value of spin only magnetic moment is equal to corresponding cyano complex
Choose the correct option.
(A) (B) (C) (D)
(1) prs qs qrs ps
(2) qps pr qrs ps
(3) pr s prs qr
(4) qs prs prs qrs
16. Match List-I with List-II and choose the correct option from the codes give bel ow.
List-I (Complex) List-II (Structure and magnetic moment)
(A) [Ag(CN)2]– (p) square planar and 1.73 BM
(B) [Cu(CN)4]3– (q) linear and zero
(C) [Cu(CN)6]4– (r) octahedral and zero
(D) [Cu(NH3)4]2+ (s) tetrahedral and zero
(E) [Fe(CN)6]4– (t) octahedral and 1.73 BM
(A) (B) (C) (D) (E)
(1) q s t p r
(2) t s p r q
(3) p r s q t
(4) s t q p r
17. Match column-I with Column-II.
Column -I (Complex compound) Column-II (Hybridisation and magnetic behaviour)
(A) [NiCl4]2– (p) sp3d2 paramagnetic
(B) [Ni(CN)4]2– (q) dsP2 diamagnetic
(C) [Ni(CO)4] (r) sp3 paramagnetic
(D) [MnCl6]3– (s) sp3 diamagnetic
Choose the correct option.
(A) (B) (C) (D)
(1) r q s p
(2) q p r s
(3) s q p r
(4) p r q s
18. Match the electronic configurations in List-I with appropriate metal complex ions in ListII and choose the correct option. [Atomic Number: Fe= 26, Mn= 25, Co= 27]
7.3 Chemistry Involved in Preparation of Compounds
7.4 Chemical Principles Involved in Experiments
The practical chemistry is to establish practical approach towards understanding the composition of various substances useful for laboratory or for our day-to-day life.
Chemical analysis is done mainly in two steps– qualitative analysis and quantitative analysis. Qualitative analysis is used to identify the sample contents. Quantitative analysis is used to determine the relative amounts of components present in the given sample. The
Table 7.1 Physical examination
Experiment
common procedure for testing any unknown sample is to make its solution and then test the solution for the ions present in it.
7.1 QUALITATIVE SALT ANALYSIS
Qualitative analysis involves the detection of cation(s) and anion(s) of a salt or a mixture of salts.
Steps involved in qualitative analysis are: preliminary tests, wet tests for acid radicals, wet tests (group analysis) for basic radicals.
7.1.1 Preliminary Tests
The identification of radicals is first done on the basis of preliminary tests; they are: physical appearance (colour and smell), dry heating test, flame test, borax bead test.
Physical appearance: Physical appearance of the radicals is the one of the preliminary tests, as shown in table 7.1
Observations
1. Colour Blue or bluish–green, Greenish
Light green Dark brown Pink
Light pink, flash colour White Cu2+ or Ni2+
Inference
2. Smell (Take a pinch of the salt between your fingers and rub with a drop of water)
Ammoniacal smell
Vinegar-like smell
Smell like that of rotten eggs
Shows the absence of Cu2+, Ni2+,Fe2+, Fe3+, Mn2+, and Co2+.
NH4+
CH3COO–
S2–
3. Density (i) Heavy (ii) Light fluffy powder
4. Deliquescence
Dry Heating Test
Salt absorbs moisture and becomes like a paste
Heat a small amount of mixture in a dry test tube. Some salts undergo decomposition,
Table 7.2 Physical Examination
Observation
Salt of Pb2+, or Ba2+ Carbonate
(i) If coloured, may be Cu(NO 3) 2, FeCl3
(ii) If colourless, may be Zn(NO3)2, chlorides of Zn2+, Mg2+ etc
evolving gases or may undergo characteristic changes in the colour of residue, as shown in Table 7.2.
1. Colourless, odourless gas is evolved that turns lime water milky CO32– or C2O42– Presents
2. Colourless gas with odour
(i) H2S gas: Smells like rotten eggs, turns lead acetate paper black Hydrated S–2
(ii) SO2 gas: Characteristics suffocating smell turns acidified potassium dichromate paper green SO32–
(iii) HCl gas: Pungent smell, white fumes with ammonia, white precipitate with silver nitrate solution. Cl–
(iv) I2 vapours: Dark violet, turns starch paper blue I-
4. Sublimate formed:
(i) White sublimate
(ii) Black sublimate accompanied by violet vapours
5. Residue
NH4+ or Hg+ I–
(i) Yellow when hot, white when cold. Zn2+
(ii) Brown when hot and yellow when cold Pb2+
(iii) White salt becomes black on heating CH3COO–
Observation Inference
(iv) Original salt blue becomes white on heating
Hydrated CuSO4
(v) Coloured salt becomes brown or black on heating. Co2+, Cu2+, Mn2+
6. Swelling: The mixture swells up into voluminous mass. PO43– , BO33–
Flame Test
A small quantity of the substance is placed on a watch glass. The substance is moistened with conc. HCl. The moistened mass is taken on the loop of a platinum wire and introduced into the non–luminous flame of the Bunsen burner into the fringe of the flame at the base. Gradually, the platinum wire is raised into the hottest part of the flame. Then, the colour of the flame is observed, as shown table 7.3
Table : 7.3 Flame test Flame
Crimson red
Golden yellow
Violet reddish
Violet blue
Brick red
Crimson red
Apple green/Yellowish green
Greenish Blue
Pale green
Flashes of green
Pale green fleeting
Borax Bead Test
This test is performed only for coloured salts. On heating borax, Na2B4O7.10H2O first loses water molecules and swells up. On further heating, it turns into a transparent liquid, which solidifies into glass-like material known as borax bead.
Na2B4O7.10H2O ∆ → Na2B4O7 + 10H2O ↑
Na2B4O7 → B2O3 + 2NaBO2
When borax is heated in a Bunsen burner
flame with metal oxides on a loop of platinum wire, different coloured M(BO 2) 2 beads are formed, as shown table 7.4
Oxidising flame: Non-luminous flame
Reducing flame : Luminous flame
Table : 7.4 Borax Bead Test
Colour of the Bead Inference In oxidising flame In reducing flame
1. Green when hot, light blue when cold Colourless when hot, opaque red when cold
2. Yellowish brown when hot, pale yellow when cold Green, hot and cold
3. Amethyst (pinkish violet) in both hot and cold Colourless, hot and cold
4. Brown when hot, pale brown when cold Grey or black when hot and opaque when cold
5. Deep blue in both hot and cold Deep blue in hot and cold
7.1.2 Analysis of Anion
or
Anions are classified into three types based on the solubility in acids.
Group A Radicals
This anion forms volatile products when reacting with dilute HCl/H 2SO4.
Carbonate Ion (CO32–)
Confirmation test for CO 3 2– ion is with the help of dilute H2SO4, lime water, MgSO4, BaCl2, phenolphthalein, methyl orange, etc.
Dilute H2SO4 test: A colourless, odourless gas is evolved with brisk effervescence.
CaCO3 + H2SO4 → CaSO4 + H2O + CO2↑
Lime water/baryta water (Ba(OH)2) test: The liberated gas can be identified by its property of rendering lime water (or baryta water) turbid
CO2 + Ca(OH)2→ CaCO3 ↑ (milky) + H2O
On prolonged passage of CO2, the milkiness disappears.
CaCO3 + CO2 + H2O→ Ca(HCO3)2(soluble)
CaCO3 ↓ H2O + CO2
Try yourself:
1. Is there any other gas like CO2 that also turns lime water milky?
milky due to the formation of insoluble calcium sulphite.
Ans: Yes, it is 2SO gas that also turns lime water
Magnesium sulphate test (for soluble carbonates): White precipitate MgCO 3 is formed.
Silver nitrate solution:White precipitate is formed CO32–+Ag+ → Ag2CO3↓
White precipitate is soluble in HNO 3 and ammonia. The precipitate becomes yellow or brown upon addition of excess reagent owing to the formation of silver oxide; the same happens if the mixture is boiled.
Ag2CO3 → Ag2O ↓+ CO2↑
Phenolphthalein is turned pink by soluble carbonates and colourless by soluble hydrogen carbonates.
Barium chloride(or calcium chloride) solution test: White precipitate of barium (or calcium) carbonate is formed.
CO32– +M2+→M C O 3↓ (M+2 = Ca+2, Ba+2)
Normal carbonate reacts; hydrogen carbonate does not. The precipitate is soluble in mineral acids and carbonic acids.
MCO3+2H+→ M+2+CO2↑ + H2O
Phenolphthalein test: Pink coloured solution is formed.
Methyl orange test: Yellow coloured solution is formed.
Sulphite Ion (SO3–2)
Confirmation test for SO 3 2– ion is with the help of dilute HCl, K 2Cr2O7, BaCl2, Zn, etc.
Dilute HCl or H2SO4 test: SO2 gas is liberated. Decomposition of salt is more rapid on warming. The gas is identified as follows: SO 2 has suffocating odour of burning sulphur.
Acidified potassium dichromate test: The filter paper dipped in acidified K2Cr2O7 turns green.
Cr2O72–+ 2H++3SO2→ 2Cr3+(green)+3SO42– + H2O.
Filter paper moistened with KIO 3 and starch solution turns to blue due to formation of iodine.
5SO2+2IO3–+4H2O → I2+5SO4–2 + 8H+
Barium chloride/strontium chloride solution: White precipitate of barium (or strontium) sulphite is obtained.
SO32– + Ba2+/Sr2+→ BaSO3/SrSO3↓ (white).
The precipitate dissolves in dilute HCl, when SO2 is evolved.
BaSO3+2H+→ Ba2++ SO2↑ +H2O
Zinc and sulphuric acid test: Hydrogen sulphide gas is evolved.
SO32– + 3Zn2+ + 8H+ → H2S + 3Zn2++3H2O
Lime water test: A white turbidity is formed.
Ca(OH)2+ SO2 → CaSO3(milky) + H2O
The precipitate dissolves on prolonged passage of the gas, due to the formation of hydrogen sulphite ions.
CaSO3 + SO2 + H2O → Ca(HSO3)2
Try yourself:
2. How can you remove SO2 from a mixture of CO2 and SO2?
Ans: The dichromate oxidises and destroys the sulphur dioxide without affecting the carbon dioxide
Lead Acetate or lead Nitrate Solution: White precipitate of PbSO3 is obtained.
SO32–+ Pb2+ → PbSO3↓
White precipitate gets soluble in dil. HNO3 on boiling. The precipitate is oxidised by atmospheric oxygen and PbSO 4 is formed.
2PbSO3+ O2→ 2PbSO4 ↓
Sulphide Ion (S2–)
Confirmation test for S2– ion with the help of dilute acid, lead acetate, sodium nitroprusside, so on etc.,
Dilute acid test: Pungent smelling gas like that of rotten egg is obtained.
S2– + 2H+→ H2S ↑
Lead acetate test: Filter paper moistened with lead acetate solution turns black.
(CH 3 COO) 2 Pb + H 2 S → PbS ↓ (black) + 2CH3COOH.
Sodium nitroprusside test: Purple colouration is obtained.
S2–+[Fe(CN)5(NO)]2–→ [Fe(CN)5NOS]4–
It is a ligand change reaction not a redox reaction. (NO+ changes to (NOS)–1)
Cadmium carbonate suspension/cadmium acetate solution: Yellow precipitate is formed.
Na2S + CdCO3 → CdS + Na2CO3
When filter paper moistened with cadmium acetate is brought in contact with evolving gas, it turns yellow.
S2–+2H+ → H2S ;
H2S+Cd2+→ CdS → +2H+
Silver nitrate solution: Black precipitate is formed, which is insoluble in cold, but soluble in hot, dilute nitric acid.
Ag++ S2–→ Ag2S
Nitrite Ion (NO2¯)
Confirmation test forNO2– ion with the help of dilute H2SO4, starch iodide, ferrous sulphate, etc.
Dilute H 2 SO 4 test: Solid nitrite in cold produces a transient pale blue liquid (due to the presence of free nitrous acid, HNO 2 or its anhydride, N2O3) first and then liberates pungent smelling reddish brown vapours of NO2. The pungent smell gas is liberated due to reaction of NO with O 2 of air.
Starch-iodide test: The addition of a nitrite solution to a solution of potassium iodide, followed by acidification with acetic acid or with dilute sulphuric acid, results in the liberation of iodine, which may be identified by the blue colour produced with starch paste. A similar result is obtained by dipping potassium iodide–starch paper moistened with a little dilute acid into the solution.
Starch+I3– Blue (starch-iodine adsorption complex)
Ferrous sulphate test (Brown ring test): When the nitrite solution is added carefully to a concentrated solution of iron(II) sulphate
acidified with dilute acetic acid or dilute sulphuric acid, a brown ring appears due to the formation of [Fe(H 2 O) 5 NO]SO 4 at the junction of the two liquids. If the addition has not been made slowly and cautiously, a brown colouration results.
NO2–+CH3COOH→ HNO2+CH3COO–→ 3 H
NO2→ H2O+HNO3+2NO ↑
Fe2++ SO42–+NO ↑ [Fe, NO]SO4
Acidified potassium permanganate solution: Pink colour of KMnO 4 is decolourised by a solution of a nitrite, but no gas is evolved.
5 NO2–+2 MnO4–+6H+→5NO3–+2Mn2++3H2O
Silver nitrate solution: White crystalline precipitate of silver nitrite from concentrated solutions. NO2–+ Ag+ → AgNO2↑
Acetate Ion (CH3COO¯)
Confirmation test for Acetate (CH3COO – ) ions is with the help of dilute H2SO4, FeCl3, silver nitrate.
Dilute H 2 SO 4 test: A vinegar like smell is obtained.
(CH3COO)2Ca+H2SO4→ 2CH3COOH+CaSO4
Neutral ferric chloride test: A deep red/ blood red colouration (no precipitate) indicates the presence of acetate.
6CH3COO–+3Fe3++2H2O → [Fe3(OH)2(CH3COO)6]++2H+
When solution is diluted with water and boiled, brownish red precipitate of basic iron (III) acetate is obtained.
Silver nitrate solution test: A white crystalline precipitate is produced in concentrated solution in the cold.
CH3COO– + Ag+ CH3COOAg ↓
Precipitate is more soluble in boiling water and readily soluble in dilute ammonia solution.
Try yourself
3. The solution obtained after dissolving in ammonia is immediately discarded why? Answer: The solution obtained after dissolving in ammonia is immediately discarded to avoid serious explosion..
Barium chloride test: No change in the presence of acetates of Ba 2+,Ca +2, and Hg +2 chloride solution
Group B Radicals
Gases or acid vapours evolved with conc. Sulphuric acid it includes Cl –, Br–, I–.
Chloride Ion (Cl¯)
Confirmation test for chloride ion(Cl–)is with the help of concentrated H2SO4, MnO2, AgCl, Na3AsO3 so on etc.,
Concentrated H2SO4 test: colourless pungent smelling gas is evolved, which gives white dense fumes of NH4Cl when a glass rod dipped in aq. Ammonia is brought in contact with evolving gas.
Cl–+H2SO4→ HCl+HSO4–
NH4OH+HCl→ NH4Cl↑ (white fumes)+H2O.
Manganese dioxide test: Heat chloride salt with small quantity of MnO2 and concentrated sulphuric acid. Greenish-yellow pungent smelling gas, turns moist starch iodide paper blue.
Silver nitrate test: White, curdy precipitate of AgCl is formed. The white precipitate
■ Insoluble in water and in dilute HNO 3.
■ Soluble in dilute ammonia solution, KCN, and in Na2S2O3solutions.
Cl–+Ag+→ AgCl(white)
White precipitate is soluble in aqueous ammonia and precipitate re-appears with HNO3.
AgCl+2NH4OH → [Ag(NH3)2]Cl (Soluble) + 2H2O
[Ag(NH3)2]Cl+2H+→ AgCl ↓ +2NH4+.
Sodium arsenite test: The white precipitate after washing with distilled water is shaken with Na3AsO3 converted into yellow precipitate.
This test is used to distinguish Cl – from Br– and I–.
3AgCl + AsO3 3– → Ag3AsO3↓ +3Cl–
Chromyl chloride test: Covalent chlorides do not give this test.
Example: Hg 2 Cl 2 , HgCl 2 , SnCl 2 , AgCl, PbCl2 SbCl3, CuCl, etc., as they are partially dissociated. This test is given generally by ionic chlorides or water soluble chlorides.
■ On heating salt with K 2Cr 2O 7 and conc. H2SO4 orange vapours of chromyl chloride (CrO2Cl2) is liberated.
■ The orange vapours, on passing through NaOH solution, give yellow solution. (Na2CrO4).
■ Acidified solution of Na2CrO4 gives yellow precipitate with lead acetate solution.
4Cl–(s)+Cr2O72–(s)+6H+(conc.) → 2CrO2Cl2(deep red vapours)+3H 2O
CrO2Cl2+4OH–→ CrO42–+2Cl–+2H2O
CrO42–+Pb+2 → PbCrO4→ (yellow)
■ Test should be carried out in a dry test tube otherwise chromic acid will be formed.
CrO2Cl2+2H2O → H2CrO4+2HCl
■ Some chlorine may also be liberated, this decreases the sensitivity of the test.
6Cl–+Cr2O7-2 +14H+ →3Cl2 +2Cr3+ +7H2O
Bromide Ion (Br¯)
Confirmation of test for bromide (Br –)ion is with the help of Concentrated H2SO4, MnO2, AgNO3, etc.
Concentrated H2SO4 test : A reddish-brown solution is formed, then reddish-brown bromine vapour accompanies the hydrogen bromide (fuming in moist air) is evolved.
2NaBr+H2SO4→ Na2SO4+2HBr
2HBr+H2SO4→ Br2↑ +2H2O+SO2↑
Manganese dioxide test: When a mixture of solid bromide, MnO2, and sulphuric acid is heated reddish brown vapour of bromine is evolved
Bromine is recognised by:
■ its powerful irritating odour
■ its bleaching of litmus paper
■ its staining of starch paper orange red
2KBr+MnO2+2H2SO4→ Br2↑+K2SO4+ MnSO4 + 2H2O
Silver nitrate test: Pale yellow precipitate is formed.
NaBr + AgNO3→ AgBr → +NaNO3
Yellow precipitate insoluble in water and in dilute HNO 3. Yellow precipitate soluble in dilute ammonia solution, KCN, and in Na2S2O3solutions
AgBr+2NH4OH → [Ag(NH3)2]Br+H2O
Lead acetate test: Bromides on treatment with lead acetate solution, give a white crystalline precipitate of lead bromide, soluble in boiling water giving colourless solution.
2Br–+Pb+2 → PbBr2↓
Chlorine water test (Organic layer test): When to a sodium carbonate extract of metal bromide containing CCl4, CHCl3 or CS2, chlorine water is added and the content is shaken and then allowed to settle down reddish brown colour is obtained in organic layer.
2Br–+Cl2→ 2Cl–+Br2↑
Br2+CHCl3/CCl4→Br2 dissolve to give reddish brown colour in organic layer.
With excess of chlorine water, the bromine is converted into yellow bromine monochloride and a pale yellow solution results.
Br2↑ +Cl2 ↑ 2BrCl
This test is used to distinguish Br 2 from I2
Starch paper test: When starch paper is brought in contact with evolving bromine gas orange red spots are produced.
Potassium dichromate and concentrated H2SO4 test: When a mixture of solid bromide, K2Cr2O7 and concentrated H2SO4 is heated and evolved vapours are passed through water, a yellowish- brown solution, containing free bromine but no chromium is obtained.
Confirmation Iodide(I–) ion is with the help of Concentrated H2SO4, Starch paper, AgNO3, so on etc.,
Concentrated H2SO4 test: Pungent smelling violet vapours of I2 are evolved.
2NaI+H2SO4→ Na2SO4+2HI
2HI+H2SO4→I2↑(dark violet)+2H2O+SO2
Evolution of dark violet fumes intensifies on adding a pinch of MnO 2
3I–+MnO2+ 2H2SO4 → I 3–+Mn2++2SO42–+ 2H2S
Starch paper test: Iodides are readily oxidised in acid solution to free iodine; the free iodine may than be identified by deep blue colouration produced with starch solution.
3I– + 2NO2– +4H+→ I 3–+ 2NO ↑ +2H2O.
Silver nitrate test: Yellow, curd-like precipitate of AgI is formed.
I– + Ag+ → AgI ↓
Yellow Precipitate is insoluble in water and in dilute HNO 3 . Yellow Precipitate very slightly soluble in concentrated ammonia solution. Readily soluble in KCN, and in Na2S2O3solutions
Chlorine water test (Organic layer test): When chlorine water is added to a solution of iodide, free iodine is liberated which colours the solution brown and on shaking with CS 2, CHCl 3 or CCl 4, it dissolves in organic layer forming a violet solution, which settles below the aqueous layer.
2NaI + Cl2 → 2NaCl+ I2
I2+CHCl3→ I2 dissolves to give violet colour in organic layer.
If excess chlorine water is added, the iodine is oxidised to colourless iodic acid.
I 3–+8Cl2↑ +9H2O → 3IO3–+16Cl–+18H+
Lead acetate solution: A yellow precipitate is formed which is soluble in hot water forming a colourless solution and yielding golden yellow plates (‘spangles’) on cooling.
2I–+Pb+2→ PbI2 ↓
Potassium dichromate and concentrated sulphuric acid: Violet vapours are liberated, and no chromate is present in distillate.
6I–+Cr2O72– + 2H2SO4 → 3I2 + Cr3+ +7SO42– + 7H2O
Copper sulphate solution test: Brown precipitate containing mixture of CuI and I 2 is formed, free iodine removed by addition of sodium thiosulphate solution or sulphurous acid, and a nearly white precipitate of copper iodine is obtained.
5I–+2Cu2+→ 2CuI ↓ +I 3 –
I 3–+2S2O32–→ 3I–+S4O62–
Nitrate Ion (NO3¯) :
Confirmation test for Nitrate(NO3–) ion is with the help of Concentrated H2SO4, Brown ring.
Concentrated H2SO4 test: Pungent smelling reddish brown vapours are evolved.
4NO3–+2H2SO4→ 4NO2 ↑+ O2+2SO42– + 2H2O
Addition of bright copper turnings or paper pellets intensifies the evolution of reddish brown gas.
2NO3–+4H2SO4+3Cu→3Cu2++2NO↑+ 4SO42–+ 4H2O ;
2NO ↑ +O2→ 2NO2 ↑
4C(paper pellet)+4HNO 3 → 2H 2 O+4NO 2 + 4CO2↑
Brown ring test: When a freshly prepared saturated solution of iron (II) sulphate is added to nitrate solution and then concentrated H2SO4 is added slowly from the side of the test tube, a brown ring is obtained at the junction of two layers.
On shaking and warming the mixture, NO escapes and a yellow solution of iron(III) ions is obtained.
Bromides and iodides interfere in brown ring test as liberated halogens obscure the brown ring. Nitrites also interfere the brown ring test and can be removed by adding a little sulphuric acid, or urea.
Group
C Radicals
These anions does not give any gas with dilute as well as concentrated H2SO4 in cold but give precipitate with certain reagents. These acid radicals are identified in inorganic salts by their individual tests as given below
Sulphate Ion (SO42–):
Confirmation test for Sulphate (SO42–)ion is with the help of Barium chloride, Lead acetate, silver nitrate so on etc.,
Barium chloride test: White precipitate is obtained when barium chloride is added to water extract containing sulphate ion
Na2SO4+BaCl2 → BaSO4 ↓ (white)+2NaCl.
White precipitate is insoluble in warm dil. HNO3 as well as HCl but moderately soluble in boiling concentrated hydrochloric acid and conc. H2SO4.
Lead acetate test: White precipitate is obtained when lead acetate is added to water extract containing sulphate ion
Na2SO4+(CH3COO)2Pb → PbSO4↓ (White) + 2CH3COONa
White precipitate soluble in excess of hot ammonium acetate and ammonium tartrate.
Cation are classified into field into different groups based on the values of their solubility product.
Basic Radicals
The basic radicals have been classified into different groups based on the values of their solubility product.
Salts with K SP values precipitates earlier so their presence is checked earlier and they fall in lower groups.
The group radicals,group reagent, and precipitate colour is given in table 7.5
■ K sp Values of chlorides of group I are low so they are precipitated while chlorides of other metals are not precipitated.
■ Depending upon if sulphides of group II are soluble in yellow ammonium sulphide (NH4)2Sx, or not they fall in group II B or Il A, respectively.
■ Ionization of H,S is small in acidic medium and more in alkaline medium. The K sp values of sulphides of group II elements is low while that of group IV is high hence group II is precipitated in acidic medium and group IV in alkaline medium.
■ NH4Cl is added to NH4OH to precipitate group III hydroxides. As their K sp values are less, the common ion NH4+from NH4Cl suppress the dissociation of NH4OH hence precipitating group III.
■ Common ion concept is applied to group V as at higher concentration of CO 3 2, MgCO3will precipitate.
I
II A. Hg2+, Cu+2,Cd+2, Pb+2, Bi+3 (Insoluble in yellow ammonium sulphide) H2S gas in presence of dil. HCl HgS (black), CuS (black), CdS(yellow), PbS(black), Bi2S3(black), As2S3(yellow)
II B As3+, Sb3+, Sn²+, Sn4+ (soluble in yellow ammonium sulphide)
Sb2S3(orange), SnS( brown) SnS2(yellow),
III Al3+, Cr3+,Fe3+ NH4OH in presence of NH4Cl Al(OH)3 (Gelatinous white ) Cr(OH)3(Green) Fe(OH)3(Reddish brown)
IV Zn2+, Ni2+, Mn2+, Co2+ H2S + NH4OH +NH4Cl ZnS(white), MnS(Buff or Pink), NiS(black),CoS(black)
V Ca+2, Sr+2 ,Ba+2 (NH4)2CO3+NH4OH CaCO3(white), SrCO3(white), BaCO3(white)
VI Mg+2 Na2HPO4+NH4OH Mg(NH4)PO4(whte)
0(Zero) NH4+ Tested indenpendently
Table 7.5 Group Radicals, Reagents and precipitate colour
Group I
Dil. HCl is added to the aqueous solution, if no precipititate, I group cations are absent.
Pb2+
Pb2+
Pb gives yellow ppt with K 2CrO4 and KI
PbCl2 + K2CrO4→ PbCrO4↓ (yellow)+2KCl
PbCrO4 + 4NaOH→ Na2PbO2+Na2CrO4+2H2O
PbCl2+2KI → PbI2↓ (yellow)+2KCl
(Soluble in hot water and reappears on cooling as golden spangles).
Sn4+ First reduced by Al to Sn+2 which gives test with HgCl2 3SnCl4+2Al →2 AlCl3+3SnCl2
Precipitate of group II A cations are insoluble in yellow ammonia sulphide.
Group II B
Precipitate of these cations are soluble in yellow ammonium sulphide.
The confirmation test of this cations are given in table 7.8
Group III
Filtrate of group ll
Precipitate
Composition of precipitate and its colour is given in Table 7.9
Table 7.9 Group III cation precipitate colours
Cation Salt colour Composition and colour of precipitate
Fe3+ Brown Fe(OH)3, Reddish Brown
Cr3+ Green Cr(OH)3, Dirty green
Al3+ Colourless Al(OH)3, Gelatinous white
Confirmation test of group III cations are given in table 7.10
Group IV
Filtrate of group III +
composition of precipitate and its colour is given in Table 7.11
Confirmation test of group IV cations are given in table 7.12
Group V
Filtrate of group IV
Flame test of white precipitate:
■ Green flame ---- Ba2+
■ Crimson red ----Sr2+
■ Brick red------Ca2+
The cation are tested in the order of Ba 2+ , Sr2+, Ca2+.
Confirmation test of group V cations are given in table 7.13
Group VI
Whiteprecipitate
Mg2++(NH4)2HPO4+NH4OH
→ MgNH4PO4↓ white + 2NH4++H2O
The precipitate is sparingly soluble in water, soluble in acetic acid and in mineral acids
Table 7.10 Group III cation confirmation and reactions
Cation Test
Brown precipitate, Fe(OH)3 is insoluble in NaOH but soluble in HCl
Fe3+ion in solution give blood red colour with KSCN
Fe3+
Cr3+
Prussian blue colour is obtained on addition of K4[Fe(CN)6] to soluble Fe3+ salts
Green precipitate, Cr(OH)3, is soluble in (NaOH+Br2 water) or Na2O2 to give yellow coloured Na 2CrO4, which gives yellow precipitate with lead acetate
Reaction
Fe(OH)3+3HCl → FeCl3+3H2O
FeCl3+3KSCN→Fe(SCN)3↓(blood red)
2FeCl3+3K4[Fe(CN)6] → K Fe[Fe(CN)6] ↓ (Prussian blue)
Br2+H2O → 2HBr+[O]
2Cr(OH)3+ 4NaOH+3[O] → 2Na2CrO4+5H2O
Na2CrO4+(CH3COO)2Pb → PbCrO4↓ (yellow)+2CH3COONa
Al3+
White gelatinous precipitate of Al(OH) 3 is solouble in NaOH.
The precipitate reappears on boiling with NH 4Cl
Table 7.11 Group IV cation precipitate colours Cation
Zn2+ colourless ZnS, white
Mn2+ light pink MnS, buff coloured
Table 7.12 Group IV cation confirmation and reactions
Cation
White precipitate ZnS soluble in dilute HCl
Zn2+
Mn2+
Ni2+
Al(OH)3+NaOH → NaAlO2+H2O
NaAlO2+NH4Cl+H2O → Al(OH)3+NaCl+NH3
and colour of precipitate
Ni2+ green NiS, black
Co2+ blackish brown CoS, black
Soluble salt gives white precipitate with NaOH which dissolves in excess of NaOH
Bluish white precipitate is obtained when the salt reacts withK4[Fe(CN)6]
Buff coloured precipitate soluble in dil.HCl
With NaOH solution initially white precipitate is form which is insoluble in excess of NaOH but rapidly oxidised on exposure to air, turning brown
Black precipitate soluble in mixture of concentrated HCl + KClO3. The mixture acts as oxidising agent
When heated with NaHCO3+Br2 water gives black precipitate
ZnS+2HCl → ZnCl2+H2S
ZnCl2+2NaOH → Zn(OH)2↓ +2NaCl
Zn(OH)2+2NaOH → Na2ZnO2+H2O
2ZnCl2+K4[Fe(CN)6] → Zn2 [Fe(CN)6] ↓ +2KCl
MnS+2HCl → MnCl2+H2S
MnCl2+2OH–→ Mn(OH)2↓
Mn(OH)2↓ + O2→ 2MnO(OH)2↓
NiS+2HCl+[O] → NiCl2+H2O+S
NiCl2+2NaHCO3→ NiCO3+2NaCl+H2O+CO2
2NiCO3+ [O] → Ni2O3↓ black +2CO2
Cation Test Reaction
Cherry red precipitate obtained when the alkaline or acid solution buffered with CH 3COONa reacts with DMG dimethyl glyoxime
Black precipitate soluble in HCl+KClO 3
Co2+
Neutral or acid solution of salt gives blue colour with NH4CNS
Yellow precipitate is formed when KNO2+CH3COOH reacts with neutral solution salt
Table 7.13 Confirmation test of group V cations Reagent
K2CrO4 yellow precipitate of BaCrO 4. Precipitate is insoluble in dilute acetic acid but readily soluble in mineral acids
(NH4)2SO4 White precipitate of BaSO4 insoluble in water, dilute acid, and in (NH4)2SO4 solution but soluble in boiling concentrated H2SO4
(NH4)2C2O4 White precipitate of BaC2O4 slightly soluble in water readily soluble in hot dilute acetic acid and mineral acids
Group VI
4 24 NHOH NaHPO Whiteprecipitate
Mg2++(NH4)2HPO4+NH4OH → MgNH4PO4↓ white + 2NH4++H2O
NiCl2+2NH4OH+2(CH3C=NOH)2
→( C 4H 7N 2O 2) 2N i ↓ +2NH4Cl+2H2O
CoS+2HCl+[O] → CoCl2+H2O+S
Co2++ 4SCN–→ [Co(SCN)4]2–
Co2++7NO2–+2H++3K+→ K3[Co(NO2)6] ↓ +NO ↑ +H2O
yellow precipitate of SrSO 4 is formed soluble in acetic acid and in mineral acid
White precipitate of SrSO4 is formed considerably soluble in water, slightly soluble in boiling HCl
White precipitate of SrC2O4 sparingly soluble in water. Acetic acid does not attack white precipitate, but dissolves in mineral acids
The precipitate is sparingly soluble in water, soluble in acetic acid and in mineral acids
A white flocculent precipitate of magnesium hydrogen phosphate is obtained in neutral solutions.
Mg2++HPO42–→ MgHPO4↓
A white gelatinous precipitate is formed with ammonium solution
Mg2++2NH4OH → Mg(OH)2↓ +2NH4+
No precipitate from dilute solution nor concentrated solution in presents of acetic acid
White precipitate soluble in water and dissolve in hot concentrated sulphuric acid
White precipitate CaC2O4 insoluble in water, acetic acid but readily soluble in mineral acids
Group Zero (Ammonium Ion)
Original salt +NaOH ∆ → smell of NH 3 ↓ HCl
■ The gas can be identified by the following characteristics/reactions NH4Cl+NaOH→ NH3+H2O+NaCl
■ The evolution of the white fumes of ammonium chloride when a glass rod dipped in dilute HCl is held in the vapour. NH3+HCl → NH4Cl ↑ (white fumes)
■ Its ability to turn filter paper moistened with CuSO 4 solution deep blue. CuSO4+4NH3→ [Cu(NH3)4]SO4
■ Its ability to turn filter paper moistened with Hg2(NO3)2 solution black.
Hg 2 (NO 3 ) 2 +2NH 3 → Hg(NH 2 )
NO3+Hg+NH4NO3
Nessler's Reagent Test
■ Alkaline solution of potassium tetra iodido mercurate (II) is Nessler's reagent.
■ Based on amount of NH 3 or NH 4 + ions brown precipitate or brown or yellow colour is obtained
■ The precipitate is known as Iodide of Millon's base.[basic mercury (II) amido iodide ]
NH 4 + +2[HgI 4 ] 2– +4OH → HgOHg(NH 2 )
I ↓ +7I–+3H2O
TEST YOURSELF
1. A brick red colour is imparted to Bunsen flame by a
(1) Ca salt (2) Sr salt
(3) Na salt (4) Co salt
2. Which of the following gives a suffocating gas when treated with dilute HCl –?
(1) Carbonate (2) Sulphite
(3) Sulphate (4) Borate
3. The carbonate of which of the following cations is insoluble in water?
(1) Na+ (2) K+
(3) NH+ (4) Ca2+
4. A mixture, when rubbed with dilute acid, smells like vinegar. It contains (1) sulphite (2) nitrate (3) nitrite (4) acetate
5. A mixture upon adding conc.H2SO4 gives deep red fumes. It may contain the anion pair
(1) Cr2O72– and Cl–(2) Br– and Cr2O72–(3) NO3– and Cl–(4) CrO42– and NO32–
6. A colourless solution of a compound gives a precipitate with AgNO 3 solution but
no precipitate with a solution of Na 2CO 3. The action of concentrated H 2SO 4 on the compound liberates a suffocating reddish brown gas. The compound is :
(1) Ba(CH3COO)2 (2) CaCl2
(3) Nal (4) NaBr
7. Which of the following statements is/are incorrect?
(1) A filter paper moistened with cadmium acetate solution turns yellow, when brought in contact with H 2S gas.
(2) Both carbonate ions as well as bicarbonate ions in the solutions, give reddish-brown precipitate with mercury(II) chloride.
(3) Sulphites in presence of zinc, reacts with diluteH2SO4 to liberate SO2 gas.
(4) A filter paper moistened with KIO3 and starch turns blue in contact with SO 2 vapours.
8. Nitrate is confirmed by ring test. The brown colour of the ring is due to formation of
(1) ferrous nitrite
(2) nitroso ferrous sulphate
(3) ferrous nitrate
(4) FeSO4.NO2.
9. For the test of sulphite, the soda extract is acidified with -
(1) dil HCl (2) dil HNO3
(3) CH3COOH (4) None of these
10. Ammonium salts on heating with slaked lime liberates a colourless gas (X). Identify the correct statement for gas (X).
(1) (X) turns red litmus blue and produces dense white fumes in contact with dilute HCl.
(2) (X) turns filter paper moistened with mercurous nitrate black and gives intense blue coloured solution with CuSO4(aq).
(3) (X) when passed through Nessler's reagent produces a brown colour precipitate.
(4) All the observations are true
11. A white crystalline substance dissolves in wa ter. On passing H 2S in this solution, a black precipitate is obtained. The black precipitate dissolves completely in hot HNO3 On adding a few drops of concentrated H2SO4, a white precipitate is obtained. This precipitate is that of (1) BaSO4 (2) SrSO4 (3) PbSO4 (4) CdSO4
12. A black sulphide is formed by the action of H2S on (1) cupric chloride (2) cadmium chloride (3) zinc chloride (4) ferric chloride
13. Intense blue precipitate of Fe 4[Fe(CN) 6] 3 and potassium hydroxide solution, when mixed, gives (1) K2Fe[Fe(CN)6]−white precipitate (2) Fe(OH)3– reddish–brown precipitate (3) Fe(CN)3– reddish–brown precipitate (4) KFe[Fe(CN6)]– Turn bull's blue
14. An aqueous solution of colourless metal sulphate M gives a white precipitate with NH 4 OH. This was soluble in excess of NH 4 OH. On passing H 2 S through this solution, a white precipitate is formed. The metal M in the salt is (1) Ca (2) Ba (3) Al (4) Zn
15. Aqueous solution of BaBr 2 gives yellow precipitate with I. K2CrO4 II. AgNO3 III. (CH3COO)2Pb (1) I and II only (2) II and III only (3) I and III inly (4) I, II, and III
16. Na2CO3 cannot be used in place of (NH 4)2 CO3 for the precipitation of V group because (1) Na+ interferes in the detection of V group (2) conc. of CO3 2− is very low (3) Na will react with acid radicals (4) Mg will be precipitated
The properties of organic compounds are the properties of the functional group present in it. The detection of functional group is most important part in the identification of a given unknown organic compound. One or more than one type of functional group may be present in the given organic compound.
7.2.1 Unsaturation
Unsaturation in the molecule can be detected by the following reactions :
Absorption of hydrogen: Absorption of one mole of hydrogen per mole of the compound indicates the presence of one mole of p-bond. Absorption of 'n' moles hydrogen per mole of the compound indicates the presence of 'n' mole of p -bonds.
Bromine water test: Unsaturated compounds discharge the colour of Br 2/CCl4, or bromine water.
The colour is discharged due to formation of vicinal dihalide or bromo hydrin.
Absorption of one mole of bromine per mole of the compound indicates the presence of one mole of p -bond.
■ If compound is aromatic having oxygen and gives light yellow precipitate with bromine water, then compound is Phenolic compound or salicylic acid.
■ If compound is aromatic amine then it gives precipitate with bromine water.
Bayer's reagent test: Cold dilute alkaline KMnO4 is Bayer's reagent.
Unsaturated compounds discharge pink colour of Bayer's reagent to brown. The brown precipitate is due to formation of MnO 2
CuCl test: Terminal alkyne gives red precipitate with ammonical cuprous chloride (Cu2Cl2+NH4OH) solution. This test is used to distinguish between terminal alkynes from internal alkynes, alkenes, and alkanes.
Ammonical silver nitrate test: Ammonical silver nitrate is known as Tollen's reagent. Terminal alkyne gives white precipitate with Tollen's reagent.
This test is used to distinguish between terminal alkynes from internal alkynes, alkenes, and alkanes.
Ozonolysis: Ozonolysis reaction indicates the number and position of double bonds and triple bonds in the unsaturated compound.
7.2.2 Hydroxyl Group
Hydroxyl group can be detected by the following reactions
Litmus test: Alcohols being neutral, do not change colour of moist blue or red litmus paper.
Reaction with active (Na) metals: Alcohols liberates hydrogen gas on treatment with sodium metal
Ester test:Alcohols on acetylation gives esters
MWof acetyl derivitative- MW of alcohol
Numberofhydroxylgroups= 42
■ Alcohols undergoes methylation when reacted with alkaline (CH 3)2SO4
MWof methyl derivitative- MW of compound
Numberofmethylgroups= 14
Ceric nitrate test: Alcoholic compounds on reaction with ceric ammonium nitrate give a red colouration due to the formation of a complex.
Lucas Test: It is used to distinguish between primary, secondary, and tertiary alcohols. This test is based on stability of intermediate carbocation
Lucas reagent is mixture of anhydrous ZnCl 2 and concentrated HCl. The alkyl chlorides produce in the reaction are insoluble, and gives turbidity, The time required for the turbidity to appear at room temperature decides the degree of alcohol.
■ Turbidity appears immediately, the alcohol is tertiary
■ The turbidity appears after five minutes or on heating, the alcohol is secondary.
■ The turbidity does not appear even after 15 minutes, the alcohol is primary.
■ Some primary alcohols gives this test because the primary carbocation formed is highly stable.
Iodoform test: Alcohols containing CH 3 CH(OH)R, R= H or alkyl group gives yellow precipitate with NaOH+I 2.
7.3.3 Phenolic (Ar-OH) Group
OH group attached directly to carbon of benzene ring is called phenolic –OH group.
Litmus test: Phenols turns blue litmus paper red.
Ferric chloride test: Phenols gives violet colour when treated with few drops of dilute aqueous FeCl3. Colour may be violet, blue, green or red depending on structure of phenol
6C6H5OH+FeCl3→[Fe(OC6H5)6]3–+3H++3HCl
Phthalein dye test: Phenol when heated with phthalic anhydride in presence of concentrated with H 2 SO 4 forms phenolphthalein which develops pink, blue, green, red etc., colours in alkaline solution
7.2.4 Aldehydic And Ketonic Groups
Aldehydes and ketones contain carbonyl group (>C = O) and are commonly known as carbonyl compounds.
These are identified by two important reactions:
■ Addition reaction on double bond of > C = O group
■ Oxidation of carbonyl group
Test given by both aldehyde's and ketones: 2,4-dinitrophenylhydrazine(2,4-DNP)
2,4-DNP is known as Brady's reagent. Aldehydes and ketones, when heated with solution of 2,4-DNP in ethanol containing small amount of H2SO4 or HCl, forms yellow, orange, or red coloured solids. If precipitate does not appear at room temperature, warm the mixture in a water bath for a few minutes and cool.
Tollen's test: Ammonical solution of silver nitrate in which silver ion is present as the ammonia complex.[Ag(NH 3)2]+
When aldehyde is heated with Tollens's reagent silver deposits on the inner walls of the test tube in the form of a mirror and aldehyde is oxidised to coresponding carboxylate ion.
■ a -hydroxy ketones, formate ion, some a-hydroxy and a-keto acids gives this test.
Fehling's test: Alkaline solution of cupric ion complexed with tartrate ion.
It is mixture of equal volume of Fehling solution I and Fehling solution II
Fehling solution I: ---aqueous CuSO 4 solution
Fehling solution II: ---Alkaline solution of sodium potassium tartarate
T his test is given by only aliphatic aldehydes.
When aliphatic aldehyde is heated with Fehling solutions red precipitate of cuprous is formed.
■ a -hydroxy ketones also give positive test.
Benedict's test: Benedict solution is alkaline solution of cupric ion complex with citric ion.
Aliphatic aldehydes reduce Benedict solution to produce a red precipitate of cupric oxide. Aromatic aldehydes do not reduce aromatic aldehydes.
Both Fehling solution and Benedict solution are deep blue coloured solutions.
Try yourself:
4. What is a significance of complexation in Fehling as well as Benedict test?
Ans: Complex formation decreases concentration the cupric ion preventing precipitation of cupric hydroxide
Schiff’s test: Aldehydes gives pink colour with Schiff’s reagent.
The reagent is prepared by decolourising aqueous solution of p–rosaniline hydrochloride dye by adding sodium sulphite or by passing SO2 gas.
Monovalent Cation(X+) Trivalent Cation(M+3)
Na+,K+,Rb+, Cs+,Tl+, Ag+, NH4+etc. Al+3, Cr+3, Fe+3, Co+3 etc.
7.2.5 Carboxylic Group
Organic compounds containing carboxyl functional groups are called carboxylic acids.
Litmus test: Carboxylic acids turn blue litmus red.
If the blue colour of the litmus paper changes to red, the presence of either a carboxylic group or a phenolic group is indicated
Sodium hydrogencarbonate test: Carboxylic acids liberates effervescence of CO2 indicates the presence of carboxyl group.
Phenols does not give this test because phenol is less acidic than carbonic acid
■ Picric acid, sulphonic acid, alkyl hydrogen sulphates, and certain enols liberate carbondioxide from NaHCO 3
Ester test: Add ethanol or methanol and 2-3 drops of concentrated sulphuric acid. Heat the reaction mixture for 10-15 minutes in a hot water bath at about 50°C. Pour the reaction mixture in a beaker containing aqueous sodium carbonate solution Sweet smell of the substance formed indicates the presence of carboxylic functional group.
Try yourself:
5. Why the mixture of carboxylic acid and alcohol after heating is poured in the beaker containing aqueous sodium carbonate solution?
Ans: To neutralise excess sulphuric acid and excess carboxylic acid
7.2.6 Amino Group
Organic compounds containing amino group are basic in nature. Both, aliphatic and aromatic amines are classified into three classes namely– primary(–NH2), secondary(NH-) and tertiary (-N<), depending upon the number of hydrogen atoms attached to the nitrogen atom. Primary amine has two hydrogen atoms, secondary has one while tertiary amine has no hydrogen atom attached to nitrogen.
Solubility test: These compounds easily react with acids to form salts, which are soluble in water.
C6H5NH2 + HCl → C6H5N+H3 Cl–
Carbylamine test: Aliphatic as well as aromatic primary amines give carbylamine test in which an amine is heated with chloroform.
In this reaaction nucleophile is primary amine and electrophile is: CCl 2
Azo dye test: Aromatic primary amines can be confirmed by azo dye test. Primary amine e.g. aniline reacts with nitrous acid generated in situ by the reaction of sodium nitrite with HCl at 0–5°C to produce diazonium salt. This couples with b -naphthol to give a scarlet red dye, which is sparingly soluble in water.
β-Naphthol azo-dye (Scarlet red)
Try yourself:
6. During diazotisation reaction the temperature is maintained always below 5°C why?
Ans: At higher temperatures diazonium salts are unstable and decomposes to liberate nitrogen gas
Try yourself:
7. Why is diazonium chloride solution always added into the alkaline solution of b -naphthol and not vice-versa?
Ans: When b -naphthol is added to diazonium chloride solution, diazotisation takes place in more than one position, giving mixture of products. To prevent this, diazonium chloride solution is added to alkaline solution of b -naphthol.
TEST YOURSELF
1. The organic compound that gives following qualitative analysis is
Test
Inference
(a) Dil.HCl Insoluble
(b) NaOH solution Soluble
(c) Br2/water Decolonisation
(1) OH (2)
(3) OH (4)
2. An organic compound A(C6H6O) gives dark green colouration with ferric chloride. On treatment with CHCl3 and KOH, followed by acidification, it gives compound B. Compound B can also be obtained from compound C on reaction with pyridinium chlorochromate (PCC). Identify A, B, and C.
(1)
A = B = C =
(2) A = B = C =
(3) A = B =
(4) A = B =
3. Tollen’s reagent is (1) ammonical cuprous chloride
5. Schiff’s reagent is (1) resorcinol solution in dilute HCl (2) rosaniline solution in water decolourised by SO2
(3) alkane CuSO 4 stabilised by Roschelle salt
(4) ammonical silver nitrate
6. Acetone and acetaldehyde are differentiated by (1) HNO2 (2) I2 (3) NaOH + I2 (4) [Ag(NH3)2]+
7. O || 65 3 CHCCH and O || 6565 CHCCH are distinguished by (1) Iodoform test (2) Tollen’s reagent (3) Fehling solution
(4) Benedict's solution
8. O 333 || CHCHOandCHC—CH are distinguished by (a) Tollen’s reagent (b) Fehling solution (c) 2,4-DNP (d) NaHSO3
(1) a and b only
(2) a, b and d only (3) b and c only (4) a, b, c, and d
9. Which of the following amine, will give the carbylamine test?
(1) NHCH3 (2) N(CH3)2
(3)
10. A positive carbylamine test is given by
(1) N, N- dimethyl aniline
(2) 2,4 dimethyl aniline
(3) N- methyl-o-methylaniline
(4) P- methyl benzylamine
11. An unknown organic compound (X) liberates H 2 gas with Na metal, gives effervescence of CO 2 with Na 2 CO 3 , and gives a fruity smelling product when it reacts with alcohol.
Compound (X) is
(1) picric acid
(2) ethyl alcohol
(3) ethanoic acid
(4) phenol
ANSWER KEY
(1) 3 (2) 1 (3) 4 (4) 2
(5) 2 (6) 4 (7) 1 (8) 3
(9) 1 (10) 4 (11) 3
7.3 CHEMISTRY INVOLVED IN PREPARATION OF COMPOUNDS
A double salt is a substance obtained by the combination of two different salts which crystallise together as a single substance but ionize as two distinct salts when dissolved in water. The constituents salts are always taken in some definite molecular proportions.
Example: Alums, Mohr's salt
Alum are double salts having general formula X2SO4.M2(SO4)3.24H2O
Mohr's Salt: Formula FeSO4.(NH4)2SO4.6H2O
Used as primary standard in volumetric analysis.
The crystals do not lose water of crystalisation by efflorescene. The crystals are not oxidised in air.
Process of crystallisation: This process involves the following steps:
■ Preparation of solution of the impure sample
■ Filtration of hot solution
■ Concentration of filtrate
■ Cooling the concentrated solution
■ Separation and drying of crystals
7.3.1 Preparation of Inorganic Compounds
Preparation of pure sample of ferrous ammonium sulphate: The chemicals required for the preparation of sample of Ferrous Ammonium sulphate crystals, Ammonium sulphate crystals, Dilute H2SO4, Distilled water
Theory: It is prepared by dissolving equimolar mixture of FeSO 4 .7H 2 O and (NH 4 ) 2 SO 4 in water containing small amount of H2SO4 The result in solution on crystallisation gives light green crystals of Mohr's salts.
FeSO4.7H2O+(NH4)2SO4→
FeSO4.(NH4)2(SO4.6H2O+H2O
Precaution:
■ Use light green ferrous sulphate crystals.
■ Add dilute sulphuric acid while dissolving ferrous sulphate in water to prevent hydrolysis of ferrous sulphate
■ Use boiling water for dissolving salt as it does not have oxygen which could cause oxidation of ferrous into ferric ions.
■ Do not heat the solution strongly and for too long It may result in oxidation of ferrous into ferric and it may also give a solid mass instead of crystals
■ Press the crystals gently between the folds of filter paper to avoid cracks in them.
Preparation of Potash Alum
C hemical Required: Potassium sulphate, aluminum sulphate, conc. H 2SO4
Theory: Potash Alum is a double salt of potassium sulphate and aluminum sulphate. It is prepared by mixing equimolecular quantities of potassium sulphate and aluminum sulphate in water. A little amount of dil. H 2 SO 4 is also added to it. The mixture solution is concentrated to crystallisation point and cooled when white crystals of potash alum separate out.
Theory: Aniline on reaction with acetic anhydride yields acetanili de. + CH3CO CH3CO O NH2 NH C O CH3 CH3COOH +
Precautions: Freshly distilled aniline should be used in order to get good results or small amount of zinc can be added to the reaction mixture.
■ Zinc reduces the coloured impurities in the aniline and also prevents its oxidation during the reaction.
■ Prolonged heating and use of excess of acetic anhydride should be avoided.
■ Reaction mixture should first be cooled and then poured in ice-cold water, otherwise hydrolysis of acetanilide may take place.
Preparation of p-Nitroacetanilide
For the preparation of p-Nitroacetanilide are the following step are involved
Chemicals required: Acetanilide, Glacial Acetic Acid, Conc. H 2 SO 4 , Fuming HNO 3 , Methylated spirit
Theory: The nitration of aniline is difficult to carry out with nitrating mixture (a mixture of conc. H2SO4 and conc. HNO3) since -NH2 group gets oxidised. The amino group is first protected by acylation to form acetanilide which is then nitrated to give p-nitroacetanilide as the major product and o-nitroacetanilide as minor product. Re crystallisation from ethanol readily removes the more soluble orthocompound and the pure p -nitroacetanilide is obtained
Theory: Aniline is diazotised, diazonium salt thus obtained is subjected to coupling reaction with 2-naphthol. Orange coloured dye separates out.
NH2 N2+Cl–
0°–5°C + +
NaNO2 + 2HCl
β-Naphthol (2-Naphthol)
Precautions:
NaCl + 2H2O
Benzene diazonium chloride
2-Naphthol aniline dye (Orange-red dye)
■ The solution of the aniline hydrochloride should be cooled to 5°C, and this temperature should be maintained throughout the addition of the sodium nitrite solution
■ Addition of sodium nitrite should be very slow because the reaction is exothermic and may cause the temperature to rise.
■ Always add diazonium chloride solution to b -naphthol solution for dye formation and not vice versa.
Iodoform:
Iodoform is used as a rubefacient. It is a topical analgesic creating heat on applied area. It also acts as antiseptic because of presence of iodine in its structure.
Iodoform is synthesized by oxidation process. Oxidation is the process in which loss of one or more electron, the reagent which undergoes oxidation is called reducing agent. KMnO4 in alkaline medium and K2Cr2O7 in acetic medium or even nitric acid is most commonly used reagent for oxidation of organic compounds. Primary alcohols are oxidized to corresponding aldehydes, which on further oxidation gives carboxylic acid group, whereas secondary alcohols are to ketones. Aromatic compounds con taining alkyl side chain, but not the group like OH & NH2 which are affected by oxidizing agents are oxidized to carboxylic acids.
Many organic compounds which contain alcoholic group or ketone group on treatment with KI and NaOCl give Iodoform.
Examples:
1 st category compounds are: Ethanol, isopropanol or lactic acid [(CH 3)2CHOH]
Iodoform can be very easily prepared by acetone by the action of ‘NaOCl’ in presence of ‘KI’]
Acetone is 1st converted into triiodo acetone which in presence of alkali is immediately converted into iodoform and sodium acetate
Procedure: Place 0.5 mL of acetone, 20 mL of 10% of KI solution and 8 mL of 10% NaOH solution in 250 mL of conical flask and then add 28 mL of NaOCl solution (5% chlorine). Mix the content well in conical flask. Yellow iodoform begins to separate, allow the mixture
7: Practical Chemistry
to stand at room temperature for 10 minutes and then filter, wash with cold water, & drain thoroughly.
Chemical reaction:
CH3COCH3+3KI+3NaOCl →
CH3COCI3+3KCl+3NaOH
CH3COCI3+NaOH → CHI3+CH3CO2Na
Crystallisation of iodoform:
■ Place the crude iodoform in a 100 ml conical flask.
■ Add small amount of rectified spirit and heat it on a water bath.
■ Add more rectified spirit slowly till the iodoform dissolves.
■ Filter the solution quickly through a fluted filter paper into a beaker.
■ Cool the solution in ice. The iodoform will crystallize rapidly.
■ Filter the crystals on a Buchner funnel, dry the crystals by pressing between the filter papers.
TEST YOURSELF
1. Which of the following is true for Mohr's salt?
a) An aqueous solution of potash alum turns blue litmus red.
b) In the preparation of ferrous ammonium sulphate, only dilute sulphuric acid is used.
c) Mohr's salt, on heating, swells up.
(1) a and b (2) a and c
(3) b and c (4) a, b and c
2. For the preparation of aniline yellow, the steps involved are
a) nitration of aniline
b) isolation of para-nitroaniline
c) reduction of para-nitroaniline
d) purification of para-phenylenediamine
The sequence of steps is
(1) a, b, c, d (2) c, a, b, d
(3) d, c, a, b (4) a, c, b, d
3. The principle involved in the preparation of acetanilinde is
(1) the replacement of one hydrogen atom of the –NH2 group of aniline by CH3CO–group in the presence of glacial acetic acid
(2) the replacement of one hydrogen atom of the – NH2 group of aniline by CH3CO–group in the presence of HCl
(3) the replacement of one hydrogen atom of the – NH2 group of aniline by CH3CO–group in the presence of alkali
(4) The addition of one hydrogen atom of the – NH2 group of aniline by CH3CO–group in the presence of glacial acetic acid
Answer Key
(1) 2 (2) 1 (3) 1
7.4 CHEMICAL PRINCIPLES INVOLVED IN EXPERIMENTS
By using calorimetric technique we can calculate enthalpy of solution of copper sulphate.
7.4.1 Enthalpy of Solution of Copper Sulphate
In this experiment, the enthalpy of dissolution is measured by the use of calorimetric techniques. A known volume of the water is taken in a polythene bottle. Its temperature is noted and then known weight of the solute is added to it. The solution is stirred gently and change in temperature is recorded. From the change in temperature, heat absorbed or evolved can be calculated. In this experiment one mole of solute is dissolved per 400 moles of water. For maintaining this ratio 7g of CuSO4.5H2O is dissolved in 200 ml of water.
Chemical required: Hydrated copper sulphate. Distilled water.
Calculations: Assuming density and specific heat of the solution to be same as that of water, heat evolved or absorbed for dissolution of wg of the solute
Q = W(t2–t1)+(200+w) (t2–t1) x 4.184 J
Heat liberated on dissolution of 1 g of copper sulphate
()() 21 21 Wtt(200w)tt4.184J w −++−× =
Heat liberated on dissolution of 1 mol (249.5 g) of copper sulphate
()() 21 21 Wtt(200w)tt4.184 249.5J w −++−× =×
D solH of copper sulphate = ()() 21 21 Wtt(200w)tt4.184 249.5J/mol w −++−× −×
7.4.2 Enthalpy of Neutralisation of HCl with NaOH
Enthalpy of neutralisation is the heat evolved when one gram equivalent of the acid is completely neutralised by a base in dilute solution
Theory: Heat is evolved during neutralisation of an acid with an alkali. Known volumes of the standard solutions of an acid and alkali are mixed and the change in temperature is observed and from this, the enthalpy neutralisation is calculated.
Chemicals required: 1.0 M hydrochloric acid and 1.0 M sodium hydroxide solution
Calculations: Heat produced during neutralisation of 100 ml of 1.0 M HCI
= (200+W)x(t2 – t1) x 4.184 joules
Heat produced during neutralisation of
1000 ml of 1 M HCl
() 21 1000 (200W)tt4.184 J 100 =+×−××
100 +×−× =
() 21 (200W)tt4.184 kJ
Since heat is liberated during neutralisation, the enthalpy of neutralisation is always negative.
() 21 (200W)tt4.184
Enthalpy of neutralisation kJ 100 +×−× =
7.4.3 Preparation of Lyophilic and Lyophobic Colloids
The terms involved in preparation of colloids are
Disperse phase: It is the component present in small proportion and consists of particles of colloidal dimensions 1-100 nm.
Dispersion medium: The medium in which colloidal particles are dispersed is called dispersion medium.
Lyophilic solutions: In this type of colloidal solutions, the disperse phase has great affinity for the dispersion medium. Examples of lyophilic solutions are gum, gelatin, starch, proteins and certain polymers in organic solvents. Such solutions are called reversible sols. If water is the dispersion medium, these are called hydrophilic solutions.
Lyophobic solutions: In this type of solutions, disperse phase has little affinity for the dispersion medium. These sols are easily precipitated (or coagulated) on the addition of small amounts of electrolytes, by heating or by shaking. Examples of lyophobic sols are sols of metals, and their insoluble compounds like sulphides and oxides. They need stabilising substances for preservation. If water is the dispersion medium, these are known as hydrophobic sols.
Preparation of Lyophilic Starch Sol
The starch sol can be prepared in the following way
Chemical required: Soluble starch (1g) and Distilled water.
Theory: Starch is a lyophilic sol when water is used as the dispersion medium. The formation of sol is accelerated by heating and the sol can be prepared by heating starch and water at about 100°C. It is quite stable and is not affected by the presence of any electrolytic impurity.
■ The apparatus used for preparing sol should be properly cleaned.
■ Distilled water should be used for preparing sols in water.
■ Addition of egg albumin should be done very slowly and with constant stirring so as to disperse the colloidal particles well in solution.
Preparation of Lyophobic Colloidal Sol of Fe(OH)3 Sol
The colloidal sol of Fe(OH)3. Prepared in the following way
Chemical required: 2 % solution of FeCl 3, prepared by dissolving 2g of pure FeCl 3 in 100 ml distilled water.
Theory: Ferric hydroxide sol is prepared by the hydrolysis of ferric chloride with boiling distilled water.
323 FeCl(aq)3HO(l)Fe(OH)(s)(Redsol)3HCl(aq)
Boil
+→+ Boil
323 FeCl(aq)3HO(l)Fe(OH)(s)(Redsol)3HCl(aq)
■ Starch should be converted into a fine paste before adding to boiling water.
■ Starch paste should be added in a thin stream to boiling water.
■ Constant stirring of the contents is necessary during the preparation of the sol.
Preparation of Lyophilic Egg Albumin Sol:
The Egg albumin sol can be prepared in the following way
Chemical required: An Egg and Distilled water
Theory: Egg albumin which is obtained from eggs forms lyophilic sol with cold water. The sol is quite stable and is not affected by the presence of traces of impurities.
Precautions:
■ Egg albumin sol is prepared in cold water because in hot water the precipitation of egg albumin takes place.
■ The yellow yolk should be separated completely from the egg albumin before using the latter in the experiment.
The hydrolysis reaction produces insoluble ferric hydroxide particles, accumulates to give bigger particles of colloidal dimensions. These particles adsorb Fe3+ions preferentially from the solution to give positive charge to the sol particles. Stability of the sol is due to the charge on the sol particles. Hydrochloric acid which is produced during hydrolysis tries to destabilize the sol and hence it must be removed from the sol by dialysis process otherwise sol will not be stable.
Precautions
■ Since ferric hydroxide sol is affected by impurities, the apparatus required for the preparation of sol should be thoroughly cleaned by steaming-out process.
■ Add ferric chloride solution drop wise.
■ Heating is continued till the desired sol is obtained.
■ Hydrochloric acid formed as a result of hydrolysis of ferric chloride is removed by dialysis process otherwise it would destablise the sol.
Preparation of Lyophobic Colloidal Solution of As2S3 Sol
The colloidal sol of AS 2 S 3 prepared in the following way
Chemical required: Solid As 2 O 3 , H 2 S gas, distilled water
Theory: Arsenious sulpide, As2S3 is a lyophobic colloid. It is obtained by the hydrolysis of As2O3with boiling distilled water, followed by passing H2S gas through the solution obtained.
As2O3+3H2 O → 2As(OH)3
2As(OH)3+3H2S → As2S3+6H2O
Precautions:
■ Use cleaned apparatus sinceAs 2 S 3 sol is affected by even traces of impurities.
■ Handle arsenious oxide with care since it is highly poisonous.
7.4.4 Kinetic Study of The Reaction of I– With H2O2
The reaction of I– with H2O2 by using kinetic study in following manner
Chemicals Required: 0.1M KI solution, 2.5
M H2SO4, Starch solution, 3% H2O2 solution, 0.05 M Na2S2O3 solution
Theory: Hydrogen peroxide oxidizes iodide ions to iodine in acidic medium
H2O2+2I–+2H+ → 2H2O+I2
The reaction is monitored by adding a known volume of sodium thiosulphate solution and starch solution to the reaction mixture. Iodine liberated at once reacts with sodium thiosulphate solution and is reduced to iodide ions
++ → Fast 22 22346 I2SOSO2I
When thiosulphate ions are completely consumed, the liberated iodine reacts with starch solution and gives blue colour
I2+ starch → Blue complex
The time elapsed before the appearance of blue colour, gives an idea about the rate of the reaction.
The rate of the reaction increases with increase in concentration of iodide ions.
Precautions.
■ Always use a freshly prepared solution of sodium thiosulphate.
■ Concentration of KI solution should be higher than the concentration of sodium thiosulphate solution.
■ Use freshly prepared starch solution.
■ Do not suck hydrogen peroxide solution with mouth but use a pipetter.
TEST YOURSELF
1. Egg albumin sol is prepared in (1) cold water because precipitation readily occurs.
(2) hot water because precipitation readily occurs.
(3) in cold water as it is stable as lyophilic sol
(4) in hot water as it is stable as lyophilic sol.
2. The enthalpy of solution of anhydrous CuSO4 is −16 kcal and that of CuSO 4 .5H 2 O is 3 kcal. Calculate the enthalpy of hydration of CuSO4.
3. 500 ml each of 0.1 M HCl solution and NaOH solution are mixed. The rise in temperature is 0.1°C. If the water equivalent of the calorimeter is 'X', then the heat of neutralisation is
■ Cations: Ammonium ion, lead ion, cupric ion, ferrous ion, ferric ion, aluminum ion, zinc ion, manganese ion etc.
■ In qualitative analysis of simple salt, identify the anion and the cation present in it.
■ Test with dilute acid for anions CO 3 2, CH3COO-,etc.
■ Test with Conc. H2SO4 for anions Cl–, Br–, I–, NO3–.
■ Individual tests for anions SO42– and PO43–.
■ Preliminary tests: Observation of the physical characteristics like colour, odour and physical state of the salt. Dry heating test (action of heat) followed by flame colour test, charcoal cavity test/cobalt nitrate test and borax bead test. These preliminary tests gives an insight into the anion and cation present in the given simple salt.
■ Flame Test: The chlorides of several metals impart characteristic colour to the flame because they are volatile in non-luminous
flame. This test is performed with the help of a platinum wire.
■ Borax bead test: This test is performed only for coloured salts. Because borax reacts with metal salts to form metal borates or metals. These metal borates or metals have characteristic colours.
■ Tests for confirmation of anions: In qualitative analysis of inorganic salts wet tests are to be carried out for confirmation of anion in the given salt.
■ Need to prepare of sodium carbonate extract: In qualitative analysis, the identification and confirmatory tests of anions are performed in their solutions. In this process, in some cases the cations interfere and forms precipitates leading to incorrect inferences. In such cases, the cations which interfere are to be eliminated by replacing them with soluble cations.
■ For carbonate anion do not use sodium carbonate extract of the salt for performing wet test as the extract contain carbonate
■ Sodium carbonate extract always contains unreacted sodium carbonate, which is alkaline. So the extract is to be neutralized with suitable dilute acid depending on the anion to be identified.
■ Analysis of cation : For analysis of cation. The preliminary examination of the given salt like colour, action of heat (dry heating test) and flame colour test gives us an indication of the cation present in the given salt. The confirmation tests for cation are to be carried out using the original solution of the salt.
■ Preparation of original solution: The following solvents used to prepared original solution (i) Distilled water (cold and hot) (ii) Dilute HCl (cold and hot)(iii) Conc. HCl (cold and hot)
Exercises
JEE MAIN LEVEL
Level - I
Qualitative Salt Analysis
Single Option Correct MCQs
1. Which of the following sulphides is yellow?
(1) ZnS (2) CdS (3) NiS (4) PbS
2. The preliminary test of which of the following ions is not conducted by dilute HCl?
(1) SO32– (2) NO2– (3) SO42– (4) S2–
3. Nitrates of all the metals, except mercury and bismuth are:
(1) coloured (2) unstable
(3) soluble in water (4) insoluble in water
4. Which of the following gives a suffocating gas when treated with dilute HCl?
(1) Carbonate (2) Sulphite
(3) Sulphate (4) Borate
5. Violet vapours are given out when________ is treated with conc. H 2SO4.
(1) bromide (2) iodide
(3) chloride (4) nitrate
6. The colour developed when sodium sulphide is added to sodium nitroprusside is (1) violet (2) yellow (3) red (4) black
7. Nessler’s reagent is: (1) K2HgI4 (2) K2HgI4 + KOH (3) K2HgI2 + KOH (4) K2HgI4 + KI
8. Turnbull’s blue is a (1) ferricyanide (2) ferrous ferricyanide (3) ferrous cyanide
(4) ferri ferrocyanide
9. How many of the following metals do not give flame test?
Li, K, Rb, Cs, Ca, Be, Sr, Mg
10. How many of the following are black precipitate group–II sulphides? CdS, As2S3, CoS, NiS
11. How many of the following compounds liberate NH 3 on heating. (NH 4 ) 2 SO 4 , (NH4)2CO3, NH4Cl, NH4NO3, (NH4)2Cr2O7
12. Test for an ester functional group is
(1) biuret test
(2) hydroxamic acid test
(3) Mulliken test
(4) Liebermann nitroso test
13. Hinsberg’s reagent which is used to test amines is
(1) benzene sulphonamide
(2) benzene diazonium chloride
(3) benzene sulphonyl chloride
(4) acetanilide
14. A reagent used to test for unsaturation of alkene is
(1) conc.H2SO4
(2) ammonical Cu2Cl2
(3) ammonical AgNO3
(4) solution of Br2 in CCl4
15. Biuret test is shown by (1) primary amine
(2) secondary amine
(3) urea
(4) nitrile group
16. Which of the following compounds will show positive test with Tollen's reagent?
(1) CH3CHO
(2) C6H5COC6H5
(3) C6H5COCH3
(4) CH3CH2OH
17. The test that distinguishes primary amines from other amines is (1) idoform test (2) Victor Meyer test (3) Lucas test (4) carbylamine test
18. Find the number of alcohols that can give positive iodoform test.
22. S O 2 + 2H 2S → 3S(Sol)+2H 2O exemplifies _____ method of preparation of colloidal solution
23. AgCl react with NH3 forms a complex (1) AgNO3 (2) AgNHH2Cl (3) [Ag(NH3)2]Cl (4) Ag mirror
24. For the test of oxalates, the soda extract is acidified with (1) dil.H2SO4 (2) dil.HNO3 (3) CH3COOH (4) dil.HCl
Chemistry involved in preparation of compounds
Single Option Correct MCQs
19. Potash alum is a (1) simple salt (2) complex salt (3) acidic salt (4) double salt
20. Which of the following is/are correct in case of Mohr’s salt?
A) It decolourises KMnO4.
B) It is primary standard titrant.
C) It is a double salt.
D) Oxidation state of Fe is +3 in the salt. (1) A only (2) B and C only (3) A, B and C only (4) A, B, C and D
Chemical principles
Single Option Correct MCQs
21. Neutralisation reactions are exothermic. Enthalpy of neutralisation is highest for (1) NH4OH vs HCN (2) NHOH vs HCN (3) NH4OH vs HCl (4) NaOH vs HCl
25. Which of the following compounds is formed in borax bead test? (1) orthoborate (2) Metaborate (3) double oxide (4) Tetraborate
26. Which compound will not give positive chromyl chloride test (1) Copper chloride CuCl2 (2) Mercuric chloride, HgCl2 (3) Zinc chloride, ZnCl2 (4) Anilinium chloride, C6H5NH+3Cl–
Numerical Value Questions
27. Consider the sulphides HgS, PbS, CuS, Sb2S3, As 2S 3, and CdS. The number of sulphides soluble in 50% HNO3 is _____.
28. How many precipitates are white in colour? AgCl, AgBr, PbCl2, Hg2Cl2, ZnS, BaCO3
Level-II
Qualitative Salt Analysis
Single Option Correct MCQs
1. If crimson flame is given when an inorganic mixture is tested by flame test, it may be due to the presence of (1) potassium (2) strontium (3) barium (4) calcium
2. The compound formed in the borax bead test of Cu2+ ion in oxidising flame is (1) Cu (2) CuBO2 (3) Cu(BO2)2 (4) Cu2(BO2)
3. The acidic solution of a salt produces blue colour with KI starch solution. The salt may be
(1) sulphite (2) bromide (3) nitrite (4) chloride
4. In the ring test for nitrates, the ring formed is due to (1) FeSO4. NO (2) FeSO4. NO2 (3) Fe(NO3)3 (4) Fe2(SO4)3.NO
5. Oxalate + MnO2+ dil. H2SO4 gas. The gas evolved is
(1) CO2 (2) CO (3) SO2 (4) O2
6. S2– and SO32– can be distinguished by using: (1) (CH3COO)2Pb (2) Na2[Fe(CN)5NO] (3) both (1) and (2) (4) dil. Ba(OH)2
7. The chromyl chloride test is meant for which of the following ions?
(1) Cl ions
(2) Both Cl and Br ions (3) I ions
(4) Cl and CrO42− ions
8. Cu2+ and Ag+ are both present in the same solution. Which of the following is added in solution so that only one of the ions is precipitated in the solution?
9. Which one of the following salts will produce clear and transparent original solution in 2 M HCl?
(1) Ag2CO3 (2) Pb(CO3)2 (3) Hg2CO3 (4) CuCO3
10. When NH 4 Cl is added to a solution of NH4OH:
(1) the dissociation of NH4OH increases.
(2) the concentration of OH– increases.
(3) the concentrations of both OH– an NH4+ increase.
(4) the concentration of OH– ion decreases.
11. The solution of sodium meta aluminate on diluting with water and then boiling with ammonium chloride, gives
(1) [Al(H2O)5OH]2+ (2) AlCl3
(3) Al(OH)3 (4) NaAl(OH)4
12. To increase significantly the concentration of free Zn2+ ion in a solution of the complex ion [Zn(NH3)4]2+,
2 2 33 aq aq 4
is added to the solution some: (1) H2O (2) HCl(aq) (3) NH3(aq) (4) NH4Cl(aq)
13. CoS (black) obtained in group IV of salt analysis is dissolved in aqua regia and is treated with an excess of NaHCO3 and then Br2 water. An apple green coloured stable complex is formed. It is ______.
(1) sodium cobaltocarbonate
(2) sodium cobaltibromide
(3) sodium cobalticarbonate
(4) sodium cobaltobromide
14. Aqueous solution of BaBr 2 gives yellow precipitate with (1) K2CrO4 (2) AgNO3
(3) (CH3COO)2Pb (4) (1) and (2) both
Numerical Value Questions
15. Among PbS, CuS, HgS, MnS, NiS, CoS, Bi2S3, and SnS2, the total number of black coloured sulphides is ______.
16. In how many of the following, white precipitate is obtained?
a) A solution of BaCl2 is treated with Na2SO3
b) A solution of NaAlO2 is heated with NH4Cl.
c) (NH4)2S is added to a solution ZnSO 4.
d) A solution of ZnSO 4 is treated with Na2CO3.
Detection of Functional Groups
Single Option Correct MCQs
17. Which of the following is not a test for phenol?
(1) FeCl3 test
(2) Br2/H2O test
(3) Liebermann nitrosation test
(4) Victor Meyer’s test
18. Test to differentiate between (CH 3 OH) (methanol) and (Ph–OH) (phenol) is/are
(1) litmus test
(2) neutral FeCl3
(3) Br2/H2O
(4) litmus test, neutral FeCl 3, and Br2/H2O
19. The amine that does not give carbylamine test is
(1) C NH C O OH H
(2) Ph–NH2
(3) CH3CH2NH2
(4) CH3CH(NH2)CH3
Numerical Value Questions
20. How many of the following tests are correctly matched?
a) Br/CCl4(or)Br2/H2O(or) alkaline KMnO4 ____ test for unsaturation.
b) Anhyd. ZnCl2 + conc.HCl ____Test for distinguishing 10, 20, 30 alcohols
c) 2,4 – DNP ____ Test for carbonyl compounds.
d) NaHCO3 ____ Test for carboxylic group
e) Carbylamine test – Identification for 10- amines
f) NaOH ____ Test for amide group
g) Hinsberg reagent test ____ Test for the separation of 10, 20, 30 amines
Chemistry Involve in Preparation of Compounds
Single Option Correct MCQs
21. Potash alum, when dissolved in water, it produces ___________.
(1) blue solution (2) alkaline solution
(3) acidic solution (4) neutral solution
22. Aniline yellow can be prepared by
(1) direct diazotisation and coupling, as followed for phenyl-azo-β-naphthol dye in acidic medium
(2) direct diazotisation and coupling, as followed for phenyl-azo-β-naphthol dye in basic medium
(3) rearrangement reaction of diazoaminobenzene with a small quantity of nitrobenezene
(4) substitution reaction of diazoaminobenzene with a small quantity of aniline hydrochloride
Numerical Value Questions
23. If potash alum is KxSO4.Aly(SO4)z .pH2O, then find the value of . 1 p xyz+++
Chemical Principles Involved in Experiments
Single Option Correct MCQs
24. Enthalpy of neutralisation of a strong acid and a strong base is
(1) –57.32 kJ mol–1
(2) –55.56 kJ mol–1
(3) –54.0 kJ mol–1
(4) –52.0 kJ mol–1
25. Enthalpy of neutralisation of all strong acids and strong bases has the same value because (1) neutralisation leads to the formation of a salt and water
(4) the net chemical change involves the combination of 1 mol of H+ ions and 1 mol OH– ions to form water
26. When a mixture of solid NaCl and solid K2Cr2O7 is heated with concentrated H2SO4, deep red vapours are obtained. This is due to the formation of :
(1) chromous chloride
(2) chromyl chloride
(3) chromic chloride
(4) chromic sulphate
27. Sodium carbonate extract is a mixture of (1) [Salt + Na2CO3 + HCl]
(2) [Salt + Na2CO3 + H2O]
(3) [Salt + CaCO3 + HCl]
(4) [Salt + Na2CO3 + HNO3]
28. Yellow ammonium sulphide solution is a suitable reagent for the separation of (1) HgS and PbS (2) PbS and Bi2S3
(3) Bi2S3 and CuS (4) CdS and Bi2S3
29. In IV group analysis, NH4OH is added before passing H2S gas because
(1) the sulphides of IV group are insoluble in NH4OH
(2) the sulphides of other metals are soluble in NH4OH
(3) the concentration of S2– ions is increased
(4) the sulphides of second group are soluble in NH4OH
30. Nitrite (NO 2 ) interferes in the ‘ring-test’ of nitrate (NO 3 ). Which of the following reagents can be used for the removal of nitrite.
I. Potassium dichromate (s)
II. CO(NH2)2(urea).
III. NH2SO3H (sulphamic acid)
IV. Zinc /sodium hydroxide
(1) I, II (2) I, II and I
(3) II, III (4) II, III and IV
31. Which of the following products are obtained wh en Na 2 CO 3 is added to a solution of copper sulphate?
(1) Basic copper carbonate [CuCO 3 Cu(OH)2], sodium sulphate, and CO 2
(2) Copper hydroxide, sodium sulphate, and CO2
(3) Copper carbonate, sodium sulphate, and CO2
(4) Copper carbonate and sodium sulphate
32. An inorganic salt solution gives a red precipitate with silver nitrate. The precipitate dissolves in dilute nitric acid. The solution contains
(1) bromide
(2) iodide
(3) phosphate
(4) chromate
33. Given reagents are HCl, NaOH, ZnCl 2 , Na 2CO 3, NH 4Cl, and Zn. Which of these reagent(s) can intensify hydrolysis of FeCl3 when added to its solution?
(1) NaOH, Na2CO3
(2) NaOH, NH4Cl, HCl
(3) Na2CO3, NaOH + HCl
(4) NH4Cl, HCl, ZnCl2
34. A gas ‘X’ is passed through water to form a saturated solution. The aqueous solution, on treatment with silver nitrate, gives a white precipitate. The saturated aqueous solution also dissolves magnesium ribbon with evolution of a colourless gas 'Y'. Identify ‘X’ and 'Y'.
(1) X = XO2, Y = Cl2
(2) X = Cl2, Y = CO2
(3) X = Cl2, Y = H2
(4) X = H2, Y = Cl2
Level - III
1. Conc.H 2SO 4 is put into two test tubes ‘A’ containing a nitrate salt and ‘B’ containing a bromide salt and the contents are heated to evolve reddish brown gases which were passed through water. Water will :
(1) turn yellow by gas coming from test tube ‘A’
(2) turn brown by gas coming from test tube ‘A’
(3) turn blood red by gas coming from test tube ‘B’
(4) turn yellow by gas coming from test tube ‘B’
2. A light yellow precipitate is formed in the second group of the qualitative analysis on passing H2S, even when no radical of second group is present. This is due to the presence of ________ in the mixture.
(1) phosphate (2) acetate (3) oxalate (4) nitrate
3. Three separate samples of a solution of a single salt gave these results. One formed a white precipitate with excess ammonia solution, one formed a white precipitate with dilute NaCl solution and one formed a black precipitate with H2S. The salt could be :
(1) AgNO3 (2) Pb(NO3)2 (3) Hg(NO3)2 (4) Mn(NO3)2
4. Consider the following observation:
Mn+ + HCl(dilute) → white precipitate ∆ → water soluble 2–CrO4 → yellow precipitate. The metal ion Mn+ will be (1) Hg2+ (2) Ag+* (3) Pb2+ (4) Sn2+
5. Sometimes, yellow turbidity appears while passing H2S gas even in slightly acidic medium the absence of group radicals. This is because
(1) sulphur is present in the mixture as impurity (2) IV group radicals are precipitated as sulphides
(3) of the oxidation of H2S gas by some acid radicals (4) III group radicals are precipitated as hydroxides
6. An original solution of an inorganic salt in dilute HCl gives a brown colouration with potassium hexacyanidoferrate (III) and reddish brown colouration with sodium acetate solution. The cation of the salt is (1) Ni2+ (2) Fe3+ (3) Cu2+ (4) Mg2+
7. To an aqueous solution containing ions such as Al+3, Zn+2, Ca+2, Fe+3, Ni+2, and Cu+2, conc. HCl was added, followed by H 2S. The total number of cations precipitated during this reaction is ______.
8. How many of the given compounds cannot be converted into anhydrous chloride by heating only? MgCl 2 .6H 2 O, AlCl 3 .6H 2 O, FeCl3.6H2O, SnCl2.2H2O, CoCl2.6H2O,
9. In Victor Meyer’s test, the colour given by 3°, 2°, 1° alcohols are respectively, (1) red, blue, colourless (2) colourless, blue, red (3) colourless, red, blue (4) red, colourless, blue
10. A compound that can give iodoform test, bromine water test, FeCl 3 test, but not Tollen’s test is (1)
(4)
11. Which of the following can be used as laboratory test for phenol
(1) It gives effervescence of CO 2 with NaHCO3.
(2) It gives purple colour with neutral FeCl3.
(3) It gives yellow precipitate with NaOH+I2
(4) It gives silver mirror with Tollen’s reagent.
12. The role of (NH4)SO4 in Mohr’s salt is
(1) to increase the solubility of FeSO4 in water
(2) to keep the FeSO 4 stable without any oxidation
(3) to provide NH4+ ions during the titration of Mohr’s salt
(4) to increase the concentration of SO 4 2–
13. p -aminoazobenzene can be prepared in a good yield by
(1) rearrangement reaction of diazoaminobenzene with a small quantity of aniline hydrochloride in the presence of aniline
(2) substitution reaction of diazoaminobenzene with a small quantity of aniline hydrochloride in the presence of aniline
(3) rearrangement reaction of diazoaminobenzene with a small quantity of nitrobenzene in the presence of aniline
(4) substitution reaction of diazoaminobenzene with a small
quantity of nitrobenzene in the presence of aniline
14. Which of the following is true about iodoform?
(1) Iodoform is a crystalline pale-yellow material that is highly flammable.
(2) Iodoform has a chemical behaviour that is nearly identical to that of chloroform.
(3) It is used as an antiseptic and in the manufacture of pharmaceuticals
(4) It is insoluble in water but soluble in ethyl alcohol and ether.
(1) 1, 2 only (2) 2, 3 only (3) 2, 3 and 4 only (4) 1, 2, 3 and 4
15. The enthalpies of solution for copper sulphate pentahydrate and anhydrous copper sulphate are −11.7 kJ mol −1 and −65.5 kJ mol−1, respectively. The hydration enthalpy of anhydrous copper sulphate is ______.
(1) –53.8 kJ mol–1 (2) +53.8 kJ mol–1
(3) –77.2 kJ mol–1 (4) +77.2 kJ mol–1
16. For kinetic study of the reaction of iodide ion with H2O2 at room temperature, (A) always use freshly prepared starch solution
(B) always keep the concentration of sodium thiosulphate solution less than that of KI solution.
(C) record the time immediately after the appearance of blue colour
(D) record the time immediately before the appearance of blue colour
(E) Always keep the concentration of sodium thiosulphate solution more than that of KI solution
Choose the correct answer from the options given below
(1) (A), (B), (C) only
(2) (A), (D), (E) only
(3) (D), (E) only
(4) (A), (B), (E) only
17. Reaction between iodide ions and hyrogen peroxide takes place in acidic medium as follows: 2I –(aq) + H 2O 2(I) + 2H + (aq) → I 2(g) + 2H2O(I); (slow reaction)
Which of the following are true?
A) In this reaction mixture if calculated amount of sodium thiosulphate (Na2S2O3) is added in the presence of starch solution as an indicator.
B) Liberated Iodine reacts with thiosulphate ions as fast as it is formed and is reduced back to iodide ions till all the thiosulphate are oxidised to tetrathionate ions.
C) Time required to consume a fixed amount of the thiosulphate ions is reproducible.
D) This reaction is also called clock reaction because time for the appearance of colour shows the accuracy of clock.
(1) A only (2) A and B only
(3) A, B and C only (4) A ,B, C and D
Multi-concept Questions
(Single Correct MCQ)
18. X: SO 2 and CO 2 both turn lime water milky, Y: SO2 also turns K2Cr2O7/H+ green.
Z: O2 is soluble in pyrogallol turning it black. These gases are to be detected in order by using these reagents. The order is
(1) (X), (Y), (Z)
(2) (Y), (X), (Z) (3) (X), (Z), (Y)
(4) The correct order cannot be predicted.
19. Which of the following is/are not correctly matched?
(A) BiI 3 → Black
(B) Cu2I2 White precipitate
(C) Pbl2 Yellow precipitate
(D) HgI2 Red precipitate
(1) A and B only (2) B and C only
(3) A, B, and C only (4) A, B, C, and D
20. Out of Fe3+, Co2+, Cu2+, which of the following ions is detected by NH4SCN?
(1) Fe3+ only
(2) Co2+ and Cu3+
(3) Fe3+and Cu2+
(4) Fe3+, Co2+, and Cu2+
21. The addition of K2CO3(aq) to the following solution is expected to produce a precipitate in every case but that one which does not produce precipitate is
(1) BaCl2(aq) (2) CaBr2(aq)
(3) Na2SO4(aq) (4) Pb(NO3)2(aq)
22. In the brown ring test for the nitrate ion, [Fe(H2O)5NO]+2 imparts the colour due to (1) π → σ transition
(2) charge transfer transition
(3) d-d transition
(4) π → σ transition
23. If Cd 2+ and Cu 2+ both are present, both form complexed with CN . When H2S gas is passed, which one of the following pairs of complexes and their relative stability enables the separation of Cu2+ and Cd2+?
(1) K 3 [Cu(CN) 4 ]: more stable and K2[Cd(CN)4]: less stable
(2) K2[Cu(CN)4]: less stable and K2[Cd(CN)4]: more stable
(3) K 2 [Cu(CN) 4 ]: more stable and K2[Cd(CN)4]: less stable
(4) K3[Cu(CN)4]: less stable and K2[Cd(CN)4]: less stable
24. Compounds A and B are treated with dilute HCl separately. The gases liberated are Y and Z, respectively. Y turns acidified dichromate paper green while Z turns lead acetate paper black. So, A and B compounds are, respectively, (1) Na2SO3, Na2S (2) NaCl, Na2SO3 (3) Na2S, Na2SO3 (4) Na2SO3, K2SO4
25. When a reagent (A) reacts with FeCl 3, the solution turns red due to the formation of a compound (B). The reagent causes no change in colour when it reacts with FeCl 2 in pure state. A and B, respectively, are
(1) K4[Fe(CN6)] and Fe4[Fe(CN)6]3
(2) NH4SCN and [Fe(SCN)3]
(3) K3[Fe(CN)6] and K2[Fe(CN)6]
(4) NH4CNS and [Fe(CNS)]2+
26. When a mixture of NaCl, K2Cr2O7, and conc. H 2 SO 4 is heated in a dry test tube, a red vapour (X) is evolved. This vapour (X) turns an aqueous solution of NaOH yellow due to the formation of Y. X and Y, respectively, are
THEORY-BASED QUESTIONS
Statement Type Questions
Each question has two statements: statement I (S-I) and statement II (S-II). Mark the correct answer as
(1) if both statement I and statement II are correct,
(2) if both statement I and statement II are incorrect
(3) if statement I is correct but statement II is incorrect,
(4) if statement I is incorrect but statement II is correct.
1. S-I : Colourless cupric metaborate is reduced to cuprous metaborate in a luminous flame.
S-II : Cuprous metaborate is obtained by heating boric anhydride and copper sulphate in a non-luminous flame.
2. S-I : Results of Victor Meyers test:
1° Red → Red colour
2° Red → Blue colour
3° Red → White or no colour
S-II : Victor Meyers test is a method for separation of 1°, 2° and 3° alcohol.
(1) CrCl3 and Na2Cr2O7
(2) CrCl3 and Na2CrO4
(3) CrO2Cl2 and Na2CrO4
(4) CrO2Cl2 and Na2Cr2O7
27. Tests on an aqueous solution of a sodium salt having an anion Xn– gave the following results:
227 KCrO/H n anion Xgreen solution + →
() 3 2 PbNO gasblack precipitate +→
Which one of the following could be X n– :
(1) I– (2) NO2–
(3) S2– (4) SO42–
3. S-I : T he replacement of one hydrogen atom of the — NH2 group of aniline by CH3CO– group in the presence of glacial acetic acid. Gives acetanilide.
S-II : In the laboratory, acetylation is usually carried out with acetic anhydride.
4. S-I : p-Nitroacetanilide is prepared by the nitration of acetanilide by using a mixture of conc. nitric acid and conc. sulphuric acid.
S-II : The mixture of the two acids releases nitronium ion(NO2+), which acts as an electrophile in the reaction.
5. S-I : Acetaldehyde responds positively with all the tests of carbonyl compounds like Tollen's test, Fehling test, 2, 4-DNP test, as well a iodoform test.
S-II : All aldehydes respond all the four tests given in statement-1.
S-II : Aldehydes do not respond positively with Benedicts reagent.
Assertion and Reason Questions
In each of the following questions, a statement of Assertion (A) is given, followed by a corresponding statement of Reason (R). Mark the correct answer as
(1) if both (A) and (R) are true and (R) is the correct explanation of (A),
(2) if both (A) and (R) are true but (R) is not the correct explanation of (A), (3) if (A) is true but (R) is false, (4) if both (A) and (R) are false.
7. (A) : In the preparation of Mohr’s salt, concentrated H 2SO 4 can be used in place of dilute H2SO4.
(R) : Conc. H 2SO 4 would oxidise ferrous ions to ferric ions.
8. (A) : On heating potash alum, it becomes light and fluffy.
(R) : On heating potash alum, it loses water of crystalisation.
9. (A) : Dilute sulphuric acid is added during the preparation of aluminium sulphate solution.
(R) : Sulphuric acid prevents the hydrolysis of aluminium sulphate.
10. (A) : CdS and As2S3 both have yellow colour and are precipitated by passing H2S through solution in dilute HCl.
(R) : Both can be separated by yellow ammonium sulphide.
(R) : Sulphur dioxide is a reducing agent. It reduces acidified potassium
JEE ADVANCED LEVEL
Multiple Option Correct MCQs
1. Which of the following statement(s) is/are true?
(1) Ag + ions give white precipitate with excess of concentrated HCl.
(2) Cu2+ ions produce a white precipitate when KCN solution is added in a small quantity and allowed to stand.
dichromate into chromic compound, which is green in colour.
12. (A) : Original solution for basic radicals can be prepared in concentrated HNO3 or concentrated H2SO4.
(R) : Both the acids dissolve majority of inorganic salts.
13. (A) : Ammonium salts evolve ammonia when heated with NaOH.
(R) : Ammonium salts are soluble in water.
14. (A) : Chloride is confirmed by chromyl chloride test.
(R) : Chlorine is evolved when chloride is heated with dil. H2SO4
15. (A) : PbCl2 and Hg2Cl2 precipitates can be separated by hot water.
(R) : Aqueous solution of PbCl2 gives yellow precipitate with K2CrO4 solution.
16. (A) : When Cl2 gas is passed into a mixture containing Br – and I – and CHCl 3, I 2 (violet) first appears in CHCl 3 layer.
(R) : In excess of Cl 2 gas, violet colour disappears and CHCl3 layer becomes brown.
17. (A) : When H2S gas is passed into aqueous solution of ZnCl2, white precipitate of ZnS appears.
(R) : ZnS is insoluble in HCl.
18. (A) : Addition of NH 4OH to an aqueous solution of BaCl2 precipitates Ba(OH)2.
(R) : Ba(OH)2 is soluble in water.
19. (A) : Sodium meta aluminate, on boiling with NH4Cl, produces white gelatinous precipitate.
(R) : Aluminium hydroxide is formed, which is not soluble in water.
(3) Hg 2+ ions give deep blue precipitate with cobalt acetate and ammonium thiocyanate.
(4) Black precipitate of BiI 3 turns white when heated with water.
2. Two metal cations P and Q, on reaction with different reagents, gave the following observation.
P(aq)
H2S(g) H2S(g) dil. H3O+ aq.NH3
Precipitate
Precipitate
Precipitate
Q(aq) clear solution clear solution
Precipitate
Possible cations P and Q respectively are (1) Ag+, Zn2+ (2) Pb+, Hg2+ (3) Cu+, Zn2+ (4) Cu+, Hg2+
3. Consider the following reactions (unbalanced).
Zn + hot conc. H2SO4 G + R + X
Zn + conc. NaOH T + Q
G + H2S + NH4OH Z( precipitate) + X + Y
Choose the correct option(s).
(1) The oxidation state of Zn in T is +1.
(2) R is a V-shaped molecule.
(3) Bond order of Q is 1 in its ground state. (4) Z is dirty white in colour.
4. A colourless water-soluble compound, on strong heating, liberates a brown coloured gas as one of the products and leaves a residue. An aqueous solution of the original solid gives a precipitate with (NH4)2S and NH4OH. The original solid could be (1) Pb(NO3)2 (2) Zn(NO3)2 (3) Ca(NO3)2 (4) Al(NO3)2
5. The pair of ions that can be separated with NH4OH in presence of NH4Cl is/are
(1) Cr+3 and Ni+2 (2) Al+3 and Zn+2
(3) Fe+3 and Ba+2 (4) Ca+2 and Sr+2
6. Select the correct characteristics of aqueous solution of Mohr’s salt.
(1) It is green in colour.
(2) pH of the solution is less than 7.
(3) It gives precipitates with BaCl 2
(4) It produces precipitate with excess of NaOH.
7. Metal 'M' is a light-red metal, which is soft, malleable, and ductile. It melts at 1038 °C. Because of its positive standard electrode potential, it is insoluble in hydrochloric acid and in dil. H2SO4, although in the presence of oxygen, some dissolution might take place. The dilute HNO 3 dissolves metal readily. If coloured solution MSO 4 obtained by dissolving metal in conc. H2SO4 is treated with KI(aq) solution, an intense brown solution of triiodide is formed. Select the correct statement regarding the above information.
(1) If a clean iron nail or a blade of a penknife is immersed in a solution of a metal salt, a red deposit of metal is obtained.
(2) Metal salt solution of M shows redox reaction with Na2S2O3, which, on further treatment with excess Na2S2O3, forms a complex.
(3) Solution obtained by reaction of metal M and AgNO3aq produces colourless solution with excess NH 4OH.
(4) Oxide of metal M (i.e. MO) dissolves in dil. HCl but not in excess KO.
8. The evolution of a reddish brown gas, on heating a salt with K2Cr2O7 and concentrated H2SO4, can arise from (1) Cl– (2) Br– (3) NO3– (4) NO2–
9. A yellow precipitate is obtained when (1) lead acetate solution is treated with K2CrO4
(2) Pb(NO3)2 solution is treated with KI
(3) AgNO3 solution treated with KI
(4) H2S is passed through a solution of CdSO4
10. When a mixture of NH 4Cl, NH 4OH, and NaH2PO4 was added to a solution containing Mg2+, a white precipitate (A) was formed. When A was heated strongly, the residue B was obtained. A is ______ and B is _____.
(1) Mg(NH4)PO4
(2) Mg2P2O7
(3) Mg2(PO4)2
(4) MgO
11. Which of the following changes do not occur when a solution containing Mn 2+ and Cr3+ is heated with NaOH and H 2O2?
(1) Hydrated manganese dioxide and Na2CrO4 are formed
(2) Mn(OH)2 and Cr(OH)3 are precipitated
(3) Na2MnO4 and Na2CrO4 are formed
(4) Na2MnO4 and Cr(OH)3 are formed
12. Which of the following combinations in an aqueous medium will give a red colour or precipitate?
(1) Fe3+ + SCN–
(2) Fe2+ + [Fe(CN)6]3–
(3) Ni2++ dimethylglyoxime + NH3 solution
(4) Co2+ + SCN–
Numerical Value Questions
13. How many of the following substance, are precipitated as hydroxides only when treated with sodium carbonate. FeSO4, FeCl3, MgCl2, AlCl3, H2SO4, CrCl3
14. Out of NiS, CoS, MnS, ZnS
A = Number of black precipitates
B = Number of sulphides which is/are precipitates by using H2S in alkaline medium.
C = Number of sulphides which is/are soluble in CH3COOH
D = Number of sulphides which is/are dissolved in aqua regia for confimatory test.
What will be the value of A + B + C + D?
15. In iodide of Million’s base how many mercury atoms are present
16. How many of the following are correct?
i) A white precipitate is obtained when a solution of BaCl2 is treated with Na2SO3
ii) A white precipitate is obtainied when a solution of NaAlO2 is heated with NH4Cl.
iii) A white precipitate is obtained when H2S is passed through a solution of ZnSO 4
iv) A white precipitate is obtained when a solution of ZnSO4 is treated with Na2CO3
17. A mixture of Zn+2, Mn+2, Ni+2, Co+2 radicals, on reaction with H2S gas, gives ‘x’ number of metal sulphide precipitates and, on further reaction with HCl, gives ‘y’ number of metal chlorides as precipitate. Find the value of (x+y).
18. How many of the following are insoluble in excess aq NaOH?
19. A mixture contained aluminium sulphate and sodium sulphate. A sample of the mixture weighing 3.36 g was dissolved in water and treated with ammonium hydroxide. The precipitate formed was incinerated to yield 0.5 g of anhydrous aluminium oxide. What is the percentage of aluminium sulphate in the mixture? (Rounding to nearest integer) (At. wt. Al=27,Na=23,S=32,O=16,N=1,H=1)
20. Find the number of basic radicals among the following cations, which can form soluble complex on adding excess of NH3 solution. Pb 2+(aq.), Ni 2+(aq.), Mn 2+(aq.), Zn 2+(aq.), Ag2+(aq.), Hg2+(aq.), Fe2+(aq.), Mg2+(aq.)
Passage-based Questions (Q: 21–22)
An aqueous salt solution containing two metal ions was treated with excess cold NaOH solution. A white ppt was obtained, which slowly turned to brown (P). To the filtrate (Q) obtained from the above solution, hydrochloric acid is added drop-wise. A white ppt was formed first, which re-dissolved in excess of HCl. The brown residue in the filter paper gave a pink colour when strong oxidising agent like PbO2/HNO3 or sodium bismuthate was added.
21. The ions present in the given solution are (1) Mg+2 and Mn+2 (2) Zn+2 and Mn+2 (3) Mg+2 and Zn+2 (4) Al+2 and Zn+2
22. The filtrate (Q) contains (1) Na2MnO2
(2) NaAlO2
(3) Na2ZnO2
(4) ZnCl2
(Q : 23–24) () () Strongly Strongly
23. How many X-O-X linkages are present in structure of A (X=central atom)?
24. If the number of tetrahedral and trigonal planar units in structure of A are x and y respectively, then x+y = _______.
(Q : 25–26)
A colourless inorganic compound (A) imparts a green colour to the flame. Its solution gives a white precipitate (B) with H2SO4. When heated with K2Cr2O7 and conc. H2SO4, a brown red vapour/gas (C) is formed. The gas/vapour, when passed through aqueous NaOH solution, turns into a yellow solution (D), which forms yellow precipitate (E) with CH 3COOH and (CH3COO)2Pb. With reference to the above information, answer the following questions.
25. The colourless inorganic compound (A) is (1) CrBr3Ba(NO3)2 (2) BaCl2 (3) CuCl2 (4) CrBr3
26. The yellow ppt. formed when (D) reacts with CH3COOH and (CH2COO)2Pb is (1) PbI2 (2) PbCrO4 (3) BaCrO4 (4) AgBr
3. (KCI + K2Cr2O7 + H2SO4) Heat Red Gas dil NaOH Yellow Solution Pb(Ac)2 X
The formula and colour of X are, respectively, (1) CrO2Cl2, red colour
(2) PbCrO4, yellow colour
(D) Its acidified salt solution decolourises pink KMnO4 solution (s) SO32–
Choose the correct option.
(A) (B) (C) (D)
(1) q,r p,r q,r,s q,s
(2) q,r p,r,s p,q,r q,r
(3) q,s p,r,s p,q,s p,q,r,s
(4) q,r q,r,s q,r p
(3) BaCrO4, green colour
(4) Cr2(SO4)3, green colour
4. Consider the following tests:
i) A white solid mixture of two salts (A) containing a common cation is insoluble in water. It dissolves in dilute HCl, producing some gases (with effervescence) that turn an acidified dichromate solution green. After the gases are passed through the acidified dichromate solution, the emerging gas turns baryta water milky.
ii) On treatment with dilute HNO 3, the white solid mixture (A) gives a solution that does not directly give a precipitate with a BaCl2 solution but the white solid mixture (A) gives a white precipitate when warmed with H2O2 and then treated with a BaCl2 solution.
iii) The solution of the mixture in dilute HCl, when treated with NH4Cl, NH4OH, and Na 2HPO 4 solution, gives a white precipitate.
Which of the following statements are correct regarding the above tests?
(1) The gases evolved in (i) are SO2 and H2S.
(2) The white precipitate obtained in (ii) indicates the presence of sulphite.
(3) The white precipitate obtained in (iii) consists of MgNH4PO4.6H2O.
(4) The white precipitate obtained in (ii) is BaSO3.
5. Consider the reaction sequence given below.
MnSO4 + NaOH + O2(air) P (Black)
(P) + NaOOC – COONa solution
24 dilute HSO
→ red colouration (Q) + R ↑ + other Product(s). If the number of optically active isomers of (Q) is ‘a’, and the atomicity of gas R is ‘b’, find the value of a + b.
6. A white crystalline solid X exhibits the following reactions:
(i) It is a binary salt.
(ii) It impart green colour to the nonluminous Bunsen burner flame
(iii) In aqueous solution of X [H3O+] = [OH ]
(iv) On addition of AgNO3 to an aqueous solution of X, yellow precipitate is obtained, which is sparingly soluble in NH3
(v) Yellow precipitate obtained in (iv) readily dissolves by adding excess of KCN or Na2S2O3
(vi) On addition of CuSO4 to an aqueous solution of X, the solution turns brown, and a white precipitate is obtained on addition of Na2S2O3
If the cationic elements belongs to group number p and the anionic element belongs to period number q in the modern periodic table, then what is the value of ( p + q)?
7. The black precipitate of nickel sulphide dissolves in aqua regia and the reaction takes place as follows: 3NiS + 2HNO 3 + 6HCl → 3NiCl2 + 2NO + 3S + 4H2O After treatment with aqua regia, nickelchloride is obtained, which is soluble in water. When dimethyl glyoxime is added to the aqueous solution of nickel chloride, made alkaline by adding NH4OH solution, a brilliant red precipitate is obtained. Find the sum of the number of Ni−N linkages and Ni−C linkages in brilliant red precipitate.
Passage 1
A gas (A) was bubbled into a solution of a common mono positive hydroxide (B) to give a solution of the salt (C). The cation of B gives a precipitate with the tetraphenyl borate ion. Heating yellow solid (D) with solution of (C) and evaporating the water gives crystals containing anion (E). Addition of iodine to a solution of anion (E) gives iodide ion and a solution of anion (F). Addition of hydrogen ion to a solution of anion (E) initially produces acid (G), which decomposes to form solid (D) and a gas (A).
8. By passing excess of gas (A) through the solution of the salt (C), a compound (X) is formed. Crystallisation of (X) by heating gives (1) Na2SO3 (2) Na2S2O3 (3) NaHSO3 (4) Na2S2O5
9. In which of the anions of following compounds, there is no S-S bond? (1) C (2) E (3) F (4) X
Passage 2
A chemist opened a cupboard and found four bottles containing water solutions, each of which had lost its label. Bottles 1, 2, and 3 contained colourless solutions, while bottle 4 contained a blue solution. The labels from the bottles were lying scattered on the floor of the cupboard. They were:
Copper (II) sulphate, Hydrochloric acid, Lead nitrate, and Sodium carbonate
By mixing samples of the contents of the bottles, in pairs, the chemist made the following observations:
Bottle 1 + Bottle 2 White precipitate is formed.
Bottle 1 + Bottle 3 White precipitate is formed.
Bottle 1 + Bottle 4 White precipitate is formed.
Bottle 2 + Bottle 3 Colourless and odourless gas is evolved.
Bottle 2 + Bottle 4 No visible reaction is observed.
Bottle 3 + Bottle 4 Blue precipitate is formed.
10. Bottle 3 contains
(1) copper (II) sulphate
FLASHBACK (Previous JEE Questions)
JEE Main
Single Option Correct MCQs
1. Element not present in Nessler’s reagent is (1) K (2) N (3) I (4) Hg
2. In the wet tests for detection of various cations by precipitation, Ba2+ cations are detected by obtaining precipitate of
(1) BaCO3
(2) Ba(OAc)2
(3) BaSO4
(4) Ba(ox) : Barium oxalate
3. Match column-I (cations) with column-II (group reagents).
Column-I (Cations)
Column-II (Group reagents)
(A) Pb2+, Cu2+ (P) H2S gas in presence of dilute HCl
(B) Al3+, Fe3+ (Q) (NH4)2CO3 in presence of NH4OH
(C) Co2+, Ni2+ (R) NH4OH in presence of NH4Cl
(D) Ba2+, Ni2+ (S) H 2S in presence of NH4OH
Choose the correct option.
(A) (B) (C) (D)
(1) S Q R P
(2) hydrochloric acid
(3) lead nitrate
(4) sodium carbonate
11. Which one of the following bottles develops deep blue colour with aqueous ammonia?
(1) Bottle 1
(2) Bottle 2
(3) Bottle 3
(4) Bottle 4
(2) P R Q S
(3) R P S Q
(4) P R S Q
4. A chloride salt solution acidified with dil. HNO3 gives a curdy white precipitate [A], on addition of AgNO3. [A], on treatment with NH4OH, gives a clear solution B. A and B are respectively,
(1) AgCl and [Ag(NH3)2]Cl
(2) AgCl and (NH4)[Ag(OH)2]
(3) H[AgCl3] and (NH4)[Ag(OH)2]
(4) H[AgCl3] and [Ag(NH3)2]Cl
5. In the wet tests for identification of various cations by precipitation, which transition element cation doesn’t belong to group IV in qualitative inorganic analysis?
(1) Ni2+ (2) Zn2+*
(3) Fe3+ (4) Co2+
6. Formula for Nessler’s reagent is
(1) KHgl3 (2) K2Hgl4
(3) KHg2l2 (4) Hgl2
7. In the flame test of a mixture of salts, a green flame with blue centre was observed. Which one of the following cations may be present in the mixture?
(1) Cu2+ (2) Sr2+
(3) Ba2+ (4) Ca2+
8. During the qualitative analysis of salt with cation Y 2+ , addition of a reagent (X) to alkaline solution of the salt gives a bright red precipitate. The reagent (X) and the cation (Y2+) present, respectively are
(1) dimethylglyoxime and Ni2+
(2) dimethylglyoxime and Co2+
(3) nessler's reagent and Hg 2+
(4) nessler's reagent and Ni 2+
9. A white precipitate was formed when BaCl2 was added to water extract of an inorganic salt. Further, a gas 'X' with characteristic odour was released when the formed white precipitate was dissolved in dilute HCl. The anion present in the inorganic salt is
(1) I– (2) SO32–
(3) S2– (4) NO2–
10. Match column–I with column-II.
Column–I (Anion)
Column–II (Gas evolved on reaction with dil. H2SO4)
(A) CO32– (P) Colourless gas that turns lead acetate paper black
(B) S2– (Q) Colourless gas that turns acidified potassium dichromate solution green.
(C) SO32– (R) Brown fumes that turn acidified KI solution containing starch blue
(D) NO2– (S) Colourless gas evolved with brisk effervescence, which turns lime water milky
Choose the correct option.
(A) (B) (C) (D)
(1) R P Q S
(2) Q P S R
(3) S P R Q
(4) S P Q R
11. Which statement is not true with respect to nitrate ion test?
(1) A dark brown ring is formed at the junction of two solutions.
(2) Ring is formed due to nitroferrous sulphate complex.
(3) The brown complex is [ Fe(H2O)5(NO)] SO4
(4) ON heating the nitrate salt with conc. H2SO4, light brown fumes are evolved.
12. Which of the following compounds is not yellow in colour?
(1) Zn2[Fe(CN)6]
(2) K3[Co(NO2)6]
(3) (NH4)3[As(Mo3O10)4]
(4) BaCrO4
13. An inorganic compound 'X', on treatment with concentrated H2SO4, produces brown fumes and gives dark brown ring with FeSO4 in presence of concentrated H 2SO 4. Also, compound 'X' gives precipitate 'Y', when its solution in dilute HCl is treated with H 2S gas. The precipitate 'Y' on treatment with concentrated HNO3, followed by excess of NH 4OH further gives deep-blue coloured solution. Compound 'X' is
(1) Co(NO3)2 (2) P(NO2)2
(3) Cu(NO3)2 (4) Pb(NO3)2t
JEE Advanced
14. The treatment of an aqueous solution of 3.74 g of Cu(NO3)2 with excess KI results in a brown solution along with the formation of a precipitate. Passing H 2S through this brown solution gives another precipitate X. The amount of X (in g) is ________. [Given: Atomic mass of H = 1, N = 14, O = 16, S = 32, K = 39, Cu = 63, I = 127]
15. The green colour produced in the borax bead test of a chromium(III) salt is due to (1) CrB (2) Cr2(B4O7)3
(3) Cr(BO2)3 (4) Cr2O3
CHAPTER TEST – JEE MAIN
1. Aq. solu tion of potash alum produces precipitate with (1) dil. HCl (2) dil. HCl + H2S (3) NH4OH + NH4OH (4) excess of KOH
2. Which one of the following gives positive 2, 4–DNP test, Tollen's test, Fehling's test but cannot give a positive haloform test?
5. A mixture when rubbed with organic acid, smells like vinegar. It contains ________. (1) sulphur (2) nitrate (3) nitrite (4) acetate
6. Which of the following combines with Fe(II) ions to form a brown complex? (1) N2O (2) NO (3) N2O3 (4) N2O4
7. Which of the following reagents can be used for making the distinction between AgCl and AgI?
(1) Sodium arsenite solution
(2) Dilute ammonia solution
(3) Both (1) and (2)
(4) Dilute HNO3
8. There are four test tubes containing dilute HCl, BaCl 2 , CdCl 2 and KNO 3 solutions. Which of the following reagents will help in the identification of BaCl 2?
(1) NaOH (2) K2CrO4
(3) AgNO3 (4) both (2) and (3)
9. Ammoni a/ammonium ion gives yellow precipitate with (1) H2PtCl6 (2) HgCl2
(3) Na3[Co(NO2)6] (4) (1) and (3) both
10. In which of the following reactions, white precipitate is not obtained as one of the reaction products?
(1) Pb2+(aq) + CO32−(aq) + H2O (l) → Products
(2) Pb2+(aq) + Br (aq) → Products
(3) Ag+(aq) + NH3(aq) + H2O(l)→ Products
(4) Ag+(aq) + Cl (aq) → Products
11. H2S in the presence of HCl precipitates II group but not IV group because (1) HCl activates H2S
(2) HCl increases concentration of Cl
(3) HCl decreases concentration of S 2−
(4) HCl lowers the solubility of H 2 S in solution
12. When excess of dilute NH4OH is added to an aqueous solution of copper sulphate, an intense blue colour is developed. This is due to the formation of
(1) [Cu(NH3)6]2+
(2) Cu(OH)2
(3) [Cu(NH3)4]2+
(4) (NH4)2SO4
13. K 4[Fe(CN) 6] can be used to detect which one or more out of Fe2+, Fe3+, Zn+, Cu2+, Ag+, Ca2+?
(1) Only Fe2+, Fe3+
(2) Only Fe3+, Zn2+, Cu2+
(3) All but not Ca2+
(4) Fe2+, Fe3+, Zn+, Cu2+, Ag+, Ca2+
14. Which one of the following ions does not give borax bead test ?
(1) Cr3+ (2) Cu2+
(3) Mn2+ (4) Zn2+
15. An aq. solution containing Hg22+, Hg2+, Pb2+, and Cd2+ ions is mixed with dil. HCl. Which will be precipitated
(1) Hg2Cl2
(2) PbCl2
(3) Both (1) and (2)
(4) CdCl2 only
16. Disodium hydrogen phoshate is used to test (1) Mg2+ (2) Na2+ (3) Ca2+ (4) Cu2+
17. Sodium extract is heated with conc. HNO3. Then,
(1) silver halides are insoluble in HNO 3
(2) Na 2S and NaCN are decomposed by HNO3
(3) Ag2S is soluble in HNO3
(4) AgCN is soluble in HNO3
Fuse this precipitate on charcoal with Na2CO3 and extract the soluble substance with H2O
18. Na2SO4 + BaCl2 White precipitate
Aqueous solution
Add dil H2SO4 and heat the solution
Gas 'G' is envolved
The gas ‘G’ will show which of the following properties?
a) It turns lead acetate filter paper black.
b) It turns acidified K2Cr2O7 filter paper green.
c) It produces a purple colouration on filter paper moistened with sodium nitroprusside already made alkaline with sodium hydroxide.
(1) Only (a)
(2) (a) and (b)
(3) (b) and (c)
(4) (a), (b) and (c)
19. Match column-I (colour) with column-II (compound).
Column I
Column II
(A) White crystalline ppt (p) Sb2S3
(B) Reddish brown ppt (q) Cr(OH)3
(C) Orange ppt (r) PbCl2
(D) Yellow ppt (s) Fe(OH)3
(E) Green ppt (t) K3[Co(NO2)6]
Choose the correct option.
(A) (B) (C) (D) (E)
(1) p r q t s
(2) r s p t q
(3) r s t p q
(4) q s r t p
20. In third group, bromine water is used to test
(1) Fe3+ ions
(2) Cr3+ ions
(3) Al3+ ions
(4) Fe2+ ions
21. Among PbCl 2, BaCl 2, ZnCl 2, ACl, HgCl 2, and CaCl 2, the total number of insoluble chlorides is _____.
22. The qualitative analysis to the salt solution addition of reagents NH 4Cl +NH4OH and ammonium carbonate give a precipitate. The oxidation state of anion in the precipitate is _______. (only magnitude)
23. Find the number of basic radicals among the following cations, which can form soluble complex on adding excess of NaOH solution. Al 3+ (aq), Ni 2+ (aq), Mn 2+ (aq), Zn 2+ (aq), Ag1+(aq), Hg2+(aq), Fe2+(aq), Mg2+(aq),
24. Adding excess of chlorine water to potassium iodide solution containing little carbon tetra chlorides first gives violet colour to carbon tetrachloride and, then, violet colour is discharged. The change in the oxidation state of iodine from initial to final compound is _____.
25. Dimethyl glyoxamate ion with Ni+2 forms rosy red colour in alkaline medium. The coordination number of metal ion in complex is _____.
CHAPTER TEST – JEE ADVANCED
1. Among PbS, CuS, HgS, MnS, Ag2S, NiS, CoS, Bi 2S 3 and SnS 2, the total number of black coloured sulphides is _____.
2. Among PbCl2, BaCl2, ZnCl2, AgCl, HgCl2, and CaCl 2, the total number of insoluble chlorides is _____.
3. Among CuSO 4 , ZnSO 4 , BaSO 4 , PbSO 4 , MnSO 4, and Na 2SO 4, the total number of insoluble chlorides is _____.
4. Among the following, the number of cations that give precipitate with Na2HPO4 solution is _____.
Mg2+, K+, Mn2+, Zn2+, Ag+, Fe3+
5. Among the following, the number of sulphides that are not soluble in yellow ammonium sulphide is _____.
As2S3, Sb2S3, CuS, Bi2S3, CdS, SnS2
6. A mixture contained aluminium sulphate and sodium sulphate. A sample of the mixture weighing 3.36 g was dissolved in water and treated with ammonium hydroxide. The precipitate formed was incinerated to yield 0.5 g of anhydrous aluminium oxide. The percentage of aluminium sulphate in the mixture is ' x ' . Find the value of 10 x (At. wt. of Al = 27, Na = 23, S = 32, O = 16, N = 14, H = 1)
7. How many of the following compounds produce(s) a gas (X) (which burns with a blue flame and reduces PdCl2(aq) to black precipitate of elemental Pd), on heating with concentrated sulphuric acid? (KCN, K 4 [Fe(CN) 6 ], K 3 [Fe(CN) 6 ], [Fe(CN) 2 ], [Fe(CN)3], Hg(CN)2, Ca(CN)2, HCOONa, CH3COONa
8. ()
Gas Y has been allowed to react with the following species in a neutral acidic medium:
(a) FeCl3
(b) CuSO4
(c) BaCl2
(d) SO2
(e) Cr2O72–
(f) CH3COONa2
(g) Hg2+
P: Number of species that undergo redox reaction with gas Y
Q : Number of species with which gas Y undergoes precipitation
R: Number of species with which gas Y produces no observable change.
Then, calculate the value of ( P + Q – R).
9. Which of the following mixtures of ions in solution can be separated by using NaOH solution?
(1) Fe3+ and Pb2+ (2) Pb3+ and Sn2+
(3) Zn3+ and Sn2+ (4) Al3+ and Cu2+
10. Which of the following statements are correct?
(1) Chromyl chloride test must not be performed in the presence of chlorates.
(2) Chromyl chloride is acidic in nature.
(3) Hydrochloric acid does not give chromyl chloride test.
(4) Chromyl chloride test should be carried out only in dry test tube.
12. Which of the following cations cannot be separated by passing H 2 S through their solutions to which NH4Cl and NH4OH have been added?
(1) Ca2+ and Ni2+ (2) Mg2+ and Mn2+
(3) Ni2+ and Mn2+ (4) Co2+ and Zn2+
13. A green coloured compound A, on heating, gives two products B and C. Metal D is deposited by passing hydrogen gas through heated B. The solution of B in HCl, on treatment with potassium ferrocyanide, gives brown-coloured precipitate. C turns lime water milky. Which of the following cannot be A?
(1) CuSO4
(2) CuCO3
(3) CuCl2
(4) Cu(NO3)2
14. Choose the correct statement (s) among the following.
(1) When aqueous KF is added to blue CuSO4 solution, a green CuF2 is precipitated
(2) If aqueous KCl is added to blue CuSO4 solution, a green [CuCl4]−2 complex is formed.
(3) If aqueous KI is added to blue CuSO4 solution a brown solution and a white precipitate are formed.
(4) When aqueous KCN is added in excess to a blue solution of CuSO4, it becomes colourless.
ANSWER KEY
JEE Main
15. Fe(OH)3 can be separated from Al(OH)3 by the addition of
(1) NaCl solution
(2) dil. HCl solution
(3) NaOH solution
(4) NH4Cl and NH4OH
16. NH4 CNS can be used to test one or more of (I)Fe3+, (II)Cu2+, (III)Co2+
(1) I only (2) II and III only
(3) I and II only (4) I, II, and III only
17. CoS (black) is dissolved in aqua regia and is treated with excess of NaHCO 3 and Br2 water. An apple-green coloured stable complex is formed. It is
(1) sodium cobaltibromide
(2) sodium cobalticarbonate
(3) sodium cobaltocarbonate
(4) sodium cobaltobromide
18. A pale green crystalline inorganic salt (A) dissolves freely in water. It gives a brown precipitate on addition of aqueous NaOH. The solution of (A) also gives a black precipitate on bubbling H 2 S in alkaline medium. An aqueous solution of (A) decolourises the pink colour of the permanganate solution. The metal in the salt solution is
