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IL Ranker Series Chemistry for JEE Grade 11 Module 4
ISBN 978-81-983575-2-6 [SECOND
EDITION]
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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.
REDOX REACTION CHAPTER 7
Chapter Outline
7.1 Classical Idea of Redox Reactions – Oxidation and Reduction Reactions
7.2 Redox reactions in Terms of Electron Transfer Reactions
7.3 Oxidation Number
7.4 Redox Reactions and Electrode Process
The subject chemistry deals with a wide variety of substances and the changes they undergo. The substances undergo different kinds of reactions. One such reaction is redox reaction. These reactions are found in our daily life, including agricultural, pharmaceutical, biological, and industrial areas.
7.1 CLASSICAL IDEA OF REDOX REACTIONS – OXIDATION AND REDUCTION REACTIONS
7.1.1 Oxidation
The term oxidation was first used to describe chemical reactions in which oxygen is added to an element or a compound. This is because of the presence of oxygen in the atmosphere. Many elements combine with it and this is the principal reason why they occur in the form of oxides.
■ Oxidation of sulphur in oxygen gives sulphur dioxide.
S+O2→ SO2
■ Oxidation of sulphite by hydrogen peroxide gives sulphate.
Na2SO3 + H2O2 → Na2SO4 + H2O
Removal of hydrogen or addition of an electronegative element or removal of an electro positive element is also known as oxidation.
■ Hydrogen is removed from hydrogen sulphide on oxidation with chlorine (or) oxygen.
H2S + Cl2 → 2HCl + S
■ Metals are oxidised by adding more electronegative fluorine or chlorine.
2Fe + 3F2 → 2FeF3 Mg+Cl2 → MgCl2
■ More electropositive iodine is removed by bromine from potassium iodide, which is oxidised to iodine.
2KI + Br2→ 2KBr + I2
I n the modern concept, loss of electron or electrons is oxidation. In other words, oxidation is de-electronation.
Zn → Zn+2 + 2e–2S2O32– → S4O62– + 2e–
A substance that undergoes reduction acts as an oxidant. An oxidant is an electron acceptor.
The strongest oxidant is fluorine. Ozone is also a good oxidant. The other important oxidising agents are: potassium permanganate, manganese dioxide, potassium dichromate, chlorine, oxygen, sodium hypochlorite, hydrogen peroxide, cupric oxide, etc.
7.1.2 Reduction
Th e term reduction is just the reverse of oxid ation. Removal of oxygen or addition of hydrogen is called reduction. Removal of an electronegative element or addition of an electropositive element is also called reduction.
■ Reduction of copper oxide with coke gives metal.
CuO + C → Cu + CO
■ Ethylene is reduced to ethane on addition with hydrogen.
C2H4 + H2 → C2H6
■ Chlorine is removed from ferric chloride by stannous chloride to give ferrous chloride.
2FeCl3 + SnCl2 → 2FeCl2 + SnCl4
■ Cupric chloride is reduced to cuprous chloride on addition with copper.
CuCl2 + Cu → Cu2Cl2
In the modern concept gain of electron or electrons is reduction. Reduction is electronation.
Al3+ + 3e– → Al
MnO4– + 8H+ + 5e– → Mn2+ + 4H2O
A substance which undergoes oxidation acts as reductant. A reductant is an electron donor. An oxidant is an electron acceptor. The redox reactions and reagents are given in the Table 7.1.
Al ka li m etals are strong reducing agents. Lithium is the strongest reducing agent. The other important reducing agents are: sodium, magnesium, aluminium, coke, hydrogen, sulphur, hydrogen sulphide, potassium iodide, sodium oxalate, lithium aluminium hydride, etc.
■ Zn + CuSO4 → Cu +ZnSO4
In the above reaction, zinc acts as a reductant and reduces copper ions of copper sulphate solution to metallic copper. Zinc transfers two electrons to copper ion. Here copper ion acts as an oxidant.
■ Cu + 2AgNO3 → 2Ag + Cu(NO3)2
In the above reaction, copper acts as a reductant and reduces silver ions of silver nitrate solution to metallic silver. Silver ion acts as oxidant.
1 2Cu2O(s) + Cu2S(s) → 6Cu(s) + SO2(g) is a redox reaction. Identify the species oxidised. What acts as an oxidant?
Oxidation Addition of oxygen or more elctronegative element; removal of hydrogen (or) removal of electropositive elements from the substance. Oxidation is loss of electrons.
Reduction Addition of hydrogen or a more electropositive element; removal of oxygen or electronegative element from a substance. (Reduction is the gain of electrons.)
In this reaction, copper is reduced from + 1 state to zero oxidation state and sulphur is oxidised from – 2 state to + 4 state. The above reaction is thus a redox reaction.
Oxidant A substance that undergoes reduction is oxidising agent. It is a substance that gains electrons.
Reductant A substance that undergoes oxidation is a reducing agent. It is a substance that loses electrons.
Table 7.1 Redox reactions and reagents
Further, Cu2O helps sulphur in Cu2S to increase its oxidation number. Therefore, Cu(I) is an oxidant; and sulphur of Cu2S helps copper both in Cu2S itself and Cu2O to decrease its oxidation number. Therfore, sulphur of Cu2S is reductant.
7.1.3 Oxidant and Reductant
Oxidising agent (or oxidant) is a reagent that can increase the oxidation number of an element in a given substance. Oxidation number of oxidising agent decreases in the redox reaction.
Ex: KMnO4, K2Cr2O7, O3, H₂O₂, F₂, Cl₂...etc
Reducing agent ( or reductant) is a reagent that can lower the oxidation number of an element in a given substance. Oxidation number of a reducing agent increases in a redox reaction.
Ex: H2S, CH4, C, CO,
The oxidising and reducing nature of compound depends upon the, oxidation number of central atom, and nature of the compound
Compounds in which element is in highest oxidation state can act as oxidising agent
Ex: H2SO4,HNO3,KMnO4,K2Cr2O7...
Compounds in which element is in lowest oxidation state can act as reducing agent.
Ex: H2S, CH4,NH3,HI....
Compounds in which element is in intermediate oxidation state, usually tends to act as both oxidant and reductant.
Ex: HNO2,H2SO3,SO2....
Try yourself:
1. In the reaction, Na2O2+SO2 → Na2SO4, which one is oxidant?
Ans: 2O2Na
TEST YOURSELF
1. M nO 2 +4HCl → MnCl 2 +Cl 2 +2H 2 O, MnO 2 acts as (1) oxidant (2) reductant
(3) both 1 and 2 (4) can’t be predicted
2. In the reaction Mg + N2 →Mg 3N2, the correct option is
(1) magnesium is reduced (2) magnesium is oxidised (3) nitrogen is oxidised
(4) both magnesium and nitrogen is oxidised
3. In C + H2O → CO + H2, H2O acts as (1) oxidising agent (2) reducing agent (3) both (1) and (2) (4) dehydrating agent
4. In a reaction between zinc and iodine in which zinc iodide is formed, what is being oxidised?
5. In the reaction 2Ag+2H2SO4→ Ag2SO4+2H2O+SO2, sulphuric acid acts as (1) oxidising agent
(2) reducing agent
(3) oxidising as well as reducing agent
(4) catalyst
6. In the reaction 2Al+N2→ 2 AlN, Al is (1) reduced (2) oxidised
(3) oxidising agent
(4) catalyst
7. In which of the following reactions, hydrogen acts as an oxidising agent?
(1) With iodine to give hydrogen iodide. (2) With lithium to give lithium hydride. . (3) With nitrogen to give ammonia. (4) With sulphur to give hydrogen sulphide.
8. Reduction does not involve, the
(1) removal of an electronegative element
(2) addition of an electropositive element
(3) removal of an electropositive element
(4) decrease in oxidation number
Answer Key
(1) 1 (2) 2 (3) 1 (4) 3
(5) 1 (6) 2 (7) 2 (8) 3
7.2 REDOX REACTIONS IN TERMS OF ELECTRON TRANSFER REACTIONS
We have learnt the redox reactions
2Na(s) + Cl2(g) → 2NaCl (s)
4Na(s) + O2(g) → 2Na2O(s)
2Na(s) + S(s) → Na2S(s) in these reactions sodium is oxidised due to the addition of oxygen or more electronegative element to sodium.
Simultaneously, chlorine, oxygen and sulphur are reduced because, the electropositive element is added. These ionic compounds formation can be shown as
Development of charges on the species produced suggests us to rewrite the reactions loss of 2e–loss of 2e–loss of 2e–
2Na(s) + Cl2(g)
2Na+ Cl– (s)
2Na(s) + O2(g)
2Na(s) + S(s) (Na+)2O2–(s) (Na+)2S2–(s) gain of 2e–gain of 2e–gain of 2e–
For convenience, each of the above processes can be considered as two separate steps, one involving the loss of electrons and the other
the gain of electrons. As an illustration, we may further elaborate one of these, say, the formation of sodium chloride.
2 Na(s) → 2 Na+(g) + 2e–Cl2(g) + 2e– → 2 Cl–(g)
Each of the above steps is called a half reaction, which shows involvement of electrons. Sum of the half reactions gives the overall reaction :
2Na(s) + Cl2 (g) → 2NaCl (s)
Reactions suggest that half reactions that involve loss of electrons are called oxidation reactions. Similarly, the half reactions that involve gain of electrons are called reduction reactions.
In above reactions sodium, which is oxidised, acts as a reducing agent because it donates electron to each of the elements interacting with it and thus helps in reducing them. Chlorine, oxygen and sulphur are reduced and act as oxidising agents because these accept electrons from sodium.
7.2.1 Competative Electron Transfer Reactions.
Electrochemical cell is a device in which electrical energy is produced at the expense of chemical energy.
When plates of two dissimilar metals are placed in a conducting liquid, such as an aqueous solution of a salt, the resulting system becomes a source of electricity. When a piece of zinc is kept in copper sulphate solution for some time, the zinc piece turns red. This is due to the deposition of copper on zinc. The reaction is given as
Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)
In this redox reaction, two electrons are transferred from zinc atom to cupric ion. However, these electrons cannot be trapped to obtain any small quantity of electrical energy because they are directly transfering from zinc to Cu+2 within the vessel. When the reactants are separated from each other and given an indirect contact, the electrical energy can be trapped.
A device in which electrical energy is gen erated by performing a redox reaction is called galvanic cell. this cell is also called voltaic cell or electrochemical cell.
So, chemical energy is converted into electrical energy in a galvanic cell. A galvanic cell can be reversed to function as an electrolytic cell. A typical galvanic cell is shown in
Zinc anode
Copper cathode
Zinc sulphate solution Coper sulphate solution
Electrochemical series or Activity series
The arrangement of electrodes in the order of increasing standard reduction potentials is called electrochemical series. This is also called activity series.
7.3 OXIDATION NUMBER
Oxidation number is defined as the residual charge, which an atom has or appears to have in a molecule when all other atoms are removed from the molecule as ions.
Oxidation number is frequently used interchangeably with oxidation state. The stock notations of oxidation numbers are based on periodic property and electronegativity.
An atom in a molecule can be assigned positive, negative, or zero oxidation number by considering its environment. In a few cases, oxidation number can even be fractional.
The rules useful for the oxidation number are stated below:
■ Each atom in an uncombined state of an element in any of its allotropic forms is assigned an oxidation number zero.
■ Oxidation number of H in hydrogen (H 2) is zero
■ Oxidation number of Au in gold is zero.
■ Oxidation number of S in sulphur S8, is zero.
■ Oxidation number of Cl in chlorine (Cl 2) is zero.
■ Oxidation number of an alkali metal in all its compounds is +1 and of an alkaline earth metal in all its compounds is +2.
■ Oxidation number of helium or neon is always zero, since they have no chemical activity. Oxidation state of a metal in an alloy and in a metal carbonyl is zero.
■ The oxidation number of fluorine in all its compounds is –1. Fluorine does not exhibit positive oxidation states. This is due to its highest electronegativity.
■ Oxidation state of hydrogen in covalent or molecular hydrides is +1 and zero in interstitial hydrides.
■ Oxidation state of H in metal hydride or ionic hydride or saline hydride is –1.
Examples: LiH, NaH, CaH 2
■ Oxidation number of oxygen in its compounds is usually –2.
Oxidation number of O in alumina is –2.
Oxidation number of O in water is –2.
However, oxidation state of oxygen in peroxides like H2O2, Na2O2, BaO2, etc., is –1 and in super oxides, like KO2, RbO 2, CsO 2, etc., it is – 1/2 . Oxygen exhibits +1 and +2 in O2F2 and OF2, respectively, because fluorine is more electronegative than oxygen.
■ The algebraic sum of oxidation numbers of all the atoms in a neutral molecule is zero. If the species is an ion, the algebraic sum of oxidation numbers of all atoms in an ion is equal to the charge of the ion.
■ Oxidation number of an atom in its charged mono-atomic ion is equal to the charge on the ion.
■ Oxidation state of Cl in chloride (Cl–) is –1.
■ Oxidation state of O in oxide (O –2) is –2.
■ Oxidation state of P in phosphide (P –3) is –3.
■ Oxidation state of Fe in ferrous ion (Fe +2) is +2.
■ Cl 2 , Br 2 , and I 2 , when combined with oxygen in oxoacids and oxoanions, have positive oxidation numbers.
■ The maximum oxidation number exhibited is equal to its group (Roman) number in the long form of the periodic table. By applying these rules, the oxidation number of any element in a molecular substance can be calculated.
It is clear that metallic elements have
positive oxidation number, while nonmetallic elements can have negative or positive oxidation states.
The highest oxidation state (+8) is known for osmium in osmium tetra oxide. The least oxidation state (–4) is known for carbon in methane.
If more atoms of an element are present in the molecular formula of a compound, the oxidation number of the element may be taken as the average of oxidation numbers of all atoms of that element.
Nitrogen can exhibit oxidation states from –3 to +5 in its compounds.
Some interesting examples of different oxidation numbers of atoms of the same element are listed in Table 7.2
Table 7.2 Examples of different oxidation number of atoms of same element
Stock Notation
The oxidation number is expressed by putting a Roman numeral in parenthesis after the symbol of the metal in the molecular formula. Thus, cuprous chloride and cupric chloride are written as Cu(I)Cl and Cu(II)Cl2. Similarly the compounds Tl2O, FeO, CuO, and MnO can be represented as Tl2(I)O, Fe(II)O, Cu(II)O, and Mn(II)O.
Oxidation and Reduction in Terms of Oxidation Number
Oxidation is a chemical change in which there is an increase in oxidation number.
Reduction is a chemical change in which there is a decrease in oxidation number.
In this reaction, oxidising agent is Cu2O and reducing agent is Cu 2S.
TEST YOURSELF
1. In which of the following compounds does sulphur exhibit least oxidation state?
(1) S2Cl2 (2) H2S2O7 (3) SOCl2 (4) Na2S2O3
2. The oxidation state(s) of Cl in CaOCl 2 (bleaching power) is/are (1) +1 only (2) –1 only (3) +1and –1 (4) –1 and –1
3. The oxidation state of nickel in [Ni(CO) 4] is (1) +2 (2) +4 (3) +5 (4) zero
4. In S 4O 6 2–, the oxidation numbers of four sulphur atoms are
(1) +4, 0, 0, and +6 (2) +4, –2, –2 and +6 (3) +5, 0, 0, and +5 (4) +5, –2, –2 and +5
5. A compound contains atoms of three elements A, B, and C. If the oxidation number of A is +2, B is +5, and that of C is –2, the possible formula of the compound is
(1) A3(BC4)2 (2) A3(B4C)2 (3) ABC2 (4) A2(BC3)2
6. Oxidation numbers of Cr in CrO 5 and K2Cr2O7, respectively, are (1) 5 and 6 (2) 10 and 6 (3) 6 and 6 (4) 6 and 5
7. Oxidation state of iron in Fe 3O4 cannot be (1) +3 (2) +2 (3) both (1) and (2) (4) +5
8. The sum of oxidation numbers of nitrogen in ammonium nitrite (NH 4NO2) is (1) +5 (2) 0 (3) 1 (4) –3
9. The oxidation number of P is + 3 in (1) H3PO3 (2) H3PO4 (3) HPO3 (4) H2P2O7
10. Oxidation number of C in CH 2Cl2 is (1) +2 (2) +4 (3) –4 (4) 0
Answer Key
(1) 1 (2) 3 (3) 4 (4) 3
(5) 1 (6) 3 (7) 4 (8) 2 (9) 1 (10) 4
7.4.1 Types of Redox Reactions
Most of the chemical reactions involve electron transfer. These reactions may be arranged in different types like combination, decomposition, displacement, disproportionation, comproportionation, etc.
Combination Reactions
Magnesium metal has zero oxidation state. When it is burnt in air, it is oxidised. It forms magnesium oxide and magnesium nitride. Oxygen is reduced to oxide and nitrogen to nitride. These are called chemical combination reactions.
2Mg+O2→ 2MgO Reduction
oxygen is oxidised. This is called a chemical decomposition reaction.
MnO2 32 2KClO2KCl3O →+
Compounds are chemically split into two or more simpler substances during decomposition reactions. Decomposition of cinnabar on heating produces mercury metal.
2HgO → 2Hg + O2
Other examples:
3Mg Mg3 N2 0 0 +2 –3
Oxidation + N2
Combustion of graphite, formation of alumina from its elements, rusting of iron, oxidation of sulphur, etc., are other familiar chemical combination reactions.
All combustion reactions in which oxygen undergoes reduction from ‘0’ to ‘–2’ are redox reactions.
C2H4(g) –2 0 +3O2(g) 2CO2(g) +2H2O(I) +4 –2
Decomposition Reaction
Oxygen is prepared in the laboratory by heating potassium chlorate in the presence of manganese dioxide catalyst. In this reaction, chlorine in chlorate is reduced and
2(I) 2HO2HO2(s)2(g) ∆ →+ (S) 2NaH2NaH2(s)2(g) ∆ →+
Displacement Reactions
Metal displacement: An active metal displaces, relatively less active metal from its salt solution.
( ) ( ) ( ) ( ) 2 s aq aq s Cu2AgCu2Ag ++ +→+ ( ) ( ) ( ) ( ) 25 ssss VO5Ca2V5CaO +→+
The above reactions are called metal displacement reactions.
Metal displacement reactions find many applications in metallurgical processes in which pure metals are obtained from their compounds or ores.
TiCl4 + 2 Mg → Ti + 2MgCl2
Displacement reactions are useful in predicting the reactivity of metals based on electrode potentials. The electron releasing tendency of the metals, Zn, Ag, Cu is in the order Zn > Cu > Ag.
Non-metal displacement: All alkali metals and some alkaline earth metals (Ca, Sr, Ba) displace hydrogen from cold water.
Example: 2Na(s)+2H2O(I) → NaOH(aq)+H2(g)
Less active metals, such as Fe, Mg, and Al, react with steam to produce H 2 gas.
2 232 2Fe3HOFeO3H ∆ +→+
Metals like Cd and Sn do not react with cold water and steam but liberate H 2 gas with dilute acids.
Sn + 2HCl → SnCl2+H2
Fluorine is the most powerful oxidising agent. It oxidises water to oxygen and potassium bromide to bromine. Chlorine, bromine, and iodine cannot oxidise water.
2F2 + 2H2O → O2 + 4HF
F2 + 2KBr → Br2 + 2KF
A lighter halogen can displace heavier halogen from its halide compound.
Cl2(g) + 2KBr(aq) → 2KCl(aq) + Br2(l)
Cl2(g) + 2KI(aq) → 2KCl(aq) + I2(s)
The above reactions are called non-metal displacement reactions.
As F 2 is the strongest oxidant, it cannot be prepared from F– by chemical oxidation, whereas the remaining halogens can be prepared by chemical oxidation of halide ion. F 2 can be prepared from F – by electrolysis only.
Disproportionation Reaction
The reactions in which the same element undergoes both oxidation and reduction simultaneously are called disproportionation reactions.
Disproportionation reactions are possible if the element exhibits a minimum of three oxidation numbers. The oxidation number of the reactant substance must not be in one of the extreme oxidation sta tes.
Some Examples of Disproportionation Reaction:
■ Reaction of white phosphorus with aqueous caustic soda solution
P4+3NaOH+3H2O → PH3+3NaH2PO2
■ Reaction of hot concentrated potash with bromine
6KOH + 3Br2 → 5KBr + KBrO3 + 3H2O
■ Hydrolysis of dimeric sulphur monochloride
2S2Cl2 + 2H2O → 4HCl + SO2 + 3S
■ Neutralisation of nitrogen dioxide with caustic soda solution
2NO2+2NaOH → NaNO2 + NaNO3 + H2O
■ Reaction of sulphur with caustic soda
S8+12NaOH → 4Na2S+2Na2S2O3+6H2O
■ Reactions of chlorine with caustic soda
Cl2(g)+ 2NaOH(g)→NaClO(aq)+ NaCl(aq)+ H2O(l)
The hypochlorite ion (ClO–) formed in the reaction oxidises the colour bearing stains of the substances to colourless compounds.
Comproportionation Reactions
In these reactions, two species with the same element in two different oxidation states form a single product. In the product, the oxidation state of the element is intermediate between those of the reactants.
Some examples of comproportionation reaction:
■ AgSO4+Ag → Ag2SO4
■ C + CO2 →2CO
■ CuCl2 +Cu →Cu2Cl2
2. Reaction between hydrogen sulphide and sulphurdioxide gives sulphur. Which type of redox reaction is this?
Sol. The reaction between hydrogen sulphide and sulphur dioxide is given as,
2H2S + SO2 → 3S + 2H2O
Oxidation state of S in H2S is –2 and is oxidised to ‘0’. Oxidation state of S in SO 2 is +4 and is reduced to ‘0’. In the product, sulphur shows zero oxidation number, which is in between –2 and +4 of the sulphur in the reactants. Therefore, this reaction is an example of comproportionation.
Try yourself:
2. 4KClO3 → 3KClO4+KCl What type of reaction is this?
Ans: Disproportionation
TEST YOURSELF
1. Which of the following is only a redox reaction but not a disproportionation reaction?
(1) 4H3PO3 → 3H3PO4+PH3
(2) 2H2O2 → 2H2O+O2
(3) P4+3NaOH+3H2O → 3NaH2PO2+PH3
(4) P4+8SOCl2→ 4PCl3+2S2Cl2+4SO2
2. Which one of the following does not undergo disproportionation in alkaline medium?
(1) F2 (2) S (3) Cl2 (4) P 4
3. Which of the following is not an example of a decomposition reaction?
(1) 2HgO → 2Hg+O2
(2) 2H2O → 2H2+O2
(3) 2KClO3→ 2KCl+3O2
(4) CH4+2O2→ CO2+2H2O
4. Which of the following is a decomposition reaction?
(1) 2HgO → 2Hg+O2
(2) CH4+2O2→ CO2+2H2O
(3) S+O2→ SO2
(4) Cl2+2KBr → 2KCl+Br2
5. 2CuI → Cu+CuI2, the reaction is (1) disproportionation
(2) neutralisation
(3) oxidation
(4) reduction
6. Which of the following is not a comproportionation reaction?
(1) Ag+2 + Ag → Ag+1
(2) SO2+H2S → S
(3) IO3–+I–→I 2
(4) H2O2 → H2O + O2
Answer Key
(1) 4 (2) 1 (3) 4 (4) 1
(5) 1 (6) 4
Balancing of redox reactions
Equations of redox reactions are balanced with respect to both the number of atoms and charge. This can be done by following a set of steps.
Oxidation Number Method
Oxidation number method is also called electron transfer method. This method is useful for balancing equations of both ionic as well as molecular reactions. The method focuses on the atoms of the elements undergoing a change in oxidation number. The different steps involved in balancing are listed below:
■ Write the skeleton equation representing a redox reaction.
■ Write the oxidation numbers of each atom at the top of it.
■ Locate the atoms undergoing changes in the oxidation numbers.
■ Find the increase and decrease in oxidation number of the atoms of the given substance
If more than one atom of the same element is involved, find the total increase
by multiplying the increase and decrease in oxidation number by the number of atoms undergoing that change.
■ Equalise the total increase in the number of units of charge to the decrease in the number of units of charge, by multiplying with suitable integers. Write them as coefficients of the corresponding substances of the equation.
■ Balance all the atoms other than H and O atoms.
■ Finally, balance O atoms by adding a suitable number of H 2 O molecules, at whichever side required.
■ In case of ionic equations balance 'O' atoms by adding H 2 O molecules to the required side of equation. Balance H-atoms by adding H+ions to the required side of equation.
■ In case the reaction is in basic medium, follow all the above steps and add OH–ions equal to the number of H+ions on both sides of the equation. Write each H ++OH– pair as H2O
■ Remove the duplication, if any. Finally, make all the stoichiometric coefficients in simplest whole number ratio.
Half Reaction Method
Half reaction method is also called ion–electron method. This method is useful for balancing ionic equations, leaving spectator ions. The total reaction is divided into two half reactions. The different steps involved in balancing are listed below:
■ Write the ionic equation representing the reaction.
■ Write the oxidation numbers of each atom at the top of it in the equation.
■ Locate the atoms undergoing changes in the oxidation number.
■ S plit the reaction into two halves, one representing oxidation and the other reduction. Then, balance each half-reaction separately, as follows:
■ Balance the atoms other than hydrogen and oxygen atoms.
■ Balance the oxygen atoms by adding the required number of water molecules to the side deficient in oxygen atoms.
To balance hydrogen atoms, follow these steps:
(i) If the reaction is in acid medium, add the required number of H+ ions to the side deficient in hydrogen atoms.
(ii) If the reaction is in basic medium, follow all the above steps. Add the number of hydroxyl ion equal to the number of H+ions on both sides of equation. Write each H + +OHpair as H 2 O. Remove duplication, if any. Finally, make the ratio of stoichiometric coefficients as simplest whole number ratio.
Add required number of electrons to balance the charges.
■ Equalise the number of electrons in the two half-reactions by multiplying each equation with the required number. Finally, add the two reactions to get the final e quation, which has no free electrons on either side and is balanced in terms of atoms as well as charges.
The following are some examples: Permanganate oxidises sulphite to sulphate in acidic solutions.
The ionic skeleton equation is written as H
Writing oxidation numbers, 742222 6 222 43 4
Locating atoms undergoing change in oxidation numbers,
2 74 6 222 43 4 MnOSOMnSO + ++ + −−+− +→+
Divide the reaction into two halves and balance in acidic medium separately.
Oxidation Half-Reaction: 22 34 SOSO →
Step 1: Balance oxygen atoms.
22 SO324 HOSO+→
Step 2: Balance hydrogen atoms in acidic medium.
22 32 4 SO HO2HSO +→++−
Step 3: Balance the charge.
22 32 4 SO HO2e2HSO +→++−+− .....(a)
Reduction Half reaction:
2 4 MnOMn−+ →
Step 1: Balance oxygen atoms.
MnO4–→ Mn2+ + 4H2O
Step 2: Balance hydrogen atoms.
MnO4–+ 8H+ → Mn2+ + 4H2O
Step 3: Balance charge.
MnO4–+ 8H+ + 5e– → Mn2+ + 4H2O ....(b)
By equalising the electrons and adding the two half reactions
eq. (a) × 5 + eq. (b) × 2, we get
5SO32–+ 5H2O → 10e– + 5 SO4-2+ 10H+
2MnO4– + 16H+ +10e– → 2Mn2+ + 8H2O
2MnO4– + 5SO32– + 6H+ → 2Mn2+ + 5 SO4-2 + 3H2O
This is a balanced equation.
■ Iodate oxidises chromic hydroxide and gives iodide and chromate in basic medium.
The ionic skeleton equation is written as OH 2 IO334 Cr(OH)ICrO + →+
■ Chromium metal in basic medium is oxidised in air to give chromic tetrahydroxide anion.
This ionic skeleton equation is written as
Writing oxidation numbers,
00321
2 4 CrOCrOH +−+ +→
Locate atoms, undergoing change in oxidation numbers.
0032
CrOCrOH +− +→
2 4
Divide the reaction into two half reactions and balance in basic medium, separately.
Oxidation Half-Reaction: Cr → [Cr(OH)4]–
Step1: Balance thee oxygen atoms
This is the balanced equation.
■ White phosphorous reacts with aqueous caustic soda to give hypophosphite and phosphine.
The ionic skeleton equation is OH 4 322 PPHHPO →+
Writing oxidation numbers 2 03111
4322 PPHHPO −+++ →+
Locating atoms, undergoing change in oxidation numbers
( ) 2 4
Cr4HOCrOH +→
Step 2: Balance the hydrogen atoms
031
4322 PPHHPO −+ →+
( ) 2 4
Cr4HO4OHCrOH4H4(OH) +− ++→++
Step 3: Balance the charge
Divide the reaction into two half reactions and balance in basic medium, separately
Oxidation half-Reaction: P 4→ H2PO2–
( ) 22 4
Cr4HO4OHCrOH4HO3e ++→++ ..(a)
Reduction half reaction: O2→[Cr(OH)4]-
Step 1: Balance the chromium atoms.
Step 1: Balance the phosphorous atoms
P 4→4 H2PO2–
Step 2: Balance the oxygen atoms
( ) 2 4
CrOCrOH +→
Step 2: Balance the oxygen atoms. ( ) 22 4
OCr2HOCrOH ++→
Step 3: Balance the hydrogen atoms. ( ) 22 4
OCr2HOCrOH ++→
Step 4: Balance the charge. ( ) 22 4
OCr2HOeCrOH +++→ ----(b)
Step 5: Equalising the electrons and adding the two half reactions, eq. (a) × 1 + eq. (b) × 3, we get ( ) ( ) 22 4 22 4 22 4
P 4 + 8H2O →4 H2PO2–
Step 3: Balance the hydrogen atoms
P4+8H2O+8OH–→4 H2PO2–+8H++8(OH-)
Step 4: Balance the charge
P4+8OH- →4 H2PO2–+4e–.....(a)
Reduction Half reaction: P 4→ PH 3
Step 1: Balance the phosphorous atoms
P 4→ 4PH 3
Step 2: Balance the hydrogen atoms
P 4 +12H++12OH- → 4PH3+12OH–12H2O+ P4 → 4PH3+12OH–
Step 3: Balance the charge
Cr4HO4OHCrOH4HO3e
3O3Cr6HO3e3CrOH
4Cr3O6HO4OH4[Cr(OH)]
++→++ +++→ +++→
P 4 + 12H2O + 12e– → 4PH 3 + 12OH–
Equalising the electrons and adding the two half reactions,
eq. (a) × 3 + eq. (b) × 1, we get
3P24HO12HPO12e
P12HO12e4PH12OH
4P12OH12HO4PH12HPO
P4+3OH–+3H2O → PH3+3H2PO2–
This is the balanced equation.
■ Acetylene is oxidised by permanganate in acidic solutions to liberate carbon dioxide.
The ionic skeleton equation is written as
Writing oxidation numbers,
Locate atoms, undergoing change in oxidation numbers
Step 2: Balance the hydrogen atoms
MnO4–+8H++5e– → Mn2++4H2O.....(b)
Equalising the electrons and adding the two half reactions.
eq. (a) × 1 + eq. (b) × 2, we get
This is a balanced equation.
Equivalent Weight
Equivalent weight of a substance is defined as ‘the number of parts by weight of it that will combine with or displace 1.008 parts by weight of hydrogen or 8 parts by weight of oxygen or 35.5 parts by weight of chlorine or 108 parts by weight of silver or the equivalent parts by weight of any other element.’
Divide the reaction into two half reactions and balance in acidic medium, separately.
Oxidation Half-Reaction: C2H2→ CO2
Step 1: Balance the carbon atoms.
C2H2→ 2CO2
Step 2: Balance the oxygen atoms.
C2H2 + 4H2O →2 CO2
Step 3: Balance the hydrogen atoms.
C2H2 + 4H2O → 2CO2 + 10H+
Step4: Balance the charge.
C2H2 + 4H2O → 2CO2 + 10H+ + 10e–---(a)
Reduction Half reaction: MnO4–→ Mn2+
Step 1: Balance the oxygen atoms
MnO4–→ Mn2++4H2O
Substances in a chemical reaction need not react in 1 : 1 ratio of their moles or masses or volumes.
The reactants in a reaction always react in equal ratio of their equivalents and the products are formed in equal ratio of their equivalents.
Equivalent weight of an element = Atomicweightofelement Valency
Sa me element may possess different equ i valent weights because valency is not a constant but a variable.
Example: Equivalent weight of carbon in CO and CO2 are 6 and 3, respectively.
The equivalent weights of some elements are given in Table 7.3
Table 7.3 Equivalent weights of some elements
Table 7.4 Equivalent weights of some cations
Equivalent weight of an ion is the number of grams of the ion which can be discharged by the loss or gain of one mole of electrons.
Equivalent weight is the ratio of ionic weight and charge magnitude of ion.
Equivalent weight of ion = Ionicweight Chargemagnitudeofion
Eq uivalent weights of important cations and important anions are respectively given in Table 7.4 and Table 7.5.
Equivalent weight of an acid is the number of grams of the acid which gives one mole of protons in water. It is the ratio of molecular weight of acid and basicity of acid.
(Number of H + ions furnished by one molecule of an acid is called basic ity).
Equivalent weight of acid = Molecularweightofacid Basicityofacid
Table 7.5 Equivalent weights of some anions
E qui valent weight of a base is the number of grams of the base which gives one mole of hydroxyl ions in water. It is the ratio of molecular weight of base and acidity of base.
(The number of OH– ions furnished by one molecule of base is called acidity).
Equivalent weight of base = Molecularweightofbase
Acidityofbase
Equivalent weight of a salt it is the ratio of formula weight of salt and total number of units of positive or negative charges of the ionic substance of one formula unit.
Equivalent weight of salt = Formulaweightofsalt
Totalnumberofpositive (ornegative)chargeunits
E quivalent weight of a salt in practice is taken as the sum of equivalent weights of ions represented in the empirical formula of salt.
Equivalent weight of salt = Equivalent weight of cation + Equivalent weight of anion
Equivalent weights of important acids, bases, and salts are given in Table 7.14, Table 7.15, and Table 7.16, respectively.
Equivalent Weights of Oxidant and Reductant
Loss of electrons is called oxidation and gain of electrons is called reduction.The reactions in which both reduction and oxidation take place are called redox reactions. In redox reactions, electrons are transferred from reductant to oxidant.
The substance that loses electrons is said to be oxidised, and it is called reductant or reducing agent. The substance that gains electrons is said to be reduced, and it is called oxidant or oxidising agent.
The weight of oxidant that gains one mole electrons is called equivalent weight of oxidant and the weight of reductant that loses one mole of electrons is called equivalent weight of reductant.
Equivalent weight of oxidant or reductant = Molecular weight
No.of electrons gained or lost per molecule
Equivalent weight of oxidant or reductant=
Molecularweight
Numberofelectronsgainedorlostpermolecule
The following examples explain calculation of equivalent weights.
■ EW of a substance undergoing disproportion:
a) EW of Cl2 with cold and dil. NaOH
Cl2+OH–→ Cl–+ClO–
ECl2 = E1+E2 = MM 71 22 +=
b) EW of Cl2 with hot and conc. NaOH: 3Cl2+6OH–→ 5Cl–+ClO3–+3H2O
The correct coefficients of the reactants, 2 MnO,CO424 and H+ for the balanced reaction, respectively, are (1) 2, 5, 16 (2) 16, 3, 12 (3) 15, 16, 12 (4) 2, 16, 5
4. 2 23 42 MnOFeHMnFeHO. ++++ ++→++
The stoichiometric coefficient associated with H+ in the stoichiometric equation is (1) 6 (2) 7 (3) 8 (4) 10
5. How many of OH ions are appearing in the given balanced equation? ( ) 2 22 2 3 OH 4 CrOHHOHOCrO ++ → (1) 2 (2) 3 (3) 4 (4) 6
6. Consider the following reaction: 22 424 22 z xMnOyCOzHxMn2yCOHO 2 −−++ ++→++
The values of x, y, and z in the reaction are, respectively, (1) 2, 5, and 8 (2) 2, 5, and 16 (3) 5, 2, and 8 (4) 5, 2, and 16
7. 222 43 4 . MnOSOHMnSO −−++− ++→+
The number of H+ ions involved is (1) 2 (2) 6 (3) 8 (4) 16
8. The value of n in 2 42 MnO8HeMn4HO −++ ++→+ n is (1) 5 (2) 4 (3) 3 (4) 2
9. The number of moles of acidified KMnO4 required that can oxidise 0.5 mole of Mohr salt is
(1) 1 (2) 2 (3) 0.1 (4) 0.2
10. The number of electrons involved in the reduction of Cr2O72- ion in acidic solution to Cr3+is
(1) 3 (2) 4 (3) 2 (4) 6
11. In the equation, 223 NOHONO2Hne +→++−+− , n stands for (1) 4 (2) 1 (3) 2 (4) 3
12. The number of electrons lost or gained during the change Fe+H2O → Fe3O4+H2 is (1) 2 (2) 4 (3) 6 (4) 8
13. Molecular weight of orthophosphoric acid is M. Its equivalent weight is (1) 3M (2) M (3) M 2 (4) M 3
14. 0.5 g of a metal, on oxidation, gave 0.79 g of its oxide. The equivalent mass of the metal is
(1) 10 (2) 14 (3) 20 (4) 40
15. The equivalent weight of Bayer’s reagent, is (alkaline KMnO4)
(1) 31.6 (2) 52.6 (3) 79 (4) 158
16. Molecular weight of KMnO 4 is ‘M’. In a reaction KMnO4 is reduced to K2MnO4. The equivalent weight of KMnO 4 is
(1) M (2) 2 Μ (3) M 3 (4) 5 Μ
17. When ferrous sulphate acts as reductant, its equivalent weight is (1) twice that of its molecular weight (2) equal to its molecular weight (3) one-half of its molecular weight (4) one-third of its molecular weight
18. 2H2O → 4e– + O2 + 4H+.
The equivalent weight of oxygen is (1) 32 (2) 16 (3) 8 (4) 4
19. In acidic medium dichromate ion oxidises ferrous ion to ferric ion. If the grammolecular weight of potassium dichromate is 294 g, its equivalent weight is (1) 294 (2) 147 (3) 49 (4) 24.5
20. The equivalent weight of hypo in the reaction [M = molecular weight]
2Na2S2O3 + I2 → 2NaI + Na2S4O6 is (1) M (2) M 2
(3) M 3 (4) M 4
21. The equivalent weight of CuSO 4 when it is converted to Cu2I2 is [M=mol.wt]
(1) M 1 (2) M 2 (3) M 3 (4) 2 M
22. The atomic weight of a metal (M) is 27 and its equivalent weight is 9. The formula of its chloride will be (1) MCl (2) MCl9 (3) M3Cl4 (4) MCl3
23. Molecular weight of Mohr’s salt is 392. Its equivalent weight when it is oxidised by KMnO4 in acidic medium is (1) 392 (2) 196 (3) 130.6 (4) 78.5
24. H3PO4 + 2KOH → K2HPO4 + 2H2O
Based on the above reaction, equivalent weight of H3PO4 is (1) 196 (2) 98
(3) 49 (4) 32.67
25. In the standardisation of Na 2 S 2 O 3 using K2Cr2O7 by iodometry, the equivalent weight of K2Cr2O7 is (1) (molecular weight)/2 (2) (molecular weight)/6 (3) (molecular weight)/3 (4) same as molecular weight
Answer Key
(1) 4 (2) 4 (3) 1 (4) 3
(5) 3 (6) 2 (7) 2 (8) 1
(9) 3 (10) 4 (11) 3 (12) 4
(13) 4 (14) 2 (15) 2 (16) 1
(17) 2 (18) 3 (19) 3 (20) 1
(21) 1 (22) 4 (23) 1 (24) 3
(25) 2
Redox Reactions as the basis of titrations
Analysis of chemical substances is generally of two types: qualitative analysis and quantitative analysis. Quantitative analysis of solutions in volumes is useful in the determination of concentration, volume of the solution, and also amount of a solute. Volumetric analysis is an experimental method of measuring the volume of a known solution required to bring about the completion of the reaction with a measured volume of unknown solution. The method is also known as titrimetric analysis and is termed as titration or titrimetry.
Titration
A titration involves two solutions. One solution whose concentration is known (standard solution) and the other solution is an unknown solution. Titration is the process
of addition of a known solution to a measured volume of solution whose concentration is not known until the reaction between the two is just complete. The volume of the known solution is determined in the titration which reacts with measured volume of unknown solution.
The substance whose solution is used to estimate the concentration of an unknown solution is called titrant. The substance whose concentration is to be estimated is called titrate or titrand.
End Point
The point at which the titrant and titrate are chemically equivalent is termed as equivalence point. The theoretical equivalence point in the volumetric titration is called stoichiometric end point or, simply, end point.
End point in a titration can be detected by observing physical change (like colour) using an indicator.
Indicator
Indicator is the auxiliary substance used for visual detection of the completion of titration. It is used to find out the end point in a titration. Indicator substance generally undergoes a change in its colour at the point of the titration.
Indicator not required in permanganometry and dichrometry as the oxidants themself exhibits different colours in oxidised and reduced form. Iodine acts as self indicator in iodimetry, which is a direct titration. However, starch is used as an indicator in indirect titrations, involving iodine and hypo, called iodometry.
Principle
At equivalence point, the number of gram equivalents of titrant and titrate are equal and the same number of gram equivalents of each
7: Redox Reaction
product will form in the reaction.
Number of gram equivalents (n) of a substance is given as : 1000 = VN n where V is the volume of solution in mL and N is the normality of solution.
At Equivalence Point
V1N1 = V2N2 where V1 is the volume of titrant solution.
V2 is the volume of titrate solution.
N1 is the normality of titrant solution.
N2 is the normality of titrate solution.
The molarity equation is also used in the direct titrations. 1122 12 = VMVM nn
Here n 1 and n 2 are the stoichiometric coefficients of titrant and titrate respectively. These coefficients are obtained by balancing the chemical reaction between solute substances of solutions employed for the titration.
Oxidation Reduction Titration
These titrations based on oxidation-reduction reactions are called redox titrations. Types of oxidation-reduction titrations are explaine below.
Reaction: K 2 Cr 2 O 7 + 7H 2 SO 4 + 6FeSO 4 → K2SO4 + Cr2 (SO4)3 + 3Fe2 (SO4)3 + 7H2O
Indicator: Diphenylamine
End point: Bluish-violet or purple colour
Iodimetric and Iodometric Titrations
The reduction of free iodine to iodide ions and oxidation of iodide ions to free iodine occurs in these titrations.
I2 + 2e– → 2I– (reduction)
2I– → I2 + 2e– (oxidation)
These are divided into two types.
Iodimetric titrations: These are the titrations in which free iodine is used. As it is difficult to prepare the solution of iodine (volatile and less soluble in water), it is dissolved in potassium iodide solution.
KI + I2 → KI 3 (Potassium tri-iodide)
This solution is first standardised before use. With the standard solution of I 2 , substances, such as sulphite, thiosulphate, arsenite, etc., are estimated.
Following examples explain iodimetric titrations.
■ I2 – Na2S2O3 . 5H2O titration.
This is iodimetry titration, since standard solution of iodine is used.
Reaction: 2Na2S2O3 + I2 → Na2S4O6 + 2Nal
Indicator: Freshly prepared starch solution.
End point: Disappearance of blue colour
■ As2O3 – I2 titration
Iodometry: Iodine is liberated during chemical reactions.
Following examples explain iodometric titrations
■ CuSO4 – Na2S2O3 . 5H2O
2CuSO4 + 4KI → Cu2I2 + K2SO4 + I2
2Na2S2O3 + I2 → Na2S4O6 + 2Nal
Indicator: Freshly prepared starch solution
End point: Disappearance of blue colour or violet
■ K2Cr2O7 – Na2S2O3 . 5H2O :
Reaction: K2Cr2O7 + 7H2SO4 + 6KI →
4K2SO4 + Cr2(SO4)3 + 7H2O + 3I2
2Na2S2O3 + I2 → 2NaI + Na2S4O6
Indicator: Freshly prepared starch solution.
End point: Disappearance of blue colour.
5. Determine the volume of M/8 KMnO4 solution required to react completely with 25.0 cm3 of M/4 FeSO4 solution in acidic medium.
Sol. The balanced ionic equation for the reaction is
MnO4–+5Fe2++8H+ → Mn2++5Fe3++4H2O
From the balanced equation, it is evident that 1 mole of KMnO4≡ 5 moles of FeSO4
Applying molarity equation to the balanced redox equation, we have, 11 22 44 12 MV MV (KMnO)(FeSO) nn = 1V1 125 8145 × =× × or 3 1 1258
Thus, the volume of M/8 KMnO4 solution required = 10.0 mL.
Try yourself:
4. A solution of ferrous oxalate has been prepared by dissolving 3.6 g L –1. Calculate the volume of 0.01 M KMnO 4 solution required for complete oxidation of 100 mL of ferrous oxalate solution in acidic medium. Ans: 150 mL
TEST YOURSELF
1. In th e titration of KMnO 4 versus Mohr's salt, the indicator used is (1) KMnO4 (self) (2) diphenyl amine (3) starch (4) methyl red
2. When KMnO4 solution is added to oxalic acid solution, the decolourisation is slow in the beginning but becomes instantaneous after some time because (1) CO2 is formed as the product (2) reaction is exothermic (3) MnO4 catalyses the reaction (4) Mn2+ acts as an autocatalyst
3. 50 mL of 0.02 M KMnO4 oxidises 50 mL of potassium iodide solution in faint alkaline medium. Molarity of potassium iodide solution is (1) 0.1 M (2) 0.01 M (3) 0.0166 M (4) 0.06 M
4. The weight of I 2 (MW=254) required to completely oxidise 1 mole of Na 2 S 2 O 3 according to the following equation (unbalanced), is
I2+Na2S2O3→ NaI+Na2S4O6 (1) 127g (2) 254 g (3) 80 g (4) 98 g
5. The amount of oxalic acid present in a solution can be determined by its titration with KMnO 4 solution in the presence of H 2SO 4. The titration gives unsatisfactory result when carried out in the presence of HCl, because HCl (1) gets oxidised by oxalic acid to chlorine (2) furnishes H+ ions in addition to those from oxalic acid (3) reduces permanganate to Mn 2+.
(4) oxidises oxalic acid to carbon dioxide and water
6. Consider the titration of potassium dichromate solution with acidified Mohr’s salt solution using dimethylamine as indicator. The number of moles of Mohr’s salt required per mole of dichromate is (1) 3 (2) 4 (3) 5 (4) 6
7. The number of moles of ferrous oxalate oxidised by one mole of KMnO 4 in acidic medium is
(1) 5 2 (2) 2 5 (3) 3 5 (4) 5 3
8. 0.8 M FeSO 4 solution requires 160 mL of 0.2 M Al 2 (Cr 2 O 7 ) 3 in acidic medium. Calculate the volume of FeSO 4 consumed.
(1) 480 mL (2) 240 mL (3) 720 mL (4) 40 mL
Answer Key
(1) 1 (2) 4 (3) 2 (4) 1
(5) 3 (6) 4 (7) 4 (8) 3
7.4 REDOX REACTIONS AND ELECTRODE PROCESSES
The experiment corresponding to reaction, can also be observed if zinc rod is dipped in copper sulphate solution. Theredox reaction takes place and during the reaction, zinc is oxidised to zinc ions and copper ions are reduced to metallic copper due to direct transfer of electrons from zinc to copper ion. During this reaction heat is also evolved. Now we modify the experiment in such a manner that for the same redox reaction transfer of electrons takes placeindirectly. This necessitates the separation of zinc metal from copper sulphate solution.
We take copper sulphate solution in a beaker and put a copper strip or rod in it. We alsotake zinc sulphate solution in another beaker and put a zinc rod or strip in it. Now reaction takes place in either of the beakers and at the interface of the metal and its salt solution ineach beaker both the reduced and oxidized forms of the same species are present.
These represent the species in the reduction and oxidation half reactions. A redox couple is defined as having together the oxidised and reduced forms of a substance taking part in an oxidation or reduction half reaction.
This is represented by separating the oxidised form from the reduced form bya vertical line or a slash representing aninterface (e.g. solid/ solution). For example in this experiment the two redox couplesare represented as Zn2+/Zn and Cu2+/Cu. In both cases, oxidised form is put before the reduced form. Now we put the beakercontaining copper sulphate solution and the beaker containing zinc sulphate solutionside y side .
We connect solutionsin two beakers by a salt bridge (a U-tubecontaining a solution of potassium chlorideor ammonium nitrate usually solidified byboiling with agar and later cooling to ajelly like substance). This provides an electriccontact between the two solutions withoutallowing them to mix with each other.
The zinc and copper rods are connected by ametallic wire with a provision for an ammeterand a switch. The set-up as shown,is known as Daniell cell. When the switch isin the off position, no reaction takes place ineither of the beakers and no current flowsthrough the metallic wire.
As soon as the switch is in the on position, we make the following observations:
i. The transfer of electrons now does not take place directly from Zn to Cu 2+ but through the metallic wire connecting the two rods as is apparent from the arrow which indicates the flow of current
ii. The electricity from solution in one beaker to solution in the other beaker flows by the migration of ions through the salt bridge. We know that the flow of current is possible only if there is a potential difference between the copper and zinc rods known as electrodes here.
The potential associated with each electrode is known as electrode potential. If the concentration of each species taking part in the electrode reaction is unity (if any gas appears in the electrode reaction, it is confined to 1 atmospheric pressure) and further the reaction
is carried out at 298K,then the potential of each electrode is said to be the Standard Electrode Potential.
Byconvention, the standard electrode potential(E 0 ) of ydrogen electrode is 0.00 volts. The electrode potential value for each electrode process is a measure of the relative tendency of the active species in the process to remainin the oxidised/reduced form. A negative E 0 means that the redox couple is a stronger reducing agent than the H +/H 2 couple. Apositive E 0 means that the redox couple is aweaker reducing agent than the H+/H2 couple.
The standard electrode potentials are very important and we can get a lot of other useful information from them. The arrangement of electrodes in the order of increasing standard reduction potentials is called electrochemical series. This is also called activity series. The negative sign of standard reduction potential indicates that when the electrode is
connected with standard hydrogen electrode, it acts as the anode and oxidation takes place. A positive sign of standard reduction potential indicates that the electrode acts as cathode when connected with a standard hydrogen electrode.
The values of standard electrode potentials for some selected half-cell reduction reactions are given. It can be seen that the standard electrode potentials for fluorine is the highest, indicating that fluorine gas (F 2 ) has the maximum tendency to get reduced to fluoride ions (F–). Therefore, fluorine is the strongest oxidising agent and fluoride ion is the weakest reducing agent.
Lithium has the lowest electrode potential, indicating that lithium ion is the weakest oxidising agent, while lithium metal is the most powerful reducing agent in an aqueous solution. It may be seen that as we go from top to bottom in Table 7.6, the standard electrode potential increase.
4.
5.
6. 3
+−+→
7. 2 Zn(aq)2eZn(s) +−+→
8. 2 Fe(aq)2eFe(s) +−+→ Fe2+/Fe – 0.44
9. 2 Co(aq)2eCo(s) +−+→ Co2+/Co – 0.28
10. 2 Ni(aq)2eNi(s) +−+→ Ni2+/Ni – 0.25
11. 2 Sn(aq)2eSn(s) +−+→ Sn2+/Sn – 0.14
12. 2 2H(aq)2eH(g) +−+→ Pt, H+/H2 0.00
13. 2 Cu(aq)2eCu(s) +−+→ Cu2+/Cu + 0.34
14. 2 I(s)2e2I(aq) +→ Pt, I2/I– + 0.54
15. Ag(aq)eAg(s) +−+→ Ag+/Ag + 0.80
Table 7.6 Comparison between the two types of cells
16. 2 2 Hg(aq)2e2Hg(l) +−+→
17. 2 Br(l)2e2Br(aq) +→
18. 2 Cl(g)2e2Cl(aq) +→
19. 3 22 O(g)2H(aq)2eO(g)HO(l) +− ++→+
20. 2 F(g)2e2F(aq) +→
Thus, the tendency to undergo the corresponding reaction (reduction) decreases, the reducing power of the particles decreases
7.2.2 Applications of Electrochemical Series
With the help of electrochemical series, one can understand the spontaneity of certain reactions and one can predict the occurrence of metal-salt solution reactions. Using these series reducing and oxidising abilities of elements can be estimated.
Chemical Reactivity of Metals
The metal that has high negative value of standard reduction potential readily loses electrons and it is converted into cations. Such a metal is said to be chemically very active. The chemical reactivity of metals decreases from top to bottom in the activity series.
Relative Strength of Oxidising and Reducing Agents
The reducing ability decreases from top to bottom in the electrochemical series. The oxidation ability increases with increase in the reduction potential.
Oxidising power ∝ reduction potential
Reducing power ∝ oxidation potential
Fluorine, with highest reduction potential, is the powerful oxidising agent. Lithium, with least reduction potential, is the powerful reducing agent in aqueous medium based on the potentials.
A given metal in activity series can reduce all the metal ions present below it. A given metal ion can oxidise all metal atoms present above it. A metal can reduce all the non-metals present below it in the activity series.
For storing a salt solution a metal container with smaller SRP value should not be used. Instead a metal container with higher SRP can be used.
Examples:
FeSO 4 solution ( ) 2 o Fe/Fe E0.44v + =− cannot be stored in Zn container ( ) 2 o Zn/Zn E0.76v + =− but it can be stored in a Cu container ( ) 2 o Cu/Cu E0.34v + =+
A non-metal can oxidise all metal atoms present above it in the activity series. It can also oxidise non-metal anions present above it.
Displacement
of Hydrogen from Dilute Acids by Metals
A metal with negative SRP can displace H+ from dilute acids to liberate H 2 gas.
A metal with positive SRP like Cu, Ag, Pt etc., cannot displace H+ from dilute acids to liberate H2.
TEST YOURSELF
1. The standard reduction potentials for the following half-reactions are given against each at 298 K. ( ) ( ) 2 aq s Zn2eZn;0.762 +−+=−
Cr3eCr;0.74V +−+=−
( ) ( ) 3 aq s
2H2eH;0.0V +−+=
( ) 2g
(3) SRP is arbitrarily taken as Zero.
(4) it is a primary reference electrode.
FeeFe;0.77V +−++=+
( ) 32 maq aq
Which is the strongest oxidising agent?
(1) Zn(s) (2) Cr(s) (3) H2(g) (4) Fe3+(aq)
2. The standard electrode potentials M/M E + ο of four metals A, B, C and D are –1.2 V, 0.6 V, 0.85 V, and –0.76 V, respectively. The sequence of deposition of metals on applying potential is
(1) A,B,C,D (2) B,D,C,A (3) C,B,D,A (4) D,A,B,C
3. Which of the following has least tendency to liberate H2 from mineral acids (1) Cu (2) Mn (3) Ni (4) Zn
4. Which of the following is true for electrochemical cell made with SHE and Cu electrode?
(Given 2 Cu/Cu
E0.36V + ο =+ )
(1) H2 is cathode and Cu is anode (2) H2 is anode and Cu is cathode (3) Reduction occurs at H2 electrode (4) Oxidation occurs at Cu electrode
5. Regarding the standard hydrogen electrode, the incorrect statement is (1) H2 gas is bubbled at 1 bar (2) 2 M HCl is electrolyte
CHAPTER REVIEW
■ Oxidation is a process in which electrons are removed from an atom or ion. Hence, de-electronation is oxidation.
■ Reduction involves addition of electrons to an atom or ion. Hence, electronation is reduction.
6. Zn gives H2 gas with H2SO4 and HCl but not with HNO3 because
(1) Zn acts as oxidising agent when reacts with HNO3.
(2) HNO3 is weaker acid than H2SO4 and HCl.
(3) in electrochemical series Zn is above H2. (4) NO3 is reduced in preference to H3O+
7. The standard reduction potentials of Cu +2 , Ag + , Hg +2 , and Mg +2 are +0.34 V, +0.80 V + 0.79 V, and –2.37 V respectively. With increasing voltage, the sequence of deposition of metals on the cathode from a molten mixture containing all those ions is (1) Ag, Hg, Mg, Cu (2) Cu, Hg, Ag, Mg (3) Ag, Hg, Cu, Mg (4) Cu, Hg, Mg, Ag
8. The chemical reaction
( ) ( ) ( ) ( ) s2g aq s 2AgClH2HCl2Ag +→+ taking place in a galvanic cell is represented by the notation.
■ In a redox reaction, electrons are transfered from the reducing agent [reductant] to the oxidising agent [oxidant].
■ In a redox reaction, the oxidant gets reduced and the reductant gets oxidised.
■ The oxidation number of the reductant increases and that of the oxidant decreases in a redox reaction.
■ Metals are reducing agents. Alkali metals are powerful reducing agents.
■ Caesium is the most powerful reducing agent in gas phase.
■ In aqueous medium, Lithium is the most powerful reducing agent.
■ Non-metals are oxidising agents. Fluorine is the most powerful oxidising agent.
■ In some compounds, all the atoms of the same element may not have the same oxidation number. Then, we calculate the average value.
■ In NCl3, the oxidation number of nitrogen is –3. In HN3 [hydrazoic acid], the oxidation number of nitrogen is –1/3.
■ An element present in its lowest oxidation state in a compound can act as a reducing agent, e.g., H2S, NH3, PH3, HCl, etc.
■ An element present in its highest oxidation state in a compound can generally act as an oxidising agent, e.g., KMnO 4, K2Cr2O7, HNO3, Conc. H2SO4, HClO4, etc.
■ An element present in its intermediate oxidation state in a compound can act as a reducing as well as oxidising agent, e.g., NaNO2, SO2
■ During redox reactions, halogens are reduced to halide ions [X – ]. Potassium permanganate is reduced to Mn 2+ in acid medium.
■ In dilute alkaline medium or neutral medium, MnO4– is reduced to MnO2. In a strong alkaline medium, MnO4– is reduced to MnO42–..
■ All combustion reactions that make use of elemental dioxygen, as well as other reactions involving elements other than dioxygen, are redox reactions.
■ All combustion reactions in which oxygen undergoes reduction from 0 to –2 are redox reactions.
■ Decompostion reactions are the opposite of combination reactions, e.g., 2HgO → 2Hg + O2
■ In the displacement reactions the place of one species in its compound is taken up by other species, e.g., Zn (s) + CuSO4 (aq)→ ZnSO4 + Cu
■ In non-metal displacement redox reactions generally hydrogen gets displaced and, rarely, oxygen.
■ Fluorine cannot be prepared from halide by chemical oxidation process, because fluorine is the strongest o xidising agent.
■ Fluorine can be prepared from F– by only electrolysis method.
■ Disproportionation reactions involve the same element in the given form to undergo both oxidation and reduction simultaneously.
■ The reacting substances in a disproportionation reaction always contain an element that can exist in at least three oxidation states.
■ The inverse of disproportionation is comproportionation.
■ In comproportionation reactions, two species with the same element in two different oxidation states form a single product in which the element is in an intermediate oxidation state.
■ All electron transfer reactions are called redox reactions.
■ In a redox reaction, number of electrons gained by one substance is equal to the
number of electrons lost by another substance.
■ Redox reactions are balanced in electron transfer method or in ion-electron method.
■ In titrimetric analysis, the substance of known concentration is called titrant and the substance being titrated is called titrand.
■ The process of adding the standard solution until the reaction is just complete is called titration.
■ The point at which the titrand just completely reacts is called equivalence point, or theoretical point, or stoichiometric end point.
■ In redox reactions, the completion of the titration is detected by a suitable method or using a reagent, known as indicator .
■ A compound added to reacting solutions that undergo an abrupt change in a physical property, usually colour, is called indicator.
■ Titrations based on oxidation-reduction reactions are called redox titrations.
■ The species participating in oxidation and reduction half reactions is called redox couple.
■ Equivalent weight of a compound = M/n f
■ Equivalent weight of an acid,
Eacid = M Basicity
■ Equivalent weight of base,
Ebase = M Acidity
■ Equivalent weight of salt, E salt = M Net change of ion
■ Equivalent weight of an element = A/Valency
■ Equivalent weight of oxidising agent, Eoxidant = M / number of electrons absorbed
■ Equivalent weight of reducing agent, Ereductant = M/Number of electrons released.
Exercises
JEE MAIN LEVEL
Level - I
Classical idea of redox reactions
Single Option Correct MCQs
1. When zinc is added to Cu SO 4 solution, copper is precipitated, it is so because of (1) reduction of Zn (2) reduction of Cu2+ (3) hydrolysis of CuSO4 (4) reduction of SO4 2–
2. PbS + 4H2O2 → PbSO4 + 4H2O. In this reaction PbS undergoes (1) oxidation (2) reduction (3) both (4) none of these
3. The largest oxidation number exhibited by an element depends on outer electronic configuration. With which of the following outer electronic configuration the element will exhibit the largest oxidation number?
(1) 3d14s2 (2) 3d34s2 (3) 3d54s2 (4) 3d54s1
4. Which of the following is an oxidation and reduction reaction?
6. The gain of oxygen is known as (1) oxidation (2) reduction (3) halogenation (4)chlorination
7. In a reaction between zinc and iodine, in which zinc iodide is formed, which one is being oxidised?
(1) Zinc ions (2) Iodide ions
(3) Zinc atom (4) Iodine
8. Which of the following is an oxidation and reduction reaction?
(1) BaO2 + H2SO4 → BaSO4 + H2O2
(2) N2O5+H2O2→2HNO3
(3) AgNO3+KI → AgI + KNO3
(4) SnCl2 + HgCl2 → SnCl4 + Hg
9. In which reaction is hydrogen acting as an oxidising agent ?
(1) With iodine to give hydrogen iodide
(2) With lithium to give lithium hydride
(3) With nitrogen to give ammonia
(4) With sulphur to give hydrogen sulphide
10. Which is not an oxidising agent?
(1) KClO3 (2) O2
(3) C6H12O6 (4) K2Cr2O7
11. Reduction does not involve in
(1) removal of an electronegative element
(2) addition of an electropositive element
(3) removal of an electropositive element (4) decrease in oxidation number
Numerical Value Questions
12. Thiosulphate reacts differently with iodine and bromine in the reactions given below:
Choose the correct statements that justifies the above equations i) Bromine is a stronger oxidant than iodine.
ii) Bromine is a weaker oxidant than iodine.
iii) Thiosulphate undergoes oxidation by bromine
iv) Thiosulphate is reduced by iodine
v) Bromine undergoes oxidation
vi) iodine undergoes reduction
13. Which of the following mainly act as oxidising agents?
O3, SO2, TeO2, PbO2
14. The Oxidation state of Nickel in [Ni (CO)4] is_____
Redox reactions in terms of electron transfer reactions
Single Option Correct MCQs
15. In the reaction: C2O42– + MnO4– + H+ → Mn2+ + CO2 + H2O, the oxidizing agent is (1) MnO4– (2) C2O4–(3) Mn2+ (4) H+
16. Which is strongest oxidizing agent?
(1) O3 (2) O2
(3) Cl2 (4) F2
17. In acidic solution, the reaction MnO4 → Mn+2 can occur through (1) oxidation by loss of 3 electrons (2) reduction by gain of 3 electrons (3) oxidation by loss of 5 electrons (4) reduction by gain of 5 electrons
18. In the reaction 2Cu2O+ Cu2S → 6Cu + SO2 the species which acts as oxidant and reductant, respectively?
(1) Cu+, O–2 (2) Cu+, S–2 (3) Cu, S–2 (4) Cu, O–2
19. The oxyacid which acts both as oxidizing and reducing agent is
(1) HNO2 (2) H3PO4 (3) H2SO4 (4) HCIO4
20. When copper is added to a solution of silver nitrate, silver is precipitated. This is due to the reduction of (1) silver
(2) copper (3) silver ion (4) cuprous ion
21. MnO2 + 4HCl → MnCl2 + Cl2 + 2H2O, MnO2 will act as (1) oxidant
(2) reductant
(3) both (4) can’t predicted
22. The correct option for a redox couple is (1) both are oxidised forms involving same element.
(2) both are reduced forms involving same element.
(3) both the reduced and oxidized forms involve same element.
(4) cathode and anode together.
23. In the following reaction, which is the species being oxidized?
(1) Fe3+ (2) I–(3) I2 (4) Fe2+
24. Which of the following can act as both oxidizing and reducing agent?
i) SO2
ii) HNO3
iii) HNO2
iv) HClO4
(1) SO2, HNO2
(2) HNO2, HNO3
(3) HNO2, HClO4
(4) HClO4, SO2
25. In which of the following reactions hydrogen is acting as an oxidizing agent?
(1) With sodium to give sodium hydride
(2) With iodine to give hydrogen iodide
(3) With nitrogen to give ammonia
(4) With sulphur to give hydrogen sulphide.
26. Which of the following species can function both as oxidising as well as reducing agent?
(1) Cl–
(2) ClO4–
(3) ClO–
(4) MnO4–
27. Which of the following is a set of reducing agents?
(1) HNO3, Fe2+, F2
(2) F–, Cl–, MnO4–
(3) I–, Na, Fe2+
(4) Cr2O72-, Cr2O42-, Na
28. Which of the following change requires a reducing agent?
(1) CrO42– → Cr2O72–
(2) BrO3 → BrO–
(3) NH3 → NF3
(4) Al(OH)3 → Al (OH)4–
29. The compound that cannot act both as oxidising and reducing agent is:
(1) H2O2
(2) HNO2
(3) H3PO4
(4) H2SO3
30. Loss of electron is termed as
(1) combustion
(2) oxidation (3) reduction
(4) neutralisation
31. In equation NO2– + H2O → NO3– + 2H+ + ne–, the value of n in balanced equation is (1) 1
(2) 0
(3) 2
(4) 3
Numerical Value Questions
32. How many compounds can act as oxidizing agents
KClO3, O3, C6H12O6, K2Cr2O7, KMnO4
33. How many electrons are involved in the following half reaction in acidic medium?
Cr2O7-2 → 2Cr+3
34. The total number of electrons transferred from 3 molecules of reductant to oxidant in the following redox process is _________?
Oxidation Number
Single Option Correct MCQs
35. Highest oxidation number that is exhibited by fluorine is (1) –1 (2) 0
(3) +1 (4) +7
36. Oxidation state of ‘S’ in S 8 molecule is (1) 0
(2) +2
(3) +4
(4) +6
37. Oxidation state of Fe in K 4[Fe(CN)6]
(1) +6
(2) +4
(3) +2
(4) +5
38. Oxidation number and valency of oxygen in OF2 are (1) +1, 2
(2) +2, 2
(3) +1, 1
(4) +2, 1
39. In which of the following the oxidation state of chlorine is +5?
(1) HClO4
(2) HClO3
(3) HClO2
(4) HCl
40. All elements commonly exhibit an oxidation state of
(1) +1
(2) –1
(3) Zero
(4) +2
41. The element that always exhibits a negative oxidation state in its compounds is
(1) Nitrogen
(2) Oxygen
(3) Fluorine
(4) Chlorine
42. Oxidation number of iron in Na2[Fe(CN)5NO] is
(1) +2
(2) +3
(3) +1
(4) 0
43. The oxidation number of phosphorus in sodium hypophosphite is
(1) +3
(2) +2
(3) +1
(4) –1
44. Which of the following reactions does not involve the change in oxidation state of metal?
(1) VO–2 → V2O3
(2) Na → Na+
(3) CrO42- → Cr2O72-
(4) Zn2+ → Zn
45. Oxidation state of oxygen in potassium superoxide is
(1) –1/2 (2) –1
(3) –2 (4) 0
46. Fluorine do not undergo disproportionation, because
(1) Fluorine is always exhibit -1 oxidation state
(2) Fluorine exhibit only two oxidation numbers
(3) Fluorine exhibit three oxidation numbers
(4) None of the above
47. 2H 2 O 2 → 2H 2 O + O 2 . This reaction is an example for
(1) decomposition
(2) combination
(3) disproportionation
(4) Both 1 and 2
48. K+Cl → KCl. This reaction is an example of (1) oxidation
(2) reduction
(3) a redox reaction
(4) none of these
49. In the reaction P4 + 3OH– + 3H2O →3H2PO2–+ PH3; phosphorus is undergoing
(1) oxidation
(2) reduction
(3) disproportionation
(4) hydrolysis
50. Which of the following is not a redox reaction?
(1) 2BaO + O2 → 2BaO2
(2) BaO2 + H2SO4 → BaSO4 + H2O2
(3) 2KClO3 → 2KCl + 3O2
(4) SO2 + 2H2S → 2H2O + 3S
51. This reaction is:
2CuI → Cu + CuI2
(1) disproportionation reaction
(2) neutralisation reaction
(3) oxidation reaction
(4) reduction reaction
52. In a reaction between zinc and iodine, in which zinc iodide is formed, the reductant is
(1) Zinc ions
(2) Iodide ions
(3) Zinc atom
(4) Iodine
53. The reaction 3ClO–(aq) → ClO3–(aq) + 2Cl–(aq) is an example of
(1) oxidation reaction
(2) reduction reaction
(3) disproportionation reaction
(4) decomposition reaction
54. In the reaction: Cl2 + H2S →2HCl + S, the oxidation number of S changes from
(1) 0 to 2
(2) 2 to zero
(3) - 2 to zero
(4) – 2 to – 1
55. Molecular weight of orthophosphoric acid is M. Its equivalent weight is (1) 3M
(2) M
(3) M/2
(4) M/3
56. Which of the following acid has the same molecular weight and equivalent weight?
(1) H3PO2
(2) H3PO3
(3) H3PO4
(4) H2SO4
57. The equivalent mass of CaCO 3 is (1) 100
(2) 50
(3) 33.3
(4) 25
58. 0.5 g of a metal on oxidation gave 0.79 g of its oxide. The equivalent mass of the metal is (1) 10
(2) 14
(3) 20
(4) 40
59. Which one among the following is not a fixed quantity?
(1) Atomic weight of an element
(2) Equivalent weight of an element
(3) Molecular weight of a compound
(4) Formula weight of a substance
60. The equivalent weight of Bayer’s reagent is (alkaline KMnO4)
(1) 31.6
(2) 52.6
(3) 79
(4) 158
61. Molecular weight of KMnO 4 is “M”. In a reaction KMnO4 is reduced to K2MnO4. The equivalent weight of KMnO 4 is
(1) M
(2) M/2
(3) M/3
(4) M/5
62. When Ferrous sulphate acts as reductant, its equivalent weight is (1) twice that of its molecular weight (2) equal to its molecular weight (3) one-half of its molecular weight (4) one-third of its molecular weight
63. 2H 2 O → 4e – + O 2 + 4H + . The equivalent weight of oxygen is (1) 32 (2) 16
(3) 8 (4) 4
64. In acidic medium dichromate ion oxidises ferrous ion to ferric ion. If the grammolecular weight of potassium dichromate is 294 g, its equivalent weight is (1) 294
(2) 147
(3) 49
(4) 24.5
65. The equivalent weight of hypo in the reaction [M = molecular weight]
2Na2S2O3 + I2 → 2NaI + Na2S4O6 is (1) M
(2) M/2
(3) M/3
(4) M/4
66. The minimum oxidation state that nitrogen exhibits is
(1) -2
(2) -3
(3) -4
(4) -5
67. In the conversion of K2Cr2O7 to K2CrO4, the oxidation number of the following changes
(1) K
(2) Cr
(3) O
(4) None
68. The number of electrons involved in the half-reaction;
Cr2O72- → 2Cr3+ + 7H2O + 14H+ is (1) 3
(2) 6
(3) 5
(4) 10
69. For the reaction, Fe2S3 + 5O2→2FeSO4 + SO2, the equivalent weight of Fe 2S3 is
(1) M/4
(2) M/16
(3) M/22
(4) M/20
70. Which of the given reactions is not an example of disproportionation reaction?
(1) 2H2O2 → 2H2O + O2
(2) 2NO2 + H2O → HNO3 + HNO2
(3) 2KMnO4 → K2MnO4 + MnO2 + O2
(4) 3MnO42−+4H+→ 2MnO4 +MnO2 +2H2O
71. For the redox reaction MnO4 + C2O4−2 + H+ → Mn+2 + CO2 + H2O the correct coefficients of the reactants for the balanced equation are
(1) 16, 5, 2
(2) 2, 5, 16
(3) 2, 16, 5
(4) 5, 16, 2
72. In the redox reaction xKMnO 4 + yNH 3 → KNO3 + MnO2 + KOH + H2O
(1) x = 4, y = 6
(2) x = 3, y = 8
(3) x = 8, y = 6
(4) x = 8, y = 3
73. The value of ‘n’ in the equation MnO 4 + 8H+ + “n” e → Mn+2 + 4H2O
(1) 4 (2) 3
(3) 5 (4) 2.5
74. On balancing the given redox react ion
T he coefficients a, b, c are found to be respectively (1) 1,3,8 (2) 3,8,1 (3) 1,8,3 (4) 8,1,3
75. Chlorine undergoes disproportionation in alkaline medium as shown below: aCl2(g)+bOH–(aq)→cClO–aq + dCl- (aq) + ne H2O(l)
The values of a, b, c and d in a balanced redox reaction are respectively: (1) 2, 2, 1 and 3 (2) 1, 2, 1 and 1 (3) 2, 4, 1 and 3 (4) 3, 4, 4 and 2
76. In the reaction the stoichiometry coefficients of 2 272 CrO,NO and H+ respectively are 23 272 32 CrONOHCrNOHO −−++− ++→++
(1) 1, 3, 8 (2) 1, 4, 8 (3) 1, 3, 12 (4) 1, 5, 12
77. x K Br + y MnO2 + z H2SO4 → MnSO4 + Br2 + K2SO4 + aH2O. Then the value of x, y and z in the balanced equation are
(1) 2, 1, 2
(2) 2, 1, 1
(3) 2, 4, 1
(4) 2, 5, 1
78. Number of moles of KMnO4 that can oxide one mole of oxalic acid in acidic media is
(1) 6.5 mole
(2) 8.3 mole
(3) 0.8 mole
(4) 0.4 mole
79. All the following are comproportionation reactions except
(1) OF2 + H2O → HF + O2
(2) H2S + SO2 → S + H2O
(3) HNO2 → HNO3 + NO
(4) Cu2+ + Cu → 2Cu+
80. The following is an example of comproportionation reaction
(1) 2 AgAg2Ag ++ +→
(2) 222 222 NaNONHCONHHC NaCNCOHO ++→ +++
(3) 22 OFHO2HFO2 +→+
(4) All
81. Number of H + ions involved in the half reaction, H 23 NONO + → is
(1) 4 (2) 1
(3) 3 (4) 2
82. 20ml of 0.1M KMnO 4 in acidic medium can completely oxidise 40 ml of Mohr’s salt solution. Molarity of Mohr’s salt solution is (1) 0.25 M
(2) 0.5 M
(3) 1 M
(4) 0.125 M
83. T he strength of an aqueous solution of I 2 can be determined by titrating the solution with standard solution of (1) Oxilic acid
(2) Sodium thiosulphate
(3) Sodium hydroxide
(4) Mohr’s salt
84. The role of starch in iodometric titrations is that,
(1) it acts an oxidant
(2) it is reducing agent
(3) it acts as indicator (4) it is a medium
85. In the standardization of Na 2S 2O 3 using K2Cr2O7 by iodometry, the equivalent weight of K2Cr2O7 is (1) (molecular weight)/2 (2) (molecular weight)/6 (3) (molecular weight)/3 (4) same as molecular weight
86. The role of permanganate in permanganometric titration’s is
(1) it acts an reductant
(2) it is a reducing agent
(3) it acts as an self-indicator and oxidant (4) it is a dehydrating agent
87. Titration of oxalic acid Vs KMnO4 is a type of (1) precipitation titration
(2) redox titration
(3) acid base titration
(4) complexometric titration
88. Titration is carried out by using KMnO4 (taken in Burette) and hot acidified oxalic acid in acidic media. The indicator used to identify the end point is (1) methyl orange
(2) methyl red
(3) phenolphthalein
(4) KMnO4 is the self-inductor
89. 0.2 moles of KMnO 4 can oxidize _____ moles of oxalic acid in acidic medium
(1) 0.3 (2) 0.4
(3) 0.2 (4) 0.5
90. What are the modes of exchange of gases (O2 and CO2) at the level of alveoli and tissues respectively?
(1) Active transport, active transport
(2) Active transport, diffusion
(3) Diffusion, active transport
(4) Diffusion, diffusion
91. In the titration of KMnO4 verses Mohr salt, the indicator used is
(1) KMnO4 (self)
(2) diphenyl amine
(3) starch
(4) methyl red
Numerical Value Questions
92. Number of moles of H + ions required by 1 mole of MnO4 to oxidise oxalate ion to CO2 is ____.
93. Total number of species from the following which can undergo disproportionation reaction is ______?
H2O2, ClO3 , P4, Cl2, Ag, Cu+1, F2, NO2, K+
94. 2MnO4 + bI + cH2O → xI2 + yMnO2 + zOH If the above equation is balanced with integer coefficients, the value of z is _______
95. In the given combination reaction, what is the value of x+y? [x and y indicate oxidation states] y x 4222 CH2OCO2HO +→+
96. Find the number of moles of K 2 Cr 2 O 7 reduced by three mole of Sn 2+ ions.
97. One mole of Fe 2+ reacts with x moles of KMnO4 is ____, in the balanced equation.
98. In the titration of KMnO4 and oxalic acid in acidic medium, the change in oxidation number of carbon at the end point is_______
99. The oxidation number of Cr in CrO 5 is +x then x is________
100. The sum of oxidation number of chlorine in bleaching powder (CaOCl 2) is_____
101. The oxidation state of sulphur in Marshalls acid (H2S2O8) is ___
Redox reactions and Electrode Processes
Single Option Correct MCQs
102. Some statements about the modified Daniel cell are given below:
(A) Cu vessel acts as a cathode.
(B) Porous pot helps to maintain electric neutrality.
(C) ZnSO4 solution is electrolyte.
(D) Anode is a negative electrode
Among these statements, correct statements are
(1) A and B only (2) A and C only (3) B, C and D only (4) A, B and D only
103. The metal that cannot displace hydrogen from dilute HCl is:
(1) Al (2) Fe (3) Cu (4) Zn
104. The standard electrode potentials of Zn, Ag and Cu are – 0.76, 0.80 and 0.34 volts respectively; then
(1) Ag can oxidize Zn and Cu
(2) Ag can reduce Zn2+ and Cu2+
(3) Zn can reduce Ag+ and Cu2+
(4) Cu can oxidize Zn and Ag
105. Consider the following reduction process:
Zn2+ + 2e- → Zn(s); E° = -0.76V
Ca2+ + 2e- → Ca(s); E° = -2.87V
Mg2+ + 2e- → Mg(s); E° = -2.36V
Ni2+ + 2e- → Ni|(s); E° = -0.25V
The reducing power of the metals increases in the order:
(1) Ca < Zn < Mg < Ni
(2) Ni < Zn < Mg < Ca
(3) Zn < Mg < Ni < Ca
(4) Ca < Mg < Zn < Ni
106. Which reaction is not feasible
(1) 2KI + Br2 → 2KBr+I2
(2) 2KBr + I2 → 2KI + Br2
(3) 2KBr + Cl2 → 2KCl + Br2
(4) 2H2O + 2F2 → 4HF + O2
107. In salt bridge normally KCl is used because (1) It is a strong electrolyte
(2) It is a good conductor of electricity
(3) K+ and Cl– ions have nearly same ionic mobility
(4) It is an ionic compound
108. Using the data given below find out the strongest reducing agent
18. In balancing the half reaction HCHO → HCOO – taking place in alkaline medium, the number of electrons(e ) to be added are (1) 2 to the reactants side (2) 2 to the products side (3) 3 to the reactants side (4) 3 to the products side
Numerical Value Questions
19. What is the value of ‘n’ in the following half equation: ( )–– 2– –42 4 CrOH+OHCrO+HO+ne →
Redox Reactions as the Basis for Titration
Single Option Correct MCQs
20. x mmol of KlO 3 reacts completely with y mmol of KI to give I2 quantitively If z mmol of hypo are required for complete titration against this I 2 , then, which statement is incorrect?
(1) z = 6x (2) 6y = 5z
(3) 5x = y (4) x + y = 2z
Numerical Value Questions
21. In iodimetric titration (where hypo is directly treated with iodine in the presence of starch), What is the average oxidation state of sulphur in the Sulphur containing product.
Multiple Concept Questions
Single Option Correct MCQs
22. The equivalent weight of phosphoric acid (H3PO4) in the reaction, NaOH+H3PO4→NaH4PO4+H2O is (1) 59 (2) 49 (3) 25 (4) 98
23. A compound contains X, Y, Z atoms. The oxidation states of X, Y and Z are +2, +2, -2 respectively. The probable formula of the compound is
(1) Y2(XZ3)2 (2) XYZ2
(3) X3(Y4Z)2 (4) X3(YZ4)2
24. Which of the following acid has the same molecular weight and equivalent weight.
(1) H3PO2 (2) H3PO3
(3) H3PO4 (4) H2SO4
25. The oxidation state of Fe in K4[Fe(CN)6] is (1) +2 (2) +6 (3) +3 (4) +4
Numerical Value Questions
26. Ionisible H atoms in H3PO4 is x and in H3PO2 is y. The ratio of x : y is_____.
Level - III
1. Which of the following is not a disproportionation reaction?
(1) 2F2+2NaOH→2NaF+OF2+H2O
(2) Cl2+2NaOH→NaCl+NaClO+H2O
(3) 6NaOH+4S→Na2S2O3+2Na2S+3H2O
(4) P4+3NaOH+3H2O→PH3+3NaOH2PO2
2. An example of comproportionation reaction
(1) 2H2O+SO2→3S+2H2O
(2) 3Cl2+6NaOH→5NaCl+NaClO3+3H2O
(3) P4→PH3+H2PO–2
(4) 2CuBr→CuBr2+Cu
3. Among P4(S), S8(S), Cl2(g), and NO2(g), how many species can undergo disproportionation in the alkaline medium?
4. How many are disproportionation reactions?
(1) 2Cu+→Cu2++Cu°
(2) 2-+4 422 3MnO+4H2MnO+MnO+2HO →
(3) 42422 2KMnOKMnO+MnO+O ∆ →
(4) -2+ + 422 2MnO+3Mn+2HO5MnO+4H →
5. – –2 +2 424 22 2MnO+ CO+ H Mn+ CO+ HO + → bcxyz if the above equation is balanced with integer co-efficients. The value of c is __.
6. In a titration, certain amount of H 2O 2 is treated with y moles of KMnO 4 in acidic medium. The left out KMnO4 when treated with X+ in basic medium oxidises X+1 to X+6 and 0.2 M, X L of X + was consumed. The mole of given H2O2 solution is:
(1) 5 yx (2) 5 2 yx
(3) ( ) 5 10 yx (4) ( ) 5 5 yx
7. 25 mL solution containing mixture of Na2CO3 and NaOH required 20 mL of 1N HCl for the phenolphthalein end point and in a separate titration 25 mL of the same
mixture required 25 mL of same HCl for methyl orange end point. Ratio of moles of Na2CO3 and NaOH taken in initial mixture and convert that into decimal form.
8. Concentrated nitric acid is act as oxidising agent. It oxidises l2 to HIO3, the n-factor of the l2 is x, and it oxidises S8 to H2SO4, then n-factor of S8 is y. What is the value of x+y. (1) 38 (2) 48 (3) 58 (4) 26
9. In the reaction the stoichiometry coefficients of 2-272 CrO,NO and H+ respectively are 2--+3+272 CrO+NO+HCrNO32 + +HO →
(1) 1, 3, 8 (2) 1, 4, 8 (3) 1, 3, 12 (4) 1, 5, 12
10. A solution contains mixture of H 2 SO 4 , H2C2O4. 20 mL of this solution requires 40 mL of M/10 NaOH for neutralisation and 20 mL of N/10 KMnO4 for oxidation. The molarity of H2C2O4,H2SO4 are
11. The oxidation states of transition metal atoms in K 2 Cr 2 O 7 , KMnO 4 and K 2 FeO 4 respectively are x, y and z. The sum of x, y and z is ____.
(1) 18 (2) 19 (3) 20 (4) 22
12. KMnO4 reacts with oxalic acid according to the equation - -2+ +2 424 22 2MnO+5CO+16H2Mn+10CO+8HO → Here 20 mL of 0.1 M KMnO4 is equivalent to (1) 120 mL of 0.25 M H2C2O4
(2) 150 mL of 0.10 M H2C2O4
(3) 25 mL of 0.20 M H2C2O4 (4) 50 mL of 0.20 M H2C2O4
13. How many of OH are present in the balanced equation?
CrOH+HOHO+CrO →
( ) OH- -2 22 24 3
(1) 2
(2) 3
(3) 4
(4) 6
14. In order to oxidise a mixture having 1 mol each of FeC2O4,Fe2(C2O4)3,FeSO4, how many mole of KMnO4 is required in acidic medium
15. Calculate number of mole of Na 2S2O3 that will react with I 2 obtained when 1 mol of K2Cr2O7 reacts with excess of KI in acidic medium as per the reaction
H2SO4 + K2Cr2O7 + KI → K2SO4 + Cr2(SO4)3 + H2O+I2
16. The number of electrons involved in the reaction of permanganate to manganese dioxide in acidic medium is ___.
17. In the balanced chemical reaction,
3 22 IO+I+HHO+I; abCd →
The sum of the coefficients of a, b, c, and d is 10+y. Then the value of y is _______.
18. How many mL of 0.3 M K2Cr2O7 required for complete oxidation of 5 mL of 0.2 M SnC2O4 solution in acidic medium. (K2Cr2O7 oxidises SnC2O4 to Sn4++CO2)
19. In basic medium, 2 4 CrO oxidises 3 2 2 SO to form 2 SO46 and itself changes into Cr(OH)4 . The volume of 0.154 M 2 4 CrO required to react with 40 mL of 0.25 M 2 SO23 is ____ mL. (Rounded off to the nearest integer)
20. 0.01 M KMnO 4 solution was added to 20.0 mL of 0.05 M Mohr’s salt solution
through a burette. The initial reading of 50 mL burette is zero. The volume of KMnO4 solution left in the burette after the end point is ____________ mL. (nearest integer)
21. Calculate the equivalent weight of phosphorous acid H 3PO3 in the following reaction.
2NaOH + H3PO3 → Na2HPO3 + 2H2O
(at. wt. H = 1, P = 31, O = 16)
22. 0.5 g of fuming H2SO4 (oleum) is diluted with water. This solution is completely neutralised by 26.7 mL of 0.4 N NaOH. The percentage of free SO3 in the sample is
(1) 30.6%
(2) 40.6%
(3) 20.6%
(4) 50%
23 2KMnO 4 + 3H 2 SO 4 + 5H 2 O 2 → K 2 SO 4 + 2 MnSO4+ 8 H2 O+5O2
Find the normality of H 2 O 2 solution, if 20 mL of it is required to react completely with 16 mL of 0.02 M KMnO 4 solution. (molar mass of KMnO4 = 158 g mol–1 )
(1) 4 × 10–2 N
(2) 2 × 10–2 N
(3) 6 × 10–2 N
(4)8 × 10–2 N
24. The oxide of a metal contains 40% of oxygen. The valency of metal is 2. What is the atomic weight of the metal?
25. 20 ml of 0.1 M FeC2O4 solution is titrated with 0.1 M KMnO4 in acidic medium. Calculate the volume (mL) of KMnO 4 solution required to oxidise FeC 2O4 completely.
THEORY BASED QUESTIONS
Statement Type Questions
Each question has two statements: statement I (S-I) and statement II (S-II). In light of the given statements, choose the most appropriate answer from the options given below.
(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 : In redox titration, the indicators used are sensitive to change in pH of the solution
S-II : In acid base titration, the indicators used are sensitive to change in oxidation potential
2. S-I : 1 mol of H2SO4 is neutralised by 2 mol of NaOH ; However, 1 equivalent of H2SO4 is neutralised by I equivalent of NaOH.
S-II : Equivalent mass of H2SO4 is half of its molecular mass, however, the equivalent mass of NaOH is equal to its molecular mass
3. S-I : 2CuCl→CuCl 2 +Cu is a disproportionation reaction.
S-II : All transition metals show disproportionation reactions.
4. S-I : Aqueous solution of K 2 Cr 2 O 7 is preferred as a primary standard in volumetric analysis over Na 2Cr 2O 7 aqueous solution.
S-II : Na 2Cr 2O 7 has a higher solubility in water than Na2Cr2O7
5. S-I : Equivalent mass of H3PO2 is equal to its molecular mass
S-II : H3PO2 is a monobasic acid
6. S-I : HNO 2 acts both as an oxidising as well as a reducing agent.
S-II : The oxidation number of nitrogen can increase above +3 and can also decrease below +3
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) just below it. Of the statement mark the correct answer.
(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) : The decomposition of H2O2 to form water and oxygen is an example of disproportionation reaction.
(R) : The oxygen of peroxide is in –1 oxidation state and it is converted to zero oxidation state in O 2 and –2 oxidation state in H2O.
8. (A) : Photosynthesis is an example of redox reactions.
(R) : In the reaction, both carbon dioxide and water are oxidised to oxygen and carbohydrates respectively.
9. (A) : Neutralisation reactions can never be redox reactions.
(R) : There is no exchange of electrons between the atoms.
10. (A) : Average oxidation state of carbon in glucose is zero.
(R) : Oxidation state of an atom in any compound is zero.
JEE ADVANCED LEVEL
Multi Option Correct MCQs
1. In acidic medium dichromate ion oxidises stannous ion as 2+ 2-+ 4+3+ 27 2 Sn+CrO+HSn+Cr+HO → xyzabc
(1) the value of x : y is 1:3
(2) the value of x+y+z is 18
(3) a : b is 3 : 2
(4) the value of z-c is 17
2. Which of the following may show fixed equivalent weight?
(1) Mg (2) Al
(3) Zn (4) Cu
3. Which statement(s) about oxidation number is (are) correct?
(1) The oxidation number is the number of electrons lost (+ve) or gained (-ve) by an atom during the formation of ionic compounds.
(2) For covalent compounds, the oxidation number is indicated by the charge that an atom of element would have acquired if the substance would have been ionic.
(3) Oxidation number may have fractional values.
(4) Oxidation number is always negative.
4. Which of the following statements(s) is (are) correct?
(1) All reactions are oxidation and reduction reactions.
(2) Oxidising agent is itself reduced.
(3) Oxidation and reduction always go side by side.
(4) Oxidation number during reduction decreases.
5. The oxidation number of Cr is +6 in
(1) FeCr2O4
(2) KCrO3Cl
(3) CrO5
(4) [Cr(OH4)]–
6. Th e different oxidation state(s) exhibited by oxygen is (are)
(1) –2 (2) –1
(3) 0 (4) –1/2
7. Which of the following reactions involve oxidation–reduction?
(1) 2Rb+2H2O→2RbOH+H2
(2) 2CuI2→2Cul+I2
(3) NH4Cl+NaOH→NaCl+NH3+H2O
(4) 4KCN+Fe(CN)2→K4[Fe(CN)6]
8. Which reactions involves iodimetric titrations?
(1) 2Na2S2O3+I2→N2S4O6+2NaI
(2) –-+ +2 4 22 2 x MnO+2I+16H2Mn+8HO+5I →
(3) Na3AsO3+I2+H2O→Na3AsO4+2HI
(4) -2+- +3 27 CrO+14H+5I2Cr+7HO+3I22 →
9. Which of the following statements are correct?
(1) In the titration of HCl vs NaOH , indicator is phenolphthalein.
(2) In the titration of KMnO4 vs H2C2O4, no external indicator is used.
(3) In the titration of I2 vs Na2S2O3, starch is the indicator.
(4) Equivalent weight of hypo is half of its molecular weight in the reaction I2+2Na2S2O3→2NaI+Na2S4O6.
10. Which of the following is/are disproportionation reaction(s)?
(1) 2H2O2(aq)→2H2O(l)+O2(g)
(2) ( ) ( ) ( ) ( ) ( ) aq aq 2322 4s l g P+3OH+3HOPH+3HPO →
(3) ( ) ( ) ( ) ( ) 2 (aq) 2 g aqaq l Cl+2OHClO+Cl+HO →
(4) ( ) ( ) ( ) ( ) 2 (aq) 2 g aq2g l 2F+2OH2F+OF+HO →
Numerical Value Questions
11. In acidic medium, dichromate ions ( ) 2–CrO27 oxidize ferrous ions to ferric ions. The molar mass of potassium dichromate is 294 g/mol. The equivalent weight is ____.
12. Methane is converted into formaldehyde. What is the ratio of molecular weight to equivalent weight of methane?
13. In the chemical reaction K2Cr2O7+xH2SO4+ ySO2→K2SO4+ySO2→K2SO4+Cr2(SO4)3+zH2O. Here sum of x,y and z is_____.
14. What will be the n–factor of the reactant in the following reaction? ( ) 4272232 2 NHCrON+CrO+4HO ∆ →
15. The oxide of an element contain 67.67% of oxygen. What is its equivalent weight?
16. 1 litre solution of KIO3 unknown molarity is given to titrate with KI in strong acidic medium 50 mL solution of KIO3 requires 10 mL of 0.1 M KI for complete reduction to I2. The molarity of KlO3 solution is x ×10–3 M. Then find the value x is _____.
17. Equivalent weight of an element ‘X’ is 3. If vapour density of volatile chloride of ‘X’ is 77, find out the number of chlorine atoms combine with element ‘X’ to form a molecule.
18. The ion An+ is oxidised to AO–3 by MnO–4 changing to Mn2+ in acidic medium. Given that 2.68×10–3 mol of An+ requires 1.61×10–3 mole of MnO–4. Thus, the value of n is
19. Find magnitude of sum of two different oxidation states of sulphur atoms in Na2S4O6 (Sodium tetra thionate).
20. What is the value of n in the following half equation?
( )- -242 4 CrOH+OHCrO+HO+e → n
21. In order to oxidise a mixture two mole of each of FeC 2 O 4 ,Fe 2 (C 2 O 4 ) 3 and FeSO 4 in acidic medium, the number of moles of KMnO4 required is:
22. In the balanced chemical equation, the value of (p+q+r)–(s+t+u) is ____. pCu2O + qMn–4 + rH2O → sMnO2 + tCu(OH2) + uOH–
23. The equivalent weight of CaCO 3 in g eq-1 is 24. 20 mL of 0.1 M FeC2O4 solution is titrated with 0.1 M KMnO 4 in acidic medium. Calculate the volume (mL) of KMnO 4 solution required to oxidise FeC 2 O 4 completely?
Passage-based Questions
Passage I: 'R' is bleaching powder 'S' is one of the component in glass SiO2 CaSiO3 s P Q R
CaCO3 CaO Ca(OH)2 CaOCl2 –CO2
25. The average oxidation state of chlorine in 'R' is ___.
26. The oxidation state of silicon in 'S' is ____.
Passage II: The oxidation number of an element in a compound decides its nature to act as oxidant or reductant. Oxidation number is defined as the residual charge which an atom has or appears to have in a molecule when all other atoms are removed the molecule as ions. Oxidation number is frequently used interchangebly with oxidation state. The stock notations of oxidation numbers are based on the periodic property electronegativity. An atom in a molecule can be assigned positive, negative or zero oxidation number by considering its environment. In few cases, oxidation number can even be fractional.
27. Calculate the oxidation number of Cr in K2Cr2O7?
28. Maximum oxidation state of Os is.
Passage III: A redox reaction involves oxidation of reductant liberations, which are then consumed by an oxidant. The sum of two half reactions give r ise to net redox change. In half reaction charge and atoms are always conserved.
29. In the reaction
As2S3+HNO3→H3AsO4+H2SO4+NO; the element oxidised is:
(1) As only (2) S only
(3) N only (4) As and S both
30. In the equation – -+223 NO+HONO+2H+e n →
n, stands for:
(1) 1 (2) 2 (3) 3 (4) 4
Matrix Matching Questions
31. Match the Column-I with Column-II
Column-I (Acid) Column-II (Nature)
(a) H3PO4 (p) mono basic
(b) H2CO3 (q) dibasic
(c) H3BO3 (r) tribasic
(d) H 4P2O7 (s) tetra basic
(a) (b) (c) (d)
(1) q r p s
(2) s p r q
(3) r q p s (4) p s q r
32. Match the Column-I with Column-II
Column-I Column-II
(a) – 2–222 2OO+O → (p) redox reaction
(b) 2–+ 4 CrO+H → (q) one of the product has trigonal planar
(c) ––+ 42 MnO+NO+H → (r) dimeric bridge tetrahedral metal ion
(d) – 2+ 224 NO+HSO+Fe → (s) disproportio nation
(a) (b) (c) (d)
(1) rq pq p r
(2) rs qr pq rs
(3) ps r pq p
(4) pq pr q rs
33. Match the Column-I and Column-II
Column–I (Reaction)
Column–II (Eq. Wt)
(a) H+ +2 4 KMnOMn → (p) M 2
(b) +2 24 MgCOMg+CO2 → (q) M 5
(c) +3 227 KCrOCr → (r) M 6
(d) +3 5 CrOCr → (s) M 3
(a) (b) (c) (d)
(1) q r p s
(2) s p r q
(3) r s p q
(4) q o r s
34. Match the redox process in List-I with n-factor for underlined species in List- II
List-I (Redox process)
List-II (n-factor for under lined species)
(a) –2–2334 AsSAsO+SO → (p) 28
(b) 23 II+IO → (q) 4/3
(c) 32333 HPOPH+2HPO → (r) 1
(d) H3PO2 + NaOH → NaH2PO2 + H2O (s) 5/3
(a) (b) (c) (d)
(1) p s q r
(2) q s r p
(3) r q p s
(4) s r p q
35. Match the following.
Column-I Column-II
(a) FeCl3 (p) 32 2KClO2KCl+3O ∆ →
(b) Redox reaction (q) reducing agent
(c) H2S (r) oxidising agent
(d) H2O2 (s) oxidising agent and Reducing agent
(a) (b) (c) (d)
(1) p q r s
(2) s q r p
(3) r p q s
(4) q r s p
36. Match the underlined element in the compound in column-I with its oxidation state in column-II
Column-I (Compound)
Column–II (Oxidation state)
(a) H2S2O8 (p) +6
(b) H2SO4 (q) +1
(c) CaOCl2 (r) –1
(d) NO2 (s) +4
(a) (b) (c) (d)
(1) p p qr s
(2) p q sq sr
(3) sq rs sq s
(4) sr qs ps pr
BRAIN TEASERS
1. How many compounds can acts as primary standard titrants?
(b) 2–+ 4 CrO+H → (q) one of the products has trigonal planar structure
37. Match the Column-I with Column-II:
Column-I Column-II
(a) +3 oxidation state (i) nitrogen
(b) +1 oxidation state (ii) nitrous oxide
(c) zero oxidation state (iii) nitrate ion
(d) +5 oxidation state (iv) hydroxylamine
(v) nitrite ion
(a) (b) (c) (d)
(1) i iv iii ii
(2) v ii iv iii
(3) iv v iii i
(4) v ii i iii
38. Match Column-I (Compounds) with Column-II (Oxidation states of Nitrogen).
Column–I Column–II
(a) NaN3 (p) +5
(b) N2H2 (q) +2
(c) NO (r) –1/3
(d) N2O5 (s) –1
(a) (b) (c) (d)
(1) r s q p
(2) p s q r
(3) p q r s
(4) s p r q
(c) ––+ 42 MnO+NO+H → (r) dimeric bridged tetrahedral metal ion
(d) - 2+ 324 NO+HSO+Fe → (s) dispropor tionation
(a) (b) (c) (d)
(1) ps r pq p
(2) p s q r
(3) s q r q
(4) p s q r
3. A 1.0 g sample of pyrolusite ore containing MnO 2 (MW=87) was dissolved in a concentrated HCl solution and liberated Cl2(g) was passed through a concentrated KI solution releasing I 2 . If the liberated iodine required 16 mL of 1.25 M sodium thiosulphate solution, (Na2S2O3 changes to Na2S4O6), mass percentage of MnO2 in the given sample is
(1) 43.5%
(2) 87%
(3) 21.75%
(4) 0.875%
4. How many milligrams of Fe 0.9 O reacts completely with 10 mL 0.1 KMnO4 solution in acidic conditions? Consider that all the iron is oxidised to iron (III) ions and manganese is reduced to +2 oxidation state. No other elements either oxidised or reduced. (Fe = 56, O= 16). Report the value after rounding off to first decimal.
5. 20 mL of x M HCl neutralises completely 10 mL of 0.1M NaHCO3 solution and a further 5 mL of 0.2 M Na2CO3 to methyl orange end point, the value of x is?
(1) 0.5 M (2) 0.01 M
(3) 0.15 M (4) 1 M
6. For the elements in the first short period of modern periodic table, maximum possible oxidation state with respect to oxygen is ______.
FLASHBACK (Previous JEE Questions)
JEE Main
1. Chlorine undergoes disproportionation in alkaline medium as shown below:
7. Given below are two statements. One is labelled Assertion (A) and the other is labelled Reason (R).
Assertion (A): When Na2CO3 is treated against HCl in presence of phenolphthalein indicator, it is converted to NaCl.
Reason (R) : Phenolphthalein shows colour change in the pH range of (3.5–4.6)
In light of the above statements, choose the correct answer from the options given below.
(1) Both (A) and (R) are true and (R) is the correct explanation of (A).
(2) Both (A) and (R) are true but (R) is not the correct explanation of (A).
(3) (A) is true but (R) is false.
(4) Both (A) and (R) are false.
8. Which relations between equivalent weight (E) and molecular weight (M) of reactions are correct for the given change?
The values of a, b, c and d in a balanced redox reaction are respectively :
(1) 1, 2, 1 and 1
2. Given below are two statements: Statement I : S 8 Solid undergoes disproportionation reaction under alkaline conditions to form S2− and 2 SO23
Statement II : 4 ClO It can undergo disproportionation reaction under acidic conditions.
In the light of the above statements, choose the most appropriate answer from the options given below:
(1) Statement I is correct but statement II is incorrect.
(2) Statement I is incorrect but statement II is correct
(3) Both statement I and statement II are incorrect
(4) Both statement I and statement II are correct
3. Iron (III) catalyzes the reaction between iodide and persulphate ions in which
a) Fe+3 oxidizes the iodide ion
b) Fe+3 oxidizes the persulphate ion
c) Fe2+ reduces the iodide ion
d) Fe2+ reduces the persulphate ion
(1) b and c only
(2) a only
(3) a and d only
(4) b only
4. Thiosulphate reacts differently with iodine and bromine in the reactions given below
5. An indicator ‘X’ is used for studying the effect of variation in concentration of iodide on the rate of reaction of iodide ion with H2O2 at room temp. The indicator ‘X’ forms blue coloured complex with compound ‘A’ present in the solution. The indicator ‘X’ and compound ‘A’ respectively are (1) starch and iodine
(2) methyl orange and H 2O 2
(3) starch and H 2O 2
(4) methyl orange and iodine
6. In neutral or faintly alkaline medium KMnO 4 being a powerful oxidant can oxidise, thiosulphate almost quantitatively, to sulphate. In this reaction overall change in oxidation state of manganese will be (1) 5 (2) 1 (3) 0 (4) 3
7. When 10 mL of an aqueous solution of KMnO4 was titrated in acidic medium, equal volume of 0.1 M of an aqueous solution of ferrous sulphate was required for complete discharge of colour.
The strength of KMnO4 in g/L is x × 10 –2 Find the value of x. [Atomic mass of K = 39, Mn = 55, O = 16]
8. The difference in oxidation state of chromium in chromate and dichromate salts is _____.
9. See the following chemical reaction:
xyz →
Which of the following statement justifies the above dual behaviour of thiosulphate?
(1) Bromine undergoes oxidation and iodine undergoes reduction in these reactions
(2) Thisulphate undergoes oxidation by bromine and reduction by iodine in these reactions
(3) Bromine is a stronger oxidant than iodine
(4) Bromine is a weaker oxidant than iodine
The sum of x, y, and z is ________
JEE Advanced
10. Consider the following molecules : Br3O8, F2O, H2S4O6, H2S5O6, and C3O2. Count the number of atoms existing in their zero oxidation state in each molecule. Their sum is ___.
CHAPTER TEST - JEE MAIN
Section-A
Single Option Correct MCQs
1. In the following reaction, which is the species being oxidised?
2. KMnO4 reacts with oxalic acid according to the equation – 2–+ 2+ 424 22
2MnO+5CO+16H2Mn+10CO+8HO →
20 mL of 0.1 M KMnO4 will react with (1) 120 mL of 0.25 M H2C2O4 (2) 150 mL of 0.10 M H2C2O4 (3) 25 mL of 0.20 M H2C2O4 (4) 50 mL of 0.20 M H2C2O4
3. When –BrO3 ion reacts with Br– in acidic medium, Br 2 is liberated. The equivalent weight of Br2 in this reaction is____
(1) 5M 8 (2) 5M 3 3) 3M 5 (4) 4M 6
4. In the reaction –2–+ 2+2–43 42 MnO+SO+HMn+SO+HO → the number of H+ ions involved is (1) 2 (2) 6 (3) 8 (4) 16
5. All elements commonly exhibit an oxidation state of (1) +1 (2) –1 (3) Zero (4) +2
6. Equivalent weight of FeC2O4 in the change: FeC2O4→Fe3++CO2 is
(1) M 3 (2) M 6 (3) M 2 (4) M 1
7. The element that always exhibits a negative oxidation state in its compounds is
(1) nitrogen (2) oxygen (3) fluorine (4) chlorine
8. Following reaction describes the rusting of iron 4Fe+3O 2 →4Fe 3+ +6O 2–Which one of the following statement is incorrect?
(1) This is an example of a redox reaction.
(2) Metallic iron is reduced to Fe 3+
(3) Fe3+ is an oxidising agent.
(4) Metallic iron is a reducing agent.
9. Consider the following reaction: 2–+ +2 424 22 MnO+CO+HMn+2CO+HO 2 → z xyzxy
(1) 2, 5 and 16 (2) 5, 2 and 8
(3) 5, 2 and 16 (4) 2, 5 and 8
10. Equivalent weight of pyrophosphoric acid is
(1) M.wt 1 (2) M.wt 2
(3) M.wt 3 (4) M.wt 4
11. Number of mole of KMnO 4 required to oxidise one mole of Fe(C 2 O 4 ) in acidic medium is
(1) 0.6 (2) 0.167
(3) 0.2 (4) 0.4
12. It is found that element 'X' forms a double salt, isomorphous with Mohr’s salt. The oxidation number of 'X' in this compound is:
(1) +3 (2) +2
(3) +4 (4) –4
13. n -factors for Cu 2 S and CuS, when they react with KMnO 4 in acidic medium are, respectively (neglecting the further oxidation of released SO2)
(1) 7,7 (2) 6,6
(3) 6,8 (4) 8,6
14. In the reaction, I2+2KClO3→2KIO3+Cl2
i) Iodine is oxidised.
ii) Chlorine is reduced. iii) Iodine displaces chlorine. iv) KClO3 is decomposed. The correct combination is (1) Only i & iv are correct. (2) Only iii & iv are correct. (3) i, ii, iii are correct. (4) All are correct.
15. Which of the following reactions does not involve the change in oxidation state of metal
(1) VO–2→V2O3 (2) K→K+ (3) Cu2+→Cus (4) Cu2+→Cu
16. The number of mole of –4 MnO and 2–CrO27
s eparately required to oxidise 1 mole of FeC2O4 each in acidic medium, respectively
17. The sum of oxidation numbers of nitrogen in ammonium nitrite (NH 4NO2) is (1) +5 (2) 0 (3) 1 (4) –3
18. The equivalent weight of potassium permanganate (M.W=158)in acid medium is
(1) 75 (2) 52.66
(3) 31.6 (4) 158
CHAPTER TEST - JEE ADVANCED
2023 P1 Model
Section-A
[Multiple Option Correct MCQs]
1. Which is/are incorrect statement?
(1) Equivalent mass of –HPO23 is 40.5
(2) Equivalent mass of –HPO24 may be equal to molar mass or less than molar mass because it depends on the reaction.
19. Which of the following is not a redox reaction?
(1) 2BaO+O2→2BaO2
(2) BaO2+H2SO4→BaSO4+H2O2
(3) 2KClO3→2KCl+3O2
(4) SO2+2H2S→2H2O+3S
20. Which one of the following can function as an oxidising agent?
(1) I– (2) H2S
(3) LiAlH 4 (4) –2CrO27
Section-B
21. For the reaction 4FeS2+11O2→2Fe2O3+8SO2
The equivalent mass of FeS 2 is x (M = Molecular mass of FeS2). Here the value of x =___.
22. -2 -+ +2 244 2 2 CO+MnO+HCO+Mn+HO → xyz
The value of (y+z)–x is ____
23. 1.0 g of metal nitrate gave 0.86 g of metal sulphate. Calculate equivalent weight of metal?
24. Calculate the number of moles of Sn + ion oxidised by 1 mol of K 2 Cr 2 O 7 in acidic medium?
25. In the compound YBa 2Cu 3O 7 that shows superconductivity, the oxidation state of Cu is x/3. (assume that the rare earth element Yttrium is in its usual +3 oxidation state) then x is
(3) KMnO4 has maximum equivalent mass in acidic medium.
(4) Oxidation state of H in MgH2 is greater than in H2O2
2. One gram equivalent of Al 2(SO4)3 contain (1) 0.33 gram ions of Al+3
(2) 2 gram atoms of oxygen
(3) 0.5 gram atoms of sulphur
(4) 0.5 gram ions of sulphate
3. When a mixture of Cu2S and CuS is titrated with Al(MnO 4 ) 3 in acidic medium the oxidation products of Cu 2 S and CuS are Cu +2 and SO 2 if the molecular weight of Cu 2S, CuS and Al(MnO 4) 3 be M 1, M 2 and M 3 respectively then which of the following statements are correct.
(1) Equivalent weight of Cu2S is M1 8
(2) Equivalent weight of CuS is 2 M 5
(3) Equivalent weight of Al(MnO 4)3 is 3 M 5
(4) Equivalent weight of Al(MnO4)3 is 3 M 15
Section-B
[Single Option Correct MCQs]
4. 2 mol N 2 H 4 loses 16 mol of electrons is being converted to a new compound x. Assuming that all of the 'N' appears in the new compound, what is the oxidation state of 'N' in compound x?
(1) –1 (2) –2 (3) +2 (4) +4
5. Oxidation states of the metal in the minerals haematite and magnetite, respectively are
(1) II, III in haematite, and III in magnetite
(2) II, III in haematite, and II in magnetite
(3) II in haematite, and II, III in magnetite (4) III in haematite, and II, III in magnetite
6. 20 mL of 0.2 M MnSO4 are completely oxidised By 16 mL of KMnO 4 of unknown normality, each forming Mn 4+ oxidation state. Find out the normality of KMnO 4 solutions.
(1) 0.5 N (2) 1 N (3) 1.5 N (4) 2 N
7. 0.59 g of the silver salt of an organic acid (mol.wt.210) on ignition gave 0.36 g of pure silver. The basicity of the acid is [At.wt of Ag=108]
(1) 1 (2) 2 (3) 3 (4) 4
Section-C
[Integer Value Questions]
8. How many of the following elements (or) ions (or) compounds will disproportionate in acidic (or) in basic medium at proper temperature?
9. 10 g of Fe3O4 is oxidised completely by 50 mL of 0.1M KMnO4 solution. The mass in grams of Fe2O3 in Fe3O4 is ___.
10. When 5.0 g of a metal is strongly heated, 9.44 g of its oxide is obtained. Then the equivalent mass of the metal is __.
11. In hot alkaline solution, Br2 disproportionates to Br– and –BrO3
2 32 3Br+6OH5Br+BrO+3HO. → The equivalent weight of Br 2 is 3M/ x (M=mol. wt.) Then the value of x is ___.
12. How many of the following show disproportionation reaction i) XeF4+H2O→ ii) XeF2+H2O→ iii) P4+NaOH+H2O→ iv) F2+NaOH→ v) Cl2+NaOH→ vi) HOCl ∆ → vii) NO2+NaOH→ viii) Zn+NaOH→ ix) S8+KOH
13. When copper is treated with a certain concentration of nitric acid, nitric oxide, and nitrogen dioxide are liberated in equal volumes according to the following equation: xCu+yHNO3→Cu(NO3)2+NO+NO2+H2O. The coefficient of x is _____.
Section-D
[Matrix Matching Questions]
14. Match the reaction in Column-I with titration method in Column-II and select answer from the answer code Column–I Column–II (a) 2 232 2SOI+→ 2 46 SO2I + (p) Redox with self indicator
15. Equivalent weight of potassium permanganate in different media.
Column–I (potassium permanganate in different media)
Column–II (Equivalent Weight)
(a) Acidic (p) 15.8
(b) Neutral (q) 52.6
(c) Strong alkali (r) 79
(d) Dilute alkali (s) 31.6
(a) (b) (c) (d)
(1) s q p q
(2) q p q s
(3) r q q p
(4) p q r s
16. Consider the redox reactions in Column-I and molar ratios of oxidising to reducing agents in Column-II, respectively. Match the items in the Columns appropriately.
Column-I (Redox Reaction)
Column-II (Molar Ratio of Oxidising to Reducing Agents)
(a) –2–MnO+CO424 MnO+CO22 → (p) 2 : 1
(b) ( ) –3 ClO+FeOH –2–4 Cl+FeO → (q) 3 : 1
(c) ( )––2 3 HO+CrOH 2––4 CrO+HO → (r) 2 : 3
(d) ( ) 24 2 NH+CuOH 2 NO+Cu → (s) 3 : 2
(a) (b) (c) (d)
(1) s q p q
(2) q p q s
(3) r q q p
(4) p q r s
17. Match the conditions if column-I with solution on right column-II.
Column-I (Medium)
Column-II (Equivalent weight)
(a) Has maximum pH at the end point when titrated against KOH (p) CH3COOH (Ka=2×10–5)
(b) Has minimum pH at the end point when titrated with KOH (q) HCN(Ka=5×10–4)
(c) Evolve maximum heat when treated with NaOH (r) HF(Ka=5×10–4)
(d) Evolve equal amount of heat when titrated with strong acid or strong base (s) NH3(Ka=2×10–5)
The elements of group 13 or group III A are boron (B), aluminium (Al), gallium(Ga), Indium (In), Thallium (Tl), and nihonium (Nh). These elements belong to p-block as differentiating electron is present in outer p-sublevel.
They possess the general outer electronic configuration, ns 2 np 1 . The electronic configuration of these elements is given in Table 8.1.
Boron is a fairly rare element, mainly occurs as orthoboric acid, (H 3BO 3), borax, Na2B4O7.10H2O and kernite (Na2B4O7.4H2O). In India, borax occurs in Puga Valley (Ladakh) and Sambhar Lake (Rajasthan).
The abundance of boron in earth crust is less than 0.0001% by mass. There are two isotopic forms of boron 10 B (19%) and 11 B (81%). Aluminium is the most abundant metal and the third most abundant element in the earth’s crust (8.3% by mass) after oxygen (45.5%) and Si (27.7%).
Bauxite, Al2O3.2H2O and cryolite, Na3AlF6 are the important minerals of aluminium. In India it is found as mica in Madhya Pradesh. Karnataka, Orissa and Jammu.
It occurs in combined state as oxides, fluorides, silicates, etc. Gallium, indium and thallium occur in traces along with the sulphide ores of zinc and lead. Abundance of elements of group 13 is given in Table 8.2.
Table 8.2 Abundance of elements of group13
Table 8.1 Electronic configuration of elements of boron family
8.1.1 Variation of Properties
Atomic radius: Atomic radius increases down the group. The sudden increase in the atomic radius of aluminium is due to greater screening effect of electrons present in the penultimate shell.
The atomic radius of Ga is less than that of aluminium. The presence of additional ten d–electrons offer only poor screening effect for the outer electrons from the increased nuclear charge in gallium. Atoms with d10 inner shells are smaller in size than would otherwise be expected.
In thallium, the presence of ‘f’ electrons in the anti-penultimate shell further affects the size.
Order of atomic size: B < Ga < Al< In < Tl
Ionisation potential: First ionisation energy of these elements corresponds to the removal of np 1 electron from ns 2 np 1 configuration while the second and third ionisation energies correspond to the removal of ns2 electrons in succession. Order of ionisation potentials is : IP1 < IP2 < IP3.
Ionisation energy decreases from B to Al significantly. This is due to sudden increase in atomic radius of Al from B. From Al to Tl, change in IP1 is less pronounced due to the poor screening caused by (n–1)d and (n–2) f electrons.
The second and third ionisation energies are considerably higher because the successive electrons are to be removed from s-orbital.
In this group, the sum of the ionisation enthalpies I1 + I2 + I 3 is very high for boron. Hence, boron is not able to exist as B 3+ ion in its compounds. However, as we move down the group, the ionisation energies decrease but not in a regular order. Ionisation enthalpies of thallium are higher, because of lanthanide contraction.
Order of I1 value: In < Al< Ga < Tl < B
Density: Density increases from boron to
thallium. Aluminium has lesser density than expected due to sudden increase in atomic size of aluminum from boron.
Order of density value: B < Al< Ga < In < Tl
Melting and boiling points: Melting points decrease from B to Ga and then increase up to thallium. The irregular trend is due to structural changes in the elements. The melting point of boron is very high because of its gaint covalent polymer structure in both liquid and solid states.
The low melting point of gallium is due to its simple molecular structure which consists of discrete Ga2 molecules. Boiling points decrease regularly from B to Tl. Gallium, with unusually low melting point (303 K), could exist in liquid state during summer. Its high boiling point (2676 K) makes it a useful material for measuring high temperatures.
Electronegativity: Electronegativity of group 13 elements decreases from boron to aluminum, and thereafter increases up to thallium. From boron to aluminum, electronegativity is decreased due to increase in atomic size.
From aluminum to gallium, electronegativity is increased due to nearly same size of aluminum and gallium. This is also due to the poor shielding effect of d–electrons in the inner shell of gallium. Since the electronegativity of Al is least in this group, it is the most metallic element.
Atomic and physical properties of elements of boron family are tabulated in Table 8.3
Oxidation states: All these elements exhibit a common oxidation state of +3 because of their outer electronic configuration. Except boron and aluminium, other elements show +1 oxidation state also. The +1 oxidation state becomes more and more stable on descending the group due to inert pair effect. B and Al atoms do not show inert pair effect and do not form B+ and Al+ cations.
Table 8.3 Atomic and physical properties of elements of boron family
–1)
(kJ mol -1)
Order of Relative Stability
With +3 state, B3+ > Al3+ > Ga3+> In3+ > Tl+3 and with +1 state, B+ < Al+ < Ga+ < In+ <Tl+, boron exhibits –3 oxidation state in borides (like in magnesium boride, Mg 3B2).
Compounds in + 1 oxidation state are ionic and, in + 3 oxidation state, are oxidising.
In trivalent state, the compounds of these elements will be electron-deficient molecules and behave as Lewis acids. Tendency to behave as Lewis acid decreases with the increase in the size of IIIA group element.
Ammonia donates lone pair of electrons to BCl3.
bond formation. For thallium, +1 oxidation state is more stable than +3 state. Thallium compounds, like TlCl3, are good oxidants, as they gain electrons easily to convert to more stable oxidation states.
8.1.2 Trends in Chemical Reactivity
■ B reacts with metals to form borides, but other elements cannot.
■ B burns in air to form B 2O3 + BN.
■ Al forms protective oxide layer. Hence, Al becomes passive.
■ Al also forms Al2O3 + AlN on burning in air at high temperature.
■ Reactivity towards oxygen of Tl > In > Ga.
Reactivity towards air: Both amorphous boron and aluminium burn in air or oxygen to form their trioxides, M2O3. B2O3 is acidic and reacts with water to form boric acid, H 3BO3. Al2O3 and Ga2O3 are amphoteric. Oxides of indium and thallium are basic.
23 2 22 3 MO MO sg s () () ()
In ert pair eff ect: It is reluctance of pair of outer s–electrons towards participation in
22 2 MN MN sg s () () ()
Boron is unreactive in the crystaline form.
Aluminium forms a thin oxide layer on surface, which prevents further attack from oxygen.
AlCl3 is a Lewis acid. It acquires stability by forming dimer.
Ti2O is more stable and more basic than Tl2O3 due to inert pair effect in Tl.
Reactivity towards halogens: The elements (except Tl) of boron family form trihalides of the formula, MX3. These are formed when the elements are heated with halogens at high temperatures.BI3 is unstable.
23 2 23 MX MX XF, Cl, Br, I () () () () sg s
Boron trihalides are covalent due to high polarising power of B3+ ion. These trihalides are electron deficient and central atom has incomplete octet. To acquire the octet, the central atom accepts a lone pair of electrons. These compounds behave as Lewis acids. BF3 is a weak Lewis acid due to back dative bonding. The order of Lewis acid strength is: BBr 3 > BCl3 > BF 3
Boron halides are readily hydrolysed, forming boric acid.
BX HO HX HBO 32 33 33
Aluminium chloride is represented as Al 2 Cl 6 up to 400°C, and above 800°C, it exists as a monomer. These halides of aluminium also function as Lewis acids. The dimeric aluminium chloride is denoted with six covalent bonds and two dative bonds. In trivalent state, most of the compounds being covalent are hydrolysed in water to form tetrahedral [M(OH) 4 ] – species; the hybridisation state of element M is sp 3 . AlCl 3, in acidified aqueous solution, forms octahedral [Al(H2O)6]3+ ion. In this complex ion, the 3d orbitals of Al are involved and the hybridisation state of Al is sp3d2. Al2Cl6 is called autocomplex. Dimeric aluminium chloride is shown in Fig. 8.1.
Reactivity towards acids and alkalies: Boron does not react with acids and alkalies even at moderate temperature; but aluminium dissolves in mineral acids and aqueous alkalies and, thus, shows amphoteric character.
Aluminium dissolves in dilute HCl and liberates dihydrogen.
26 26 3 3 2 Al HClAlClH ()saqaqaqg () () () ()
However, concentrated nitric acid renders aluminium passive by forming a protective oxide layer on the surface.
Aluminium also reacts with aqueous alkali and liberates dihydrogen.
22 6 2 Al NaOH HO ()saql () ()
2342 Na Al OH H aq g [( )] () ()
Sodium tetrahydroxoaluminate (III)
8.1.3 Anamolous Properties of Boron
Certain important trends can be observed in the chemical behaviour of group 13 elements. The tri-chlorides, bromides, and iodides of all these elements, being covalent in nature, are hydrolysed in water. Species like tetrahedral [M(OH)4]– and octahedral [M(H2O)6]3+, except in boron, exist in aqueous medium.
The monomeric trihalides, being electron deficient, are strong Lewis acids. Boron trifluoride easily reacts with Lewis bases, such as NH3, to complete octet around boron.
It is due to the absence of d orbitals that the maximum covalence of B is 4. Since the
Fig. 8.1. Dimeric aluminium chloride
d orbitals are available with Al and other elements, the maximum covalence can be expected beyond 4. Most of the other metal halides are dimerised through halogen bridging. The metal species completes its octet by accepting electrons from halogen in these halogen bridged molecules.
Boron shows anamolous behaviour due to (i) small size, (ii) high IP, (iii) high EN, (iv) nonavailability of vacant d-orbitals, (v) different number of electrons in penultimate shell.
Boron shows some differences in some of the properties with aluminium, which are given in Table 8.4.
8.1.4 Compounds of Boron (Advance)
Boron is a non-metal.It is not found in native state.It is found in combined state. Borax, boric acid, and diborane are some important compounds of boron.
Borax (Advance)
The most common metaborate is borax. Its molecular formula is Na 2 B 4 O 7 .10H 2 O (or) Na2[B4O5(OH)4].8H2O.
Borax is available in nature as (i) tincal (or) crude borax, (ii) kernite (or) rasorite Na2B4O7.4H2O from these two minerals, the decahydrate is obtained by extracting with hot water and concentrating the extract.
Borax contains the tetranuclear units [B4O5(OH)4]2–, as shown in Fig. 8.2.
Properties of Borax
B orax exists in crystalline forms namely prismatic borax (Na2B4O7.10H2O), octahedral borax (Na2B4O7.5H2O), crystallized at 370 K. When borax is heated to above its melting point, it forms anhydrous sodium tetraborate, called borax glass (Na 2B4O7).
1. Boron is a rare element and a non-metal. Aluminium is the most abundant metal and the third largest available element in nature.
2. It is a bad conductor of heat and electricity. It is a good conductor.
3. It exhibits allotropy. It does not exhibit allotropy.
4. It does not react with dilute HCl or H2SO4 It evolves H2 gas with dilute HCl or H2SO4
5. It forms acidic oxide. It forms amphoteric oxide.
6. It forms stable borates. It forms unstable aluminates.
7. It forms borides with metals. It forms alloys with metals.
8. It forms stable covalent hydrides. It forms unstable hydrides.
9. B(OH)3 is an acid. Al(OH)3 is amphoteric.
10. It does not form cation. It does not form anion.
11. It can exhibit–3 oxidation state. It cannot exhibit–3 oxidation state.
12. It forms only covalent compounds. It forms both covalent and ionic compounds.
13. Its maximum covalency is 4. Its maximum covalency is 6.
Fig. 8.2 Tetranuclear units of borax
Table 8.4 Difference between boron and aluminum
Boron
Aluminium
Borax is sparingly soluble in cold water but fairly soluble in hot water.
Borax dissolves in water to give an alkaline solution due to anionic hydrolysis.
Na BO HO 2NaOH+ 4H BO 24 72 33 7
Borax Bead Test
The test is useful in the identification of basic radicals. The test is carried out as follows.
■ A small loop is made at the end of platinum wire. It is heated in Bunsen flame till it becomes red hot.
■ It is then dipped in powdered borax and again heated in the flame. Then, it loses water of crystallisation, forming a colourless, transparent, glass-like bead, called borax bead.
Na BO HO Na BO NaBo BO Sodium meta borate Boric anh 24 72 24 72 23 10 2 y ydride Boraxbead
■ The bead is allowed to cool and dipped into the sample (transition metal salts) to be tested. The bead with the adhering substance is heated.
The metaborates of various trasition metal ions form coloured beads.
BO CoOCoBO blue bead 23 ()22
BO Cr OCrBO green bead 23 23 22 2 ()
Uses of Borax
■ Borax is used as flux in soldering, welding, and in certain metallurgical processes.
■ Borax bead test is used to identify coloured basic radicals in qualitative analysis.
■ Borax is used in making optical glass and pyrex glass.
■ Borax is used as food preservative.
■ Borax is used in leather industry for cleaning hides and skin.
Boric Acid (Advance)
Boron trioxide will give various boric acids with water. These are given below:
Orthoboric acid: H3BO3 (or) B2O3.3H2O
Metaboric acid: HBO2 (or) B2O3.H2O
Tetraboric acid: H2B4O7 (or) 2B2O3.H2O
Pyroboricacid: H6B4O9 (or) 2B2O3.3H2O
Boric acid is trivial name of ortho boric acid. It is prepared from powdered colemanite, by dissolving in hot water and passing SO 2 until saturation.
It can also be prepared by acidifying an aqueous solution of bor ax.
Na BO HClH ONaClB OH 24 72 3 25 24 ()
Properties of Boric Acid
H3BO3 is a white cystalline solid which is soapy to touch. H3BO3 is a weak monobasic acid. It loses water on heating. Orthoboric acid above 370 K forms metaboric acid HBO2, which, on further heating, yields boric oxide, B2O3.
H3BO3 H2B4O7 B2O3 573K 370K HBO2 ∆ ∆
Bori c acid has layer structure, in which planar BO 3 units are joined by hydrogen bonds. The hydrogen atoms constitute covalent bond with one unit and hydrogen bond with the other unit, as shown in Fig. 8.3. The dotted lines represent hydrogen bonds.
Fig. 8.3 Structure of boric acid (the dotted lines represent hydrogen bonds)
BOH HOH BOHH O () [( 34)] 3 2
Uses of Boric Acid
■ Boric acid solution is a weak antiseptic for eyes.
■ Boric acid is used in enamel and glass industries.
Diborane (Advance)
Boron forms many borohydrides, but diborane is the simplest and most important. Boranes are high-potential energy fuels, in view of their high heat of combustion. These are better fuels than the hydrocarbons. Heat of combustion of diborane is much higher than ethane.
Preparation of Diborane
■ Industrially, diborane is prepared by the reaction of boron trifluoride with ionic hydrides, like sodium hydride or lithium hydride.
2BF+ 6LiH BH +6LiF 3 450K 26
It is prepared on a large scale by the reduction of BF3 with LiAlH4 .
4BF+ LiAlH2BH +2LiF+ 3AlF 34 26 3 →
■ In laboratory, diborane is obtained when sodium borohydride is oxidised with iodine 22 42 26 2 NaBH IB HNaI +H
Properties of Diborane
■ Diborane is a colourless toxic gas. It is stable at low temperatures in the absence of grease and moisture. At high temperature, it is decomposed to give other hydrides of boron. Diborane catches fire spontaneously upon exposure to air. It burns in oxygen, releasing an enormous amount of energy (1976 KJmol–1).
BH OB OH O 26 22 3332
■ It readily reacts with water, giving boric acid and hydrogen.
BH HO HBOH 26 23 32 62 6
■ It dissolves in strong alkalies to produce metaborates and hydrogen.
BH KOH HO KBOH 26 22 2 22 26
■ It combines with carbon monoxide at 1000 C and 20 atmospheres to form borane carbonyl.
BH CO BH CO 26 3 22[. ]
Addition compound is formed due to dative bonding between BH 3 and CO.
■ Diborane undergoes cleavage reactions with Lewis bases (L) to give borane adducts, BH3.L.
BH N(CH )BHN CH 26 33 33 3 22 .( )
■ Diborane reacts with metal hydrides in ether to give hydridoborates.
BH MH MBH 26 4 22 []
M is lithium or sodium.
It undergoes spontaneous combustion in air due to strong affinity of boron with oxygen, forming boric anhydride. Large amount of heat is evolved, because of which, it is considered a potential rocket fuel.
■ Reaction of ammonia with diborane gives, initially at 120°C, an addition product, diammoniate of diborane, which can be formulated as [BH2(NH3)2]+ [BH4]–. This, on further heating to 200°C, gives borazine or borazole. This is structurally similar to benzene. Hence it is called inorganic benzene. However, borazole is polar and more reactive than benzene. Structure of borazole is given in Fig. 8.4.
Fig. 8.4. Structure of borazole
36 3 26 32 32 4
BH NH BH NH BH [( )] []
Heat BN 21HH 2 33 62 At higher temperatures, it decomposes to give boron nitride and hydrogen.
nB NH 3(BN)+ 3nH 33 6 ” n2
In B 3 N 3 H 6 , both B and N atoms are sp 2 hybridised.
Structure
■ In B2H6, total number of valence electrons is 12. In the similar compound, C 2H6, the total number of valence electrons is 14. This indicates that B2H6 is electron deficient.
■ Methylation of B2H6 gives B2H2Me4. Four H-atoms of B2H6 are replaced by methyl groups but remaining two H-atoms are not methylated. This indicates that there are two different sets of H-atoms are present in B2H6.
■ In B 2H 6, boron uses sp 3 hybrid orbitals. Out of four hybrid orbitals, three contain one unpaired electron in each and one is a vacant orbital.
■ Two hybrid orbitals of B, each containing one unpaired electron, are used in the formation of two terminal B–H bonds.
■ 1s orbital of each bridge H-atom overlaps with an sp 3 orbital containing unpaired electron of boron and also with vacant sp3 orbital of second boron. Thus, three-centre two electrons bonds is formed. Two such B–H–B bonds are formed in B2H6. They are also known as banana bonds or tau bonds.
■ The bridge hydrogen atoms are present in a plane perpendicular to the rest of the molecule.
The distance between the two non-bonded boron atoms is 177 pm.
8.1.5 Alums (Advance)
■ Alums are double salts having general formula, X2SO4Y2(SO4)324H2O. Here, X+ is monovalent ion and Y+3 is trivalent cation.
X+ion = Na+, K+,NH4+...
Y+3ion = Cr+3, Al+3, Fe+3...
Examples:
Potash alum – K2SO4Al2(SO4)324H2O
Soda alum– Na2SO4Al2(SO4)324H2O
Ammonium alum–(NH4)2SO4Al2(SO4)324H2O
Chrome alum – K2SO4Cr2(SO4)324H2O
Ferric alum– (NH4)2SO4Fe2(SO4)324H2O
■ The alums are isomorphous (they crystallise in similar forms).
■ They exist with the given composition only in solid state. Once they dissolve in water, they split into ions and behave as a mixture of various constituent ions.
■ Their aqueous solutions are acidic in nature (they undergo cationic hydrolysis in water).
■ In alum, each cation is surrounded by six H2O molecules. Among the alkali metals, lithium does not form alum, due to its small size.
Preparation of Alum
■ Equimolar quantities of Al 2 (SO 4 ) and K2SO4 salts are taken and dissolved them in minimum quantity of water.
■ The resultant solution is subjected to evaporation till crystallisation point is reached.
Fig 8.5. Structure of diborane
Fig 8.6. Diborane molecule and formation of banana bands.
■ The resultant solution, upon cooling, forms crystals of alum.
8.1.6 Uses of B and Al
Boron being extremely hard refractory solid of high melting point, low density and very low electrical conductivity, finds many applications. Boron fibres are used in making bullet-proof vest and light composite material for aircraft.
The boron (10B) isotope has high ability to absorb neutrons and, therefore, metal borides are used in nuclear industry as protective shields and control rods.
The main industrial application of borax and boric acid is in the manufacture of heat resistant glasses (e.g., Pyrex), glass-wool and fibreglass.
Borax is also used as a flux for soldering metals, for heat, scratch and stain resistant glazed coating to earthenwares and as constituent of medicinal soaps. An aqueous solution of orthoboric acid is generally used a mild antiseptic.
Aluminium is bright silvery-white metal, with high tensile strength. It has a high electrical and thermal conductivity. On a weight-to-weight basis, the electrical conductivity of aluminium is twice that of copper.
Aluminium is used extensively in industry and every day life. It forms alloys with Cu, Mn, Mg, Si and Zn.
Aluminium and its alloys can be given shapes of pipe, tubes, rods, wires, plates or foils and, therefore, find uses in packing, utensil making, construction, parts of aeroplane and transportation industry.
The use of aluminium and its compounds for domestic purposes is now reduced considerably because of t heir toxic nature.
TEST YOURSELF
1. Which of the following statements is correct regarding 13th group?
(1) Boron has low melting point.
(2) Gallium has high Boiling point than In.
(3) The stable oxidation state of thallium is +3.
(4) Atomic radius of Gallium is greater than that of aluminium.
2. Which of the following set of elements is correct with respect to least melting point and least boiling point?
(1) Tl, Ga (2) Ga, Tl
(3) Al, Ga (4) B, Al
3. Choose the correct stability order of group 13 elements in their +1 oxidation state.
(1) Al < Ga < In < Tl
(2) Tl < In < Ga < Al
(3) Al < Ga < Tl < In
(4) Tl < Al < In < Ga
4. Maximum covalency of boron is X and that of aluminium is Y. Then, X and Y, respectively are
(1) 4 and 6 (2) 6 and 7
(3) 6 and 6 (4) 4 and 4
5. Bond order of B–F in BF3 is (1) 1.33 (2) 0.5
(3) 1 (4) 0.75
6. Hybridisation of ‘Al’ in the compound [AlH2O6]3+ ion is
(1) sp3 (2) dsp2
(3) sp3d3 (4) sp3d2
7. The mineral of aluminium which does not contain oxygen is
(1) bauxite (2) cryolite
(3) clay (4) diaspore
8. H3BO3 is
(1) monobasic and weak Lewis acid
(2) monobasic and strong Lewis acid
(3) monobasic and weak Bronsted acid
(4) tribasic and weak Bronsted acid
9. T he covalency and oxidation state, respectively, of boron in [BF 4]– are (1) 4 and 3 (2) 3 and 5
(3) 3 and 4 (4) 4 and 4
10. The 13th group element with least ionisation potential is
(1) In (2) Tl
(3) Ga (4) Al
11. Which one of the properties in 13th group shows a regular decrease from top to bottom?
(1) Atomic radius
(2) Electronegativity
(3) Melting point
(4) Boiling point
12. The correct order of atomic radii in group 13 elements is
(1) B < Al < In < Ga < Tl
(2) B < Al < Ga < In < Tl
(3) B < Ga < Al < Tl < In
(4) B < Ga < Al < In < Tl
13. The oxidation state of most metallic element present in cryolite is
(1) +3 (2) +1
(3) –1 (4) 0
14. Which element reacts with acids as well as alkalies?
(1) Mg (2) Si
(3) Al (4) Ca
Answer Key
(1) 2 (2) 2 (3) 1 (4) 1
(5) 1 (6) 4 (7) 2 (8) 1
(9) 1 (10) 1 (11) 4 (12) 4
(13) 2 (14) 3
8. 2 GROUP 14 ELEMENTS –THE CARBON FAMILY
The elements of group 14 or group IV A are ca rbon (C), silicon (Si), germanium (Ge), Tin (Sn), lead (Pb), and flerovium (Fl). These elements belong to p-block and possess the general outer electronic configuration, ns2np2 In the penultimate shell, carbon contains two electrons, silicon contains eight electrons, and other elements contain eighteen electrons each. Due to this difference, carbon differs from silicon in some properties and these two differ from the rest of the members of the group. The electronic configuration of these elements is given in Table 8.5.
Table 8.5 Electronic configuration of elements of carbon family
8.2.1
Occurrence and Abundance
Carbon and silicon are widely distributed in nature, compared to tin and lead.
Carbon is the seventeenth most abundant element by mass in the earth’s crust. It is widely distributed in nature in free as well as in the combined state. Carbon is the most abundant element in the human body. It is the most versatile element in the world. It is an essential constituent of all living organisms. Organic chemistry is a branch of chemistry that deals with compounds containing carbon. In the elemental state, carbon is available as coal, diamond, and graphite. In the combined state, carbon exists as metal carbonates, hydrocarbons, and carbon dioxide (0.03% in air). Naturally occurring carbon contains two stable isotopes; C12 and C13. In addition to these, third isotope C14 is also present. It is a radioactive isotope with half-life 5770 years
and is used for radiocarbon dating. Silicon is the second most abundant element (27.7%) by mass in the earth’s crust and is present in nature in the form of silica and silicates. Silicon is a very important component of ceramics, glass, and cement. Germanium is only found in traces. The abundance of tin and lead is comparatively low, but they occur as concentrated ores. Tin occurs mainly as cassiterite (SnO2) and lead as galena (PbS). Ultra pure forms of germanium and silicon are useful in making transistors and semiconductor devices.
The atomic number of flerovium is 114, atomic mass is 289 g mol -1 , and electronic configuration is [Rn] 5f 14 6d 10 7s 2 7p 2 . It has been prepared only in a small amount. Its half life is short and its chemistry has not been established yet.
Atomic and physical properties of elements of group 14 are given in Table 8.6.
(20°C) (ohm cm) (diamond)
Table 8.6 Atomic and physical properties of carbon family
8.2.2 Variation of Properties
There is no regular gradation in properties of group 14 elements. Some of them are discussed as follows.
Covalent Radius
Atomic or covalent radius gradually increases from carbon to lead. There is a considerable increase from carbon to silicon. Thereafter from silicon to lead, a small increase in radius is observed. This is due to the presence of completely filled inner ‘d’ and ‘f’ orbitals in heavier members.
Order of atomic radius: C < Bi < Ge < Sn < Pb
Ionisation Energy
The first ionisation energy of elements of carbon family is higher than the corresponding elements of boron family. The ionisation energies decrease down the group gradually but not systematically as in alkali metals and alkaline earth metals. The value of lead is slightly higher than expected due to lanthanide contraction. Large decrease in ionisation potential from carbon to silicon is due to sudden increase in size of silicon atom. The very high ionisation energy of carbon is due to smaller size and high electronegativity.
Order of I1: C > > Si > Ge > Pb > Sn
Density, Melting Point, and Boiling Point
Except for carbon, the densities increase with an increase in the atomic number. Carbon has an extremely high melting point. Silicon melts appreciably at lower temperature than carbon, but the values of silicon and germanium are still high. They all have the very stable diamond type lattice. Melting involves breaking the strong covalent bonds in the lattice, and so, it requires a lot of energy. The melting points decrease on descending the group because the M–M bonds become weaker as the atoms increase in size. Tin and lead have much lower melting points. They do not use all four outer electrons for metallic bonding.
Nature of Elements
C and Si are non-metals, Ge is metalloid, Sn is metalloid but more metallic in nature, and Pb is a metallic element.
Electronegativity
Due to small size, the elements of this group are slightly more electronegative than elements of group 13. Carbon is the most electronegative element of this group while silicon, germanium, tin, and lead possess nearly the same value. This is due to filling of d-orbitals in tin and germanium and also f-orbitals in lead.
Oxidation States
The group 14 elements have four electrons in the outermost shell. The common oxidation states exhibited by these elements are +4 and +2. Due to greater electronegativity, carbon exhibits negative oxidation states also. Since the sum of the first four ionisation potentials is very high, compounds in +4 oxidation state are generally covalent in nature.
In heavier elements, the tendency to show +2 oxidation state increases in the sequence: Ge < Sn < Pb. It is due to the inability of ns2 electrons of valence shell to participate in bonding. This is called inert pair effect. The relative stabilities of these two oxidation states vary down the group.
Due to inert pair effect, the formation of compounds having these elements in +2 oxidation state is limited to certain elements.
Carbon and silicon mostly show +4 oxidation state. Carbon and silicon do not show +2 oxidation state in their stable compounds. In CO, the oxidation state of carbon is +2, but due to the presence of dative bond from oxygen to carbon, the + 2 state is doubtful. Germanium forms stable compounds in +4 state and only few compounds in +2 state. Tin forms compounds in both oxidation states (Sn in +2 state is a red ucing agent). Lead
compounds in +2 state are stable and in +4 state are strong oxidising agents. Since the inert pair effect increases from germanium to lead, the stability of M4+ ions decreases and that of M2+ ions increases. Thus, the stability of these ions follow the following order:
Ge2+ < Sn2+ < Pb2+ ; Ge4+ > Sn4+ > Pb4+
Being electron precise molecules, these tetravalent compounds are normally not expected to act as electron acceptor or electron donor species. Although carbon cannot exceed its covalency more than four, other elements of this group can do so because of the presence of vacant d-orbitals in them. Due to this, their halides undergo hydrolysis and have tendency to form complexes by accepting electron pairs from donor species. For example, the species like [SiF6]2–, [GeCl6]2–, and [Sn(OH)6]2– exist where the hybridisation of the central atom is sp3d2
8.2.3
Chemical Properties
Group 14 elements are relatively unreactive but reactivity increases down the group. Carbon, silicon, and germanium are unaffected by water. Tin reacts with steam to form SnO2 and hydrogen.
Sn HO 22SnOH 22 2
Lead is not affected by water due to an oxide film on the surface.
Halides
The members of this group form two types of halides. They are dihalides (MX 2 ) and tetrahalides of the type MX 4, except PbBr 4 and PbI4. The halides are covalent and formed by sp 3 hybridisation. Exceptions are SnF 4 and PbF4, which are ionic in considering the thermal stability, GeX 4 is more stable than GeX2, whereas PbX2 is more stable than PbX4. PbI4 does not exist. This is because Pb–I bond formed does not release enough energy to unpair 6s2 electrons and excite them to higher orbitals to have four unpaired electrons in lead.
The thermal stability of halides of different elements with a common halogen decreases with increasing atomic number. The thermal stability of tetrahalides of the same element decreases with increase in molecular mass of the tetrahalide. Except carbon tetra halides, other halides are readily hydrolysed by water. The trend towards hydrolysis, however, decreases down the group. In case of elements other than carbon, vacant d-orbitals are present in the valency shell. Hence, water molecules can coordinate with these elements and so, are easily hydrolysed by water.
T he tetrahalides of silicon, germanium, tin and lea d act as strong Lewis acids. Heavier members, germanium to lead, are able to form halides of formula MX2. Stability of dihalides increases down the group.
GeF4 and SiCl4 act as Lewis acids because Ge and Si have vacant d-orbitals in valence shell. CCl4 is not a Lewis acid as carbon does not have d-orbitals in valence shell.
Oxides
All members, when heated in oxygen, form oxides. There are mainly two types of oxides, monoxide and dioxide of formula MO and MO 2 , respectively. SiO exists only at high temperature. Oxides in higher oxidation states of elements are generally more acidic than those in lower oxidation states. The dioxides CO 2 , SiO 2 , and GeO 2 are acidic, whereas SnO 2, and PbO 2 are amphoteric in nature. GeO is distinctly acidic whereas SnO and PbO are amphoteric; but CO is neutral among monoxides.
8.2.4 Anomalous Behaviuor of the Carbon
The first element, carbon, shows dissimilarities with other members of group 14. It is due to its smaller size, higher electronegativity, higher ionisation enthalpy, and unavailability of d-orbitals. They are as follows.
■ Carbon is widely distributed in nature. Carbon occurs in the free state whereas other elements are almost not available in nature.
■ The electronic configuration of carbon is 1s 2 2s 2 2p 2 . Carbon has no available d-orbitals while other elements have d-orbitals. Therefore, expansion of the octet can take place in other elements. The coordination number of the element is increasing from 4 to 6.
Example : SiF4 + 2F–→ [SiF6]2–
■ Carbon is non-metallic and, being a small atom, forms compounds with greater covalent nature.
■ The maximum covalency for C is 4 while it is possible to have 6 for other elements.
■ Carbon differs from the other elements in its unique ability to form chains. This is because C–C bond energy is very high (348 kJmol–1) compared to the bond energy in the other elements. The chain length is infinitely long while the longest chain that could be formed is eight atoms in silicon.
■ Carbon alone forms multiple bonds amongst themselves and with other elements of small size and high electronegativity readily, like C = C, C ≡ C, C = O, C = S, and C ≡ N.
Catenation
The linking of identical atoms with each other to form long chains is called catenation. Except lead all the elements of this group have the property of catenation. However, this property decreases from carbon to ti n.
These chains may be linear or cyclic. When an element is divalent or polyvalent, catenation is possible. Carbon is tetravalent and forms single or multiple bonds in the linear chain or in cyclic rings.
Carbon has the maximum property of catenation.
Down the group, the size increases and electronegativity decreases and thereby, tendency to show catenation decreases. This can be clearly seen from the bond energy values which are given in Table 8.7.
8.2.5 Allotropes
of Carbon
The property due to which an element exists in two or more forms which differ in their physical properties and some of the chemical properties is known as allotropy.
The various forms are called allotropes or allotropic modifications. Carbon occurs in many allotropic modifications. Three of the allotropes of carbon are crystalline and the remaining are amorphous as shown in Fig.8.7
Amorphous Coal, coke, wood, charcoal, animal charcoal, lamp black, carbon black, gas carbon, and petroleum coke
Fig.8.7 Allotropes of carbon
C >> Si > Ge = Sn
Table 8.7 Bond energies of elements of carbon family
Carbon
Crystalline
Graphite
Fullerenes
Diamond
Diamond
In diamond, each carbon atom is in sp3 hybridisation and is linked tetrahedrally to four other carbon atoms. This results in the formation of giant molecule. C–C bond lengths are equal to 1.54A° and bond angle is 109°28’. As a result, in diamond, there is a three dimensional network of strong covalent bonds as shown in Fig.8.8 and it is the hardest substance on the earth due to difficulty in breaking of extended covalent bonding. It has exceptionally high refractive index. 154 pm
bonds with three other carbon atoms, as shown in Fig.8.9. All these carbon atoms lie in the same plane. The fourth electron present in the pure ‘p’ orbital is free and, hence, graphite is a good conductor of electricity. The fourth electron forms a pi bond and pi electrons are delocalised. The carbon–carbon bond length in graphite is 1.42 A°
142 pm
335 pm
Fig. 8.8. Structure of Diamond
Uses of Diamond
■ Diamond is used as a precious stone in jewellery. Its purity is measured in carat (1 carat = 200 mg of diamond).
■ Due to its extreme hardness, diamond is used for making tools for cutting and grinding other hard materials and is used as abrasive.
■ Because of its extraordinary sensitivity to heat rays, it is used in high precision thermometers.
■ Diamond is used in the manufacture of tungsten filaments for electric light bulbs and also in making dyes.
Graphite
The structure of graphite is different from that of diamond. Graphite has two-dimensional sheet-like structure. It consists of a series of layers in which there are hexagonal rings made up of carbon atoms. Each carbon atom is in sp 2 hybridisation and forms three covalent
8.9. Structure of graphite
The different sheets of carbon atoms are held by weak van der Waals forces and are separated by a distance of 3.35 A°. These layers can easily slide over one another, which explains the slippery nature of graphite.
Properties of Graphite
■ Graphite is soft and slippery, with melting point 3500°C and density 2.22g cm –3 .
■ It is a good electrical and thermal conductor due to delocalised pi electrons.
■ On heating strongly in air, it burns, giving carbon dioxide.
■ Graphite is thermodynamically more stable than diamond.
Fig.
Uses of Graphite
■ Graphite is used as a lubricant. Graphite cleaves easily between the layers and, therefore, it is very soft and slippery. So, it can be used as a dry lubricant in machines running at high temperatures where oil cannot be used as a lubricant.
■ Mixed with clay, graphite is used in lead pencils. Graphite embedded in plastic material forms high-strength, light-weight composites. These are used in products such as aircrafts, cancos, fishing rods, and tennis rockets.
■ It is used for making carbon electrodes in electrolytic cells and dry cells.
■ It is used in electrotyping and electroplating.
■ It is used for making graphite crucibles, which are refractory and also inert to dilute acids and alkalies.
■ It is used as a moderator in nuclear reactors. The differences between diamond and graphite are listed in Table 8.8.
S.No.
1. Diamond is very hard. It has high density (3.5 gcc–1).
2. It is bad conductor of heat and electricity due to the absence of free electrons.
3. It possesses three-dimensional network structure as each carbon is linked to 4 other carbon atoms.
4. Chemical reactivity is low.
5. Carbon undergoes sp3 hybridisation. C–C bond length is 154 pm and bond angle is 109 028 ’ .
6. Thermodynamically, diamond is less stable than graphite
7. It is transparent to X-rays with high refractive index (2.45).
Fullerenes
Fullerenes are newly discovered crystalline allotropes of carbon. Fullerenes are made by heating graphite in an electric arc in the presence of inert gases, such as helium or argon. The sooty material formed by condensation of vapourised Cn small molecules consists of mainly C60 with smaller quantity of C70 and traces of fullerenes consisting of even number of carbon atoms upto 350 and above. Fullerenes are the only pure form of carbon because they have smooth structure without having ‘dangling’ bonds. Dangling bond is an unsatisfied valence on an immobilised atom. Fullerenes are large cage-like spheroidal molecules. C 60 molecule has a shape like a soccer ball, is as shown in Fig.8.10. It is called buckminister fullerene.
Graphite is soft. It has relatively low density (2.2 gcc–1).
It is good conductor of heat and electricity as it has free electrons.
It possesses a sheet type structure as each carbon is linked to 3 carbon atoms.
Chemical reactivity is high.
Carbon undergoes sp2 hybridisation. C–C bond length is 142 pm and bond angle is 120 0 The distance between the layers is 3.35 A 0
Thermodynamically, graphite is more stable.
It is layer lattice and layers are slippery due to weak van der Waals forces.
Fig. 8.10 Buckminister fullerene
Table 8.8 Difference between diamond and graphite
C60 molecule consists of twenty six-membered rings and twelve five-membered rings. A six-membered ring is fused with six or fivemembered ring but a five-membered ring can only fuse with six-membered ring. All the carbon atoms are equal and they undergo sp2 hybridisation. Each carbon atom forms three sigma bonds with other three carbon atoms. The remaining electron at each carbon atom is delocalised in molecular orbitals, which, in turn, give aromatic character to molecule. This soccer ball-shaped molecule has 60 vertices and each one is occupied by one carbon atom. It also contains both single and double bonds with C–C distances 1.435 A° and 1.383 A°, respectively. Spherical fullerenes are also called bucky balls in short. C 70 molecule consists of twenty five six-membered rings and twelve five-membered rings.
However, the pattern of arrangement is same as that in C 60 molecule. The structure closes to acquire the shape of rugby ball.
Fullerenes are covalent, hence they are soluble in organic solvents.
D fH° graphite is taken as zero because it is t hermodynamically most stable allotrope of carbon, whereas D f H° of diamond and fullerene, C 60 , are 1.90 and 38.1 kJ/mole, respectively.
All the allotropic forms of carbon exhibit same chemical properties but they differ in their chemical reactivity. Amorphous forms of carbon are more reactive than crystalline forms due to their large surface area.
Other forms of elemental carbon, like carbon black, coke, and charcoal are all impure forms of graphite or fullerenes. Carbon black is obtained by burning hydrocarbon in a limited supply of air. Charcoal and coke are obtained by heating wood or coal, respectively, at high temperature in the absence of air.
Uses of Carbon
■ Carbon is used in the manufacture of water gas, producer gas, and carbon disulphide.
■ Activated charcoal is used in adsorbing poisonous gases. Charcoal is also used in water filters to remove organic contaminators.
■ Coke is used as a fuel and largely as a reducing agent in metallurgy.
■ In the form of graphite, it is used for electrodes in batteries and industrial electrolysis.
■ Carbon black is used in black ink and as a filler in automobile tyres.
8.2.6 Oxides of Carbon (Advance)
Carbon forms more number of oxides than other elements present in the group. Of these, CO and CO2 are important stable oxides.
Carbon Monoxide
It is neutral oxide of carbon and is water insoluble.
Preparation of Carbon Monoxide
■ Carbon monoxide is formed by incomplete combustion of carbon or carbon containing compounds in the limited supply of oxygen or air.
■ 2C(g)+O2(g)→2CO(g)
■ On small scale, pure carbon monoxide is prepared by dehydration of formic acid with concentrated sulphuric acid at 100°C.
HCOOH HO CO 2
On commercial scale, it is prepared by the passage of steam over hot coke.A mixture of carbon monoxide and hydrogen, thus produced is known as ‘water gas’ or ‘synthesis gas’ or ‘syn gas’.
C(s)+H2O(g) → CO(g)+H2(g)
■ When air is used instead of steam, a mixture of carbon monoxide and nitrogen is produced, which is called ‘producer gas’.
2C(s)+O2(g)+4N2(g)→ 2CO(g)+4N2(g)
Properties of Carbon Monoxide
■ Carbon monoxide is a colourless, odourless gas water in soluble.
■ Carbon monoxide is neutral in nature and does not effect litmus
■ It burns with a blue flame and forms carbon dioxide.
■ Its density is nearly equal to that of air.
■ It is highly poisonous in nature due to its ability to form a complex with haemoglobin, which is about 300 times more stable than the oxygen haemoglobin complex. This prevents RBC to function as oxygen carrier.It is a powerful reducing agent. It reduces almost all metal oxides other than those of the alkali and alkaline earth metals, aluminium and a few transition metals. This property is used in extraction of metals.
Fe2O3+3CO → 2Fe+3CO2
ZnO + CO → Zn + CO2
■ It combines with chlorine in the presence of sunlight to give phosgene or carbonyl chloride, which is a poisonous gas.
CO + Cl2 → COCl2
■ In CO, carbon atom undergoes ‘sp’ hybridisation. In CO molecule, a dative bond is present between C and O. Its structure can be represented as : C O :
■ In carbon monoxide molecule, there are one sigma and two pi bonds between carbon and oxygen. Because of the presence of a lone pair of electrons on carbon, it acts as a donor and reacts with certain metals, when heated, to form metal carbonyls.
Ni + 4CO → Ni(CO)4
■ Action of NaOH: NaOH CO HCOONa Sodium formate
Uses of Carbon Monoxide
■ Carbon monoxide is an important industrial fuel in the form of water gas (CO + H2) and producer gas (CO + N 2).
■ It is used as a reducing agent in many metallurgical processes.
■ It is used in the purification of nickel in Mond’s process.
■ It is used in the manufacture of methanol, formic acid, phosgene gas, etc.
Carbon Dioxide
It is acidic oxide of carbon and is partly water soluble.
Preparation of Carbon Dioxide
■ Carbon dioxide is prepared by complete combustion of carbon and carbon containing fuels in excess of air.
C(s)+O2(g)→ CO2(g)
CH4(g)+2O2(g)→ CO2(g)+2H2O(g)
■ It is prepared by the action of dilute mineral acids on carbonates or bicarbonates in the laboratory.
MgCO3(s)+H2SO4(aq)→
MgSO4(aq)+H2O(l) + CO2(g)
NaHCO3(s)+HCl(aq) → NaCl(aq)+ CO2(g)+H2O(l)
■ On commercial scale, it is obtained by heating limestone.
CaCO3(s)→ CaO(s)+CO2(g)
■ Limestone is strongly heated with silica
CaCO +SiO CaSiO+ CO 3(s) 2(s) 3(s) 2
Properties
■ Carbon dioxide is a colourless, odourless gas.
■ It is heavier than air.
■ Unlike carbon monoxide, it is not poisonous but it does not support life in animals and humans.
■ It acts as an oxidising agent.
■ It is acidic in nature. Its low solubility in water makes it of immense biochemical and geochemical importance. With water, it forms carbonic acid, H2CO3, which is a weak dibasic acid and dissociates in two steps:
Carbonic acid forms two series of salts, the bicarbonates and carbonates. Aqueous solutions of both bicarbonates and carbonates are alkaline due to anionic hydrolysis. H2CO3/ HCO3– buffer system helps to maintain pH of blood between 7.26 to 7.42.
Uses of Carbon Dioxide
■ Gaseous carbon dioxide is extensively used to carbonate soft drinks.
■ Solid carbon dioxide is obtained in the form of dry ice by allowing the liquefied carbon dioxide to expand rapidly. Dry ice is used as a refrigerant for ice-cream and frozen food.
■ Being heavy and non-supporter of combustion, it is used as a fire-extinguisher.
■ A substantial amount of carbon dioxide is used to manufacture urea.
CO2 + 2NH3 NH2COONH4→ NH 2CONH2 + H2O
■ Carbogen, a mixture of oxygen and carbondioxide (5–10%), is used for artificial respiration.
■ Green plants convert atmospheric CO 2 into carbohydrates, such as glucose. By this process, plants make food for themselves as well as food for animals and human beings.
■ Carbon dioxide, which is normally present to the extent of ~ 0.03% by volume in the atmosphere, is removed from it by the process known as photosynthesis. It is the process by which green plants convert atmospheric CO2 into carbohydrates, such as glucose. The overall chemical change can be expressed as:
6CO2+12H2O → C6H12O6+6O2+6H2O
By this process, plants make food for themselves as well as for animals and human beings. Unlike CO, it is not poisonous. But the increase in combustion of fossil fuels and decomposition of limestone for cement manufacture in recent years seem to have increased the CO 2 content in the atmosphere. This may lead to increase in greenhouse effect and thus, raise the temperature of the atmosphere, which might have serious consequences.
In CO2 molecule, carbon atom undergoes sp hybridisation. Two sp hybridised orbitals of carbon atom overlap with two p orbitals of oxygen atoms to make two sigma bonds while other two electrons of carbon atom are involved in p – p bonding with oxygen atom. This results in its linear shape [with both C–O bonds of equal length (115 pm) with no dipole moment. The resonance structures shown below: OC OO CO OC O
Resonance structures of carbon dioxide
8.2.7 Compounds of Silicon (Advance)
The second-most abundant element in earth’s crust is silicon, found mostly in the form of silicon dioxide and silicates.
Silicon Dioxide
95% of the earth is crust is made up of silica and silicates. Silicon dioxide is commonly known as silica. It exhibits polymorphism— crystalline as well as amorphous forms. Quartz,
tridymite, and crystobalite are the crystalline forms and can be interconvertable at suitable temperature. whereas agate, jaspar, and onyx are the amorphous forms.
Properties of Silicon Dioxide
■ Silica is insoluble in water but dissolves to a small extent when heated under high pressure. Silicon resists attack of halogens, hydrogen, metals, and acids at high temperature.
■ Silicon in its normal form is almost nonreactive. This is because of high Si–O bond energy. It is insoluble in all acids, except hydrofluoric acid. It reacts with hydrofluoric acid to give silicon tetra fluoride and hydrofluoro silicic acid.
SiO2+4HF → SiF4+2H2O
2HF+SiF4→ H2SiF6
■ Silica is acidic oxide. It reacts with alkalies, metal oxides and metal carbonates to give silicates.
SiO2+2NaOH → Na2SiO3+H2O
SiO2+ CaO → CaSiO3
SiO2+Na2CO3→ Na2SiO3+CO2
Na2O+SiO2→ Na2SiO3
SiO2+CaCO3→ CaSiO3+CO2
■ When a mixture of silica and carbon is heated in an electric furn ace, silicon
carbide is formed. Silico n carbide is c alled ‘carborundum’. It has diamondlike structure.
■ S ilica, on heating to 1600°C, changes to quartz glass, which is used in light experiments.
Uses of Silicon Dioxide
■ Silica is used as a building material.
■ It is used as acidic flux in metallurgy.
■ Quartz glass is used in light experiments with UV radiations and in making glass articles.
■ Coloured quartz is used as gems and transparent quartz is for lenses.
■ Bricks made from a mixture of sand, clay, and lime are used in the lining of furnaces in steel manufacture.
■ Quartz is extensively used as a piezo electric material. It has made it possible to develop extremely accurate clocks, modern radio, and television broadcasting.
■ Silica gel is used as drying agent and support for chromatographic material and as a catalyst.
■ Kieselghur is an amorphous form of silica and is used in filtration plants.
Comparison between CO2 and SiO2 is given in Table 8.9.
1. CO2 is linear molecule with ‘sp’ hybridisation of carbon and has 2 s and 2 delocalised p bonds. It has polymeric three-dimensional structure with ‘Si’ undergoing ‘sp3’ hybridisation.
2. C and O are bonded by double bonds. Si and O are bonded by single bonds.
3. It is a gas at room temperature. It is a solid at room temperature.
4. It exits as individual molecules due to weak van der Waals forces between them. The atoms are linked by covalent bonds. It exists as a covalent solid.
5. It is an acidic oxide. It is weakly acidic in nature.
6. It has low melting and boiling point. It has high melting and boiling point.
Table 8.9 Comparison between carbon dioxide and silicon dioxide
S.No Carbon dioxide Silicon dioxide
Structure of Silicon Dioxide
Like diamond, silica is a giant molecule with three-dimensional structure. Each silicon is tetrahedrally linked to four oxygen atoms by covalent bonds. Each corner of tetrahedron is shared with another tetrahedron. Each silicon atom shares only half of the four oxygen atoms bonded to it, so the ratio of silicon to oxygen is 1 : 2. So, the formula is SiO2. Structure of silica is shown in Fig.8.11.
such as CH 3 SiCl 3 , (CH 3 ) 2 SiCl 2 , (CH 3 ) 3 SiCl, and small amount of (CH3)4Si. (CH3)2SiCl2, on hydrolysis followed by polymerisation, yields straight chain polymer.
The chain length of silicones can be controlled by adding (CH3)3SiCl, which blocks the ends. Silicones may be long chain linear compounds or cyclic and branched chain silicones.
Fig. 8.11. Structure of silica
Si – O – Si bonds in silica are weaker than the C – C bonds in diamond. So, silica is not as hard as diamond. Melting point of silica is less than that of diamond. Silica has low temperature a -form and high temperature b modifications. These are interconvertible at a suitable temperature.
Silicones (Advance)
Silicones are synthetic organo silicon compounds containing Si – O – Si linkages with R2SiO as repeating units. So, the general formula is (R2SiO)n, where R is alkyl or aryl group. These are a group of organosilicon polymers.
Commercial silicones are generally methyl d erivatives and, to a lesser extent, phenyl derivatives. The general formula of the starting material is R n SiCl (4 – n) . These are prepared when methyl chloride reacts with silicon in the presence of copper as a catalyst at 300°C. Various types of organosilanes are formed,
Uses of Silicones
Silicones, being surrounded by non-polar alkyl groups, are water repelling in nature. They have high thermal stability, high dielectric strength, and resistance to oxidation and chemicals. So, they are used as follows:
■ In the preparation of water-proof clothes and papers, as alkyl groups are water repelling in nature.
■ To prepare grease, lubricants in aeroplanes, as their viscosities change little with temperature.
■ In paints and enamels, as they can withstand high temperature and chemical inertness
■ Used as insulating material for electrical motors and also as sealants.
■ In surgical and cosmetic plants as they are bio-compatible.
Silicates (Advance)
Silicates are metal derivatives of silicic acid. Granite, slates, bricks, cement, glass, feldspar, zeolites, mica, asbestos, and ceramics are examples of silicates. The Si–O bonds in silicates are very strong. Silicates do not dissolve in any of the common solvents. The silicates are mainly divided into six types, depending on the manner in which different SiO 4 4– units are linked together. Silicates are formed by heating metal oxide or metal carbonates with sand.
Na2CO3 +SiO2→ Na2SiO4 + CO2
A large number of silicate minerals exist in nature. The basic structural unit of silicates is SiO44–, in which silicon atom is sp3 hydridised and bonded to four oxygen atoms in tetrahedral fashion, as shown in Fig.8.12. In
silicates, either the discrete unit is present or a number of such units are joined together via corners by sharing 1, 2, 3, or 4 oxygen atoms per silicate unit. When silicate units are linked together, they form chain, ring, sheet, or three dimensional structures. Negative charge on silicate structure is neutralised by positively charged metal ions. If all the four corners are shared with other tetrahedral units, three -dimensional network silicates are formed.
Two important man-made silicates are glass and cement. Various types of silicates are compared in Table 8.10.
Fig. 8.12. (a) SiO44– ion tetrahedral structure (b) Representation of SiO44– unit
S.No. Types Units present Number of oxygen atoms shared
If aluminium ions (Al3+) replace a few silicon ions (Si 4+) in three-dimensional network of silicon dioxide, overall structure acquires a negative charge and is known as alumino silicate. Cations, such as Na + , K + ,or Ca 2+ , balance the negative charge. Examples are feldspar and zeolites.
In sheet silicates, one-fourth or one-half of silicon ions are replaced by Al3+ ions, resulting in zeolite. They trap Ca2+ ions from hard water and replace Na + ions, hence acting as ion exchanger. The structure of zeolite permits the formation of cavities of different sizes. Molecules like water, NH3, CO2, and ethanol are trapped in these cavities. Hence, zeolites act as molecular sieves.
Zeolites are widely used as a catalyst in petrochemical industries for cracking of hydrocarbons and isomerisation. A type of zeolite, ZSM–5, is used to convert alcohols directly into gasoline.
Hydrated zeolites are used as ion exchangers in softening of hard water.
TEST YOURSELF
1. Correct trend of electronegativity (on Pauling scale) in 14th group elements is
(1) C > Pb > Si > Ge > Sn
(2) C > Pb > Si = Ge = Sn
(3) C > Pb = Si = Ge > Sn (4) C > Si > Ge > Sn > Pb
2. Strongest oxidant among the following is
(1) C+4 (2) Pb+4
(3) Si+4 (4) Ge+4
3. The catenation tendency of C, Si, and Ge is in the order Ge < Si < C. The bond energies (in kJ mol–1) of C–C, Si–Si and Ge–Ge bonds are, respectively,
7. An example of network solid is (1) SiO2 (2) MgO (3) CaF2 (4) ZnS
8. Thermodynamically, most stable allotrope of carbon is (1) diamond (2) graphite (3) coal (4) coke
9. T he dioxides and monoxides of elements X and Y are amphoteric in nature. X and Y are, respectively, (1) C, Si (2) Si, Ge (3) Sn, Pb (4) Ge, Pb
10. CCl4 does not show hydrolysis but SiCl4 is readily hydrolysed because of (1) presence of vacant 3d orbitals in carbon (2) absence of vacant 3d orbitals in silicon (3) presence of vacant 4d orbitals in silicon (4) presence of vacant 3d orbitals in silicon
11. Correct melting point order for IVA group elements, is
(1) C > Si > Ge > Pb > Sn
(2) C > Si > Ge > Sn > Pb
(3) C > Ge > Sn > Si > Pb
(4) C > Si > Pb > Ge > Sn
12. The repeating unit of silicones is (1) RSiO (2) R2SiO (3) RSiO2 4) R2SiO2R2Si
13. The covalency of silicon and oxygen in SiO2 respectively, are (1) 2, 4 (2) 4, 4 (3) 4, 2 (4) 4, 6
14. In Buckminister fullerene, each carbon atom is (1) sp-hydridised (2) sp2-hydridised
CHAPTER REVIEW
■ Elements of group 13 are: B, Al, Ga, In and Tl. The general electronic configuration is ns2np1.
■ Aluminium is the most abundant metal and the 3rd most abundant element in earth crust.
■ Atomic radius increases suddenly from boron to aluminium due to greater screening effect of electrons of penultimate shell of Al.
■ The atomic radii of Al and Ga are same. This is due to poor shielding effect of d-electrons in Ga.
■ The decreasing order of electronegativity of elements of group 13 is B > Tl > In > Ga > Al.
■ Due to poor shielding effect of ‘d’ or ‘f’ orbitals, the electronegativity values of Tl, In, and Ga are more than that of Al.
■ The melting and boiling points of boron are very high, since boron exists as a giant covalent polymer in liquid and in solid states.
■ Gallium is called summer liquid.
■ Order of densities of group 13 elements is B < Al < Ga < In < Tl. Al has relatively low density due to its large atomic volume.
(3) sp3-hydridised (4) pure p-orbitals involved
15. Water gas is (1) CO+N2 (2) CO+H2 (3) CO2+N2 (4) CO2+H2
Answer Key
(1) 2 (2) 2 (3) 1 (4) 1 (5) 4 (6) 4 (7) 2 (8) 1
(9) 2 (10) 3 (11) 4 (12) 1 (13) 3 (14) 2 (15) 2
■ O rder of ionisation potential values of group 13 group is B > Tl > Ga > Al > In.
■ Order of SRP values of group 13 elements is Al < B < Ga < In < Tl.
■ The inability of ns 2 electrons in the participation of chemical bond is called inert pair effect.
■ Thallium exhibits stable +1 oxidation state in its compounds, due to inert pair effect.
■ Boron does not displace hydrogen from acids and never appears as cation. It always forms covalent compounds.
■ B and Al can react with N2 to form nitrides, which on hydrolysis, gives ammonia.
2223 33 2 BN BN NH HBO HO
2223 3 2 Al NAlN NH Al OH HO ()
■ Borax is re presented by the formula Na2B4O7.10H2O or Na2 [B4O5(OH)4].8H2O. Impure borax is called tincal.
■ Borax contains tetrahedral nuclear units [B4O5(OH)4]2–
■ The aqueous solution of borax is alkaline in nature. This is due to anionic hydrolysis.
■ Borax bead test is useful in the identification of basic radicals
■ Borax is used as a flux, in making optical pyrex glass, in the leather industry for cleaning hides and skin, and as a preservative
■ Boric acid is a weak monobasic acid.
■ Boric acid does not act as a protonic acid but behaves as a Lewis acid by accepting a pair of electrons from OH – ion.
■ H3BO3 loses water on heating. The reaction takes place depending on temperature, finally giving B2O3
■ Boric acid is used as an antiseptic, in enamel and glass industries.
■ Boric acid has a layer structure in which planar BO33– units are joined by hydrogen bonds.
■ Tetrahydridoborates of Li and Na are used as reducing agents.
■ Diborane is prepared, by the reduction of BCl 3 with LiAlH 4 or hydrogen in the presence of Cu–Al.
■ Diborane burns in oxygen to give B 2 O 3 and energy.
■ Boranes are better fuels than hydrocarbons, because the heat of combustion of diborane is very much higher.
■ Diborane diammonate, on heating at 200°C, gives borazole B 3N3H6.
■ Borazole or borazine is known as inorganic benzene.
■ Borazole is polar and is more reactive than benzene.
■ Diborane molecule contains only 12 bonding electrons and is an electron deficient molecule. It has no B–B bond.
■ Diborane has two coplanar BH 2 groups. There are 4 terminal and 2 bridge H atoms.
■ B–H–B bond or tau (or) banana (or) hydrogen bridge bond is formed by sp3–s–sp3 orbital overlapping.
■ Banana bond is 3 centered -2 electron bond. There are two B–H–B banana bonds in a diborane molecule.
■ Aluminium is used in thermite welding, in making foils, and as mordant in the form of alums.
■ Aluminium alloys are used in making utensils, and aeroplanes and in the construction and transport industry.
■ Al(OH)3 is a white gelatinous ppt.
■ AlF 3 is ionic and other halides are covalent.
■ MnSO 4 .Al 2 (SO 4 ) 3 .24H 2 O and FeSO 4 Al2(SO4)3. 24H2O are called pseudo alums. It is also prepared by heating alumina with chlorine in the presence of coke.
Al OC Cl AlCl CO C 23 2 100 3 33 23 0
It i s also formed when ‘Al’ metal (or) Al(OH)3 is dissolved in dilute HCl.
26 2332 Al HClAlClH
Al(OH)3+3HCl → AlCl3+3H2O
HCl gas is circulated through the solution to obtain crystals of hydrated AlCl 3
■ Anhydrous AlCl 3 is a white solid, deliquescent, and fumes in moist air. etc.
■ It is dimer in non-polar solvents and exists as [Al(H 2 O) 6 ]Cl 3 in water. It ionises in water due to high heat of hydration.
Elements of group 14 are C, Si, Ge, Sn, and Pb.
■ The general valence shell configuration of elements of group 14 is ns2 np2 and usual valency is 4.
■ The atomic radius difference between Sn and Pb is small. This is due to lanthanide contraction.
■ Melting point and boiling points decrease from carbon to lead.
■ High values of melting of C and Si is due to the tendency of these elements to form giant molecules.
■ Order of electronegativity of group 14 elements is C > Si = Ge = Sn < Pb.
■ Order of M–M bond dissociation energy is C–C >> Si–Si > Ge–Ge> Sn–Sn > Pb–Pb.
■ Carbon and silicon exhibit – 4 state in their compounds with metals.
■ The stability of +2 state increases down the group. This is due to inert pair effect. Pb2+ is more stable than Pb 4+ .
■ C, Si, and Ge are unaffected by water. Sn reacts with steam to give SnO 2 and H2.
■ Lead is not affected by water due to an oxide film on the surface.
■ Maximum covalency of C is 4, but Si is six.
■ CCl 4 does not undergo hydrolysis, due to the absence of vacant d orbitals in the valency shell.
■ SiCl 4 undergoes hydrolysis due to the presence of vacant 3d-orbitals in its valency shell.
■ SnCl2 has high melting point than SnCl4 because SnCl2 has more ionic nature. SnCl4 is more covalent due to high oxidation state of Sn4+ .
■ Availability of an element in two or more forms is termed allotropy.
■ Diamond has three-dimentional network structure.
■ Diamonds are used in making dies used for drawing thin wires and for cutting glass and drilling rocks.
■ Graphite is soft and is thermodynamically more stable than diamond.
■ Graphite has layer lattice structure.
■ Diamond is an insulator, but graphite is an electrical conductor.
■ Gra phite is used as a lubricant, in the manufacture of lead pencils, and in the manufacture of refractory crucibles.
■ Coal, coke, charcoal, and carbon black are the amorphous forms of carbon.
■ The amorphous forms of carbon are more reactive due to their increased surface area.
■ CO is an important component of fuel gases. It is a good ligand in the metallurgy of Ni by Mond’s process.
■ Producer gas is a mix of N2 and CO( 2 : 1).
■ Water gas is a mix of H 2 and CO .
■ Solid CO2 is dry ice or cardice.(refrigerant).
■ Carbon dioxide is used in making urea, inert atmosphere, and as a fire extinguisher.
■ Silica has two allotropic forms, namely crystalline and amorphous forms.
■ Crystalline forms of silica are quartz, tridymite, and crystoballite.
■ Amorphous forms of silica are agate, jaspar, and onyx.
■ Silica melts at 1600ºC to form quartz glass. It is used in making optical instruments.
■ Silica is soluble in hydrofluoric acid, forming hydrofluorosilicic acid, H 2SiF6.
■ Silica is acidic in nature.
■ Silica is a giant molecule with three dimensional structure. Hence, it exists as a solid. In SiO2, each Si atom is tetrahedrally surrounded by 4 oxygen atoms.
■ Silicones are organosilicon polymers containing R2SiO repeating units. These are polymeric and are formed by the hydrolysis of chlorosilanes.
■ Silicones have high thermal stability, high dielectric strength, and are resistant to oxidation and chemicals.
■ The water repellency of silicones is due to the presence of organic side chain groups.
■ Silicones are used to prepare grease, lubricants in aeroplanes, insulators in electrical motors and tools, and in paints and enamels.
■ Glass and cement are man-made silicates.
■ The basic structural unit in silicates is SiO4–4 tetrahedron.
■ In silicates, either the discrete unit is present or a number of such units are
joined together through corners and form a chain, ring, sheet, or three-dimensional structures.
■ Clay minerals are used for absorbing chemicals and sheet silicates are used for electrical insulation.
■ Zeolites are silicates in which some of Si4+ is replaced by Al3+, with the incorporation of other cations, like Na+, K+, and Ca2+, for maintaining charge balance.
Exercises
JEE MAIN LEVEL
Level - I
Introduction to p-block Elements
Single Option Correct MCQs
1. How many groups of p-block elements are present in the periodic table?
(1) 5 (2) 6
(3) 7 (4) 8
2. Which group in the p-block has the general electronic configuration ns 2np5?
(1) Group 13
(2) Group 15
(3) Group 16
(4) Group 17
3. What is the maximum covalence exhibited by second-period p-block elements?
(1) 3
(2) 4
(3) 5
(4) 6
4. What type of bonding is prominent in compounds formed between non-metals with small differences in electronegativity?
(1) Ionic bonding
(2) Covalent bonding
(3) Metallic bonding
(4) Hydrogen bonding
5. Which of the following is attributed to the presence of vacant d-orbitals in heavier p-block elements?
(1) Inert pair effect
(2) Ability to expand covalence
(3) Low ionization enthalpy
(4) High electronegativity
6. Whi ch element can form [EF 4 ] −, but not [EF6]3− due to the absence of d-orbitals?
(1) Aluminium
(2) Boron
(3) Silicon
(4) Phosphorus
Numerical Value Questions
7. The maximum covalency of Boron is
8. P-block elements starts from IUPAC group X, and ends with IUPAC group Y. The value of X+Y is
9. Sum of the maximum oxidation states of all groups of p-block elements will be ________
10. The number of p-block elements, which do not have availability of empty d orbitals ________?
Group 13 Elements: The Boron Family
Single Option Correct MCQs
11. Which of the following are correct statements
I) The abundance of boron in early earth crust is less than 0.01% by mass
II) Bauxite and cryolite are not most important minerals of aluminium
III) Kernite 2472 NaBO4HO
IV) Nihonium has symbol Nh atomic number
113
(1) I, II, IV
(2) II, IV
(3) II, III
(4) III, IV
12. Synthetically prepared radioactive element of Boron family is
(1) Bh
(3) Nb
(2) Nh
(4) Fl
13. T he order of abundance of IIIA group elements is
(1) Al > Ga > B
(2) B > Ga > Al
(3) B > Al > Ga
(4) Ga > Al > B
14. ‘Be’ and ‘Al’ exhibit many properties which are similar, but the two elements differ in
(1) Forming covalent halides
(2) Forming polymeric hydrides
(3) Exhibiting maximum covalency in compounds
(4) Exhibiting amphoteric nature in their oxides
15. The relative stability of +1 oxidation state of group 13 elements follows the order
(1) Al < Ga < Tl < In
(2) Tl < In < Ga < Al
(3) Al < Ga< In < Tl
(4) Ga < Al < In < Tl
16. The element with least first ionization enthalpy [IP1] is
(1) Al
(2) Ga
(3) Tl
(4) ln
17. 13th group element with max negative
3 0 M/M E + value is
(1) Tl
(2) B
(3) ln
(4) Al
18. Some statements are given below:
(A) The atomic radius of Ga is greater than Tl
(B) The melting point of Ga is greater than In
(C) First ionization enthalpy of Ga is greater than In
(D) The Electronegativity of Tl is greater than B
Among these incorrect statements are
(1) A, and B only
(2) A,B, and D only
(3) A, B, and C only
(4) A, B, C ,D
19. IIIA group element with highest density is (1) B
(2) Al (3) ln
(4) Tl
20. Which of the following statement(s) is/are incorrect?
1) Boron shows +1 and +3 oxidation states
2) Sn+2 is more stable than Sn +4
3) Tl+3 is more stable than Tl+1 due to inert pair effect
4) Pb+4 is a stronger oxidising agent than Pb+2
(1) 1,2 and 4 (2) 1,2 and 3 (3) 1,3 and 4 (4) 1,2,3 and 4
21. Which element reacts with acids as well as alkalis.
(1) Mg
(2) Na
(3) Al
(4) Ca
22. The main factor responsible for weak acidic nature of BF3 is
(1) large electro negativity of F
(2) three centered two electron bonds in BF 3
(3) pπ−pπ back bonding
(4) small size of B atom
23. 2Al (s)+2NaOH (aq)+6H 2O (l) → 2X (aq) +nH 2(g). Identify ‘X’ and ‘n’ in the balanced equation?
(1) x – [Al (OH)4]+ n –3
(2) x – [Al (OH)4]+3 n –3
(3) x – [Al (OH)4]–3 n –3
(4) x – [Al (OH)4]– n –3
24. AlCl 3 is an electron deficient compound, but AlF3 is not because.
(1) Atomic size of F is smaller than Cl which makes AlF3 more covalent
(2) AlCl3 is a covalent compound while AlF3 is an ionic compound
(3) Al in AlF 3 is sp2 hybridized while in AlCl3 it is sp3 hybridized
(4) AlCl3 exists as dimer, but AlF 3 is not
25. Which of the following does not undergo hydrolysis?
(1) BCl3
(2) BBr 3
(3) BF 3
(4) BI 3
26. In Al2Cl6, the covalency of aluminium is (1) 6
(2) 4
(3) 3
(4) 2
27. Which of the following sublimes on heating?
(1) Al2O3
(2) Al(OH)3
(3) (AlH3)n
(4) (AlCl3)n
28. Which one gives methane on hydrolysis
(1) Be2C
(2) Al4C3
(3) Both (a) and (b)
(4) AlN
29. Which among the following is the most basic oxide?
(1) Al2O3
(2) Ga2O3
(3) l2O3
(4) B2O3
30. Aluminium chloride exist as dimer Al2Cl6 in solid state as well as in solution of nonpolar solvent such as C6H6. When dissolved in water it gives
(1) Al2O3+ 6HCl
(2) [Al(H2O)6]3+ + 3Cl–
(3) [Al(OH)6]3–+3HCl
(4) Al3+ 3Cl–
31. In the reaction between boron and sodium hydroxide to liberate hydrogen gas, boron acts as
(1) an oxidizing agent
(2) a reducing agent
(3) a precipitating agent
(4) a deoxidizer
32. The hybridisation of ‘Al’ in [Al (H2O)6]3+ is
(1) sp3d2
(2) d2sp3
(3) dsp3
(4) sp3
33. Which metal forms a protective oxide layer to prevent corrosion?
(1) Au
(2) Cu
(3) Al
(4) Ag
Numerical Value Questions
34. How many Elements are more Electronegative, than Boron in Be, Al, Ga, In, Tl
35. Aluminium is the _______ most abundant element in nature
36. The stable oxidation state of thallium, a IIIA group element is _____
37. Number of f- electrons in the electronic Configuration of Thallium (Tl) are ______?
38. The number of amphoteric compounds among the following is _________
BeO, Al2O3, Al(OH)3, Ba(OH)2, Be(OH)2
39. The no. of moles of dihydrogen liberated when one mole of Aluminium dissolves in dilute HCl
Important trends and anomalous properties of B
Single Option Correct MCQs
40. Boron exhibits diagonal relationship with (1) Si (2) C
(3) Al (4) Be
41. Which of the following cannot liberate H 2 gas with acids?
(1) Tl (2) Ga
(3) Al (4) B
42. The correct statement regarding Boron is/ are
I) Boron has a maximum covalency of 4.
II) Boron does not form cations in aqueous solution.
III) The oxide of boron, B2O3 is acidic in nature.
IV) Boron is a metal.
(1) I, II, III, IV
(2) Only II, III, IV
(3) Only I, II, III
(4) I, III, IV
43. Boron does not form B³+ ions whereas Al forms Al+3 ions. This is because
(1) The size of B atom is smaller than that of Al
(2) The sum of IE1 + IE2 + IE3, B is much higher than that of Al
(3) The sum of IE1 + IE2 + IE 3 of Al is much higher than that of B
(4) Both (a) and (b)
44. Which of the following complex ions does not exist?
(1) ( ) 3 2 6 BHO +
(2) ( ) 3 2 6 AlHO +
(3) ( ) 3 2 6 GaHO +
(4) ( ) 3 2 6 InHO +
45. Which one is a non-metal in group 13 (III A)?
(1) B (2) Al
(3) Ga (4) ln
Numerical Value Questions
46. Boron has 2 isotopes, B-x, B-y. x, y is mass numbers. x+y is ______
47. The boron-x (xB) isotope has high ability to absorb neutrons. What is x?
Some Important Compounds of Boron
Single Option Correct MCQs
48. Which of the following statement is incorrect regarding the structure of Borax?
(1) No. of B-B bonds are zero
(2) Hybridisation of each B- atom: sp 2
(3) No. of B-O-B bonds:5
(4) Borax contains two different types of B-O bonds
49. The correct statement(s) for orthoboric acid is/are?
(1) It behaves as Lewis acid by accepting a lone pair of electrons from OH – ion in water
(2) H3BO3 on heating at 100°C gives HBO 2
(3) H3BO3 on heating at 160°C gives H2B4O7
(4) All are correct.
50. The blue colour produced in the borax bead test of a salt due to
(1) Mn(BO2)2
(2) Co(BO2)2 (3) Ni(BO2)2 (4) Cr(BO2)2
51. The structure of diborane (B 2H6) contains (1) four 2c-2e bonds are two 3c-2e bonds
(2) two 2c-2e bonds and four 3c-2e bonds
(3) two 2c-2e bonds and two 3c-2e bonds
(4) four 2c-2e bonds and four 3c-2e bonds
52. Borax is converted into crystalline boron by the following steps: XY 3323 BoraxHBOBOB ∆ ∆ →→→
X and Y respectively.
(1) HCl, Mg
(2) HCl, C
(3) C, Al
(4) HCl, Al
53. Borax bead test is used to identify the
(1) Anion in coloured salt
(2) Cation in coloured salt
(3) Anion in white salt
(4) Cation in white salt
54. Consider following statements about borax:
a) Each boron atom has four B – O bonds
b) Each boron atom has three B – O bonds
c) Two boron atoms have four B – O bonds while other two have three B – O bonds
d) Each boron atom has one – OH group
Select correct statement (s):
(1) a, b
(2) b, c
(3) c, d
(4) a, c
55. 2472 NaBO10HOXYZ ∆∆ ⋅→→+
X, Y and Z in the reaction are
(1) X = Na2B4O7, Y = B2O3, Z = H3BO3
(2) X = Na2B4O7, Y = NaBO2, Z = B2O3
(3) X = B2O3, Y = NaBO2, Z = B(OH)3
(4) X = NaBO2, Y = B2O3, Z = B(OH)3
56. What is the nature of aqueous borax solution?
(1) Neutral
(2) Acidic
(3) Alkaline
(4) Amphoteric
57. During the borax bead test with CuSO 4 , blue green colour of the bead was observed oxidizing flame due to the formation of
(1) Cu3B2
(2) Cu
(3) Cu(BO2)2
(4) CuO
58. Which of the following molecule can cleave diborane unsymmetrically?
(1) NH3 (2) CO
(3) N (4) R3N
59. B orax is made of two tetrahedra and two triangular units joined together and should be written as: Na2B4O5(OH)4 8H2O Consider the following statements about borax:
(P) Each boron atom has four B−O bonds. (Q) Each boron atom has three B−O bonds?
(R) Two boron atoms have four B−O bonds while other two have three B−O bonds.
(S) Each boron atom has one -OH groups.
Select correct statement(s):
(1) P and Q
(2) Q and R (3) R and S
(4) P and R
Numerical Value Questions
60. In borazole, ‘x’ be the number of sigma bonds and ‘y’ be the number of Pi bonds. Then (x+y)2 will be?
61. Basicity of ortho boric acid is?
62. In Borax, ‘x’ be the number of Sp 3hybridized B – atoms and ‘y’ be the number of Sp2 - hybridized B-atoms the ( ) 3 xy 2 +× will be _____
63. Borax is Na x B y O z10H2O then x+y+z =?
Uses of Boron and Aluminium and their compounds
Single Option Correct MCQs
64. Which is used as disinfectant?
(1) Boric acid
(2) Sulphuric acid
(3) Phosphorus acid
(4) Phosphoric acid
65. Which compound can make fireproof clothes?
(1) Aluminium sulphate
(2) Ferrous sulphate
(3) Magnesium sulphate
(4) Cuprous sulphate
66. Boron carbide, B4C is widely used for (1) making acetylene (2) making plaster of Paris (3) as hardest substance after diamond (4) making boric acid
67. In metallurgy, the substance which can act as de-oxidizer is
(1) B (2) Al2O3
(3) AlN (4) Al
68. Aluminium powder is used in (1) The extraction of gold
(2) Calico-printing
(3) Sizing paper
(4) In flash bulbs
Numerical Value Questions
69. In K2SO4 ⋅Al2(SO4)3 ⋅24H2O (Potash alum), each metal ion is surrounded by _______ number of H2O molecules
Group 14 Elements: The Carbon Family
Single Option Correct MCQs
70. Lowest density of 14th group element is (1) Graphite (2) Diamond
(3) Silicon
(4) Germanium
71. The element that shows greater ability to form Pπ−Pπ multiple bonds is
(1) C (2) Ge
(3) Sn (4) Si
72. The correct order of catenation is
(1) C > Sn > Si = Ge
(2) C > Si > Ge = Sn
(3) Si > Sn > C > Ge
(4) Ge > Sn > Si > C
73. Which one of the following compounds of group-14 elements is not known?
(3) Carbon atoms have tendency to link with one another through covalent bonds to form Chains and rings
(4) Heavier elements of Si, Ge to form pπ−pπ bonds
76. The order of catenation power is
(1) C > Si > Ge > Sn
(2) Si > C > Ge > Sn
(3) Sn > Ge > Si > C
(4) Ge > Sn > C > Si
77. Correct trend of electro negativity (on Pauling scale) in 14th group elements is
(1) C > Pb > Si > Ge > Sn
(2) C > Pb > Si = Ge = Sn
(3) C > Pb > Si = Ge > Sn
(4) C > Si > Ge > Sn > Pb
78. The relative stability of +2 oxidation state of G-14 elements follows the order (1) Pb < Sn < Ge < Si
(2) Ge < Si < Sn < Pb
(3) Si < Ge < Sn < Pb
(4) Sn < Ge < Si < Pb
79. Among the following, which of the following elements is a metalloid?
(1) C (2) S (3) Ge (4) Pb
80. The 14th group element with least ionization potential is (1) Ge
(2) Pb
(3) Si
(4) Sn
81. The following bond has highest bond energy?
(1) Si - Si
(2) C - C
(3) Sn - Sn (4) b - Pb
82. Silicon generally has high (i) thermal stability (ii) dielectric strength (iii) resistance to chemicals (1) (i) and (ii) (2) (i) and (iii) (3) (ii) and (iii) (4) (i) (ii) and (iii)
83. On strong heating lead nitrate gives (1) PbO, NO, O2 (2) PbO, NO, NO2 (3) PbO2, PbO, NO2 (4) PbO, NO2, O2
84. Which one of the following does not exist? (1) PbF4
(2) PbI4
(3) SnF4
(4) SiCl4
85. The dioxides and monoxides of elements X and Y are amphoteric in nature. X and Y are, respectively
(1) C, Si (2) Si, Ge (3) Sn, Pb (4) Ge, Pb
86. Hydrolysis of SiCl4 gives (1) Si(OH)2 and HCl (2) Si(OH)4 and HCl
(3) H2[SiCl6] and HCl (4) H2[SiCl4] and HCl
87. CCl4 does not show hydrolysis but SiCl4 is readily hydrolysed because (1) of carbon is higher than that of silicon
(2) electronegativity of carbon is higher than that of silicon
(3) carbon cannot expand its octet but silicon can expand
(4) carbon forms double and triple bonds but not silicon
88. Which of the following does not exists?
(1) PbF4 (2) SnF4
(3) CCl4 (4) PbI4
89. The correct order of melting point is
(1) C > Si > Ge > Sn > Pb
(2) C < Si < Ge < Sn < Pb
(3) C > Si > Ge > Pb > Sn
(4) C > Si > Pb > Sn > Ge
90. Evaluate the following statements related to group 14 elements
(A) Covalent radius decreases down the group from C to Pb in a regular manner.
(B) Electro negativity decreases from C to Pb down the group gradually.
(C) Maximum covalence of C is 4 whereas other elements can expand their covalence due to presence of d orbitals.
(D) Heavier elements do not form pπ- pπ bonds.
(E) Carbon can exhibit negative oxidation states.
Choose the incorrect answer from the options given below:
(1) (C), (D) and (E) Only
(2) (A) and (B) Only
(3) (A), (B) and (C) Only
(4) (C) and (D) Only
Numerical Value Questions
91. Number of penultimate electrons in silicon?
92. Atomic number of elements Flerovium (Fl) is,
93. The number of ions from the following that are expected to behave as oxidizing agent is ______Sn4+,Sn+2,Pb+2,Tl3+,Pb+4,Tl+
94. How many of the following have density more than that of water? Carbon, Silicon, Germanium, Tin, lead
Important Trends and Anomalous Behaviour of Carbon
Single Option Correct MCQs
95. The element that shows greater ability to form pπ-pπ multiple bounds is
(1) C (2) Ge
(3) Sn (4) Si
96. Which of the following cannot form complex compounds?
(1) C (2) Si
(3) Ge (4) Al
97. Carbon forms many compounds due to:
(1) Tetravalency
(2) Variable valency
(3) Large chemical affinity
(4) property of catenation
98. Maximum covalency exhibited by Carbon and Silicon respectively are
(1) 4, 6
(2) 4, 4
(3) 6, 6
(4) 4, 8
Numerical Value Questions
99. The number of pure atomic orbitals at each carbon atom present in graphite is _____
100. Graphite has layered structure, Layers are held by van der Waals forces and distance between two layers (in A °) is,
101. The number of d orbitals available for bonding in the case of carbon is _____
Allotropes of Carbon
Single Option Correct MCQs
102. The correct statement about allotropes of carbon is / are
i) Diamond has 3D polymeric structure.
ii) Graphite is thermally stable.
iii) C60 molecule contains 12 five membered and 20 six membered rings.
iv) In Diamond C – C bond lengths is 1.54 A0, and in graphite C – C bond length is 1.34 A0.
(1) All
(2) Only ii and iii
(3) i, ii and iii
(4) Only i and iii
103. The hybridisation of carbon atoms in Diamond, Graphite and fullerenes are
(1) sp2,sp3,sp2
(2) sp2, sp2, sp3
(3) sp3, sp2, sp2
(4) sp2, sp2, sp2
104. The following are some statements about graphite
I) C-C bond length is 1.42 A°
II) distance between two layers is 3.35 A°
III) bond angle is 60°
The correct combination is
(1) all are correct
(2) only I and II are correct
(3) only II is correct
(4) all are incorrect
105. Thermodynamically most stable allotrope of carbon is (1) Diamond (2) Graphite (3) Coal (4) Coke
106. Graphite is similar to (1) B
(2) B4C
(3) B2H6 (4) (BN)n
107. Which of the following has similar structure to graphite?
(1) BN
(2) N
(3) B4C (4) B2H6
Numerical Value Questions
108. C60 Fullerene is called Buckminster fullerene that contains six-member as well as fivemember rings. If a five-member ring is fused with X six- member And Y five-member rings, what is the value of X+Y?
109. The number of hybrid orbitals used by each carbon atom in graphite and diamond are x and y respectively. The value of x+y =
110. Covalency of carbon in Diamond is _____
Some Important Compounds of Carbon and Silicon
Single Option Correct MCQs
111. CO2 is a gas but SiO2 is solid because (1) SiO2 has continuous tetrahedral structure while CO2 is discrete covalent molecule (2) SiO2 is heavier than CO2 (3) SiO2 is less acidic than CO 2 (4) Melting point of SiO2 is very high
112. The covalency of Silicon and oxygen in SiO2 respectively (1) 2, 4 (2) 4, 4
(3) 4, 2 (4) 4, 6
113. Which is the crystalline form of silica?
(1) Agate (2) Jaspar (3) Onyx (4) Cristobalite
114. Dry ice is:
(1) Solid NH3 (2) Solid SO2
(3) Solid N2 (4) Solid CO2
115. The element which forms neutral as well as acidic oxide is ____
(1) Si (2) Sn
(3) C (4) Pb
116. The products of the following reaction are:
2 SiOC ∆ +→
(1) SiC and CO2
(2) SiO and CO
(3) SiC and CO
(4) SiC and CO2
117. Which of the following is a pyro silicate.
(1) Sc2Si2O7
(2) Zn2SiO4
(3) Ca2 Si3O9 (4) Be3Al2Si6O18
118. The compound which has three-dimensional structure
(1) graphite
(2) SiO2
(3) CO2
(4) Sheet – silicate
119.
0 2 HO 1000C 4 SiClXY →→
X, Y respectively are (1) SiO2 and Si (2) H4SiO4 and SiO2
(3) H2SiCl4 and SiO2 (4) H4SiO4 and Si
120. The repeating unit of silicones
(1) RSiO
(2) R2SiO
(3) RSiO2
(4) R2SiO2R2Si
121. The basic structural unit of feldspar, zeolites, mica, and asbestos is:
(1) ( )2 SiO3
(2) SiO2
(3) ( ) 4 4 SiO
(4)
122. Exhausted permutit it does not contain ______ ion
(1) Na+
(2) Mg+2
(3) Al+3
(4) Si+4
123. The incorrect statement about the silicones is
(1) They are thermally unstable because of Si – C bond
(2) They are insoluble in water
(3) They are organosilicon polymer
(4) They have stable silica – like skeleton
(–Si – O – Si – O – Si –)
124. Zeolite is a silicate of two elements X and Y. What are X and Y?
(1) Na,Ca
(2) Mg,Al
(3) Na,Al
(4) Mg,Zn
125. Which of the following pairs of ions represent in cyclic and double chain silicates?
(1) 2 SiO27 and ( )2n 3 3 SiO
(2) 6 SiO39 and ( )6n 411 n SiO
(3) SiO27 and ( )2n 25 n SiO
(4) 7 SiO27 and ( )2n 3 3 SiO
Numerical Value Questions
126. Total number of pi (π) bonds in CO2, C3O2 molecules _____________
127. Bond order in carbon monoxide is?
128. Covalency of Si in hydrofluoric silicic acid is?
129. Oxidation state of silicon in silicic acid is
130. The oxidation state of ‘C’ in CO is x and the oxidation state of ‘C’ in COCl2 is y. Then y-x is
131. How many of the following are amphoteric oxides
CO,GeO,SnO,PbO,PbO2,SiO2,GeO2,As2O3,Sb2O3
Level - II
Group-13
Single Option Correct MCQs
1. Aluminium exhibits diagonal relationship with (1) C (2) Si (3) Be (4) Ge
2. The relative stability of +3 oxidation state of group 13 elements follows the order (1) Al<Ga<Tl<In (2) Tl<In<Ga<Al (3) Al<Ga<In<Tl (4) In<Ga<Tl<Al
3. Choose the correct stability order of group 13 elements in their +1 oxidation state (1) Al<Ga<In<Tl (2) Tl<In<Ga<Al (3) Al<Ga<Tl<ln (4) Al<Tl<Ga<ln
4. The metal that has very low melting point and its periodic position is closer to a metalloid is
(1) Al (2) Ga
(3) Se (4) In
5. Which of the following is an electron deficient molecule?
(1) LiH
(2) B2H6
(3) LiBH 4
(4) B3N3H6
6. The Lewis acid character of boron tri halides follows the order:
(1) BCl3>BF3>BBr3>BI3
(2) BI 3 >BBr3>BCl3>BF3
(3) BBr 3 >BI3>BCl3>BF3
(4) BF3>BCl3>BBr3>BI3
7. Boric acid is an acid because its molecule
(1) contains replaceble H+ ion
(2) gives up proton
(3) accepts OH– from water releasing proton
(4) combines with proton from water molecule
Numerical Value Questions
8. Aluminium is the _______ most abundant element in nature.
9. Maximum covalency of boron________
Group-14
Single Option Correct MCQs
10. Select the element that belongs to 14th group
(1) Fl (2) Nh
(3) Lv (4) Rf
11. Two stable isotopes of carbon present in naturally occurring carbon is
(1) 12C,14C
(2) 13C,14C
(3) 12C,13C
(4) 11C,13C
12. Even though carbon and silicon are non -me tals, they have higher melting points than others because
(1) they exist as 3 dimensional covalent solids
(2) the bonds in their molecules are strong
(3) they exhibit multiple bonding
(4) they are highly electronegative
13. The general trend in the properties of elements of carbon family shows that, with the rise in atomic number,
(1) the tendency towards catenation increases
(2) the tendency to show +2 oxidation state increases
(3) the metallic character decreases
(4) the tendency to form complexes with covalency higher than four decreases
14. Which of the following can decompose steam into hydrogen and gives dioxide?
(1) Sn (2) C
(3) Si (4) Ge
15. Reason for the highest catenation of carbon is
(1) C is more electronegative
(2) C has higher ionisation potential value
(3) C has only one stable isotope
(4) C–C bond is strong
16. Which of the following structure is similar to graphite?
(1) B (2) B4C
(3) B2H6 (4) BN
Numerical Value Questions
17. The covalency of carbon in diamond is
18. The number of penultimate electrons in silicon is______
19. The total number of electrons that a molecule of carbondioxide has ______
Multiple Concept Questions
Single Option Correct MCQs
20. Ionisation enthalpy for the elements of group-13 follows the order:
(1) B > Al > Ga > In > Tl
(2) B < Al < Ga < In < Tl
(3) B < Al > Ga < In > Tl
(4) B > Al < Ga > In < Tl
21. Stability of monovalent and trivalent cations of Ga, In, Tl lie in the following sequence.
i) Ga3+<In3+>Tl3+
ii) Ga3+>In3+>Tl3+
iii) TlInGa ⊕⊕⊕ >>
iv) GaInTl ⊕⊕⊕ >>
(1) i,ii (2) ii,iii
(3) i,iv (4) iii,iv
22. Group-13 elements react with O 2 to form oxides of type M 2O 3(M=element). Which among the following is the least basic oxide?
(1) Ga2O3
(2) Al2O3
(3) B2O3
(4) Tl2O3
23. Which one of the following does not exist?
(1) BH 3
(2) H2F2
(3) SbH3
(4) N2H4
24. Which of the following is used as black pigment in black ink?
(1) Coke
(2) Carbon black
(3) Germanium
(4) Graphite
25. Which of the following does not represent the correct resonance structure of carbon monoxide?
(1) : C : : : O :
(2) : C ≡ O :
(3) : C ← O:
(4) :C → O :
26. An acidic flux among the following is
(1) CaO
(2) MgO
(3) SiO
(4) CaH2
Numerical Value Questions
27. The total number of 2–centre–2–electron and 3–centre–2–electron bonds in B2H6is-----------.
28. The number of acidic oxides in 13 th group is_____.
Level -III
1. Which of the following is the correct statement?
(1) Be exhibits coordination number of 6.
(2) Chloride of both Be and Al have bridged structure in solid phase.
(3) Boric acid is a protonic acid.
(4) B 2 H 6 .2NH 3 is known as ‘inorganic benzene’.
2. Aluminum chloride exist as dimer, Al2Cl6 in solid phase as well as in solution of non-polar solvents such as benzene. When dissolved in water it gives
(1) [Al(OH)6]–3+ 3HCl
(2) [Al(H2O)6]+3+ 3Cl–
(3) Al3+ 3Cl–
(4) Al2O3+ 6HCl
3. Maximum covalency of boron and aluminum are a and b respectively. Then 2 ab + is
4. Heating orthoboric acid above A K forms metaboric acid, what is the value of A?
5. The reaction that gives CO2 as one of the products is
(1) 23 FeO3C ∆ +→
(2) 3C4HNO3 ∆ +→
(3) 2 SnO2C+→
(4) 6NaOH2C+→
6. Silica is insoluble in (1) HF (2) NaOH (3) KOH (4) HNO3
7. A piece of graphite has 10 layers and each layer consisting of 40 carbons. If it is a good conductor of electricity, then the number of unpaired electrons present in that piece is
8. Which of the following is used in filtration plants?
(1) quartz
(2) kieselghur
(3) sylica gel
(4) Acetylene
9. Thallous chloride is more stable than thallic chloride because of
(1) more ionic character
(2) larger size of Tl+ ion
THEORY BASED QUESTIONS
Statement Type Questions
Each question has two statements: statement I (S-I) and statement II (S-II). In light of the given statements, choose the most appropriate answer from the options given below.
(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.
(3) high hydration energy of Tl + ion
(4) inert pair effect
10. SiO 2 is a solid ,while CO 2 is a gas. This is because
(1) SiO 2 contains weak van der Waals attraction, while CO2 contains strong covalent bonds.
(2) Solid SiO2 has a three dimensional network structure, whereas CO2 contains discrete molecules.
(3) Both contain strong covalent bonds
(4) Both contain weak van der Waals attractions
11. 13 th group elements show maximum + x oxidation state and +y oxidation state (with inert pair effect) then x y is------.
12. B2H6+NH3→X (ionic compound) heated 2 Y H →+↑ . Find the total number of sp2 hybridised atoms per Y molecule.
13. The covalency of silicon and oxygen respectively in SiO2 is x and y. The product of x and y is _________
14. The number of five-membered carbon rings in C60 fullerence is ____.
(4) if statement I is incorrect, but statement II is correct.
1. S-I : Organic substance that has –OH groups on adjacent carbon atoms in C is configuration enhances the acidic properties of H3BO3.
S-II : Such type of compound shift the equilibrium ( ) -+ 332 4 HBO+HO[BOH]+H in forward direction.
2. S-I : H3BO3 is used as antiseptic
S-II : In B2H6, each boron is sp2 hybridized
3. S-I : Boric acid (H3BO3) and fluroboric acid (HBF4) both are mono basic acid in water.
S-II : Both the acids are OH – acceptors rather than proton donors.
4. S-I : Increase in the atomic radius from B to Al is more than that of consecutive elements of the same group.
S-II : Electrons in penultimate shell of aluminium have greater screening effect.
5. S-I : Graphite is used as dry lubricant, as it possesses slippery nature.
S-II : Diamond is thermodynamically more stable than graphite.
6. S-I : Beryllium halide is IIA group metallic dihalide, which is covalent in nature.
S-II : Beryllium chloride has dimer state in solid state but it exhibit chain structure in vapour 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). 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).
JEE ADVANCED LEVEL
Multiple Option Correct MCQs
1. Compounds that readily undergo hydrolysis are
(1) CCl4
(2) BCl3
(3) SiCl4
(4) CF4
(3) if (A) is true but (R) is false.
(4) if both (A) and are (R) false.
7. (A) : Compounds of 13th group are act as Lewis acids.
(R) : Molecules with incomplete octet are Lewis acids.
8. (A) : Even though diamond is covalent, it has a high melting point.
(R) : Diamond is a three dimensional gaint molecule. The C–C bonds in it are very strong.
9. (A) : Gallium is used in high temperature thermometer.
(R) : There is large difference between melting point and boiling point of gallium
10. (A) : Graphite is used as a lubricant.
(R) : Graphite has a layer like structure and the attractive forces between layers are weak.
11. (A) : GeF4 and SiCl4 act as Lewis bases.
(R) : Ge and Si have d-orbitals to accept electrons.
12. (A) : Al liberates hydrogen gas with both NaOH and HCl.
(R) : Al is amphoteric metal.
13. (A) : Beryllium has less negative value of reduction potential compared to the other alkaline earth metals.
(R) : Beryllium has large hydration energy due to small size of Be2+ but relatively large value of atomisation enthalpy.
2. When metal carbides react with H 2O, the correct equations are
(1) Al4C3+H2O→CH ≡ CH
(2) CaC2+H2H→CH ≡ CH
(3) Mg2C3 +H2O→H3C ≡ CH
(4) Be2C+H2O→CH4
3. Which of the following is true about ‘CO’?
(1) CO forms a volatile compound with nickel.
(2) CO and Cl 2 forms phosgene gas in presence of sunlight.
(3) CO is absorbed by ammonical solution of Cu2Cl2.
(4) Producer gas is a mixture of CO + N 2.
4. Among the following, methanides are (1) Al4C3 (2) Be2C (3) Mg2C3 (4) Both 1 and 3
5. Which of the following are basic?
(1) B2O3 (2) Tl2O3
(3) In2O3 (4) Al2O3
6. Which of the following does not exhibit inert pair effect?
(1) B (2) Al
(3) Tl (4) Sc
7. Select the correct statements about diborane.
(1) B2H6 has three centre two electron bond.
(2) Each boron atom lies in sp3 hybrid state.
(3) H t ....B.....H t bond angle is 122°.
(4) All hydrogens in B2H6 lie in the same plane.
Numerical Value Questions
8. TK 3 26 2BF+ 6NaH BH+6NaF → the value of T is
9. In vapour phase of AlCl 3 , Al atom is surrounded by _______ number of Clatoms.
10. How many of the following statements are correct?
(I) ‘B’ has a smaller first ionisation enthalpy than Be.
(II) It is easier to remove 2p electron than 2s electron.
(III) 2p electron of ‘B’ is more shielded from the nucleus by the inner core of electrons than 2s electrons of Be.
(IV) 2s electron has more penetration power than 2p electron.
(V) Atomic radius of B is more than Be.
11. The distance between two layers in graphite is x A° and C–C bond length in graphite is y A°, then x–y is _____A°(round off )
12. The number of moles of dihydrogen liberated when one mole of aluminium dissolves in dilute HCl
13. In the structure of anionic part of beryl, if the number of oxygen atoms shared per tetrahedral unit = x, number of tetrahedral units involved in the cyclic structure = y , magnitude of charge on the anionic part = z, then x+y+z =-------------.
14. The value of ‘n’ in the molecule formula Be n Al2Si6O18 is ____.
15. 3 mole of B2H6 are completely reacted with methanol. The number of mole of boron containing product formed is __?
Integer Value Questions
16. How many of the following are basic oxides CO, CO2, SiO2, Al2O3, PbO, In2O3, Tl2O, SnO, SnO2
∆ →+ then the value of (x–y+z) is
18. How many properties are common in diamond and graphite?
(1) Electrical conductivity
(2) Relative atomic weight
(3) Crystal structure
(4) Density
(5) Allotropes of carbon
19. How many of the following are non-metals among B, Al, Ga, In, Tl, C, Si, Ge, Sn, Pb.?
20. Diamond is formed by fusion of several carbon tetrahedrons, in which carbon atom form homocyclic ring. The number of carbon atoms in each ring is ___.
21. Basicity of orthoboric acid is___.
Passage-based Questions
Passage I: A large number of silicate minerals exist in nature. The basic structural unit of silicate is 4 4 SiO in which silicon atom is bonded to four oxygen atoms in tetrahedron fashion. In silicates either the discrete unit is present (or) a number of such units are joined together via corners by sharing 1, 2, 3 or 4 oxygen atoms per silicate units
22. Number of oxygen atoms sharing per silicate unit in sheet silicates is
23. Total negative charge present on trimer of cyclic silicate is (give only magnitude)
Passage II: Anhydrous aluminium chloride exist as a dimer at below 625 K and as a monomer at above 1025 K. In between temperatures , it is an equilibrium mixture of AlCl3 and Al2Cl6. The structure of dimer is
24. The correct order of bond angles in Al 2Cl6 is
(1) x>y>z
(2) z>x>y
(3) y>x>z
(4) z>y>x
25. Pi ck out correct statement(s) from the following:
I. On methylation Al2Cl6 forms (CH3)6Al2 and B2H6 form (CH3)4B2H2.
II. On methylation both Al2Cl6_and B2H6 forms hexamethyl derivatives((CH2)6M2).
III. In Al2Cl6, Al–Cl–Al bridge is 3c–4e bond, while in B2H6, B–H–B bridge is 3c–2e bond.
IV. Both Al2Cl6, and B2H6 act as Lewis acids.
V. In gaseous state, both compounds exist as a mixture of monomer and dimer.
(1) II, III, IV
(2) I, III, IV
(3) I, III, IV, V
(4) II, III, IV, V
Passage III: The heavier members of 13 and 14 groups besides the group oxidation state also show another oxidation state that is two unit less than the group oxidation state. Down the group decreasing, the stability state of higher oxidation state and that of lower oxidation state increases. The concept which is commonly called inert pair effect has been used to explain many physical and chemical properties of the element of these groups.
26. Which among the following is the strongest reducing agent?
(1) GaCl
(2) InCl
(3) BCl3
(4) TlCl
27. Which among the following is the strongest oxidising agent?
(b) Single chain silicate (q) SiO bonds are 50% ionic and 50% covalent
(c) Pyro silicates (r) General formula is ( )2n3 n SiO
(d) Sheet silicates (s) Two oxygen atoms per (two dimensional) tetrahedron are shared
(a) (b) (c) (d)
(1) pqrs pqrs pq pq
(2) pq rs pq rs
(3) rs p qs rs
(4) pqrs rqs pq pq
BRAIN TEASERS
1. Match Column-I with Column-II.
Column–I Column–II
(a) Graphite (p) Layered structure
(b) Boric acid (q) Delocalisation of electrons
(c) Borazole (r) Electrical conductor
(d) Boron nitride (s) Hydrogen bonds
(a) (b) (c) (d)
(1) pqr ps q pq
(2) pr q s p
(3) pq r q s
(4) ps p q r
2. The solubility of anhydrous AlCl 3 and hydrated AlCl 3 in diethyl ether are S 1 and S2, respectively. Then,
(1) S1=S2
(2) S1>S2
(3) S1<S2
(4) S1<S2 but not S1=S2
Passage I : Boric acid B(OH) 3 is weak monobasic acid reacts with alkali to form borates. The most common borate of boric acid is borax represented as Na2(B4O5(OH)4. 8H2O
34. Match Column-I with Column-II.
Column–I Column–II
(a) Graphite (i) sp3
(b) Fullerene (ii) sp
(c) Ethyne (iii) sp2
(d) Diamond (iv) both sp2 and sp2
(a) (b) (c) (d)
(1) ii iii ii i
(2) iii iv ii i
(3) ii iv ii i
(4) i iv iv i
that is made up of two tetrahedral and two triangular units. On dissolution in water, these tetrahedral and triangular units are separated. Borax is useful primary standard for titration against acids.
3. Oxidation state of boron atom in borax is/ are
(1) +3 only
(2) three atoms +3 and one atom +2
(3) +2 only
(4) two atoms +3 and two atoms +4
4. The number of B–O–B linkage in borax is/ are
(1) 2 (2) 5
(3) 4 (4) 6
5. 0 f H∆ values of graphite, diamond and fullerene are respectively (in kJ mol –1)
(1) 1.9, 38.1 and 0
(2) 0, 1.9 and 38.1
(3) 0, 38.1 and 1.9
(4) 1.9, 0 and 38.1
6. Correct statements among a to d regarding silicones are:
(a) They are polymers with hydrophobic character.
(b) They are biocompatible.
(c) In general, they have high thermal stability and low dielectric strength.
(d) Usually, they are resistant to oxidation and used as greases.
(1) (a), (b), and (c) only (2) (a), (b), (c), and (d)
(3) (a) and (b) only (4) (a), (b), and (d) only
7. Given below are two statements. One is labelled as Assertion (A) and the other is labelled as reason (R).
Assertion (A) : Metal borides are used in nuclear industry as protective shields and controlling rods.
FLASH BACK (Previous JEE Questions)
JEE Main
1. Sum of valencies of nitrogen & boron in borazole is ________. (2023)
2. Consider the elements Mg, Al, S, P and Si, the correct increasing order of their first ionization enthalpy is (2021)
(1) Mg < Al < Si < S < P
(2) Mg < Al < Si < P < S
(3) Al < Mg < S < Si < P
(4) Al < Mg < Si < S < P
3. If the formula of borax is Na 2B 4O x (OH) y .zH2O x + y + z = _____.
4. Among the following, the number of halide(s) that is /are inert to hydrolysis is ___ (2023)
(A) BF 3 (B) SiCl4 (C) PCl5 (D) SF6
JEE Advanced
5. Given below are two statements, one is labeled as Assertion A and the other is labelled as Reason R (2021)
Reason (R) : 10B isotope has ability to absorb neutrons.
In the light of the given statements, choose the correct answer from the options given.
(1) Both (A) and (R) are true and (R) is the correct explanation of (A).
(2) Both (A) and (R) are true but (R) is not the correct explanation of (A).
(3) (A) is true but (R) is false.
(4) Both (A) and (R) are false.
8. The total number of σ & π bonds are present in carbon suboxide is ________.
9. The low atomic radius of Ga than Al can be explained by the (1) presence of d-electrons
(2) amphoteric property of Ga
(3) higher Zeff of Ga than Al
(4) high ionization energy of Ga
Assertion (A) : Carbon forms two important oxides CO and CO2, CO is neutral whereas CO2 is acidic in nature.
Reason (R) : CO2 can combine with water in a limited way to form carbonic acid, while CO is sparingly soluble in water.
(1) Both A and R are correct and R is the correct explanation of A.
(2) A is not correct but R is correct.
(3) A is correct but R is not correct.
(4) Both A and R are correct but R is NOT the correct explanation of A.
6. Aluminium is usually found in +3 oxidation state. In contrast, thallium exists in +1 and +3 oxidation states. This is due to.
(1) lanthanoid contraction
(2) diagonal relationship
(3) lattice effect
(4) inert pair effect
7. Formula of kernite is Na2BxO7.4H2O. What is the value of x? (2019)
CHAPTER TEST - JEE MAIN
Section-A
1. Which of the following elements have positive standard reduction potential for M+3/M transformation?
(1) Al (2) Tl
(3) Ga (4) In
2. Which element reacts with acids as well as alkalies?
(1) Mg (2) Si
(3) Al (4) Cu
3. Carbon cannot expand its valency beyond 4, because (1) it has only 4 electrons. (2) it has only 4 shells.
(3) it lacks valence p-orbitals.
(4) it lacks valence d-orbitals.
4. Maximum covalency exhibited by carbon and silicon respectively are
(1) 4, 6 (2) 4, 4
(3) 6, 6 (4) 4, 8
5. The correct order of melting points of IIIA group elements is
(1) B > Al > Tl > In > Ga
(2) B > Al > Ga > In > Tl
(3) B > Al > Tl > Ga > In
(4) B > Al > In > Tl > Ga
6. Incorrect statement regarding CO 2 is (1) Cardice is the solid CO2.
(2) Dry ice is used as refrigerant.
(3) CO2 is used in making urea.
(4) CO2 is insoluble in water.
7. CCl4 is used as a fire extinguisher because (1) Its m.p. is high.
(2) It forms covalent bond.
(3) Its b.p. is low.
(4) It gives noncombustible vapours.
8. A solid element (symbol Y) conducts electricity and forms two chlorides YCl n (a colorless volatile liquid) and YCl n-2 (a colorless solid). To which one of the following groups of the periodic table does Y belong?
(1) 13 (2) 14
(3) 15 (4) 16
9. Diborane on hydrolysis gives (1) BC3 (2) H3BO3
(3) HNO2 (4) B3N3H6
10. Zeolites are used as
a) ion exchangers
b) molecular sieves
c) water softener
The correct uses are
(1) a and b only (2) b and c only
(3) a and c only (4) a, b and c
11. The statements regarding ‘B’ and ‘Al’ are I) Boron is a bad conductor of heat and electricity.
II) Aluminium hydrides are stable.
III) Maximum covalency of Boron is 4.
The correct statements are
(1) Only I is correct.
(2) I and III are correct.
(3) All are correct.
(4) only III is correct.
12. L 1 is the length between two adjacent carbon atoms in a layer and L2 is the length in between two layers of graphite. The approximate ratio between L 1 and L2 is (1) 1 : 1 (2) 2 : 5
(3) 5 : 2 (4) 1 : 5
13. IIIA group element with highest density is (1) B (2) Al (3) In (4) Tl
14. The covalency and oxidation state respectively of boron in [BF 4]-, are (1) 4 and 3 (2) 4 and 4 (3) 3 and 4 (4) 3 and 5
15. The number of pentagons and hexagons, respectively in C60 fullerene are (1) 10 & 20 (2) 30 & 30 (3) 20 & 10 (4) 12 & 20
16. The maximum number of atoms present in the same plane in diborane molecule is (1) 2 (2) 6 (3) 4 (4) 3
17. The most reactive form of carbon is (1) charcoal (2) graphite (3) diamond (4) coal
18. Electronegativity is least for (1) Tl (2) Al (3) Ga (4) B
19. Which of the following halides is least stable and has doubtful existence (1) CCl4
CHAPTER TEST - JEE ADVANCED
2019 P2 Model
Section-A
[Multiple Option Correct MCQ]
1. Which of the following order(s) is/are correct?
(1) BCl3>AlCl3: Lewis acidic nature
(2) BF3<BCl3<BBr3: Lewis acidic nature
(3) SiF 4>SiCl 4 : Ease of hydrolysis under normal conditions
(4) Diamond > graphite: Thermal conductivity
(2) GeI4
(3) SnI4
(4) PbI4
20. The correct statements among the following I) Fullerene has both single and double bonds.
II) C–C bond lengths in fullerene are 143.5 pm and 138.3 pm.
III) Fullerene has one electron at each carbon which is delocalised into molecular orbitals.
IV) Fullerene is aromatic in nature.
(1) I and II only (2) I, II, and III only
(3) I, II, III, and IV (4) I, II, and IV only
Section-B
21. Number of BB bonds in diborane
22. Graphite has layered structure, Layers are held by van der Waals forces and distance between two layers(in A 0) is,
23. How many elements are more electronegative, than Boron in Be, Al, Ga, ln, Tl?
24. What is the value of ‘ n’ in the following silicate ion? [Si6O18]n–
25. The stable oxidation state of thallium, a IIIA group element is _____.
2. Which halides of group 13 exist as dimer in vapour state?
(1) Al (2) B
(3) Ga (4) In
3. The non-existence of PbI 4 is due to (1) highly oxidising nature of Pb 4+
(2) highly reducing nature of Pb 4+
(3) sufficiently large covalent character
(4) highly reducing nature of I –4 ions
4. Stability of monovalent and trivalent cations of Ga, In, Tl lie in the following sequence
(1) Ga3+<In3+>Tl3+
(2) Ga3+>In3+>Tl3+
(3) TiInGa ⊕⊕⊕ >>
(4) GaInTl ⊕⊕⊕ >>
5. Orthoboric acid (H3BO3) and metaboric acid (HBO2) differ in respect of
(1) Basicity
(2) Structure
(3) Melting point
(4) Oxidation
6. Which of the following statements are true for H3BO3?
(1) It is mainly monobasic acid and a Lewis acid.
(2) It does not act as a proton donor but acts as an acid by accepting hydroxyl ions.
(3) It has a layer structure in which BO3 units are joined by hydrogen bonds.
(4) It is obtained by treating borax with conc. H2SO4.
7. Which of the following statements is/are correct for CO?
(1) CO is an important fuel.
(2) CO is poisonous gas and a neutral oxide.
(3) It can be prepared by dehydrating formic acid with conc.H2SO4.
(4) CO is acidic in nature.
8. Which of the following statement is correct?
(1) Tl (III) salt undergoes disproportionation.
(2) CO is used as a reducing agent.
(3) CO2 is a greenhouse gas.
(4) SiO2 is a covalent solid.
Section-B
[Numerical Value Questions]
9. Among the given oxides:
B 2O 3, Al 2O 3, In 2O 3, CO 2, SiO 2, CO, SnO 2, PbO2 SnO, PbO, Tl2O
x= Acidic oxides,y = Basic oxides, z = Amphoteric oxides, m = neutral oxide
Then, xyz m ++ is-----------.
10. In borax the number of sp3 hybridised boron = x & sp2 hybridised boron = y, find (x + y)× 2/3 .
11. The ratio of six membered rings in 70 60 CFullerene CFullerene
12. In borazole, if the number of σ bonds are x, and number of π bonds are y, then (x + y)2 is_____.
13. Beryl is a cyclic silicate containing 12 18 n SiO The value of n is-------.
14. The oxidation state of ‘C’ in CO is x and the oxidation state of ‘C’ in COCl2 is y. Then y-x is __________
Section-C
[Passage-based Questions]
Passage I: Given below two lists: List-I and List-II. Analyze both the lists and answer the given questions:
List–I List–II
(a) AlCl3 (p) Planar geometry
(b) BF 3 (q) Layered structure and have delocalised electrons
(c) Graphite (r) Lewis acid
(d) (BN)x (s) Non-polar molecule
(t) Sublimes on heating
15. Which of the following has the correct combination considering List-I and List-II?
(1) (IV), (Q)
(2) (II), (P)
(3) (I), (Q)
(4) (III), (Q)
16. Which of the following has the correct combination considering List-I and List-II?
(1) (III), (R)
(2) (I), (T)
(3) (II), (S)
(4) (IV), (P)
Passage II: Consider List-I and List-II:
List–I List–II
(a) Pb (p) Unaffected by water
(b) C (q) Releases hydrogen gas with water
ANSWER KEY
(c) Sn (r) Stable in +2 oxidation state
(d) Si (s) Does not show catenation
17. Which of the following has the correct combination considering List-I and List-II?
(1) (A)-(S) (2) (B)-(P)
(3) (C)-(S) (4) (D)-(Q)
18. Which of the following has the correct combination considering List-I and List-II?
