Published by Rankguru Technology Solutions Private Limited
IL Achiever Series Chemistry for JEE Module 2
ISBN 978-81-985634-0-8 [SECOND
EDITION]
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Key Features of the Book
Chapter Outline
1.1 Importance of Chemistry
1.2 Nature of Matter
This outlines topics or learning outcomes students can gain from studying the chapter. It sets a framework for study and a roadmap for learning.
Solved Examples
Specific problems are presented along with their solutions, explaining the application of principles covered in the textbook.
Try yourself:
1. How many significant f igures are there in ‘ p ’?
Ans: This is a non-terminating and nonrecurring value. Hence, the number of significant figures in ‘p ’ is infinity.
Solved example
1. A jug contains 2 L of milk. Calculate the volume of the milk in m3 .
Sol. Since 1 L = 1000 cm3 and 1 m = 100 cm, 1m100cm =1= 100cm1 m
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.
TEST YOURSELF
1. Which of the following pairs can be cited as an example to illustrate the law of multiple proportions?
(1) Na2O, K2O (2) CO, CO2
(3) SO2, SO3 (4) Both 2 and 3
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.
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
FLASHBACK
CHAPTER TEST
This comprehensive test is modelled after the JEE exam 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 needing 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.
CLASSIFICATION OF ELEMENTS CHAPTER 4
Chapter Outline
4.1 Genesis of Periodic Classification
4.2 Modern Periodic Law and Present form of The Periodic Table
4.3 Nomenclature of Elements with Atomic number >100
4.4 Electronic configuration of Elements and the Periodic Table
4.5 Electronic Configuration and Types of Elements: s,p,d,f-blocks
4.6 Periodic Trends in Properties of Elements
4.1 G ENESIS OF PERIODIC CLASSIFICATION
■ By 1865, the number of known elements increased from 31 to 63; 88 occur naturally, while 26 are man-made, with atomic weight becoming a key property after Dalton’s atomic theory.
Dobereiner's Law of Triads
■ John Dobereiner identified triads, sets of three elements with similar chemical properties, where the middle element’s atomic weight is approximately the average of the other two.
Dobereiner’s triads
Telluric Helix
■ In 1862, A.E.B. de Chancourtois arranged elements by increasing atomic weight in a spiral cylindrical table called the Telluric Helix.
■ It did not apply to all known elements.
Newland’s Law of Octaves
■ John Newlands arranged elements by increasing atomic weight and observed that every eighth element showed similarities, calling it the Law of Octaves.
■ Drawback: It worked only for lighter elements (up to calcium) and failed for heavier elements.
CHAPTER 4: Classification of Elements
Lothar Meyer’s Work
■ Lothar Meyer classified by plotting atomic volume, melting point, and boiling point against atomic weight, showing a periodic pattern which is shown in Fig.4.1.
Solved Example
1. What would be the group and period of the element with atomic number 80?
Sol. The element with Z value 80 is mercury (Hg). It is present in period 6 and group II B.
TEST YOURSELF
1. Beryllium follows Newland’s law of octaves. What is its eighth similar element in the classification? (1) K (2) Mg (3) P (4) Si
2. The number of gaseous elements available is (1) 5 (2) 11 (3) 12 (4) 15
3. Which of the following is a Dobereiner's triad? (1) Li, Na, K (2) Fe, Co, Ni (3) Ru, Rh, Pd (4) Os, Ir, Pt
Answer Key (1) 2 (2) 2 (3) 1
4.2 MODERN PERIODIC LAW AND PRESENT FORM OF THE PERIODIC TABLE
■ Periodic Law – "The physical and chemical properties of elements are periodic functions of their atomic weights."
■ Table Structure – Elements are arranged in horizontal rows (periods) and vertical columns (groups), with each group divided into subgroups A and B.
■ Short & Long Periods – The first three periods are short, while the remaining long periods contain two rows (series).
■ Arrangement Criteria – Elements were arranged by atomic weight and chemical properties.
■ Prediction of Missing Elements – Mendeleev left gaps in the table, predicting the properties of undiscovered elements.
Fig. 4.1 Lother Meyer’s curves
Table 4.1 Comparison of properties – Mendeleef’s Eka Elements
Property
of the element
K
Eka Aluminium Eka Silicon As Predicted As Observed As Predicted As Observed
Formula of oxide (EkaAl)2O3 Ga2O3 (EkaSi)O2 GeO2
Formula of chloride (EkaAl)Cl3 GaCl3 (EkaSi)Cl4 GeCl4
Isolation of the elements - - -
■ Predicted Elements – Eka-boron, Eka-aluminium, and Eka-silicon were later identified as Scandium (Sc), Gallium (Ga), and Germanium (Ge).
■ Corrected Atomic Weights – Revised atomic weights of Be, In, U, using Atomic weight = Equivalent weight × Valency.
■ Anomalous Pairs – Some elements were arranged against atomic weight order:
■ Ar (40) & K (39), Co (59) & Ni (58), Te (128) & I (127), Th (232) & Pa (231).
■ Transition Triads (Group VIII) – Elements placed horizontally due to similar properties:
■ Fe, Co, Ni; Ru, Rh, Pd; Os, Ir, Pt.
■ Limitations –Zero group elements were unknown.
■ Cu, Ag, Au were placed with K, Rb, Cs despite different properties.
■ Atomic structure was unknown at Mendeleev’s time.
■ The electronic configuration provided fundamental basis for the properties of elements.
■ Moseley’s Experiments – Bombarding elements with cathode rays produced X-rays with characteristic frequencies. He showed that X-ray frequency is linked to the atomic nucleus charge (atomic number). ()aZbυ=−
■ Here, υ is the frequency of X-rays, Z is the atomic number, a and b are constants for a selected type of line. A plot of υ against ‘Z’ gives a straight line, as sh own in Fig.4.2 Y o x Z 0 a
oa Intercept of Y axis is ‘ab’ ν
Fig. 4.2 Plot of υ and atomic number (Z)
■ As atomic number increases, the frequency of characteristic X-rays increases. Hence, atomic number is concluded as the fundamental quantity of element,
original periodic table
TEST YOURSELF
1. Considering the chemical properties, atomic weight of the element ‘Be’ was corrected based on (1) valency (2) configuration (3) density (4) atomic volume
2. Eka silicon is now known as (1) scandium (2) gallium (3) germanium (4) boron
3. The element ‘Sc’ was known long back as (1) eka-aluminium (2) eka-boron (3) eka-silicon (4) eka-mercury
4. Anomalous pair among the following is (1) Boron – Silicon (2) Beryllium – Indium (3) Aluminium – Gallium (4) Cobalt – Nickel
5. Choose the triad not present in group VIII of Mendeleef’s table. (1) Li, Na, K (2) Fe, Co, Ni (3) Ru, Rh, Pd (4) Os, Ir, Pt
6. The frequency of the characteristic X-ray of line of metal target ‘M’ is 2500 cm–1 and the graph between ν vs ‘Z’ is as follows. The atomic number of M is
(1) 49 (2) 50 (3) 51 (4) 25
4.2.1 MODERN PERIODIC LAW
■ In 1913, Moseley demonstrated that atomic number is a more fundamental property than atomic weight, determining an element’s position in the periodic table and explaining anomalous pairs.
■ He stated: "The physical and chemical properties of elements are periodic functions of their atomic numbers."
■ Atomic number (Z) = Nuclear charge = Number of electrons in a neutral atom.
■ Electronic configuration governs element properties, forming the basis of the modern periodic law.
■ Modern Periodic Law: "The physical and chemical properties of elements are periodic functions of their atomic numbers or electronic configurations."
■ The modern periodic table retains the same number of horizontal rows (periods) as Mendeleev’s table.
CHAPTER 4: Classification of Elements
Solved Examples
2. How would you justify the presence of 18 elements in the 5th period of the periodic table?
Sol. When n = 5, l= 0, 1, 2, 3. The order in which the energy of the available orbitals 4d, 5s, and 5p increases is 5s < 4d < 5p. The total number of orbitals available is 9. The maximum number of electrons that can be accommodated is 18 and, therefore, 18 elements are there in the 5 th period.
3. Why is there a break in the third period elements of the long form of the periodic table?
Sol. In the third period, 3s and 3p orbitals are only filled successively. Four orbitals together can hold eight electrons.
TEST YOURSELF
1. The basis of modern periodic law is (1) atomic number (2) atomic size (3) atomic volume (4) atomic mass
2. Which of the following pairs of elements are from the same group of the peri odic table? (1) Mg, Cs (2) Mg, Sr (3) Mg, Cl (4) Na, Cl
3. Elements of a vertical group have (1) same atomic number (2) same electronic configuration (3) same number of valence electrons (4) same number of core electrons
4. The first element of fifth period is (1) K (2) Rb (3) Kr (4) Xe
5. As per the modern periodic law, the physical and chemical properties of elements are periodic functions of their (1) atomic volume (2) electronic configuration (3) atomic weight (4) atomic size
6. The period that contains only gaseous elements is (1) 1 (2) 2 (3) 3 (4) 4
7. The first element and last element in the largest period in modern periodic table are (1) Rb and Xe (2) Cs and I (3) Cs and Rn (4) Fr and Kr
8. Which of the following has both members from the same period of the periodic table? (1) Na, F (2) Mg, Ca (3) Na, Cl (4) Be, Al
Answer Key
(1) 1 (2) 2 (3) 3 (4) 2 (5) 2 (6) 1 (7) 3 (8) 3
4.2.2 LONG FORM OF PERIODIC TABLE
■ Bohr’s Periodic Table – Arranged elements by electronic configuration based on modern periodic law.
Periodic Table of the Elements (Long form) (Representing electronic configurations)
Atomic number Symbol Valence-shell configuration
Main group Elements S-subshell is
Main-Group Elements p-subshell is gradually filled up IA Group 0 (zero) Transition Elements d-subshell is gradually filled up 1 H 1s 1 Inner -Transition Elements f-subshell is gradually filled up
7 Fr Ra AC ** Rf Db Sg Bh Hs Mt Ds Rg Uub Uut
Table 4.7 A modern form of the periodic table (long form)
Salient Features
■ Based on electronic configuration and arranged by increasing atomic number, following the Aufbau principle, as shown in the table 4.3.
■ Consists of 7 periods (horizontal rows) and 18 groups (vertical columns).
■ Periods correspond to the principal quantum number of the outermost orbit, starting with alkali metals and ending with noble gases.
■ First element in a period has its differentiating electron in the s-orbital, while the last element has it in the p-orbital.
■ Fourteen elements from the 6th and 7th periods (3rd group) are placed separately at the bottom are listed in Table 4.4.
Table 4.3 Sub-energy levels filled in periods
Table 4.4 Number of elements present in different periods
7 (20) (Incomplete)
TEST YOURSELF
1. The number of elements present in 2 nd, 3rd, 4th, and 5th periods of the modern periodic table, respectively, are (1) 2, 8, 8, and 18 (2) 8, 8, 18, and 32 (3) 8, 8, 18, and 18 (4) 8, 18, 18, and 32
2. Outer shell octet configuration is observed for the elements of the group (1) 2 (2) 8 (3) 18 (4) 32
3. The element with Z=117 and Z=120 belong to ____and______family, respectively. (1) halogen family, alkaline earth metals (2) nitrogen family, alkali metals (3) halogen family, alkali metals (4) chalcogens family, alkali metals
4. The following statements are related to elements in the periodic table. Which of the following is true?
(1) All the elements in group-17 are gases.
(2) The group-13 elements are all metals.
(3) Elements of group-16 have lower ionisation enthalpy values compared to those of group-15 in the corresponding periods.
(4) For group-15 elements, the stability of +5 oxidation state increases down the group.
5. In a period, elements are arranged in a strict sequence of
(1) decreasing charges in the nucleus (2) increasing charges in the nucleus (3) constant charges in the nucleus (4) equal charges in the nucleus
6. Which of the following pairs has elements containing the same number of electrons in the outermost orbit?
(1) N, O (2) Na, Cl (3) Ca, Cl (4) Cl, Br
Answer Key
4.3 NOMENCLATURE OF ELEMENTS WITH ATOMIC NUMBER >100
■ Element 104 was disputed between American (Rutherfordium) and Soviet (Kurchatovium) scientists.
■ To prevent conflicts, IUPAC introduced a systematic naming system based on numerical roots for atomic numbers (Z > 100).
■ IUPAC Naming Rules: Each digit of the atomic number is replaced with its corresponding root. The roots are combined in sequence, followed by "-ium". are shown in Table 4.5.
Table 4.5 IUPAC nomenclature of heavy elements Atomic
Unniltrium Unt Lawrencium Lr
Unnilquadium Unq Rutherfordium Rf
Unnilpentium Unp Dubnium Db
Unnilhexium Unh Seaborgium Sg
Unnilseptium Uns Bohrium Bh
Unniloctium Uno Hassium Hs
Unnilennium Une Meitnerium Mt
Ununnilium Unn Darmstadtium Ds
Unununnium Uuu Roentgenium* Rg*
Ununquadium Uuq
Ununpentium Uup
Official name and symbol yet to be announced by IUPAC
TEST YOURSELF
1. From the following select, the elements belonging to the same group. (1) Z = 12, 38, 4, 88 (2) Z = 9, 16, 3, 35 (3) Z = 5, 11, 27, 19 (4) Z = 24, 47, 42, 55
2. Rare earths are generally (1) actinides (2) all f-block elements (3) all inner transition elements (4) lanthanides
4. In the periodic table, transition elements begin with (1) scandium (2) zinc (3) copper (4) mercury
5. The general electronic configuration (n-1)d3ns2 indicates that the particular element belongs to the group (1) VB (2) VA (3) IVB (4) IIB
6. In the sixth period, the orbitals being filled with electrons are (1) 5s, 5p, 5d (2) 6s, 6p, 6d, 6f (3) 6s, 5f, 6d, 6p (4) 6s, 4f, 5d, 6p
7. The representative elements get the nearest inert gas configuration by (1) losing electrons (2) gaining electrons (3) sharing electrons (4) losing or gaining or sharing electrons
8. The period number and group number in which maximum number of elements are placed are, respectively, (1) 4th and IA (2) 5th and zero (3) 7th and IIIA (4) 6th and IIIB
9. The formula of the compound formed by the pair of elements Al and S is (1) AlS (2) Al2S3 (3) Al3S2 (4) AlS2
10. An element has 18 electrons in the outer- most shell. The element is a/an (1) transition metal (2) rare earth metal (3) alkaline earth metal (4) alkali metal
11. Match the columns.
Column-I Column II
A. Polonium I) Liquid metal
B. Mercury II) Liquid non-metal
C. Bromine III) Diamond
D. Carbon IV) VIA group
Choose the correct answer from the options given below.
12. The electronic configuration of an element ‘X’, is 1s2 2s2 2p6 3s2 3p3. What is the atomic number of the element which is just below ‘X’ in the periodic table? (1) 33 (2) 34 (3) 31 (4) 49
Answer Key
(11) 1 (12) 1
4.4 ELECTRONIC CONFIGURATION OF ELEMENTS AND THE PERIODIC TABLE
4.4.1
Electronic Configuration in Periods
1. Period Length & Orbitals – The number of elements in each period is twice the number of atomic orbitals available in that energy level.
2 Period Number & Shells – The period number corresponds to the outermost shell of the atom.
3. Successive Filling – Each period fills the next higher principal energy level.
Period-wise Breakdown:
1st Period (2 Elements) –
■ Fills 1s (K-shell), includes H (1s¹) and He (1s2).
2nd Period (8 Elements) –
■ Starts with Li (2s¹), ends at Ne (2s2 2p⁶).
3rd Period (8 Elements) –
■ Starts with Na (3s¹), ends at Ar (3s2 3p⁶).
4th Period (18 Elements) –
■ Begins with K (4s¹), ends at Kr (4s 24p⁶).
■ Includes 10 transition elements (Sc-Zn) due to 3d filling.
■ Exception: Cr and Cu have one electron in 4s orbital.
5th Period (18 Elements) –
■ Starts with Rb (5s¹), ends at Xe (5p⁶).
■ Includes 10 transition elements (Y-Cd) due to 4d filling.
6th Period (32 Elements) –
■ Starts with Cs (6s¹), ends at Rn (6p⁶).
■ Includes Lanthanides (Ce-Lu, 4f series).
CHAPTER 4: Classification of Elements
7th Period(Incomplete, 32 Elements)
■ Starts with Fr (7s¹).
■ Actinides (Th-Lr, 5f series) included.
■ Expected to end at atomic number 118.
■ Special Placements
■ Hydrogen is placed separately due to its unique properties.
■ Lanthanides (4f) & Actinides (5f) are placed separately to maintain periodic table structure.
4.4.2
Electronic Configuration in Groups
■ Purpose of Classification – Groups elements with similar chemical properties based on their electronic configuration.
■ Groupwise Electron Configuration:
Group 1 (Alkali Metals) – ns¹ (1 electron in outermost shell).
Group 2 (Alkaline Earth Metals) – ns2 (2 electrons in outermost shell).
Comparison of Covalent Radii with Van der Waals' Radii
■ Covalent radius is about 20% shorter than the theoretical atomic radius due to orbital overlap in covalent bonding, which reduces internuclear distance.
■ Van der Waals radius is 40% larger than the covalent radius since non-bonded atoms are held by weak forces.
■ Inert gases are monoatomic and rarely form bonds.
■ For monoatomic elements, the van der Waals radius is considered as the atomic radius.
■ The covalent and vander Waals radii of some non-metallic elements are compared in Table.4.7
Table 4.7 Covalent and van derwaals radii of some elements
Variation of Atomic Radius in Groups
■ Atomic radius increases down a group due to an increase in the number of shells, despite increasing nuclear charge.
■ The effect of additional shells outweighs nuclear attraction, leading to larger atomic size.
■ Shielding effect increases, further reducing nuclear attraction.
■ Metallic (crystal) radius is greater than covalent radius.
■ Hydrogen has the smallest atomic radius, while Caesium has the largest among available elements.
■ The increase in the atomic radius in groups, with increasing atomic number, is shown diagrammatically in Fig.4.4.
Atomic number (Z)
Fig. 4.4 Atomic radii in group
number (Z)
Fig. 4.5 Atomic radii in period
Variation of Atomic Radius in Period
■ Atomic radius decreases across a period from left to right up to noble gases.
■ Reason 1: Increasing atomic number raises effective nuclear charge, pulling electrons closer to the nucleus, reducing atomic size.
■ Reason 2: From halogens to noble gases, atomic radius increases because:
Noble gases use van der Waals radius, which is larger.
Completely filled subshells in noble gases cause inter-electronic repulsions, slightly increasing size.
Radii of second period elements are listed in Table 4.8.
■ The trend in the decrease of atomic radius is valid in any period from alkali element to halogen.
Table 4.8 Radii of 2nd Period Elements
CHAPTER 4: Classification of Elements
■ The decrease in the atomic radius in periods, with increase in atomic number.
■ Every period starts with an electron entering s-sub shell of a new orbit. When the next electron enters in the same s-sub shell, the resulting decrease in the atomic radius is significant.
■ But the decrease in the radius with the p, d, and f-sub shells are being filled is normal.
Variation of Atomic Radius in Transition and Inner Transition Elements
■ Transition elements: Atomic radius decreases slowly across a period.
■ Effective nuclear charge per electron increases, pulling electrons closer to the nucleus.
■ Cation radius is smaller than its parent atom due to greater nuclear attraction.
■ This is illustrated by the data in the Table 4.9.
Table 4.9 Atomic and cation radii of some elements
Anionic Radius
■ Anion is a negatively charged ion formed by the gain of electrons by a neutral atom.
■ Effective nuclear charge per electron decreases, reducing nuclear attraction on electrons.
■ Anion radius is larger than its parent atom due to weaker attraction.
■ Cation and anion radii increase down a group due to added electron shells.
Nuclear Charge
■ Anion forms by electron gain in a neutral atom.Lower nuclear attraction increases anion size.
■ Anion radius > Parent atom due to weaker attraction.Cation & anion radii increase down a group due to added shells.
Radii of Isoelectronic Species
■ Ions that have equal number of electrons are called isoelectronic ions.
■ In such ions with decrease of nuclear charge radius increases, Table 4.10.
Table 4.10 Atomic and ionic radii of some isoelectronic series
Solved Examples
6. Compare the radii of H atom, H + ion and H– ion
Sol. H+ is the nucleus of H atom. Its radius is very small.
H– ion has number of electrons more than number of protons. Its size is more than that of H atom.
The radius is in the order: H + < < H < H–
7. Which is a bigger ion among Na +, F–, O2– and Mg2+? Why?
Sol. O2– is bigger ion among the given four. Among isoelectronic ions, the more the negative charge on the ion, the more is its size.
The order is O–2 > F–1 > Na+ > Mg+2
Try yourself:
2. If the van der Waals radius of hydrogen is 120 pm, what should be the intermolecular distance in solid Hydrogen?
Ans:240 pm
Ionisation Enthalpy
Electrons in an atom are attracted by the positively charged nucleus. To remove an electron from an atom, energy has to be supplied in order to overcome the attractive forces. This energy is known as ionisation enthalpy or ionisation energy or ionisation potential. Energy is always required to remove electrons from an atom and, hence, ionisation enthalpies [ D H] are always positive
()() +−+→+ gg MIEMe
■ The minimum energy required to remove electron from an isolated gaseous atom to convert it into a gaseous ion is called ionisation energy.
Here, M is the isolated atom, I is the ionisation, potential and M + is the cation formed by the loss of one electron from M.
The units of ionisation enthalpy = kJmol -1 or kcal mol–1
1eV/atom= 23.06 k cal mol –1
1eV/atom = 96.43 kJ mol –1
1eV/atom = 1.602 × 10 –19 J atom–1
■ IE1, IE2, IE3, IE4, IE5, etc., are collectively known as successive ionisation potential values.
■ In general, the increasing trend in these values is : IE 1 < IE2 < IE 3 ...... IE n .
■ The IE1, IE2, and IE3 values of aluminium are, respectively, 578, 1820, and 2750 kJ/mole.
Table 4.11 Ionisation potential of second period elements
Factors Influencing Ionisation Potential
■ Atomic Radius → Ionisation potential (IP) is inversely proportional to atomic radius.
■ Oxygen and Sulphur have lower IE than their preceding group 15 elements due to less stable electron configurations.
Solved Examples
8. The successive ionisation enthalpies of an element M are 5.98, 18.82, 28.44, 119.96, 153.77....eV/atom. What is the formula of chloride of M?
Sol. Observing the IE1, IE2, IE3, IE4, IE5, .... it is noticed that there is a sudden jump from IE 3 and IE4. This observation gives the idea that the element has 3 electrons in the outer most shell, as there is a great difference between 3rd and 4th ionisation enthalpies.
M3+ state is stable and valency is 3.
Formula of chloride of M is MCl 3.
9. The ionisation enthalpy of sodium is 5.14 eV. How many kcal of energy is required to ionise all atoms present in one gram of gaseous Na atoms?
CHAPTER 4: Classification of Elements
Sol. 1 eV atom–1 = 23 kcal mol–1
Energy required to ionise all atoms of 23 g (one mole) of gaseous Na atoms = 23 × 5.14 kcal
Energy required for ionisation of all atoms present in one gram of gaseous Na atoms = 5.14 kcal.
Try yourself:
3. The 1st, 2nd and 3rd ionisation enthalpies of Aluminium are 577, 1816 and 2744 kJmol –1. What is the amount of energy required to convert one gaseous Aluminium atom to gaseous Al 3+?
Ans:8.53 × 10–18 J/atom..
Electron Gain Enthalpy
■ In the process of addition of electron to neutral isolated gaseous atom, a certain amount of enthalpy change is involved. This is called Electron gain enthalpy ( D Heg)
XeX,HH +→∆=∆
()() atomion ggeg
■ Units: Expressed in kJ/mol or kcal/mol.
■ Negative ΔHeg: Energy is released during electron gain (exothermic).
■ Halogens have highly negative Δ Heg due to their strong tendency to gain electrons.
■ Positive ΔHeg: Energy is required for electron addition (endothermic).
■ Atoms reluctant to accept electrons show positive Δ Heg
■ Electron affinity & ionisation potential are defined at absolute zero, but at other temperatures, heat capacities of reactants and products must be considered.
5 2, ∆=−− ege HART where eg H∆ = electron gain enthalpy A e = electron affinity
Similarly, 5 HI.PRT 2 ∆=+ , where H∆ is ionisation enthalpy
■ The ionisation potential of a neutral atom A is equal in magnitude with the electron affinity of A+ ion. However, they have opposite signs.
AIAe:AeAE +−+− +→++→+
■ In the above equations, I and E are numerically same but the sign is opposite.
Successive Electron Affinities
■ Successive addition of electrons to atomic species involve energy changes.
■ The utilisation of E1 results in the formation of uninegative ion.
()() gg1 XeXE +→+
■ The utilisation of E2 results in the formation of dinegative ion.
()() 2 2 ++→ gg XeEX
■ In a similar manner, the third electron affinity and the fourth electron affinity are defined. E1, E2, E3 etc., are collectively known as successive electron affinity values.
11
OeO142kJmole +→−
121
OeO702kJmole +→+
Two successive electron affinities of group 16 elements are listed in Table 4.14.
■ Sec ond Electron Gain Enthalpy is positive as energy is required to overcome repulsion between the uninegative ion and added electron.
Table 4.14 First and second electron affinity values of group 16 elements
Element E1 (in eV/atom) E2 (in eV/atom) Oxygen – 1.46 + 8.09
■ Atomic Radius → Larger radius → Lower electron affinity.
■ Electronic Configuration → Half-filled & fully filled sub-shells are stable, so they have low electron gain enthalpy.
■ Group Trend → Electron gain enthalpy decreases down a group
■ Variation in Groups
■ In a group, from top to bottom, electron gain enthalpy gradually decreases.
■ Trend in the electron affinity in Table 4.15.
Table 4.15 Electron affinity of halogens
■ The element with most negative electron gain enthalpy is chlorine, and the one with the least negative electron gain enthalpy is phosphorus.
Electron Gain Enthalpy Trends & Anomalies
■ Down a group: F (–328 kJ/mol) < Cl (–349 kJ/mol) due to higher repulsion in F’s small size; O < S follows the same trend.
■ Across a period: Electron affinity increases as nuclear charge increases and atomic radius decreases.
■ Anomalies:
1. Inert gases have high positive enthalpy due to a stable octet.
2. Group 15 elements have lower electron affinity; N has positive gain enthalpy due to half-filled stability.
3. Alkali metals have low negative values, while Be, Mg have positive enthalpy, approximated as zero.
The trend in the electron gain enthalpy values in a period can be observed from the data in Table 4.16.
Table 4.16 Electron gain enthalpy values of some elements (kJ mol–1)
– 46
Solved Examples
10. Write the descending order of electron affinity va lues of chalcogens.
Sol. Decreasing order of electron affinity values of chalcogens: S > Se > Te > O. Electron affinity of oxygen is less because oxygen has small atomic size and the added electron experiences greater repulsion on oxygen atom.
11. Process (A) : F2(g)+ 2e– → 2F–(g)
Process (B) : Cl2(g) + 2e– → 2Cl–(g) Which of these processes is easier? Why?
Sol. F2(g) + 2e– → 2F–(g) is easy. Though electron gain enthalpy of Cl (g) to give Cl–(g) is more than that of F(g) to give F–(g), the bond dissociation of F2(g) is very less, compared to that of Cl2(g).
IL ACHIEVER SERIES FOR JEE CHEMISTRY
Try yourself:
4. The quantity of energy released when one million gaseous iodine atoms are converted into I –(g) ions is 5.0 × 10–16 kJ. In this question: I (g) + e– → I–(g) + Q kJ mol–1 What is the value of Q?
Ans: 301
Electronegativity
■ Electronegativity is the relative tendency of an atom to attract bonded electrons.
■ Defined as an atom’s ability to pull shared electron pairs toward itself.
■ Not directly measurable, so quantitative scales are used for calculation.
Mulliken scale: According to Mulliken, Electronegativity = I.EE.A 2 + ; here, both IE and EA are in eV unit.
Pauling scale: Linus Pauling determined electronegativities with the help of bond dissociation energies.
The Pauling’s equation for the determination of electronegativity is written as,
AB XX0.208−=∆ where D is in k.cals.mol-1
AB XX0.1017−=∆ where D is in kJ.mol-1
■ Here, XA is electronegativity of element A, XB is electronegativity of element B and D is the bond stabilisation or resonance energy
D = Experimental bond energy - Calculated bond energy ()1/2 ABAABB 1 EEE 2 ∆=−+
■ EA–A, EB–B and EA–B are the bond dissociation energy values of the bonds A–A, B–B and A–B, respectively.
■ Pauling arbitrarily assigned a value of 2.1 to hydrogen and determined the electronegativity of fluorine as 4.
■ Based on this electronegativity values of other elements are calculated.
■ Pauling’s values of electronegativity are listed in Table 4.17
■ Penetration Ability → Greater penetration → Higher electronegativity (Order: s > p > d > f).
■ s-Character in Hybrid Orbitals → More s-character → Higher electronegativity (Order: sp > sp2 > sp3).
Table 4.17 Pauling’s electronegativity values of representative elements
■ Variation in Groups → Electronegativity decreases down a group due to increased atomic radius and shielding effect.
■ Variation in Periods → Electronegativity increases across a period (left to right) up to halogens due to decreasing atomic radius and increasing nuclear charge.
■ Variation of values of electronegativity with atomic number of elements is shown graphically in Fig.4.10.
4.10 Variation of electronegativity
Fig.
■ Halogens have the highest electronegativity in a period.
■ Fluorine (4.0) is the most electronegative, while Cesium (0.7) is the least on Pauling’s scale.
■ Applications of Electronegativity
■ Indicates non-metallic nature → Elements with electronegativity ≥ 2 are generally non-metals.
■ Related to reactivity → Fluorine is the most reactive, while Helium is the least reactive.
■ Elements with similar electronegativity have similar chemical behavior.
= 16 (XA – XB ) + 3.5(XA – XB)2, where XA and XB are the electro-negativit ies of two atoms A and B.
Solved Examples
12. How is the nature of covalent bond between two atoms predicted?
Sol. Nature of the covalent bond between two atoms is predicted based on the difference in electronegativity.
If there is no difference in the electronegativity of bonded atoms, the bond is pure covalent.
If electronegativity difference arises between bonded atoms, the bond is polar covalent, and further, increase of difference of electronegativity leads to the formation of ionic bond.
13. Bond energies of H2, Cl2, and HCl are, respectively, 104, 58, and 100 kcal mol –1. Calculate Pauling’s electronegativity of chlorine.
Sol. Average of bond energies of H 2 and Cl2 is the calculated bond energy of HCl
10458 2 + = = 81 kcal mol–1
Experimental bond energy of HCl = 100 kcal mol –1
D = Bond (resonance) stabilisation energy = 100 - 81 = 19 kcal mol–1
X1– X2 = 0.208 = 0.208 = 0.208 × 4.358 = 0.90
Since Pauling’s electronegativity of hydrogen is 2.1, that of chlorine = 2.1 + 0.9 = 3.0.
Try yourself:
5. Is it correct to say that pauling electronegativity of carbon is 2.5 in every compound?
Ans: no
Valency
■ Valency is the combining capacity of an element.
■ Compared to hydrogen, chlorine, or oxygen in bonding.
■ Defined as: Number of hydrogen or chlorine atoms an element can bond with.
■ Twice the number of oxygen atoms it can combine with.
■ Valencies of elements in some compounds are listed in Table 4.18
Table 4.18 Valencies of some elements
Valenc y of group elements is generally same.
■ The periodicity of valency of typical elements is provided in Table 4.21 in the form of hydrides, and oxides
■ Valency remains the same within a group.
■ Valency is zero for uncombined elements and Group 18 (noble gases) under normal conditions.
■ Valency electrons (outermost electrons) determine valency in representative elements.
■ For elements with ≤ 4 valence electrons, valency = number of valence electrons.
■ For elements with ≥ 4 valence electrons, valency = 8 – number of valence electrons.
■ Maximum valency corresponds to group number (in Roman numerals) and is never more than 8.
■ Osmium in OsO₄ and Xenon in XeO₄ have a valency of 8.
■ Valency in a period (for hydrides): Increases from 1 to 4 (Group 1 to 4), then d ecreases back to 1 (Group 5 to 7).
■ Periodic valency trends are evident in hydrides, oxides, and fluorides across periods.
Oxidation Number
■ Oxidation state is the charge present or appearing to be present on an atom in a species.
■ Oxidation states can be positive, negative, zero, or fractional.
■ Examples:
■ Alkali metals (Group IA): Always +1 (lose one electron).
■ Alkaline earth metals (Group IIA): Always +2.
■ Halogens: Always –1 (gain one electron).
■ p-block elements:
■ Oxidation number = Group number or Group number – 8.
■ Example: Group VA elements show +5 and –3 states.
■ Inert Pair Effect: Lower oxidation states become more stable down the group due to the reluctance of ns electrons to bond.
■ Example:
■ Tl (IIIA Group): +1 is more stable than +3.
■ Pb (IVA Group): +2 is more stable than +4.
■ Bi (VA Group): +3 is more stable than +5.
■ d-block elements:
■ Show variable oxidation states (+1 to +8) due to configuration ns 0-2, (n–1)d1-10
■ Common oxidation state = +2.
■ Examples:
■ Mn exhibits +2, +3, +4, +6, +7 states.
■ Cr, Cu, Ag, Au, Hg show +1 state.
■ f-block elements:
■ Exhibit +2, +3, and +4 oxidation states.
■ Lanthanides: +3 is most common.
■ Maximum oxidation state (+8) is shown by Ru, Os, and Xe in RuO₄, OsO₄, and XeO₄.
Variation of oxidation number in transition elements is shown in Table 4.19
Solved Examples
14. Using the periodic table, predict the formula of compound formed between an element X of group 13 and another element Y of group 16.
Sol. The valency of X (group 13) = 3
The valency of Y (group 16) = 2
The compound has 2 atoms of X and 3 of Y.
Hence, the formula = X2Y3
Table 4.19 Variation of oxidation states in transition elements
15. What are the valencies of K in K2O and S in H2S? What should be the formula of compound of K and S in which the above valencies are reflected?
Sol. Two K atoms are in combined state with one oxygen atom. One K-atom combines with 1/2 atom of oxygen. Hence, valency of K is one. One S-atom combined with two H atoms. Hence, valency of S is two. Hence, the formula of the compound is K 2S.
Metallic Nature
■ Electropositivity & Metallic Nature
■ Electropositivity is the opposite of electronegativity, indicating metallic nature.
■ Defined as an element’s tendency to lose electrons and form cations (M →Mⁿ ++ne-).
■ Higher electropositivity means greater metallic nature and lower ionization enthalpy.
■ Electropositivity Trends & Properties
■ Alkali metals are highly electropositive due to:
■ No hydrolysis of their ions.
■ Stable solid bicarbonates (except LiHCO₃).
■ Strong reducing ability.
■ Highly soluble oxides forming strong bases.
■ s-block metals are the most electropositive, followed by p-block metals.
■ All d-block and f-block elements are metals.
■ Metallic nature increases down a group.
■ Metalloids exhibit both metallic and non-metallic properties:
■ Group 14: Arsenic (As) and Antimony (Sb).
■ Group 15: Selenium (Se) and Tellurium (Te).
Nature of Oxides(four types)
■ Generally, metallic oxides are basic. They neutralise acids and they dissolve in water to give bases. e.g., K2O, MgO, Tl2O, etc.
Na2O + H2O → 2NaOH
CaO + H2O → Ca(OH)2
■ Generally, non-metallic oxides are acidic. They neutralise bases like sodium hydroxide. They dissolve in water to give acids. They are called acid anhydrides, e.g., SO 2, P4O10, CO2, etc.,
2223 COHOHCO +→
24223 NOHOHNOHNO +→+
■ Oxides of metalloids are generally amphoteric. They react with both acids as well as bases e.g., GeO2, Sb4O6, TeO2, As2O3,etc.
■ Some of the metallic oxides are also amphoteric, e.g., ZnO, Al 2O3, SnO2, etc.,
ZnO2HClZnClHO +→+
ZnO2NaOHNaZnOHO +→+
2332 AlO6HCl2AlCl3HO +→+
2322 AlO2NaOH2NaAlOHO +→+
■ Some of the non-metallic oxides are neutral. They do not react with acids as well as with bases e.g., CO, N2O, NO, etc.
■ Basic nature of the oxides increases generally with an increase in the electropositivity of metal forming oxide.
■ Oxides of all elements of group1 are basic and of group 17 are acidic. Down the group, basic nature of oxides increases and acidic nature decreases, as shown in Table 4.20.
Table 4.20 Nature of trioxides of group-15
Element Nature of the element Formula of trioxide Nature of oxide
N non-metal N2O3 acidic
P non-metal P4O6 acidic
As metalloid As4O6 weakly acidic
Sb metalloid Sb4O6 amphoteric
Bi metal Bi2O3 basic
■ Trends in Oxide Nature Across a Period
■ Basic nature of oxides decreases, while acidic nature increases across a period.
■ Oxide acidity increases with higher oxygen content in multiple oxides of the same element.
■ Most basic oxide: Caesium oxide (Cs₂O) → Forms the strongest base, CsOH.
■ Most acidic oxide: Chlorine heptoxide (Cl₂O₇) → Forms the strongest acid, HClO₄ (Perchloric acid).
Periodic Trends and Chemical Reactivity
■ Periodic Trends & Chemical Reactivity
■ Chemical properties are determined by electronic configuration.
■ Across a period (left to right): Atomic radius decreases → Ionization enthalpy increases → Electron gain enthalpy becomes more negative.
■ Reactivity Trends: Alkali metals (Group 1) are highly reactive due to low ionization potential, forming cations (electropositive, good reductants).
■ Halogens (Group 17) are highly reactive due to high electron affinity, forming anions (nonmetallic, good oxidants).
■ Reactivity peaks at period extremes:
■ Left (alkali metals) → React by losing electrons (cation formation, electropositivity).
■ Right (halogens, not noble gases) → React by gaining electrons (anion formation, nonmetallic nature).
■ The periodic trends in the properties of elements are diagramatically given in Fig.4.11.
Solved Examples
16. Is hydrogen electropositive?
Sol. Hydrogen has electropositivity. It is evidenced by the formation of proton. H → e – + H+
However, hydrogen is not a metal. It is a common non-metal.
17. In aqueous solutions lithium is the best reductant. Why?
Sol. Lithium cation is small and its hydration ability is high. The stronger reduction ability of lithium is also reflected in least standard potential of Li +/Li.
Anamolous properties of second period elements
■ Second-period elements show anomalous behavior due to:
■ Small atomic size.
■ High electronegativity.
■ Large charge-to-radius ratio.
■ Absence of d-orbitals for bonding.
■ Second-period elements have a maximum covalency of four.
■ First p-block elements readily form multiple bonds.
■ Oxide behaviour of third period elements is shown in table 4.21.
Fig.4.11 Periodic trends of elements
Table 4.21 Nature of oxides of elements of third period
Group IA Group IIA Group IIIA
VIA Group VIIA
Na2O MgO Al2O3 SiO2 P4O10 SO3 Cl2O7
Strong base Basic Amphoteric Weak acid
Moderately acidic Strong acid Strong acid
NaOH Mg(OH)2 Al(OH)3 H2SiO3 H3PO3 H2SO4 HClO4
Strong base Basic Amphoteric Weak acid
Examples:
1) CC,CC,NN,NN =≡=≡
2) CO,CN,CN,NO ==≡=
Moderately acidic Strong acid Strongest acid
■ Lithium (IA) & Beryllium (IIA) form covalent compounds, unlike heavier group members.
■ Boron forms [BF4]–, while heavier group members expand their valence shells (e.g., [AlF6]3–).
■ Chemical similarities exist in groups, but diagonal relationships also occur.
■ Diagonal relationship: occurs when lighter second-period elements resemble third-period elements diagonally placed in the periodic table.
■ Diagonal relationship of elements is shown in Table 4.22.
■ This similarity is called diagonal relationship.
■ The phenomenon of diagonal relationship does not appear after group 4 and also below third period of the periodic table.
■ Diagonal relationship arises due to similar electronegativity of elements.
■ Elements showing diagonal relationships have similar polarizing power.
■ Polarizing power is the cation’s ability to attract an anion’s charge cloud.
■ Formula: Polarizing po wer = Ionic charge /(Ionic radius)2.
Polarising power = () Chargeofion 2 Ionicradius
■ Across a period: Charge increases, size decreases, leading to higher polarizing power.
■ Down a group: Size increases, causing polarizing power to decrease.
■ Be2+ and Al3+ have similar polarizing power, leading to similar properties.
■ Both dissolve in caustic soda, form amphoteric oxides, and their carbides hydrolyze to produce methane.
Be2C + 4H2O → 2Be(OH)2 + CH4
Al4C3 + 12H2O → 4Al(OH)3 + 3CH4
■ Diagonal similarities are most prominent among lighter electropositive elements.
■ The metal-nonmetal boundary in the periodic table also follows a diagonal t rend.
Table 4.22 Diagonal relationship of elements
Solved Examples
18. Compare the oxidation ability of sulphur and chlorine.
Sol. 2 CleCl
S2eS +→ +→
Chlorine is better oxidant than sulphur. Electron gain enthalpy is more for chlorine. Chlorine accepts electron easily and becomes stable chloride.
19. Lithium is monovalent. Magnesium is divalent. But Li and Mg are diagonally related pair of elements. Why?
Sol. Lithium and magnesium have certain similarities in their properties. Hence, they are called a diagonally related pair of elements.
The reasons are:
(a) Li and Mg have similar electronegativity.
(b) Li+ and Mg2+ have similar polarising power.
TEST YOURSELF
1. The correct order of van der Waals radius of F, Cl, and Br is (1) F > Br > Cl (2) Br > Cl > F (3) F > Cl > Br (4) Br > F > Cl
2. The correct arrangement of O, P, and N in order of increasing radii is (1) O < N < P (2) P < O < N (3) O < P < N (4) N < O < P
3. Covalent bond length of chlorine molecule is 1.98 Å. Covalent radius of chlorine is (1) 1.98 Å (2) 1.7 Å (3) 2.05 Å (4) 0.99 Å
4. The covalent and van der Waals radii of chlorine, respectively, are (1) 1.80 Å and 0.99 Å (2) 0.99 Å and 1.80 Å (3) 1.80 Å and 1.80 Å (4) 0.99 Å and 0.99 Å
5. Electrons with the highest penetrating power are (1) p-electrons (2) s-electrons (3) d-electrons (4) f-electrons
6. The species with largest ionisation potential (1) Li+ (2) Mg+ (3) Al+ (4) Ne
7. Second ionisation energy is higher than first ionisation energy for an element. This is because (1) nuclear charge is high in cation (2) size of cation is higher than neutural atom (3) effective nuclear charge is more for cation (4) bond energy changes with charge
8. In modern periodic table, the groups table that possesses the highest and lowest ionisation energies, respectively, are (1) IA, VIIA (2) zero, IA (3) IA, IIA (4) VIIA, IA
9. In the lithium atom, screening effect of valence shell electron is caused by (1) electrons of K and L shell (2) electrons of K shell (3) two electrons of 1st and one of 2nd shell (4) electrons of L-shell
10. First four ionisation energy values of an element are 191, 578, 872, and 5972 kcals. The number of valence electrons in the element is (1) 4 (2) 3 (3) 1 (4) 2
11. The element with highest electron affinity is (1) fluorine (2) cesium (3) helium (4) chlorine
12. Ionisation of energy of F – is 320 kJ mol–1. The electron gain enthalpy of fluorine would be (1) – 320 kJ mol–1 (2) –160 kJ mol–1 (3) + 320 kJ mol–1 (4) 160 kJ mol–1
13. Which of the following is an endothermic process? (1) First electron affinity of chlorine (2) Second electron affinity of oxygen (3) Formation of NaCl from gaseous ions (4) Hydration of MgCl2
14. In a period from left to right, electron affinity (1) increases (2) decreases (3) remains constant (4) first increases and then decreases
15. Configuration that shows the highest energy released, when an electron is added to the atom, is (1) 1s2 2s2 2p3 (2) 1s2 2s2 2p4 (3) 1s2 2s2 2p5 (4) 1s2 2s2 2p6
16. Pauling’s electronegativity is based on (1) electron affinity (2) ionisation potential (3) both IP and EA (4) bond energies
17. The electronegativity value of chlorine and bromine are, respectively 3 and 2.8. Formula of a binary compound is best represented as (1) BrCl (2) ClBr3 (3) ClBr (4) ClBr5
18. Reference element for Pauling’s electronegativity is (1) H (2) C (3) Cl (4) He
19. What is the correct order of electronegativity?
(1) M+1 < M+2 < M+3 < M+4
(2) M+1 > M+2 > M+3 > M+4
(3) M+1 < M+2 > M+3 < M+4
(4) M+4 < M+2 < M+3 < M+1
20. In a period, electronegativity is highest for (1) chalcogen (2) halogen (3) inert gas (4) alkali metal
21. All the following elements show both positive and negative oxidation states, except (1) N (2) H (3) O (4) F
22. An element with electronic arrangement as 2, 8, 18, 1 will exhibit the following oxidation states (1) + 2 and + 4 (2) + 1 and + 2 (3) + 2 only (4) + 1 only
23. Oxidation state and covalency of Al in [AlCl(H 2O)5]2+ are (1) +2, 6 (2) +3, 6 (3) +2, 4 (4) +3, 4
24. Basic nature of the oxides of a period from left to right (1) increases (2) decreases (3) remains constant (4) first increases and then decreases
25. Oxide that is most acidic is (1) Cl2O7 (2) SO3 (3) P4O10 (4) N2O5
26. Generally, the nature of non-metal oxides is (1) basic (2) acidic (3) amphoteric (4) neutral
27. Most acidic oxide in the periodic table is formed by an element in (1) 2nd period, group VI A (2) 4th period, group VII A (3) 3rd period, group VI A (4) 3rd period, group VII A
28. The correct increasing order of metallic nature of Si, Be, Mg, Na, P is
(1) P < Si < Be < Mg < Na
(3) Si < P < Mg < Be < Na
(2) Si < P < Be < Mg < Na
(4) Na < Mg < Be < Si < P
29. The diagonal relationship phenomenon is not observed after (1) I A group
(2) II A group (3) III A group
(4) IV A group
30. Which of the following is not correct in the case of Be and Al?
(1) Both are rendered passive by conc. HNO 3
(2) Carbides of both give methane on hydrolysis.
(3) Both give hydroxides that are basic.
(4) Both give covalent chlorides.
31. The correct order of polarisability of ion is
(1) Cl– > Br– > I– > F–
(2) F– > I– > Br– > Cl–
(3) I– > Br– > Cl– > F– (4) F– > Cl– > Br– > I–
32. The chemistry of lithium is very similar to that of magnesium, even though they are placed in different groups. This is because
(1) both are found together in nature (2) both have nearly the same size
(3) both have similar electronic configuration
(4) the ratio of charge and size is nearly same
33. Beryllium and aluminium exhibit many properties that 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
Answer Key
3 (32) 4 (33) 3
# EXERCISES
JEE MAIN
Level I
Genesis of periodic classification
Single Option Correct MCQs
1. Which of the following is Dobereiner triad
(1) Li, Na, K
(2) Fe, Co, Ni
(3) Ru, Rh, Pd
(4) Os, Ir, Pt
2. Which is not a Dobereiner’s triad?
(1) Fe, Co, Ni
(2) Li, Na, K
(3) Ca, Sr, Ba
(4) Cl, Br, I
3. Law of Octave is not applicable to (1) Li,Na,K
(2) Be,Mg,Ca
(3) B,Al,Ga
(4) All of the above
4. The Newland’s law of octaves for the classification of elements was found to be applicable only up to the element
(1) Potassium
(2) Calcium
(3) Cobalt
(4) Phosphorus
5. Elements which occupied position in the Lother Meyer curve, on the peaks, were:
(1) alkali metals
(2) highly electropositive elements
(3) elements having large atomic volume
(4) all of the above
6. Lothar Meyer obtained the curve for the known elements by plotting their atomic volumes against (1) atomic numbers
(2) atomic masses
(3) densities
(4) ionisation energies
7. Considering the chemical properties, atomic weight of the element ‘Be’ was corrected based on
(1) Valency
(2) Configuration
(3) Density
(4) Atomic volume
8. Eka silicon is now known as (1) Scandium
(2) Gallium
(3) Germanium
(4) Boron
9. Anomalous pair among the following is
(1) Boron - Silicon
(2) Beryllium - Indium
(3) Aluminium - Gallium
(4) Cobalt - Nickel
10. Number of short periods in short form of periodic table
(1) 3 (2) 2
(3) 4 (4) 6
11. According to Mendeleev’s periodic law
(1) the properties of the middle element were in between those of the other two members
(2) three elements arranged according to increasing weights have similar properties.
(3) the properties of the elements are a periodic function of their atomic weights
(4) the elements can be grou ped in the groups of six elements.
12. The number of elements known at that t ime when Mendeleev arranged them in the periodic table was :
(1) 63 (2) 60
(3) 70 (4) 65
Numerical Value Questions
13. The group of Mendeleev’s periodic table consisting of maximum elements is ____
14. The elements A, B and C form a Dobereiner’s Triad. If the sum of atomic mass of A, B and C is 180, then atomic mass of B is:
[The order of atomic masses is A < B < C]
[Divide your answer by 10]
15. Number of elements present in second series of Mendeleev's periodic table are ___
Modern periodic law and the present form of the periodic table
Single Option Correct MCQs
16. Henry Moseley plotted a graph between ν and Z, where ν was the frequency of X–ray emitted by an atom and Z was its atomic number. This graph showed that
(1) the atomic mass is a fundamental property of an element.
(2) the atomic number is a fundamental property of an element.
(3) Both (1) and (2)
(4) the frequency (ν) was independent of atomic number.
17. Henry Mosely studied characteristic X− ray spectra of elements. The graph which his observation correctly is (1) υ
18. The frequency of the characterstic X ray of K α line of metal target ‘M’ is 2500 cm−1 and the graph between v Vs ‘z’ is as follows, then atomic number of M is (1) 49 (2) 50 (3) 51 (4) 25
19. According to Moseley, a straight line graph is obtained on plotting: Where, ν is frequency and Z is atomic number.
(1) ν vs Z (2) ν2 vs Z
(3) v vs Z (4) 1 v vs Z
Numerical Value Questions
20. The moseleys expression for iron metal would be v = a (X – b), the value of X is ___.
21. The atomic number of metal preceding nickel metal based on Moseley's experiment will be _________
Nomencl ature of elements with atomic number more than 100
Single Option Correct MCQs
22. The IUPAC name of an element with atomic number 119 is
(1) unnilennium (2) unununium
(3) ununoctium (4) ununennium
23. Identify the incorrect match
Name IUPAC official name
A. unnilunium I) Mendelevium
B. unniltrium II) Lawrencium
C. unnilhexium III) Seaborgium
D. unununium IV) Darmstadtium
(A) (B) (C) (D)
(1) IV I II III
(2) III I IV II
(3) III I II IV
(4) III IV I II
24. Ununseptium is the systematic name for element having atomic number
(1) 113 (2) 115
(3) 117 (4) 119
25. The name of the element having atomic no. 104 is/are
(1) Rutherfordium (2) Unnilquadium
(3) Kurchatovium (4) All of the above
26. The IUPAC nomenclature of an element with electronic configuration [Rn]5f146d107s2 is
(1) ununbium
(2) Unnilunium
(3) Unnilquadium
(4) Unniltrium
27. Identify the incorrect statement from the following
(1) The IUPAC name of an element with atomic number 101 is Unnilunium
(2) The IUPAC name of an element with atomic number 102 is Ununseptium
(3) The IUPAC name of an element with atomic number 103 is Unniltrium
(4) The IUPAC name of an element with atomic number 104 is unnilquadium
Numerical Value Questions
28. If IUPAC name of an element is “Unununnium” then the element belongs to nth group of periodic table. The value of n is______
29. The atomic number of Unnilunium is
30. Sum of atomic numbers of Unununium and Unnilennium are?
Electronic configurations of elements and the periodic table
Single Option Correct MCQs
31. The element with ns2np4 as outer electron configuration is a ____________
(1) Alkali Metal (2) Chalcogen
(3) Noble gas (4) Halogen
32. Match column I (electronic configuration) with column II (group)
Column I Electronic configuration
Column II Group name
A. ns2np3 I) Halogens
B. ns2np6 II) Noble gases
C. ns1 III) Group 15 elements
D. ns2np5 IV) Transition elements
V) Alkali metal
(A) (B) (C) (D)
(1) III II V I
(2) III I V IV
(3) II III V I
(4) III IV I V
33. Elements A, B, C, D and E have the following electronic configurations:
A. 1s2, 2s22p1
B. 1s2, 2s22p6, 3s23p1
C. 1s2, 2s22p6, 3s23p3
D. 1s2, 2s22p6, 3s23p5
E. 1s2, 2s22p6, 3s23p6
Which among these will belong to the same group in the periodic table?
(1) A and C
(2) A and D
(3) A and B (4) A and E
34. Match List I with List II
List I (Block) List II (General electronic configuration)
A. Most reactive metals I) ns2np6
B. P-block II) ns2(n−1)d1−10
C. Transition metals III) ns1–2
D. Inner transition metals IV) (n–2)f1–14 (n–1) d0 or 1 ns2
(A) (B) (C) (D)
(1) II III IV I
(2) III II III IV
(3) II III I IV
(4) III I II IV
35. Match the following in view of period and the orbitals being filled
List I
List II
A. Second period I) s, f, d
B. Fourth period II) s, f, d, p
C. Sixth period III) s, p
D. First period IV) s V) s, d, p
(A) (B) (C) (D)
(1) iv i iii v (2) iii v ii iv
(3) ii iii iv v (4) iii i ii v
36. Number of unpaired electrons in Gd(Z =64) and the net electrons spin are (1) 7, 3.5 (2) 8, 3 (3) 6, 3 (4) 8, 4
37. The electronic configuration of an element ‘X’, is 1s2 2s2 2p6 3s2 3p3. What is the atomic number of the element which is just below ‘X’ in the periodic table (1) 33 (2) 34 (3) 31 (4) 49
38. Identify the element that has the following electronic configuration 1s22s22p63s23p64s23d104p65s24d105p66s24f2 (1) Ba (2) At (3) Ce (4) Pr
39. In the long form of the periodic table, the valence shell electronic configuration of 5s25p4 corresponds to the element present in:
(1) Group 16 and period 5
(2) Group 17 and period 6
(3) Group 17 and period 5
(4) Group 16 and period 6
Numerical Value Questions
40. The electronic configuration of an element is 1s22s22p6 5s25p3. What is the atomic number of the element?
41. The element with the lowest atomic number that has a ground state electronic configuration of (n−1)d5ns2 is located in___ period
42. The number of electrons in the valency shell of non-metallic liquid element in the periodic table is _____
Electronic configurations and types of elements s,p,d,f blocks
Single Option Correct MCQs
43. The element having electronic configuration (Xe)4f05d16s2 belongs to (1) d-block (2) f-block (3) p-block (4) s-block
List I (Atomic number)
List II
A. 53 I) d-block
B. 55 II) p-block
C. 57 III) f-block
D. 62 IV) s-block
(A) (B) (C) (D)
(1) I II III IV
(2) II IV I III
(3) II I III IV
(4) I III II IV
45. Match List I with List II.
List I (Atomic number)
List II (Block of periodic table)
A. 37 I) p-block
B. 78 II) d-block
C. 52 III) f-block
D. 65 IV) s-block
Choose the correct answer from the options given below.
(A) (B) (C) (D)
(1) II IV I III
(2) IV III II I
(3) IV II I III
(4) I III IV II
46. The period in which s-block, p-block and d-block elements not present are
(1) 4 (2) 6 (3) 7 (4) 3
47. Match List I with List II.
List I List II
A. s block I) RareEarths
B. d block II) NobleGases
C. f block III) StrongReducing agents
D. p block IV) n-1s2p6d10
V) present from fourth period
The correct match is (A) (B) (C) (D)
(1) II I IV III
(2) V III I II (3) III V I II (4) I V II III
48. Which of the following sets of atomic number belong to that of alkali metals?
49. Element with atomic number 56 belongs to which block?
(1) s (2) p (3) d (4) f
50. Representative elements mainly belongs to (1) s- and p-blocks
(2) p- and d-blocks (3) f-block only (4) d- and f-blocks
Numerical Value Questions
51. The number of elements among the following atomic numbers that are p block elements is ______ 83, 79, 42, 64, 37, 54, 34
52. Find out total number of representative elements in the given elements:
Cd, Nb, Ta, Te, Ra, Mo, Po, Pd, Tc.
53. Number of f-electrons present in the electronic configuration of Thallium(Tl) are ______
Periodic trends in properties of elements
Single Option Correct MCQs
54. The correct order of radii is (1) N < Be < B (2) F < O2− < N3− (3) Na < Li < K (4) Fe3+ <Fe2+ <Fe4+
55. Which of the following is correct order of size of the given species?
(1) I > I > I+
(2) I+ > I > I
(3) I > I+ > I (4) I > I > I+
56. Atomic radius depends upon (A) Number of bonds formed by the atom (B) Nature of the boding (C) Oxidation state of the atom
(1) A, B (2) B, C
(3) A, C (4) A, B, C
57. The correct order of van der waal radius of F, Cl, Br is
(1) F > Br > Cl (2) Br > Cl > F
(3) F > Cl > Br (4) Br > F > Cl
58. Atomic radii of fluorine and neon in Angstrom units are respectively given by:
(1) 0.72, 1.60
(2) 1.60, 1.60
(3) 0.72, 0.72
(4) None of these
59. Covalent radius of Cl is 99pm. Select best representation for Cl2 molecule.
60. The electronic configuration with the highest ionization enthalpy is
(1) [Ne]3s23p1
(2) [Ne]3s23p2
(3) [Ne]3s23p3
(4) [Ar]3d104s24p3
61. Second ionization energy is higher than first ionization energy for an element. This is because
(1) Nuclear charge is low in cation
(2) Size of cation is higher than neutral atom
(3) Effective nuclear charge is more for cation
(4) Bond energy changes with charge
62. The I 1 ,I 2 ,I 3 ,I 4 values of an element ‘M’ are 120, 600, 1000, and 8000 kJ/mole respectively then the formula of its sulphate.
(1) MSO4
(2) M2(SO4)3
(3) M2SO4
(4) M3(SO4)2
63. What is the value of electron gain enthalpy of Na+ if IE1 of Na is 5.1 ev ?
(1) –10.2 ev
(2) –5.1 ev
(3) +2.55 ev
(4) +10.2 ev
64. The correct increasing order of ionization enthalpy of He, Li+ and Be+2 is
(1) He < Li+ < Be+2 (2) Li+ < Be+2 < He
(3) Be+2 < Li+ < He (4) Be+2 < He < Li+
65. Electron affinity of oxygen is less than that of sulphur because
(1) Electronegativity of oxygen is more (2) repulsions with incoming electron (3) sulphur is a stronger oxidant (4) Bond dissociation energy of O 2 is less
66. The electron affinity of chlorine is 3.7eV.
1 g of chlorine is completely converted to Cl ion in the gaseous state (1 eV=23.06 k cal mol −1 ). The energy released in the process is
(1) 7.2 K.Cal (2) 4.8 K.Cal
(3) 8.2 K.Cal (4) 2.4 K.Cal
67. The first ionization potential of K is 3.1 eV, the value of electron gain enthalpy of K+ will be
(1) +1.5 eV (2) –1.5 eV (3) –3.1eV (4) –9.3eV
68. Inert gases have positive electron gain enthalpy. Its correct order is
(1) Xe < Kr < Ne < He
(2) He < Ne < Kr < Xe
(3) He < Kr < Xe < Ne
(4) He < Xe < Kr < Ne
69. 1st electron affinity (EA1) is positive for (1) O (2) F
(3) C (4) N
70. Which of the following property increases down the group
(1) Metalic bond strength
(2) Ionization energy
(3) Electropositivity
(4) Electron affinity
71. Which among the following has high electron affinity?
(1) O (2) S
(3) Se (4) Te
72. Which of the following has the lowest electron gain enthalpy?
(1) F (2) Cl
(3) I (4) Br
73. Which group elements have almost zero affinity for electrons?
(1) VIIA (2) VIA
(3) VA (4) VIIIA
74. The element with highest electronegativity is
(1) O (2) F
(3) Cl (4) Br
75. The correct option with respect to the Pauling electronegativity values of the elements is:
(1) Ga > Ge (2) Si < Al
(3) P > S (4) Te > Se
76. Which of the following factors does not affect electronegativity?
(1) Effective nuclear charge
(2) Screening effect of inner electrons
(3) Type of hybridization of the atom
(4) Strength of bond in which it is participated
77. Fluorine has the highest electro negativity among the group on the pauling scale, but the electron affinity of fluorine is less than that of chlorine because.
(1) The atomic number of fluorine is less than that of chlorine.
(2) Fluorine being the first member of the family behaves in an unusual manner.
(3) Chlorine can not accommodate an electron better than fluorine by utilizing its vacant 3d – orbital.
(4) Small size, high electron density and an increased electron repulsion make addition of an electron to fluorine less favourable than that in the case of chlorine.
78. Pair of elements with equal values of electro negativity.
(1) Be, Al
(2) Mg, Al
(3) Mg, Ca
(4) F, Ne
79. EN of Fluorine in Mulliken scale is (1) 4 (2) 1.428 (3) 11.2 (4) 0.7
80. Which of the following is not a measurable quantity?
84. Among the following basic oxide is (1) SO3 (2) SiO2 (3) CaO (4) Al2O3
85. All the following are neutral oxides except (1) N2O (2) NO2 (3) NO (4) CO
86. The oxidation number and covalency of sulphur, in molecule are (1) 0 and 2 (2) +6 and 8 (3) 0 and 8 (4) +6 and 2
87. Which of the following pair is metalloid (1) Na, Al (2) Si, C (3) Ge, As (4) Sb, P
88. The effect of lanthanoid contraction in the lanthanoid series of element by and large means:
(1) Increase in both atomic and ionic radii
(2) Decrease in atomic radii and increase in ionic radii
(3) decrease in both atomic and ionic radii
(4) Increase in atomic radii and decrease in ionic radii
89. Which of the following metals does not show inert pair effect?
(1) Tl (2) Ga (3) ln (4) Al
90. Similarity in the radius of Zr and Hf is explained on the basis of
(1) Lanthanide contraction
(2) Inert pair effect
(3) Same outershell configuration
(4) Anomalous configuration
91. The order of metalic character is
(1) P < Mg < Si < Na
(2) P < Si < Mg < Na
(3) Mg < Na < Si < P
(4) Si < P < Na < Mg
92. Which pair of symbols identifies two elements that are metalloids?
(1) As and Ge
(2) Mg and Si
(3) P and As
(4) Ti and V
93. Lanthanide contraction means:
(1) Contraction of atom of lanthanum element due to poor shielding d-subshell electron
(2) Contraction of atom of lanthanum element due to high shielding of d-subshell electron
(3) Contraction of atom of elements after lanthanum due to poor shielding of f-subshell electron
(4) Contraction of atom of elements before lanthanum due to poor shielding of f-subshell electron
94. A compound contains atoms X, Y and Z. The oxidation number of X is +3, Y is +5 and Z is –2. The possible formula of the compound is
(1) XYZ2 (2) Y2(XZ3)2 (3) X(YZ4) (4) X3(Y4Z)2
95. Among the following which element exhibits highest number of oxidation state in its compounds.
(1) Osmium (2) Nitrogen
(3) Carbon (4) Chlorine
Numerical Value Questions
96. The C-C single bond length is 1.54 A° and that of Cl−Cl is 1.98A°. If the electronegativity of Cl and C are 3.0 and 2.5 respectively, the Cl−Cl bond – length will be equal to ________A°
97. If 1st Ionization enthalpy of K is 419 kJ/mol then K+ + 1e → K. ΔH = −x. x is equal to
98.The five successive ionization enthalpies of an element are 800, 2427, 3658, 5024 and 32824 kJ mol −1 . The number of valence electrons in the element is:
99. I.E. and E.A. of an element are 13 eV and 3.8 eV respectively. The electronegativity of the element on Pauling’s scale is.
100. The values of electronegativity of atom A and B are 1.20 and 4.0 respectively. The percentage of ionic character of A – B bond is nearly
103. How many oxides are Amphoteric in nature CO2, CO, Na2O, Al2O3, PbO, SnO, ZnO,
CHAPTER 4: Classification of Elements
Level II
Genesis of Periodic Classification
Single Option Correct MCQs
1. Which of the following sets of elements follows Newland’s octave rule?
(1) Be, Mg, Ca (2) F, Cl, Br
(3) Na, K, Rb (4) B, Al, Ga
Modern Periodic Law and Present Form of the Periodic Table
Single Option Correct MCQs
2. The atomic weights of ‘Be’ an d ‘In’ were corrected by Mendeleef using the formula
(1) () v aZb=−
(2) 2 nh mvr π =
(3) Atomic weight = Equivalent weight × valency
(4) Equivalent weight = Atomic weight × valency
3. Eka-silicon is now known as (1) Scandium
(2) gallium
(3) germanium
(4) Boron
4. As per the Mendeleef’s periodic table the physical and chemical properties of elements are periodic functions of their (1) atomic mass
(2) atomic number
(3) atomic volume
(4) atomic radii
5. As per the modern periodic law, the physical and chemical properties of elements are periodic functions of their (1) atomic volume
(2) electronic configuration
(3) atomic weight
(4) atomic size
Long Form of Periodic Table
Single Option Correct MCQs
6. In the periodic table, the elements are arranged in the periods following the (1) Hund’s rule of maximum multiplicity (2) Pauli’s exclusion principle (3) Aufbau principle (4) Both (1) and (2)
7. The long form of the periodic table consists of (1) 8 horizontal and 7 vertical series (2) 7 horizontal and 18 vertical series (3) 7 horizontal and 7 vertical series (4) 8 horizontal and 8 vertical series
8. The position of the element with Z = 106 in the periodic table is (1) d-block (2) s-block (3) f-block (4) p-block
9. Atomic number of nitrogen is 7. The atomic number of the third member in the same family is (1) 23 (2) 15 (3) 33 (4) 51
10. Atomic numbers of actinides are (1) 57 to 71 (2) 80 to 103 (3) 58 to 71 (4) 90 to 103
11. Which of elements, with the following atomic numbers belong to the same group? (1) 9, 16, 35, 3 (2) 12, 20, 4, 38 (3) 11, 19, 27, 5 (4) 24, 47, 42, 55
12. The element that belongs to 3rd period and IVA group of the periodic table is (1) silicon (2) carbon (3) germanium (4) tin
13. An element belongs to group 17 with atomic number 17. What is the atomic number of the element belonging to the same group and present in the fifth period? (1) 25 (2) 33 (3) 35 (4) 53
14. The sub-shells filled one by one for 4 th period elements are (1) 3d, 4s, and 4p (2) 4s, 4p, and 4d (3) 4s, 3d, and 4p (4) 3d, 4p, and 4s
Numerical Value Questions
15. According to Moseley’s equation: ()aZbν=− the graph between vs Z ν is given below, the frequency v (in s–1) for atomic number (Z)=52 is _____.
Ο Z 45° 1
Nomenclature of Elements with Atomic number >100
Single Option Correct MCQs
16. As per IUPAC rule, the atomic number for the element unbiunium is:
(1) 120 (2) 121
(3) 112 (4) 122
17. IUPAC official symbol of the element with the symbol Unu is
(1) No (2) Md (3) Lr (4) Fm
Electronic Configuration of elements and Periodic Table
Single Option Correct MCQs
18. Which group and period does an element belong to if the electronic configuration of an element in its – 2 oxidation state is 1s22s22p63s23p6?
(1) Period 3, group 16
(2) Period 3, group 17
(3) Period 4, group 16
(4) Period 4, group 17
19. The 79th electron of an element ‘X’ with an atomic number 79 enters into
Electronic Configuration of elements and Types of elements
Single Option Correct MCQs
20. An element has 18 electrons in the outer most shell. The element is a/an
(1) transition metal
(2) rare earth metal
(3) alkaline earth metal
(4) alkali metal
21. Which of following is not correctly matched?
(1) d-block element: Electronic configuration is ns0–2(n-1)d–10
(2) p-block element: Electronic configuration is ns1–2np1–6
(3) s-block element: Electronic configuration is ns1–2
(4) Cerium(Ce): The first member in the f-block in modern periodic table.
22. The outer most electronic configuration of the most electronegative element is
(1) ns2np3
(2) ns2np6(n-1)d2
(3) ns2np5
(4) ns2np6
23. Two elements ‘X’ and ‘Y’ have the following configuration X=1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 , Y=1s 2 2s 2 3s 2 3p 5 . The compound formed by the combination of ‘X’ and ‘Y’ will be (1) XY2 (2) X 3Y2 (3) X2Y3 (4) XY 5
Numerical Value Questions
24. Find the total number of 6th period elements from the given atomic numbers : 81, 63, 80, 50, 54, 48, 86
25. Number of elements present in shortest period is ___.
26. The VII B group of Mendeleef’s periodic table consists of ____ elements.
27. When 4f level is completely filled with electrons, the next electron will enter into a subshell whose (n+l) is equal to ____.
28. The longest period in which s-block, p-block, d-block, and f-block elements are present is ___.
29. The total number of gaseous elements is ___.
Periodic Trends in Properties of Elements
Single Option Correct MCQs
30. The correct increasing order of the ionic radii is
31. The correct order of atomic radii of the elements O, N, S, and P is (1) N < P < S < O (2) N < O < P < S (3) O < N < P < S (4) O < N < S < P
32. Atomic radii of fluorine and neon in angstrom units are respectively (1) 0.72, 1.62 (2) 0.72, 0.72 (3) 1.2, 1.2 (4) 1.62, 0.72
33. Screening by inner electrons will be more effective in (1) Mg (2) K (3) Sr (4) Cs
34. The first ionisation enthalpies of the elements C, N, P, and Si are in the order of (1) C < N < Si < P (2) N < Si < C < P (3) Si < P < C < N (4) P < Si < N < C
35. The correct option for the 1 st ionisation energy (kJ/mol) of Li, Na, K, and Cs, respectively is (1) 496, 520, 419, 374 (2) 374, 419, 496, 520 (3) 520, 496, 419, 374 (4) 374, 419, 520, 496
36. The correct order of increasing first ionisation energy is
(1) Ca < K < Ne < P < F
(2) F < Ca < Ne < P < K
(3) K < Ca < P < F < Ne (4) Ne < F < P < Ca < K
37. Which of the following is the correct order of ionisation energy?
i) Be+ > Be ii) Be > Be+ iii) C > Be iv) B > Be
(1) ii, iii (2) iii, iv (3) i, iii (4) i, ii
38. Which of the order for ionisation energy is correct?
39. The five successive ionisation energies of an element ‘X’ are 800, 1427, 2658, 25024, and 32824 kJ mol−1, respectively. The valency of ‘X’ is
(1) 1 (2) 2 (3) 3 (4) 4
40. The first, second, and third ionisation energies are 737, 1045, and 7733 kJ/mol, respectively. The element can be (1) Na (2) B (3) Al (4) Mg
41. The element with highest IP1 in the modern periodic table is (1) Cl (2) He (3) N (4) O
42. The group of elements with highest second ionisation energy is
(1) II A (2) VIII A (3) VII A (4) I A
43. The correct order of electron gain enthalpy is
(1) O > S > Se > Te (2) Te > Se > S > O (3) S > O > Se > Te (4) S > Se > Te > O
44. Which of the following is the correct order of electron affinity?
(1) I > Br > F > Cl (2) F > Cl > Br > l
(3) Br > I > F > Cl (4) Cl > F > Br > l
45. Which of the following has the highest electron gain enthalpy?
(1) [Ne]3s23p3 (2) [Ne]3s23p4
(3) [Ne]3s23p5 (4) [Ne]3s23p2
46. The correct order of electron affinity is (1) Be < B < C < N (2) Be < N < B < C
(3) Be < N < B < C (4) N < C < B < Be
47. Energy is liberated in which of the following processes?
(1) Cl (g)→ Cl+ (g) +e–
(2) HCl(g) → H+ (g) +Cl– (g)
(3) Cl (g) + e– → Cl– (g)
(4) O– (g) + e– → O2– (g)
48. What is the value of electron gain enthalpy of Na+ if IE1 of Na is 5.1 eV?
(1) –10.2 eV (2) –5.1 eV
(3) +2.55 eV (4) + 10.2 eV
49. 3.55 g of Cl (g) liberates Q kJ of heat by gaining electrons. Electron gain enthalpy of chlorine is ______ kJ/mol.
(1) –5 Q (2) –10 Q
(3) –20 Q (4) –2.5 Q
50. The electronegativity of the following elements increases in the order
(1) S < P < N < O (2) P < S < N < O
(3) N < O < P < S (4) N < P < S < O
51. Which list includes elements in order of increasing metallic character?
(1) Si, P, S (2) As, P, N
(3) Al, Ge, Sb (4) Br, Se, As
52. Maximum valency with respect to hydrogen shown by a third period element is (1) 3 (2) 4
(3) 7 (4) 6
53. Among a) Na2O, b) MgO, c) Al2O3 , d) P2O5 , and e) Cl2O7 the most basic, most acidic,
56 and amphoteric oxides, respectively, are (1) a, b, c (2) b, e, c (3) a, e, c (4) e, c, a
54. Among Al 2 O 3 , SiO 2 , P 2 O 3 , and SO 2 , the correct order of acid strength is (1) SO2< P2O3< SiO2< Al2O3 (2) SiO2 < SO2< Al2O3 < P2O3 (3) Al2O3 < SiO2< SO2< P2O3 (4) Al2O3< SiO2< P2O3< SO2
55. In which pair of oxides, both are acidic in nature?
(1) CaO, SiO2 (2) B2O3, SiO2 (3) B2O3, CaO (4) N2O, BaO
56. The diagonal relationship phenomenon is not observed after
(1) I A group (2) II A group (3) III A group (4) IV A group
57. Diagonal relationship is present between the lighter elements of ____ and ___periods.
59. The IP1, IP2, IP3, IP4 , and IP5 of an element are 7.1, 14.3, 103.4, 166.8, and 262.2 eV, respectively. The element belongs to ___ group.
60. The element with highest electron affinity belongs to period number _____.
61. The period with maximum electronegativity belongs to _____ .
CHAPTER 4: Classification of Elements
Multiple Concept Questions
Single Option Correct MCQs
62. Which of the following is an incorrect set?
Group Period
(1) Z = 38 IIA 5th
(2) Z =58 IIIB 6th
(3) Z = 25 VIIB 4th
(4) Z = 22 VIB 4th
63. The element with atomic number 56 is likely to have the same outer shell configuration as the element with atomic number .
(1) 12 (2) 19
(3) 14 (4) 22
64. The electronic configuration of seventh element in lanthnoid series is (1) [Xe]4f35d56s2 (2) [Xe]4f75d26s1 (3) [Xe]4f75d16s2 (4) [Xe]4f35d66s2
65. The element having electronic configuration [Xe]4f145d06s2 belongs to (1) d-block, 12th group (2) f-block, 3rd group (3) s-block, 2nd group (4) f-block, 12th group
66. Atomic radius depends upon i) number of bonds formed by the atom ii) nature of the bonding iii) oxidation state of the atom
(1) i and ii (2) ii and iii
(3) i and iii (4) i, ii, and iii
67. The pair that has similar atomic radii is ? (1) Sc and Ni (2) Ti and Hf (3) Mo and W (4) Mn and Rc
68. A sudden jump between the values of second and third ionisation energies of an element is associated with configuration (1) 1s22s22p63s1 (2) 1s22s22p63s23p1 (3) 1s22s22p63s23p2 (4) 1s22s22p63s2
69. Which of the following elements have electron affinity greater than fluorine?
(1) O (2) S (3) Se (4) Cl
70. The correct order of acidic strength is (1) K2O>CaO>MgO (2) CO2>N2O5>SO3 (3) Na2O>MgO>Al2O3 (4) Cl2O7>SO2>P4O10
72. From the given compounds, number of compounds that give acidic solution when dissolved in water is __.
CaO, SO 2, SO 3, Fe 2O 3, Cl 2O 7, CO 2, NaO, NO2
73. An element ‘X’ belong to 4th period and 4th group in the modern periodic table. Then, the atomic number of ‘X’ is ___.
74. How many of the following exist with +2 as the most common oxidation state?
Al, Si, Pb, B, Zn, Ca, Ba, Cr, Sc, and N
75. Common oxidation state of inner transition elements is ___.
76. Atomic number of fifth member of noble gas is ____.
77. The number of elements present in the group-3 in the modern periodic table is ____.
78. CO2, N2O, BaO, Cl2O7, Al2O3, N2O3, NO, CaO in above oxides, the number of amphoteric oxides ____.
79. In a period, the element with largest atomic volume belongs to which of the groups
Level III
Single Option Correct MCQs
1. Which of the elements whose atomic numbers are given below, cannot be accommodated in the present set up of the long form of the periodic table?
(1) 107 (2) 118
(3) 126 (4) 102
2. The period number in the long form of the periodic table is equal to
(1) magnetic quantum number of any element of the period
(2) atomic number of any element of the period
(3) maximum principal quantum number of any element of the period
(4) maximum azimuthal quantum number of any element of the period
3. What is the position of the element in the periodic table satisfying the electronic configuration (n-1)d1ns2 for n=4?
(1) 3rd period and 3rd group
(2) 4th period and 4th group
(3) 3rd period and 2nd group
(4) 4th period and 3rd group
4. Choose the statement that is not correct for periodic classification of elements is:
(1) The properties of elements are periodic function of their atomic numbers.
(2) Non-metallic elements are Iess in number than metallic elements.
(3) For transition elements, the 3d-orbitals are filled with electrons after 3p-orbitals and before 4s-orbitals.
(4) The first ionisation enthalpies of elements generally increase with increase in atomic number as we go along a period.
5. The first ionisation potentials of Na, Mg, and Si, respectively, are 496, 737, and 786 kJ/mol. The first ionisation potential of ‘Al’ in kJ/mol is
(1) 577 (2) 487
(3) 856 (4) 768
6. Consider the following changes:
A(g)→A+(g) + e– is E1 and A+(g)→A2+(g)+e– is E2
The energy required to pull out the two electrons are E 1 and E 2, respectively. The correct relationship between two energies would be
(1) E1<E2 (2) E1>E2
(3) E1=E2 (4) E1≥ E2
7. Elements X, Y, and Z have atomic numbers 19, 37, and 55, respectively. Which of the following statements is true about them?
(1) Their ionisation potential would increase with increasing atomic number.
(2) Y would have an ionisation potential between those of X and Z.
(3) Z would have the highest ionisation potential.
(4) Y would have the highest ionisation potential.
8. Choose the correct statement about an element having electronic configuration 1s22s22p63s23p63d54s2
(1) It belongs to s-block.
(2) Its highest oxidation state is +7 in its compounds.
(3) It is a non-metal.
(4) It belongs to 3 rd period of modern periodic table.
9. Ionisation energy values of an atom are 495, 767, 1250, and 4540 kJ mol-1. The formula of its sulphate is (1) MSO4 (2) M2SO4 (3) M2(SO4)3 (4) M(SO4)2
10. Which of the following orders is not correct?
(1) IE(I) of Be > IE(I) of B but IE(II) of Be < IE(II) of B
(2) IE(I) of Be < IE(I) of B but IE(II) of Be < IE(II) of B
(3) IE(II) of O > IE(II) of N
(4) IE(I) of Mg > IE(I) of Al
11. Screening effect influences
A) atomic radius
B) ionisation enthalpy
C) electron gain enthalpy
(1) A, B only (2) B, C only
(3) A, C only (4) A, B, and C
12. Which of the following is incorrect?
(1) Cesium is the most electropositive element while F is the most electronegative element.
(2) Chlorine has the highest –ve electron gain enthalpy of all the elements.
(3) Electron gain enthalpy of N as well as that of noble gases is positive.
(4) In any period, the atomic radius of the noble gas is lowest.
13. Electronic configuration of four elements
A, B, C, and D are given below:
(A) 1s22s22p6 (B) 1s22s22p4
(C) 1s22s22p63s1 (D) 1s22s22p5
Which of the following is the correct order of increasing tendency to gain electron?
(1) A < C < B < D (2) A < B < C < D
(3) D < B < C < A (4) D < A < B < C
14. Fluorine has the highest electronegativity among the ns 2np 5 group on the Pauling scale, but the electron affinity of fluorine is less than that of chlorine because
(1) the atomic number of fluorine is less than that of chlorine
(2) fluorine, being the first member of the family, behaves in an unusual manner.
(3) fluorine can accommodate an electron better than chlorine by utilising its vacant 3d–orbital
(4) small size, high electron density and an increased electron repulsion makes addition of an electron to fluorine less favourable than in the case of chlorine in isolated state
15. A, B, and C are hydroxyl compounds of the elements X, Y, and Z, respectively. X, Y and Z, are in the same period of the periodic table. A gives an aqueous solution of pH less than seven. B reacts with both strong acids and strong alkalies. C gives an aqueous solution which is strongly alkaline.
Which of the following statements is/are true?
I: The three elements are metals.
II: The electronegativity decreases from X to Y to Z.
III: The atomic radius decreases in the order X, Y, and Z.
IV: X, Y, and Z could be phosphorus, aluminium, and sodium, respectively.
(1) I, II, and III are correct.
(2) I and III are correct.
(3) II and IV are correct.
(4) II, III, and IV are correct.
16. Which of the following statements are not correct?
1) The electron gain enthalpy of ‘F’ is more negative than that of ‘Cl’
2) Ionisation enthalpy decreases in a group of periodic table.
3) The electronegativity of an atom depends upon the atoms bonded to it.
4) Al 2 O 3 and NO are examples of amphoteric oxides.
(1) 1, 3, and 4
(2) 1, 2, and 3
(3) 3 only
(4) 1, 2, 3, and 4
17. Which of the following is not true about Pauling scale electronegati vity
(1) ()0.1017/ AB XXkJmol −=∆
(2) 2.8 mulliken pauling X X =
(3) reference element for determination of EN is hydrogen
(4) ()0.208/ AB XXkJmol −=∆
18. Percentage of ionic character in the covalent bond A-B is (given XA= 2, XB= 3)
(1) 12.5% (2) 30%
(3) 19.5% (4) data is insufficient
19. In which of the following arrangements the order is not according to the property indicated against it?
(1) Al3+<Mg2+<Na+<F– increasing ionic size
(2) B<C<O<N increasing first ionisation energy
(3) I<Br<F<Cl increasing electronegativity
(4) Li<Ca<Al<Si valency with respect to hydrogen
20. Which of the following statements is incorrect?
(1) H+ is the smallest cation in the periodic table.
(2) van der Waals radius of chlorine is more than its covalent radius.
(3) Ionic mobility of hydrated Li+ is greater than that of hydrated Na +
(4) He has the highest ionisation enthalpy in the periodic table.
21. Consider the following points:
(a) Cs is the strongest reducing agent among IA group elements (in aqueous solution).
(b) Be(OH)2 is amphoteric.
(c) The density of potassium is less than sodium.
(d) Among the alkali metals Li, Na, K, and Rb, lithium has the maximum value of melting point.
Choose the correct statements
(1) (a), and (b) are correct
(2) (a), (b), and (c) are correct
(3) (b), and (c) are correct
(4) (b), (c), and (d) are correct
22. Which of the following statements about 4BF and 3 6AlF are correct?
(a) B and Al differ in their oxidation states.
(b) B, Al differ in their covalency.
(c) B obeys octet rule.
(d) B and Al are in diagonal relationship.
(1) a, b (2) b, c, d
(3) a, b, c (4) b, c
23. Which of the following properties is true regarding diagonal relationship between Li and Mg?
(1) Li and Mg salts do not impart colour in oxidising flame.
(2) Li or Mg nitrate liberates only one paramagnetic gas on heating.
(3) Their bicarbonates are highly stable.
(4) Their carbonates decompose on heating to form a gas whose molecule has net zero dipole moment.
Numerical Value Questions
24. Calculate the electronegativity of fluorine from the following data. Express your answer to the nearest possible integer.
(E = bond dissociation energy and X= electronegativity)
EH–H=104.2 k cal/mol
EF–F=36.6 k cal/mol
EH–F=13.4 k cal/mol
XH=2.1
25. How many of the following factors influence the electronegativity of the atom of an element?
i) Effective nuclear charge
ii) Screening effect of inner electrons
iii) Size of the atom.
iv) Oxidation state of the atom in bond
v) Type of hybridisation of the atom
vi) Strength of bond in which it is participated
vii) Number of surrounding atoms on the other atom with which it is bonded
viii) Nature of bond
26. How many of the following orders/ statements are correct?
a) IP1of nitrogen is not less than that of oxygen.
CHAPTER 4: Classification of Elements
b) IP1 of Pb is less than that of Sn.
c) Electronegativity order.
d) IP2 order B > C > Be.
e) Electron gain enthalpy order Ne > Ar > Kr > Xe > He > Rn.
f) Electron affinity of Be+4 is 217.6 eV/ion nearly.
g) Electronegativities of K and Rb are same
h) Order of sizes Tl > In > Al > Ga > B
i) Official IUPAC name of element with atomic number 104 is Unnilquadium
27. A metal has electronic configuration [Ar]183d74s2. On the basis of this electronic configuration, find out the group number of metal.
28. The ionisation energy and electron affinity of an element are 12.0 eV and 3.8 eV, respectively. Its electron negativity on Milliken scale is __.
29. Among the following species, how many have their ionic size greater than O 2–?
Se2–, F–, N3–, P3–
THEORY-BASED QUESTIONS
Statement Type Questions
Each question has two statements. Statement I and statement 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 correct but statement II is incorrect.
(4) if statement I is incorrect but statement II is correct.
1. S-I : 5th period of the long form periodic table is the longest period.
S-II : The longest period of the periodic table includes 32 elements.
2. S-I : van der Waals’ radius of an element is always larger than its c ovalent radius.
S-II : Two molecules can never be closer than two atoms in a molecule.
3. S-I : In Cl2 molecule, the covalent radius is double the atomic radius of chlorine.
S-II : Radius of anionic species is always greater than their parent atomic radius.
4. S-I : Ionisation enthalpy difference from B to Al is more than that of Al to Ga.
S-II : Ga has completely filled d-orbital.
5. S-I : Electron gain enthalpy becomes less negative as we go down the group.
S-II : Size of atom increases on going down the group and the addition of electron would be farther from the nucleus.
6. S-I : Li and Mg show similar chemical properties.
S-II : Li + and Mg 2+ have nearly same polarising power.
7. S-I : For transition elements, on moving from left to right in a transition series, ionisation energy increases.
S-II : As the atomic number increases, the effective nuclear charge also increases.
8. S-I : Metallic or electropositive character of elements increases as the value of ionisation potential decreases.
S-II : In a group, moving from top to bottom, metallic or electropositive character increases.
9. S-I : The first IP of nitrogen is greater than oxygen, while the reverse is true for the second IP values.
S-II : Oxygen unipositive ion has a halffilled electronic configuration.
Assertion and Reason Type Questions
In each of the following questions, a statement of Assertion (A) is given, followed by a corresponding statement of
Reason (R). Mark the correct answer as
(1) if both (A) and (R) are true and (R) is the correct explanation of (A)
(2) if both (A) and (R) are true but (R) is not the correct explanation of (A)
(3) if (A) is true but (R) is false
(4) if both (A) and (R) are false
10. (A) : According to modern periodic law, the physical and chemical properties of elements are periodic functions of their atomic numbers.
(R) : Atomic number is equal to the number of electrons present in any species (atom or ion).
11. (A) : ‘He’ and ‘Be’ both have the same outer electronic configuration, like ns2 type.
(R) : Both are chemically inert.
12. (A) : Mg2+ and Al3+ are isoelectronic but ionic radius of Al3+ is less than that of Mg2+
(R) : The effective nuclear charge on the outer shell electrons in Al3+is more than that in Mg2+
13. (A) : The first ionisation enthalpy of 3d series elements is more than that of group 2 metals.
(R) : In 3d series of elements, successive filling of d-orbitals takes place.
14. (A) : The first ionisation enthalpy for oxygen is lower than that of nitrogen.
(R) : The four electrons in 2p orbitals of oxygen experience more electron–electron repulsion due to small size.
15. (A) : The first ionisation enthalpy for oxygen is lower than that of nitrogen.
(R) : The four electrons in 2p orbitals of oxygen experience more electron–electron repulsion.
16. (A) : Helium has the highest value of ionisation energy among all the elements known.
(R) : Helium has the highest value of electron affinity among all the elements known.
17. (A) : The first electron gain enthalpies for all the elements is exothermic.
(R) : Electron gain enthalpy is the amount of energy accompanied when an electron is added to neutral isolated atom.
18. (A) : Lithium, having maximum negative E° (SRP) value, is the strongest reducing agent among all alkali metals in solution.
(R) : Lithium is the lightest metal in the periodic table.
19. (A) : Both Be and Al exhibit similar properties in their compounds.
(R) : Due to almost the same polarising power of cations, some pairs of elements, like Be and Al, are diagonally related in their properties of compounds.
20. (A) : Tl, in its +1 oxidation state, is more stable than its +3 oxidation state.
(R) : Tl exhibits inert pair effect.
21. (A) : The decreasing order of acidic character of CO2, N2O5, SiO2, and is SO3>N2O5>CO2>SiO2
(R) : As the electronegativity difference between the central atom and oxygen decreases, the acidic character of the oxide increases.
JEE ADVANCED LEVEL
Multiple Option Correct MCQs
1. The electron affinity of nitrogen is very low, which can be attributed to
(1) ‘N’ atom has stable filled 2p 3 configuration
(2) addition of electron to 2p orbital increase spin paired repulsion
(3) energy released due to attraction between the nucleus and incoming electron is offset by the inter-electron repulsion by pairing
(4) due to small size, inter-electron repulsions are more than the other elements in the group
2. Consider the following ionisation steps:
()() ;100; gg MMeHeV +− →+∆=+
()() 2 2;250 gg MMeHeV +− →+∆=+
select correct statement(s):
(1) IE1 of M(g) is 100 eV
(2) IE1 of M+ (g) is 150 eV
(3) IE2 of M (g) is 250 eV
(4) IE2 of M (g) is 150 eV
3. The correct order of radii is (1) Mg+2<Li+2<N–3 (2) Li<Na<K (3) Fe+2<Fe+3<Fe+4 (4) Be>B>N
4. Which of the following statements is/are incorrect?
(1) IE 2 of group VA<IE 2 of group VI elements.
(2) IE2 of group IA<IE2 of group II elements.
(3) IE2 of group VIA<IE2 of group VII A elements.
(4) IE2 of 4d series <IE1 of 5d series.
5. Which of the following have a diagonal relationship?
(1) Li→Mg (2) B→Mg (3) Be→Al (4) Be→Na
6. In which of the following arrangements, the order is not correct according to the property indicated against it?
(1) Increasing size: Cu2+<Cu+<Cu
(2) Increasing IE1: B<C<N<O
(3) Increasing IE1: Na<Al<Mg<Si
(4) Increasing IE1: Li<Na<K<Rb
7. Which of the following orders of atomic / ionic radius is/are correct?
(1) I– > I >I+ (2) Mg2+>Na+>F–(3) P5+<P3+ (4) Li>Be>B
9. Generally, the ionisation potential in a period increases but there are some exceptions. Such as (1) Be and B (2) N and O (3) Mg and Al (4) Na and Mg
10. Select the correct statement(s)
(1) Sulphur has a lower electron affinity than chlorine.
(2) Iodine has a lower electron affinity than bromine.
(3) Boron has a lower first ionisation energy than Be.
(4) Sulphur has a lower first ionisation energy than phosphorus.
11. Which of the following sequences is/are incorrect with respect to the property indicated?
(1) Oxidising power: F2<Cl2>Br2
(2) Bond energy: F2>Cl2>Br2
(3) Electronegativity: F>Cl>Br
(4) Electron affinity: F>Cl>Br
12. If 0 2 N atoms of X(g) are converted into X+(g) by absorbing energy E1, 0 2 N atoms of X are converted to X–, and E2 energy is released, then,
(1) ionisation potential of X would be 1 0 2 E N
(2) ionisation potential of X would be 2E 1
(3) electron affinity of X would be 2 2 0 E N
(4) electron affinity of X would be 2E 2
13. Consider three elements with the following abbreviated electronic configurations: X=[Ar]4s2; Y=[Ne]3s23p4; Z=[Ar]3d104s24p4
Then, which of the following statements is/are correct?
(1) X is metal; Y is non-metal.
(2) The element having highest atomic size is X.
(3) The element having highest ionisation energy is Z.
(4) The element having highest electron affinity is Y.
14. The covalent radius of an element depends on
(1) number of bonds formed between atoms of that element
(2) type of hybridisation involved by its atoms in the covalent bond formation
(3) ionic character of the covalent bond formed by its atom with the atoms of other element
(4) oxidation state of the atom in its covalent compounds
15. Ionisation energy of atoms A and B are 350 and 250 kcal/mol, respectively. If the electron affinity of these atoms are 70 kcal/ mol and 90 kcal/mol, respectively, then
(1) electron cloud is more attracted by A
(2) electron cloud is more attracted by B
(3) electronegativity of A is more than B
(4) electronegativity of A is less than B
16. Which of the following is/are true order(s)?
(1) B+<B<B– → size
(2) I<Br<Cl<F→ electron affinity
(3) O2–<O–<O+→Zeffective
(4) Na<Al<Mg<Si→ ionisation potential
17. The first( D r H1 ) and the second ( D r H2 ) ionisation enthalpies (in kJ mol–1), and the ( D egH) electron gain enthalpy (in kJmol –1) of a few elements are given below:
Elements D r H1 D r H1 D eg H
I 520 7300 –60
II 419 3051 –48
III 1681 3374 –328
IV 1008 1846 –295
V 2372 5251 +48
VI 738 1451 –40
Choose the correct option(s):
(1) The most reactive metal is II.
(2) III and IV are both non-metals, III being the most reactive non-metal.
(3) VI is a metal, and it can form a stable binary halide of the formula MX (X=halogen).
(4) V is the most unreactive element.
Numerical Value Questions
18. The electronic configuration of thorium is [Rn]5fx6dy7sz. What is the value of y+z–2x?
19. IP1 and IP2 of Mg are 178 and 348 kcal mole-1 , The energy required for the reaction. Mg → Mg2++2e– is ____ kcal.
20. The amount of energy released when 10 6 atoms of iodine in vapour state are converted to I – ions is 4.9 ×10 –13 J. The electron affinity of iodine in eV per atom is___.
21. Among the following, the total number of orders that are correct with respect to the property indicated against each is
(i) Mg>Al>Si>P: covalent radius
(ii) Na+<O2-<F-<N3-: ionic size
(iii) Al3+<Mg2+<Li+<K+: ionic size
(iv) C<Si>P>N: electron affinity value
(v) N<C<O<F: electron affinity value
(vi) F>Cl>Br>I: electron affinity value
(vii) Si>Mg>Al>Na: first ionisation energy
(viii) O>F>N>C: second ionisation energy
(ix) N>P>Sb>As: third ionisation energy
CHAPTER 4: Classification of Elements
22. The number of species having higher first ionisation energy than Ca from the following is ___.
Ge, Ga, Br, Se, Kr, As, K
23. The number of pairs, in which electron affinity of the second element is more than that of the first element is ____.
(F, Cl) (C, N) (O, N) (F, Ne) (B, C), (O, S)
24. If the atomic radius of non-metal bromine is 1.14 Ao , its covalent radius is ____ A o .
25. Zeff for the last electron in carbon atom is __________.
26. Number of possible ionisation potential values for C-atom is ____.
Integer Value Questions
27. Covalent bond length of chlorine molecule is 2 Ao. The covalent radius of chlorine is __ Ao
28. The first four ionisation energy values of an element are 191, 578, 872, and 8982 kcal. The number of chlorine atoms per molecule of the compound formed by it is ___.
29. Ionisation potential and electron affinity of an element are 17.92 eV and 4.48 eV, respectively. The electronegativity of the element is ___.
30. Common oxidation state of d-block elements is +x. The value of x is ____.
31. In how many pairs, first species has lower ionisation energy than second species?
(i) N and O
(ii) Br and K
(iii) Be and B
(iv) I and I–
(v) Li and Li+
(vi) O and S
(vii) Ba and Sr
32. How many of the following elements have positive or zero electron gain enthalpy values?
Be, Mg, B, N, He, Ne, O, P
33. The bond energies of H–H; X–X, and H–X are 104 kcal, 38 kcal, and 138 kcal, respectively. The electronegativity of X i s
Y. Then, find Y 3.8. Given: () 67=8.18
Passage-based Question
(Q: 34-35)
In the modern periodic table, the elements are placed in order of increasing atomic number. There have been numerous designs of the table over the years but the most common is the long form of periodic table. The long form of periodic table shows all the elements in numerical order.
34. The atomic number of second element in period-3 is
(1) 10 (2) 20 (3) 12 (4) 13
35. The number of d-electrons in copper atom is
(1) 10 (2) 12 (3) 9 (4) 4
(Q: 36-38)
In the modern periodic table, the elements are placed in order to increasing atomic number. There have been numerous designs of the table over the years but the most common is the long form of periodic table. The long form of periodic table shows all the elements in numerical order.
36. What is the atomic number of the (as yet undiscovered) alkali earth metal after radium?
(1) 120 (2) 121 (3) 124 (4) 118
37. In which of the following inert gases electrons are occupying in 4f-orbitals but
no electron in 6d-orbitals in ground state electronic configuration?
(1) Kr (2) Xe (3) Rn (4) Uuo
38. Total number of S-block elements in the periodic table is ___ (1) 14 (2) 12 (3) 13 (4) 18
(Q: 39-40)
Comprehension given below is followed by some multiple choice questions. Each question has one correct option. Choose the correct option.
Ionisation energies of three hypothetical elements are given below (in kJ/mole): I II III X 122 340 1890 Y 99 931 1100 Z 118 1220 1652
39. What is the value (magnitude only) of ()()() 2 gg ZeZkJ +−+ +→ ?
40. Energy (in kJ/mol) required for the process ()() 2 2 gg ZZe+−→+ will be _____.
(Q: 41-42)
The tendency of the atom of an element to attract the shared electron pairs, more towards itself in a hetero nuclear diatomic molecule or in a polar covalent bond is called electronegativity. Pauling calculated the electronegativities of different elements from the bond energies using the equation:
0.208/ AB xxkcalmol −=∆
41. Calculate the electronegativity of chlorine from the bond energies of ClF(66 kcal/mol), F2(40 kcal/mol), and Cl2(62.5 kcal/mol).
42. From the bond energies of HF, H 2, and F2 are 135, 100, and 36 kcal/mol, respectively. Calculate the extra ion resonance energy in the HF molecule.
Matrix Matching Questions
43. Match the Column I (element) with Column II (group to which the element belongs)
Column I (Element) Column II (Group to which element belongs to)
A. N I) VI A group
B. F II) VA group
C. O III) VII A group
D. C IV) IV A group
Choose the correct answer from the options given below.
(A) (B) (C) (D)
(1) II III IV I
(2) II III I IV
(3) III II I IV
(4) I `II III IV
44. Match Column I (electronic configuration) with Column II (element)
Column I (Electronic configuration) Column II (Element)
A. 1s2,2s2,2p2,3s2,3p6,4s1 I) d-block element
B. 1s2,2s2,2p2,3s2,3p6 II) Halogen
C. 1s2,2s2,2p2,3s2,3p6, 3d6,4s2 III) Alkali metal
D. 1s2,2s2,2p5 IV) Noble gas
Choose the correct answer from the options given below.
(A) (B) (C) (D)
(1) I II III IV
(2) III IV I II
(3) I III II IV
(4) II IV III I
45. Match the Column I(atomic number) with Column II(period).
Column I (Atomic number)
Column II (Period)
A. 31 I) 5
B. 50 II) 3
C. 56 III) 4
D. 14 IV) 6
Choose the correct answer from the options given below.
(A) (B) (C) (D)
(1) I II III IV
(2) II I IV II
(3) III IV I II
(4) III I IV II
46. Match Column I(element) with Column II (atomic radii).
Column I (Element)
Column II (Atomic radii in pm)
A. O I) 88
B. C II) 74
C. B III) 66
D. N IV) 77
Choose the correct answer from the options given below.
(A) (B) (C) (D)
(1) IV III II I
(2) I IV III II
(3) III IV I II
(4) II I IV III
47. Match the Column I(Oxide) with Column II (Nature)
Column I (Oxide)
Column II (Nature)
A. MgO I) Acidic
B. BeO II) Neutral
C. Cl2O7 III) Basic
D. CO IV) Amphoteric
Choose the correct answer from the options given below:
(A) (B) (C) (D)
(1) II IV I III
(2) II III I IV
(3) II III IV I
(4) III IV I II
48. Match the List I (oxide) with List II (nature)
List I (Oxide) List II (Nature)
A. N2O5 I) Amphoteric
B. BaO II) Basic
C. BeO III) Neutral
D. NO IV) Acidic
Choose the correct answer from the options given below.
(A) (B) (C) (D)
(1) IV III I II
(2) IV II I III
(3) II IV III I
(4) I II III IV
49. Match Column I (Element) with Column II (Characteristic property)
Column I (Element) Column II (Characteristic property)
A. Li I) Forms acidic oxide
B. P II) Has diagonal relation with Mg
C. Be III) Has highest electronegativity among all elements and exhibits only one oxidation state
D. F IV) Electronegativity is equal to Al
Choose the correct answer from the options given below.
(A) (B) (C) (D)
(1) II I IV III
(2) II III I IV
(3) III I II IV
(4) II I IV III
FLASH BACK (Previous JEE Questions)
JEE Main
1. Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R: (2024)
Assertion (A) : The first ionisation enthalpy decreases across a period.
Reason (R) : The increasing nuclear charge outweighs the shielding across the period.
In the light of the above statements, choose the most appropriate from the options given below:
(1) Both A and R are true and R is the correct explanation of A
(2) A is true but R is false
(3) A is false but R is true
(4) Both A and R are true but R is NOT the correct explanation of A
2. The element having the highest first ionization enthalpy is (2024)
(1) Si (2) Al
(3) N (4) C
3. Given below are two statements:
Statement I : Fluorine has most negative electron gain enthalpy in its group.
Statement II : Oxygen has least negative electron gain enthalpy in its group.
In the light of the above statements, choose the most appropriate from the options given below.
(1) Both Statement I and Statement II are true
(2) Statement I is true but Statement II is false
(3) Both Statement I and Statement II are false
(4) Statement I is false but Statement II is true
4. Given below are two statements: (2024)
Statement I : Along the period, the chemical reactivity of the element gradually increases from group 1 to group 18.
Statement II : The nature of oxides formed by group 1 element is basic while that of group 17 elements is acidic.
In the the light above statements, choose the most appropriate from the questions given below:
(1) Both Statement I and Statement II are true
(2) Statement I is true but Statement II is false
(3) Statement I is false but Statement II is true
(4) Both Statement I and Statement II are false
5. The first ionisation enthalpies of Be, B, N, and O follow the order (2022)
(1) O < N < B < Be (2) O < N < B < Be
(3) B < Be < N < O (4) B < Be < O < N
6. The IUPAC nomenclature of an element with electronic configuration [Rn]5f146d17s2 is (2022)
(1) unnilbium
(2) unnilunium
(3) unnilquadium
(4) unniltrium
7. Given below are two statements
Statement I : In Cl2 molecule the covalent radius is double the atomic radius of chlorine.
Statement II : Radius of anionic species is always greater than their parent atomic radius.
In light of the above statements, choose the most appropriate answer from the options given below:
(2022)
(1) Both statement I and statement II are correct.
(2) Both statement I and statement II are incorrect.
(3) Statement I is incorrect but statement II is correct.
(4) Statement I is correct but statement II is incorrect.
8. The electronic configuration of Pt (atomic number 78) is
(1) [Xe]4f145d96s1
(2) [Kr]4f145d10
(3) [Xe]4f145d10
(4) [Xe]4f145d86s2
9. Give below are two statements.
One is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A) : The ionic radii of O2- and Mg2+ are same.
Reason (R) : Both are isoelectronic species.
In light of the above statements, choose the correct answer from the options given below.
(2022)
(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) (A) is false but (R) is true
10. Which of the following statements are not correct?
A. The electron gain enthalpy of F is more negative than that of Cl.
B. Ionisation enthalpy decreases in a group of periodic table.
C. The electronegativity of an atom depends upon the atoms bonded to it.
D. Al 2 O 3 and NO are examples of amphoteric oxides.
Choose the most appropriate answer from the options given below. (2022)
(1) A, B, C, and D
(2) A, C, and D only
(3) A, B, and D only
(4) B and D only
11. Given below are two statements:
Statement I: The decrease in first ionisation enthalpy from B to Al is much larger than that from Al to Ga.
Statement II : The d orbitals in Ga are completely filled. (2022)
In light of the above statements, choose the most appropriate answers from the options given below.
(1) Statement I is incorrect but statement II is correct.
(2) Both the statements I and II are incorrect.
(3) Both the statements I and II are correct.
(4) Statement I is correct but statement II is incorrect .
CHAPTER TEST – JEE MAIN
Section - A
Single Option Correct MCQ's
1. The triad not present in group VIII of Mendeleef table is
(1) Li, Na, K (2) Fe, Co, Ni (3) Ru, Rh, Pd (4) Os, Ir, Pt
2. The IUPAC symbol for the element with atomic number 119 would be (1) Unh (2) Uue (3) Uun (4) Une
3. The number of elements present in 2 nd , 3rd, 4th, and 5th periods of modern periodic table, respectively are
(1) 2, 8, 8, and 18 (2) 8, 8, 18, and 32
(3) 8, 8, 18, and 18 (4) 8, 18, 18, and 32
4. Following are some statements about modern periodic table. Pick the correct statement(s).
i) It consists of s, p, d, and f- blocks.
ii) The energy levels filling order in 6th period is 6s, 4f, 5d, and 6p.
iii) IIIA group contains maximum number of elements.
(1) i and ii (2) only i (3) ii and iii (4) i, ii, and iii
5. Which one of the following is correct order of the size?
(1) I–>I>I+ (2) I>I–>I+ (3) I>I+>I– (4) I+>I–>I
6. In which of the following sets, elements have nearly same atomic radii?
(1) Li, Be, B (2) Mg, Ca, Sr (3) Fe, Co, Ni (4) O, S, Se
7. The lanthanoid contraction is responsible for the fact that
(1) Zr and Y have about the same radius
(2) Zr and Nb have similar oxidation state
(3) Zr and Hf have about the same radius
(4) Zr and Zn have same oxidation state
8. Consider the following ionisation reaction: I.E (kJ/mol) I.E (kJ/mol) A(g)→A+(g)+e–,A1 B(g)→B+(g)+e–,B1 B+(g)→B+2(g)+e–,B2 C(g)→C+(g)+e–,C1 C+(g)→C+2(g)+e–,C2 C+2(g)→C+3(g)+e–,C3
If uni-positive ion of A, di-positive ion of B
and tri-positive ion of C have zero electron. Then, incorrect order of corresponding IE is
(1) C3>B2>A1 (2) B1<A1>C1
(3) C3<C2>B2 (4) B2<C2>A1
9. In which of the following arrangement the order is incorrect according to the property indicated against it?
(1) Al+3<Mg2+<Na+<F– = increasing ionic size
(2) B<C<N<O= increasing first ionisation enthalpy
(3) I<Br<F<Cl= increasing electron gain enthalpy (with negative sign)
(4) Li<Na<K<Rb= increasing metallic radius
10. Which one of the following arrangements of the incorrect representation of the property indicated with it?
(1) Br<Cl<F: electronegativity
(2) F<Br<Cl: electron affinity
(3) F2<Br2<Cl2: bond energy
(4) Br2<Cl2<F2: oxidising strength
11. The elements ‘X’, ‘Y’, and ‘Z’ form oxides, which are acidic, basic, and amphoteric, respectively.The correct order of their electronegativity is
(1) X > Y > Z (2) Z > Y > X
(3) X > Z > Y (4) Y > X > Z
12. Calculate the percentage ionic character for molecule AB when electronegative difference is 2.0.
(1) 46% (2) 36% (3) 30% (4) 50%
13. Which of the following oxides is amphoteric?
(1) CrO (2) Cr2O3 (3) CrO3 (4) CrO5
14. Which of the following orders presents the correct sequence of the increasing basic nature of the given oxides?
CHAPTER 4: Classification of Elements
(1) Al2O3<MgO<Na2O<K2O
(2) MgO<K2O<Al2O3<Na2O
(3) Na2O<K2O<MgO<Al2O3
(4) K2O< Na2O<Al2O3<MgO
15. An element A of group-1 shows similarity with an element B belonging to group-2. If A has maximum hydration enthalpy in group-1, then B is ___.
(1) Mg (2) Be
(3) Ca (4) Sr
16. The ionisation potential and electron affinity of fluorine are 17.42 and 3.45 eV, respectively. Calculate the electronegativity of fluorine in Mulliken scale
(1) 3.726 (2) 4.726
(3) 2.726 (4) 1.726
17. The electron affinity of chlorine is 3.7 eV. How much energy in kcal is released when 2 g of chlorine is completely converted to C in ion in a gaseous state? (1 eV = 23.06
kcal /mol)
(1) 48 kcal (2) 2.8 kcal
(3) 8.4 kcal (4) 4.8 kcal
18. The correct order of ionic radius is (1) Ti4+<Mn7+
(2) 35Cl– > 37Cl–(3) K+ > Cl–(4) P3+ > P5+
19. Which of the following processes do not involve absorption of energy?
20. Mar k the correct statement out of the following.
(A) Helium has the highest first ionisation enthalpy in the periodic table.
(B) Fluorine has less negative electron gain enthalpy than chlorine.
(C) In any period, the atomic radius of the noble gas is the highest.
(D) Hg and Br are liquids at room temperature.
(1) (A), (B)only (2) (A), (C) only (3) (A), (B), (C) only (4) (A), (B), (C), and (D)
Section-B
Numeric Value Questions
21. Calculate the electronegativity of fluorine from the following data. Express your answer to the nearest possible integer.
(E= bond enthalpy and X = electronegativity)
EH–H = 104.2 kcal/mol
EF–F = 36.6 kcal/mol
EH–F = 13.4 kcal/mol
XH = 2.1
22. From the given compounds, number of compounds acidic in water is ____.
CaO, SO2, SO3, Fe2O3, Cl2O7, CO2, Na2O, No2
23. How many of the following factors influence the electronegativity of the atom of an element?
i) Effective nuclear charge
ii) Screening effect of inner electrons
iii) Size of the atom
iv) Oxidation state of the atom in bond
v) Type of hybridisation of the atom
vi) Strength of bond in which it is participated
vii) Number of surrounding atoms on the other atom with which it is bonded
viii) Nature of bond
24. An element ‘X’ belongs to 4th period and 4th group in the modern periodic table. Then, the atomic number of ‘X’ is ___.
25. How many of the following orders/ statements are correct?
a) IP1 of nitrogen is not less than that of oxygen.
b) IP1 of Pb is less than that of Sn.
c) Electronegativity order of B > Tl > In > Ga > Al.
d) IP2 of order B > C > Be.
e) Electron gain enthalpy order Ne < Ar < Kr < Xe < He < Rn.
f) Electron affinity of Be+4 is 217.6 eV/ ion.
g) Electronegativities of K and Rb are same.
h) Order of sizes is Tl > In > Al > Ga > B.
i) Official IUPAC name of element with atomic number 104 is unnilquadium.
CHAPTER TEST – JEE ADVANCED
2022 P-1 Model Section-A
[Numerical Value Questions]
1. How many of the following statements are correct?
(i) Each period consists of a series of elements having the same valence shell.
(ii) Each period corresponds to a particular principal quantum number of the valence shell present in it.
(iii) Each period starts with an alkali metal having the outermost electronic configuration ns1 .
(iv) Each period ends with a noble gas with the outermost electronic configuration ns2np6 ,except helium, having outermost electronic configuration 1s 2 .
(v) Each period starts with the filling of a new energy level.
(vi) The number of elements in each period is twice the number of atomic orbitals available in energy level that is being filled.
2. How many of them are metalloids in the modern periodic table?
3. The bond energies of H–H, X–X, and H–X are 104 kcal, 38 kcal, and 138 kcal, respectively. The electronegativity of X is p. Then, p–2.8 is ____. Given: () 67=8.18
4. How many of the following elements have positive or zero electron gain enthalpy values?
Al, Be, Mg, B, N, He, Ne, O, P, F
5. The common oxidation state of the first transition series is +x. The value of x is ___.
6. If covalent bond length of fluorine molecule is 20 pm, the covalent radius of chlorine is ____ pm.
7. Among CO 2, CO, SO 2, Al 2O 3, SO 3, P 4O 6, B2O3, BeO, CaO, BaO, N2O5, and NO
the number of acidic oxides = x
the basic oxides = y
the amphoteric oxides = z
the neutral oxides = w. What is the value of xy zw + + ?
8. Among the following number of oxides which is/are more basic as compared to Na2O is _____.
Li2O, K2O, Cs2O, Rb2O, MgO, CaO, Al2O3
Section-B
[Multiple Option Correct MCQs]
9. Select the correct statements about the element Uuo.
CHAPTER 4: Classification of Elements
(1) This element belongs to period-7.
(2) This element belongs to group-18.
(3) This is a p-block element.
(4) This is a metal.
10. Stability of ions of Ge, Sn, and Pb will be in the order
(1) Ge+2<Sn+2<Pb+2
(2) Ge+4<Sn+4<Pb+4
(3) Sn+4>Sn+2
(4) Pb+2>Pb+4
11. Following statements regarding the periodic trends of chemical reactivity of alkali metals and halogens are given. Which of these statements gives the correct picture?
(1) Chemical reactivity increases with increase in atomic number down the group in both alkali metals and halogens.
(2) In alkali metals, the reactivity increases, but in the halogens, it decreases with increases in atomic number down the group.
(3) The reactivity decreases in alkali metals but increases in halogens with increases in atomic number down the group.
(4) In both alkali metals and halogens, chemical reactivity decreases with increase in atomic number down the group.
12. In which of the following arrangements, the order is correct according to property indicated?
(1) Al 3+<Mg 2+<Na +<F –: increasing ionic size
(2) B<C<O<N: increasing first ionisation enthalpy
(3) CI<F<Br<I: increasing electron affinity
(4) Li<Na<K<Rb: increasing metallic radius
13. Which of the following sets of elements have more similarities in their properties?
(1) S, Na (2) Li, Mg
(3) Be, Al (4) Zr, Hf
14. Which of the following statements are true?
(1) In a multi-electron atom, the effect of nuclear charge experienced by the outermost electron is less than the theoretical value of the nuclear charge (Z).
(2) On moving from second to third transition series in a group [except Y(39) and La (57)] electronegativity increases due to the increase of +18 units in nuclear charge.
(3) Pauling assumed that the electronegativity value of fluorine is 4 and calculated the electronegativity values of other elements from this value.
(4) Higher ionisation potential and higher electron affinity values implies lower electronegativity value.
Section -C
[Matrix Matching Questions]
15. Match the elements in List I with Lis t II.
List I (Elements) List II (Classification)
A. He, Ne, Ar I) Representative elements
B. Fr, Ra, At II) Lanthanoids
C. Ce, Gd, Yb III) Noble gases
D. Rb, Ga, Cl IV) Radioactive elements
Choose the correct answer from the options given below:
(A) (B) (C) (D)
(1) III IV II I
(2) IV III II I
(3) I IV II III
(4) III IV I II
16. Match the electronic configuration in Column I with the ionisation energy in Column II ( kJ mole)
Column I
Column II
A. ns2 I) 2100
B. ns2np1 II) 1400
C. ns2np3 III) 800
D. ns2np6 IV) 900
Choose the correct answer from the options given below.
(A) (B) (C) (D)
(1) III I II IV
(2) IV I III II
(3) IV III I II
(4) IV III II I
17. Match the List I with List II
List I List II
A. I>Br>Cl>F I) First ionisation energy
B. F>CI>Br>I II) Atomic radii
C. CI>F>Br>I III) Electron affinity
D. O>F>N>C IV) Electronegativity V) Second ionisation energy
Choose the correct match:
(A) (B) (C) (D)
(1) II I,IV,V V V
(2) II I,V III V
(3) II I,V IV,III V
(4) II,IV I,V III V
18. Match Column I(element) with Column II (group to which element belongs)
Column I (Element)
Column II (Group to which element belongs)
A. P I) VI A group
B. Br II) VA group
C. Se III) VII A group
D. La IV) IIIB group
Choose the correct answer from the options given below.
■ Chemical bonds are the attractive forces holding two or more atoms or ions together in a molecule or ion.
■ When a pair of atoms combines to form a bond, energy is released.
■ The electronic theory of valency—also known as chemical bond theory or the modern theory of valency—was proposed by Kossel and Lewis.
■ Kossel explained the formation of electrovalent bonds.
■ Lewis explained the formation of covalent bonds.
Valence and Core Electrons:
■ Electrons in the outermost shell are called valence electrons; those in inner shells are core electrons.
■ The nucleus, along with the core electrons, forms the positively charged kernel.
Example: A sodium atom has one valence electron and ten core electrons.
Chemical Reactivity:
■ Valence electrons drive chemical activity, while core electrons generally do not participate in reactions.
■ Stability and the Octet Rule:
■ Elements in the zero group have a stable electronic configuration.
CHAPTER 5: Chemical Bonding and Molecular Structure 78
■ Most noble gases (except helium) have eight electrons (an octet) in their outermost shells, which accounts for their inertness.
■ Although helium has only two electrons, it is stable and inert because its only orbital (1s²) is completely filled.
■ Atoms outside the zero group are typically reactive because they have fewer than eight electrons in their outermost energy level.
■ The tendency of atoms to acquire eight electrons in their outermost shell for stability is known as the octet rule.
■ The valence electrons, in this system, are represented by dots (one for each electron). These notations are called Lewis symbols.
Examples:
Li, Be, B , C , N , O , F , Ne
■ Octet configuration is acquired in two types: By the transfer of electrons (ionic bond) or by mutual sharing of electrons (covalent bond).
Octet Rules:
■ Lewis formulated the octet rule, stating that an atom must have eight electrons in its outermost energy level for stability.
■ This rule applies even when electron pairs are shared in molecules (see Table 5.3 for examples).
Exceptions to the Octet Rule
Incomplete Octet:
■ Some molecules have fewer than eight electrons around the central atom (e.g., BeCl₂, BF₃, AlCl₃).
The Expanded Octet:
■ Some compounds feature more than eight valence electrons around the central atom.
■ Elements in the third period can utilize 3d orbitals (in addition to 3s and 3p) for bonding (e.g., SCl₄, PCl₅, SF₆, IF₇, H₂SO₄)
Odd Electron Molecules
■ Some molecules contain an odd number of electrons. Examples are, NO, NO2, ClO2, etc. Octet configuration has even number of electrons.
■ The octet rule clearly can never be satisfied for molecules with odd number of electrons.
Cations
■ Some of the transition metal ions are known in which the octet is not complete. Examples are Fe2+, Fe3+, Co3+, Ni2+, Cu2+ etc.
Guidelines to Write the Lewis Dot Formula
■ The total number of electrons for the structure is the sum of the valence electrons from the combining atoms.
■ Example: CH₄ has 8 valence electrons (4 from C and 4 from the four H atoms).
Adjusting for Overall Charge:
■ For anions, add electrons equal to the overall negative charge.
Example: In CO₃²–, electrons = 4 (from C) + 18 (3 × 6 from O) + 2 (charge) = 24.
■ For cations, subtract electrons equal to the overall positive charge.
Example: For an ion with N and four H atoms, electrons = 5 (from N) + 4 (from H’s) – 1 (charge) = 8.
Distributing Electrons as Bond Pairs:
■ Write the skeletal structure of the compound, placing the least electronegative atom at the center.
Example: In NF₃, nitrogen is the central atom and fluorines are bonded atoms.
■ Construct single bonds first using bond pairs.
■ Use remaining electrons for forming multiple bonds or as lone pairs so that each bonded atom (except hydrogen) attains an octet.
Example – Methyl Nitrite (CH₃ONO):
■ Step 1: Total valence electrons = 24, which equals 12 electron pairs.
For Anions:
■ Add electrons equal to the overall negative charge to the total from step (a).
Example: In CO₃²–, electrons = 4 (from C) + 18 (3×6 from O) + 2 (charge) = 24.
Some examples of Lewis dot formulae
Oxygen(O2)
Nitrogen (N2)
Carbon dioxide (CO2)
Carbon monoxide (CO)
Boron trichloride (BCL3)
(CH4)
Ethylene (C2H4)
Ammonia(NH3)
Ammonium cation (NH4+)
Carbonate anion (CO32–)
Nitrite anion (NO2–)
Nitric acid (HNO3)
Phosphorus Pentachloride (PCl5)
Sulphur hexafluoride(SF6)
Nitrogen trifluoride (NF3)
CHAPTER 5: Chemical Bonding and Molecular Structure
For Cations:
■ Subtract electrons equal to the overall positive charge from the total from step (a).
Example: For an ion with N and four H's, electrons = 5 (from N) + 4 (from H's) – 1 (charge) = 8.
Distributing Electrons as Bond Pairs:
■ Write the skeletal structure of the compound, making the least electronegative atom the central atom.
■ In NF₃, nitrogen serves as the central atom while the fluorines are bonded atoms.
■ Construct single bonds using bond pairs first.
■ Then, assign any remaining electron pairs as either multiple bonds or lone pairs.
■ Ensure that every bonded atom (except hydrogen) ac hieves an octet of electro ns.
■ Let us consider methyl nitrite, CH 3ONO.
Step 1: Total number of valency electrons = 24, the number of elect ron pairs = 12
Step 2:
Step 3: 6 electron pairs are used as bond pairs. Now, the remaining six electron pairs are to be placed as lone pairs on the atoms in the descending order of electro-negativity.
Nitrogen does not satisfy the octet rule.
Step 4: Convert one lone pair on terminal oxygen as π bond.
Now, all the atoms in the molecule are satisfying the octet rule. Thus, it is the correct structure.
Formal
Charge
■ A molecule’s overall charge is zero; for polyatomic ions, the net charge resides on the ion, not on any single atom, though a formal charge can be assigned to each atom.
IL ACHIEVER SERIES FOR JEE CHEMISTRY
Definition:
■ The formal charge of an atom is the difference between its free-state valence electrons (N A) and the electrons assigned to it in the Lewis structure (N M).
Formula:
Qf = NA – NM = NA – (NLp + ½ NBp)
■ NA: Number of valence electrons in the free atom.
■ NLp: Number of electrons in lone pairs.
■ NBp: Number of electrons in bond pairs.
Assumption:
■ In counting electrons, each shared bond contributes one electron to the atom, while both electrons in a lone pair are fully assigned to it.
Illustrative Examples
■ PH 3 molecule: The Lewis dot structure is HPH H or HPH H
Formal charge of first N atom is Q f = 5 – 4 – 1/2(4) = –1. Formal charge of second N atom is Qf = 5 – 0 – 1/2(8) = +1
Formal charge of O atom is Q f = 6 – 4 – 1/2(4) = 0
CHAPTER 5: Chemical Bonding and Molecular Structure
Note:
■ Formal charge of central oxygen atom is +1 only and terminal oxygen atoms may be 0 or –1
Importance of Formal Charge:
■ Formal charges do not represent actual ionic charges.
■ They help in comparing various Lewis structures of a molecule.
■ The most stable (lowest energy) structure is the one with the smallest formal charges on the atoms.
Covalency
■ It is the number of electrons an atom shares (or electron pairs it forms) to achieve a noble gas configuration.
■ For non-metallic elements (except hydrogen), covalency is given by (8 – G), where G is the group (Roman) number (e.g., oxygen in H₂O is 2, carbon in CH₄ is 4; common valences: IVA = 4, VA = 3, VIA = 2, VIIA = 1).
■ Generally, an element’s covalency equals the total number of unpaired electrons in its valence s and p orbitals (e.g., N = 3, O = 2, F = 1).
■ Elements with vacant d orbitals can exhibit variable covalency in excited states by increasing unpaired electrons—a shift not possible for H, N, O, and F due to the absence of d orbitals.
Examples:
■ Phosphorus: Ground state ([Ne] 3s²3p³) has 3 unpaired electrons (covalency = 3); excited state ([Ne] 3s¹3p³3d¹) has 5 unpaired electrons (covalency = 5).
■ Sulphur: Exhibits covalency of 2, 4, and 6 in its ground, first excited, and second excited states respectively.
■ Chlorine: Exhibits covalency of 1, 3, 5, and 7 in its ground, first excited, second excited, and third excited states respectively.
Properties of Covalent Compounds
■ Physical State: Typically gases or volatile liquids; high-mass compounds may be soft solids.
■ Melting/Boiling Points: Low, due to weak Van der Waals forces.
■ Crystal Structure: Loosely packed neutral molecules; network solids (e.g., diamond) are an exception.
■ Mechanical Nature: Soft and non-brittle.
■ Solubility: Generally insoluble in polar solvents; soluble in non-polar ones; hydrogen bonding can aid water solubility.
■ Isomerism: Rigid, directional bonds allow for isomers.
■ Electrical Conductivity: Generally non-conductive (graphite is an exception).
■ Reaction Rates: Slow molecular reactions, often needing catalysts.
■ Transfer of electrons yields electrovalency; sharing of electrons yields covalency.
■ Formed between low electronegativity (metal) atoms.
Bond Strength & Properties:
■ Metallic bond strength increases with more valence electrons and decreases with larger atom size.
■ Stronger metallic bonds lead to higher enthalpy of atomization and melting points.
Solved Examples
1. Will the formal charges of an atom in a molecule remain the same ? Why or why not?
Sol. No. the formal charge on an atom may not be same. It changes with the structural environment of the atom in the molecule. In resonance structures, the electronic environment changes. Hence, the formal charges also may change.
Example:
N2O has the resonance structures and
(1) (1) (2) (2)
The formal charges in structure I were of N (1), N(2) are –1 and +1. In structure (II) the formal charges of N(1), N(2) and O are 0, +1 and – 1, respectively.
2. In which excited state sulphur forms six covalent bonds in SO 3.
Sol. Electron configuration of S = 1s 2 2s2 2p6 3s2 3p4
In 1st excited state it has four unpaired electrons (1s2 2s2 2p6 3s2 3p3 3d1). In 2nd excited state its configuration is 1s2 2s2 2p6 3s1 3p3 3d2. In this excited state it forms six bonds.
TEST YOURSELF
1. In molecule of COCl2, between carbon and oxygen there are four dots in Lewis dot formula. What does it indicate?
(1) Four non-bonding electrons
(2) A double bond between carbon and oxygen
(3) Between carbon and oxygen the bond formed is odd electron bond.
(4) Valency shown by carbon is 4.
2. Which of the following has odd electron?
(1) SO2 (2) ClO3 (3) N2O4 (4) H2O
3. Which of the following molecules obeys octet rule?
(1) BCl3 (2) BeCl2 (3) NCl3 (4) SF4
4. Which is the total number of valence electrons involved in bonding by all atoms in the compound, H2CO3?
(1) 16 (2) 12 (3) 24 (4) 10
5. In drawing Lewis dot formula of CN – ion, how many number of dots is expected to appear? (1) 10 (2) 11 (3) 9 (4) 13
6. What are the formal charges of carbon in 2 3 CO and CN– ions, respectively? (1) Zero and +1 (2) Zero and –1 (3) +1 and –1 (4) +1 and –4
7. The bond between two identical non-metal atoms has a pair of electrons (1) unequally shared between the two (2) transferred fully from one atom to another (3) with identical spin (4) equally shared between them
8. Covalent compounds are generally soluble in (1) polar solvents (2) non-polar solvents (3) concentrated acids (4) all solvents
9. Which of the following boils at higher temperature? (1) CCl4 (2) CO2 (3) C6H12O6 (4) KCl
10. CCl4 is insoluble in water because (1) H2O is non-polar (2) CCl4 is non-polar (3) they form intermolecular H-bonding (4) they form intramolecular H-bonding
11. Which of the following is very volatile? (1) Diamond (2) Sodium chloride (3) Calcium (4) Dry ice
12. The following are some statements about the characteristics of covalent compounds.
a) The combination of a metal and non-metal must give a covalent compound.
b) All covalent substances are bad conductors of electricity. c) All covalent substances are gases at room temperature. Choose the correct option.
(1) All are correct. (2) Only a and b are correct. (3) Only b and c are correct. (4) All are wrong.
■ Formed by the electrostatic attraction between oppositely charged ions created when electrons are transferred from a metal to a nonmetal.
■ Occurs between atoms with low ionization energy (forming cations) and high electron affinity (forming anions) when the electronegativity difference exceeds 1.7.
■ Alkali and alkaline earth metals generally form cations, while halogens and chalcogens form anions.
■ Ionic bonds form readily between alkali metals and halogens (e.g., CsF).
Factors Favourable for the Formation of Ionic Bond
■ Only when cation and anion are formed, they will have ionic bond. The following are the factors which favour their formation.
Formation of Cation
■ Electron Loss: Metals lose electrons to form cations.
■ Low Ionisation Energy: Atoms with lower ionisation energy lose electrons more readily (e.g., K+ forms easier than Na +; order: Na+ < K+ < Rb+ < Cs+).
■ Low Charge: Lower positive charges are easier to form; higher charges require losing more electrons (order: Al³ + < Mg²+ < Na+).
■ Large Atomic Size: Larger metal atoms (Li < Na < K < Rb < Cs) have a weaker nuclear hold on valence electrons.
■ Inert Gas Configuration: Cations with ns² np⁶ (or pseudo inert gas with 18 electrons) are more stable and favor ionic compound formation.
Formation of Anion
■ Electron Gain: Non-metals gain electrons to form anions.
■ High Electron Affinity: Atoms with high electronegativity and electron affinity form anions readily (order: Cl– > Br– > I–).
■ Small Atomic Size: Smaller non-metals favor stable anion formation.
■ Inert Gas Configuration: Anions with ns² np⁶ configuration (eight electrons in the outer shell) are especially stable.
■ Fig.5.1 shows the variation of the nature of the bond, i.e. ionic nature of the bond, with the difference in electronegativities of the bonded atoms.
Fig. 5.1 Variation of Ionic Nature of the Bond Versus Difference in Electronegativities
■ Large E.N difference: Leads to electron transfer and ionic bonds.
■ Small E.N difference: Leads to electron sharing and covalent bonds. .
■ The distinction between ionic and covalent bonds is presented in the Table.
Differences between ionic and covalent bonds
1. Ionic bond is formed by the transfer of one or more electrons.
1. Covalent bond is formed by sharing of electrons.
2. It is formed between metal and non-metal. 2. It is formed between non-metals.
3. It is also called electrovalent bond and is due to electro-valency.
3. It is called electron pair bond and is due to covalency.
4. Ionic bond consists of electrostatic force of attraction between the oppositely charged ions. 4. Covalent bond consists of shared pair or pairs of electrons which are attracted by both the nuclei.
5. Ionic bond is non rigid and non directional. 5. Covalent bond is rigid and directional except for H2
6. Ionic bond is polar in nature.
Nature of Bond
6. Covalent bond may be polar or non polar.
■ Generally, bonds between a metal and a non-metal are ionic, though covalent bonds can also form.
■ Polarisation in Ionic Bonds:
■ In ionic bonds, a smaller cation attracts the outer electrons of a larger anion and repels its nuclear charge, causing the anion’s electron cloud to distort toward the cation.
■ Polarising Power & Polarisability:
■ Polarising power is the cation’s ability to distort an anion’s electron cloud.
■ Polarisability is the tendency of the anion to be distorted.
■ Resulting Bond Character:
■ This polarisation leads to partial electron sharing, giving the bond some covalent character.
■ Cation with higher charge and smaller size has high polarising power, which gives more covalent character to the compound.
i.e., Polarising power of cation charge size α
■ Cation Polarising Power & Anion Polarisability:
■ Cation polarising power decreases down a group and increases across a period.
■ Anions with higher charge and larger size are more easily polarised, thus increasing covalent character; for a given cation, anion polarisability increases down the group and decreases across a period.
Fajan’s Rules:
1. Cation Size: Larger cations increase ionic character (Li +< Na+< K+< Rb+< Cs+; hence, LiCl is less ionic than CsCl).
2. Anion Size: Larger anions favour covalent character (F– < Cl– < Br– < I–; CaF₂ is more ionic than CaI₂).
3. Charge Magnitude: High charges on cations and/or anions promote covalent bonds (e.g., NaCl is more ionic than AlCl₃ due to Al³ +’s higher charge).
4. Electronic Configuration: Cations with inert gas configurations form ionic compounds, while those with pseudo inert gas configurations favour covalent bonds (e.g., NaCl is ionic; CuCl is more covalent).
Ionic Bond Conditions:
■ Large cation, small anion, and low magnitudes of charge on both ions favor ionic bonding.
Examples: LiI is the least ionic among lithium halides; AgCl is less ionic than NaCl; LiF is ionic while BeF₂ is more covalent.
■ Order of increasing covalent character: NaCl < MgCl₂ < AlCl₃ < SiCl₄ (as cation charge increases).
Energy Changes in ionic Bond Formation
■ Ions arrange into a close-packed, regular pattern, releasing energy (crystal lattice energy) and stabilizing the system.
■ In forming NaCl, Na atom ionizes (absorbing energy) and Cl atom accepts an electron (releasing energy), making the formation of gaseous Na +Cl– endothermic (+147.1 kJ/mol).
■ However, the exothermic lattice formation of solid NaCl (–788 kJ/mol) more than compensates for this energy input, driving the overall reaction.
■ Thus, the stability of an ionic compound is best gauged by its lattice enthalpy rather than just achieving an octet in the gaseous state.
■ Sodium chloride is formed by the union of Na and Cl2. To form an ionic solid, the energy changes are summarized as follows:
Na ionizes to Na+ ions by absorbing energy. 1 (s)(g)1 NaNa;H108.7kJmol. →∆=+
■ It can be readily concluded from equation (3) that energy is absorbed during the reaction to form Na+Cl–(g)
Lattice Energy
■ Lattice is a systematic three dimensional arrangement of oppositely charged ions. This arrangement leads to the stability of the ionic substances.
■ The lattice energy of an ionic crystal is the amount of energy released when one mole of crystal is formed from the oppositely charged gaseous ions. This is denoted by “U”.
■ It can also be defined as the amount of energy needed to dissociate one mole of ionic crystal into isolated constituent gaseous ions.
■ Lattice energies of ionic crystals are not measured directly, instead, they are derived from data using the Born-Haber cycle.
■ The magnitude of lattice energy is described by coulomb’s law, as ZZ UK d +− −=×
■ The value of constant K depends on the geometrical arrangement of ions.
■ Within compounds with the same anion, lattice energy increases as cation size decreases (e.g., LiF > NaF > KF).
■ For compounds with the same cation, lattice energy increases as anion size decreases (e.g., LiF > LiCl > LiBr > LiI).
■ Compounds with higher charged ions have greater lattice energies (e.g., NaF < MgO < AlN).
Properties of Ionic Compounds
■ Crystalline nature: All ionic compounds are crystalline solids.
■ State is due to close packing of oppositely charged ions in the crystal lattice.
■ As a result no isolated, discrete molecule exists in the crystal lattice.
■ NaCl crystal lattice is described as face centred cubic lattice (FCC), as shown in Fig. 5.2.
Fig. 5.2 Unit cell of NaCl
■ One unit cell of NaCl is associated with 27 ions. However, the stoichiometric ratio of constituent ions in sodium chloride crystal is 1:1.
■ Hardness & Brittleness: Ionic compounds are hard due to strong electrostatic forces but brittle as crystal layers shift under external force.
■ Melting & Boiling Points: High due to strong interionic attraction; melting points follow the order NaF > NaCl > NaBr > NaI.
■ Solubility: Ionic compounds dissolve in polar solvents like water but are insoluble in nonpolar solvents; solubility depends on hydration energy being greater than lattice energy.
■ Electrical Conductivity: Ionic solids do not conduct electricity in solid state but do in molten or aqueous states due to free ion movement.
■ Ionic Reactions: Occur instantly in solution with high reaction rates as they involve freemoving ions.
■ Space Isomerism (Stereo Isomerism): Ionic compounds do not exhibit space isomerism because ionic bonds are non-directional.
■ Differences between ionic and covalent substances is shown in the table below.
Property
Ionic substances
Physical state Ionic compounds are crystalline solids at room temperature.
Melting and boiling points.
They possess high melting and high boiling points
Solubility Freely soluble in polar solvents like water and liquid ammonia
Conductivity Bad conductors of electricity in solid state, but good conductors in molten state and in aqueous solution
Nature of reactions Isomerism
They undergo ionic reactions. The ionic reactions are fast and instantaneous. Ionic compounds do not exhibit isomerism, as electrovalent bond is non directional.
Covalent substances
Covalent substances are usually gases, low boiling liquids and soft solids under ordinary conditions.
Possess low melting and low boiling points with the exception of gaint molecules
Generally, insoluble in water but soluble in nonpolar solvents
Bad conductors of electricity, with a few exceptions, like graphite
they undergo molecular reactions. The reactions are slow and the rates are low. Covalent substances exhibit isomerism, as the covalent bonds are directional.
Solved Example
3. Lattice energy in sodium chloride is x kJ. Assuming the same interionic distance, what will be the lattice energy of magnesium sulphide?
Sol. The charge magnitudes of ions in NaCl are 1 and 1, respectively. The product of q1 and q2 = 1 × 1 = 1
The charge magnitudes of ions in magnesium sulphide are 2 and 2 respectively.
The product of q1 and q2 = 2 × 2 = 4
Hence, lattice energy of MgS = 4 x kJ.
TEST YOURSELF
1 Ease of formation of anion is favoured by (1) lower value of ionisation energy (2) lower value of electronegativity (3) lower value of electron affinity (4) higher value of electron affinity
3. Which of the following is a favourable condition for the formation of ionic bond?
(1) Small cation with small charge (2) Small anion with large charge (3) Large difference in the electronegativity (4) Small cation with high charge
4. Number of electrons transferred from one Al atom during bond formation in aluminium fluoride is (1) 1 (2) 2 (3) 3 (4) 4
5. Which of the following is least ionic? (1) CaF2 (2) CaBr2 (3) CaCl2 (4) CaI2
6. The strongest ionic bond is present in (1) LiF (2) NaF (3) RbF (4) CsF
7. The compound with highest melting point is (1) NaF (2) NaCl (3) NaBr (4) NaI
8. CuCl has more covalent character than NaCl because (1) Na+ has more polarising power than Cu + (2) Cu+ has more polarising power than Na + (3) Cl– has pseudo inert gas electron configu-ration (4) Na+ has pseudo inert gas electron configuration
9. The electronic structure of four elements a, b, c and d are (a) 1s2 (b) 1s2, 2s2, 2p2 (c) 1s2, 2s2, 2p5 (d) 1s2, 2s2, 2p6 The tendency to form electrovalent bond is greatest in (1) a (2) b (3) c (4) d
10. If Na+ ion is larger than Mg2+ ion and S2– ion is larger than Cl– ion, which of the following will be least soluble in water? (1) NaCl (2) Na2S (3) MgCl2 (4) MgS
11. Lattice energy of NaCl is ‘X’. If the ionic size of A+2 is equal to that of Na+, and B–2 is equal to Cl–, then lattice energy associated with the crystal AB is
(1) X (2) 2 X (3) 4 X (4) 8 X
12. For which of the following sets are all the compounds ionic? (1) NaF, BF3, MgF2 (2) NaBr, MgBr2, MgO (3) Al2O3, MgO, SO3 (4) NCl3, BeCl2, AlCl3
13. Among the following compounds, the one with greatest ionic character is (1) NaCl (2) KCl (3) CsCl (4) RbCl
Chemical bonds are identified with certain measurable properties. They are also follows
Bond Length
■ Bond Length: The average equilibrium distance between the nuclei of two covalently bonded atoms.
■ Reason for Variation: Atoms vibrate due to simultaneous attractive and repulsive forces.
■ Measurement: Determined using spectroscopy, X-ray, or electron diffraction.
■ Units: Expressed in Angstroms (Å) or picometers (pm).
Fig. 5.3 The Bond Length in a Covalent Molecule AB
bond between two atoms, the bond length is taken as the sum of their covalent radii, as shown in Fig.5.3. Bond lengths of some molecules are listed in Table 5.15.
Bond length A–B = ra + rb where r a and rb are the covalent radii of the atoms A and B respectively.
Important Features
■ Bond Length Consistency: The bond length between identical atoms remains the same across different molecules (e.g., O–H in H₂O, H₂O₂, and C₂H₅OH = 96 pm).
■ Effect of Bond Type: Bond length varies when the bond type between specific atoms changes.
CHAPTER 5: Chemical Bonding and Molecular Structure
Bond lengths in some molecules
■ Bond Length and Atomic Size: Increases with the size of bonded atoms (e.g., F–F < Cl–Cl < Br–Br < I–I, H–F < H–Cl < H–Br < H–I).
■ Homonuclear Diatomic Molecules: Bond length = 2 × covalent radius (e.g., H covalent radius = 37 pm, H–H bond length = 74 pm).
■ Bond Order & Length: More bonds → stronger overlap → shorter bond length (Single > Double > Triple).
■ Examples: C–C (154 pm, ethane) > C=C (134 pm, ethylene) > C ≡ C (120 pm, acetylene).
■ Average bond lengths for some single, double and triple bonded molecules are given in the table below.
Average bond lengths for some molecules
■ Effect of Hybridization: Bond length decreases with increasing s character and decreasing p character in hybrid orbitals.
■ Polar Bonds: Have shorter bond lengths than theoretical values (sum of covalent radii).
Bond Enthalpy
Diatomic Molecule
■ Bond Enthalpy (Bond Energy/Bond Strength): Energy required to break one mole of diatomic molecules in the gaseous state into free atoms.
■ Units: Expressed in kJ mol –¹ at room temperature.
■ Bond Length and Strength: Shorter bond length corresponds to greater bond strength.
()() 1 gg HH2H;H436kJmol −→∆=
(H–H bond length, 74 pm)
ClCl2Cl;H242kJmol −→∆=
()() 1 gg
(Cl–Cl bond length, 198 pm).
■ Bond Dissociation Enthalpy (ΔH): Always positive, as energy is required to break a bond.
■ Bond Formation: Releases energy, so ΔH is negative.
■ Bond energies of some covalent bonds are listed in table below.
■ Polyatomic Molecules: Bond enthalpy differs from bond dissociation enthalpy; bond energy is the average of dissociation energies of similar bonds in the molecule.
Example:
■ In CH4, the bond dissociation energies of the four C–H bonds a re also follows:
CHCHH;→+D H = 426 kJ mol–1
()()() 4g3gg
CHCHH;→+D H = 349 kJ mol–1
()()() 3g2gg
CHCHH;→+D H = 451 kJ mol–1
()()() 2ggg
CHCH;→+D H = 347 kJ mol–1
()()() ggg
CHC4H;→+D H ≅ 1663 kJ mol–1
()()() 4ggg
■ The average bond dissociation energies (e.g., C–H = 1663/4 = 416 kJ mol –¹).
Table 5.18 Bond energies
■ Heterolytic bond cleavage requires more energy than homolytic cleavage.
■ Bond enthalpy varies within a molecule due to different chemical environments (e.g., O–H bonds in H₂O require different energies to break).
2(g)(g)(g)a1 HOHOHH kJ mol ;5021→+∆=
■ Bond Energy & Stability: Higher bond energy increases bond stability and decreases chemical reactivity (e.g., N₂ is highly stable due to its high bond energy of 944 kJ mol –¹).
■ Bond Type Strength: Sigma (σ) bonds are stronger than pi (π) bonds.
■ Multiple Bonds: More bonds between the same atoms increase overall bond strength.
Bond Energy Order
The bond energies of similar type of bonds gradually decrease down the group. HF > HCl > HBr > HI
As the number of lone pairs of electrons on bonded atoms increases, the bond energy decreases.
C O O C N > > N (341 kJ) (163 kJ) (146 kJ)
The heat of reaction can be estimated from the values of bond energy.
D H = Heat of reaction = total bond energy of reactants – total bond energy of products.
Among the halogens, chlorine has highest bond energy. Cl 2 > Br2 > F2 > I2.
Bond Angle
■ Bond Angle: The average angle between two adjacent atoms bonded to the central atom, measured in degrees.
■ Determination: Measured using X-ray diffraction and spectroscopy.
■ Stability & Shape: Larger bond angles increase molecular stability and help determine molecular shape.
Factors Affecting Bond Angle:
■ Molecular Geometry: Bond angle depends on the shape of the molecule.
■ Hybridization: More s character in hybrid orbitals increases bond angle.
■ Lone Pairs: Reduce bond angle (e.g., NH₃ has 107° despite sp³ hybridization).
Electronegativity:
Lower central atom electronegativity → smaller bond angle.
For the same central atom, lower surrounding atom electronegativity → larger bond angle.
Examples: Bond angles in OF₂ and Cl₂O are 103° and 111°, respectively.
Solved Examples
4. If the C=O bond length is 121 pm, what is the distance between the nuclei of oxygen atoms in carbondioxide molecule?
Sol. Carbon dioxide is a linear molecule with two C=O bonds opposite to each other. O = C = O 121pm 121pm
The distance between the nuclei of oxygen atoms in CO 2 molecule is 121 + 121 = 242pm.
5. The As–Cl bond distance in AsCl3 is 2.20A°. Estimate the single bond covalent radius of Arsenic. (Covalent radius of Cl is 0.99A°).
Sol. Internuclear distance - radius of chlorine atom = radius of arsenic atom 2.20 – 0.99 = 1.21 A°
Covalent radius of As = 1.21A°.
TEST YOURSELF
1. Bond energy is least in (1) HF (2) HCl (3) HBr (4) HI
2. Bond length of H2 is 0.074 nm, bond length of Cl 2 is 1.98A°. Bond length of HCl is (1) 2.72 A° (2) 136 pm (3) 1.027 nm (4) 0.136 A°
3. The O-H bond length in H2O is xA°. The O-H bond length in H2O2 is (1) < x A° (2) x A° (3) > x A° (4) 2x A°
4. Bond energy is highest in the molecules of (1) F2 (2) Br2 (3) I2 (4) Cl2
5. Bond energy of C C bond is highest in (1) H3C –CH3 (2) H2C = CH2 (3) CH ≡ CH (4) C2H5Cl
6. The bond dissociation energy of the molecules A2, B2, and C2 are 498, 158, and 945 kJ/ mole, respectively. So, the correct decreasing order of their bond orders is (1) A2, B2, C2 (2) C2, B2, A2 (3) C2, A2, B2 (4) B2, C2, A2
7. The order of bond length of (O – O) in O 2, O3, and H2O2 is (1) O2 > H2O2 > O3 (2) O3 > H2O2 > O2 (3) H2O2 > O3 > O2 (4) O2 > H2O2 > O3
8. The correct order of increasing bond angles is (1) PF 3 < PCl3 < PBr 3 < PI 3 (2) PF 3 < PBr 3 < PCl3 < PI 3 (3) PI 3 < PBr 3 < PCl3 < PF 3 (4) PF 3 > PCl3 < PBr 3 < PI 3
9. The correct order of increasing C-O bond length of CO, 2 CO3 and CO2 is (1) 2 CO3 < CO2 < CO (2) CO2 < CO< 2 CO3 (3) CO < 2 CO3 < CO2 (4) CO < CO2 < 2 CO3
10. Which of the following has least bond energy?
(1) F2 (2) H2 (3) N2 (4) O2
Answer Key
1
Bond Order
■ Bond Order: Number of bonds between two atoms.
Examples: H₂ (1), O₂ (2), N₂ (3), CO (3). H H H Bond order =1 or H, –
■ Single shared electron pair in H 2. O O , O O = Bond order =2 or
■ Two shared electron pairs in O 2. N N N N , Bond order =3 or ≡
■ Three shared electron pairs in N 2. C C O O , Bond order =3 or ≡
■ Isoelectronic Species: Have identical bond orders (e.g., N₂, CO, NO+ each have 14 electrons and a bond order of 3).
■ Resonance & Bond Order: Calculated as x/y, where x is the total number of bonds, and y is the number of canonical structures.
■ Bond Order Effects: Higher bond order → increased bond energy & decreased bond length.
■ Calculation Method: Now determined using Molecular Orbital Theory (MOT).
■ Bond orders in some molecules and ions are listed in the adjacent table Bond orders are now calculated by using molecular orbital theory.
IL ACHIEVER SERIES FOR JEE CHEMISTRY
Bond orders in some molecules and ions
Solved Examples
6. Explain the structure of CO 2 molecule.
Sol. CO₂ Bond Length: Experimentally determined as 115 pm, lying between a typical C=O (121 pm) and C ≡ O (110 pm) bond length.
■ Resonance in CO₂: The structure is a hybrid of multiple canonical forms, leading to a n intermediate bond length ⊕⊕
7. What is the nitrogen-oxygen bond order in NO 3 – ion?
Sol. NO3– The structure of NO3 – ion is: It has four bonds and three canonical structures. Bond order = 1.33.
TEST YOURSELF
1 How many resonance structures are possible for NO 3 – ion? (1) Five (2) Three (3) Four (4) Six
2. According to the concept of resonance, among the ions OCl–, ClO2–, ClO3– , and ClO4–, the most stable ion is (1) OCl– (2) ClO2– (3) ClO3– (4) ClO4–
3. Based on the resonance structures of CO3 –2 incorrect observation would be (1) every structure has the same number of lone pairs. (2) i one structure central atom is oxygen. (3) in all the structures one C=O exists per structure. (4) every structure has –2 units of charge.
4. C–O bond order in CO2 molecule is (1) 3 (2) 4 (3) 2 (4) 1
Answer Key (1) 2 (2) 4 (3) 2 (4) 3
BOND POLARITY AND DIPOLE MOMENT
■ Polar Covalent Bonds: Form when bonding electrons are shared unequally between two atoms but not completely transferred.
■ Nature: They exist between purely covalent and purely ionic bonds.
■ A formula was proposed to calculate the percentage of ionic character in an A–B bond based on the electronegativity values of atoms A and B.
Percentage of ionic chara cter =
XA = electronegativity of the atom A
XB = electronegativity of the atom B
Dipole Moment (μ)
■ Measures the polarity of a molecule.
■ Definition: The product of the magnitude of charge on either pole and the distance between the poles.
■ Formula: μ = q × d, where q is charge and d is bond length.
■ The electric charge is in the order of 10 –10 esu and the distance is in the order of 10 –8 cm.
■ 10–18 esu cm is known as Debye unit. 1 Debye = 3.336 × 10–30 C m. The SI unit of dipole moment is coulomb meter, (C-m)
■ Every bond between two hetero atoms has a definite dipole moment called bond moment. Dipole moment is a vector. It has both magnitude and direction.
■ Dipole moment is a vector quantity and it is depicted by a small arrow with tail on the positive centre and head pointing towards the negative centre.
Dipole moment of HX as in the Table.
Dipole moments of hydrogen halides
100
May be represented as H X
CHAPTER 5: Chemical Bonding and Molecular Structure
■ Electron Density Shift: Represented by a crossed arrow in Lewis structures, indicating the direction of electron movement.
■ Dipole Moment in Polyatomic Molecules:
■ Depends on bond dipoles and spatial arrangement of bonds.
■ The dipole moment of the molecule is the vector sum of individual bond dipoles.
■ Zero Dipole Moment:
■ Symmetrical molecules (linear, trigonal, tetrahedral) have zero dipole moment.
■ Distorted shapes (angular, pyramidal, see-saw) result in a net dipole moment. For example: H2O, SO2, H2S, NH3, SF4
■ In H2O molecule, which has a bent structure, the two O–H bonds are oriented at an angle of 104.5º. Net dipole moment of 6.17 × 10 –30 C m (1D = 3.33564 10–30 C m) is the resultant of the dipole moments of two O – H bonds.
Bond dipole Resultant dipole moment H
H (a) (b)
Net Dipole moment, m = 1.85 D
= 1.85 × 3.33564 × 10 –30 Cm = 6.17 × 10–30 Cm
■ The dipole moment in case of BeF2 is zero. This is because the two equal bond dipoles point in opposite directions cancel the effect of each other.
F F + Be
BeF2 molecule
■ Dipole Moment in Tetraatomic Molecules (e.g., BF₃):
■ Although B–F bonds are polar and oriented at 120°, their bond moments cancel out.
■ The vector sum of any two bond moments equals and opposes the third, resulting in a net dipole moment of zero .
BF3 Molecule; Representation of (a) Bond Dipoles and (b) Total Dipole moment
Resultant dipole moment in NH 3=4.9 × 10–30 Cm Resultant dipole moment in NF 3=0.80×10–30Cm
Calculation of Resultant Bond Moments
■ Let AB and AC have two polar bonds of a molecule inclined at an angle q . Let m 1 and m 2 have their dipole moments, respectively. The resultant dipole moment, may be obtained by vectorial method.
R12122cos µ=µ+µ+µµθ
When q = 0, the resultant is maximum. (cos 0 = 1)
R12 µ=µ+µ
When 180º, the resultant is minimum. (cos 180 = – 1)
R12 µ=µ−µ
■ Dipole moments of some covalent molecules are listed in the table below.
Dipole moments of some molecules
Applications of Dipole Moment
■ Used to determine the percentage of ionic character in a covalent bond.
■ Calculated from the ratio of the observed dipole moment to the dipole moment expected for complete electron transfer.
Percentage ionic character × Observeddipolemoment100
Dipolemomentforcompleteioniccharacter
■ The bond angles can be calculated and inference can be drawn about the symmetry of the molecule. If q is the angle of AB2 molecule, 2(bondmoment)cos2 θ µ=××
■ To distinguish cis- and trans- forms of geometrical isomers: For example, the following two molecules have the same molecular formula C 2H2Cl2, the same type and number of bonds, but different molecular structures.
Distinguishing Isomers:
Cis-dichloroethene is polar with a dipole moment of 1.89D, while trans-dichloroethene is non-polar (μ = 0).
Dipole moment measurements help differentiate them.
Determining Substituent Orientation in Benzene Rings:
■ Dipole moment helps determine substituent positioning in disubstituted benzene rings.
■ General Order: Ortho > Meta > Para (in terms of dipole moment m agnitude).
Order of Dipole Moment:
■ NH3 > NCl3 > NF3
■ CH3Cl > CH2Cl2 > CHCl3 > CCl4
■ CH3Cl > CH3F > CH3Br > CH3 I
and
For and m≠0 (non-linear angular shape)
■ Dipole moment of PCl3F2 is 0
Solved Examples
6. The dipole moments of SO2 and CO2 are 5.37 × 10–30 Cm and zero, respectively. What can be said about the shapes of the two molecules?
Sol. Oxygen is considerably more electronegative than either sulphur or carbon. The sulphur-oxygen and the carbon-oxygen bond should be polar with a net negative charge residing on the oxygen.
SO2 is angular, as the S = O bond moments do not cancel.
CO2 is linear. Though C = O bonds are polar, the bond moments cancel each other.
7. The dipole moment of HBr is 2.60 × 10–30 C m and the interatomic spacing is 1.41A°. What is the percent ionic character of HBr?
Sol. The dipole moment of 100% assumed ionic molecule, HBr at the given internuclear distance
= (1.60 × 10–19C) (1.41 × 10–10m)
= 2.26 × 10–29 C m
The actual dipole moment is less.
The percent ionic character = 30 29 2.6010Cm 10011.5% 2.2610Cm × ×= ×
2. Carbon tetrachloride has no dipole moment because of (1) its regular tetrahedral structure (2) its planar structure (3) similar sizes of carbon and chlorine atoms (4) similar electron affinities of carbon and chlorine
3. Molecule with dipole moment among the following is (1) SF6 (2) SF4 (3) CCl4 (4) CH4
4. The molecule having zero dipole moment is (1) CHCl3 (2) CH2Cl2 (3) CCl4 (4) CH3Cl
5. The dipole moment of CO2 is zero, because its bond angle is (1) 120° (2) 180° (3) 130° (4) 90°
6. SI unit for dipole moment is (1) esu-cm (2) coulomb-cm (3) coulomb-metre (4) esu - metre
7. Molecule with zero dipole moment is (1) BCl3 (2) BeCl2 (3) CCl4 (4) All of these
8. Molecule with dipole moment among the following is (1) SF6 (2) PCl5 (3) CCl4 (4) BF 3
9. In which of the following pairs, both molecules possess dipole moment? (1) CO2, SO2 (2) BCl3, PCl3 (3) H2O, SO2 (4) CO2, CS2
10. The dipole moment of HX is 1.2 D. If the percentage ionic character of the bond is 25%, then its bond length is
(1) 10 A°
(3) 10–8 m
11. Which of the following has m =0?
(1) CH2Cl2
(3) NH3
12. Which bond is most polar?
(1) Cl – F
(3) I – F
(11) 2 (12) 3
RESONANCE
(2) 10–10 m
(4) 10–6 m
(2) SO3
(4) H2O
(2) Br – F
(4) F – F Answer Key
■ Resonance Concept: Applied when a single Lewis structure cannot fully explain a molecule’s properties.
■ Resonance Hybrid: The actual structure of a molecule that is a blend of two or more alternate Lewis structures.
■ For example, the ozone molecule, O 3 can be equally represented by the structures I and II, as shown in Fig. 5.4.
Fig. 5.4 Resonance in Ozone Resonance in Ozone (O₃):
■ O₃ has two possible Lewis structures, each with O–O (148 pm) and O=O (121 pm) bonds.
■ Experimental Bond Length: Both O–O bonds in O₃ are 128 pm, intermediate between a single and double bond.
■ Resonance Hybrid: The actual structure is a blend of canonical forms (I & II), represented as structure III.
■ Definition: Resonance is the phenomenon of writing two or more equivalent structures for a molecule when a single Lewis structure cannot fully represent it.
CHAPTER 5: Chemical Bonding and Molecular Structure
Other examples of resonance structures are those for the carbonate ion Fig. 5.5
Fig. 5.5 Resonance in CO3–2
Resonance in Carbonate Ion (CO₃²–):
■ Experimental evidence shows all C–O bonds are equivalent, meaning resonance structures must be considered together to describe its properties accurately.
Resonance in Benzene:
■ If benzene existed as a single resonance form, two different bond lengths would be present.
■ Actual Bond Length: 139 pm, intermediate between C–C (154 pm) and C=C (134 pm).
■ Benzene’s structure is often represented as a hexagonal skeleton without explicitly showing C and H atoms.
Effect of Resonance:
■ Stabilization: The resonance hybrid has lower energy than individual canonical structures, making the molecule more stable.
■ Bond Averaging: Resonance averages bond characteristics, leading to equalized bond lengths and strengths.
Resonance Energy
■ Resonance Energy (E): The difference between the actual bond energy of a molecule (E₀) and the most stable canonical form's energy (E₃).
■ Formula: E = E₃ – E₀
■ Stability: Higher resonance energy means greater stability of the molecule or ion.
Fig.5.6 Resonance Energy
■ It may be noted that the canonical forms have no real existence. The real molecule cannot be depicted by a single Lewis structure.
5.4 VSEPR THEORY [VALENCE SHELL ELECTRON PAIR REPULSION THEORY]
■ Lewis concept is not able to explain the shapes of molecules.
■ This VSEPR theory provides a procedure to predict the shapes of different polyatomic molecules.
■ Proposed by sidgwick and powell. this was fur ther developed by Nyholm and Gillespic.
■ A central atom of a covalent molecule has electron pairs in its valence shell.(bp,lp).
■ These electrons pairs are oriented in space with least repulsions. Thus the molecular assumes a particular shape.
■ When central atoms of molecules have no lone pairs, the shapes predicted are as follows in the Table.
Postulates of VSEPR Theory
■ Molecular Shape: Determined by repulsions between all electron pairs in the valence shell of the central atom.
Lone Pair vs. Bond Pair:
■ Lone pairs (LP) occupy more space than bond pairs (BP) since they are attracted to only one nucleus.
■ Repulsion order: LP–LP > BP–LP > BP–BP.
Effect on Bond Angles:
■ Lone pairs distort bond angles by reducing the actual bond angle between bonded atoms.
Electronegativity Influence:
■ Stronger electronegativity differences between the central and bonded atoms affect bond pair repulsions.
Effect of Multiple Bonds:
■ Triple bonds > Double bonds > Single bonds in terms of repulsion strength.
■ However, multiple bonds between two atoms count as a single direction of repulsion in determining molecular shape.
■ In studying molecular geometry it is convenient to divide different molecules into two categories:
108
Shapes based on VSEPR model Number of electron pairs
CHAPTER 5: Chemical Bonding and Molecular Structure
and 900
i) Molecules in which the central atom has no lone pairs
ii) Molecules in which the central atom has one or more lone pairs
Molecules with no Lone Pairs on Central Atom
■ The molecules have the general formula AB x, where x is an integer 2,3,4,5,6 and so on.
AB2 (Beryllium Chloride)
■ BeCl2 has two bond pairs. Thus the ClBeCl angle in BeCl2 is 180° and the molecule is linear.
AB3(Boron Trifluoride)
■ BF 3 contains three covalent bonds or bond pairs. Thus, each of the three FBF angles in BF 3 is 120° and all four atoms lie in the same plane. The geometry of BF 3 is trigonal planar.
AB4 (Methane)
■ Since there are four bonding pairs, the geometry of CH 4 is tetrahedral. This is a three dimensional shape with each angle 109°28΄.
AB5 (Phosphorus Pentachloride)
■ PCl 5 has five bond pairs, shape is trigonal bipyramid. The two Cl atoms are in the axial position and three in equatorial positions.
AB6 (Sulphur Hexafluoride)
■ SF6 has six bond pairs and has octahedral arrangement. In SF6 molecule in which each of the FSF angles is 90° and 180°.
■ Iso-electronic species usually have the same structure.
■ Thus, BF4, and NH4+ are all tetrahedral. CO32–, NO3– and SO3 are all trigonal planar and CO2, N3–, BO2– and NO2 + are all linear.
Molecules with Lone Pairs on the Central Atom
■ We designate molecules with lone pairs as AB x Ey. Here, x and y are integers.
AB2E (Sulphur dioxide):
■ The sulphur dioxide molecule has three electron pairs on the sulphur atom.
■ The overall arrangement is trigonal planar. the SO2 molecule has a bent shape. Due to lone pair and bonding pair repulsion, the two sulphur- oxygen bonds are pushed together slightly and OSO angle is less than 120°. By experiment, OSO bond angle is found to be 119.5°.
AB3E (Ammonia)
■ The ammonia molecule contains three bond pairs and one lone pair. But in NH3 one of the electron pairs is a lone pair, so the geometry of NH 3 is trigonal pyramidal.
■ The HNH angle in ammonia 107°is smaller than, the ideal tetrahedral angle of 109 .5°.
AB2E2(Water)
■ A water molecule contains two bonding pairs and two lone pairs.
■ These lone pairs tend to be as far from each other as possible. Consequently, the two O–H bond pairs are pushed towards each other, so that we can predict an even greater deviation form the tetrahedral angle tha n in NH3. This prediction is correct and HOH bond angle is 104.5°.
■ Structures of some molecules with lone pairs on the central atom are given in the table.
CHAPTER 5: Chemical Bonding and Molecular Structure
Structures of some molecules and ions with lone pairs on central atom
8. SnCl2 is angular, but HgCl2 is linear. Why? Write their VSEPR notations.
Sol. SnCl2 has two bond pairs and one lone pair.
Hence, it is angular. HgCl2 has 2 bond pairs and no lone pair. Hence, it is linear. SnCl2 is denoted as AB2E. HgCl2 is denoted as AB2
9. In SF4 molecule, lone pair of electrons occupies equatorial position but not axial position Why?
Sol.
In (a) the lp is present at axial position so there are three lp – bp repulsions at 90°. In (b) the lp is present at equatorial position, and there are only two lp – bp repulsions at 90°. Hence arrangement (b) is more stable i.e lp occupies equatorial position.
IL ACHIEVER SERIES FOR JEE CHEMISTRY
TEST YOURSELF
1. Maximum number of planar atoms in SF6 molecule is (1) 5 (2) 4 (3) 6 (4) 7
2. The angle between two covalent bonds is minimum in (1) H2O (2) CO2 (3) NH3 (4) CH4
3. Octahedral molecule among the following is (1) SO3 (2) CHCl3 (3) SF6 (4) PCl5
4. In PCl5, bond angle in plane is (1) 90° (2) 120° (3) 180° (4) 109°.28΄
5. The shape of water molecule and bond angle in water molecule, according to hybridisation concept, respectively, are (1) angular, 90° (2) angular, 104°30΄ (3) tetrahedral, 109°28΄ (4) tetrahedral, 104°30΄
6. The geometry of H3O+ ion is (1) planar (2) triangular (3) pyramidal (4) tetrahedral
7. The shape of AB3E type molecule is (1) pyramidal (2) tetrahedral (3) angular (4) linear
8. Which one of the following is a correct set? (1) H2O, sp3 hybridisation, angular (2) H2O, sp2 hybridisation, linear (3) NH4+, dsp2 hybridisation, square planar (4) CH4, dsp2 hybridisation, tetrahedral
9. In which of the following molecules, sigma bonds formed by the overlap of sp 3d and p-orbitals are absent?
(1) PCl5 (2) ClF3 (3) SbCl5 (4) HClO4
10. Which of the following is not tetrahedral? (1) BF 4 – (2) NH4+ (3) CO32– (4) SO42–
11. Which of the following has square pyramidal structure? (1) BrF 5 (2) ClF3 (3) IF 7 (4) ClF
12. Structure of ICl2 – is (1) trigonal (2) octahedral (3) square planar (4) linear
13. The lone pairs in ClF3 and SCl4 are present respectively, at (1) axial, axial (2) axial, equatorial (3) equatorial, axial (4) equatorial, equatorial
14. Which has triangular planar shape? (1) CH3+ (2) CIO2– (3) H3O+ (4) CIO3–
5.5 VALENCE BOND THEORY
■ Valence Bond Theory (VBT): Proposed by Heitler & London to explain covalent bond formation using Schrödinger’s wave equation.
■ Extension by Pauling & Slater: Expanded VBT to explain molecular shapes and bond directionality.
Postulates
■ Covalent Bond Formation: Occurs by the overlap of atomic orbitals in the valence shell of two atoms.
■ Overlap Condition: Only half-filled orbitals can participate in bonding.
■ Electron Density: The shared electron pair is concentrated along the internuclear axis.
■ Spin Pairing: The bond consists of a pair of electrons with opposite spins.
■ Strength of Bond: Greater overlap → Stronger covalent bond.
■ Bond Directionality: Bonds form in the direction where electron clouds are concentrated.
Attraction & Stability:
■ Bonding electrons are attracted by both nuclei, reducing repulsions and increasing stability.
■ The electron cloud is symmetric, ensuring a stable structure.
Bond Energy:
■ Energy released per mole during orbital overlap stabilizes the system.
■ The molecule has lower energy and is more stable than isolated atoms.
Formation of H2 Molecule
The orbital overlap present in hydrogen molecule is s-s overlap.
Formation of Cl2 molecule
The orbital overlap present in chlorine molecule is p-p overlap.
Formation of HCl Molecule
The orbital overlap present in hydrogen chloride molecule is s-p overlap.
1s3ps-poverlap
In a given energy level, the overlapping efficiency is in the order: s-s < s-p < p-p.
■ Due to homoatomic overlap, homoatomic molecule is formed. eg. H2, F2, Cl2, Br2, I2, O2, N2 etc.
■ The overlap between atoms of different elements is called heteroatomic overlap.The molecule formed is called heteroatomic or heteronuclear molecule. Eg. HF, HCl, IC.
■ Homoatomic molecules are non-polar but hetrodiatomic molecules are polar.
■ Based on type of overlapping, covalent bonds are of two types sigma (s) bond and pi (p) bond.
Sigma Bond
■ Formation: End-to-end (axial/head-on) overlap of half-filled orbitals along their axes.
■ Strength: Stronger due to maximum electron cloud overlap.
■ Symmetry: Electron cloud is symmetrical around the internuclear axis.
■ Types of Overlap: s–s, s–p, p–p (coaxial overlapping).
■ First Bond: Always s is the first bond formed betw een bonded atoms.
Pi (π) Bond:
■ Formation: Sidewise/lateral/parallel overlap of half-filled orbitals.
■ Orbital Alignment: Overlapping orbitals must be parallel to each other and perpendicular to the internuclear axis.
Sidewise
Overlaping p- orbitals
π bonded electroncloud (or)
Pi (π) Bond Structure:
■ Consists of two electron clouds above and below the plane of bonded atoms.
■ Formed by p or d orbitals (not s-orbitals).
Strength & Rotation:
■ σ bond is stronger due to greater overlap; π bond is weaker with limited overlap.
■ σ bond allows free rotation, while π bond restricts rotation.
Formation & Molecular Shape:
■ σ bond forms independently, but π bond always exists with a σ bond.
■ Molecular shape is determined by σ bonds, not π bonds.
Reactivity:
■ σ bond-only substances are less reactive.
■ π bond-containing substances are more reactiv e.
■ Differences between sigma bond and pi bond is shown in the table below.
1. Sigma bond is formed by the overlap of orbitals along their axes. It includes s-s, s-p and p-p overlapping.
2. s bond is relatively stronger bond.
3. Electron cloud is symmetrical about the inter nuclear axis.
CHAPTER 5: Chemical Bonding and Molecular Structure
1. Pi bond is formed by the side-wise or lateral overlap of orbitals.
2. p bond is relatively weaker bond.
3. Electron cloud is unsymmetrical, but present in two lobes are in equidistant to the axis.
4. There can be free rotation of the atoms around the bond. 4. Free rotation is not possible around the bond.
5. Substances with bonds are less reactive. 5. Substances with p bonds are more reactive.
6. The shape of the molecule is described by bonds. 6. p bonds do not affect the shape of the molecule.
7. s electrons are referred to as localised electrons. 7. p electrons are referred to as mobile
8. s bond has independent existence.
Hybridization and Bonding:
8. p bond exists along with the bond.
■ Hybrid orbitals and s orbitals form σ bonds, not π bonds.
■ Single bonds consist of only one σ bond, formed by sharing one pair of electrons.
■ Double Bond: Consists of one σ bond and one π bond, formed by sharing two electron pairs.
■ Triple Bond: Consists of one σ bond and two π bonds, formed by sharing three electron pairs.
■ Multiple Bonds: Double and triple bonds are collectively referred to as multiple bonds.
■ Without forming a s bond between two atoms, a pi bond can never be formed. The difference between a s bond and a p bond are given in Table 5.6.
Localised Orbital
■ An orbital with bonded electron cloud between the nuclei of the bonded atoms is called localised orbital.
Formation of Oxygen Molecule
■ The half-filled 2p z orbitals on both the oxygen atoms are parallel to one another. They undergo lateral overlap to form a weak p bond, as shown in Fig. 5.7
Fig. 5.7 A Double Bond in O2 Molecule
Sigma (s ) bond
(p ) bond
Formation of Nitrogen Molecule
■ The half filled 2p x orbitals of two nitrogen atoms overlap along the same axis to form a strong s bond.
■ The remaining half filled 2py and 2pz orbitals to form two p bonds as shown in Fig. 5.8.
Nitrogen atom Nitrogen atom
Fig. 5.8 A triple bond in N2 molecule
Formation of H2O Molecule
■ Bond Formation in H₂O: Two O–H sigma (σ) bonds form by the overlap of oxygen’s 2py and 2pz orbitals with hydrogen’s 1s orbitals.
■ Theoretical Bond Angle: Expected to be 90° since 2p y and 2pz orbitals are perpendicular.
■ Experimental Bond Angle: 104°30', explained by VSEPR theory, which accounts for lone pair repulsions that slightly reduce the bond angle.
Formation of CH4 Molecule
■ Tetravalence of Carbon: Explained using excited state electronic configuration: 1s² 2s¹ 2p x ¹ 2py¹ 2pz¹.
■ Energy Source for Excitation: Comes from C–H bond formation energy.
■ One C–H bond is s–s,three C–H bonds are s–p.
■ One bond is non-directional, and the remaining three are at 90° angles.
Experimental Observations:
■ All four C–H bonds in CH₄ are identical with a bond angle of 109°28'.
■ Explanation: Later justified by hybridization and VSEPR theory.
Directional Property of Covalent Bond
■ Hydrogen Bonding: Formed by non-directional 1s–1s overlap, making it the simplest covalent bond.
■ Directional Covalent Bonds: All other covalent bonds exhibit directionality in simple and polyatomic molecules.
CHAPTER 5: Chemical Bonding and Molecular Structure
■ Types of Overlap in Bond Formation:
■ Positive Overlap: Leads to covalent bond formation.
■ Negative Overlap: Results in weak bonding or instability.
■ Zero Overlap: No bond formation occurs.
■ No bond forms between p x and py orbitals.
■ No bond forms between an s orbital and a px orbital if aligned along the y-axis or z-axis.
■ Positive or negative overlaps lead to the bond formation, as shown in table below.
Positive or negative overlapping
Drawbacks
■ Explains bond strength and directionality but fails to justify identical valencies of Be, B, and C.
■ Does not explain the shape and bond angles of molecules like CH₄, NH₃, and H₂O.
■ Cannot predict actual molecular geometries despite accounting for bond formation.
Solved Examples
10. What kind of a bond is formed when the orbitals of two atoms A and B undergo (i) s-s overlap (ii) s-p overlap?
Sol. In both (i) and (ii), only s -bonds are formed. Because s-orbitals are nondirectional in nature, always overlap along the internuclear axis only. So s-orbitals form s -bonds only. The s -bonds in (i) and (ii) are denoted as and , respectively.
11. Can px overlap a py orbital? State the reason?
Sol. No. They cannot overlap because their orientations and sy mmetries are not same.
Try yourself
1. On the basis of valence bond theory predict the bond angle in NH 3(experimental bond angle is 107°)
Answer: 90°
TEST YOURSELF
1. Most important concept of valence bond theory is (1) overlap of atomic orbitals results in the bond (2) sharing of odd number of electrons for bonding (3) sharing of electrons follows the octet rule (4) transfer of electrons follows the octet rule
2. The type of overlap present in the bonds of hydrogen sulphide molecule is (1) s–p (2) s–s (3) p–p (4) sp3–s
3. Iodine monochloride molecule is formed by the overalap of (1) s-s orbitals (2) s-p orbitals (3) p-p orbitals end to end (4) p-p orbitals sideways
4. Choose the molecule that has only one pi bond . (1) CH2 = CH – CH = CH2 (2) CH2 = CHCOOH (3) CH3CH = CH2 (4) CH ≡ CH
5. Double bonds are present in (1) CO2 (2) BeCl2 (3) HgCl2 (4) MgO
6. According to VB theory, the bonds in methane are formed due to the overlapping of (1) 1 s s–s, 3 s s–p (2) 1 s s–p, 3 s s–s (3) 2 s s–s, 2 s s–p (4) σ 3 sp 4
7. The bond between chlorine and bromine in BrCl is (1) ionic (2) non-polar (3) polar with negative end on Br (4) polar with negative end on Cl
8. Some statements about valence bond theory are given below. a) The strength of bond depends upon extent of overlapping. b) The theory explains the directional nature of covalent bond. c) According to this theory oxygen molecule is paramagnetic in nature. Choose the correct option. (1) All are correct. (2) Only a and c are correct. (3) Only a and b are correct. (4) All are wrong.
9. In the electronic structure of acetic acid, there are (1) 16 shared and 8 unshared valence electrons
(2) 8 shared and 16 unshared valence electrons
(3) 12 shared and 12 unshared valence electrons
(4) 18 shared and 6 unshared valence electrons
Answer Key
5.6 HYBRIDISATION
■ Pauling introduced hybridization.
■ Hybridization: Intermixing of orbitals with similar energy to form an equal number of equivalent orbitals.
■ Hybrid Orbitals: Newly formed orbitals with identical energy and shape.
Common hybridisations and corresponding shapes
S. No. Pure orbitals involved in hybridisation Resulting hybrid orbitals (their no.) Shape of the molecules as a result Bond angles in them. Examples
1. one s; one p sp (2) Linear 1800; BeCl2; C2H2
2. one s; two p sp2 (3) Trigonal planar 1200; BCl3; C2H4
3. one s; three p sp3 (4) Tetrahedral 109028'; CH4 ; [NiCl4]2–, MnO4–(NH3 and H2O bond angle decreases)
4. one s; two p one d dsp2 (4) Square planar 90 0 and 180 0; [Cu(NH 3) 4] 2+; [PdBr 4] 2–, [Ni (CN)4]2–,
5. one s; three p one d sp3d (or) dsp3 (5) Trigonal bipyramidal 900; 1200; 1800 PCl5, PF5, SbCl5
6. one s; three p; two d sp 3 d 2 (or) d2sp3 (6) Square pyramidal (or) Octahedral 900; 1800; BrF5, SF6, TeF6, PF6–, [Co(NH3)6]3+, XeOF4, CrF63–
7. one s; three p; three d sp3d3 Pentagonal bipyramidal 900; 720; 1800 IF 7
Rules of Hybridisation
■ Hybridization occurs only within orbitals of the same atom, not between different atoms.
■ Participating orbitals must have nearly equal energy levels.
■ Half-filled, completely filled, and empty orbitals can hybridize.
■ Outer orbitals are not necessarily required to hybridize.
■ Number of hybrid orbitals = Number of atomic orbitals involved.
■ Hybrid orbitals arrange symmetrically for maximum stability.
■ Bond angles are determined by equal spacing between hybrid orbitals.
■ Electron distribution follows Pauli’s exclusion principle and Hund’s rule.
■ Formation: One s, three p, and one d orbital mix to form five equivalent sp³d hybrid orbitals.
■ Examples: PCl₅ (Phosphorus Pentachloride, with equatorial and axial bonds)
■ The axial bonds (2.19A0) have been found to be larger than equatorial bonds(2.04A 0). This is because axial P–Cl bonds experience greater repulsion from equatorial P-Cl bonds. PCl 5 is highly reactive and in the solid state exists as [PCl 4]+ and [PCl6]– ions
sp3d2 Hybridisation
■ sp³d² Hybridization (Octahedral, 90°)
■ Formation: One s, three p, and two d orbitals mix to form six equivalent sp³d² hybrid orbitals.
■ Examples: SF₆ (Sulfur Hexafluoride, completely regular structure with 90° bond angles)
■ This structure is completely regular with bond angles of 90° in Figure.
CHAPTER 5: Chemical Bonding and Molecular Structure
■ Formation: One s, three p, and three d orbitals mix to form seven equivalent sp³d³ hybrid orbitals.
■ Examples: IF₇ (Iodine Heptafluoride, bond angles 72° and 90°)
sp3d3 Hybridisation
■ The intermixing of one ‘s’, three ‘p’ and three ‘d’ orbitals to form seven equivalent sp3d3 hybrid orbitals is called sp 3d3 hybridisation.
■ Hybrid orbitals are orientated towards the corners of a pentagonal bipyramid and angles between them are 72° and 90°. Eg.IF 7. In this bond angle between equitorial F–I–F bonds is 72°, bond angle between equitorial and axial bond is 90°.
Solved Examples
12. Discuss the hybridisation of carbon atoms in allene, C 3 H4, and show the orbital overlaps.
Sol. Allene is : CH2 = C = CH2
I II III
Carbon atoms I and III are sp2 hybridised, while carbon II is sp hybridised. The two unhybridised orbitals of carbon II overlap sidewise with unhybridised p orbitals of carbon I and III to form bonds.
13. Calculate the ratio of number of pure and hybrid orbitals used for bonding in an acetylene molecule.
Sol. Acetylene has two ‘C’ atoms and two ‘H’ atoms.
Carbon undergoes sp hybridisation.
Number of hybrid orbitals = 2 × 2 = 4
Number of pure orbitals = 4 (from C) + 2 (from H) = 6
Ratio of number of pure and hybrid orbitals = 4:6 = 2:3
TEST YOURSELF
1. The percentage of ‘p’ character in hybrid orbital of the central atom of water molecule is (1) 25% (2) 50% (3) 75% (4) 33.3%
2. Which hybridisation is found in NH3 and H2O? (1) sp3 (2) dsp2 (3) sp (4) sp2
3. Octahedral shape is due to the hybridisation (1) sp3d (2) sp3d2 (3) sp3 (4) sp
4. The type of hybrid orbitals used by the central atom in perchlorate ion is (1) sp3 (2) sp2 (3) sp (4) dsp2
5. The hybrid orbitals have a bond angle of 109.50. The percentage of p-character in the hybrid orbital is nearly (1) 25% (2) 33% (3) 50% (4) 75%
6. In BCl3 molecule, the Cl-B-Cl bond angle is (1) 90° (2) 120° (3) 109°.28' (4) 180°
7. The hybridisation of oxygen in OF2 molecule is (1) sp (2) sp2 (3) sp3 (4) dsp2
8. In hydrazine (N2H4), the hybridisation of nitrogen is (1) sp (2) sp2 (3) sp3 (4) dsp2
9. Increasing order of size of hybrid orbitals is (1) sp, sp2, sp3 (2) sp3, sp2, sp (3) sp2, sp3, sp (4) sp2, sp, sp3
10. A square planar complex is formed by hybridization of which atomic orbitals? (1) s, px, py, 2 z d (2) s, px, py, 22 xy d (3) s, px, py, dyz (4) s, px, pz, d xy
11. The type of overlapping not observed in the formation of ethylene molecule is (1) 22 spsp σ (2) 2 spp σ (3) 2 sps σ (4)
12. The d-orbitals involved in SP 3d2 hybridisation of central atom are (1) 222 zxy d,d (2) 22 xy,yz,zx xy d,ddd (3) 2 xy,yz,zx z
■ Molecular Orbital Theory (MO Theory): Developed by Hund and Mulliken.
Postulates:
■ Electrons in Molecules: Just like atomic orbitals hold electrons in atoms, molecular orbitals hold electrons in molecules.
■ Formation of Molecular Orbitals: Created by the combination of atomic orbitals with similar energies and proper symmetry.
Electron Influence:
■ Atomic Orbital: Electron is influenced by one nucleus (monocentric).
■ Molecular Orbital: Electron is influenced by two or more nuclei (polycentric).
■ Number of Molecular Orbitals: Equals the number of combining atomic orbitals.
Types of Molecular Orbitals:
■ Bonding Molecular Orbital (BMO): Lower energy, greater stability.
■ Anti-Bonding Molecular Orbital (ABMO): Higher energy, less stability.
■ Electron Probability Distribution:
■ Atomic Orbital (AO): Around a single nucleus.
■ Molecular Orbital (MO): Around multiple nuclei.
Main differences between atomic and molecular orbitals
S.No. Atomic orbitals
CHAPTER 5: Chemical Bonding and Molecular Structure
Molecular orbitals
1. They belong to one specific atom only They belong to all the atoms in a molecule
2. They are the internal characteristic of an atom They result when atomic orbitals of similar energies combine
3. They have simple shapes of geometries They have complex shapes
4. The atomic orbitals are named as s, p, d, f, .... etc. The molecular orbitals are named as s,p,d , . . . etc.
5. The stabilities of these orbitals are less than bonding and more than the anti-bonding orbitals
Filling of Molecular Orbitals:
The stabilities of these orbitals are either more or less than the atomic orbitals.
■ Follows Aufbau Principle, Pauli’s Exclusion Principle, and Hund’s Rule.
Notation:
■ Like atomic orbitals (s, p, d), molecular orbitals are denoted by Greek letters (σ, π, δ).
Formation of Molecular Orbitals
■ Formation of Molecular Orbitals: Occurs by the combinatio n of atomic orbitals of bond ed atoms.
Wave Functions and Schrödinger Equation:
■ Atomic orbitals are expressed as wave functions (Ψ) derived from Schrödinger’s equation.
■ Solving Schrödinger’s equation for molecules is complex, so the Linear Combination of Atomic Orbitals (LCAO) method is used as an approximation.
LCAO Method in H₂ Molecule:
■ Hydrogen atoms A and B have 1s atomic orbitals, represented by wave functions ΨA and ΨB. Molecular Orbitals (MOs) Formation:
■ Addition of Wave Functions: Forms a bonding molecular orbital (Ψ = ΨA + ΨB).
■ Subtraction of Wave Functions: Forms an anti-bonding molecular orbital (Ψ = ΨA - ΨB).
Therefore two molecular orbitals s and s* are formed as shown in Fig. 5.18
s = Y A + Y B; s * = Y A – Y B
■ Bonding Molecular Orbital (σ): Formed by the addition of atomic orbitals (Ψ = ΨA + ΨB), resulting in lower energy and greater stability.
■ Anti-Bonding Molecular Orbital (σ*): Formed by the subtraction of atomic orbitals (Ψ = ΨA - ΨB), leading to higher energy and reduced stability.
Bonding Molecular Orbital (BMO):
■ Formed by constructive interference, where electron waves reinforce each other.
■ Lower energy than parent atomic orbitals, increasing stability.
Anti-Bonding Molecular Orbital (ABMO):
■ Formed by destructive interference, where electron waves cancel each other.
■ Higher energy than parent atomic orbitals, reducing stability.
Energy Conservation:
■ The total energy of the bonding and anti-bonding orbitals equals the energy of the original atomic orbitals.
Energy Order of Orbitals:
■ Bonding < Non-Bonding < Anti-Bonding.
Conditions for the combinations
1. Energy Compatibility: The combining atomic orbitals must have the same or nearly the same energy.
2. Symmetry Matching: The orbitals must have the same symmetry about the molecular axis (z-axis by convention).
3. Maximum Overlap: The greater the overlap, the higher the electron density between the nuclei, leading to stronger molecular orbitals.
4. Orbital Approach:
■ s s–s and p–p orbitals can combine when aligned along the z-axis.
■ Sigma molecular orbitals are formed as shown in Fig. 5.18.
Fig. 5.16 Order of Energies of Bonding and Anti-bonding Orbital
■ Sidewise Overlap: When atomic orbitals overlap laterally, they form molecular orbitals with a nodal plane.
■ Orbital Lobes: Appear on either side of the internuclear axis.
■ Bond Symmetry: The resulting bond lacks cylindrical symmetry and is called a π-bond.
■ p –bonds due to p-p overlaps are shown in Fig. 5.19 and 5.20.
Orbitals of 1st atom Orbitals of 2nd atom
CHAPTER 5: Chemical Bonding and Molecular Structure
Fig. 5.17 Sigma bonding and anti-bonding molecular orbits
■ Main difference between sigma and pi molecular orbitals are given in the table below.
Differences between s and p Molecular Orbitals.
S.No.
s -molecular orbitals
p -molecular orbitals
1. Formed by the end on overlap along the inter nuclear axis. Formed by the side wise overlap, perpendicular to the internuclear axis.
2 Overlapped region is very large. Overlapped region is small.
3. Rotation about the internuclear axis is symmetrical. Rotation about the internuclear axis is unsymmetrical.
4. Strong bonds are favoured. Weak bonds are favoured.
■ The sequence of increasing order of energy of various molecular orbitals for O 2 and F2 is s 1s < s* 1s < s 2s < s* 2s < s2 pz< ()() xyxy 2p2p*2p*2p π=π<π=π < s2 pz,
Elec tronic Configuration and Molecules Behaviour
Stability of Molecules
■ Let Nb = Number of electrons in bonding orbitals, N a = Number of electrons in anti-bonding orbitals.
1. If Nb > Na, the molecule is stable.
2. If Nb < N a or Nb = Na, the molecule is unstable.
Bond Order
■ The bond order is defined as one half of the difference between the number of electrons present in bonding and in anti bonding orbitals.
Bond order ba 1 NN 2 =−
■ Bond Order & Stability:
■ Positive bond order → Stable molecule.
■ Negative bond order → Unstable molecule.
■ Bond Order & Bond Type:
■ Bond Order = 1 → Single Bond.
■ Bond Order = 2 → Double Bond.
■ Bond Order = 3 → Triple Bond.
Bond Length
■ Higher bond order → Shorter bond length.
■ Lower bond order → Longer bond length.
■ Bond order is an approximate meas ure of bond length.
Magnetic Behaviour
■ Diamagnetic Substance: All molecular orbitals are fully occupied (paired electrons).
■ Paramagnetic Substance: One or more molecular orb itals have unpaired electrons.
Bonding in Some Homonuclear Diatomic Molecules
In this section we shall discuss bonding in some homonuclear diatomic molecules as follows.
Hydrogen Molecules (H2)
■ Electronic configuration is H 2 : ( s 1s2)
■ Bond order NNba 20 1 22 ===
■ Hydrogen Molecule (H₂): The two hydrogen atoms are bonded together by a single covalent bond.
■ Bond Dissociation Energy: 438 kJ/mol.
■ Bond Length: 74 pm.
■ Magnetic Nature: Since no unpaired electrons are present, H₂ is diamagnetic.
Helium Molecule (He2)
■ The electronic configuration of He2 is 2*2 2 He:(1s)(1s) σσ
Bond order of He2 is 1/2 (2–2) = 0
■ Therefore unstable and does not exist.
■ Be2 bond order is zero, hence does not exist.
Lithium Molecule (Li2)
■ 2*22 2 Li:(1s)(1s)(2s) σσσ
■ Its bond order, therefore, is 1/2(4–2) = 1. It means that Li 2 molecule is known to exist in the vapour phase.
CHAPTER 5: Chemical Bonding and Molecular Structure
Carbon Molecule (C2)
■ is 2*22*222 2y C:(1s)(1s)(2s)(2s)(2p2P) σσσσπ=π x or 2*222 xy KK(2s)(2s)(2p2p) σσπ=π
■ Bond Order of C₂: ½(8–4) = 2, indicating a double bond.
■ Magnetic Nature: C₂ is diamagnetic as all electrons are paired.
■ Bond Composition: The double bond in C₂ consists of two π bonds, due to the presence of four electrons in two π molecular orbita ls.
Nitrogen Molecule (N2)
■ The electronic configuration of N 2 molecule, therefore, is
So, in nitrogen molecule, atoms are held by a triple bond. N 2 is diamagnetic because of no unpaired electron.
Oxygen Molecule (O2)
■ The electronic configuration of oxygen molecule is KK 22222 zxy (2s)(*2s)(2p)(2p)(2p) σσσπ≡π 11 xy (*2p)(*2p) π≡π
Bond order 106 2 =
■ The electrons in p *2px and p *2py orbitals remain unpaired. In oxygen molecule, the atoms are held by a double bond. It should be paramagnetic.
■ The bond dissociation energy of oxygen is 495 kJ mol –1. Bond length in O2 is 121 pm.
■ The electronic configuration of O 2 + is KK ( s 2s)2 ( s* 2s)2 ( s 2pz)2 ( p 2px)2 ( p 2pz)2 ( p* 2px)1
Bond order 83 2.5 2 ==
■ The bond strength of O2 + is more than that of O2 and bond length of O2 + is less than that of O2. O2+ is paramagnetic.
The electronic configuration of O 2 – is
KK ( s 2s)2 ( s* 2s)2 ( s 2pz)2 ( p 2px)2 ( p 2py)2 ( p 2px)2 ( p* 2px)2 ( p* 2py)1
Bond order 853 1.5 22 ===
■ The bond order of O2 – is less than that of O2. The bond in O2 – is weaker than in O2 and the bond length of O2 – is larger than that of O 2. O2– is paramagnetic.
■ The electronic configuration of O 2 2– is KK ( s 2s)2 ( s* 2s)2 ( s 2pz)2 ( p 2px)2 ( p 2py)2 ( p 2px)2 ( p* 2px)2 ( p* 2py)2
Bond order 86 1 2 ==
■ Peroxide Ion (O₂²–):
■ Bond Order: 1 (Weaker bond than O₂).
■ Bond Strength: Weaker than O₂ due to lower bond order.
■ Bond Length: Longer than O₂.
Order of bond energies is: O 2 + > O2 > O2– > O22–
Order of bond lengths is: O 2 2– > O2– > O2 > O2+
Configuration and magnetic properties of some diatomic molecules and their ions are in given Table 5.14. Electronic arrangement in molecules are in given Table 5.15.
Table 5.14 Configuration and Magnetic Properties of Diatomic Molecules and Ions
Molecule
O2 s 1s2 s* 1s2 s 2s2 s* 2s2 s2 p2z p2 p2x p2 p2y p* 2p1x p 2p1y 2 2 paramagnetic
CHAPTER 5: Chemical Bonding and Molecular Structure
Table 5.15 Electronic Arrangement in Molecules Diamagnetic Diamagnetic with 2s-2p mixing Bond energy/k.rmol
without 2s-2p mixing
Bond order Magnetic Properties valence electron configuration Paramagnetic Paramagnetic Diamagnetic
Molecular Orbital Theory for Heteronuclear Diatomic Molecules
■ NO molecule [Total number of electrons = 15]
Electron Configuration:
KK: xyy 22222*1 2s2sz2p2p2p *2p σσσπππ
Bond order in NO = 105 2.5 2 =
Bond order in NO + = 104 3 2 =
Bond order NO–= 106 2 2 =
■ CN– ion [number of electrons = 14]
Electron configuration:
KK xyy 22222*1 2s2sz2p2p2p *2p σσσπππ
Bond order in CN– = 104 3 2 =
■ BN molecule (Total number of electrons = 12)
The energy order of its molecular orbitals will be the same as that of molecules having electrons less than those in N 2.
Electron configuration, KK xy 2*222 2s2s2p2p σσππ
Bond order 62 2 2 ==
Note:
Bond order of CO + is higher than that of CO, i.e., Bond length of CO + is shorter than that of CO. This is explained by suggesting that 2s orbital of C-atom. When they mix to form s 2s and s 2s orbitals, the latter exceeds the energy of s 2pz as well as x 2pπ and y 2pπ
Thus, electron configuration
■ CO: KK xyz 2222*2 2s2p2p2p2s σππσσ
■ CO + : KK yz x 222*2*1 2s 2p2p2s 2p σππσσ
■ Bond order of CO = 82 3 2 =
■ Bond order of CO + 81 3.5 2 == .
Solved Examples
14. By the use of molecular orbital consideration, account for the fact that oxygen is paramagnetic.
Sol. Electronic configuration of molecular orbitals of O 2 is ( s 1s)2 ( s* 1s)2 ( s 2s)2 ( s 2pz)2 ( p 2px)2 ( p 2py)2 ( p* 2px)1 ( p* 2pz)1
The last two electrons are in p*2px and p*2py anti bonding orbitals so as to maximize pi electrons in accordance with Hund’s rule. These two unpaired electrons confer paramagnetism in oxygen.
15. Use molecular orbital theory to explain why Be 2 molecule does not exist.
Sol. The atomic no.(z) of Be is 4. This means that 8 electrons are to be filled in the M.O. of Be 2
The configuration is : s 1s2 s* 1s2 s 2s2 s*2 s2; ba 11 B.O.NN440 22 −=−=
As bond order of Be2 is zero, the molecule of Be2 does not exist.
TEST YOURSELF
1. The paramagnetic nature of oxygen is best explained by (1) VB theory (2) hybridisation (3) MO theory (4) VSEPR theory
3. Number of anti-bonding electrons in O2 molecule are (1) 10 (2) 6 (3) 4 (4) 2
4. In which pair, the stronger bond is found in the first species? a) O,O22 b) N,N22+ c) NO,NO+− (1) a only (2) b only (3) a and c only (4) b and c only
5. Bond energy is maximum in (1) F2 (2) N2 (3) O2 (4) Br2
6. Bond order is 2.5 in
a) O2+ b) N2+ c) NO (1) a, b only (2) b, c only (3) a, c only (4) a, b, and c
7. Which of the following molecular orbitals has the lowest energy for O 2 molecule? (1) s 2pz (2) x 2p σ (3) x 2pπ (4) x 2p * σ
8 The molecular orbital shown in the diagram can be described as (1) s (2) s * (3) p (4) p *
9. In O2 molecule, the correct order of energy of molecular orbitals is (1) yz 2p2pπ>π (2) yz 2p2p = ππ (3) 2s2s * σ<σ (4) x 2s2p **σ>σ
10. Maximum number of electrons that can be present in any molecular orbital is (1) 3 (2) 6 (3) 8 (4) 2
11. The molecular electronic configuration of B 2 is
(1) 2*211 xy KK(2s)(2s)(2p)(2p) σσππ
(2) 2*22 xKK(2s)(2s)(2p) σσπ
(3) 2*22 KK(2s)(2s)(2p) σσπ
(4) 2*211 KK(2s)(2s)(2p)(2p) σσσπ
12. When N2 goes to N2+, the N–N bond distance _____ and when O2 goes to O2+, the O–O bond distance ______.
13. Which combination of atomic orbitals is not allowed according to MO theory?
(1) px – px
(2) px – py
14. A bonding molecular orbital is produced by (1) destructive interference of wave functions (2) constructive interference of wave functions
(3) pairing of electrons with opposite spins
(4) combination of +ve and –ve wave functions
(3) py – py
(4) pz – pz
15. If Z-axis is the molecular axis, then pi-molecular orbitals are formed by the overlap of (1) s + pz
(3) pz + pz
(2) px + py
(4) px + px
16. Which one of the following is correct regarding s molecular orbital?
(1) The rotation along the inter nuclear axis is symmetric.
(2) It is formed by the partial overlap of atomic orbitals at right angle to inter-nuclear axis. (3) It is very weak bond.
(4) It has less overlapping region.
17. p 2px differs from p 2py molecular orbital in which of the following?
(1) Number of nodal planes
(3) Symmetry
(2) Energy
(4) Shape
18. In the formation of a homo-diatomic neutral molecule, if ‘N’ atomic orbitals combine, then the total number of bonding molecular orbitals formed is
(1) 2N
(3) N/2
(2) N
(4) N/4
Answer Key
5.10 HYDROGEN BONDING
■ Hydrogen Bonding: Introduced by Latimer and Rodebush to explain molecular association in liquids like water, ammonia, hydrogen fluoride, and formic acid.
■ Dipole Formation: The bond electron pair is strongly attracted toward the electronegative atom, creating a dipole.
■ Intermolecular Attraction: When such molecules come close, the positive end of one molecule attracts the negative end of another, forming a weak electrostatic force (hydrogen bond).
CHAPTER 5: Chemical Bonding and Molecular Structure
■ Thus, these molecules associate together to form a cluster of m olecules. XH...XH....XH....XH
■ Hydrogen Bond: A weak electrostatic attraction between a partially positive hydrogen atom and a highly electronegative atom (F, O, or N) in the same or different molecule.
■ Requirements: The atom bonded to hydrogen must be small and highly electronegative (F, O, or N).
■ Strength & Physical State Dependence:
■ Strongest in solids, weakest in gases.
■ Hydrogen bond energy: 12–50 kJ/mol, lower than a covalent bond.
■ O···H hydrogen bond length: ~1.76 Å (longer than covalent bonds).
■ Note : Deuterium also forms H.bonds.
5.10.1 Types of Hydrogen Bonding (two types)
Intermolecular Hydrogen Bonding
■ Intermolecular Hydrogen Bond: Formed between a partially positive hydrogen atom and an electronegative atom of different molecules within the same or different substances.
Hydrofluoric acid
Water
Ammonia
Ammonia in water
Formicacid
Parafluorophenol
Metanitrophenol
Para hydroxy benzaldehyde H–C=O H–O O-H O=C-H
Effects of Intermolecular H.Bonding:
■ Increases boiling point due to molecular association.
■ Enhances solubility in water.
■ Increases viscosity; stronger hydrogen bonding leads to higher viscosity..
Intramolecular Hydrogen Bonding
■ Intramolecular Hydrogen Bonding:
■ Formed within the same molecule between a partially positive hydrogen and an electronegative atom.
■ Leads to chelation, preventing molecular association.
■ No significant increase in boiling point; substances remain more volatile
Salicylaldehyde
Orthonitrophenol
Orthofluorophenol
Orthonitroaniline
Salicylic acid
Consequences of Hydrogen Bonding:
Abnormal Physical Properties:
■ Hydrogen-bonded substances exhibit unexpected physical states.
■ Example: Water should be a gas at room temperature but remains liquid due to hydrogen bonding.
Higher Melting & Boiling Points:
■ Water, ammonia, and hydrogen fluoride have higher boiling points than expected due to hydrogen bonding.
Solubility Enhancement:
■ Lower alcohols (e.g., CH₃OH, C₂H₅OH) dissolve in water due to hydrogen bonding.
■ Comparison of Para and Ortho Hydroxybenzaldehyde:
Para-hydroxybenzaldehyde:
■ Exhibits intermolecular hydrogen bonding → high boiling point, soluble in water.
■ Cannot be purified by steam distillation.
Ortho-hydroxybenzaldehyde:
■ Exhibits intramolecular hydrogen bonding → low boiling point, less soluble in water.
■ Volatile in steam, can be purified by steam distillation.
Viscosity & Density:
■ Sulfuric acid is denser and more viscous due to molecular association via hydrogen bonding.
■ Boiling Point Comparison of Water and HF:
■ Water has a higher boiling point than HF, despite HF having stronger hydrogen bonds.
■ Reason: Water forms more hydrogen bonds per molecule than HF, leading to greater overall intermolecular attraction.
■ In vapor phase, HF exists as (HF) n polymers, without breaking all hydrogen bonds.
Boiling Point of Ammonia vs. HCl:
■ Ammonia (NH₃) has a higher boiling point than HCl, even though N and Cl have similar electronegativity.
■ Reason: NH₃ forms hydrogen bonds, while HCl does not (due to the larger size of Cl).
Existence of Complex Salts:
■ KHF₂ (Potassium hydrogen fluoride) exists due to hydrogen bonding, but KHCl₂ does not (no hydrogen bonding in HCl).
■ Factors Affecting Hydrogen Bonding:
■ Electronegativity of the hydrogen-bonding atom.
■ Number of hydrogen bonds formed per molecule.
■ Size of the electronegative atom (smaller size favors stronger hydrogen bonding).
Group Trends in Hydrides:
■ Hydrogen bonding is absent in Group 14 hydrides (e.g., CH₄, SiH₄).
■ Hydrogen bonding is present in hydrides of Groups 15, 16, and 17 (NH₃, H₂O, HF).
■ The influence of hydrogen bonding on the boiling points of these hydrides is shown graphically in Fig. 5.20.
Fig. 5.18 Boiling Points of Hydrides
5.19 Pi bonding and anti-bonding molecular orbits
CHAPTER 5: Chemical Bonding and Molecular Structure
Fig.
Solved Examples
16. Why is salicylaldehyde less soluble in water?
Sol. Salicylaldehyde has chelate ring formed due to intramolecular hydrogen bonding.
Salicylaldehyde
Solute-solute attractions are stronger. Hence, solubility in water is less.
17. Why is orthofluorophenol more volatile than its meta and para isomers?
Sol. Intramolecular hydrogen bonding is present in orthofluorophenol, which does not lead to association of molecules.
Hence, it is volatile.
Intermolecular hydrogen bonding is present in meta and para isomers, which leads to molecular association.
TEST YOURSELF
1. Intermolecular hydrogen bonding is absent in (1) H2O (2) NH3 (3) C2H5OH (4) CH4
2. Which among the following has the highest volatility? (1) H2O (2) H2S (3) H2Se (4) H2Te
3. Among t he following, which has the highest boiling point? (1) NH3 (2) PH 3 (3) AsH3 (4) CH4
4. Abnormal boiling point of a compound is due to (1) van der Waals forces
(2) covalent bonding
(3) intermolecular hydrogen bonding
(4) intramolecular hydrogen bonding
5. The pair of substances that has hydrogen bonding between them is (1) C2H4, H2O (2) CH3CHO, H2O (3) CH3Cl, H2O (4) C6H6, H2O
6. In vapour phase, acetic acid dimerises due to (1) inter-molecular hydrogen bonding
(2) covalent bonds
(3) dative bonds
(4) intra-molecular hydrogen bonding
7. Maximum number of hydrogen bonds that one water molecule is capable of forming in ice is (1) 1 (2) 2 (3) 3 (4) 4
8. Which of the following hydrogen bonds is relatively weaker?
■ A type of covalent bond where a shared pair of electrons is donated by only one atom.
■ Sidgwick's Extension of Lewis Concept: Explained the mechanism of coordinate bond formation.
■ Electron Pair Donor: The atom that contributes the electron pair.
■ Electron Pair Acceptor: The atom that accepts the electron pair.
■ Notation: Represented by an arrow (→) pointing from the donor to the acceptor.
■ Condition for a Donor Atom: Must have at least one lone pair in its valence shell.
Example: NH3 H2O SCl one lone pair two lone pairs one lone pair Acceptor must have an empty orbital to accommodate the lone pair of electrons. Example: ☐H+ ☐BF3 ☐AlCl3 empty orbital empty orbital empty orbital
■ Dative (Coordinate) Bond Formation: Occurs when a lone pair donor shares electrons with an acceptor.
unshared pair
unshared pair
ion
■ Hydronium Ion Formation: H₃O+ forms when HCl dissolves in water, via a coordinate bond.
■ Dative vs. Covalent Bond: Once formed, a coordinate bond is indistinguishable from a covalent bond, differing only in mode of formation.
■ NH₃ and BF₃ Reaction: NH₃ donates a lone pair to BF₃, forming a dative bond as B lacks a full octet.
■ Orbital Overlap: sp³ hybrid orbital of N (with lone pair) overlaps with vacant p-orbital of B.
■ Mechanism: NH₃ donates a lone pair to H+, forming a coordinate covalent b ond.
Some other examples
HNO3 H O N=O
[Cu(NH3)4]2+ H3N H3N NH3 NH3 Cu 2+
[Ag(CN)2]– CN–→ Ag+ ← CN–
I3 I–I –I O3 O O O SO2 S O
CHAPTER 5: Chemical Bonding and Molecular Structure
according to octet rule)
SO3 s O O O (according to octet rule)
■ Inter med iate Properties: Coordinate co mpounds exhibit properties between ionic and covalent compounds.
■ Physical State: Exist as gases, liquids, or solids under normal conditions.
■ Melting & Boiling Points: Higher than covalent compounds but lower than ionic compounds.
■ Solubility: Less soluble in polar solvents (e.g., water) but readily soluble in non-polar solvents (e.g., benzene, CCl₄).
■ Stability: As stable as covalent compounds.
■ Electrical Conductivity: Poor conductors, like covalent compounds.
■ Reaction Rate: Undergo molecular reactions, which are slow.
■ Bond Strength & Nature: Dative bond is strong and directional.
TEST YOURSELF
1. In the molecule, HNO3, which atom acts as electron pair donor?
(1) Oxygen (2) Nitrogen (3) Hydrogen (4) NO atom
2. In NH3BF3 addition compound formation, boron is electron pair acceptor because (1) boron has five valence electrons
(2) boron is a non-metal
(3) boron has vacant orbital and is electron deficient
(4) NH3 can donate electron pair
3. Which one of the following has no coordinate bond?
(1) Ni(CO)4 (2) 4 NH + (3) CCl4 (4) CO
4. Which one of the following is not the property of coordination compounds?
(1) They have directional bond.
(2) They are soluble in polar solvents.
(3) They have polar bonds.
(4) They are not good conductors of electricity.
Answer Key
(1) 2 (2) 0 (3) 3 (4) 2
# EXERCISES
JEE MAIN
Level I
Kossel-Lewis approach to Chemical Bonding
Single Option Correct MCQ's
1. The number of electron-deficient molecules among PH 3 , CCl 4 , NH3, BeCl2, and BF 3 is (1) 0 (2) 1 (3) 2 (4) 3
2. The octet rule is not obeyed in (1) CO2 (2) BCl3 (3) PCl5 (4) (B) and (C) both
3. In Lewis formula of O3, there are (1) 2σ, 1π, and 4 lone pairs of electrons (2) 1σ, 2π, and 1 lone pairs of electrons (3) 2σ, 1π, and 6 lone pairs of electrons (4) 2σ, 2π, and 3 lone pairs of electrons
4. In drawing the Lewis dot formula of CN- ion, how many dots are expected to appear?
(1) 10 (2) 11 (3) 9 (4) 13
5. According to the Lewis dot structure for ozone, what is the formal charge on the central oxygen atom?
0 = 0—0
(1) –2 (2) –1 (3) 0 (4) +1
6. What is the formal charge on the nitrogen atom in HNO3?
H—O —N = O O
(1) 0 (2) +1
(3) +3 (4) +5
7. Covalent bonds are (1) rigid and directional
(2) rigid and non-directional (3) neither rigid nor directional (4) non-rigid and directional
Numerical Value Questions
8. The number of molecules among the following that do not follow the octet rule is ________.
C2H6, CO, PCl5, PCl3, AlCl3, SF6, BF3, NH3
9. The total number of electrons that takes part in forming the bonds in N2 is _____.
10. In the Lewis dot structure for NO2–, the total number of valence electrons around nitrogen is _______________.
Ionic or Electrovalent bond
Single Option Correct MCQ's
11. Lattice enthalpy and enthalpy of solution of NaCl are 788 kJ mol-1 and 4 kJ mol-1, respectively. The hydration enthalpy of NaCl is:
(1) 784 kJ mol-1
(2) 780 kJ mol-1
(3) –784 kJ mol-1
(4) –780 kJ mol-1
12. The compound with maximum ionic character is formed from (1) Na and Cl (2) Cs and F (3) Cs and l (4) Na and F
13. Correct order of Ionic nature is (1) CaF2 < CaCl2 < CaBr2 < CaI2
(2) CaF2 > CaCl2 > CaBr2 > CaI2
(3) CaCl2 > CaF2 > CaBr2 > CaI2
(4) CaI2 > CaCl2 > CaF2 > CaCl2
14. In which of the following species are the bonds non-directional?
(1) NCl3 (2) RbCl
(3) BeCl2 (4) BCl3
15. The compound with the highest melting point is
(1) NaF (2) NaCl
(3) NaBr (4) NaI
16. Which of the following is a favourable condition for the formation of ionic bonds?
(1) Small cation with small charge
(2) Small anion with large charge
(3) Large difference in electronegativity
(4) Small cation with high charge
17. AB is an ionic solid. The ionic radii of A+ and B are, respectively, rc and r a . The lattice energy of AB is proportional to
(1) c a r r (2) () ca rr +
(3) a c r r (4) ca 1 rr +
Numerical Value Questions
18. How many of the following compounds contain(s) ionic bond(s)?
BaCl2, HNO3, NH3, NaCl, O3, BF3, H2O
19. The values of electronegativity of atoms A and B are 1.2 and 4, respectively. The percentage of ionic character of A-B bond is ________. (Round off to nearest integer)
20. The lattice energy of NaCl is y kJ/mole. The value of the lattice energy of MgS is ny kJ/mole. If the ionic radii of Mg+2 and S−2 are same as Na+ and Cl , respectively, then the value of ‘n’ is _____.
Bond parameters
Single Option Correct MCQ's
21. Which among the following species has unequal bond lengths?
(1) XeF4 (2) SiF4 (3) PCl5 (4) BF –4
22. What is the correct order of increasing C–O bond length of 2 CO,CO,CO32?
24. Highest bond length between carbons is observed in (1) CH4 (2) C2H2 (3) C2H4 (4) C2H6
25. Which one of the following statements is true about the structure of CO2–3 ion?
(1) Out of the three C–O bonds, two are longer and one is shorter.
(2) It has three sigma and pi bonds.
(3) All three C–O bonds are equal in length, with a bond order between 1 and 2.
(4) It can be explained by considering sp 3 hybridisation.
26. Species having maximum ‘Cl-O’ bond order is
(1) ClO3 (2) ClO3 (3) ClO2 (4) ClO2
27. In 3 4 PO , P–O bond order is (1) 1.25 (2) 2 (3) –0.75 (4) –3
28. Bond order of SO2 is (1) 8 (2) 4
(3) 2 (4) 1.5
29. Correct order of bond angles in NH 3 , H2O, and BCl3 is (1) H2O > NH3 > BCl3
(2) NH3 > BCl3 > H2O
(3) NH3 > H2O > BCl3 (4) BCl3 > NH3 > H2O
30. Consider the O–N–O bond angles in the following species:
A. NO2⊕ B. NO2
C. NO2
(1) A < B < C (2) B < C < A
(3) A < C < B (4) B < A < C
31. Bond angle is least in (1) H2O (2) CH4 (3) CO2 (4) NH3
32. The correct increasing bond angle among BF3, PF3, and ClF3 follows the order (1) BF 3 < PF 3 < ClF3 (2) PF 3 < BF 3 < ClF3 (3) ClF3 < PF 3 < BF 3 (4) BF 3 = PF 3 = ClF3
33. Which of the following compounds has the smallest bond angle in its molecule? (1) NH3 (2) SO2 (3) H2O (4) H2S
34. According to the concept of resonance, among the ions 234 OCl, ClO, ClO and ClO,the most stable ion is
(1) OCl (2) ClO2
(3) ClO3 (4) 4 ClO
35. How many resonance structures are possible for NO3 ion? (1) Five (2) Three (3) Four (4) Six
36. Identify the incorrect statement about resonance from the following. (1) Canonical structures are hypothetical
and, individually, do not represent any real structure.
(2) There exists a dynamic equilibrium between the canonical structures.
(3) Due to resonance, the position of nuclei does not change.
(4) The energy of resonance hybrid is less than the most stable resonance structures.
37. The resonance hybrid of nitrate ion is (1) –1/2 –1/2 –1/2 N O O O (2) –5/3 –5/3 –5/3 N O O O (3) –1/3 –1/3 –1/3 N O O O (4) –2/3 –2/3 –2/3 N + O O O
38. Which one among the following, which molecule has zero dipole moment?
(1) C = C Cl Cl H H (2) C = C Cl Cl H H (3) C = C H Cl Cl H (4) Both (2) and (3)
39. Dipole moment is highest for (1) CHCl22 (2) 4 CH (3) 3 CHCl (4) 4 CCl
40. Which of the following compounds has non-zero resultant dipole moment? (1) BF2
(2) NO2
Numerical Value Questions
45. Find the number of shorter bonds in IF7
46. Bond energy of CH4 is 396 kcal and that of C2H6 is 680 kcal. The C–C bond energy in C2H6 (in kcal) is ____.
47. The number of species with bond angle more than 100° but less than 180° is _____.
C2H2; CO2; SO2; NH3; H2O; C2H4
(3) CN CN (4) OH OH
41. Which has the highest value of dipole moment?
(1) 3 CHCl (2) CHCl22 (3) 3 CHCl (4) 4 CCl
42. The observed dipole moment of HI is 0.38 D. The percentage of ionic character of the molecule having H-I bond distance of 1.61 is
(1) 16.4% (2) 12% (3) 5% (4) 20%
43. The dipole moment of HA molecule is 0.42 × 10−18 esu.cm, and the interatomic distance is 1.75 A 0. The percentage of ionic character is (1) 27 (2) 5 (3) 15 (4) 50
44. For which of the following molecules, μ≠0?
Cl i) ii) iii) iv)
(1) Only (i) (2) (i) and (ii) (3) Only (iii) (4) (iii) and (iv)
48. Based on VSEPR theory, the number of 90-degree bond angles in SF6 is ______.
49. The bond order of Cl–O bond on ClO3 is X. Then, 3X is ______. (Nearest integer)
50. The total number of molecules with zero dipole moment, among CH4, BF3, H2O, HF, NH3, CO2, and SO2, is _________.
51. The dipole moment of a certain diatomic molecule X−Y is 0.38 D. If the X–Y distance is 158 pm, the percentage of electronic charge developed on X–atom is ______.
52. The dipole moment of trans-dichloro ethene is ___________.
53. A molecule X−Y has a dipole moment of 2.88 D. If its bond length is 1.2 A 0, then find the percentage ionic character in X−Y.
VSEPR Theory
Single Option Correct MCQ's
54. Among the following species, identify the isostructural pairs.
33333 NF,NO,BF,HO,NH −+
(1) NF,NOand3333BF,HO −+
(2) 33 33 NF,NH and NO,BF
(3) 33 33 NF,HOand NO,BF+−
(4) 33 33 NF,HOand NH,BF +
55. The numbers of lone pairs of electrons in the central atom of 2463 XeF,XeF,XeF,XeO respectively, are (1) 0, 1, 2, and 2 (2) 3, 2, 1, and 1 (3) 1, 2, 2, and 0 (4) 2, 1, 3, and 0
56. Which of the following species has shape similar to XeOF4?
(1) XeO3 (2) IOF4+
(3) PCl5 (4) XeF5 ⊕
57. The maximum number of 180° angles possible between X-M-X bond for compounds SF6 and PCl5 are, respectively, (1) 3, 3 (2) 3, 1 (3) 1, 2 (4) 3, 0
58. The central atom in a molecule has three bond pairs and one lone pair of electrons. The shape of the molecule is (1) triangular (2) pyramidal (3) linear (4) tetrahedral
59. The geometry of ClO3 ion, according to the valence shell electron pair repulsion (VSEPR) theory, will be (1) planar triangular (2) pyramidal (3) tetrahedral (4) square planar
Numerical Value Questions
60. Among the following, the number of species having linear shape is ____________.
2332322263 XeF,I,CO,I,CO,BeCl,SO,XeF,BF +−
61. The number of species having nonpyramidal shape, among the following, is ______.
SO3, NO3 , PCl3, CO32−, SF6, BF3
62. Based on VSEPR theory, the number of 90° F–Br–F angles in BrF5 is _____.
63. Find the number of molecules that are planar.
(i) SF4 (ii) XeF6
(iii) XeF2 (iv) XeF4
(v) H2O (vi) NH3
(vii) PH3 (viii) PCl5
(ix) PCl3
64. The total number of lone pairs of electrons on oxygen atoms of ozone is ________.
VBT
Single Option Correct MCQ's
65. What is the property of oxygen molecule that cannot be explained by valence bond theory?
(1) Formation of double bond
(2) Formation of pi bond
(3) Strength of double bond
(4) Paramagnetic nature
66. Which of the following is called negative overlap?
(1) + + – –
(2) + + – –
(3) + + – –
(4) + + –
67. Indicate the wrong statement, according to valence bond theory.
(1) A sigma bond is stronger than π-bond.
(2) p-orbitals always have only sidewise overlapping.
(3) s-orbitals never form π-bonds.
(4) There can be only one sigma bond between two atoms.
Numerical Value Questions
68. The bond angle in H 2 O, according to valency bond theory, is _____ degrees.
CHAPTER 5: Chemical Bonding and Molecular Structure
69. What is the ratio of dπ−pπ bonds and pπ−pπ bonds in sulphur trioxide?
70. What is the number of zero overlaps among the following?
75. For which hybridisation are there two unequal bond angles?
(1) sp3 (2) sp2 (3) sp (4) sp3d
76. The percentage of ‘s’ character is least in the hybrid orbitals of which of the following molecules?
(1) H2O (2) BF 3
(3) SF6 (4) PCl5
77. sp3d2 hybridisation is not displayed by (1) SF6 (2) PF 5 (3) IF 5 (4) BrF 5
Hybridisation
71. Which of the following compounds has sp, sp2, and sp3 hybrid carbon atoms?
(1) C = C = C
(2) 32 CHCHCHCHCH −=−=
(3) 3 CHCCCCH −≡−≡
(4) 2 CHCHCCH =−≡
72. The structure of the molecule with 25% s-character in hybrid orbital is (1) plane triangular (2) linear (3) tetrahedral (4) octahedral
73. Which of the following has been arranged in increasing order of size of the hybrid orbitals?
74. Choose the pair of compounds that have different hybridisation but same shape.
(1) SO3, ClF3 (2) BF3, PCl3
(3) XeF2, CO2 (4) XeF4, SF4
78. The hybridisations of central atoms in CO2, SiO2, and SO2 are (1) sp2, sp3, sp (2) sp, sp3, sp2 (3) sp3, sp , sp2 (4) sp2, sp, sp3
79. The hybrid orbital of carbon in diamond, graphite, and ethyne is, respectively, (1) sp3, sp2, sp (2) sp2,sp3, sp (3) sp, sp2, sp3 (4) sp, sp3, sp2
80. The hybridisations of atomic orbitals of nitrogen in 234 NO,NOandNHare +−+ (1) SP, SP3, and SP2, respectively (2) SP, SP2, and SP3, respectively (3) SP2, SP, and SP3, respectively (4) SP2, SP3, and SP, respectively
Numerical Value Questions
81. The total number of species from the following, with the central atom utilising sp2 hybrid orbitals for bonding, is ____.
101. The strongest hydrogen bonds can be formed by (1) HF (2) H2O (3) NH3 (4) HCl
102. Intramolecular hydrogen bond is exhibited by (1) COOH HO
(2) CH3 OCH3
(3) CH3 H3CO
(4) COOH OH
103. Intramolecular hydrogen bonding is present in (1) meta nitrophenol (2) ortho salicylic acid (3) hydrogen chloride (4) benzophenone
104 Hydrogen bonds are present even in vapour state of (1) H2O (2) HF (3) p-hydroxy benzaldehyde (4) C2H5OH
105. Intermolecular hydrogen bonding increases the enthalpy of a liquid due to the (1) decrease in the attraction between molecules (2) increase in the attraction between molecules
CHAPTER 5: Chemical Bonding and Molecular Structure
(3) decrease in the molar mass of unassociated liquid molecules (4) increase in the effective molar mass of hydrogen-bonded molecules
106. The type(s) of bonds present in 42 CuSO5HO ⋅ (blue vitriol) is/are (1) electrovalent and covalent (2) electrovalent, covalent, co-ordinate, and hydrogen (3) covalent and co-ordinate covalent (4) electrovalent
107. KF combines with HF to form KHF2. The compound contains the species (1) K+, F–, and H+ (2) K+, F–, and HF
(3) K+ and [HF2]–1 (4) [KHF]+ and F–
108. Among the three isomers of nitrophenol, which is the least soluble in water?
(1) ortho isomer (2) para isomer (3) meta isomer (4) All are insoluble.
Numerical Value Questions
109.The number of hydrogen bonds water can form is ____.
110.The number of water molecules not coordinated to copper ion directly in 42 CuSO5HO is _____.
111.How many of the following molecules are known to exist?
KHF2, KHCl2, FBr3, KI3, ICl3
Level II
Kossel–Lewis Approach to Chemical Bonding
Single Option Correct MCQs
1. Two elements ‘X’ and ‘Y’ have the following configuration:
X = 1s2 2s2 2p6 3s2 3p6 4s2
Y = 1s2 2s2 2p6 3s2 3p5
The compound formed by the combination of ‘X’ and ‘Y’ will be
(1) XY2 (2) X 3Y2
(3) X2Y3 (4) XY 5
2. The molecule that deviates from octet rule is
(1) CCl4 (2) BF 3
(3) MgO (4) NCl3
3. Which of the following has highest lattice energy?
(1) AlF 3
(2) NaF
(3) MgF2
(4) NaCl
Numerical Value Questions
4. How many compound are ionic, among sodium hydride, sulphur dioxide, potassium oxide, and calcium carbide?
Covalent Bond
Single Option Correct MCQs
5. Which combination will give strongest ionic bond?
(1) Na+ and Cl
(2) Mg2+ and Cl–
(3) Na+ and O2–
(4) Mg2+ and O2–
6. In the given molecule, the formal charges on oxygen atoms 1, 2, and 3 are, respectively, O 1 2 3 O O
(1) –1, 0, +1 (2) 0, –1, +1
(3) 0, +1, –1 (4) +1, 0, –1
Coordinate Covalent Bond
Single Option Correct MCQs
7. The type of bond formed by CO with haemoglobin of blood is
(1) covalent
(2) dative
(3) ionic bond
(4) van der Waals forces
Valence Bond Theory
Single Option Correct MCQs
8. Which of the following represents zero overlap?
(1) + + – –
(2) + + – –
(3) + + – –
(4) + + –
9. Some statements about valence bond theory are given below
i) The strength of bond depends upon extent of overlapping.
ii) The theory explains the directional nature of covalent bond.
iii) According to this theory, oxygen molecule is paramagnetic in nature.
(1) All statements are correct.
(2) Only statements i and iii are correct
(3) Only statements i and ii are correct
(4) All statements are wrong
Hybridisation
Single Option Correct MCQs
10. There will be no change in hybridisation when
(1) AlH 3 is converted to [AlH4]–
(2) BF 3 combines with F–
(3) SnCl2 combines with Cl2
(4) SO2 combines with SO3
11. In which of the following sets, the central atom is involved in sp3 hybridisation?
(1) BF–4 , ClO–4
(2) BF3, ClO–3
(3) XeF4, XeO3
(4) SO3, XeO3
12. Hybridisation of I in lCl+ 2
(1) sp (2) sp2 (3) sp3 (4) dsp2
Numerical Value Questions
13. The number of species involving sp 3d hybridisation among the following, is____.
XeF2, SCl4, PCl5, CIF3, ICI–2, XeF4, XeOF2.
VSEPR theory
Single Option Correct MCQs
14. A central atom in a molecule has two lone pairs of electrons and forms three single bonds. The shape of the molecule is (1) T-shaped (2) see-saw
(3) trigonal pyramidal (4) planar triangular
15. Shape of BrF5 molecule is (1) pentagonal ipyramidal
(2) trigonal pyramidal
(3) square pyramidal
(4) trigonal bipyramidal
16. In which of the following molecules, all the atoms lie in one plane?
(1) CH4 (2) BF 3
(3) PF 5 (4) NH3
Numerical Value Questions
17. Maximum number of atoms in one plane in SF6 molecule is ____.
18. How many of the following having regular tetrahedral geometry?
CCl4, CHCl3, SO3, SiF4, BF4–, NH3, H2O, SiO44−
19. Find the number of planar species out of SF2, SF4, SF6, SO2, SO3
20. The number of σ bonds in naphthalene molecule is ____.
CHAPTER 5: Chemical Bonding and Molecular Structure
Molecular Orbital Theory
Single Option Correct MCQs
21. The correct order of bond orders of C22–, N22–, and O2 2– is
(1) 2-2-2222 C<N<O
(2) 2-2-2222 O<N<C
(3)
(4)
2-2-2222 C<O<N
2-2-2222 N<C<O
22. In which of the following transformations, the bond order has increased and the magnetic behaviour has changed?
(1) + 22CC →
(2) NO+ → NO
(3) + 22OO →
(4) + 22NN →
Numerical Value Questions
23. The spin-only magnetic moment value of B2+species is ___ × 10–2 BM.
[Given 3 = 1.73]
24. Among H2, O2, N2,O2+, O2–, B2, F2, C2, and N2+, the number of diamagnetic species is ___.
Hydrogen Bonding
Single Option Correct MCQs
25. All the following molecules involve in intra- molecular hydrogen bonding, except
(1) o-nitro phenol
(2) salicylic acid
(3) p-nitro aniline
(4) o-hydroxy benzaldehyde
Numerical Value Questions
26. The number of water molecules that are coordinate to copper ion directly in CuSO4. 5H2O is ____.
Bond Parameters
Single Option Correct MCQs
27. What is the correct order of bond angles in the following?
29. Bond enthalpy is maximum for (1) Br2 (2) F2 (3) Cl2 (4) I2
30. Which of the following molecule has the highest value of dipole moment?
(1) PF2Cl3 (2) PBr3Cl2 (3) PCl5 (4) PCl2F3
31. Which of the following has fractional bond order?
(1) O2+2 (2) O2–2 (3) F2–2 (4) H2–
Multiple Concept Questions
Single Option Correct MCQs
32. The hybrid orbitals of carbon in diamond, graphite, and ethyne, respectively, are (1) sp3, sp2, sp (2) sp2,sp3, sp (3) sp, sp2, sp3 (4) sp, sp3, sp2
33. sp3d2 hybridisation is not displayed by (1) SF6 (2) PF 5 (3) IF 5 (4) BrF 5
34. A, B, and C are atoms of elements with atomic number Z , Z +1, and Z +2, respectively. If ‘B’ has octet configuration, the bond formed between A and C
predominantly is (1) covalent bond (2) ionic bond (3) dative bond (4) hydrogen bond
35. Peroxo bond is present in (1) H2S2O7 (2) HClO4 (3) H2S4O6 (4) H2S2O8
36. PH 3 and BF 3 form an adduct readily because they form (1) co-ordinate covalent bond (2) a covalent bond (3) an ionic bond (4) a hydrogen bond
37. If the values of Madelung constants of the following compounds are equal, then their lattice energy values decrease in the order
(1) NaF > KCl > CaO > Al2O3 (2) Al2O3 > CaO > KCl > NaF (3) Al2O3 > CaO > NaF > KCl (4) KCl > NaF > CaO > Al2O3
38. Which of the following is not correct? (1) Low ionisation potential is a favourable condition for the formation of cation (2) Coordination number of Na in NaCl is 6 (3) Ionic bond is directional. (4) Ionic compounds have high melting points.
39. Among LiCl, RbCl, BeCl 2, and MgCl 2, the compounds with greatest and least ionic character, respectively, are (1) LiCl, RbCl (2) RbCl, BeCl2 (3) RbCl, MgCl2 (4) MgCl2, BeCl2
40. The species having pyramidal shape according to VSEPR theory is (1) SO3 (2) BeF2 (3) SiO32– (4) OSF2
41. The correct pair of species that are not iso-structural is (1) PF6– and SF6 (2) IO3– and XeO3 (3) BH 4 – and NH4+ (4) BrF 3 and XeF4
42. Among the following, which species has same number of σ and π bonds?
(1) C7H8 (2) C2(CN)4
(3) C2H4 (4) HC ≡ CH
43. Number of σ and π bonds in acrylonitrile are
(1) 6σ and 3π (2) 7σ and 2π (3) 7σ and 3π (4) 2 p and 6σ
44. In which species, the N-atom is in sp hybridisation?
(1) NO2– (2) NO3–(3) NO2 (4) NO2+
45. In which of the following pairs, both the species have the same hybridisation?
(I) SF4, XeF4 (II) I3–, XeF2 (III) ICl4+, SiCl4 (IV) ClO3–, PO43–(1) I, II (2) II, III (3) II, IV (4) I, II, III, and IV
46. The ion that has sp 3d hydridisation for the central atom is
(1) [ICl4] (2) [ICl2] (3) [IF6] (4) [BrF4]
47. If the magnetic moment of a dioxygen species is 1.73 BM, then the ion may be (1) O2– or O2+ (2) O2 or O2+ (3) O2 or O2– (4) O2, O2–, or O2+
48. The total number of antibonding electrons in C2 and B2 molecules respectively is (1) 4,4 (2) 5,6 (3) 8,6 (4) 4,6
49. Among the following molec ules or ions
-2-2-2
2222 C,N,O,O , which one is diamagnetic and has the shortest bond length? (1) O2 (2) N2−2 (3) O2–2 (4) C2–2
50. A diatomic molecule has a dipole moment of 1.2 D. If the bond distance is 1.0 Ao, the fraction of an electronic charge on each atom is
(1) 0.25 (2) 0.33 (3) 0.66 (4) 0.90
51. Which of the following hydrogen halide is most volatile?
(1) HF (2) HCl
(3) HBr (4) HI
Numerical Value Questions
52. Find out the total number of species where the central atom is sp2 hybridised.
ClO2+, NO2, SO2,ClO3, . C –H3, C3O2, (CN)2, N2H4
53. Maximum number planar atoms in XeF4 molecule ___.
54. Number of 90° bond angles in the molecule of PCl5(g) is ___.
Level III
Single Option Correct MCQ's
1. The order of relative ease of formation of various ions is
(1) F– > O–2 > N–3 (2) N–3 > O-2 > F-
(3) O–2 > N–3 > F– (4) F– > N–2 > O–
2. For NaCl, MgCl 2, AlCl3, and SiCl4 , the increasing order of covalent character is
(1) NaCl > MgCl2 > AlCl3 > SiCl4
(2) NaCl < MgCl2 > AlCl3 > SiCl 4
(3) NaCl < MgCl2 < AlCl3 < SiCl4
(4) NaCl > AlCl3 > MgCl2 > SiCl4
3. Which one of the following is correct regarding σ bond?
(1) Free rotation of atoms about the bond axis is permitted.
(2) It is formed by the partial overlap of atomic orbitals at right angle to internuclear axis.
(3) It is very weak bond.
(4) It has less overlapping regions
4. Identify the pair in which the geometry of the species is T-shape and squarepyramidal, respectively
(1) ClF3 and IO 4 (2) ICI2– and ICI5
(3) XeOF2 and XeOF4 (4) IO3– and IO2F2–
5. Intra-molecular hydrogen bond is present in
(1) ortho-hydroxy benzoic acid
(2) para-hydroxy benzoic acid
(3) ethyl alcohol
(4) hydrogen fluoride
6. For which of the following molecules, the resultant dipole moment μ≠0?
(ii) (iii) (iv)
(1) only (i) (2) (i) and (ii)
(3) only (iii) (4) (iii) and (iv)
7. When these species are arranged in order of increasing bond energy, what is the correct sequence?
(1) N2, O2, F2
(2) F2, O2, N2
(3) O2, F2, N2
(4) O2,N2, F2
8. Wrong match among the following is
(1) dipole moment: NF3 < NH3
(2) bond length: N2 < N–2
(3) bond angle: NH3 < PH 3
(4) ionic character: CuCl <KCl
9. Of the following sets which one does not contain isoelectronic species?
(1)
-3-2444 PO,SO,ClO
(2) --2 22CN,N,C
(3)
(4)
-2-2333 SO,CO,NO
-3-2333 BO,CO,NO
10. The follo wing are some statements about the characteristics of covalent compounds:
(i) The combination of a metal and nonmetal must give a covalent compound.
(ii) All covalent substances are bad conductors of electricity.
(iii) All covalent substances are gases at room temperature.
(1) Only i, and iii are correct.
(2) Only i and ii are correct.
(3) Only ii and iii are correct.
(4) i, ii, and iii are incorrect.
11. Find the correct order of given property from the following:
(A) K2CO3 > Na2CO3 > Li2CO3: thermal stability order
(B) CaO > CaF2: extent of polarisation order
(C) BaO > BaF2: melting point order
(1) A and B only (2) A and C only
(3) B and C only (4) A, B and C
12. In which of the following solvents should KCl be more soluble at 25 °C ? (D = Dielectric constant value)
(1) C6H6[D = 0]
(2) CH3COCH3[D = 2]
(3) CCl4[D = 0]
(4) CH3OH[D = 32]
13. The correct order of solubility of the sulphates of group II elements would be
(1) BeSO4 < MgSO4 < CaSO4 < SrSO4 < BaSO4
(2) MgSO4 < CaSO4 < SrSO4 < BaSo4 < BeSO4
(3) BeSO4 > MgSO4 > CaSO4 > SrSO4 > BaSO4
(4) MgSO4 <BeSO4 < CaSO4 < SrSO4 < BaSO4
14. Choose the correct statement regarding molecules SF4, CF4, and XeF4 .
(1) 2, 0, and 1 lone pairs on central atom, respectively (2) 1, 0, and 1 lone pairs on central atom, respectively (3) 0, 0, and 2 lone pairs on central atom, respectively (4) 1, 0, and 2 lone pairs on central atom respectively,
15. Which of the following are iso-structural pairs?
A) SO42– and CrO42– B) BiCl4 and TiCl4
C) NH3 and NO3– D) BCl3 and BrCl3
(1) C and D only (2) B and C only (3) A and C only (4) A and B only
16. Molecular shapes of CIF3, I3–, and XeO3 , respectively, are (1) T-shape, linear, pyramidal (2) planar, linear, tetrahedral (3) T-shape, planar, pyramidal (4) trigonal bipyramidal, linear, tetrahedral
17. The correct order of increasing s-character (in percentage) in the hybrid orbitals of following molecules/ions is I) CO32– II) XeF4 III) I 3 – IV) NCl3 V) BeCl2
(1) II < III < IV < I < V
(2) II < IV < III < V < I
(3) III < II < I < V < IV
(4) II < IV < III < I < V
18. If E is the energy of atomic orbitals involved to form molecular orbitals, Eb and E a are the energies of bonding and antibonding molecular orbitals formed respectively, then
(1) Eb > E a > E
(2) E a >E > Eb
(3) E > E a > Eb
(4) E a > Eb > E
19. If the bond order in CO is 3, then what is the bond order in CO+?
(1) 3 (2) 3.5
(3) 2.5 (4) 2
20. In which of the given molecules, the C–H bond length is the longest?
(1) C2H6
(2) C2H4
(3) C2H2
(4) Cannot be determined
21. The correct order of bond angle in the given molecules is
(1) + 3324 PH<NH<HO<NH
CHAPTER 5: Chemical Bonding and Molecular Structure
(2) + 3234 PH<HO<NH<NH
(3) + 3324 NH<PH<HO<NH
(4) + 4323 NH<NH<HO<PH
22. A molecule is described by three Lewis structures having energies E 1 , E 2 , and E 3, respectively . The energies of these structures follow the order E1 > E2 > E3. If resonance hybrid energy of the molecule is E0, the resonance energy is
(1) E0– E1 (2) E0– E2 (3) E0– E 3 (4) E 0 – (E1 + E2 + E3)
23. The resonance hybrid of nit rate ion is (1) –1/2 –1/2
–1/2 N O O O
(2)
–5/3 –5/3 –5/3 N O O O
(3) –1/3 –1/3 –1/3 N O O O
(4) –2/3 –2/3 –2/3 N + O O O
Numerical Value Questions
24. Coordination number of Na+ ion in NaCl crystal lattice is x , and coordination number of Cl in CsCl crystal lattice is y.
Then, x+y−10 =?
25. Number of lone pairs on the central atom of XeF2, XeF4, XeO4, and XeOF4 are x, y, z, and w. Then the value of x+y+z+w is
26. For XeOF4 molecule, if x= number of σ bond pairs and y = number of lone pairs on central atoms, then calculate the value of x-y.
27. How many of the following compounds have odd number of electrons?
Cs2O, MgO, Cl2O, ClO2, NO, NO2, N2O4, N2O5, CaOCl2, N2O, CO2
28. Assuming there is no 2s–2p mixing, the number of paramagnetic species amongst the following is _____
29. How many bond electron pairs are present in IF 7 molecule?
30. The value of electronegativity of atoms ‘A’ and B are 1.2 and 4.0, respectively. The percentage of ionic character of A-B bond is ___.
31. When hybridisation involving d-orbitals are considered, all the five orbitals are not degenerate; rather 222 , xyz dd and d xy , d yz and d zx form two different sets of orbitals, and orbitals of appropriate set are involved in the hybridisation. The number of species where 22 xy d set of orbitals participate in hydrisation is ___
XeF6, SF6, IF7, XeO3, PCl5, PF5
32. The number of paramagnetic species among the following is _____.
B2, Li2, C2, C2–, O22–, O2+, He2+
33. The number of molecules having dipole moment greater than zero is ____.
O3, PF2Cl3, NO2, N2O, CFCl3, ClF3, XeOF4, SF4
THEORY-BASED QUESTIONS
Statement Type Questions
Each question has two statements: statement I (S-I) and statement II (S-II). Mark the correct answer as
(1) if both statement I and statement II are correct,
(2) if both statement I and statement II are incorrect,
(3) if statement I is correct, but statement II is incorrect,
(4) if statement I is incorrect, but statement II is correct
1. S–I : NaF is more ionic than Na2O.
S–II : Formation of O –2 from O –1 is exothermic.
2. S–I : Carbon sub-oxide is bent molecule.
S-II : Each carbon atom is in sp-hybrid state.
3. S–I : Hyperconjugation is a permanent effect.
S–II : Hyperconjugat ion in ethyl cation + 32CH-CH
inv olves the overlapping of 21s sp C-H bond with empty 2p orbitals of other carbon.
4.
S–I: In C 2 vapours, the bond order is two with one sigma bond and one pi bond.
S–II: Formal charge on terminal nitrogen atom of NNO == is '+1'.
5. S–I: Bond energy of 22NN−+ <
S–II: Antibonding electrons are more in N–2 than in N+ 2
6. S–I: Dipole moment is a vector quantity and, by convention, it is depicted by a small arrow with tail on the negative centre and head pointing towards the positive centre.
S–II: The crossed arrow of the dipole moment symbolises the direction of the drift of electrons in the molecu les.
7. S–I: CO2 has no dipole moment where as SO2 and H2O have dipole moment.
S–II: SnCl2 is ionic, whereas SnCl4 is covalent.
8. S–I: The actual structure of benzene is said to be a resonance hybrid.
S-II: Alternative structures are referred to as resonance structures or canonical forms of benzene.
Assertion and Reason Questions
In each of the following questions, a statement of Assertion (A) is given, followed by a corresponding statement of Reason (R). Mark the correct answer as
(1) if both (A) and (R) are true and (R) is the correct explanation of (A),
(2) if both (A) and (R) are true but (R) is not the correct explanation of (A),
(3) if (A) is true but (R) is false,
(4) if both (A) and (R) are false.
9. (A) : Formal charge on the two terminal nitrogen atoms in N –3 (azide ion) is the same in all different possible Lewis structures.
(R) : Formal charge on an atom is independent of Lewis structure.
10. (A) : POF3 exists but NOF3 does not exist.
(R) : Both ‘P’ and ‘N can form five bonds by expanding their octet.
11. (A) : Water is a good solvent for ionic compounds but a poor one for covalent compounds.
(R) : Hydration energy of ions releases sufficient energy to overcome lattice energy and break hydrogen bonds in water, while covalent bonded compounds interact so weakly that even van der Waals forces between molecules of covalent compounds cannot be broken.
12. (A) : Zero orbital overlap is an out of phase overlap.
(R) : It results due to different orientation/ direction of approach of orbitals.
13. (A) : The shape of N(SiH 3 ) 3 is trigonal planar.
(R) : In N(SiH3)3, the hybridisation of ‘N’ is sp2
14. (A) : [I(CN)2]– is planar, whereas [I(CH3)2]–is not planar.
(R) : Iodine atom is sp hybridised in both the species.
15. (A) : Diamagnetic C 2 molecule has been detected in vapour phase, in which double bond consists of both Pi bonds.
(R) : In the above C 2 molecule, four electrons are present in two Pi molecular orbitals.
16. (A) : In the bonding molecular orbital (MO) of H 2, electron density is increased between the nuclei.
(R) : The bonding MO is ψ A + ψ B, which shows destructive interference of the combining electron waves.
17. (A) : Bond order in a molecule can assume any value, positive or negative, integral or fractional, when number of electrons in bonding molecular orbitals is greater than that in antibonding orbitals.
(R) : Bond order does not depend on the number of electrons in the bonding and anti-bonding orbitals.
18. (A) : Among the two O−H bonds in H 2O molecule, the energy required to break the first O−H bond and the other O−H bond is different.
(R) : This is because the electronic environment around oxygen is different even after breakage of one O−H bond.
19. (A) : The dipole moment helps to predict whether a molecule is polar or nonpolar.
(R) : The dipole moment helps to predict the geometry of molecules.
20. (A) : The electronic structure of azide ion (N3–) is [::] NNN ←≡ .
(R) : N N = N –is not a resonating structure of azide ion, because the position of atoms cannot be changed.
JEE ADVANCED LEVEL
Multiple Option Correct MCQs
1. Reso nating structures of a molecule should have
(1) identical arrangement of atoms
(2) nearly the same energy content
(3) the same number of paired electrons
(4) the same number of unpaired electrons
2. Choose the correct statement/s among the following.
(1) Hydride (H ) ion is similar in size to fluoride (F ) ion.
(2) In solid state, BeH2 and BeCl2 are structurally similar but bonding scheme in these compounds is not similar.
(3) Super oxides are less stable than peroxides due to lesser lattice energy with a particular cation.
(4) Liquid water can act both as oxidising and reducing agent.
3. Which of the following is/are correct relation? (in general)
(1) Bond energy ∝( Polarity of the bond)-1
(2) Bond energy ∝ (s-character of hybrid orbitals)
(3) Bond energy ∝( Atomic radius)–1
(4) Bond energy ∝( Bond order)1
4. Each of the following options contains a set of four molecules. Identify the option(s), where the molecule possesses permanent dipole moment at room temperature.
(1) BeCl2, CO2, CHCl3, CH4, SF6, IF7
(2) NO2, NH3, POCl3, CH3Cl
(3) BF3, O3, SF6, XeF6, CO2, XeF2, SO3
(4) SO2, C6H5, H2Se, BrF5
5. Which of the following molecules involves intramolecular H-bond?
(1) ortho-nitrophenol
(2) Chloralhydrate
(3) para-nitrophenol
(4) Benzoic acid
6. Which of the following statements are correct based on Fajan's rule?
(1) The smaller the cation the higher is its polarising power.
(2) Cation with pseudo noble gas configuration (ns 2 np 6 nd 10 ) having relatively high polarising power then those with nobel gas configuration (ns2np6 ).
(3) The larger the size of the anion, the higher is its polarisability.
(4) The larger the size of the cation the higher is its polarising power.
7. Which among the following are isostructural?
(1) XeO2F2, SF4 (2) CO2, I3–
(3) SO32–, CO32– (4) CIF3, XeF2
8. Consider the shape of PCl3F2, and identify the correct statement(s).
(1) No face contains two atoms of fluorine.
(2) There are two chlorine atoms per face.
(3) There is one fluorine atom per face.
(4) Only one face contains three chlorine atoms.
9. If the internuclear axis is z-axis, correct combination(s) of orbitals overlapping and type of bond formed is/are
(1) s-orbital and s-orbital σ−bond
(2) d xz -orbital and d xz -orbital p −bond
(3) d xy -orbital and d xy -orbital p −bond
(4) pz -orbital and pz -orbital σ−bond
10. In which combination of compounds, their geometry and hybridisation are correct?
(1) XeOF4 – Square pyramidal, sp3d2
(2) BF 4–– Tetrahedral, sp3
(3) SF4, – Tetrahedral sp3d
(4) ICl–2 – Linear, sp3d
11. Which of the following molecular orbitals has two nodal planes?
(1) π2py (2) π*2px
(3) σ*2pz (4) π*2py
12. Which of the following molecule(s) has/ have dπ−pπ bonding?
(1) SO32– (2) H2PO2–
(3) ClO2– (4) NO2+
13. Select the correct statements regarding σ and π bonds.
(1) σ-bond lies on the line joining the nuclei of bonded atoms.
(2) π-electron cloud lies on either side to the line joining the nuclei of bonded atoms.
(3) The order of the strengths of different types will be σ>π>δ.
(4) σ-bond has primary effect to decide direction of covalent bond, whereas π-bond has no primary effect in direction of bond.
14. Choose the correct statements according to molecular orbital theory for a diatomic molecule (LCAO method).
(1) π-bonding molecular orbital is not symmetrical around bond axis.
(2) σ-bonding molecular orbital is symmetrical around bond axis.
(3) π*-antibonding molecular orbital has a node between nuclei of bonding atoms.
(4) σ*-antibonding molecular orbital has an antinode between nuclei of bonding atoms.
CHAPTER 5: Chemical Bonding and Molecular Structure
15. Which of the following is correctly arranged in decreasing order of bond angle?
(1) NO2+ > NO2 > NO2–
(2) ClO2– > ClO2+ > ClO2
(3) NH3 > NF3 > F2O
(4) Cl2O > H2O > F2O
Numerical Value Questions
16. Calculate the electron gain enthalphy of O atom to O–2 ion from the following data:
I. fH ∆ [MgO(S)] =–600 kJ mol–1
II. u H ∆ [MgO(S)] =–3860 kJ mol–1
III. IE1 +IE2 of Mg(g) = 2170 kJ mol–1
IV. diss H ∆ [O2(g)] = +494 kJ mol–1
V. D dissH°[Mg(s)] = +150 kJ mol–1
17. The molecule ML x is planar with 6 pairs of electrons around M in the valence shell. The value of x is ___.
18. Among the following, how many are with same hybridisation, same number of lone pairs, and same structure?
NI3, I3–, SO3–2, NH4+, ClF3, SF4
19. The number of (dπ−pπ) bonds in the species SO 3 2– , SO 4 2– , and SO 3 are, respectively, x , y, and z . Then, find the value of xy z + .
20. In how many of the following reactions, paramagnetic gas is liberated?
21. A covalent compound MX has bond length 2.22 A o. The observed dipole moment of MX is equal to 2.4 debye. The percentage of ionic character of MX is ___.
22. The ∠CNC bond angle in CH3NCS should be greater than _____.
23. Out of the given six species, find the number of species that have intramolecular hydrogen bonding.
26. Among [I3]+, [SiO4]4–, SO2Cl2, XeF2, SF4, CIF 3, Ni(CO) 4, XeO 2F 2, [PtCl 4] 2-, XeF 4, and SOCl2, the total number of species having sp 3 hybridised central atom is _____.
27. How many possible 109°28' bond angles can be observed in a molecule of methane?
Passage-based Questions
Q(28-29)
CH(COOH)
(b)
(a) COOH COOH
CH(COO–) cis :
(c) H3BO3(s)
(d) o-nitrophenol
(e) p-hydroxybenzaldehyde
(f) Tetra methyl ammonium hydroxide
Integer Value Questions
24. Consider the following in the liquid form: N2, HCl, BF3, CO2, H2S, SF4, SiCl4 Cl Cl Cl , Cl
When a charged comb is brought near their following stream, how many of them show deflection as per the following figure?
When hybridisations involving d-orbitals are considered, then all the five orbitals are not degenerate, rather, 222 x-yz d,d and d xy , dyz and d zx form two different sets of orbitals and orbitals of appropriate set is involved in the hybridisation.
28. The number of molecules/ ions in which central atom used its 2 z d orbitals in hybridisation among the following species is XeF 4 , SOF 4 , NO 3 Θ, ClO 3 Θ , BrF3, IF 4 Θ , COCl2, PCl6 Θ
29. The number of species where 22 x-y d orbitals participates in hybridisation is/ are
XeF4,SF6,IF7,XeO3,PCl5, PF 5
Q(30-31)
When hybridisation involving d-orbitals are considered, all the five d-orbitals are not degenerate, rather 222 x-yz d,d and d xy , dyz, d zx form two different sets of orbitals and orbitals of appropriate set is involved in the hybridisation.
30. In sp 3 d 2 hybridisation, which sets of d-orbitals is involved?
(1) 222 x-yz d,d (2) 2xy z d,d
25. Number of lone pair and bond pair repulsion at 90° are p in ICl2 –
Number of lone pair and bond pair repulsion at 90° are q in ICl4 – Find the difference of q and p.
(3) d xy , dyz (4) 22xyx-y d,d
31. In sp3d3 hybridisation, which orbitals are involved?
(1) 222xyx-yz d,d,d (2) d xy , dyz, dzx
(3) 22xyxzx-y d,d,d (4) 2yzzx z d,d,d
CHAPTER 5: Chemical Bonding and Molecular Structure 162
Q(32-33)
In the polyatomic molecules, the bond energy of similar bond may differ due to the difference in chemical environment around the central atom. Bond order is given by the number of bonds between the two atoms in a molecule. Iso-electronic molecules and ions will have identical bond orders. In general, as bond order increases, the bond energy increases and bond length decreases.
32. In the following, the number of species that have less bond order than O2 + and are paramagnetic is____.
CN , CO, N2 + , N2 , O2, O2 , O2−2, F2, CO+
33. Among the following, the number of species with fractional bond order is ____.
H 2 + , H 2, H 2 , He 2, He 2 + , B 2, C 2,O 2, O 2 + , O 2 , O 3, BF3, CO3−2, C6H6 (Only C–C bond)
Q(34-35)
Dipole moment is equal to the product of charge separation, q and the bond length d for the bond. The unit of dipole moment is debye. One debye is equal to 10–18 esu. cm. Dipole moment is a vector quantity. It has both magnitude and direction. Dipole moment of molecules depends upon the relative orientation of the bond dipoles, but not on the polarity of the bonds alone.
34. The diatomic molecule has a dipole moment of 1.2 D. If the bond length is 1.0 × 10 –8 cm, what fraction of charge exists on each atom?
35. The dipole moment of HBr is 2.6 × 10–30 coulomb-metre and inter-atomic spacing is 1.4 Ao. The percentage ionic character of HBr is ___.
Matrix Matching Questions
36. Match the following and choose the correct option
List-I (order)
List-II (property)
A. LiI < LiBr < LiCl <LiF I) lattice energy
B. LiCl < BeCl2 < BCl3 < CCl4 II) ionic nature
C. NaCl < KCl < RbCl < CsCl III) covalent character
D. RbI < NaBr < MgF2 < CaO IV) electrical conductivity in fused state
V) solubility in water
(A) (B) (C) (D)
(1) IV III II I
(2) II T IV III
(3) I II T IV (4) III I T II
37. Match Column-I with Column-II. Choose the correct answer from the options given below.
Column-I (Molecule)
Column-II (Hybridisation; shape)
A. XeO3 I) sp3d ; linear
B. XeF2 II) sp3; pyramidal
C. XeOF4 III) sp3d3; distorted octahedral
D. XeF6 IV) sp3d2; square pyramidal
(A) (B) (C) (D)
(1) II I IV III
(2) II IV III I
(3) IV II III I
(4) IV II I III
38. M atch the orbital overlap figures shown in Column I with the description given in Column II and select the correct answer using the codes given below.
Column I Column II
A. I) p–d π anti-bonding
B. II) d–d σ bonding
C. III) p–d π bonding
D. IV) d–d σ bonding
(A) (B) (C) (D)
(1) II I III IV
(2) IV III I II
(3) II III I IV
(4) IV III I II
FLASHBACK (Previous JEE Questions)
JEE Main
1. The molecule/ion with square pyramidal shape is (2024)
(1) [Ni(CN)4] (2) PCl5
(3) BrF5 (4) PF5
2. Given below are two statements (2024)
Statement I : Since fluorine is more electronegative than nitrogen, the net dipole moment of NF3 is greater than NH3.
Statement II : In NH3, the orbital dipole due to lone pair and the dipole moment of NH bonds are in opposite direction, but in NF3 the orbital dipole due to lone pair and dipole moments of N-F bonds are in same direction.
I n t he light of the above statements. Choose the most appropriate from the options given below.
(1) Statement I is true but Statement II is false.
(2) Both Statement I and Statement II are false.
(3) Both statement I and Statement II is are tru
(4) Statement I is false but Statement II is are true.
3. The linear combination of atomic orbitals to form molecular orbitals takes place only when the combining atomic orbitals (2024)
A. have the same energy
B. have the minimum overlap
C. have same symmetry about the molecular axis
D. have different symmetry about the molecular axis
Choose the most appropriate from the options given below:
(1) A, B, C only
(2) A and C only
(3) B, C, D only
(4) B and D only
4. Which of the following is least ionic? (2004)
(1) BaCl2 (2) AgCl
(3) KCl (4) CoCl2
5. The pair from the following pairs having both compounds with net non-zero dipole moment is (2023)
6. Which one of the following pairs is an example of polar molecular solids? (2023) (1) SO2( s), NH3(s) (2) HCl(s), AlN(s) (3) MgO(s), SO2(s) (4) SO2( s), CO2(s)
7. The bond order and magnetic property of acetylide ion are same as that of (2023)
(1) NO+ (2) O2–
(3) N2+ (4) O2+
8. In which of the following processes, the bond order increases and paramagnetic character changes to diamagnetic one? (2023)
(1) NO → NO+ (2) O2 → O22–
(3) N2→ N2+ (4) O2 → O2+
9. Given below are two statements.
Statement–I : SO2 and H2O both possess V-shaped structure
Statement–II : The bond angle of SO2 is less than that of H2O.
In the light of the above statements, choose the most appropriate answer from the options given below. (2023)
(1) Statement I is correct but statement II is incorrect.
(2) Both Statement I and statement II are incorrect.
(3) Both Statement I and statement II are correct.
(4) Statement I is incorrect but statement II is correct.
10. Consider the following statements (2023)
(A) NF 3 molecule has a trigonal planar structure.
(B) Bond length of N2 is shorter than O2.
(C) Isoelectronic molecules or ions have identical bond order.
(D) Dipole moment of H2S is higher than that of water molecule.
(1) (A) and (D) are correct.
(2) (B) and (C) are correct.
(3) (A) and (B) are correct.
(4) (C) and (D) are correct.
11. Order of covalent bond: (2023)
A) KF > KI; LiF > KF
B) KF < KI; LiF > KF
C) SnCl4 > SnCl2; CuCl > NaCl
D) LiF > KF; CuCl < NaCl
E) KF < KI; CuCl > NaCl
Choose the correct answer from the options given below:
(1) A, B only (2) B, C only
(3) B, C, E only (4) C, E only
12. What is the number of unpaired electron(s) in the highest occupied molecular orbital of the following species
N2, N2+, O2 , O2+? (2023)
(1) 2, 1, 2, 1 (2) 2, 1, 0, 1
(3) 0, 1, 0, 1 (4) 0, 1, 2, 1
13. Given below are two statements.
Statement I : Dipole moment is a vector quantity and, by convention, it is depicted by a small arrow with tail on the negative centre and head pointing towards the positive centre.
Statement II : The crossed arrow of the dipole moment symbolises the direction of the shift of charges in the molecules.
In light of the above statements, choose the most appropriate answer from the options given below: (2023)
(1) Statement I is incorrect but statement II is correct.
(2) Both Statement I and statement II are correct.
(3) Statement I is correct but statement II is incorrect.
(4) Both statement I and statement II are incorrect.
14. The magnetic behaviour of Li2O, Na2O2, and KO2, respectively, are (2023)
(1) paramagnetic, diamagnetic, and paramagnetic
(2) diamagnetic, diamagnetic, and paramagnetic
(3) diamagnetic, paramagnetic, and diamagnetic
(4) paramagnetic, paramagnetic, and diamagnetic
15. According to MO theory the bond orders for O22−, CO, and NO+ , respectively, are (2023)
(1) 1, 3, and 3 (2) 2, 3, and 3
(3) 1, 2, and 3 (4) 1, 3, and 2
16. Match list-I with list-II. (2023)
List I (Molecules/ ions)
List II (Number of lone pairs of e- on central atom)
A. IF 7 I) Three
B. ICI4 II) One
C. XeF6 III) Two
D. XeF2 IV) Zero
Choose the correct answer from the options given below.
(A) (B) (C) (D)
(1) IV III II I
(2) II I IV III
(3) IV I II III
(4) IV I III II
17. Resonance in carbonate ion (CO32-) is (2023)
18. The correct order of bond enthalpy (kJ mol–1) is (2023)
(1) Si–Si > C–C > Sn–Sn > Ge–Ge
(2) Si–Si > C–C > Ge–Ge> Sn–Sn
(3) C–C> Si–Si > Ge–Ge > Sn–Sn
(4) C–C > Si–Si > Sn–Sn > Ge–Ge
19. The number of species from the following that have square pyramidal structure is PF5, BrF4 , IF5, BrF5, XeOF4, ICl4 (2023)
20. In an ice crystal each water molecule is hydrogen bonded to neighboring molecules (2023)
21. The number of species having a square planar shape from the following is XeF4, SF4, SiF4, BF4 , BrF4 , [Cu(NH3)4]+2, [FeCl4]−2,[PtCl4]2− (2023)
22. Which are the factors that affect the percent covalent character of the ionic bond? (2023)
A) Polarising power of cation
B) Extent of distortion of anion
C) Polarisability of the anion
D) Polarising power of anion
23. The number of species from the following, carrying a single lone pair on central atom xenon, is ____.
29. The correct order of bond orders of C22–, N22–and O22– is, respectively, (2022)
(1) C2–2 < N2–2 < O22–
(2) O22– < N2–2 < C2–2
(3) C2–2 < O22– < N2–2
(4) N2–2 < C2–2 < O22–
30. Match List-I with List-II. (2022) List-I List-II
(A) [PtCl4]2– (I) sp3d
(B) BrF 5 (II) d2sp3
(C) PCl5 (III) dsp2
(D) [Co(NH3)6]3+ (IV) sp3d2
Choose the correct option.
(A) (B) (C) (D)
(1) II IV I III
(2) III IV I II
(3) III I IV II
(4) II I IV III
31. Arrange the following in increasing order of their covalent character. (2022)
(A) CaF2
(B) CaCl2
(C) CaBr2
(D) CaI2
Choose the correct answer from the options given below. (2022)
(1) B < A < C < D
(2) A < B < C < D
(3) A < B < D < C
(4) A < C < B < D
32. Given below are two statements. One is labelled Assertion (A) and the other is labelled Reason (R).
Assertion (A) : LiF is sparingly soluble in water.
Reason (R) : The ionic radius of Li + ion is smallest among its group members, hence has least hydration enthalpy.
In light of the above statements, choose the most appropriate answer from the options given below: (2022)
(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) (A) is false but (R) is true.
33. Match List-I with List-II (2022)
List I (Compound) List II (Shape)
A. BrF 5 I) Bent
B. [CrF6]3– II) Square pyramidal
C. O3 III) Trigonal bipyramidal
D. PCl5 IV) Octahedral
Choose the correct answer from the options given below.
(A) (B) (C) (D)
(1) I II III IV
(2) IV III II I
(3) II IV I III
(4) III IV II I
34. Match List-I with List-II (2022)
List-I
List-II
A. ΨMO = ΨA−ΨB I) Dipole moment
B. μ = Q × r II) Bonding molecular orbital
C. 2 baNN III) Anti-bonding molecular orbital
D. ΨMO = ΨA + ΨB IV) Bond order
C hoose the correct answer from the options given below.
(A) (B) (C) (D)
(1) II I IV III
(2) III IV I II
(3) III I IV II
(4) III IV II I
35. The sum of number of lone pairs of electrons present on the central atoms of XeO 3, XeOF 4, and XeF 6 is _______. (2022)
36. Among the following species, N2, N2+, N2 , N22−, O2, O2+, O2 , O22−, the number of species showing diamagnetism is______. (2022)
37. Amongst BeF2,BF3, H2O, NH3, CCl4 and HCl, the number of molecules with nonzero net dipole moment is__________. (2022)
38. Amongst SF 4, XeF 4, CF 4, and H 2O, the number of species with two lone pairs of electrons on central atom is____ . (2022)
JEE Advanced
39. The correct molecular orbital diagram for F2 molecule in the ground state is ____ (2023) (1)
40. For diatomic molecules, the correct statement(s) about the molecular orbitals formed by the overlap of two 2pz orbitals is (are) (2022)
(1) σ orbital has a total of two nodal planes
(2) σ*orbital has one node in the x z -plane containing the molecular axis
(3) π orbital has one node in the plane which is perpendicular to the molecular axis and goes through the center of the molecule
(4) π* orbital has one node in the xy-plane containing the molecular axis
CHAPTER TEST – JEE MAIN
Section - A
Single Option Correct MCQ's
1. Which of the following contains unshared electron on cental atom?
(1) NO2
(2) CO2
(3) NO2
(4) CN
2. The formal charges on nitrogen atoms 1 and 2, respectively, are 1 ..2.. NNN ==
(1) −1, −1 (2) –1, +1
(3) +1, –1 (4) +1, +1
3. Dative bond is absent in
(1) BF 4 (2) NH4+
(3) O3 (4) CO–2 3
4. Oxidation numbers of A, B, and C are +2, +5, and –2 respectively. Possible formula of compound is
(1) A2(BC2)2 (2) A3(BC4)2
(3) A3(BC2)2 (4) A2(B2C)2
5. The lattice energies of KF, KCl, KBr and KI follow the order
(1) KF > KCl > KBr > Kl
(2) Kl > KBr > KCl > Kl
(3) KF > KCl > Kl > KBr
(4) Kl > KBr > KF > KCl
CHAPTER 5: Chemical Bonding and Molecular Structure
6. Given below are two statements. One is labelled Assertion (A) and the other is labelled Reason (R).
Assertion (A): many ionic solids are more soluble in water than in common organic solvents.
Reason (R): dielectric constant of water is more when compared with common organic solvents.
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.
7. If Na+ ion is larger than Mg2+ ion and S2–ion is larger than Cl– ion, which of the following will be least soluble in water?
(1) NaCl (2) Na2S
(3) MgCl2 (4) MgS
8. For central atom ‘A’, let ‘E’ be the lone pair.
Match the Column I with Column II
Column I Column II
A. AB2E I) T-shape
B. AB 3E2 II) square pyramid (Distorted)
C. AB 5 E III) bent
D. AB2E2 IV) see-saw V) non-zero dipole moment
Choose the correct option.
(A) (B) (C) (D)
(1) III,V IV IV V
(2) III IV,V IV,V I,V
(3) III,V I,V II,V III,V
(4) I,V II,V II,V III
9. Which of the following has zero overlapping of orbitals, if z-axis is internuclear axis?
(1) s−pz
(2) pz–pz
(3) px–px
(4) s–px
10. Match Column-I with Column-II and choose correct option.
Column I Column II
A. C2H2 I) sp3d2 hybridisation
B. SF6 II) sp3d3 hybridisation
C. SO2 III) sp hybridisation
D. IF 7 IV) sp2 hybridisation
(A) (B) (C) (D)
(1) I III II IV
(2) III I IV II
(3) II III I IV
(4) IV I III II
11. Given below are two statements. One is labelled Assertion (A) and the other is labelled Reason (R).
Assertion (A) : Xe in XeF 2 is d 2 sp 3 hybridised.
Reason (R) : XeF 2 molecule does not follow octet rule.
In light of the above statements, choose the most appropriate 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.
12. Given below are two statements. One is labelled Assertion (A) and the other is labelled Reason (R).
Assertion (A) : According to MOT, the process
N2(g) → N2 + (g)+ e− is more endothermic than the process
O2(g) → O2++ e
Reason (R) : Electron is removed from bonding molecular orbital in N 2 whereas electron is removed from antibonding molecular orbital in O2.
In light of the above statements, choose the most appropriate 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 and (R) is not the correct explanation of (A)
(3) (A) is true but (R) is false.
(4) Both (A) and (R) are false.
13. In which of the following processes, the value of magnetic moment does not change?
(i) 22NN →
(ii) 22NN + →
(iii) + 22OO →
(iv) 22OO→+
(v) + 22NN →
(vi) 22OO →
(1) (i) and (ii) (2) (iii) and (v)
(3) (iii) and(iv) (4) (v) and (ii)
14. The bond order and magnetic behavior of O2−2 ion , respectively, are (1) 1.5 and paramagnetic
(2) 1.5 and diamagnetic
(3) 2 and diamagnetic
(4) 1 and diamagnetic
15. When the molecules N2, N2O, and N2O4 are arranged in order of decreasing N–N bond length, which order is correct?
(1) N2O4, N2O, N2 (2) N2, N2O, N2O4
(3) N2O, N2, N2O4 (4) N2, N2O4, N2O
16. The correct order of increasing bond angle is
(1) PF 3 < PCl3 < PBr 3 < PI 3
(2) PF 3 < PBr 3 < PCl3 < PI 3
(3) PI 3 < PBr 3 < PCl3 < PF 3
(4) PF 3 > PCl3 < PBr 3 < PI 3
17. Which of the following orders is correct against the given property?
(1) H2O > NH3 > NF3: Dipole moment
(2) O2–2, > O2 > O+2: Bond order
(3) AlF 3 > MgF2 > NaF: Ionic character
(4) BF 3 < BCl3 < BBr3: Bond angle
18. The electronegativity difference between N and F is greater than electronegativity difference between N and H. Yet, the dipole moment of NH3(1.5 D) is larger than that of NF3(0.2 D). This is because
(1) in NH 3, the atomic dipole and bond dipole are in the same direction, whereas in NF3 , these are in opposite directions
(2) in NH 3 as well as NF 3 , the atomic dipole and bond dipole are in opposite directions
(3) in NH3 , the atomic dipole and bond dipole are in opposite directions, whereas in NF3, these are in the same direction
(4) in NH 3 as well as in NF 3, the atomic dipole and bond dipole are in the same direction
19. Identify the species having only one π -bond and three sigma bonds, and show canonical forms, is
(1) CO32- (2) O2 (3) SO2 (4) SO3
20. Which of the following hydrogen bonds is the strongest?
21. What is the sum of bond orders of the following species?
O2, O2–, O2–2, O+ 2
22. Number of sigma bonds in B2 molecule is __.
23. The number of nodal planes of σ*2pz is __.
24. How many of the following molecules is/ are known to exist?
KHF2, KHCl2, KHBr2, KHI2, KHAt2
25. If the total number of bonds in IF7 that are at 72°is x, then (x/2) = ?
CHAPTER TEST – JEE ADVANCED
2022 P2 Model
Section-A
[Integer Value Questions]
1. Calculate the value of P+Q for XeOF4 ( P = number of σ bond pairs and Q = number of lone pairs on central atom)
2. The value of electronegativity of atom 'X' and 'Y' are 1.20 and 4.0, respectively. The percentage of ionic character of X-Y bond is ____.
3. Dipole moment of toluene and chlorobenzene are 0.4 D and 2.6 D. The dipole moment of p -chloro toluene in debye units is ___.
4. When hybridisation involving d-orbitals are considered, then all the five orbitals are not degenerate, rather 222 x-yz d,d and d xy,dyz and d zx form two different sets of orbitals and orbitals of appropriate set are involved in the hybridisation. The number of species where 22 x-y d orbitals participate in hybridisation is/are XeF6 , SF6 , IF7 , XeO3, PCl5, PF 5
5. Lattice energy of sodium chloride is x kJ. Assuming the same inter-ionic distance, the lattice energy of MgS is y kJ. The value of y is ___.
6. A list of species having the formula XZ4, is given below:
XeF4, SF4, SiF4, BF 4 , and BrF 4
Defining shape on the basis of the location of X and Z atoms, the total number of species having a square planar shape is
7. How many non-axial d-orbitals are involved in the hybridisation of CrO2Cl2?
8. In how many of the following species, the N−O bond length is greater than that in NO+?
N2O, NO2–, NO3–, NOCl, NO2 + , NO2Cl
Section-B
[Multiple Option Correct MCQs]
9. Most ionic compounds have
(1) high melting points and low boiling points
(2) high melting points and non-directional bonds
(3) high solubilities in polar solvents and low solubilities in non-polar solvents
(4) three-dimensional network structures, and they are good conductors of electricity in the molten state
10. The linear structure is assumed by (1) SnCl2 (2) CO2 (3) NO2(+) (4) CS2
11. CO2 is iso-structural with (1) HgCl2 (2) SnCl2 (3) C2H2 (4) NO2
12. Which of the following statements is/are correct?
(3) sp2-hybridised orbitals have equal 's' and 'p' character.
(4) Usually, hybridised orbitals form bonds.
13. Which of the following species is/are paramagnetic?
(1) N2 (2) O22+
(3) NO+ (4) B2+
14. Which of the following have identical bond order?
(1) CN (2) O2–
(3) NO+ (4) CN+
Section-C
[Single Option Correct MCQs]
15. Electronic configurations of four elements E1, E2, E3, and E4 are, respectively, 1s2, 1s2 2s2 2p2, 1s2 2s2 2p5, and 1s2 2s2 2p6. The element that is capable of forming ionic as well as covalent bonds is (1) E1 (2) E2 (3) E 3 (4) E 4
16. The incorrect statement regarding the formation of ionic bond is
(1) it involves the electrostatic attraction
(2) it is a redox process
(3) it is an exothermic process
(4) it involves the absorption of energy
17. Polarisation is the distortion of the shape of an anion by an adjacently placed cation. Which of the following statements is correct?
(1) Maximum polarisation is brought about by a cation of high charge.
(2) Minimum polarisation is brought about by a cation of low radius.
(3) A large cation is likely to bring about a large degree of polarisation.
(4) A small anion is likely to undergo a large degree of polarisation.
18. Which among the following species has unequal bond lengths?
■ Found as orthoboric acid (H₃BO₃), borax (Na₂B₄O₇·10H₂O), kernite (Na₂B₄O₇·4H₂O).
■ Indian sources: Puga Valley (Ladakh), Sambhar Lake (Rajasthan).
■ Abundance: <0.0001% in Earth’s crust.
■ Isotopes: ¹⁰B (19%) and ¹¹B (81%).
■ Occurrence of Aluminium:
■ Third most abundant element (8.3% by mass) after Oxygen (45.5%) and Silicon (27.7%).
■ Found in bauxite (Al₂O₃·2H₂O), cryolite (Na₃AlF₆).
■ Found as oxides, fluorides, and silicates.
■ Gallium, Indium, and Thallium: Found in trace amounts with zinc and lead sulfide ores.
■ Abundance of elements o f group 13 is given in Table 6.1.
Table 6.1 Abundance of elements of group13
B 3 ×10–4 Al 8.13
Ga 1.3 × 10–3 In 1.0 × 10–5 Tl 10–4–10–5
Table 6.2 Electronic configuration of elements of boron family
Variation of Properties
■ Atomic radius: Increases down the group, but Ga is smaller than Al due to poor shielding by d-electrons.
■ Order: B < Ga < Al < In < Tl.
■ Ionization Energy:
■ First ionization energy (IP₁) < Second (IP₂) < Third (IP₃).
■ Boron has the highest ionization enthalpy, making B³ + formation difficult.
■ Ionization energy decreases irregularly down the group due to poor d- and f-electron screening.
■ Order of IP₁: In < Al < Ga < Tl < B.
■ Density: Increases down the group; Al is unexpectedly less dense due to its larger atomic size.
■ Order: B < Al < Ga < In < Tl.
■ Melting & Boiling Points:
■ Melting point decreases from B to Ga, then increases up to Tl.
■ Boron has a high melting point due to its giant covalent polymer structure.
■ Gallium has a low melting point (303 K) but a high boiling point (2676 K), making it useful for high-temperature measurements.
■ Boiling points decrease regularly from B to Tl.
■ Electronegativity:
■ Decreases from B to Al, then increases up to Tl due to poor d-electron shielding in Ga.
■ Al has the lowest electronegativity, making it the most metallic.
■ Oxidation States:
■ All elements exhibit +3 oxidation state due to ns²np¹ configuration.
■ Ga, In, Tl also show +1 oxidation state due to the inert pair effect, which increases down the group.
■ B and Al do not show the +1 state (no iner t pair effect).
Order of Relative Stability
■ Oxidation State Trends:
■ +3 stability: B³+ > Al³+ > Ga³+ > In³+ > Tl³+ .
■ +1 stability: B+ <Al+ <Ga+ <In+ < Tl+ (increases due to inert pair effect).
■ Boron also shows –3 oxidation state in borides (e.g., Mg₃B₂).
■ Ionic & Oxidizing Nature:
■ +1 oxidation state compounds are ionic.
■ +3 oxidation state compounds are oxidizing.
■ Lewis Acid Behavior:
■ Trivalent compounds are electron-deficient, making them Lewis acids.
■ Lewis acidity decreases down the group as atomic size increases.
■ Example: NH₃ donates a lone pair to BCl₃, forming a coordinate bond.
Inert pair effect:
■ Reluctance of outer s-electrons to participate in bonding.
■ Stability of Oxidation States: Tl (+1) is more stable than Tl (+3) due to the inert pair effect.
■ Oxidizing Nature of Tl³ + Compounds: TlCl₃ and similar compounds are strong oxidants,
Table 6.3 Atomic and physical properties of elements of boron family
Third
CHAPTER 6: The p-block Elements-I
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, M 2O3. B2O3 is acidic and reacts with water to form boric acid, H 3BO3. Al2O3 and Ga2O3 are amphoteric. Oxides of indium and tha llium are basic.
23 2 22 3 MO MO sg s () () ()
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 Tl 2O3 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.BI 3 is unstable.
2M(s)+3X2(g) → 2MX3(s) (X = F, Cl, Br, I)
■ Boron trihalides are covalent due to high polarising power of B 3+ ion. These trihalides are electron deficient and central atom has incomplete octet.
■ Lewis Acid Behavior: Group 13 trivalent compounds are electron-deficient and act as Lewis acids by accepting lone pairs.
■ BF₃ Weak Lewis Acidity: Due to back dative bonding (p-π–p-π interaction), reducing its tendency to accept electrons.
■ Order of Lewis Acid Strength: BBr₃ > BCl₃ > BF₃ (due to decreasing back dative bonding from Br to F).
■ Boron halides are readily hydrolysed, forming boric acid.
BX HO HX HBO 32 33 33
■ Aluminium Chloride Forms Dimers: Al₂Cl₆ exists up to 400°C, and above 800°C, it dissociates into monomeric AlCl₃.
■ Lewis Acid Nature: AlCl₃ acts as a Lewis acid, accepting electron pairs.
■ Dimeric Structure: Al₂Cl₆ contains six covalent bonds and two dative bonds.
Hydrolysis in Water:
■ Trivalent aluminium compounds (covalent) hydrolyze to form tetrahedral [M(OH)₄] – (sp³ hybridized).
■ AlCl₃ in acidified aqueous solution forms octahedral [Al(H₂O)₆]³ + (sp³d² hybridized, involving 3d orbitals).
■ Al₂Cl₆ as an Autocomplex: Al₂Cl₆ is called an autocomplex due to its self-association into a dimeric structure.
■ Dimeric aluminium chloride is shown in Fig. 6.1.
Reactivity towards acids and alkalies:
■ Boron’s Inertness: Boron does not react with acids or bases at moderate temperatures.
■ Aluminium’s Amphoteric Nature: Aluminium dissolves in both mineral acids and aqueous alkalies, showing amphoteric behavior.
■ Reaction with Dilute HCl: Aluminium reacts with dilute HCl, liberating dihydrogen (H₂) gas.
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)
General Chemical properties:
Hydrolysis of Trihalides:
■ Trichlorides, bromides, and iodides of Group 13 elements are covalent and hydrolyze in water.
■ Tetrahedral [M(OH)₄] – and octahedral [M(H₂O)₆]³ + species form in aqueous solutions (except for boron).
Lewis Acidity of Trihalides:
■ Monomeric trihalides (BF₃, AlCl₃, etc.) are electron-deficient and act as strong Lewis acids.
■ BF₃ reacts with Lewis bases (e.g., NH₃) to complete boron’s octet.
Fig. 6.1 Dimeric aluminium chloride
CHAPTER 6: The p-block Elements-I 180
Covalence Limitations:
■ Boron’s maximum covalence is 4 (due to the absence of d orbitals).
■ Aluminium and heavier elements exceed a covalence of 4 (due to the availability of d orbitals).
Dimerization of Metal Halides:
■ Most Group 13 metal halides dimerize through halogen bridging, completing the octet by accepting electrons from halogen atoms.
Anamolous Properties of Boron
■ Small atomic size
■ High ionization potential (IP)
■ High electronegativity (EN)
■ No vacant d orbitals
■ Different electron configuration in the penultimate shell.
■ Boron shows some differences in some of the properties with aluminium, which are given in Table 6.4.
Table 6.4 Difference between boron and aluminum Boron Aluminium
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.
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) razorite Na2B4O7.4H2O from these two minerals, the decahydrate is obtained by extracting with hot water and concentrating the extract.
■ Borax contains the tetranuclear units [B 4O5(OH)4]2–, as shown in Fig. 6.2
Properties of Borax
■ Borax exists in crystalline forms namely prismatic borax (Na 2B4O7.10H2O), 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).
■ Borax is sparingly soluble in cold water but fairly soluble in hot water.
■ Borax dissolves in water to give an alkaline solut ion 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.
Fig. 6.2 Tetranuclear units of borax
CHAPTER 6: The p-block Elements-I
■ 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 borax.
Na2B4O7+2HCl+5H2O → 2NaCl+4B(OH)3
Properties of Boric Acid
■ H3BO3 is a white cystalline solid which is soapy to touch. H 3BO3 is a weak monobasic acid. It loses water on heating. Orthoboric acid above 370 K forms metaboric acid HBO 2, which, on further heating, yields boric oxide, B 2O3
■ Boric acid has layer structure, in which planar BO3 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. 6.3. The dotted lines represent hydrogen bonds.
Fig. 6.3 Structure of boric acid (the dotted lines represent hydrogen bonds)
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 h ydride.
2BF+ 6LiH BH +6LiF 3 450K 26
It is prepared on a large scale by the reduction of BF 3 with LiAlH 4
4BF+ LiAlH2BH +2LiF+ 3AlF 34 26 3 →
■ In laboratory, diborane is obtained when sodiu m 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 hydrog en.
BH HO HBOH 26 23 32 62 6
■ It dissolves in strong alkalies to produce metabor ates and hydrogen.
BH KOHH OKBO H 26 22 2 22 26
■ It combines with carbon monoxide at 1000 C and 20 atmospheres to form borane carbonyl.
BHCOBHCO 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, BH 3.L.
BH NCHBHN CH 26 33 33 3 22 () .( )
■ Diborane reacts with metal hydrides in ether to giv e hydridoborates.
BH MH MBH 26 4 22 []
M is lithium or sodium.
CHAPTER 6: The p-block Elements-I
■ 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 fue l.
■ Reaction of ammonia with diborane gives, initially at 120°C, an addition product, diammoniate of diborane, which can be formulated as [BH(NH )] [BH] 23 2 + 4.
■ 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 b orazole is given in Fig. 6.4.
At higher temperatures, it decomposes to give boron nitride and hydrogen.
nB NH 3(BN)+ 3nH 33 6 ” n2 In B3N3H6, both B and N atoms are sp 2 hybridised.
Structure
■ In B2H6, total number of valence electrons i s 12. In the similar compound, C 2H6, the total number of valence electrons is 14. This indicates that B 2H6 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 B 2H6.
■ In B2H6, boron uses sp3 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 sp 3 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.
Fig. 6.4 Structure of borazole
IL ACHIEVER SERIES FOR JEE CHEMISTRY 185
■ The bridge hydrogen atoms are present in a plane perpendicular to the rest of the molecule. (Fig. 6.5 and 6.6)
6.6
■ The distance between the two non-bonded boron atoms is 177 pm.
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 H 2O molecules. Among the alkali metals, lithium does not form alum, due to its small size.
Preparation of Alum
■ Equimolar quantities of Al2(SO4) 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. KSOAlSOwater KSOAlSOH O aq aq 24 42 42 43 2 24 () ()() ()
■ The resultant solution, upon cooling, forms crystals of alum.
Fig 6.5 Structure of diborane
Fig
Diborane molecule and formation of banana bands.
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, f or heat, scratch and stain resistant glazed coating to earthenwares and as co nstituent 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 their toxic nature.
TEST YOURSELF
1. Which of the following statements is correct r egarding 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 [Al(H 2O)6]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. The 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 13 th 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
(11) 4 (12) 4 (13) 2 (14) 3
6.2 GROUP 14 ELEMENTS – THE CARBON FAMILY
■ Carbon (C), Silicon (Si), Germanium (Ge), Tin (Sn), Lead (Pb), Flerovium (Fl).
■ p-Block Elements: General electronic configuration: ns²np².
■ Penultimate Shell Electron Count:
■ Carbon (C): 2 electrons
■ Silicon (Si): 8 electrons
■ Ge, Sn, Pb, Fl: 18 electrons each
■ Differences in Properties
■ C arbon differs from Silicon due to its small size, high electronegativity, high ionization energy, and lack of d-orbitals.
■ Both Carbon and Silicon differ from heavier elements (Ge, Sn, Pb, Fl) due to the presence of d and f orbitals in heavier elements, leading to different chemical behavior.
■ The electronic configuration of these elements is given in Table 6.5.
Table 6.5 Electronic configuration of elements of carbon family Name
Occurrence and Abundance
■ Carbon (C):
■ 17th most abundant element in Earth’s crust.
■ Most abundant element in the human body and essential for life.
■ Exists free (coal, diamond, graphite) and in combined forms (carbonates, hydrocarbons, CO₂ in air ~0.03%).
■ Isotopes: ¹²C, ¹³C (stable), and ¹⁴C (radioactive, half-life 5770 years, used for radiocarbon dating).
Silicon (Si):
■ 2nd most abundant element in Earth’s crust (27.7%).
■ Found as silica (SiO₂) and silicates.
■ Used in ceramics, glass, and cement.
Germanium (Ge):
■ Found only in traces.
■ Ultra-pure Ge is used in transistors and semiconductors.
Tin (Sn) & Lead (Pb):
■ Less abundant, but occur in concentrated ores.
■ Tin (Sn): Found as cassiterite (SnO₂).
■ Lead (Pb): Found as galena (PbS).
Flerovium (Fl):
■ Atomic number: 114, Atomic mass: 289 g/mol.
■ Electronic configuration: [Rn] 5f¹⁴6d¹⁰7s²7p².
■ Short half-life, synthesized in small amounts, and chemistry is not well established.
■ Atomic and physical properties of elements of group 14 are given in Table 6.6.
Table 6.6 Atomic and physical properties of carbon family
2.22(graphite)
(Pauling scale)
Electronegativity (Pauling scale)
(20°C) (ohm cm) (diamond)
Variation of Properties
■ There is no regular gradation in properties..
Covalent Radius
■ Gradual increase from Carbon to Lead (C → Pb).
■ Significant jump from Carbon to Silicon, then smaller increases from Si to Pb due to d- and f-orbital shielding.
■ Heavier elements (Ge, Sn, Pb) have completely filled inner d- and f-orbitals, which reduce effective nuclear attraction, causing only a slight increase in atomic size.
■ Order of Atomic Radius:
■ C < Si < Ge < Sn < Pb.
Ionisation Energy
■ Higher First Ionization Energy than Group 13 due to greater nuclear charge.
■ Ionization energy decreases down the group but not systematically as in alkali and alkaline earth metals.
■ Large drop from C to Si due to the sudden increase in atomic size.
■ Pb has slightly higher ionization energy than expected due to lanthanide con traction.
■ Carbon has the highest ionization energy due to its small atomic size and high electronegativity.
■ Order of I1: C > > Si > Ge > Pb > Sn
190
Density, Melting Point, and Boiling Point
Density Trend:
■ Increases down the group (except for Carbon).
CHAPTER 6: The p-block Elements-I
■ Carbon has lower density than expected due to its unique structure.
Melting & Boiling Point Trend:
■ Carbon has an extremely high melting point due to its strong covalent bonding in a diamondtype lattice.
■ Si and Ge also have high melting points, but lower than Carbon.
■ Melting points decrease down the group as M–M bonds weaken with increasing atomic size.
■ Sn and Pb have significantly lower melting points because they do not use all four valence 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
■ Slightly higher electronegativity than Group 13 due to smaller atomic size.
■ Carbon is the most electronegative in the group.
■ Si, Ge, Sn, and Pb have nearly the same electronegativity because of:
■ d-orbital filling in Ge and Sn.
■ f-orbital filling in Pb, which reduces effective nuclear charge.
Oxidation States
■ Valence Electrons: ns²np² configuration → Common oxidation states: +4 and +2.
Carbon’s Unique Behavior:
■ Can show negative oxidation states due to high electronegativity.
■ +4 state is more common, +2 state is doubtful (e.g., in CO due to dative bonding from O to C).
Stability of Oxidation States:
■ Heavier elements show increasing stability of +2 state due to the inert pair effect.
■ Order of +2 stability: Ge² + < Sn²+ < Pb²+ .
■ Order of +4 stability: Ge⁴ + > Sn⁴+ > Pb⁴+ .
■ Pb²+ is more stable, while Pb⁴ + is a strong oxidizing agent.
■ Sn²+ acts as a reducing agent, while Sn⁴ + is more stable.
Covalency and Complex Formation:
■ Carbon cannot exceed a covalency of 4, while Si, Ge, Sn, and Pb can due to vacant d-orbitals.
■ Tendency to form complexes → Accepting electron pairs from donor species.
■ Hydrolysis of halides occurs for Si, Ge, Sn, but not for C-halides.
■ Examples of Complexes: [SiF₆]² –, [GeCl₆]²–, [Sn(OH)₆]²– (sp³d² hybridization).
Chemical Properties
■ General Reactivity: Relatively unreactive, but increases down the group.
Reaction with Water:
■ Carbon, Silicon, and Germanium → Unreactive wi th water.
■ Tin (Sn) → Reacts with steam, forming SnO₂ and H₂.
Sn HO 22SnOH 22 2
■ Lead (Pb) → Unaffected by water due to the formation of a protective oxide film.
Halides
Types of Halides:
■ Dihalides (MX₂) and Tetrahalides (MX₄) (except PbBr₄ and PbI₄, which do not exist).
■ Formed by sp³ hybridization, making them covalent, except SnF₄ and PbF₄, which are ionic.
Thermal Stability:
■ GeX₄ is more stable than GeX₂, while PbX₂ is more stable than PbX₄ due to the inert pair effect.
■ PbI₄ does not exist as the Pb–I bond does not release enough energy to excite 6s² electrons for hybridization.
■ Tetrahalides’ thermal stability decreases with increasing molecular mass.
Hydrolysis Trend:
■ Carbon tetrahalides (CX₄) do not hydrolyze, while other tetrahalides readily hydrolyze in water.
■ Hydrolysis tendency decreases down the group due to the decreasing availability of vacant d-orbitals.
■ Elements (except carbon) contain vacant d-orbitals, allowing water molecules to coordinate, making them prone to hy drolysis.
■ Tetrahalides (MX₄) of Si, Ge, Sn, Pb act as strong Lewis acids due to vacant d-orbitals in heavier elements.
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CHAPTER 6: The p-block Elements-I
■ Stability of Dihalides (MX₂) Increases Down the Group → GeX₂ < SnX₂ < PbX₂ (due to the inert pair effect).
Lewis Acidity of Halides:
■ GeF₄ and SiCl₄ act as Lewis acids (can accept electron pairs) due to available d-orbitals.
■ CCl₄ is NOT a Lewis acid as carbon lacks vacant d-orbitals.
Oxides
■ Formation: All members form oxides (MO & MO₂) when heated in oxygen.
Monoxides (MO):
■ CO is neutral.
■ GeO is acidic.
■ SnO and PbO are amphoteric.
Dioxides (MO₂):
CO₂, SiO₂, GeO₂ are acidic.
SnO₂ and PbO₂ are amphoteric.
■ SiO exists only at high temperatures.
■ Higher oxidation state oxides (+4) are more acidic than lower oxidation state oxides (+2).
Anomalous Behaviuor of the Carbon
Reasons for Dissimilarity:
■ Small atomic size
■ High electronegativity
■ High ionization energy
■ Absence of d-orbitals (no octet expansion possible)
Unique Properties of Carbon:
■ Exists freely in nature, while other elements are mostly found in combined states.
■ Max covalency = 4, whereas Si, Ge, Sn, Pb can expand their octet to covalency 6 (e.g., SiF₄ + 2F– → [SiF₆]²–).
■ Forms highly covalent compounds due to small size.
■ Strong C–C bond (348 kJ/mol) allows it to form long chains (catenation), whereas Si forms chains only up to 8 atoms.
■ Readily forms multiple bonds (e.g., C = C, C ≡ C, C = O, C = S, and C ≡ N.) unlike heavier elements.
Catenation (Self-Linking of Atoms)
■ Decreases down the group due to increasing atomic size & decreasing electronegativity.
■ Order: C >> Si > Ge = Sn (Pb does not show catenation).
■ Carbon exhibits the strongest catenation, forming linear, cyclic, and multiple-bonded structures.
■ This can be clearly seen from the bond energy values which are given in Table 6.7.
Table 6.7 Bond energies of elements of carbon family
Allotropes of Carbon
■ Definition: Allotropy is the existence of an element in two or more different forms, differing in physical and some chemical properties.
■ Forms of Carbon: Carbon exhibits allotropy and occurs in multiple modifications.
Types of Allotropes:
■ Crystalline forms (e.g., Diamond, Graphite, Fullerenes).
■ Amorphous forms (e.g., Charcoal, Coke, Soot).
■ Three of the allotropes of carbon are crystalline and the remaining are amorphous as shown in Fig.6.7
Carbon
Crystalline
Diamond
Graphite
Fullerenes
Amorphous
Coal, coke, wood, charcoal, animal charcoal, lamp black, carbon black, gas carbon, and petroleum coke
Fig.6.7 Allotropes of carbon Diamond
■ Structure: sp³ hybridized carbon atoms, each bonded tetrahedrally to four other carbon atoms, forming a giant 3D covalent network.
■ Bond Length: C–C = 1.54 Å
■ Bond Angle: 109°28’ Properties:
■ Hardest natural substance due to strong covalent bonding.
■ Exceptionally high refractive index
154 pm
CHAPTER 6: The p-block Elements-I
Uses of Diamond
■ Jewelry: Used as a precious stone, purity measured in carats (1 carat = 200 mg).
■ Cutting and Grinding: Used in tools & abrasives due to extreme hardness.
■ Heat Sensitivity: Used in high-precision thermometers.
■ Industry: Used in manufacturing tungsten filaments for light bulbs and in making dyes.
Graphite
■ Structure: Layered, 2D sheet-like structure with hexagonal rings of sp² hybridized carbon atoms.
Bonding:
■ Each carbon forms three covalent bonds in a plane.
■ C–C bond length = 1.42 Å.
■ Fourth electron remains delocalized, forming π bonds.
■ Electrical Conductivity: Good conductor due to delocalized π-electrons.
■ Layered Structure of Graphite
■ Interlayer Spacing: 3.35 Å between carbon sheets.
■ Forces: Weak van der Waals forces hold the layers together.
■ Sliding Property: Layers slide easily, making graphite slippery.
142 pm 335 pm
■ 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.
Fig. 6.8 Structure of Diamond
Fig. 6.9 Structure 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.
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 6.8.
Fullerenes
■ Newly discovered crystalline allotropes of carbon.
■ Produced by heating graphite in an electric arc with inert gases (He or Ar).
Table 6.8 Difference between diamond and graphite
S.No. Diamond Graphite
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.
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.
4. Chemical reactivity is low. Chemical reactivity is high.
5. Carbon undergoes sp3 hybridisation. C–C bond length is 154 pm and bond angle is 109 028 ’ .
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
6. Thermodynamically, diamond is less stable than graphite Thermodynamically, graphite is more stable.
7. It is transparent to X-rays with high refractive index (2.45).
It is layer lattice and layers are slippery due to weak van der Waals forces.
■ Composition: Mainly C₆₀, with smaller amounts of C₇₀ and traces of higher fullerenes (C n up to C₃₅₀ and beyond).
CHAPTER 6: The p-block Elements-I
■ Purest form of carbon, having smooth structure without dangling bonds.
■ Structure: Large cage-like spheroidal molecules.
■ C60 molecule has a shape like a soccer ball, is as shown in Fig.6.10. It is called buckminister fullerene.
Fig. 6.10 Buckminister fullerene
■ Structure & Properties of Fullerenes
■ C₆₀ Structure:
■ Composed of 20 six-membered rings and 12 five-membered rings.
■ Five-membered rings fuse only with six-membered rings.
■ All carbon atoms are sp² hybridized and form three sigma bonds.
■ Delocalized π-electrons give aromatic character.
■ Soccer ball-shaped molecule (buckyball) with 60 vertices, each occupied by a carbon atom.
■ Contains both single and double bonds: C–C (1.435 Å) & C=C (1.383 Å).
C₇₀ Structure:
■ 25 six-membered rings & 12 five-membered rings.
■ Same ring arrangement as C₆₀, but rugby ball-shaped.
■ Solubility: Fullerenes are soluble in organic solvents due to covalent bonding.
■ Thermodynamic Stability (Δ fH°):
■ Graphite = 0 kJ/mol (most stable allotrope).
■ Diamond = 1.90 kJ/mol.
■ Fullerene (C₆₀) = 38.1 kJ/mol.
■ Other Carbon Allotropes
■ Reactivity: Amorphous forms are more reactive than crystalline forms due to larger surface area.
Impure Forms of Carbon:
■ Carbon black → Burning hydrocarbons in limited air.
■ Charcoal & Coke → Heating wood /coal in absence of air.
Uses of Carbon
■ Industrial Gases: Used in the production of water gas, producer gas, and carbon disulfide.
■ Activated Charcoal: Adsorbs poisonous gases; used in water filters to remove organic contaminants.
■ Coke: Fuel & reducing agent in metallurgy.
■ Graphite: Used in electrodes for batteries & industrial electrolysis.
■ Carbon Black: Used in black ink and as a filler in automobile tires.
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
■ Solubility: Insoluble in water.
■ Nature: Neutral, does not affect litmus.
■ Combustion: Burns with a blue flame, forming CO₂.
■ Density: Nearly equal to air.
■ Toxicity: Highly poisonous → Forms a stable complex with hemoglobin (300× stronger than oxygen-hemoglobin complex), preventing oxygen transport by RBCs.
CHAPTER 6: The p-block Elements-I
■ Reducing Agent: Reduces most metal oxides (except alkali, alkaline earth metals, aluminum, & some transition metals) → Used in metal extraction.
■ 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.
■ CO₂, being heavy & non-combustible, is used in fire extinguishers.
■ CO₂ are used in urea production.
CO2 + 2NH3 NH2COONH4→ NH 2CONH2 + H2O
■ Carbogen (O₂ + 5–10% CO₂) is used for artificial respiratio n.
■ Green plants convert CO₂ into carbohydrates (glucose) via photosynthesis.
■ CO₂ (~0.03% in the atmosphere) is removed through photosynthesis.
■ The overall chemical change can be expressed as:
6CO2+12H2O → C6H12O6+6O2+6H2O
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CHAPTER 6: The p-block Elements-I
■ Plants produce food for themselves, animals, and humans via photosynthesis.
■ CO₂ is non-poisonous, unlike CO, but its increase due to fossil fuel combustion & cement production raises greenhouse effect & global temperatures.
■ CO₂ molecule:
■ Carbon undergoes sp hybridization.
■ Two sigma (σ) bonds form by overlap of sp orbitals of C with p orbitals of O.
■ Two pi (π) bonds form by p–p overlap with oxygen.
■ Linear structure with C–O bond length = 115 pm, and n o dipole moment.
■ Resonance structures exist.
OC OO CO OC O
Resonance structures of carbon dioxide
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
■ Silica & Silicates make up 95% of Earth’s crust.
■ Silicon dioxide (SiO₂) is commonly known as silica.
■ Exhibits polymorphism with crystalline & amorphous forms.
■ Crystalline forms: Quartz, Tridymite, Cristobalite (interconvertible at suitable temperatures).
■ Amorphous forms: Agate, Jasper, Onyx.
Properties of Silicon Dioxide
■ Silica & Silicates make up 95% of Earth’s crust.
■ Silicon dioxide (SiO₂) is commonly known as silica.
■ Exhibits polymorphism with crystalline & amorphous forms.
■ Crystalline forms: Quartz, Tridymite, Cristobalite (interconvertible at suitable temperatures).
■ Amorphous forms: Agate, Jasper, Onyx.
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
■ Silicon carbide (SiC) or ‘Carborundum’ is formed by heating silica (SiO₂) with carbon in an electric furnace.
■ SiC has a diamond-like structure.
■ Silica (SiO₂) heated to 1600°C forms quartz glass, used in light experiments.
Uses of Silicon Dioxide
■ Silica is used as a building material and acidic flux in metallurgy.
■ Quartz glass is used in UV light experiments & glass articles.
■ Coloured quartz is used as gems, while transparent quartz is used for lenses.
■ Bricks (sand, clay, lime) are used in furnace linings for steel manufacturing.
■ Quartz is a piezoelectric material, enabling accurate clocks, modern radio, and TV broadcasting.
■ Silica gel is used as a drying agent, chromatographic support, and catalyst.
■ Kieselguhr (amorphous silica) is used in filtration plants.
Comparison between CO 2 and SiO2 is given in Table 6.9.
Table 6.9 Comparison between carbon dioxide and silicon dioxide
S.No Carbon dioxide Silicon dioxide
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.
Structure of Silicon Dioxide
■ Silica (SiO₂) has a giant 3D structure, similar to diamond.
■ Each silicon atom is tetrahedrally bonded to four oxygen atoms via covalent bonds.
■ Each oxygen is shared between two tetrahedra, maintaining a Si:O ratio of 1:2.
■ So, the formula is SiO 2. Structure of silica is shown in Fig.6.11.
■ 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.
CHAPTER 6: The p-block Elements-I
■ Silica exists in two forms: Low-temperature (α-form) and high-temperature (β-form). These forms are interconvertible at suitable temperatures.
Silicones (Advance)
■ Silicones are synthetic organosilicon polymers with Si–O–Si linkages and (R₂SiO)n as the repeating unit.
■ R = Alkyl or Aryl group.
■ Commercial silicones are mainly methyl derivatives and, to a lesser extent, phenyl derivatives.
■ Starting material formula: R nSiCl(4 – n).
■ Preparation: Methyl chloride reacts with silicon at 300°C in the presence of copper catalyst.
■ Organosilanes formed: CH₃SiCl₃, (CH₃)₂SiCl₂, (CH₃)₃SiCl, and (CH₃)₄Si.
■ (CH₃)₂SiCl₂ undergoes hydrolysis, followed by polymerization, to form a straight-chain silicone polymer.
■ Chain length of silicones is controlled by adding (CH₃)₃SiCl, which blocks the polymer ends.
■ Silicones can be linear, cyclic, or branched-chain compounds.
Fig. 6.11 Structure of silica
Uses of Silicones
■ Silicones are water-repellent due to non-polar alkyl groups.
■ Properties: High thermal stability, dielectric strength, oxidation & chemical resistance.
Uses:
■ Waterproofing clothes & paper.
■ Grease & lubricants (stable viscosity in temperature changes).
■ Paints & enamels (heat & chemically resistant).
■ Electrical insulation & sealants.
■ Surgical & cosmetic implants (biocompatible).
Silicates (Advance)
■ Silicates are metal derivatives of silicic acid.
■ Aluminosilicates form when Al³+ replaces Si⁴+ in a 3D silicate network, creating a negatively charged structure.
■ Cations (Na+, K+, Ca²+) balance the charge; examples include feldspar and zeolites.
■ Zeolites trap Ca²+ from hard water, replacing it with Na +, acting as ion exchangers.
■ Zeolite structures have cavities that trap H₂O, NH₃, CO₂, ethanol, making them molecular sieves.
■ Zeolites are catalysts in petrochemical industries for hydrocarbon cracking and isomerization.
■ ZSM-5 zeolite converts alcohols into gasoline.
■ Hydrated zeolites are used for softening hard water.
TEST YOURSELF
1. Correct trend of electronegativity (on Pa uling 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, (1) 348, 297, 260 (2) 297, 348, 260 (3) 348, 260, 297 (4) 260, 297, 348
4. Which of the following has maximum ionic character? (1) PbF4 (2) PbI4 (3) PbCl4 (4) PbBr4
5. The element that shows greater ability to form Pπ−Pπ multiple bonds is (1) Ge (2) Si (3) Sn (4) C
6. The stability of dihalides of Si, Ge, Sn and Pb follows the sequence (1) SiX2 < GeX2 < PbX2 < SnX2 (2) SiX2 < GeX2 < SnX2 < PbX2 (3) PbX2 < SnX2 < GeX2 < SiX2 (4) GeX2 < SiX2 < SnX2 < PbX2
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 SiCl 4 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
CHAPTER 6: The p-block Elements-I
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 SiO 2 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
(3) sp3-hydridised (4) pure p-orbitals involved
15. Water gas is (1) CO+N2 (2) CO+H2 (3) CO2+N2 (4) CO2+H2
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− duetotheabsenceofd-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 maximum 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)+6H2O(l) → 2X(aq) +nH2(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. AlCl3 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) AlCl 3 is a covalent compound while AlF 3 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 (1) and (2)
(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 Al 2Cl6 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 H2 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 (1) and (2)
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 +
CHAPTER 6: The p-block Elements-I
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 inwater
(2) H3BO3 on heating at 100°C gives HBO2
(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 (B2H6) 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 CuSO4, 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 o r ax 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 Bor ax, ‘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
CHAPTER 6: The p-block Elements-I
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?
(1) [SiF6]−2 (2) [GeCl6]−2
(3) [SiCl6]−2 (4) [Sn(OH)6]−2
74. Which of the following is correct?
(1) Boiling Point: - Si > Ge > Sn > Pb
(2) Ionisation enthalpy: - C > Si > Ge > Sn > Pb
(3) Atomic radius: - Sn > Pb > Ge > Si > C
(4) Density: - Si>Ge>Sn>Pb
75. Which one is incorrect
(1) Catenation power of Ge=Sn (nearly)
(2) Lead cannot show catenation
(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) Pb - 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 _____
CHAPTER 6: The p-block Elements-I
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 fivemember 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) SiO 2 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. T he basic structural unit of feldspar, zeolites, mica, and asbestos is:
(1) ()2 SiO3 (2) SiO2
(3) () 4 4 SiO
(4)
122. Exha usted 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
CHAPTER 6: The p-block Elements-I
131. How many of the following are amphoteric oxides
22. Group-13 elements react with O 2 to form oxides of type M2O3(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, Al 2Cl6 in solid phase as well as in solution of non-polar solvents such as benzene. When
CHAPTER 6: The p-block Elements-I
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
(3) high hydration energy of Tl + ion
(4) inert pair effect
10. SiO2 is a solid ,while CO2 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 Y2 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 ____.
T HEORY-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 : 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 (HBF 4) 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).
(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.
JEE ADVANCED LEVEL
Multiple Option Correct MCQs
1. Compounds that readily undergo hydrolysis are
(1) CCl4
(2) BCl3
(3) SiCl4
(4) CF4
2. When metal carbides react with H2O, the correct equations are
(1) Al4C3+H2O→CH ≡ CH
(2) CaC2+H2O→CH ≡ CH
(3) Mg2C3 +H2O→CH3C ≡ 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 Cl –atoms.
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
17. ()()
∆ →+ 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
222 of AlCl3 and Al2Cl6. The structure of dimer is
24. The correct order of bond angles in Al2Cl6 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 Al 2 Cl 6_ and B 2 H 6 forms hexamethyl derivatives((CH2)6M2).
III. In Al 2Cl 6, 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
CHAPTER 6: The p-block Elements-I
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
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
FLASHBACK (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 Na2B4Ox (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)
Assertion (A) : Carbon forms two important oxides CO and CO 2, 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 6: The p-block Elements-I
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 so lid 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) 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 (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 _____.
CHAPTER TEST - JE E 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) SiF4>SiCl4 : Ease of hydrolysis under normal conditions
(4) Diamond > graphite: Thermal conductivity
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
