CLASS 11 · CHEMISTRY · CHAPTER 4

Chemical Bonding and Molecular Structure

Simple solutions to four worked examples and all 40 end exercises on Lewis structures, molecular shape, bonding and molecular orbitals.

44 worked answersComplete chapter answers

Read the NCERT questions ↗← All chapters

44 worked answers · short steps first, explanations when you need them

Worked Questions from the Chapter

Textbook pages 104–110

Question Problem 4.1

How do we draw a Lewis structure for CO?

Steps only

C contributes 4 and O contributes 6 valence electrons, total 10. Draw a triple bond C≡O and one lone pair on each atom; formal charges are C⁻ and O⁺.

Answer

C contributes 4 and O contributes 6 valence electrons, total 10. Draw a triple bond C≡O and one lone pair on each atom; formal charges are C⁻ and O⁺.

Question Problem 4.2

How do we draw the Lewis structure of NO₂⁻?

Steps only

Total valence electrons=5+2(6)+1=18. Draw O–N=O with one lone pair on N; the double bond can be on either O, giving two equivalent resonance forms.

Answer

Total valence electrons=5+2(6)+1=18. Draw O–N=O with one lone pair on N; the double bond can be on either O, giving two equivalent resonance forms.

Question Problem 4.3

How does resonance describe CO₃²⁻?

Steps only

Three equivalent Lewis forms shift the C=O bond among the three O atoms. The real ion is a hybrid with three equal C–O bonds.

Answer

Three equivalent Lewis forms shift the C=O bond among the three O atoms. The real ion is a hybrid with three equal C–O bonds.

Question Problem 4.4

How is CO₂ described by resonance in this chapter?

Steps only

The chapter describes three canonical forms to account for the measured C–O length between typical double and triple bond lengths. The linear molecule is a resonance hybrid, not a structure switching among forms.

Answer

Chapter 4 Exercises

Textbook pages 133–135

Question 4.1

Exercise 4.1: Explain why atoms form chemical bonds.

Steps only

Atoms bond when the bonded arrangement has lower energy and greater stability than separate atoms. They do this by transferring or sharing valence electrons.

Answer

Atoms bond when the bonded arrangement has lower energy and greater stability than separate atoms. They do this by transferring or sharing valence electrons.

Question 4.2

Exercise 4.2: Write Lewis dot symbols for Mg, Na, B, O, N and Br.

Steps only

Show valence electrons as dots: Mg 2; Na 1; B 3; O 6; N 5; Br 7. Put one dot on each side before pairing.

Answer

Show valence electrons as dots: Mg 2; Na 1; B 3; O 6; N 5; Br 7. Put one dot on each side before pairing.

Question 4.3

Exercise 4.3: Write Lewis symbols for S, S²⁻, Al, Al³⁺, H and H⁻.

Steps only

S has 6 dots; S²⁻ has 8 in brackets with 2−. Al has 3 dots; Al³⁺ is bracketed with 3+ and no dots. H has 1 dot; H⁻ has 2 dots in brackets with −.

Answer

S has 6 dots; S²⁻ has 8 in brackets with 2−. Al has 3 dots; Al³⁺ is bracketed with 3+ and no dots. H has 1 dot; H⁻ has 2 dots in brackets with −.

Question 4.4

Exercise 4.4: Describe Lewis structures for H₂S, SiCl₄, BeF₂, CO₃²⁻ and HCOOH.

Steps only

H₂S: H–S–H with two lone pairs on S. SiCl₄: four Si–Cl single bonds; each Cl has three lone pairs. BeF₂: F–Be–F, linear; each F has three lone pairs. CO₃²⁻: three equivalent resonance forms, each with one C=O and two C–O⁻. HCOOH: H–C(=O)–O–H; each O has two lone pairs.

Answer

Question 4.5

Exercise 4.5: Define the octet rule and state its importance and limitations.

Steps only

Atoms tend to gain, lose or share electrons to reach eight valence electrons (H seeks two). It helps predict simple bonding. Exceptions include incomplete octets (BF₃), odd-electron species (NO), and expanded valence shells (SF₆); it does not predict shape or stability by itself.

Answer

Question 4.6

Exercise 4.6: What conditions favour formation of an ionic bond?

Steps only

Ionic bonding is favoured when cation formation needs little ionisation energy, anion formation is favourable, and a large amount of lattice energy is released.

Answer

Ionic bonding is favoured when cation formation needs little ionisation energy, anion formation is favourable, and a large amount of lattice energy is released.

Question 4.7

Exercise 4.7: Using VSEPR, state shapes of BeCl₂, BCl₃, SiCl₄, AsF₅, H₂S and PH₃.

Steps only

BeCl₂ linear; BCl₃ trigonal planar; SiCl₄ tetrahedral; AsF₅ trigonal bipyramidal; H₂S bent; PH₃ trigonal pyramidal. Lone pairs repel more strongly and compress bond angles.

Answer

BeCl₂ linear; BCl₃ trigonal planar; SiCl₄ tetrahedral; AsF₅ trigonal bipyramidal; H₂S bent; PH₃ trigonal pyramidal. Lone pairs repel more strongly and compress bond angles.

Question 4.8

Exercise 4.8: Why is the H–O–H angle smaller than the H–N–H angle although both geometries are distorted tetrahedral?

Steps only

Oxygen in H₂O has two lone pairs; nitrogen in NH₃ has one. Lone-pair repulsion is stronger than bond-pair repulsion, so H₂O is compressed more (104.5°) than NH₃ (about 107°).

Answer

Oxygen in H₂O has two lone pairs; nitrogen in NH₃ has one. Lone-pair repulsion is stronger than bond-pair repulsion, so H₂O is compressed more (104.5°) than NH₃ (about 107°).

Question 4.9

Exercise 4.9: How is bond strength related to bond order?

Steps only

Higher bond order generally means a stronger, shorter bond: triple > double > single.

Answer

Higher bond order generally means a stronger, shorter bond: triple > double > single.

Question 4.10

Exercise 4.10: Define bond length.

Steps only

Bond length is the equilibrium distance between the nuclei of two bonded atoms.

Answer

Bond length is the equilibrium distance between the nuclei of two bonded atoms.

Question 4.11

Exercise 4.11: Explain resonance in CO₃²⁻.

Steps only

CO₃²⁻ has three equivalent Lewis forms, with the double bond on a different O in each. The real ion is their hybrid: all three C–O bonds are equal, with average bond order 1⅓.

Answer

CO₃²⁻ has three equivalent Lewis forms, with the double bond on a different O in each. The real ion is their hybrid: all three C–O bonds are equal, with average bond order 1⅓.

Question 4.12

Exercise 4.12: Can two shown H₃PO₃ structures be resonance forms? Explain.

Steps only

No. Valid resonance forms keep the same positions of atoms and differ only in electron placement. The two proposed structures change atom connectivity or hydrogen attachment, so they are different structures, not resonance forms.

Answer

Question 4.13

Exercise 4.13: Write resonance forms for SO₃, NO₂ and NO₃⁻.

Steps only

SO₃: three equivalent forms with the S=O position shifting. NO₂: two forms with one N=O and one N–O bond; it has an odd electron. NO₃⁻: three equivalent forms with one N=O and two N–O⁻ bonds.

Answer

Question 4.14

Exercise 4.14: Use Lewis symbols to show electron transfer in (a) K+S (b) Ca+O (c) Al+N.

Steps only

Metals lose electrons, and non-metals gain them.

Answer

(a) K and S: K₂S (2K⁺ and S²⁻).
(b) Ca and O: CaO (Ca²⁺ and O²⁻).
(c) Al and N: AlN (Al³⁺ and N³⁻).

Question 4.15

Exercise 4.15: Why is H₂O bent but CO₂ linear? Explain using dipole moments.

Steps only

CO₂ has two opposite C=O bond dipoles that cancel in a linear shape. H₂O is bent, so its two O–H dipoles do not cancel; the net dipole is non-zero.

Answer

CO₂ has two opposite C=O bond dipoles that cancel in a linear shape. H₂O is bent, so its two O–H dipoles do not cancel; the net dipole is non-zero.

Question 4.16

Exercise 4.16: Give uses or significance of dipole moment.

Steps only

It measures molecular polarity; helps infer shape, distinguish isomers, estimate percentage ionic character, and compare charge separation.

Answer

It measures molecular polarity; helps infer shape, distinguish isomers, estimate percentage ionic character, and compare charge separation.

Question 4.17

Exercise 4.17: Define electronegativity and distinguish it from electron-gain enthalpy.

Steps only

Electronegativity is an atom’s relative ability to attract shared bonding electrons. Electron-gain enthalpy is an energy change when an isolated gaseous atom gains an electron.

Answer

Electronegativity is an atom’s relative ability to attract shared bonding electrons. Electron-gain enthalpy is an energy change when an isolated gaseous atom gains an electron.

Question 4.18

Exercise 4.18: Explain a polar covalent bond with an example.

Steps only

In HF, both atoms share electrons, but F attracts the pair more strongly. This creates partial charges Hδ⁺ and Fδ⁻.

Answer

In HF, both atoms share electrons, but F attracts the pair more strongly. This creates partial charges Hδ⁺ and Fδ⁻.

Question 4.19

Exercise 4.19: Arrange LiF, K₂O, N₂, SO₂ and ClF₃ by increasing ionic character.

Steps only

N₂ < ClF₃ < SO₂ < K₂O < LiF. N₂ is non-polar; between ClF₃ and SO₂, the S–O bond has the slightly larger electronegativity difference. Ionic character increases as the difference increases.

Answer

Question 4.20

Exercise 4.20: Correct the Lewis structure of acetic acid, CH₃COOH.

Steps only

CH₃–C(=O)–O–H. The carbonyl carbon is double-bonded to one O and single-bonded to OH; the CH₃ group bonds to that carbon. Each O has two lone pairs.

Answer

CH₃–C(=O)–O–H. The carbonyl carbon is double-bonded to one O and single-bonded to OH; the CH₃ group bonds to that carbon. Each O has two lone pairs.

Question 4.21

Exercise 4.21: Why is methane tetrahedral rather than square planar?

Steps only

Tetrahedral sp³ bonds are 109.5° apart, maximising separation of four bonding pairs. A square plane has 90° angles and much stronger pair repulsion.

Answer

Tetrahedral sp³ bonds are 109.5° apart, maximising separation of four bonding pairs. A square plane has 90° angles and much stronger pair repulsion.

Question 4.22

Exercise 4.22: Why does BeH₂ have zero dipole moment although Be–H bonds are polar?

Steps only

BeH₂ is linear H–Be–H. Its equal Be–H bond dipoles point in opposite directions and cancel.

Answer

BeH₂ is linear H–Be–H. Its equal Be–H bond dipoles point in opposite directions and cancel.

Question 4.23

Exercise 4.23: Which has greater dipole moment, NH₃ or NF₃? Why?

Steps only

NH₃ has the larger dipole. In NH₃, N–H bond dipoles point broadly with the lone-pair direction; in NF₃, N–F bond dipoles oppose the lone-pair direction and largely cancel.

Answer

NH₃ has the larger dipole. In NH₃, N–H bond dipoles point broadly with the lone-pair direction; in NF₃, N–F bond dipoles oppose the lone-pair direction and largely cancel.

Question 4.24

Exercise 4.24: Define hybridisation and describe sp, sp² and sp³ hybrid orbital shapes.

Steps only

Hybridisation mixes atomic orbitals on one atom to form equivalent directional orbitals. sp: two linear orbitals (180°); sp²: three trigonal-planar orbitals (120°); sp³: four tetrahedral orbitals (109.5°).

Answer

Question 4.25

Exercise 4.25: Describe the change in Al hybridisation in AlCl₃ + Cl⁻ → AlCl₄⁻.

Steps only

In AlCl₃, Al is trigonal planar (sp²). Cl⁻ donates a lone pair to form tetrahedral AlCl₄⁻, where Al is sp³.

Answer

In AlCl₃, Al is trigonal planar (sp²). Cl⁻ donates a lone pair to form tetrahedral AlCl₄⁻, where Al is sp³.

Question 4.26

Exercise 4.26: Does B or N hybridisation change in BF₃ + NH₃ → F₃B←NH₃?

Steps only

B changes from sp² in BF₃ to sp³ in F₃B←NH₃ when it accepts a lone pair. N remains sp³: it has three N–H bonds and one donated lone pair, now used in the coordinate bond.

Answer

B changes from sp² in BF₃ to sp³ in F₃B←NH₃ when it accepts a lone pair. N remains sp³: it has three N–H bonds and one donated lone pair, now used in the coordinate bond.

Question 4.27

Exercise 4.27: Describe how C–C double and triple bonds form in C₂H₄ and C₂H₂.

Steps only

C₂H₄: each C is sp²; one C–C σ bond and one sideways p–p π bond make the double bond. C₂H₂: each C is sp; one σ plus two perpendicular π bonds make the triple bond.

Answer

C₂H₄: each C is sp²; one C–C σ bond and one sideways p–p π bond make the double bond. C₂H₂: each C is sp; one σ plus two perpendicular π bonds make the triple bond.

Question 4.28

Exercise 4.28: Count sigma and pi bonds in (a) C₂H₂ (b) C₂H₄.

Steps only

(a) C₂H₂ (H–C≡C–H): 3 σ and 2 π. (b) C₂H₄ (H₂C=CH₂): 5 σ and 1 π.

Answer

(a) C₂H₂ (H–C≡C–H): 3 σ and 2 π. (b) C₂H₄ (H₂C=CH₂): 5 σ and 1 π.

Question 4.29

Exercise 4.29: With x as internuclear axis, which pair cannot form a sigma bond: (a) 1s/1s (b) 1s/2pₓ (c) 2pᵧ/2pᵧ (d) 1s/2s? Explain.

Steps only

(c) 2pᵧ and 2pᵧ cannot form a σ bond when x is the internuclear axis; their side-on overlap forms a π bond. The other listed pairs can overlap head-on.

Answer

(c) 2pᵧ and 2pᵧ cannot form a σ bond when x is the internuclear axis; their side-on overlap forms a π bond. The other listed pairs can overlap head-on.

Question 4.30

Exercise 4.30: Name carbon hybridisation in (a) CH₃–CH₃ (b) CH₃–CH=CH₂ (c) CH₃–CH₂–OH (d) CH₃–CHO (e) CH₃COOH.

Steps only

(a) CH₃–CH₃: both C sp³. (b) CH₃–CH=CH₂: first C sp³, alkene C atoms sp². (c) CH₃–CH₂–OH: both C sp³. (d) CH₃–CHO: methyl C sp³, carbonyl C sp². (e) CH₃COOH: methyl C sp³, carboxyl C sp².

Answer

Question 4.31

Exercise 4.31: Distinguish bond pairs from lone pairs and give an example of each.

Steps only

A bond pair is shared between two atoms, e.g. the C–H pair in CH₄. A lone pair stays on one atom, e.g. a non-bonding pair on O in H₂O.

Answer

A bond pair is shared between two atoms, e.g. the C–H pair in CH₄. A lone pair stays on one atom, e.g. a non-bonding pair on O in H₂O.

Question 4.32

Exercise 4.32: Distinguish sigma and pi bonds.

Steps only

A σ bond forms by head-on orbital overlap, has electron density along the internuclear axis, and is usually stronger. A π bond forms by side-on overlap with density above/below that axis; it occurs alongside a σ bond in multiple bonds.

Answer

Question 4.33

Exercise 4.33: Explain H₂ bond formation using valence-bond theory.

Steps only

Two singly occupied 1s orbitals overlap head-on. Opposite-spin electrons pair in the shared region; attraction to both nuclei lowers energy and forms a σ(1s) bond.

Answer

Two singly occupied 1s orbitals overlap head-on. Opposite-spin electrons pair in the shared region; attraction to both nuclei lowers energy and forms a σ(1s) bond.

Question 4.34

Exercise 4.34: What conditions must atomic orbitals meet to combine into molecular orbitals?

Steps only

Atomic orbitals combine effectively when they have similar energies, compatible symmetry about the molecular axis, and appreciable overlap. The number of molecular orbitals formed equals the number of atomic orbitals combined.

Answer

Question 4.35

Exercise 4.35: Use molecular-orbital theory to explain why Be₂ is unstable.

Steps only

Each Be contributes two 2s electrons. They fill σ2s² and σ*2s²; bond order=½(2−2)=0. With no net bond, Be₂ is unstable.

Answer

Each Be contributes two 2s electrons. They fill σ2s² and σ*2s²; bond order=½(2−2)=0. With no net bond, Be₂ is unstable.

Question 4.36

Exercise 4.36: Compare O₂⁺, O₂ and O₂⁻ for stability and magnetic behaviour.

Steps only

Bond orders: O₂⁺=2.5, O₂=2, O₂⁻=1.5, so stability increases O₂⁻ < O₂ < O₂⁺. All are paramagnetic: O₂⁺ has one unpaired electron, O₂ two, O₂⁻ one.

Answer

Bond orders: O₂⁺=2.5, O₂=2, O₂⁻=1.5, so stability increases O₂⁻ < O₂ < O₂⁺. All are paramagnetic: O₂⁺ has one unpaired electron, O₂ two, O₂⁻ one.

Question 4.37

Exercise 4.37: What do plus and minus signs represent in orbital diagrams?

Steps only

Plus and minus show opposite phases (signs) of the orbital wavefunction, not positive or negative electric charge. Same-phase overlap is constructive; opposite-phase overlap is destructive.

Answer

Plus and minus show opposite phases (signs) of the orbital wavefunction, not positive or negative electric charge. Same-phase overlap is constructive; opposite-phase overlap is destructive.

Question 4.38

Exercise 4.38: Describe PCl₅ hybridisation and explain why axial bonds are longer than equatorial bonds.

Steps only

P uses sp³d hybridisation: three equatorial bonds and two axial bonds in a trigonal bipyramid. Axial bonds have more 90° interactions, so their electron pairs are repelled more and the bonds are longer/weaker.

Answer

Question 4.39

Exercise 4.39: Define hydrogen bonding and compare its strength with van der Waals forces.

Steps only

A hydrogen bond is attraction between H covalently attached to N, O or F and a lone pair on N, O or F. It is stronger than ordinary van der Waals forces, but weaker than a covalent bond.

Answer

A hydrogen bond is attraction between H covalently attached to N, O or F and a lone pair on N, O or F. It is stronger than ordinary van der Waals forces, but weaker than a covalent bond.

Question 4.40

Exercise 4.40: Define bond order and calculate it for N₂, O₂, O₂⁺ and O₂⁻.

Steps only

MO bond order=½(Nbonding−Nantibonding). N₂=3; O₂=2; O₂⁺=2.5 (remove an antibonding electron); O₂⁻=1.5 (add one antibonding electron).

Answer

MO bond order=½(Nbonding−Nantibonding). N₂=3; O₂=2; O₂⁺=2.5 (remove an antibonding electron); O₂⁻=1.5 (add one antibonding electron).