Chemical Bonding and Molecular Structure Class 11 Notes

• Favoured by low ionisation enthalpy, high electron gain enthalpy and high lattice energy.
• Fajan's rules: small cations and large, polarisable anions increase covalent character (e.g. LiCl is more covalent than NaCl).
• Ionic compounds conduct electricity in molten or aqueous state, not as solids.

• Valence pairs arrange themselves to minimise repulsion; lone pairs repel more strongly than bond pairs, squeezing bond angles.
• Linear (2 pairs, 180°) — BeCl₂ and CO₂.
• Trigonal planar (3 pairs, 120°) — BF₃.
• Tetrahedral (4 pairs, 109.5°) — CH₄; with one lone pair NH₃ becomes trigonal pyramidal (107°), with two lone pairs H₂O is bent (104.5°).

• CH₄, NH₃, H₂O — sp³ with SN = 4.
• BCl₃ — sp² with SN = 3.
• BeCl₂ and C₂H₂ (acetylene) — sp with SN = 2.
• Greater s-character shortens and strengthens the bond: sp bonds are shorter than sp² bonds, which are shorter than sp³ bonds.

Example: Draw the Lewis structure of ${\text{CO}}_2$ and state its molecular shape.

Solution: Carbon shares four electrons — a double bond with each oxygen. No lone pairs on carbon, so the shape is linear (sp hybridised) with bond angle 180°.

Example: Why is the H₂O bond angle 104.5° and not 109.5°?

Solution: Water is sp³ hybridised (SN = 4) with two lone pairs on oxygen. Lone-pair — lone-pair repulsion is stronger than bond-pair repulsion, so the two O–H bonds are pushed closer from 109.5° to 104.5°.

Exercises (NCERT Q4.1–Q4.40)

• 4.1) Explain the formation of a chemical bond.
• 4.2) Write Lewis dot symbols for atoms of the following elements: Mg, Na, B, O, N and Br.
• 4.3) Write Lewis symbols for the following atoms and ions: S and S²⁻; Al and Al³⁺; H and H⁻.
• 4.4) Draw the Lewis structures for H₂S, SiCl₄, BeF₂, CO₃²⁻ and HCOOH.
• 4.5) Define the octet rule. Write its significance and limitations.
• 4.6) Write the favourable factors for the formation of an ionic bond.
• 4.7) Discuss the shapes of BeCl₂, BCl₃, SiCl₄, AsF₅, H₂S and PH₃ using the VSEPR model.
• 4.8) Although the geometries of NH₃ and H₂O are described as distorted tetrahedral, the bond angle in water is less than that in ammonia. Discuss.
• 4.9) How do you express bond strength in terms of bond order?
• 4.10) Define bond length.
• 4.11) Explain the important aspects of resonance with reference to the CO₃²⁻ ion.
• 4.12) H₃PO₃ can be represented by structures I and II shown in the source. Can these be taken as canonical forms of the resonance hybrid representing H₃PO₃? If not, give reasons.
• 4.13) Write the resonance structures for SO₃, NO₂ and NO₃⁻.
• 4.14) Use Lewis symbols to show electron transfer between (a) K and S, (b) Ca and O and (c) Al and N so that cations and anions form.
• 4.15) Although both CO₂ and H₂O are triatomic molecules, H₂O is bent while CO₂ is linear. Explain this on the basis of dipole moment.
• 4.16) Write the significance and applications of dipole moment.
• 4.17) Define electronegativity. How does it differ from electron gain enthalpy?
• 4.18) Explain, with a suitable example, a polar covalent bond.
• 4.19) Arrange the bonds in order of increasing ionic character in LiF, K₂O, N₂, SO₂ and ClF₃.
• 4.20) The skeletal structure of CH₃COOH is correct, but some bonds are shown incorrectly. Write the correct Lewis structure for acetic acid.
• 4.21) Apart from tetrahedral geometry, another possible geometry for CH₄ is square planar, with four H atoms at the corners and C at the centre. Explain why CH₄ is not square planar.
• 4.22) Explain why BeH₂ has zero dipole moment although its Be–H bonds are polar.
• 4.23) Which of NH₃ and NF₃ has the higher dipole moment, and why?
• 4.24) What is meant by hybridisation of atomic orbitals? Describe the shapes of sp, sp² and sp³ hybrid orbitals.
• 4.25) Describe the change, if any, in the hybridisation of the Al atom in AlCl₃ + Cl⁻ → AlCl₄⁻.
• 4.26) Is there any change in the hybridisation of B and N as a result of BF₃ + NH₃ → F₃B·NH₃?
• 4.27) Describe the formation of a double bond in C₂H₄ and a triple bond in C₂H₂.
• 4.28) What is the total number of σ and π bonds in C₂H₂ and C₂H₄?
• 4.29) Taking the x-axis as the internuclear axis, which pair will not form a σ bond: (a) 1s and 1s, (b) 1s and 2pₓ, (c) 2pᵧ and 2pᵧ, or (d) 1s and 2s?
• 4.30) Which hybrid orbitals are used by carbon in (a) CH₃–CH₃, (b) CH₃–CH=CH₂, (c) CH₃–CH₂–OH, (d) CH₃–CHO and (e) CH₃COOH?
• 4.31) What are bond pairs and lone pairs of electrons? Give one example of each type.
• 4.32) Distinguish between a σ bond and a π bond.
• 4.33) Explain the formation of the H₂ molecule on the basis of valence-bond theory.
• 4.34) Write the important conditions required for the linear combination of atomic orbitals to form molecular orbitals.
• 4.35) Use molecular-orbital theory to explain why the Be₂ molecule does not exist as a stable discrete molecule.
• 4.36) Compare the relative stability of O₂, O₂⁺, O₂⁻ and O₂²⁻ and indicate their magnetic properties.
• 4.37) What is the significance of a plus sign and a minus sign shown while representing orbitals?
• 4.38) Describe the hybridisation in PCl₅. Why are the axial bonds longer than the equatorial bonds?
• 4.39) Define a hydrogen bond. Is it weaker or stronger than van der Waals forces?
• 4.40) What is meant by bond order? Calculate the bond order of N₂, O₂, O₂⁺ and O₂⁻.

Answers at a glance

1. 1
   Q4.1 → A chemical bond forms when electron rearrangement and attraction between the bonded particles produce a lower-energy arrangement, usually approaching a stable noble-gas valence configuration.
2. 2
   Q4.2 → The Lewis symbols contain respectively 2, 1, 3, 6, 5 and 7 valence-electron dots for Mg, Na, B, O, N and Br.
3. 3
   Q4.3 → S and S²⁻ show 6 and 8 valence electrons; Al and Al³⁺ show 3 and 0; H and H⁻ show 1 and 2, respectively.
4. 4
   Q4.4 → H₂S has 2 S–H bonds and 2 S lone pairs; SiCl₄ has 4 Si–Cl bonds; BeF₂ has 2 Be–F bonds; CO₃²⁻ has three C–O connections with the charges stated above; HCOOH has H–C(=O)–O–H connectivity with 2 lone pairs on each O.
5. 5
   Q4.5 → The octet rule is a useful Lewis guideline for attaining a noble-gas valence configuration, but it has exceptions for incomplete octets, odd-electron species, expanded octets and noble-gas compounds and cannot alone predict shape.
6. 6
   Q4.6 → Ionic bond formation is favoured by low metal ionisation enthalpy, a large exothermic electron-gain magnitude for the non-metal and high lattice enthalpy, particularly with small, highly charged ions.
7. 7
   Q4.7 → BeCl₂ is linear, BCl₃ trigonal planar, SiCl₄ tetrahedral, AsF₅ trigonal bipyramidal, H₂S bent and PH₃ trigonal pyramidal.
8. 8
   Q4.8 → Both have steric number 4, but the two lone pairs on oxygen repel more strongly and compress the O–H bonds more, so H₂O has the smaller bond angle.
9. 9
   Q4.9 → A higher dimensionless bond order corresponds to a stronger bond because there is a greater excess of bonding over antibonding electrons.
10. 10
    Q4.10 → Bond length is the equilibrium distance between bonded nuclei, normally reported in pm or Å; 1 Å = 100 pm = 10⁻¹⁰ m.
11. 11
    Q4.11 → CO₃²⁻ is a resonance hybrid of three equivalent contributors, one for each possible C=O position, so all three C–O bonds are identical and have bond order 1⅓.
12. 12
    Q4.12 → No. Structures I and II change the positions or connectivity of atoms, whereas resonance contributors must retain the same nuclear arrangement and differ only in electrons.
13. 13
    Q4.13 → SO₃ and NO₃⁻ each have three equivalent resonance contributors with the double bond on a different O; NO₂ has two equivalent contributors and one unpaired electron.
14. 14
    Q4.14 → The transfers are K⁺ + S⁻ (and K₂S), Ca²⁺ + O²⁻ (CaO), and Al³⁺ + N³⁻ (AlN).
15. 15
    Q4.15 → The two equal C=O dipoles cancel in linear CO₂ (0 D), whereas the two equal O–H dipoles do not cancel in bent H₂O, whose dipole moment is 1.84 D.
16. 16
    Q4.16 → Dipole moment is μ = qr in C·m and is used to test polarity, infer molecular geometry, compare isomers and estimate the ionic character of bonding.
17. 17
    Q4.17 → Electronegativity is a relative bonded-atom ability to attract electron density, whereas electron gain enthalpy is the measurable enthalpy change for adding an electron to a gaseous atom.
18. 18
    Q4.18 → In HCl, χ(Cl) > χ(H), so the shared pair shifts toward chlorine and the bond is polar covalent with Hδ+ and Clδ− ends.
19. 19
    Q4.19 → N₂ < SO₂ < ClF₃ < K₂O < LiF.
20. 20
    Q4.20 → Acetic acid is CH₃–C(=O)–O–H, with 2 lone pairs on each oxygen and zero formal charge on every atom.
21. 21
    Q4.21 → CH₄ has steric number 4 and four sp³ orbitals, so tetrahedral bonding at 109.5° is more stable than a 90° square-planar arrangement.
22. 22
    Q4.22 → Linear BeH₂ has two equal and opposite Be–H bond dipoles, so its net dipole moment is 0 D.
23. 23
    Q4.23 → NH₃ has the higher dipole moment, 1.46 D, compared with 0.24 D for NF₃, because the N–H bond-resultant reinforces the lone-pair contribution whereas the N–F resultant opposes it.
24. 24
    Q4.24 → Hybridisation produces equivalent directed orbitals: 2 linear sp orbitals at 180°, 3 sp² orbitals at 120° and 4 sp³ orbitals at 109.5°.
25. 25
    Q4.25 → Al changes from sp² trigonal planar in AlCl₃ to sp³ tetrahedral in AlCl₄⁻ because coordination raises its steric number from 3 to 4.
26. 26
    Q4.26 → B changes from sp² to sp³ on adduct formation, while N remains sp³ hybridised.
27. 27
    Q4.27 → C₂H₄ forms a C=C double bond from one σ and one π overlap; C₂H₂ forms a C≡C triple bond from one σ and two π overlaps.
28. 28
    Q4.28 → C₂H₂ contains 3 σ and 2 π bonds; C₂H₄ contains 5 σ and 1 π bond.
29. 29
    Q4.29 → The 2pᵧ–2pᵧ pair cannot form a σ bond because overlap perpendicular to the internuclear axis forms a π bond.
30. 30
    Q4.30 → (a) both C sp³; (b) C₁ sp³ and C₂, C₃ sp²; (c) both C sp³; (d) CH₃ carbon sp³ and CHO carbon sp²; (e) CH₃ carbon sp³ and COOH carbon sp².
31. 31
    Q4.31 → C₂H₆ has 7 bond pairs and no carbon lone pairs; H₂O has 2 bond pairs and 2 lone pairs on oxygen.
32. 32
    Q4.32 → σ bonds are head-on, axially symmetric and generally stronger; π bonds are lateral, nodal about the internuclear axis, weaker and rotation-restricting.
33. 33
    Q4.33 → The two H 1s orbitals overlap head-on, opposite-spin electrons pair in the resulting σ molecular orbital and the internuclear attraction produces an H–H bond at its minimum-energy separation.
34. 34
    Q4.34 → Effective LCAO requires similar orbital energies, compatible symmetry and orientation, and substantial overlap; these conditions generate bonding and antibonding molecular orbitals.
35. 35
    Q4.35 → Be₂ has bond order 0 because its bonding and antibonding electrons cancel, so it does not form a stable discrete molecule.
36. 36
    Q4.36 → Stability: O₂⁺ > O₂ > O₂⁻ > O₂²⁻; O₂⁺, O₂ and O₂⁻ are paramagnetic, while O₂²⁻ is diamagnetic.
37. 37
    Q4.37 → The signs denote the algebraic phase of the wavefunction; plus-like overlap is constructive and minus-like overlap is destructive, not positive and negative charge.
38. 38
    Q4.38 → PCl₅ is sp³d hybridised and trigonal bipyramidal; its axial bonds are longer because each undergoes three 90° interactions with equatorial bonds.
39. 39
    Q4.39 → A hydrogen bond is an X–H···Y attraction with N, O or F; it is stronger than van der Waals forces but generally weaker than covalent or ionic bonds.
40. 40
    Q4.40 → The dimensionless bond orders are N₂ = 3, O₂ = 2, O₂⁺ = 2.5 and O₂⁻ = 1.5.

MCQ Questions

1 Mark Questions

1. 1
   State the octet rule. (Ans: atoms tend to acquire eight electrons in their valence shell during bond formation)
2. 2
   What is a polar covalent bond? (Ans: a covalent bond between atoms of different electronegativity sharing electrons unequally, giving partial charges)
3. 3
   Define bond enthalpy. (Ans: the energy needed to break one mole of a particular covalent bond in gaseous state)
4. 4
   What is the shape and bond angle of CH₄? (Ans: tetrahedral, 109.5°)

2 Mark Questions

1. 1
   Draw the Lewis structure of SO₂ and state its shape.
2. 2
   Distinguish between sigma (σ) and pi (π) bonds on two points. (Ans: σ — head-on overlap, single bond, free rotation; π — sidewise overlap of p orbitals, present in multiple bonds, restricts rotation)
3. 3
   Give two examples of molecules that do not obey the octet rule. (Ans: BeCl₂ (4 e⁻ on Be), BF₃ (6 e⁻ on B), SF₆, PCl₅ (expanded octet))

3 Mark Questions

1. 1
   Explain the VSEPR theory with the prediction of the shapes of NH₃ and H₂O from their lone-pair–bond-pair repulsions. (Ans: NH₃ − 1 lone pair, pyramidal; H₂O − 2 lone pairs, bent; bond angles 107° and 104.5°)
2. 2
   Describe the shapes of sp, sp² and sp³ hybrid orbitals and give one molecule for each. (Ans: sp — linear (BeCl₂); sp² — trigonal planar (BF₃); sp³ — tetrahedral (CH₄))
3. 3
   What are the factors that determine the polarity of a molecule? Show with examples how bond polarity and shape combine to give a polar or non-polar molecule. (Ans: electronegativity difference and molecular shape/symmetry; CO₂ non-polar, H₂O polar)

5 Mark Questions

1. 1
   Explain the valence bond theory with reference to the formation of H₂ and Cl₂ molecules, and describe orbital overlap (sigma and pi) for the formation of CH₂=CH₂. (Ans: overlap of half-filled orbitals stabilises the bond; sp² carbons form a σ framework with a p–p π bond)
2. 2
   Describe the molecular orbital theory: LCAO, bonding and antibonding MOs, bond order, and apply it to show that O₂ is paramagnetic (with its MO configuration and bond order = 2).