Chapter 4: Chemical Bonding and Molecular Structure

4.1 Introduction

Atoms combine to achieve a stable noble gas electronic configuration. The attractive force that holds atoms together in a molecule or crystal is called a chemical bond.

Lewis Symbols (Electron Dot Structures): The valence electrons of an atom are represented as dots around its symbol.

4.2 Types of Chemical Bonds

Ionic Bond (Electrovalent Bond)

Formed by the transfer of electrons from a metal to a non-metal.

Example: Formation of NaCl

\[ \text{Na} \rightarrow \text{Na}^+ + e^- \quad (\text{loses 1 electron}) \] \[ \text{Cl} + e^- \rightarrow \text{Cl}^- \quad (\text{gains 1 electron}) \]

Factors favouring ionic bond formation:

• Low ionization enthalpy of the metal
• High electron gain enthalpy of the non-metal
• High lattice enthalpy

Lattice Enthalpy: The energy required to completely separate one mole of an ionic solid into gaseous ions.

\[ \text{NaCl}(s) \rightarrow \text{Na}^+(g) + \text{Cl}^-(g) \quad \Delta_{\text{lattice}}H^\circ = +786;\text{kJ/mol} \]

Covalent Bond

Formed by the sharing of electrons between two atoms (usually non-metals).

Lewis Structures represent bonded atoms showing shared and lone pairs.

Formal Charge

\[ \text{Formal charge} = (\text{Valence electrons}) - (\text{Lone pair electrons}) - \frac{1}{2}(\text{Bonded electrons}) \]

Bond Parameters

\[ \text{Bond order} = \frac{1}{2}(N_b - N_a) \]

where \(N_b\) = number of bonding electrons, \(N_a\) = number of antibonding electrons (in MO theory).

4.3 The Octet Rule and Its Limitations

Octet Rule: Atoms tend to achieve 8 electrons in their valence shell (or 2 for H and He) — a duet.

Limitations:

1. Incomplete octet: BF₃ (B has 6 electrons), BeCl₂ (Be has 4)
2. Expanded octet: PCl₅ (P has 10 electrons), SF₆ (S has 12), possible for elements in Period 3+ due to available d-orbitals
3. Odd-electron molecules: NO, NO₂ have an odd number of electrons

4.4 Polar and Nonpolar Covalent Bonds

When bonded atoms have different electronegativities, the shared pair is displaced towards the more electronegative atom, creating a polar covalent bond.

Dipole moment (\(\mu\)):

\[ \mu = q \times d \]

where \(q\) = magnitude of charge, \(d\) = distance of separation. Unit: Debye (D), \(1;\text{D} = 3.336 \times 10^{-30};\text{C·m}\)

• \(\mu = 0\) for symmetrical molecules (e.g., CO₂, CCl₄, BF₃)
• \(\mu \neq 0\) for asymmetrical molecules (e.g., H₂O, NH₃, CHCl₃)

4.5 Fajan’s Rules — Covalent Character of Ionic Bonds

Ionic bonds develop covalent character when:

1. Small cation with high charge (high polarising power)
2. Large anion with high charge (high polarisability)

4.6 VSEPR Theory (Valence Shell Electron Pair Repulsion)

The shape of a molecule is determined by the repulsion between electron pairs (bonding and lone pairs) in the valence shell of the central atom.

Order of repulsion: lp–lp > lp–bp > bp–bp

4.7 Valence Bond Theory (VBT)

• A covalent bond is formed when half-filled orbitals of two atoms overlap.
• The greater the overlap → stronger the bond.
• Types of overlap:
  • Sigma (σ) bond: Head-on overlap (s-s, s-p, p-p along bond axis)
  • Pi (π) bond: Lateral/sideways overlap (p-p perpendicular to bond axis)

A single bond consists of one σ bond. A double bond = 1σ + 1π. A triple bond = 1σ + 2π.

4.8 Hybridization

Hybridization is the intermixing of atomic orbitals of similar energies to form new, equivalent hybrid orbitals.

Important: Lone pairs also occupy hybrid orbitals:

• NH₃: sp³ hybridized, 3 bp + 1 lp → trigonal pyramidal
• H₂O: sp³ hybridized, 2 bp + 2 lp → bent

4.9 Molecular Orbital Theory (MOT)

Key Concepts

1. Atomic orbitals of comparable energy combine to form molecular orbitals (MOs)
2. Number of MOs formed = Number of atomic orbitals combined
3. Bonding MOs (lower energy) and Antibonding MOs (higher energy)
4. Filling follows Aufbau, Pauli, and Hund’s rules

MO Energy Level Diagram for Homonuclear Diatomic Molecules

For O₂ and F₂ (\(Z \geq 8\)):

\[ \sigma 1s < \sigma^* 1s < \sigma 2s < \sigma^* 2s < \sigma 2p_z < \pi 2p_x = \pi 2p_y < \pi^* 2p_x = \pi^* 2p_y < \sigma^* 2p_z \]

For Li₂ to N₂ (\(Z < 8\)):

\[ \sigma 1s < \sigma^* 1s < \sigma 2s < \sigma^* 2s < \pi 2p_x = \pi 2p_y < \sigma 2p_z < \pi^* 2p_x = \pi^* 2p_y < \sigma^* 2p_z \]

Bond Order:

\[ \text{Bond order} = \frac{N_b - N_a}{2} \]

4.10 Resonance

When a single Lewis structure cannot explain all properties of a molecule, we write resonance structures (canonical forms). The actual structure is a resonance hybrid.

Example: Carbonate ion \(\text{CO}_3^{2-}\) has three equivalent resonance structures. All C–O bond lengths are equal (127 pm), intermediate between C=O (122 pm) and C–O (143 pm).

4.11 Hydrogen Bond

A type of electrostatic attraction between a hydrogen atom bonded to a highly electronegative atom (F, O, N) and a lone pair on another electronegative atom.

Types:

• Intermolecular: Between different molecules (e.g., H₂O, HF)
• Intramolecular: Within the same molecule (e.g., o-nitrophenol)

Consequences:

• High boiling point of H₂O, NH₃, HF compared to hydrides of their group members
• Ice is less dense than liquid water (hydrogen bonding creates an open structure)

Practice Questions

Multiple Choice Questions (MCQs)

1. The correct order of bond angles is:

 (a) \(\text{H}_2\text{O} \gt \text{NH}_3 \gt \text{CH}_4\)

 (b) \(\text{CH}_4 \gt \text{NH}_3 \gt \text{H}_2\text{O}\)

 (c) \(\text{NH}_3 \gt \text{H}_2\text{O} \gt \text{CH}_4\)

 (d) \(\text{CH}_4 \gt \text{H}_2\text{O} \gt \text{NH}_3\)

2. The hybridization of Xe in XeF₄ is:

 (a) sp³

 (b) sp³d

 (c) sp³d²

 (d) dsp²

3. O₂ is paramagnetic because it has:

 (a) unpaired electrons in bonding MOs

 (b) unpaired electrons in antibonding MOs

 (c) no unpaired electrons

 (d) more antibonding electrons than bonding electrons

4. The bond order in NO⁺ is:

 (a) 2

 (b) 2.5

 (c) 3

 (d) 1.5

5. Which molecule has the shortest bond length?

 (a) O₂

 (b) O₂⁻

 (c) O₂²⁻

 (d) O₂⁺

Short Answer Questions (2–3 Marks)

6. Using VSEPR theory, predict the shapes of BrF₅ and XeOF₂.

7. Draw the resonance structures of ozone (O₃). What is the O–O bond order in ozone?

8. Explain why the bond angle in H₂O (104.5°) is less than in NH₃ (107°).

9. Define hydrogen bonding. Why is the boiling point of HF much higher than that of HCl?

10. Use MO theory to explain why He₂ does not exist but He₂⁺ does.

Long Answer Questions (5 Marks)

11. (a) What is hybridization? Explain sp, sp², and sp³ hybridization with examples.

 (b) What is the hybridization of each carbon atom in CH₂=C=CH₂?

12. (a) Draw the MO energy level diagram for N₂. Calculate its bond order and predict its magnetic behaviour.

 (b) Compare the bond order and stability of N₂ and N₂⁺.

13. (a) Predict the shape and hybridization of the following: SF₆, ClF₃, ICl₂⁻.

 (b) Explain why PCl₅ exists but NCl₅ does not.

Assertion-Reason Questions

14. Assertion (A): The bond angle in PH₃ is less than in NH₃.

Reason (R): P is less electronegative than N.

15. Assertion (A): CO₂ has zero dipole moment though it has two polar C=O bonds.

Reason (R): CO₂ is a linear molecule and the bond dipoles cancel each other.

Answer Key