NCERT Solutions for Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure
NCERT Solutions for Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure covers octet rule and Kössel-Lewis approach, ionic and covalent bonds, Lewis symbols, formal charge, dipole moment, Resonance, VSEPR Theory, Valence Bond Theory (VBT), orbital overlap, Hybridisation (sp, sp2, sp3, sp3d, sp3d2), Molecular Orbital Theory (MOT), Bond Order, magnetic properties, and Hydrogen Bonding.
Overview of Topics & Subtopics (Click to Jump)
1. Kössel-Lewis Approach, Octet Rule, Formal Charge & Bond Parameters
Short Note:
• Chemical Bond: Attractive force holding constituent atoms/ions together in different chemical species.
• Octet Rule: Atoms gain, lose, or share electrons to attain 8 electrons in their valence shell (noble gas configuration). Limitations: Incomplete octet (e.g., LiCl, BeH2, BCl3), odd-electron molecules (NO, NO2), and expanded octet (PF5, SF6, H2SO4).
• Formal Charge (F.C.): F.C. = [Total valence e- in free atom] - [Total non-bonding (lone pair) e-] - 1/2 [Total bonding (shared) e-].
• Dipole Moment (μ): Product of magnitude of charge (Q) and distance of separation (r): μ = Q × r (expressed in Debye, D). Vector quantity, indicates covalent character polarity.
2. VSEPR Theory & Prediction of Molecular Geometries
Short Note:
• VSEPR Theory (Valence Shell Electron Pair Repulsion): Shape of a molecule depends upon the number of valence shell electron pairs (bonded and non-bonded) around the central atom.
• Order of Repulsive Interactions: Lone Pair - Lone Pair (lp - lp) > Lone Pair - Bond Pair (lp - bp) > Bond Pair - Bond Pair (bp - bp).
• Common Geometries:
- 2 bp, 0 lp: Linear (e.g., BeCl2, CO2, 180°)
- 3 bp, 0 lp: Trigonal Planar (e.g., BF3, 120°)
- 2 bp, 1 lp: Bent / V-shaped (e.g., SO2, <120°)
- 4 bp, 0 lp: Tetrahedral (e.g., CH4, 109.5°)
- 3 bp, 1 lp: Trigonal Pyramidal (e.g., NH3, 107°)
- 2 bp, 2 lp: Bent / V-shaped (e.g., H2O, 104.5°)
- 5 bp, 0 lp: Trigonal Bipyramidal (e.g., PF5)
- 6 bp, 0 lp: Octahedral (e.g., SF6)
3. Valence Bond Theory (VBT) & Concept of Hybridisation
Short Note:
• Valence Bond Theory: Chemical bond forms by pairing of electrons with opposite spins through orbital overlap along internuclear axis.
- Sigma (σ) Bond: Head-on (axial) overlap along internuclear axis. Stronger bond.
- Pi (π) Bond: Lateral / sideways overlap perpendicular to internuclear axis. Weaker bond.
• Hybridisation: Intermixing of atomic orbitals of slightly different energies to produce a set of equivalent hybridized orbitals equal in number to the mixed atomic orbitals.
- sp: Linear (180°)
- sp2: Trigonal Planar (120°)
- sp3: Tetrahedral (109.5°)
- sp3d: Trigonal Bipyramidal (axial bonds longer than equatorial)
- sp3d2: Octahedral
4. Molecular Orbital Theory (MOT) & Hydrogen Bonding
Short Note:
• Molecular Orbital Theory (MOT): Linear Combination of Atomic Orbitals (LCAO) forms Bonding Molecular Orbitals (BMO, lower energy) and Antibonding Molecular Orbitals (ABMO, higher energy).
• Bond Order (B.O.): B.O. = 1/2 (Nb - Na). Higher B.O. means greater bond stability and bond energy, and shorter bond length.
- If B.O. > 0, molecule is stable. If B.O. = 0, molecule does not exist (e.g., He2).
- Magnetic Behaviour: Unpaired electrons = Paramagnetic; All electrons paired = Diamagnetic.
• Hydrogen Bonding: Electrostatic force of attraction between a hydrogen atom covalently bonded to a highly electronegative atom (F, O, N) and another electronegative atom.
- Intermolecular H-bond: Between different molecules (e.g., H2O, HF, NH3) - increases boiling point.
- Intramolecular H-bond: Within the same molecule (e.g., o-nitrophenol).
Intext & Conceptual Questions
(i) BeCl2: Central atom Be has 2 valence electrons, forms 2 bond pairs with Cl and has 0 lone pairs. Geometry: Linear (180°).
(ii) BCl3: Central atom B has 3 valence electrons, forms 3 bond pairs with Cl and has 0 lone pairs. Geometry: Trigonal Planar (120°).
(iii) SiCl4: Central atom Si has 4 valence electrons, forms 4 bond pairs with Cl and has 0 lone pairs. Geometry: Tetrahedral (109.5°).
(iv) AsF5: Central atom As has 5 valence electrons, forms 5 bond pairs with F and has 0 lone pairs. Geometry: Trigonal Bipyramidal.
(v) H2S: Central atom S has 6 valence electrons, forms 2 bond pairs with H and has 2 lone pairs. Geometry: Bent / V-shaped (approx 92.2° due to lone pair repulsion).
Textbook Chapter End Exercise Questions
• Al (Aluminum, Z=13): Valence electrons = 3. Lewis structure: •Al: (or three dots around Al).
• Al3+: Loses 3 electrons. Symbol: [Al]3+ (no dots).
• O (Oxygen, Z=8): Valence electrons = 6. Lewis structure: :O: with 2 lone pairs and 2 single dots.
• O2-: Gains 2 electrons, completing octet. Symbol: [:O:]2- (8 dots inside brackets).
• N (Nitrogen, Z=7): Valence electrons = 5. Lewis structure: :N: with 1 lone pair and 3 single dots.
• N3-: Gains 3 electrons, completing octet. Symbol: [:N:]3- (8 dots inside brackets).
Total electrons in O2 = 16.
MO Electronic Configuration for O2 (16 e-):
σ1s2 σ*1s2 σ2s2 σ*2s2 σ2pz2 (π2px2 = π2py2) (π*2px1 = π*2py1)
Calculation of Bond Orders:
• O2 (16 e-): Nb = 10, Na = 6 → Bond Order = 1/2 (10 - 6) = 2.0
• O2+ (15 e-): Remove 1 e- from ABMO → Nb = 10, Na = 5 → Bond Order = 1/2 (10 - 5) = 2.5
• O2- (17 e-): Add 1 e- to ABMO → Nb = 10, Na = 7 → Bond Order = 1/2 (10 - 7) = 1.5
• O22- (18 e-): Add 2 e- to ABMO → Nb = 10, Na = 8 → Bond Order = 1/2 (10 - 8) = 1.0
Order of Stability:
Stability is directly proportional to Bond Order:
O2+ (2.5) > O2 (2.0) > O2- (1.5) > O22- (1.0)
Read More & Chapter Resources
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