Chemical Bonding & Molecular Structure
Class 11 CBSE | ISC | WBCHSE – Detailed Notes, Modern Theories, MCQs with Answers
1. Introduction
Chemical Bond: The attractive force that binds atoms together in a molecule or compound, responsible for the structure and properties of matter.
Chemistry’s central theme: Why and how atoms link to form molecules; what determines their shapes, strengths, and behavior.
2. Types of Chemical Bonds
(a) Ionic (Electrovalent) Bond
- Electrons transferred from metal to non-metal. Results in cations/anions.
- Example: NaCl (Na → Na⁺ + e⁻; Cl + e⁻ → Cl⁻)
Ionic solids: High melting point, brittle, conduct electricity in molten/solution.
(b) Covalent Bond
- Shared pair(s) of electrons between two non-metals.
- Examples: H₂, O₂, N₂
Cl₂: Cl· + ·Cl → Cl:Cl
Covalent compounds: Soft, low melting, poor conductors.
(c) Coordinate (Dative) Bond
- Both shared electrons from one atom (donor).
- Examples: [NH₄]⁺, SO₂, O₃
3. Octet Rule and Its Limitations
- Stable configuration: 8 electrons in outer shell (2 for H, He).
- Limitations: Odd-electron species (NO); Expanded octet (PF₅); Incomplete octet (BF₃).
[Lewis structures: BF₃, PF₅, NO]
4. Lewis Structures and Resonance
- Lewis structure: Shows valence electrons as dots/lines, lone pairs/double bonds.
- Formal charge: Used to predict the most stable structure.
- Resonance: When multiple valid Lewis structures, actual molecule is a hybrid.
[CO₃²⁻ resonance structures]
5. Theories of Chemical Bonding
5.1 Valence Bond Theory (VBT)
- Atomic orbitals overlap; electrons shared/pairs localized between two atoms.
- Sigma (σ) bond: Head-on overlap; Pi (π) bond: Sidewise overlap.
[Sigma and Pi bond formation]
5.2 Hybridization
- Mixing of atomic orbitals to create new, equal-energy hybrid orbitals.
- sp = linear, sp² = trigonal planar, sp³ = tetrahedral…
Examples: CH₄ – sp³; BF₃ – sp²; BeCl₂ – sp
5.3 Molecular Orbital Theory (MOT)
MOT: Developed by Hund & Mulliken – explains the formation of molecular orbitals (MOs) spanning entire molecule.
- Atomic orbitals combine to form MOs: same number as original orbitals.
- Bonding (lower energy, more stability, denoted σ, π) and antibonding (higher energy, less stable, σ*, π*) orbitals.
- Electrons fill MOs in increasing energy (Aufbau) with 2 electrons per orbital.
Bond Order (BO) = ½ × [No. in bonding MOs – No. in antibonding MOs]
Order of energy for diatomics Z ≤ 7: σ1s < σ*1s < σ2s < σ*2s < (π2px=π2py) < σ2pz …
For O₂, F₂: σ2pz below π2p MOs.
For O₂, F₂: σ2pz below π2p MOs.
[Molecular Orbital diagram for O₂, N₂]
- O₂: 2 unpaired e⁻ in π* MOs ⇒ paramagnetic (verified by experiment).
- N₂: BO = 3 ⇒ strong triple bond, diamagnetic.
- He₂: BO = 0 ⇒ molecule doesn’t exist.
Bond Order examples:
O₂ (8 in bonding, 4 in antibonding) → BO = 2.
O₂⁻: BO = 1.5 (one extra e⁻ in antibonding).
[MO electron population for O₂, O₂⁺, O₂⁻]
6. VSEPR Theory (Valence Shell Electron Pair Repulsion)
Explains molecular shapes based on minimizing electron pair repulsions in the valence shell of the central atom.
- Both lone and bond pairs repel; Lone pair-lone pair > lone pair-bond pair > bond pair-bond pair.
- Molecule adopts shape that minimizes these repulsions.
Steps in VSEPR Application:
- Write Lewis structure and count total electron pairs (lone + bond pairs) around central atom (use AXmEn notation).
- Predict basic geometry (electron pair geometry): Linear, trigonal planar, tetrahedral, trigonal bipyramidal, octahedral, etc.
- Account for lone pairs to get actual shape (molecular geometry).
| AXmEn | Total electron pairs | Example | Electron Geometry | Molecular Shape | Bond Angle (°) |
|---|---|---|---|---|---|
| AX₂ | 2 | BeCl₂, CO₂ | Linear | Linear | 180 |
| AX₃ | 3 | BF₃, SO₃ | Trigonal planar | Trigonal planar | 120 |
| AX₂E | 3 | SO₂ | Trigonal planar | Bent | ~118 |
| AX₄ | 4 | CH₄ | Tetrahedral | Tetrahedral | 109.5 |
| AX₃E | 4 | NH₃ | Tetrahedral | Trigonal pyramidal | 107 |
| AX₂E₂ | 4 | H₂O | Tetrahedral | Bent | 104.5 |
| AX₅ | 5 | PCl₅ | Trigonal bipyramidal | Trigonal bipyramidal | 90, 120 |
| AX₄E | 5 | SF₄ | Trigonal bipyramidal | Seesaw | 101-119 |
| AX₃E₂ | 5 | ClF₃ | Trigonal bipyramidal | T-shaped | 87-102 |
| AX₂E₃ | 5 | XeF₂ | Trigonal bipyramidal | Linear | 180 |
Key Points:
– Lone pairs compress bond angles.
– The shape depends only on positions of atoms, not lone pairs.
– Results in a rich variety of molecular shapes (trigonal planar, tetrahedral, pyramidal, bent, etc).
[VSEPR geometries: CH₄, NH₃, H₂O, BF₃, XeF₂, SF₄, ClF₃]
7. Formal Charge
Helps assign the most likely arrangement of electrons in Lewis structures; indicates the “electron bookkeeping” on each atom.
Formal charge = [Valence electrons] – [Nonbonding electrons] – ½ [Bonding electrons]
Calculation Steps:
- Write the Lewis structure of the molecule/ion.
- Count valence electrons of atom in free state.
- Count non-bonding electrons (lone pairs).
- Assign half of bonding electrons (shared pairs) to the atom.
Example: Ozone (O₃) – For central O atom:
- Valence electrons: 6
- Lone pair electrons: 2
- Bonding electrons: 3 bonds × 2 = 6
- Formal charge = 6 – 2 – ½×6 = 6 – 2 – 3 = +1
Importance:
– The structure with the lowest (closest to zero) formal charges and with negative charge on the most electronegative atom is favored.
– Useful in choosing most appropriate resonance structure.
– The structure with the lowest (closest to zero) formal charges and with negative charge on the most electronegative atom is favored.
– Useful in choosing most appropriate resonance structure.
8. Lattice Enthalpy
Lattice enthalpy (∆Hlattice): The energy released when 1 mole of an ionic crystalline compound is formed from its gaseous ions, or the energy required to separate 1 mole of solid into gaseous ions.
- Exothermic (formation): M⁺(g) + X⁻(g) → MX(s) + ∆Hlattice
- Endothermic (dissociation): MX(s) → M⁺(g) + X⁻(g)
- Larger lattice energy = higher melting point, less solubility in non-polar solvents, greater bond strength
- Increases with: Smaller ionic radii, higher ionic charge.
Trends:
– Lattice enthalpy (NaF) > (NaCl) > (NaBr) > (NaI) – as anion size increases, lattice energy decreases.
– MgO has greater lattice energy than NaCl (higher charge, smaller ions).
– Lattice enthalpy (NaF) > (NaCl) > (NaBr) > (NaI) – as anion size increases, lattice energy decreases.
– MgO has greater lattice energy than NaCl (higher charge, smaller ions).
9. Born-Haber Cycle
A thermochemical cycle that analyzes the steps of formation of an ionic compound from its elements, allowing the calculation of lattice enthalpy via Hess’s Law.
Key Steps in NaCl Formation:
- Sublimation of Na(s) → Na(g): ∆Hsubl (endothermic)
- Ionization: Na(g) → Na⁺(g) + e⁻: IE (endothermic)
- Dissociation of Cl₂(g): ½Cl₂(g) → Cl(g): ½ bond energy (endothermic)
- Electron gain: Cl(g) + e⁻ → Cl⁻(g): EA (exothermic)
- Formation of NaCl(s): Na⁺(g) + Cl⁻(g) → NaCl(s): ∆Hlattice (exothermic)
∆Hf (formation) = ∆Hsubl + IE + ½ × bond energy + EA + ∆Hlattice
Example for NaCl:
- ∆Hf (NaCl) = ∆Hsubl(Na) + IE₁(Na) + ½ D(Cl₂) + EA(Cl) + ∆Hlattice
- Allows indirect measurement of lattice energy, as it cannot be measured directly.
[Born-Haber cycle step diagram for NaCl formation]
Summary: The Born-Haber cycle demonstrates how the energetics of all component steps combine to yield the overall enthalpy of formation, critical for understanding stability of ionic solids.
10. Practice MCQs (Answer & Explanation Below)
| MCQ | Options |
|---|---|
| 1. Which of the following forms an ionic bond? | a) H₂ b) O₂ c) NaCl d) Cl₂ |
| 2. The number of lone pairs in NH₃ is: | a) 0 b) 1 c) 2 d) 3 |
| 3. Shape of BF₃ molecule is: | a) Tetrahedral b) Trigonal planar c) Bent d) Linear |
| 4. Which is a polar molecule? | a) CO₂ b) CH₄ c) H₂ d) H₂O |
| 5. The bond order of O₂ molecule is: | a) 1 b) 2 c) 3 d) 4 |
| 6. Bond angle in water (H₂O) is: | a) 90° b) 104.5° c) 120° d) 180° |
| 7. Which has a coordinate bond? | a) H₂ b) NH₄⁺ c) O₂ d) H₂O |
| 8. Which molecule has an expanded octet? | a) BH₃ b) PF₅ c) CO₂ d) CH₄ |
| 9. Valence electrons in BeCl₂? | a) 4 b) 6 c) 8 d) 10 |
| 10. Resonance occurs in: | a) CH₄ b) SO₃ c) NH₃ d) BeCl₂ |
| 11. Which is not explained by VSEPR? | a) Lone pairs effect b) d-p π bonding c) Hybridization d) Bond length |
| 12. Which is paramagnetic? | a) N₂ b) O₂ c) CO₂ d) HCl |
| 13. Molecule with highest bond order: | a) O₂ b) N₂ c) F₂ d) H₂ |
| 14. Which hybridization leads to linear shape? | a) sp b) sp² c) sp³ d) d²sp³ |
| 15. Bond angle in CH₄ is: | a) 90° b) 104.5° c) 120° d) 109.5° |
| 16. The term ‘Lewis acid’ refers to: | a) Proton donor b) Electron donor c) Electron pair acceptor d) None |
| 17. Example of molecule with bent shape: | a) CO₂ b) H₂O c) BeF₂ d) BF₃ |
| 18. Which does not obey the octet rule? | a) CH₄ b) NH₃ c) BeCl₂ d) H₂O |
| 19. The correct electron-dot structure for NO₃⁻: | a) Single N−O bond only b) Double bonds only c) Resonance hybrid d) None |
| 20. Bond order for C≡N in HCN: | a) 1 b) 2 c) 3 d) 4 |
| 21. Presence of two unpaired electrons in O₂ is explained by: | a) Lewis theory b) VBT c) MOT d) VSEPR |
| 22. Which has maximum number of lone pairs on central atom? | a) NH₃ b) H₂O c) BeCl₂ d) PF₅ |
| 23. CO₂ is nonpolar because: | a) C has no electronegativity b) Shape is linear c) Bonds not polar d) None |
| 24. A triple bond consists of: | a) 3 σ bond b) 2 σ + 1 π c) 1 σ + 2 π d) 3 π |
| 25. The formal charge on O in O₃ (central O): | a) +1 b) 0 c) -1 d) +2 |
| 26. The maximum covalency of nitrogen is: | a) 3 b) 4 c) 5 d) 6 |
| 27. The energy required to break a bond is: | a) Bond energy b) Ionization enthalpy c) Electron affinity d) Lattice energy |
| 28. Molecule with one lone pair and three bond pairs: | a) NH₃ b) CH₄ c) H₂O d) BF₃ |
| 29. Which is a planar molecule? | a) NH₃ b) CH₄ c) BeCl₂ d) H₂O |
| 30. Back bonding is seen in: | a) C₂H₄ b) BF₃ c) H₂O d) NH₃ |
Answers & Explanations to MCQs
- c) NaCl – Ionic bond; electrons transfer from Na to Cl.
- b) 1 – One lone pair on N in NH₃.
- b) Trigonal planar – BF₃ is AX₃, sp² hybridized.
- d) H₂O – Bent shape, net dipole.
- b) 2 – O₂, by MOT calculations.
- b) 104.5°
- b) NH₄⁺ – N→H⁺ forms coordinate bond.
- b) PF₅ – P has 10 outer electrons (expanded octet).
- a) 4 – Be(2) + Cl(1 each).
- b) SO₃ – Has resonance.
- b) d-p π bonding – Not a focus of VSEPR.
- b) O₂ – Two unpaired e⁻ in MO diagram.
- b) N₂ – BO=3 (triple bond).
- a) sp – Linear, 180°.
- d) 109.5°
- c) Electron pair acceptor – Lewis acid.
- b) H₂O – Two lone pairs, bent.
- c) BeCl₂ – Be has incomplete octet.
- c) Resonance hybrid – NO₃⁻.
- c) 3 – C≡N is triple bond.
- c) MOT – O₂ paramagnetism explained by MOT.
- b) H₂O – Two lone pairs.
- b) Shape is linear – CO₂, dipoles cancel.
- c) 1 σ + 2 π – Triple bond.
- a) +1 – Central oxygen in O₃.
- c) 5 – Nitrogen can form up to five bonds.
- a) Bond energy
- a) NH₃ – 3 bond pairs, 1 lone pair.
- c) BeCl₂ – Linear planar molecule.
- b) BF₃ – Exhibits back bonding.