3.6 Valence Bond Theory (VBT)
Introduction to VBT
Valence Bond Theory explains covalent bond formation through orbital overlap. It considers a molecule as a combination of atoms with overlapping atomic orbitals.
- A covalent bond is formed when half-filled orbitals in valence shells of two atoms with similar energy overlap.
- Greater overlap results in a stronger bond.
- Covalent bonds are directional (determined by shape and mode of overlapping orbitals).
Memorization Tip
VBT = Valence Bond Theory = Overlap Creates Bonds
Remember: “Overlap determines bond strength and direction.”
3.6.1 Formation of Sigma (σ) Bond
A sigma bond is formed by linear (head-on) overlap of two half-filled atomic orbitals on adjacent atoms.
Two s orbitals overlap linearly to form H₂ molecule
s orbital of H overlaps with p orbital of Cl
Two p orbitals overlap linearly to form Cl₂
- Formed by head-on (end-to-end) overlap
- Electron density is concentrated along the internuclear axis
- Strongest type of covalent bond
- Free rotation around sigma bond is possible
Fig 3.8: Overlap between s orbitals of two hydrogen atoms to form the H₂ molecule
Formation of Pi (π) Bond
A pi bond is formed by sideways (parallel) overlap of two half-filled p-orbitals.
Example: O₂ molecule formation
Oxygen atoms have two unpaired electrons in perpendicular p-orbitals (p_x and p_y):
- p_x orbitals overlap end-to-end → σ bond
- p_y orbitals overlap sideways → π bond
Key Difference: σ vs π Bonds
σ Bond: Head-on overlap, stronger, electron density along axis
π Bond: Sideways overlap, weaker, electron density above/below axis
Fig 3.12: Formation of double bond (σ and π) between oxygen atoms
π bonds have electron density above and below the internuclear axis (not between nuclei), making them more diffuse and weaker than σ bonds.
Solved MCQs – VBT
3.7 Atomic Orbital Hybridization
Introduction to Hybridization
Hybridization is the process of mixing atomic orbitals of slightly different energies and shapes to form a new set of equivalent orbitals with same energy and shape.
- Explains molecular geometries that can’t be explained by simple orbital overlap
- Accounts for equivalent bond lengths and angles in molecules
- Explains bonding in molecules like CH₄ (tetrahedral), BF₃ (trigonal planar), etc.
3.7.1 Types of Hybridization
| Hybridization | Orbitals Mixed | Geometry | Bond Angle | Examples |
|---|---|---|---|---|
| sp | 1s + 1p | Linear | 180° | BeCl₂, CO₂, C₂H₂ |
| sp² | 1s + 2p | Trigonal planar | 120° | BF₃, C₂H₄, SO₃ |
| sp³ | 1s + 3p | Tetrahedral | 109.5° | CH₄, NH₃, H₂O |
| sp³d | 1s + 3p + 1d | Trigonal bipyramidal | 90°, 120° | PCl₅, SF₄ |
| sp³d² | 1s + 3p + 2d | Octahedral | 90° | SF₆, [AlF₆]³⁻ |
sp Hybridization (Linear Geometry)
In sp hybridization, one s orbital and one p orbital mix to form two equivalent sp hybrid orbitals oriented at 180°.
Be (4): 1s² 2s² → excited state: 1s² 2s¹ 2p¹ → hybridized: 1s² (sp)¹ (sp)¹
Two sp hybrid orbitals of Be overlap with p orbitals of two Cl atoms → Linear molecule (180°)
Fig 3.13: Mixing of s and p orbitals to give two hybrid sp orbitals and formation of BeCl₂
sp² Hybridization (Trigonal Planar Geometry)
In sp² hybridization, one s orbital and two p orbitals mix to form three equivalent sp² hybrid orbitals oriented at 120°.
B (5): 1s² 2s² 2p¹ → excited state: 1s² 2s¹ 2p¹ₓ 2p¹ᵧ → hybridized: 1s² (sp²)¹ (sp²)¹ (sp²)¹
Three sp² hybrid orbitals of B overlap with p orbitals of three F atoms → Trigonal planar (120°)
Fig 3.14: Formation of sp² hybrid orbitals from atomic orbitals
sp³ Hybridization (Tetrahedral Geometry)
In sp³ hybridization, one s orbital and three p orbitals mix to form four equivalent sp³ hybrid orbitals oriented at 109.5° (tetrahedral).
C (6): 1s² 2s² 2p² → excited state: 1s² 2s¹ 2p¹ₓ 2p¹ᵧ 2p¹₂ → hybridized: 1s² (sp³)¹ (sp³)¹ (sp³)¹ (sp³)¹
Four sp³ hybrid orbitals of C overlap with s orbitals of four H atoms → Tetrahedral (109.5°)
Fig 3.16: Formation of sp³ hybrid orbitals of C in methane
Hybridization Identification Trick
Count σ bonds + lone pairs on central atom:
- 2 = sp (linear, 180°)
- 3 = sp² (trigonal planar, 120°)
- 4 = sp³ (tetrahedral, 109.5°)
- 5 = sp³d (trigonal bipyramidal)
- 6 = sp³d² (octahedral)
Solved MCQs – Hybridization
3.8 Valence Shell Electron Pair Repulsion Model (VSEPR)
Introduction to VSEPR Model
The VSEPR model predicts molecular shapes based on electron pair repulsions around central atoms. Developed by Sidgwick and Powell (1940).
- Both lone pairs and bond pairs participate in determining geometry
- Electron pairs arrange to maximize distance (minimize repulsion)
- Lone pairs occupy more space than bond pairs
- Repulsion order: lp-lp > lp-bp > bp-bp
- Multiple bonds treated as single electron pair for geometry
VSEPR Memory Aid
“Lone Pairs Push More”
Remember: Lone pair-lone pair repulsion > Lone pair-bond pair > Bond pair-bond pair
VSEPR Shapes and Geometries
| Steric Number | Bond Pairs | Lone Pairs | Electron Geometry | Molecular Shape | Bond Angle | Examples |
|---|---|---|---|---|---|---|
| 2 | 2 | 0 | Linear | Linear | 180° | BeCl₂, CO₂ |
| 1 | 1 | Linear | Linear | 180° | N₂O, NO⁺ | |
| 0 | 2 | Linear | Linear | 180° | XeF₂, I₃⁻ | |
| 3 | 3 | 0 | Trigonal Planar | Trigonal Planar | 120° | BF₃, SO₃ |
| 2 | 1 | Trigonal Planar | Bent/V-shaped | <120° | SO₂, SnCl₂ | |
| 4 | 4 | 0 | Tetrahedral | Tetrahedral | 109.5° | CH₄, CCl₄ |
| 3 | 1 | Tetrahedral | Trigonal Pyramidal | <109.5° | NH₃, H₃O⁺ | |
| 2 | 2 | Tetrahedral | Bent/V-shaped | <109.5° | H₂O, OF₂ | |
| 5 | 5 | 0 | Trigonal Bipyramidal | Trigonal Bipyramidal | 90°, 120° | PCl₅, PF₅ |
| 6 | 6 | 0 | Octahedral | Octahedral | 90° | SF₆, [AlF₆]³⁻ |
AB₂ Type Molecules (Linear Geometry)
Example: BeCl₂, CO₂
Two electron pairs around central atom arrange at 180° to minimize repulsion.
Fig 3.17: Linear shape of the BeCl₂ molecule
AB₃ Type Molecules (Trigonal Planar)
Example: BF₃, SO₃
Three electron pairs arrange at 120° (trigonal planar).
Fig 3.18: BF₃ is an AB₃ system having triangular planar geometry
With 2 bond pairs + 1 lone pair → Bent/V-shaped structure (<120°)
AB₄ Type Molecules (Tetrahedral)
Example: CH₄ (Tetrahedral)
Four electron pairs arrange at 109.5° (tetrahedral).
Fig 3.22: Distorted geometry and pyramidal shape of the NH₃ molecule
- CH₄: 4 bond pairs, 0 lone pairs → Perfect tetrahedron (109.5°)
- NH₃: 3 bond pairs, 1 lone pair → Trigonal pyramidal (107°)
- H₂O: 2 bond pairs, 2 lone pairs → Bent/V-shaped (104.5°)
Why NF₃ (102°) has smaller bond angle than NH₃ (107°)?
F is more electronegative than H, so N-F bond pairs are pulled toward F, reducing repulsion between bond pairs. This allows lone pair to compress bond angle more.
AB₅ and AB₆ Type Molecules
AB₅ Example: PCl₅ (Trigonal Bipyramidal)
Five electron pairs: 3 equatorial (120°), 2 axial (90° to equatorial).
AB₆ Example: SF₆ (Octahedral)
Six electron pairs at 90° angles.
Trigonal bipyramidal
Octahedral
Square planar (AB₆ with 2 lone pairs)
Fig 3.27: Lewis structure and octahedral shape of SF₆ with each angle of 90°
Solved MCQs – VSEPR
3.9 Molecular Orbital Theory (MOT)
Introduction to MOT
Molecular Orbital Theory explains bonding through formation of molecular orbitals that belong to the entire molecule (not individual atoms).
- Atomic orbitals combine to form molecular orbitals (MOs)
- Number of MOs formed = Number of AOs combined
- Two types of MOs: Bonding (lower energy) and Antibonding (higher energy)
- Electrons fill MOs following Aufbau principle, Pauli exclusion, Hund’s rule
- Bond order = ½ (No. of bonding electrons – No. of antibonding electrons)
MOT vs VBT
VBT: Localized electrons, bonds between specific atoms
MOT: Delocalized electrons, molecular orbitals cover entire molecule
Formation of Molecular Orbitals
Bonding Molecular Orbitals (σ, π): Formed by in-phase overlap (+ with + or – with -), lower energy than parent AOs.
Antibonding Molecular Orbitals (σ*, π*): Formed by out-of-phase overlap (+ with -), higher energy than parent AOs.
Fig 3.30: Formation of bonding and antibonding orbitals for the H₂ molecule
MO Diagrams of Diatomic Molecules
| Molecule | Electron Configuration | Bond Order | Magnetic Property | Stability |
|---|---|---|---|---|
| H₂ | (σ1s)² | 1 | Diamagnetic | Stable |
| He₂ | (σ1s)²(σ*1s)² | 0 | Diamagnetic | Not stable |
| N₂ | KK(σ2s)²(σ*2s)²(π2p)⁴(σ2p)² | 3 | Diamagnetic | Very stable |
| O₂ | KK(σ2s)²(σ*2s)²(σ2p)²(π2p)⁴(π*2p)² | 2 | Paramagnetic | Stable |
| F₂ | KK(σ2s)²(σ*2s)²(σ2p)²(π2p)⁴(π*2p)⁴ | 1 | Diamagnetic | Stable |
σ1s < σ*1s < σ2s < σ*2s < π2p_y = π2p_z < σ2p_x < π*2p_y = π*2p_z < σ*2p_x
Energy Order for MOs (O₂ and beyond):σ1s < σ*1s < σ2s < σ*2s < σ2p_x < π2p_y = π2p_z < π*2p_y = π*2p_z < σ*2p_x
Nitrogen Molecule (N₂)
Why N₂ is diamagnetic and has triple bond:
Electronic Configuration: KK(σ2s)²(σ*2s)²(π2p)⁴(σ2p)²
All electrons are paired → Diamagnetic
Fig 3.34: Molecular orbital diagram of N₂ Molecule
Oxygen Molecule (O₂)
Why O₂ is paramagnetic and has double bond:
Electronic Configuration: KK(σ2s)²(σ*2s)²(σ2p)²(π2p)⁴(π*2p)²
Two unpaired electrons in π*2p orbitals → Paramagnetic
Why MOT is Superior to VBT for O₂?
VBT predicts all electrons paired in O₂ (wrong). MOT correctly predicts 2 unpaired electrons (paramagnetism), explaining experimental observations.
Fig 3.35: Molecular orbital diagram of O₂ Molecule
Helium Molecule (He₂)
Why He₂ doesn’t exist:
Electronic Configuration: (σ1s)²(σ*1s)²
Bonding electrons = 2, Antibonding electrons = 2 → Net bond order = 0
He₂ molecule is not stable because antibonding electrons cancel the effect of bonding electrons.
Solved MCQs – MOT
Solved Exercise Questions
Textbook Exercise MCQ Solutions
| Q.No. | Question | Correct Answer | Explanation |
|---|---|---|---|
| iii | Which molecule has zero dipole moment? | D | BF₃ has symmetrical trigonal planar structure → dipole moments cancel |
| iv | Number of σ and π bonds in N₂ | B | N₂ has 1 σ + 2 π bonds (triple bond: 1σ + 2π) |
| v | Species with unpaired electrons in antibonding MOs | B | N₂²⁻ has electronic configuration with unpaired electrons in π* orbitals |
| vi | Shape of ICl₃ | D | ICl₃: 3 bond pairs + 2 lone pairs → T-shaped |
| vii | Molecule with dipole moment | C | SO₂ is bent (V-shaped) → net dipole moment |
| viii | Valence electrons of P in PO₄³⁻ | B | P has 10 electrons in valence shell (expanded octet) |
| ix | Extra electrons in I in ICl₅ | C | I has 12 electrons (4 extra beyond octet) |
| x | Type and shape of [ICl₄]⁻ | D | AB₆ type (4 bond pairs + 2 lone pairs) → Square planar |
| xi | Molecule with sp³ hybridization & tetrahedral geometry | C | CCl₄ has sp³ hybridization, tetrahedral geometry |
| xii | Species with dative bond | C | NH₄⁺ has dative bond from N to H⁺ |
Important Short Questions Solved
H₂O: O (sp³ hybridized) with 2 bond pairs (O-H σ bonds) and 2 lone pairs. V-shaped structure with ~104.5° bond angle.
N₂: Triple bond between N atoms: 1 σ bond (p-p head-on overlap) + 2 π bonds (p-p sideways overlap). Linear molecule.
F₂: p-p head-on overlap of half-filled p orbitals → σ bond.
HF: s-p head-on overlap: H(1s) with F(2p) → σ bond (polar due to electronegativity difference).
sp hybridization: Mixing of one s and one p orbital to form two equivalent sp hybrid orbitals oriented at 180°.
Example: BeCl₂ (linear, 180°), CO₂ (linear), C₂H₂ (acetylene).
H₂O: O is sp³ hybridized (tetrahedral electron geometry, bent molecular shape).
CO₂: C is sp hybridized (linear molecule).
- Electron pairs (both bond pairs and lone pairs) arrange to maximize distance.
- Lone pairs occupy more space than bond pairs (lp-lp > lp-bp > bp-bp repulsion).
Lone pairs are attracted by only one nucleus (more localized electron density), while bond pairs are shared between two nuclei (more diffuse). This makes lone pairs bulkier with greater repulsive effect.
Both have 3 electron domains (2 bond pairs + 1 lone pair) → Trigonal planar electron geometry, bent/V-shaped molecular geometry (<120°).
PbCl₄ has 4 bond pairs + 0 lone pairs → Tetrahedral geometry (109.5°).
H₂O has strong hydrogen bonding (O-H bonds are highly polar), while H₂S has weaker dipole-dipole forces (S is less electronegative). Hydrogen bonding raises boiling point of water.
Bonding MO: Formed by in-phase overlap, electron density between nuclei, lower energy.
Antibonding MO: Formed by out-of-phase overlap, electron density away from nuclei, higher energy, node between nuclei.
He₂ would have electronic configuration: (σ1s)²(σ*1s)². Bond order = (2-2)/2 = 0 → No net bonding → He₂ doesn’t exist as stable molecule.
All electrons in N₂ are paired in molecular orbitals: KK(σ2s)²(σ*2s)²(π2p)⁴(σ2p)² → No unpaired electrons → Diamagnetic.
O₂ has electronic configuration: KK(σ2s)²(σ*2s)²(σ2p)²(π2p)⁴(π*2p)² with two unpaired electrons in π* orbitals → Paramagnetic.