Chemistry Physical How to Solve: Chemical Bonding – VSEPR, Hybridisation, Molecular Orbital Theory (Bond Order) | IIT JEE Guide

How to Solve: Chemical Bonding – VSEPR, Hybridisation, Molecular Orbital Theory (Bond Order) | IIT JEE Guide

Introduction

Mastering VSEPR, hybridisation, and MOT unlocks 10–15 marks in IIT JEE—enough to push you from a 90 to a 100+ percentile. These concepts explain why water is bent, why oxygen is paramagnetic, and how to predict bond strengths in seconds. If you can solve these, you can solve any bonding question in the exam.

WHAT YOU NEED TO KNOW FIRST

1. Lewis structures – How to draw them and count valence electrons.
2. Electron configuration – s, p, d orbitals and their shapes.
3. Basic molecular geometry – Linear, trigonal planar, tetrahedral, etc.

(If you’re shaky on these, pause and review them first.)

KEY TERMS & FORMULAS

1. VSEPR Theory (Valence Shell Electron Pair Repulsion)

Key Terms: - Steric number (SN): Number of atoms bonded to central atom + number of lone pairs on central atom. - Electron pair geometry: Arrangement of all electron pairs (bonding + lone pairs). - Molecular geometry: Shape considering only bonded atoms (ignores lone pairs).

Formulas: - Steric Number (SN) = (Number of bonded atoms) + (Number of lone pairs on central atom) - MEMORISE THIS – Used to predict shape. - Bond angle deviations: - Lone pairs repel more than bonding pairs → compress bond angles. - Double/triple bonds repel more than single bonds → slightly larger angles.

VSEPR Chart (MEMORISE THIS): | SN | Electron Pair Geometry | Molecular Geometry (0 lone pairs) | Molecular Geometry (1 lone pair) | Molecular Geometry (2 lone pairs) | |--------|----------------------------|---------------------------------------|--------------------------------------|----------------------------------------| | 2 | Linear | Linear (180°) | – | – | | 3 | Trigonal Planar | Trigonal Planar (120°) | Bent (<120°) | – | | 4 | Tetrahedral | Tetrahedral (109.5°) | Trigonal Pyramidal (<109.5°) | Bent (<109.5°) | | 5 | Trigonal Bipyramidal | Trigonal Bipyramidal (90°, 120°) | See-saw (<90°, <120°) | T-shaped (<90°) | | 6 | Octahedral | Octahedral (90°) | Square Pyramidal (<90°) | Square Planar (90°) |

2. Hybridisation

Key Terms: - Hybridisation: Mixing of atomic orbitals to form new hybrid orbitals. - Hybrid orbitals: sp, sp², sp³, sp³d, sp³d² (MEMORISE these). - Sigma (σ) bond: Head-on overlap (single bonds). - Pi (π) bond: Sideways overlap (double/triple bonds).

Formulas: - Hybridisation = Steric Number (SN) - SN = 2 → sp - SN = 3 → sp² - SN = 4 → sp³ - SN = 5 → sp³d - SN = 6 → sp³d² - MEMORISE THIS – Directly linked to VSEPR.

Exceptions (MEMORISE THESE): - NH₃ (SN = 4, sp³ hybridised) – But bond angle is 107° (not 109.5°) due to lone pair repulsion. - H₂O (SN = 4, sp³ hybridised) – Bond angle is 104.5°. - CO₂ (SN = 2, sp hybridised) – Linear, but has two double bonds (π bonds).

3. Molecular Orbital Theory (MOT) & Bond Order

Key Terms: - Molecular orbitals (MOs): Formed by combining atomic orbitals (AOs). - Bonding MOs (σ, π): Lower energy, stabilise the molecule. - Antibonding MOs (σ, π): Higher energy, destabilise the molecule. - Bond order (BO): Indicates bond strength and stability.

Formulas: - Bond Order (BO) = ½ [(Number of electrons in bonding MOs) – (Number of electrons in antibonding MOs)] - MEMORISE THIS – Given on exam sheet, but you must know how to apply it. - Magnetic properties: - Paramagnetic: Unpaired electrons present. - Diamagnetic: All electrons paired.

MO Diagrams (MEMORISE THESE): | Molecule | MO Order (for Z ≤ 7) | MO Order (for Z ≥ 8) | |--------------|--------------------------|--------------------------| | B₂, C₂, N₂ | σ1s² < σ1s² < σ2s² < σ2s² < π2pₓ = π2pᵧ < σ2p_z | – | | O₂, F₂ | σ1s² < σ1s² < σ2s² < σ2s² < σ2p_z < π2pₓ = π2pᵧ < π2pₓ = π2pᵧ | Same as left |

Key Observations (MEMORISE): - O₂ is paramagnetic (2 unpaired electrons in π orbitals). - N₂ has BO = 3 (triple bond, very stable). - He₂ does not exist (BO = 0).

STEP-BY-STEP METHOD

Step 1: Draw the Lewis Structure

• Count total valence electrons.
• Place least electronegative atom in the center (except H).
• Form bonds, then distribute remaining electrons as lone pairs.
• Check octet rule (exceptions: H, Be, B, Al, expanded octets for P, S, Cl).

Step 2: Determine Steric Number (SN)

• SN = (Number of bonded atoms) + (Number of lone pairs on central atom).
• Use SN to predict electron pair geometry (from VSEPR chart).

Step 3: Predict Molecular Geometry

• Ignore lone pairs → molecular geometry (from VSEPR chart).
• Adjust bond angles if lone pairs or multiple bonds are present.

Step 4: Assign Hybridisation

• Hybridisation = Steric Number (SN).
• Match SN to hybridisation type (sp, sp², sp³, etc.).

Step 5: For MOT Questions – Draw MO Diagram

• Write electron configuration of atoms.
• Fill MOs in order (memorise the order for Z ≤ 7 vs. Z ≥ 8).
• Count bonding and antibonding electrons.
• Calculate Bond Order (BO).

Step 6: Check Magnetic Properties

• If any unpaired electrons → paramagnetic.
• If all electrons paired → diamagnetic.

WORKED EXAMPLES

Example 1 – Basic: NH₃ (Ammonia)

Question: Predict the shape, hybridisation, and bond angle of NH₃.

Step 1: Lewis Structure - N (5 valence e⁻) + 3 H (1 each) = 8 valence e⁻. - N forms 3 bonds with H → 6 e⁻ used. - Remaining 2 e⁻ → 1 lone pair on N.

Step 2: Steric Number (SN) - Bonded atoms = 3 (H). - Lone pairs = 1. - SN = 3 + 1 = 4.

Step 3: Molecular Geometry - SN = 4 → Tetrahedral electron pair geometry. - 1 lone pair → Trigonal pyramidal molecular geometry.

Step 4: Hybridisation - SN = 4 → sp³ hybridised.

Step 5: Bond Angle - Ideal tetrahedral angle = 109.5°. - Lone pair repulsion → 107° (actual angle).

What we did and why: - We used VSEPR to predict shape and hybridisation. - Lone pair repulsion compresses the bond angle below 109.5°.

Example 2 – Medium: CO₂ (Carbon Dioxide)

Question: Predict the shape, hybridisation, and magnetic properties of CO₂.

Step 1: Lewis Structure - C (4 valence e⁻) + 2 O (6 each) = 16 valence e⁻. - C forms 2 double bonds with O → 8 e⁻ used. - No lone pairs on C.

Step 2: Steric Number (SN) - Bonded atoms = 2 (O). - Lone pairs = 0. - SN = 2 + 0 = 2.

Step 3: Molecular Geometry - SN = 2 → Linear molecular geometry.

Step 4: Hybridisation - SN = 2 → sp hybridised.

Step 5: Magnetic Properties (MOT) - C (1s² 2s² 2p²) + 2 O (1s² 2s² 2p⁴) → Total 16 e⁻. - MO order for Z ≥ 8: σ1s² < σ1s² < σ2s² < σ2s² < σ2p_z² < π2pₓ² = π2pᵧ² < π2pₓ⁰ = π2pᵧ⁰ < σ2p_z⁰. - Bonding e⁻ = 8 (σ2s² + σ2p_z² + π2pₓ² + π2pᵧ²). - Antibonding e⁻ = 4 (σ1s² + σ2s²). - BO = ½ (8 – 4) = 2. - All electrons paired → Diamagnetic.

What we did and why: - CO₂ is linear due to sp hybridisation. - MOT confirms BO = 2 (double bond) and diamagnetism.

Example 3 – Exam-Style: NO (Nitric Oxide)

Question: Predict the bond order and magnetic properties of NO. Is NO stable?

Step 1: Lewis Structure (Not needed for MOT, but good practice) - N (5 e⁻) + O (6 e⁻) = 11 valence e⁻. - Forms a double bond (N=O) with 1 unpaired e⁻ on N.

Step 2: MOT Approach - Total e⁻ = 11 (N: 7, O: 8 → 15 e⁻, but NO has 15 – 4 (from 2s²) = 11 in valence MOs). - MO order for Z ≤ 7: σ1s² < σ1s² < σ2s² < σ2s² < π2pₓ² = π2pᵧ² < σ2p_z¹ < π2pₓ⁰ = π2pᵧ⁰. - Bonding e⁻ = 8 (σ2s² + π2pₓ² + π2pᵧ² + σ2p_z¹). - Antibonding e⁻ = 3 (σ1s² + σ2s²). - BO = ½ (8 – 3) = 2.5. - Unpaired e⁻ in σ2p_z → Paramagnetic.

Step 3: Stability Check - BO = 2.5 > 0 → Stable molecule.

What we did and why: - NO has a fractional BO (2.5) due to an unpaired electron. - Paramagnetism is confirmed by MOT.

COMMON MISTAKES

EXAM TRAPS

1-MINUTE RECAP (Night Before Exam)

"Listen up—this is your 60-second VSEPR, hybridisation, and MOT crash course.

1. VSEPR: Count steric number (SN) = bonded atoms + lone pairs. Match SN to the VSEPR chart for shape. Lone pairs squish bond angles (e.g., H₂O = 104.5°, not 109.5°).
2. Hybridisation: SN = hybridisation type (2=sp, 3=sp², 4=sp³, 5=sp³d, 6=sp³d²). No exceptions—just memorise.
3. MOT: For diatomic molecules, draw the MO diagram. Z ≤ 7 (N₂, B₂): π2p before σ2p. Z ≥ 8 (O₂, F₂): σ2p before π2p. Bond order = ½ (bonding e⁻ – antibonding e⁻). Unpaired e⁻? Paramagnetic.
4. Exam traps: O₂ is paramagnetic (2 unpaired e⁻). N₂ has BO=3 (triple bond). XeF₄ is sp³d² (SN=6).

You’ve got this. Now go crush that exam."