Chemistry Inorganic - How to Solve: Chemical Bonding – Formal Charge, Hybridisation, VSEPR, MOT (Bond Order, Magnetic Character)

How to Solve: Chemical Bonding – Formal Charge, Hybridisation, VSEPR, MOT (Bond Order, Magnetic Character)

For IIT JEE (Main + Advanced)

Introduction

"Mastering formal charge, hybridisation, VSEPR, and Molecular Orbital Theory (MOT) doesn’t just get you 10–15 marks in JEE—it’s the key to predicting whether a molecule is stable, polar, or even magnetic. One question on bond order or hybridisation can decide your rank!

WHAT YOU NEED TO KNOW FIRST

1. Lewis Structures – How to draw them (octet rule, exceptions).
2. Electron Configuration – s, p, d orbitals and valence electrons.
3. Basic Bonding – Sigma (σ) and pi (π) bonds, lone pairs.

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

KEY TERMS & FORMULAS

1. Formal Charge (FC)

Formula: FC = (Valence electrons in free atom) – (Non-bonding electrons) – ½(Bonding electrons) - Valence electrons in free atom = Group number (e.g., N = 5, O = 6). - Non-bonding electrons = Lone pair electrons on the atom. - Bonding electrons = Total electrons in bonds (each bond = 2 electrons).

MEMORISE THIS – Used to determine the most stable Lewis structure.

2. Hybridisation

Key Rules: - sp³ → 4 electron domains (tetrahedral, 109.5°). - sp² → 3 electron domains (trigonal planar, 120°). - sp → 2 electron domains (linear, 180°). - dsp³ → 5 electron domains (trigonal bipyramidal). - d²sp³ → 6 electron domains (octahedral).

MEMORISE THIS – Count σ bonds + lone pairs to determine hybridisation.

3. VSEPR Theory (Valence Shell Electron Pair Repulsion)

Steps:
1. Draw Lewis structure.
2. Count electron domains (σ bonds + lone pairs).
3. Use VSEPR chart to predict shape and bond angles.

MEMORISE THIS TABLE (Common Shapes): | Electron Domains | Lone Pairs | Shape | Bond Angle | |----------------------|----------------|--------------------|----------------| | 2 | 0 | Linear | 180° | | 3 | 0 | Trigonal planar | 120° | | 3 | 1 | Bent | <120° | | 4 | 0 | Tetrahedral | 109.5° | | 4 | 1 | Trigonal pyramidal | <109.5° | | 4 | 2 | Bent | <109.5° | | 5 | 0 | Trigonal bipyramidal| 90°, 120° | | 6 | 0 | Octahedral | 90° |

MEMORISE THIS – Lone pairs reduce bond angles.

4. Molecular Orbital Theory (MOT)

Key Formulas:
1. Bond Order (BO): BO = ½ [(Number of bonding electrons) – (Number of antibonding electrons)] - MEMORISE THIS – Higher BO = stronger bond. - BO = 0 → Molecule does not exist.

1. Magnetic Character:
2. Paramagnetic → Unpaired electrons present.
3. Diamagnetic → All electrons paired.

MEMORISE THIS ORDER (for B₂, C₂, N₂, O₂, F₂): σ1s² < σ1s² < σ2s² < σ2s² < π2pₓ² = π2pᵧ² < σ2p_z² < π2pₓ¹ = π2pᵧ¹ < σ2p_z¹

(For O₂ and F₂, σ2p_z comes before π2pₓ/π2pᵧ.)

STEP-BY-STEP METHOD

Step 1: Formal Charge (FC) – Find the Best Lewis Structure

1. Draw all possible Lewis structures for the molecule.
2. Calculate FC for each atom in each structure.
3. Choose the structure with:
4. FC closest to zero on all atoms.
5. Negative FC on more electronegative atoms (e.g., O, F).

Step 2: Hybridisation – Determine the Central Atom’s Hybridisation

1. Count σ bonds (single bonds = 1 σ, double/triple bonds = 1 σ).
2. Count lone pairs on the central atom.
3. Add them → Total electron domains.
4. Match to hybridisation:
5. 2 domains → sp
6. 3 domains → sp²
7. 4 domains → sp³
8. 5 domains → dsp³
9. 6 domains → d²sp³

Step 3: VSEPR – Predict Shape & Bond Angles

1. Draw Lewis structure (from Step 1).
2. Count electron domains (σ bonds + lone pairs) around the central atom.
3. Use VSEPR table to find shape and bond angles.
4. Adjust for lone pairs (they compress bond angles).

Step 4: MOT – Bond Order & Magnetic Character

1. Write electron configuration of the molecule (e.g., O₂ = 16 electrons).
2. Fill MOs in order (use the correct sequence for the molecule).
3. Count bonding & antibonding electrons.
4. Calculate bond order (BO = ½ [bonding – antibonding]).
5. Check for unpaired electrons → Paramagnetic (if yes), Diamagnetic (if no).

WORKED EXAMPLES

Example 1 – Basic: Formal Charge & Hybridisation (CO₂)

Question: Find the formal charge on each atom in CO₂ and determine the hybridisation of C.

Step 1: Draw Lewis Structure - C (4 valence) + 2 O (6 each) = 16 valence electrons. - Structure: O=C=O (double bonds, no lone pairs on C).

Step 2: Calculate Formal Charge - C: FC = 4 – (0) – ½(8) = 0 - O (each): FC = 6 – (4) – ½(4) = 0

Step 3: Hybridisation of C - σ bonds = 2 (one in each C=O). - Lone pairs = 0. - Total domains = 2 → sp hybridisation.

What we did and why: - We confirmed the most stable structure (FC = 0 on all atoms). - We used σ bonds + lone pairs to find hybridisation.

Example 2 – Medium: VSEPR & Bond Angles (NH₃)

Question: Predict the shape and bond angle of NH₃.

Step 1: Draw Lewis Structure - N (5 valence) + 3 H (1 each) = 8 valence electrons. - Structure: N with 3 single bonds and 1 lone pair.

Step 2: Count Electron Domains - σ bonds = 3 (N-H). - Lone pairs = 1. - Total domains = 4 → sp³ hybridisation.

Step 3: VSEPR Shape - 4 domains, 1 lone pair → Trigonal pyramidal. - Bond angle = <109.5° (lone pair compresses it to ~107°).

What we did and why: - We used VSEPR to predict shape and bond angle distortion due to lone pairs.

Example 3 – Exam-Style: MOT (O₂⁺ Ion)

Question: Find the bond order and magnetic character of O₂⁺.

Step 1: Electron Configuration - O₂ has 16 electrons → O₂⁺ has 15 electrons. - MO order: σ1s² < σ1s² < σ2s² < σ2s² < σ2p_z² < π2pₓ² = π2pᵧ² < π2pₓ¹ = π2pᵧ¹

Step 2: Fill Electrons - Total electrons = 15. - Configuration: (σ2s)² (σ2s)² (σ2p_z)² (π2pₓ)² (π2pᵧ)² (π2pₓ)¹

Step 3: Count Bonding & Antibonding - Bonding electrons = 2 (σ2s) + 2 (σ2p_z) + 4 (π2pₓ/π2pᵧ) = 8 - Antibonding electrons = 2 (σ2s) + 1 (π2pₓ) = 3

Step 4: Calculate Bond Order - BO = ½ (8 – 3) = 2.5

Step 5: Magnetic Character - Unpaired electron (π2pₓ¹) → Paramagnetic.

What we did and why: - We used MOT to find bond order and magnetic properties, crucial for JEE questions.

COMMON MISTAKES

1. Mistake: Counting π bonds in hybridisation. Why it happens: Students confuse σ and π bonds. Correct approach: Only σ bonds + lone pairs determine hybridisation.

2. Mistake: Forgetting lone pairs in VSEPR. Why it happens: Students only count bonds. Correct approach: Always count σ bonds + lone pairs for electron domains.

3. Mistake: Wrong MO order for O₂/F₂. Why it happens: Using the same order as N₂. Correct approach: For O₂/F₂, σ2p_z comes before π2pₓ/π2pᵧ.

4. Mistake: Incorrect formal charge calculation. Why it happens: Misidentifying bonding/non-bonding electrons. Correct approach: FC = Valence – (Non-bonding + ½ Bonding).

5. Mistake: Ignoring electronegativity in formal charge. Why it happens: Not checking if negative FC is on the right atom. Correct approach: Negative FC should be on more electronegative atoms.

EXAM TRAPS

1. Trap: "Which structure is correct?" with multiple Lewis structures. How to spot it: Examiner gives 2–3 structures with similar FC. How to avoid it: Choose the one where negative FC is on the most electronegative atom.

2. Trap: Hybridisation of ions (e.g., NH₄⁺). How to spot it: Central atom has a charge. How to avoid it: Count electrons correctly (NH₄⁺ has 8 valence electrons, not 9).

3. Trap: MOT for heteronuclear molecules (e.g., CO). How to spot it: Different atoms → MO order changes. How to avoid it: Use the same order as N₂ (σ2s < σ2s < π2p < σ2p_z).

1-MINUTE RECAP (Night Before Exam)

"Listen up—this is your 60-second cheat sheet for Chemical Bonding in JEE:

1. Formal Charge: FC = Valence – (Non-bonding + ½ Bonding). Pick the structure with FC closest to zero, negative on the most electronegative atom.
2. Hybridisation: Count σ bonds + lone pairs. 2 = sp, 3 = sp², 4 = sp³.
3. VSEPR: Count electron domains → Use the table. Lone pairs compress angles.
4. MOT: Fill MOs in order (N₂ order for B₂/C₂/N₂, O₂ order for O₂/F₂). Bond order = ½ (bonding – antibonding). Unpaired electrons? Paramagnetic. Paired? Diamagnetic.

One last tip: For O₂, remember it’s paramagnetic—examiners love this! Now go crush that exam."