IB Chemistry How to Solve: IB Chemistry – Bonding (VSEPR, Hybridisation, Formal Charge, Resonance)

How to Solve: IB Chemistry – Bonding (VSEPR, Hybridisation, Formal Charge, Resonance)

Complete Guide

Introduction

Mastering VSEPR, hybridisation, formal charge, and resonance unlocks 6–8 marks in IB Chemistry Paper 2 and Paper 3—enough to boost your grade by a full band. These concepts explain why molecules have specific shapes, reactivity, and even drug interactions in real-world chemistry.

WHAT YOU NEED TO KNOW FIRST

1. Lewis structures – How to draw them (octet rule, exceptions).
2. Electron domains – Counting bonding pairs and lone pairs.
3. Bond types – Single, double, triple bonds and their electron counts.

KEY TERMS & FORMULAS

1. VSEPR (Valence Shell Electron Pair Repulsion) Theory

• Key Idea: Electron pairs (bonding + lone) arrange themselves to minimise repulsion.
• Electron Domain Geometry vs. Molecular Geometry:
• Electron Domain Geometry = Shape considering all electron pairs (bonding + lone).
• Molecular Geometry = Shape considering only atoms (ignores lone pairs).
• MEMORISE THIS: VSEPR chart (2–6 electron domains).

2. Hybridisation

• Key Idea: Atomic orbitals mix to form new hybrid orbitals for bonding.
• MEMORISE THIS: Hybridisation types based on electron domains:

3. Formal Charge

• Formula: Formal Charge (FC) = (Valence electrons) – (Non-bonding electrons) – ½(Bonding electrons)
• What it means:
• FC = 0 → Most stable structure.
• Lowest FC (closest to 0) → Preferred Lewis structure.
• MEMORISE THIS: Always calculate FC for all possible structures in resonance.

4. Resonance

• Key Idea: When multiple valid Lewis structures exist, the actual structure is a hybrid.
• Rules:
• Only π bonds and lone pairs move (σ bonds stay fixed).
• Octet rule must be satisfied (except for expanded octets).
• Formal charges must be minimised (closest to 0).

STEP-BY-STEP METHOD

Step 1: Draw the Lewis Structure

1. Count total valence electrons (group number × atoms).
2. Draw skeleton structure (central atom = least electronegative, except H).
3. Place single bonds between atoms.
4. Distribute remaining electrons as lone pairs (start with outer atoms).
5. Check octet rule (exceptions: H, Be, B, Al, expanded octets for Period 3+).
6. If octet is incomplete, form double/triple bonds.

Step 2: Determine Electron Domain Geometry (VSEPR)

1. Count electron domains around the central atom (each lone pair = 1 domain, each bond = 1 domain regardless of bond order).
2. Use the VSEPR chart to find electron domain geometry.

Step 3: Determine Molecular Geometry

1. Count lone pairs on the central atom.
2. Use the VSEPR chart to find molecular geometry (ignoring lone pairs).

Step 4: Assign Hybridisation

1. Match electron domains to hybridisation (e.g., 4 domains = sp³).

Step 5: Calculate Formal Charges (If Needed)

1. Apply the formal charge formula to each atom.
2. Choose the structure with FC closest to 0 (most stable).

Step 6: Check for Resonance (If Applicable)

1. If multiple valid Lewis structures exist (same atom positions, different electron placement), draw all resonance structures.
2. Calculate formal charges for each.
3. The actual structure is a hybrid of all resonance forms.

WORKED EXAMPLES

Example 1 – Basic: CO₂ (VSEPR & Hybridisation)

Question: Predict the molecular geometry, bond angle, and hybridisation of CO₂.

Step 1: Lewis Structure - Valence electrons: C (4) + O (6) × 2 = 16 e⁻. - Skeleton: O–C–O. - Single bonds: 2 × (2 e⁻) = 4 e⁻ used. - Remaining: 16 – 4 = 12 e⁻ (6 pairs). - Place lone pairs on O: 3 pairs each (O now has octet). - C has only 4 e⁻ (incomplete octet) → Form double bonds (C=O). - Final structure: O=C=O (all atoms have octet).

Step 2: Electron Domain Geometry - Central atom (C) has 2 electron domains (2 double bonds, 0 lone pairs). - Electron domain geometry = Linear.

Step 3: Molecular Geometry - 0 lone pairs → Molecular geometry = Linear.

Step 4: Hybridisation - 2 electron domains → sp hybridisation.

Step 5: Bond Angle - Linear = 180°.

Answer: - Molecular geometry: Linear - Bond angle: 180° - Hybridisation: sp

What we did and why: We drew the Lewis structure first to count electron domains. Then, we used VSEPR to predict shape and hybridisation. Double bonds count as 1 domain in VSEPR.

Example 2 – Medium: SO₂ (VSEPR, Resonance, Formal Charge)

Question: Draw the Lewis structure, resonance forms, and predict molecular geometry of SO₂.

Step 1: Lewis Structure - Valence electrons: S (6) + O (6) × 2 = 18 e⁻. - Skeleton: O–S–O. - Single bonds: 2 × (2 e⁻) = 4 e⁻ used. - Remaining: 18 – 4 = 14 e⁻ (7 pairs). - Place lone pairs on O: 3 pairs each (O has octet). - S has 4 e⁻ (incomplete octet) → Form double bond (S=O). - Now, S has 6 e⁻ (octet). - But we have 1 lone pair left on S → Resonance possible.

Step 2: Resonance Structures - Structure 1: O=S–O (lone pair on S, double bond to left O). - Structure 2: O–S=O (lone pair on S, double bond to right O). - Both are valid → Resonance hybrid.

Step 3: Formal Charges - Structure 1: - S: FC = 6 – 2 – ½(6) = +1 - Left O (double bond): FC = 6 – 4 – ½(4) = 0 - Right O (single bond): FC = 6 – 6 – ½(2) = -1 - Structure 2: Same as Structure 1 (just flipped). - Both have FC = +1, 0, -1 → Equally valid.

Step 4: Electron Domain Geometry - Central atom (S) has 3 electron domains (1 lone pair + 2 bonds). - Electron domain geometry = Trigonal planar.

Step 5: Molecular Geometry - 1 lone pair → Molecular geometry = Bent.

Step 6: Bond Angle - Bent (1 lone pair) ≈ 119° (less than 120° due to lone pair repulsion).

Answer: - Lewis structure: Resonance between O=S–O and O–S=O. - Molecular geometry: Bent - Bond angle: ~119°

What we did and why: We drew two resonance structures because the double bond can be on either side. Formal charges helped confirm stability. The lone pair on S bends the molecule.

Example 3 – Exam-Style: NO₃⁻ (Resonance, Formal Charge, Hybridisation)

Question: For the nitrate ion (NO₃⁻): a) Draw all resonance structures. b) Calculate formal charges for each. c) Predict molecular geometry and hybridisation.

Step 1: Lewis Structure - Valence electrons: N (5) + O (6) × 3 + 1 (negative charge) = 24 e⁻. - Skeleton: O–N–O (central N). - Single bonds: 3 × (2 e⁻) = 6 e⁻ used. - Remaining: 24 – 6 = 18 e⁻ (9 pairs). - Place lone pairs on O: 3 pairs each (O has octet). - N has 6 e⁻ (incomplete octet) → Form double bond (N=O). - Now, N has 8 e⁻ (octet). - Resonance possible (double bond can be on any O).

Step 2: Resonance Structures - Structure 1: O=N–O⁻ (double bond to top O, single bonds to others). - Structure 2: O–N=O (double bond to right O). - Structure 3: ⁻O–N=O (double bond to left O). - All three are equivalent.

Step 3: Formal Charges - Structure 1: - N: FC = 5 – 0 – ½(8) = +1 - Double-bonded O: FC = 6 – 4 – ½(4) = 0 - Single-bonded O (with 3 lone pairs): FC = 6 – 6 – ½(2) = -1 - Single-bonded O (with 2 lone pairs + negative charge): FC = 6 – 6 – ½(2) = -1 - Total FC = +1 + 0 – 1 – 1 = -1 (matches ion charge). - Structures 2 & 3: Same as Structure 1 (just rotated).

Step 4: Electron Domain Geometry - Central atom (N) has 3 electron domains (3 bonds, 0 lone pairs). - Electron domain geometry = Trigonal planar.

Step 5: Molecular Geometry - 0 lone pairs → Molecular geometry = Trigonal planar.

Step 6: Hybridisation - 3 electron domains → sp² hybridisation.

Answer: a) Resonance structures: Three equivalent forms (double bond rotates). b) Formal charges: N = +1, double-bonded O = 0, single-bonded O = -1 (each). c) Molecular geometry: Trigonal planar Hybridisation: sp²

What we did and why: We drew three resonance structures because the double bond can be on any O. Formal charges confirmed stability (total = -1, matching the ion). The trigonal planar shape comes from 3 bonding domains.

COMMON MISTAKES

EXAM TRAPS

1-MINUTE RECAP

"Listen up—this is your 60-second bonding cheat sheet for the IB exam.

1. VSEPR: Count electron domains (bonds + lone pairs) around the central atom. 2 domains = linear, 3 = trigonal planar, 4 = tetrahedral, 5 = trigonal bipyramidal, 6 = octahedral. Lone pairs bend the shape (e.g., 3 domains + 1 lone pair = bent).
2. Hybridisation: Match domains to orbitals. 2 = sp, 3 = sp², 4 = sp³, 5 = sp³d, 6 = sp³d².
3. Formal Charge: Use the formula FC = valence e⁻ – non-bonding e⁻ – ½(bonding e⁻). The lowest FC (closest to 0) wins.
4. Resonance: If you can move double bonds or lone pairs without changing atom positions, draw all structures and pick the one with lowest FC.
5. Exam tricks: Double bonds = 1 domain, lone pairs change shape, and always check FC in resonance.

Now go crush that exam—you’ve got this!