AQA A-Level Chemistry Paper 3, June 2018: Question 30

Determining Molecular Shapes & Lone Pairs

📋 What this question tests

This question assesses your ability to apply Valence Shell Electron Pair Repulsion (VSEPR) theory to determine the number of bonding pairs and lone pairs of electrons around a central atom in neutral molecules and charged polyatomic ions.

Deducing the number of valence electrons on the central atom (including adjustments for positive and negative charges).
Distinguishing between electron-pair geometry and final molecular shape.
Recognising how lone pairs repel more strongly than bonding pairs and influence bond angles and geometry.

Question 30: Analysis & Solution

Multiple Choice: Identifying Lone Pairs on a Central Atom

✅ Correct Answer

B: ClF₃

Award [1 mark] for selecting option B.

Chlorine has 7 outer electrons and forms 3 single covalent bonds with fluorine atoms. This leaves 4 unbonded electrons, which form 2 lone pairs. The shape is T-shaped (derived from a trigonal bipyramidal electron arrangement), which is strongly influenced by these two lone pairs.

📐 Step-by-Step Species Deduction

Species	Valence e⁻ Calculation	BP / LP	Shape
A: AlCl₃	Group 3: 3 e⁻ − 3 (in bonds) = 0	3 BP, 0 LP	Trigonal planar (120°)
B: ClF₃	Group 7: 7 e⁻ − 3 (in bonds) = 4 e⁻	3 BP, 2 LP	T-shaped (<90°)
C: IF₆⁺	Group 7: (7 − 1) = 6 e⁻ − 6 = 0	6 BP, 0 LP	Octahedral (90°)
D: PCl₆⁻	Group 5: (5 + 1) = 6 e⁻ − 6 = 0	6 BP, 0 LP	Octahedral (90°)

💡 Key Knowledge: Counting Electron Pairs

Step 1: Find the periodic group of the central atom to get its starting number of valence electrons.
Step 2: Adjust for overall ionic charge:
• For a positive charge (+), subtract an electron (e.g., I has 7, so I⁺ has 6).
• For a negative charge (−), add an electron (e.g., P has 5, so P⁻ has 6).
Step 3: Each terminal halogen (F, Cl) forms a single bond by sharing 1 electron. Subtract 1 electron per bond to find non-bonding electrons.
Step 4: Divide remaining non-bonding electrons by 2 to determine the number of lone pairs (LP).

🧠 Exam Technique & Strategy

Spot the odd one out quickly: Notice that IF₆⁺ and PCl₆⁻ both have 6 bonding atoms attached to an atom with an effective count of 6 valence electrons, creating symmetrical octahedral species with zero lone pairs.
Beware of Group 3 electron deficiency: Students often mistakenly give Al an octet (4 pairs). Remember, AlCl₃ is electron-deficient with only 3 pairs (6 electrons in outer shell), giving 0 lone pairs.
VSEPR hierarchy: Remember that lone pair–lone pair repulsion > lone pair–bonding pair repulsion > bonding pair–bonding pair repulsion. In ClF₃ , the two lone pairs occupy equatorial positions, bending the axial F–Cl–F axis to produce a slightly distorted T-shape.

❌ Common Errors & Misconceptions

Forgetting to adjust for ion charge: Overlooking the + on IF₆⁺ leads to assuming Iodine still has 7 valence electrons, incorrectly calculating (7 + 6)/2 = 6.5 pairs or an extra lone electron.
Assuming 6 bonds always means lone pairs are present: Expanded octets occur in Period 3 and below (like P and I), allowing 6 bonding pairs with 0 lone pairs when charged appropriately.
Confusing central atom lone pairs with ligand lone pairs: The question specifically states "around the central atom". Fluorine and chlorine ligands have multiple lone pairs, but these do not dictate the primary VSEPR geometry of the central skeleton.

AQA A-Level Chemistry Paper 3, June 2018: Question 30

Question

Mark scheme

How to answer it