AQA A-Level Chemistry Paper 3, June 2025: Question 27

AQA A-Level Chemistry • Paper 1 Multiple Choice

Shapes of Species: Molecules and Complex Ions

What this question tests

This question tests your ability to deduce molecular geometries and transition metal complex shapes. You need to distinguish between 4-coordinate transition metal complexes (which typically adopt tetrahedral geometry when bulky halide ligands like Cl⁻ are coordinated) and main-group p-block species that expand their octet to form see-saw or square planar shapes due to the presence of lone pairs.

Question 27 Analysis

Which species has a tetrahedral shape? [1 mark]

✅ Correct Answer

B: FeCl₄⁻

Award: 1 mark (AO2)

In [FeCl₄]⁻ , iron is in the +3 oxidation state. Chloride ( Cl⁻ ) is a relatively large ligand, meaning only four can pack around the Fe³⁺ ion due to steric repulsion. In first-row transition metal complexes with four monodentate chloride ligands, maximum separation yields a tetrahedral geometry with bond angles of 109.5°.

📐 Systematic Deduction of All 4 Options

Using VSEPR theory and coordination chemistry principles:

Species	Electron Pairs / Ligands	Molecular Shape
A: SF₄	6 (S valence) + 4 (from F) = 10 e⁻ → 4 bonding pairs, 1 lone pair	See-saw
B: FeCl₄⁻	4 large Cl⁻ ligands coordinated to Fe³⁺ → coordination number = 4	Tetrahedral
C: XeF₄	8 (Xe valence) + 4 (from F) = 12 e⁻ → 4 bonding pairs, 2 lone pairs	Square planar
D: ICl₄⁻	7 (I valence) + 1 (charge) + 4 = 12 e⁻ → 4 bonding pairs, 2 lone pairs	Square planar

💡 Key Knowledge

Transition Metal Complex Shapes:
- Small ligands (e.g. H₂O, NH₃) typically achieve a coordination number of 6 (octahedral).
- Larger ligands (e.g. Cl⁻) only permit a coordination number of 4. First-row complexes such as [CuCl₄]²⁻ , [CoCl₄]²⁻ , and [FeCl₄]⁻ are tetrahedral.
- Square planar (coordination number 4) in A-Level is primarily associated with Pt²⁺ and Pd²⁺ (e.g. cisplatin, Pt(NH₃)₂Cl₂ ).
VSEPR Lone Pair Rule: Lone pairs repel more strongly than bonding pairs. For 6 pairs (4 bonding, 2 lone), the lone pairs position opposite each other (180° apart) to minimise repulsion, producing a square planar shape.

❌ Common Errors

Assuming "4 surrounding atoms = tetrahedral": A fatal trap! SF₄, XeF₄, and ICl₄⁻ all have 4 attached atoms, but none are tetrahedral because the central atom possesses unshared lone pairs.
Miscounting valence electrons on ions: In ICl₄⁻ , students often forget to add the extra electron for the negative charge (7 + 1 + 4 = 12 electrons = 6 pairs).
Overcomplicating the transition metal: Trying to draw a simple Lewis dot-and-cross diagram for FeCl₄⁻ as if it were a p-block covalent molecule rather than recognising it as a standard complex ion.

🧠 Exam Technique: Elimination Strategy

Check the non-metal options first using the VSEPR counting method:
Total pairs = (Central group valence + bonded atoms - charge) ÷ 2
For XeF₄: (8 + 4) ÷ 2 = 6 pairs. 4 bonds + 2 lone pairs = Square planar. Eliminate C.
For ICl₄⁻: (7 + 4 + 1) ÷ 2 = 6 pairs. 4 bonds + 2 lone pairs = Square planar. Eliminate D.
For SF₄: (6 + 4) ÷ 2 = 5 pairs. 4 bonds + 1 lone pair = See-saw. Eliminate A.
By elimination (and direct specification recall of chloride transition metal complexes), B must be correct.