AQA A-Level Chemistry Paper 3, June 2022: Question 19

Transition Metal Complexes: Ligand Substitution & Geometry

📋 What this question tests

This question assesses your ability to deduce coordination numbers, ligand substitution reactions, and 3D geometries of transition metal complexes. Specifically, it tests recognition of when ligand size (e.g. large chloride ions) causes a change in coordination number from 6 to 4, yielding a tetrahedral complex.

Question 19 Analysis

Identifying the Reagent Pair Forming a Tetrahedral Complex

✅ Correct Option: D

FeCl₃(aq) and concentrated HCl(aq)

In aqueous solution, iron(III) exists as the octahedral hexaaqua ion [Fe(H₂O)₆]³⁺ . Addition of concentrated hydrochloric acid provides a high concentration of chloride ions ( Cl⁻ ), which displace the smaller neutral water ligands to form the tetrachloroferrate(III) ion:

[Fe(H₂O)₆]³⁺ + 4Cl⁻ ⇌ [FeCl₄]⁻ + 6H₂O

Because chloride ligands are relatively large and charged, ligand-ligand repulsions prevent 6 chlorides from fitting around the central Fe³⁺ ion. Only 4 chloride ligands can coordinate, giving a tetrahedral geometry with bond angles of approximately 109.5°.

💡 Key Knowledge: Ligand Sizes & Complex Shapes

Small, uncharged ligands: H₂O and NH₃ are similar in size. Complex formation usually leads to a coordination number of 6 and an octahedral geometry (e.g. [Co(NH₃)₆]²⁺ ).
Large, charged ligands: Cl⁻ is much larger than H₂O . When chloride ions coordinate to metal ions like Fe³⁺ , Cu²⁺ , or Co²⁺ , the coordination number drops to 4, forming a tetrahedral complex (e.g. [FeCl₄]⁻ , [CuCl₄]²⁻ , [CoCl₄]²⁻ ).
Bidentate ligands: Ethanedioate ( C₂O₄²⁻ ) forms 2 coordinate bonds per ligand. Three ethanedioate ligands coordinate to give a coordination number of 6 (octahedral).

🧠 Exam Technique: Systematic Option Breakdown

Option	Complex Formed	Shape
A	[Co(NH₃)₆]²⁺	Octahedral
B	[Cu(NH₃)₄(H₂O)₂]²⁺	Octahedral (distorted)
C	[Cu(C₂O₄)₃]⁴⁻	Octahedral
D	[FeCl₄]⁻	Tetrahedral

❌ Common Misconceptions & Traps

Confusing CoCl₂ with the reagent adding Cl⁻: Option A includes aqueous cobalt(II) chloride, which forms the pink octahedral hexaaqua ion [Co(H₂O)₆]²⁺ in solution. Adding concentrated NH₃ produces the octahedral hexammine complex. If concentrated HCl had been added to CoCl₂ , it would form the blue tetrahedral [CoCl₄]²⁻ ion. Do not mix up the added ligand with the spectator counter-ion!
Misinterpreting the 4:2 ratio in copper-ammonia: In Option B, adding concentrated ammonia replaces four water molecules to form deep blue [Cu(NH₃)₄(H₂O)₂]²⁺ . Many students assume four ammonia ligands means coordination number 4 (tetrahedral or square planar). In reality, two water molecules remain coordinated at the axial positions, maintaining a coordination number of 6 and an octahedral shape.
Forgetting bidentate denticity: In Option C, ethanedioate ( en -like or ox²⁻ ) is bidentate. Even if fewer ligands bind, each donates two lone pairs, resulting in 6 coordination sites and an octahedral shape.

Mark Scheme Breakdown: D (AO3) = 1 Mark. Awarded for correctly selecting the reaction between FeCl₃(aq) and concentrated HCl(aq) that produces the 4-coordinate tetrahedral [FeCl₄]⁻ complex.

AQA A-Level Chemistry Paper 3, June 2022: Question 19

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