AQA A-Level Chemistry Paper 3, June 2025: Question 23

1 mark · Medium difficulty · Multiple Choice

Identify which reaction produces an octahedral complex ion from the given options.

Practise this question

Question

Multiple-choice question asking: 'Which reaction produces an octahedral complex ion?' Four options are given: A: The addition of excess aqueous ammonia to a solution containing [Co(H2O)6]2+ ions; B: The addition of excess aqueous sodium carbonate to a solution containing [Fe(H2O)6]2+ ions; C: The addition of excess concentrated hydrochloric acid to a solution containing [Cu(H2O)6]2+ ions; D: The addition of excess aqueous sodium carbonate to a solution containing [Fe(H2O)6]3+ ions.
Question text

23 Which reaction produces an octahedral complex ion?

[1 mark]

The addition of excess aqueous ammonia to a solution containing

A 2+

[Co(H2O)6] ions.

The addition of excess aqueous sodium carbonate to a solution

B 2+

containing [Fe(H2O)6] ions.

The addition of excess concentrated hydrochloric acid to a solution

C 2+

containing [Cu(H2O)6] ions.

The addition of excess aqueous sodium carbonate to a solution

D 3+

containing [Fe(H2O)6] ions.

Mark scheme

Show the mark scheme Mark scheme table row for question 23, showing the correct answer is A, worth 1 mark (AO1), corresponding to 'The addition of excess aqueous ammonia to a solution containing [Co(H2O)6]2+ ions.'

23 A 1 (AO1) The addition of excess aqueous ammonia to a solution

containing [Co(H O) ]2+ ions.

How to answer it

AQA A-Level Chemistry • Paper 1 • Multiple Choice

Formation of Octahedral Complex Ions

What this question tests

Recall and understanding of transition metal reactions in aqueous solution (Topic 3.2.5 / 3.2.6): distinguishing between ligand substitution and precipitation, identifying coordination numbers (6 for octahedral vs 4 for tetrahedral), and recognising whether products are charged complex ions or neutral precipitates.

Question 23 Analysis

Identifying the reaction forming an octahedral complex ion

1 Mark [AO1] • Multiple Choice

✅ Correct Answer: Option A

Reaction: Addition of excess aqueous ammonia to [Co(H₂O)₆]²⁺

[Co(H₂O)₆]²⁺ + 6NH₃ → [Co(NH₃)₆]²⁺ + 6H₂O

  • Shape: Octahedral (coordination number = 6).
  • Nature of product: Soluble complex ion (charge of 2+).
  • Ammonia replaces water ligands completely in excess to form hexaamminecobalt(II).

💡 Key Knowledge: Shapes & Ligands

  • Octahedral (Coordination No. = 6): Formed with small, neutral unidentate ligands like H₂O and NH₃, or bidentate ligands (e.g. en, C₂O₄²⁻).
  • Tetrahedral (Coordination No. = 4): Formed with larger, negatively charged ligands like Cl⁻ due to ligand-ligand electrostatic repulsion and steric hindrance.
  • Complex Ion vs Precipitate: A complex ion must carry an overall charge (e.g. [Co(NH₃)₆]²⁺, [CuCl₄]²⁻). Neutral species like FeCO₃ or [Fe(H₂O)₃(OH)₃] are insoluble solid precipitates.

Option-by-Option Breakdown

Why the distractors are incorrect

❌ Why Option B is Incorrect

[Fe(H₂O)₆]²⁺ + CO₃²⁻ → FeCO₃(s) + 6H₂O

With Fe²⁺ (a 2+ aqua ion), carbonate ions cause simple precipitation of insoluble iron(II) carbonate, FeCO₃(s) . This forms a green precipitate, not a complex ion.

❌ Why Option C is Incorrect

[Cu(H₂O)₆]²⁺ + 4Cl⁻ ⇌ [CuCl₄]²⁻ + 6H₂O

This reaction forms a complex ion ( [CuCl₄]²⁻ ), but because chloride ligands are large and charged, only 4 can fit around the copper ion. The shape is tetrahedral, not octahedral.

❌ Why Option D is Incorrect

2[Fe(H₂O)₆]³⁺ + 3CO₃²⁻ → 2[Fe(H₂O)₃(OH)₃](s) + 3CO₂ + 3H₂O

Because Fe³⁺ is more polarising (higher charge density), it acts as an acid. The reaction produces a brown precipitate of neutral [Fe(H₂O)₃(OH)₃] and CO₂ gas. It is not an ion.

Exam Technique & Examiner Tips

🧠 Exam Technique: Two-Step Screening

When an exam question specifies an octahedral complex ion, filter the options using two distinct criteria:

  1. Is it an ion? Eliminate options that produce neutral precipitates (e.g. metal carbonates or neutral metal hydroxides like FeCO₃ and Fe(H₂O)₃(OH)₃).
  2. Is it octahedral? Check the coordination number. If 4 large ligands like Cl⁻ are substituted, the shape is tetrahedral.

❌ Common Pitfalls to Avoid

  • Confusing Fe²⁺ and Fe³⁺ with CO₃²⁻: Remember that M²⁺ ions simply precipitate MCO₃, while M³⁺ ions undergo hydrolysis to give M(H₂O)₃(OH)₃ and CO₂ gas. Neither gives a complex ion.
  • Ignoring the word "ion": Many students correctly identify that Fe(H₂O)₃(OH)₃ has 6-coordinate iron, but forget that it has no charge (neutral precipitate), so it fails the "complex ion" criterion.
  • Forgetting coordination change with conc. HCl: Addition of excess conc. HCl always changes coordination number from 6 to 4 with Cu²⁺ and Co²⁺.

Topics

Inorganic Chemistry · 3.2.5 Transition Metals · 3.2.6 Reactions of Ions in Aqueous Solution

Question and mark scheme from the AQA A-Level Chemistry examination, Paper 3, June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.