AQA A-Level Chemistry Paper 3, 2019: Question 18

✅ Correct Answer: D

The solution changes colour and no precipitate forms.

💡 Key Knowledge

The reaction occurring is a complete ligand substitution:

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

• Initial complex: [Cu(H₂O)₆]²⁺ is blue and octahedral (coordination number 6).
• Final complex: [CuCl₄]²⁻ is yellow-green (or yellow) and tetrahedral (coordination number 4).
• Both starting and product complexes are charged ions fully dissolved in water, meaning no solid precipitate ever forms.
• No redox or acid-carbonate reactions occur, so no gas is evolved.

🧠 Exam Technique: Systematic Elimination

• Check for gas evolution (Eliminates A & B): Adding acid to CuSO₄ does not involve a carbonate, sulfite, or redox couple that yields a gas. No effervescence can occur.
• Check for precipitate formation (Eliminates A & C): Adding OH⁻ or dilute NH₃ produces a Cu(OH)₂(H₂O)₄ precipitate. However, adding Cl⁻ undergoes a simple substitution forming the soluble complex ion [CuCl₄]²⁻. No precipitate forms at any stage.
• Conclusion: Only D correctly describes a homogeneous colour change (blue to green/yellow) with no precipitate or gas.

❌ Common Misconceptions & Traps

• Confusing Cl⁻ with NH₃ additions (Option C): With aqueous ammonia, copper(II) first forms a blue precipitate of Cu(OH)₂, which then redissolves in excess NH₃ to form deep blue [Cu(NH₃)₄(H₂O)₂]²⁺. Students often mistakenly apply this "precipitate then dissolve" pattern to concentrated HCl.
• Thinking green means an insoluble salt: When adding concentrated HCl dropwise, the solution passes through an intermediate green stage (a mixture of blue [Cu(H₂O)₆]²⁺ and yellow [CuCl₄]²⁻). Some students mistake this colour darkening for precipitation.
• Confusing with carbonate tests: Seeing an acid added to a solution tempts some students into guessing effervescence occurs (Options A and B).