Edexcel A-Level Chemistry Paper 3, June 2017: Question 1

✅ Correct Answer

Cr(OH)₃(s) + 3H⁺(aq) → [Cr(H₂O)₆]³⁺(aq)

Alternative accepted: Using 3H₃O⁺(aq) yielding 6H₂O(l) on the product side, or including water molecules coordinated in the reactant formula ( Cr(OH)₃(H₂O)₃ ).

💡 Key Knowledge

• Solid chromium(III) hydroxide acts as a base (amphoteric).
• In acidic conditions, protons coordinate to form the hexaaquachromium(III) complex ion.
• State symbols (s) and (aq) are strictly required for the mark.

❌ Common Errors

• Omitting or misplacing state symbols (e.g. leaving out (s) or (aq) ).
• Failing to balance charges or stoichiometry (forgetting the coefficient 3 in front of H⁺).
• Writing incorrect complex ion coordination numbers (chromium(III) typically forms hexacoordinated complexes).

✅ Correct Answer

1. Green solid / grey-green solid disappears (or dissolves/forms precipitate).

2. Forms a green solution.

🧠 Exam Technique

Always distinguish clearly between a solid/precipitate and a solution. Examiners penalize vague terms like "it turns green" without stating whether the solid dissolved or what state the resulting mixture is in.

✅ Correct Answer

Cr(OH)₃(s) + 3OH⁻(aq) → [Cr(OH)₆]³⁻(aq)

Alternative accepted: Cr(OH)₃(s) + 3OH⁻(aq) → [Cr(OH)₆]³⁻(aq) + 3H₂O(l) (if starting from hexaaqua formulation).

💡 Key Knowledge

Because chromium(III) hydroxide is amphoteric, it reacts with excess sodium hydroxide (alkali) to form a soluble complex anion, dissolving the precipitate.

❌ Common Errors

• Writing a lower coordination number like [Cr(OH)₄]⁻ without balancing the rest of the equation properly (though [Cr(OH)₄]⁻ is accepted if balanced with water molecules, [Cr(OH)₆]³⁻ is the standard A-Level expectation for excess alkali).

✅ Correct Answer

A green solution.

🧠 Exam Technique

Keep your description concise. The mark scheme accepts "green solution" and ignores colour qualifiers like 'dark' or 'pale', but you must explicitly state it is a solution (not a solid).

✅ Correct Answer

1. Blue solution initially forms a pale blue precipitate.

2. The precipitate dissolves in excess ammonia to form a dark / deep / royal blue solution.

💡 Key Knowledge

This is a two-stage process: first, dropwise addition causes a precipitation/deprotonation reaction forming copper(II) hydroxide. Second, excess ammonia causes a ligand substitution reaction where NH₃ replaces H₂O and OH⁻ ligands.

❌ Common Errors

Describing the precipitate as "dark blue" or referring to "blue crystals". The initial precipitate is pale blue, and the final solution is deep/royal blue.

✅ Correct Answer

[Cu(H₂O)₆]²⁺ + 4NH₃ → [Cu(NH₃)₄(H₂O)₂]²⁺ + 4H₂O

Note: State symbols are not required for this specific part as per the question stem.

💡 Key Knowledge

Ammonia acts as a neutral ligand (NH₃). Four ammonia molecules substitute four of the water molecules in the octahedral copper(II) complex, resulting in a complex with formula [Cu(NH₃)₄(H₂O)₂]²⁺ .

❌ Common Errors

• Forgetting to balance the equation (missing coefficients 4 for NH₃ and H₂O).
• Writing an incorrect formula for the ammine complex (e.g. forgetting the remaining coordinated water molecules in the tetrammine complex).