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A2 · Practice questions · Ligand exchange

Swap the ligands.

Six original Cambridge-style questions on ligand exchange and hydroxide precipitates.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.
Keep these straight

Ammonia, chloride, hydroxide.

01
Recall
[3 marks]

Describe what is seen, and write an equation for each stage, when aqueous ammonia is added dropwise until in excess to aqueous copper(II) sulfate.

  • Pale blue precipitate: [Cu(H2O)6]2+ + 2OH− → [Cu(OH)2(H2O)4] + 2H2O. ✓
  • The precipitate dissolves in excess to give a deep blue solution. ✓
  • [Cu(OH)2(H2O)4] + 4NH3 → [Cu(NH3)4(H2O)2]2+ + 2OH− + 2H2O. ✓
02
Explain
[3 marks]

Concentrated hydrochloric acid is added to aqueous cobalt(II) chloride. Describe the colour change and explain the change in shape of the complex.

  • Pink to blue. ✓
  • [Co(H2O)6]2+ + 4Cl− ⇌ [CoCl4]2− + 6H2O. ✓
  • Cl− ions are larger than water molecules, so only four fit: octahedral (6) becomes tetrahedral (4). ✓
03
Analysis
[3 marks]

Solutions X, Y and Z contain Fe2+, Fe3+ and Cr3+ (not in that order). Describe how aqueous NaOH alone would identify each.

  • Fe2+: green precipitate, insoluble in excess, turning brown at the surface on standing. ✓
  • Fe3+: red-brown precipitate, insoluble in excess. ✓
  • Cr3+: grey-green precipitate that dissolves in excess NaOH to give a dark green solution. ✓
04
Explain
[2 marks]

Explain why the coordination number of copper does not change when excess ammonia reacts with [Cu(H2O)6]2+.

  • NH3 and H2O are similar in size and both are neutral. ✓
  • Four NH3 replace four H2O, so there are still six ligands: [Cu(NH3)4(H2O)2]2+, octahedral. ✓
05
Recall
[2 marks]

Write an equation for the reaction of [Cr(OH)3(H2O)3] with excess hydroxide ions, and state the colour of the product.

  • [Cr(OH)3(H2O)3] + 3OH− → [Cr(OH)6]3− + 3H2O. ✓
  • Dark green solution. ✓
06
Analysis
[2 marks]

The yellow-green solution formed from copper(II) ions and concentrated HCl turns blue when diluted with water. Explain.

  • [Cu(H2O)6]2+ + 4Cl− ⇌ [CuCl4]2− + 6H2O is an equilibrium. ✓
  • Adding water (and lowering [Cl−]) shifts it to the left, re-forming blue [Cu(H2O)6]2+. ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.