Ligand exchange and preparing a complex (Core Practical 14)

Ligand exchange (spec 17.24)

• Ammonia with copper(II): pale blue [Cu(H2O)6]2+ -> blue precipitate Cu(OH)2(H2O)4 -> deep blue [Cu(NH3)4(H2O)2]2+ in excess. Overall: [Cu(H2O)6]2+ + 4NH3 -> [Cu(NH3)4(H2O)2]2+ + 4H2O. The coordination number stays 6.
• Chloride with copper(II): add concentrated hydrochloric acid: [Cu(H2O)6]2+ + 4Cl- ⇌ [CuCl4]2- + 6H2O. Blue -> (green mixture) -> yellow. Octahedral (6) becomes tetrahedral (4).
• Chloride with cobalt(II): [Co(H2O)6]2+ + 4Cl- ⇌ [CoCl4]2- + 6H2O. Pink -> blue. Adding water reverses it.

Core Practical 14 (spec 17.33)

One option is tetraamminecopper(II) sulfate: dissolve copper(II) sulfate crystals in a little water, add concentrated ammonia (in a fume cupboard) until the precipitate redissolves to a deep blue solution, then add ethanol, in which the complex is less soluble, so it crystallises. Filter under reduced pressure, wash with a little ethanol and dry. Another option is iron(II) ethanedioate, FeC2O4.2H2O, from acidified iron(II) ammonium sulfate and ethanedioic acid.

Calculate the theoretical yield from the limiting reagent and the percentage yield. Losses come from crystals left in solution and on the apparatus.

Worked example

2.50 g of CuSO4.5H2O gives 1.80 g of [Cu(NH3)4]SO4.H2O. Calculate the percentage yield. (Cu = 63.5, S = 32.1, O = 16.0, H = 1.0, N = 14.0)

1. M(CuSO4.5H2O) = 249.6; n = 2.50 / 249.6 = 1.002 x 10^-2 mol
2. M([Cu(NH3)4]SO4.H2O) = 245.6; theoretical mass = 2.460 g
3. % yield = 1.80 / theoretical x 100

Answer: 73.2%

Check yourself

1. What colour is [CuCl4]2-?
   • Deep blue
   • Yellow
   • Pink
   • Violet
   Show answer

   Yellow. Adding concentrated HCl to blue copper(II) solution gives green then yellow.

2. Which equation shows the reaction of aqueous cobalt(II) ions with concentrated hydrochloric acid?
   • [Co(H2O)6]2+ + 6Cl- ⇌ [CoCl6]4- + 6H2O
   • [Co(H2O)6]2+ + 2Cl- ⇌ CoCl2 + 6H2O
   • [Co(H2O)6]3+ + 4Cl- ⇌ [CoCl4]- + 6H2O
   • [Co(H2O)6]2+ + 4Cl- ⇌ [CoCl4]2- + 6H2O
   Show answer

   [Co(H2O)6]2+ + 4Cl- ⇌ [CoCl4]2- + 6H2O. Four large Cl- ligands replace six water ligands.

3. Water is added to a blue solution of [CoCl4]2-. What happens?
   • It turns deep blue
   • It turns pink as the equilibrium shifts back to [Co(H2O)6]2+
   • A precipitate forms
   • No change
   Show answer

   It turns pink as the equilibrium shifts back to [Co(H2O)6]2+. Adding water shifts [Co(H2O)6]2+ + 4Cl- ⇌ [CoCl4]2- + 6H2O to the left.

4. What is the coordination number of copper in [Cu(NH3)4(H2O)2]2+?
   • 4
   • 2
   • 6
   • 8
   Show answer

   6. Four NH3 and two H2O ligands, all monodentate.

5. In Core Practical 14, 2.50 g of CuSO4.5H2O could give at most 2.46 g of [Cu(NH3)4]SO4.H2O. A student obtains 1.85 g. Calculate the percentage yield.
   Show answer

   75.2 %. 1.85 / 2.46 x 100 = 75.2%.

6. In the preparation of tetraamminecopper(II) sulfate, why is ethanol added?
   • Ethanol is a ligand
   • Ethanol oxidises the copper
   • The complex is less soluble in ethanol, so it crystallises out
   • Ethanol neutralises the ammonia
   Show answer

   The complex is less soluble in ethanol, so it crystallises out. Lowering the solubility makes the crystals form.

7. How do the coordination number and shape change when [Cu(H2O)6]2+ forms [CuCl4]2-?
   • 6 to 4, octahedral to tetrahedral
   • 6 to 6, no change of shape
   • 4 to 6, tetrahedral to octahedral
   • 6 to 4, octahedral to square planar
   Show answer

   6 to 4, octahedral to tetrahedral. Chloride ligands are larger, so only four fit.

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Sources

• Pearson Edexcel IAL Chemistry (2018) specification
• Pearson Edexcel IAL Chemistry (2018) scheme of work
• Pearson IAL Chemistry teacher practical guide

Last updated 7 November 2026. Spotted a mistake? Tell us.