Topic 17 · page 14 of 20

Ligand exchange and colour

Replacing water ligands by ammonia or by chloride ions changes the colour. Ammonia keeps six ligands round copper; chloride gives four, so the shape changes too.

11 min read17.24(i)17.24(ii)17.24(iii)

Read first: Eight ions with ammonia

Copper(II) with ammonia 17.24(i)

Add ammonia solution to blue copper(II) sulfate solution a little at a time.

  1. Pale blue solution of [Cu(H₂O)₆]²⁺.
  2. A few drops: a pale blue precipitate of [Cu(OH)₂(H₂O)₄]. Ammonia acts as a base (deprotonation).
  3. Excess: the precipitate dissolves to a deep blue solution of [Cu(NH₃)₄(H₂O)₂]²⁺. Ammonia acts as a ligand.
[Cu(HX2O)X6]X2+→[Cu(OH)X2(HX2O)X4]→[Cu(NHX3)X4(HX2O)X2]X2+ \begin{aligned} &\ce{[Cu(H2O)6]^2+ -> [Cu(OH)2(H2O)4]} \\ &\ce{-> [Cu(NH3)4(H2O)2]^2+} \end{aligned}

The last step is ligand exchange: four NH₃ ligands replace the two OH⁻ ligands and two of the four H₂O ligands. Starting from the aqua ion, the overall change is that four H₂O are replaced by four NH₃:

[Cu(HX2O)X6]X2++4 NHX3→[Cu(NHX3)X4(HX2O)X2]X2++4 HX2O \ce{[Cu(H2O)6]^2+ + 4NH3 -> [Cu(NH3)4(H2O)2]^2+ + 4H2O}

The coordination number stays 6 and the shape stays octahedral. Water and ammonia are similar in size and both uncharged, so the exchange does not change the number of ligands that fit. The colour changes from pale blue to deep blue because the new ligands split the d-orbitals by a different amount.

Copper(II) and cobalt(II) with chloride 17.24(ii)17.24(iii)

Chloride ions are larger than water molecules and carry a charge. Add concentrated hydrochloric acid and they replace the water ligands. Only four fit round the metal ion, so the coordination number falls from 6 to 4 and the shape changes from octahedral to tetrahedral.

Copper(II). Pale blue [Cu(H₂O)₆]²⁺ turns green and then yellow as [CuCl₄]²⁻ forms:

[Cu(HX2O)X6]X2++4 ClX−⇌[CuClX4]X2−+6 HX2O \ce{[Cu(H2O)6]^2+ + 4Cl- <=> [CuCl4]^2- + 6H2O}

The green is a mixture of blue [Cu(H₂O)₆]²⁺ and yellow [CuCl₄]²⁻.

Cobalt(II). Pink [Co(H₂O)₆]²⁺ turns blue as [CoCl₄]²⁻ forms:

[Co(HX2O)X6]X2++4 ClX−⇌[CoClX4]X2−+6 HX2O \ce{[Co(H2O)6]^2+ + 4Cl- <=> [CoCl4]^2- + 6H2O}

Both are equilibria. Adding water dilutes the chloride and shifts the position back to the left, to the octahedral aqua ion and its original colour.

Three rows of test tubes with arrows. Row one, copper(II) with ammonia: pale blue solution, then a pale blue precipitate with a little ammonia, then deep blue solution with excess. Row two, copper(II) with concentrated HCl: blue, then green mixture, then yellow. Row three, cobalt(II) with concentrated HCl: pink, then blue.
Ligand exchange in three reactions. The colours are schematic.
Reaction Ligand change Coordination number Shape Colour change
Cu²⁺ with excess NH₃ 4 H₂O → 4 NH₃ 6 → 6 octahedral → octahedral pale blue → deep blue
Cu²⁺ with conc. HCl 6 H₂O → 4 Cl⁻ 6 → 4 octahedral → tetrahedral blue → yellow
Co²⁺ with conc. HCl 6 H₂O → 4 Cl⁻ 6 → 4 octahedral → tetrahedral pink → blue

PracticalTesting ligand exchange with chloride

  1. Put a few cm³ of aqueous copper(II) sulfate into a test tube. Note the colour.
  2. In a fume cupboard, add concentrated hydrochloric acid a few drops at a time, shaking after each addition. Note the colours.
  3. Add water slowly. Note what happens to the colour.

Concentrated hydrochloric acid is corrosive and gives off fumes: wear eye protection and gloves.

Check yourself

  1. Describe what you would see when ammonia solution is added dropwise to copper(II) sulfate solution until in excess.
  2. Write the equation for the formation of [Cu(NH₃)₄(H₂O)₂]²⁺ from [Cu(H₂O)₆]²⁺.
  3. Write the equation for the formation of [CoCl₄]²⁻ from [Co(H₂O)₆]²⁺ and state the colour change.
  4. Give the change in coordination number and shape in question 3. Why does it happen?
  5. What would you see on adding a large volume of water to blue [CoCl₄]²⁻ solution, and why?
  6. Explain why the mixture of the copper(II) aqua and chloro complexes looks green.

Answers cover these until you have tried the questions

  1. A pale blue precipitate forms first. In excess ammonia it dissolves to give a deep blue solution.
  2. [Cu(HX2O)X6]X2++4 NHX3→[Cu(NHX3)X4(HX2O)X2]X2++4 HX2O\ce{[Cu(H2O)6]^2+ + 4NH3 -> [Cu(NH3)4(H2O)2]^2+ + 4H2O}
  3. [Co(HX2O)X6]X2++4 ClX−⇌[CoClX4]X2−+6 HX2O\ce{[Co(H2O)6]^2+ + 4Cl- <=> [CoCl4]^2- + 6H2O}. Pink to blue.
  4. Coordination number 6 to 4, octahedral to tetrahedral. Chloride ions are relatively large, so only four fit round the cobalt ion.
  5. The solution turns pink again. Water dilutes the chloride ions, so the equilibrium shifts to the left and [Co(H₂O)₆]²⁺ re-forms.
  6. The green is a mixture of the blue [Cu(H₂O)₆]²⁺ and the yellow [CuCl₄]²⁻. A mixture of blue and yellow looks green.