Ligand exchange and the chelate effectEdexcel International A Level Chemistry: Revision notes

In a ligand exchange (substitution) reaction one or more ligands in a complex ion are replaced by different ligands. Ligands form dative covalent bonds using lone pairs, so the metal ion is a Lewis acid and each ligand a Lewis base.

Water and ammonia are similar in size and both are uncharged, so replacing H₂O by NH₃ does not change the coordination number or the shape (octahedral stays octahedral). Chloride ions are larger, so fewer fit around the metal ion: the coordination number falls from 6 to 4 and the shape changes from octahedral to tetrahedral.

Most ligand exchanges are reversible, so the position of equilibrium depends on the concentrations of the ligands.

[Cu(H₂O)₆]²⁺ is pale blue. Adding a few drops of ammonia solution gives a pale blue precipitate. Ammonia acts as a Brønsted–Lowry base and removes H⁺ from water ligands:

[Cu(H₂O)₆]²⁺ + 2NH₃ → Cu(OH)₂(H₂O)₄ + 2NH₄⁺

Adding excess ammonia dissolves the precipitate to give a deep blue solution. Ammonia now acts as a ligand and replaces four water ligands:

Cu(OH)₂(H₂O)₄ + 4NH₃ → [Cu(NH₃)₄(H₂O)₂]²⁺ + 2H₂O + 2OH⁻

The two remaining water ligands stay because the ion remains octahedral.

Adding concentrated hydrochloric acid replaces water with chloride ligands:

[Cu(H₂O)₆]²⁺ + 4Cl⁻ ⇌ [CuCl₄]²⁻ + 6H₂O (blue to yellow; a mixture looks green)

[Co(H₂O)₆]²⁺ + 4Cl⁻ ⇌ [CoCl₄]²⁻ + 6H₂O (pink to blue)

Both products are tetrahedral and the coordination number falls from 6 to 4, because Cl⁻ ions are bigger than H₂O molecules. Adding water reverses each change by shifting the equilibrium to the left, so the original colour returns.

A monodentate ligand forms one dative bond (H₂O, NH₃, Cl⁻). A bidentate ligand forms two, for example 1,2-diaminoethane (en) and ethanedioate. EDTA⁴⁻ is hexadentate and forms six.

Substituting monodentate ligands with a bidentate or hexadentate ligand gives a more stable complex. This is the chelate effect. For example:

[Ni(NH₃)₆]²⁺ + 3en → [Ni(en)₃]²⁺ + 6NH₃ (4 particles → 7 particles)

[Cu(H₂O)₆]²⁺ + EDTA⁴⁻ → [Cu(EDTA)]²⁻ + 6H₂O (2 particles → 7 particles)

The number of particles increases, so ΔS_system is positive (greater disorder). The number and type of dative bonds are similar, so ΔH is about zero, and the positive ΔS_system makes ΔS_total more positive, so the product is favoured.

[Ni(NH₃)₆]²⁺ + 3en → [Ni(en)₃]²⁺ + 6NH₃ has ΔH = −12 kJ mol⁻¹ and ΔS_system = +145 J K⁻¹ mol⁻¹ at 298 K.

ΔS_surroundings = −ΔH/T = 12000 ÷ 298 = +40.3 J K⁻¹ mol⁻¹

ΔS_total = ΔS_system + ΔS_surroundings = 145 + 40.3 = +185 J K⁻¹ mol⁻¹

ΔS_total is positive, so the reaction is feasible. Remember to convert ΔH from kJ to J before dividing by T.

A typical preparation is tetraamminecopper(II) sulfate monohydrate, [Cu(NH₃)₄]SO₄·H₂O.

1. Dissolve copper(II) sulfate in a little water and add concentrated ammonia solution until the deep blue solution forms.
2. Add ethanol. The complex salt is less soluble in ethanol, so deep blue crystals form.
3. Filter under reduced pressure, wash with a little ethanol to remove soluble impurities without dissolving the product, and dry between filter papers.

Do not heat the crystals, because the complex loses ammonia. Percentage yield = actual mass ÷ theoretical mass × 100.