Ligand Substitution Reactions

Ligand substitution (or ligand exchange) occurs when one ligand in a complex ion is replaced by another. This is a key topic in AQA 3.2.5. You need to know the specific examples involving cobalt(II), copper(II), chromium(III), and iron(III) complexes, and understand the role of stability constants and Le Chatelier's principle.

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General Principle

In a ligand substitution reaction, one type of ligand replaces another in the coordination sphere of the metal ion. The reaction can be represented as an equilibrium:

[M(H₂O)₆]^(n+)(aq) + 6L ⇌ [ML₆]^(n+)(aq) + 6H₂O(l)

The position of equilibrium depends on the relative strength of the bonds formed between the metal ion and the incoming ligand compared to the outgoing ligand.

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Key Substitution Reactions

Cobalt(II) Complexes

Water to chloride substitution:

[Co(H₂O)₆]²⁺(aq) + 4Cl⁻(aq) ⇌ [CoCl₄]²⁻(aq) + 6H₂O(l) Pink → Blue

• The coordination number changes from 6 to 4 (Cl⁻ is a larger ligand than H₂O).
• The shape changes from octahedral to tetrahedral.
• This reaction is endothermic in the forward direction.
• Adding concentrated HCl (which provides Cl⁻) shifts the equilibrium to the right (blue).
• Adding water dilutes the Cl⁻ concentration, shifting equilibrium to the left (pink).
• Heating favours the endothermic forward reaction (blue); cooling favours the reverse (pink).

Exam Tip: The cobalt(II) chloride test can be used as a thermochromic indicator — blue when hot, pink when cold. Cobalt(II) chloride paper turns from blue (anhydrous) to pink (hydrated) in the presence of water.

Copper(II) Complexes

Water to chloride substitution:

[Cu(H₂O)₆]²⁺(aq) + 4Cl⁻(aq) ⇌ [CuCl₄]²⁻(aq) + 6H₂O(l) Blue → Yellow/green

• Coordination number changes from 6 (octahedral) to 4 (tetrahedral).
• A mixture of the blue and yellow/green complexes often appears green.

Water to ammonia substitution (partial then full):

Step 1 — Adding a small amount of NH₃(aq):

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

• Ammonia is acting as a base (accepting H⁺ from coordinated water molecules).
• A pale blue precipitate of copper(II) hydroxide forms.

Step 2 — Adding excess NH₃(aq):

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

Or overall from the hexaaqua complex:

[Cu(H₂O)₆]²⁺(aq) + 4NH₃(aq) ⇌ [Cu(NH₃)₄(H₂O)₂]²⁺(aq) + 4H₂O(l) Pale blue → Deep blue (royal blue)

• Only 4 of the 6 water ligands are replaced by ammonia.
• The coordination number remains 6 (octahedral shape).
• The deep blue colour of [Cu(NH₃)₄(H₂O)₂]²⁺ is diagnostic for copper(II).

Chromium(III) Complexes

Water to ammonia substitution:

[Cr(H₂O)₆]³⁺(aq) + 6NH₃(aq) ⇌ [Cr(NH₃)₆]³⁺(aq) + 6H₂O(l)

• Excess ammonia replaces all 6 water ligands (NH₃ and H₂O are similar in size).
• Coordination number stays at 6 (octahedral).
• Green/violet [Cr(H₂O)₆]³⁺ → purple [Cr(NH₃)₆]³⁺.

Water to hydroxide (with dilute NaOH):

[Cr(H₂O)₆]³⁺(aq) + 3OH⁻(aq) → Cr(OH)₃(H₂O)₃ + 3H₂O(l)

• Green precipitate of chromium(III) hydroxide.
• With excess NaOH: Cr(OH)₃(H₂O)₃ + 3OH⁻(aq) → [Cr(OH)₆]³⁻(aq) + 3H₂O(l)
• The precipitate dissolves to form a green solution of [Cr(OH)₆]³⁻ (showing amphoteric behaviour of Cr(OH)₃).

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EDTA Substitution

EDTA⁴⁻ is a hexadentate ligand that can replace all six monodentate water ligands: