6.4 Ligand Substitution Reactions | The Transition Metals | Chemistry 12

Ligand substitution (also called ligand exchange) reactions occur when one or more ligands in a complex ion are replaced by different ligands. These reactions are a key feature of transition metal chemistry and often result in a change in colour and sometimes a change in coordination number.

What is Ligand Substitution?

In a ligand substitution reaction, the overall oxidation state of the central metal ion does not change. Only the identity of the surrounding ligands changes. The driving force is usually the formation of a more stable complex.

$[M (L)_{n}]^{x +} + m L^{'} \to [M (L)_{n - m} (L^{'})_{m}]^{x +} + m L$

Ligand Substitution in Copper(II) Complexes

1. $[C u (H_{2} O)_{6}]^{2 +}$ with Ammonia

When excess aqueous ammonia is added to the pale blue hexaaquacopper(II) ion, four water ligands are replaced by four ammonia ligands:

$[C u (H_{2} O)_{6}]^{2 +} + 4 N H_{3} \to [C u (N H_{3})_{4} (H_{2} O)_{2}]^{2 +} + 4 H_{2} O$

Complex	Formula	Colour
Hexaaquacopper(II)	$[C u (H_{2} O)_{6}]^{2 +}$	Pale blue
Tetraamminediaquacopper(II)	$[C u (N H_{3})_{4} (H_{2} O)_{2}]^{2 +}$	Deep blue

The deep blue colour is due to the stronger ligand field of $N H_{3}$ compared to $H_{2} O$ , which changes the energy gap $Δ E$ between the split d-orbitals.

2. $[C u (H_{2} O)_{6}]^{2 +}$ with Hydroxide Ions

When $N a O H (a q)$ is added to $[C u (H_{2} O)_{6}]^{2 +}$ , hydroxide ions replace water ligands:

$[C u (H_{2} O)_{6}]^{2 +} + 2 O H^{-} \to [C u (H_{2} O)_{4} (O H)_{2}] + 2 H_{2} O$

A pale blue precipitate of $[C u (H_{2} O)_{4} (O H)_{2}]$ forms. If excess $N H_{3}$ is then added, this precipitate dissolves to give the deep blue solution of $[C u (N H_{3})_{4} (H_{2} O)_{2}]^{2 +}$ .

3. $[C u (H_{2} O)_{6}]^{2 +}$ with Chloride Ions

When concentrated $H C l$ is added to $[C u (H_{2} O)_{6}]^{2 +}$ , chloride ions replace all six water ligands:

$[C u (H_{2} O)_{6}]^{2 +} + 4 C l^{-} \to [C u C l_{4}]^{2 -} + 6 H_{2} O$

Complex	Formula	Colour	Geometry
Hexaaquacopper(II)	$[C u (H_{2} O)_{6}]^{2 +}$	Pale blue	Octahedral
Tetrachlorocuprate(II)	$[C u C l_{4}]^{2 -}$	Yellow	Tetrahedral

Why does the coordination number change from 6 to 4? The $C l^{-}$ ion is much larger than $H_{2} O$ . Due to steric hindrance (ligand-ligand repulsion), only four $C l^{-}$ ions can fit around the $C u^{2 +}$ centre.

This reaction is reversible. Adding water shifts the equilibrium back to the blue $[C u (H_{2} O)_{6}]^{2 +}$ . The equilibrium mixture often appears green (a mixture of blue and yellow species).

Ligand Substitution in Cobalt(II) Complexes

1. $[C o (H_{2} O)_{6}]^{2 +}$ with Ammonia

When excess aqueous ammonia is added to the pink hexaaquacobalt(II) ion:

$[C o (H_{2} O)_{6}]^{2 +} + 6 N H_{3} \to [C o (N H_{3})_{6}]^{2 +} + 6 H_{2} O$

Complex	Formula	Colour
Hexaaquacobalt(II)	$[C o (H_{2} O)_{6}]^{2 +}$	Pink
Hexaamminecobalt(II)	$[C o (N H_{3})_{6}]^{2 +}$	Straw/yellow

2. $[C o (H_{2} O)_{6}]^{2 +}$ with Hydroxide Ions

When $N a O H (a q)$ is added to $[C o (H_{2} O)_{6}]^{2 +}$ :

$[C o (H_{2} O)_{6}]^{2 +} + 2 O H^{-} \to [C o (H_{2} O)_{4} (O H)_{2}] + 2 H_{2} O$

A blue-green precipitate of $[C o (H_{2} O)_{4} (O H)_{2}]$ forms.

3. $[C o (H_{2} O)_{6}]^{2 +}$ with Chloride Ions

When concentrated $H C l$ is added to $[C o (H_{2} O)_{6}]^{2 +}$ , the coordination number changes from 6 to 4:

$[C o (H_{2} O)_{6}]^{2 +} + 4 C l^{-} \to [C o C l_{4}]^{2 -} + 6 H_{2} O$

Complex	Formula	Colour	Geometry
Hexaaquacobalt(II)	$[C o (H_{2} O)_{6}]^{2 +}$	Pink	Octahedral
Tetrachlorocobaltate(II)	$[C o C l_{4}]^{2 -}$	Blue	Tetrahedral

Summary Table: Key Ligand Substitution Reactions

Metal Ion	Ligand Added	Product Complex	Colour Change
$C u^{2 +}$	$N H_{3}$ (excess)	$[C u (N H_{3})_{4} (H_{2} O)_{2}]^{2 +}$	Pale blue → Deep blue
$C u^{2 +}$	$O H^{-}$	$[C u (H_{2} O)_{4} (O H)_{2}]$	Pale blue → Pale blue precipitate
$C u^{2 +}$	$C l^{-}$ (conc.)	$[C u C l_{4}]^{2 -}$	Pale blue → Yellow
$C o^{2 +}$	$N H_{3}$ (excess)	$[C o (N H_{3})_{6}]^{2 +}$	Pink → Straw/yellow
$C o^{2 +}$	$O H^{-}$	$[C o (H_{2} O)_{4} (O H)_{2}]$	Pink → Blue-green precipitate
$C o^{2 +}$	$C l^{-}$ (conc.)	$[C o C l_{4}]^{2 -}$	Pink → Blue

What is Ligand Substitution?

$[M (L)_{n}]^{x +} + m L^{'} \to [M (L)_{n - m} (L^{'})_{m}]^{x +} + m L$

Ligand Substitution in Copper(II) Complexes

1. $[C u (H_{2} O)_{6}]^{2 +}$ with Ammonia

When excess aqueous ammonia is added to the pale blue hexaaquacopper(II) ion, four water ligands are replaced by four ammonia ligands:

$[C u (H_{2} O)_{6}]^{2 +} + 4 N H_{3} \to [C u (N H_{3})_{4} (H_{2} O)_{2}]^{2 +} + 4 H_{2} O$

Complex	Formula	Colour
Hexaaquacopper(II)	$[C u (H_{2} O)_{6}]^{2 +}$	Pale blue
Tetraamminediaquacopper(II)	$[C u (N H_{3})_{4} (H_{2} O)_{2}]^{2 +}$	Deep blue

The deep blue colour is due to the stronger ligand field of $N H_{3}$ compared to $H_{2} O$ , which changes the energy gap $Δ E$ between the split d-orbitals.

2. $[C u (H_{2} O)_{6}]^{2 +}$ with Hydroxide Ions

When $N a O H (a q)$ is added to $[C u (H_{2} O)_{6}]^{2 +}$ , hydroxide ions replace water ligands:

$[C u (H_{2} O)_{6}]^{2 +} + 2 O H^{-} \to [C u (H_{2} O)_{4} (O H)_{2}] + 2 H_{2} O$

3. $[C u (H_{2} O)_{6}]^{2 +}$ with Chloride Ions

When concentrated $H C l$ is added to $[C u (H_{2} O)_{6}]^{2 +}$ , chloride ions replace all six water ligands:

$[C u (H_{2} O)_{6}]^{2 +} + 4 C l^{-} \to [C u C l_{4}]^{2 -} + 6 H_{2} O$

Complex	Formula	Colour	Geometry
Hexaaquacopper(II)	$[C u (H_{2} O)_{6}]^{2 +}$	Pale blue	Octahedral
Tetrachlorocuprate(II)	$[C u C l_{4}]^{2 -}$	Yellow	Tetrahedral

Ligand Substitution in Cobalt(II) Complexes

1. $[C o (H_{2} O)_{6}]^{2 +}$ with Ammonia

When excess aqueous ammonia is added to the pink hexaaquacobalt(II) ion:

$[C o (H_{2} O)_{6}]^{2 +} + 6 N H_{3} \to [C o (N H_{3})_{6}]^{2 +} + 6 H_{2} O$

Complex	Formula	Colour
Hexaaquacobalt(II)	$[C o (H_{2} O)_{6}]^{2 +}$	Pink
Hexaamminecobalt(II)	$[C o (N H_{3})_{6}]^{2 +}$	Straw/yellow

2. $[C o (H_{2} O)_{6}]^{2 +}$ with Hydroxide Ions

When $N a O H (a q)$ is added to $[C o (H_{2} O)_{6}]^{2 +}$ :

$[C o (H_{2} O)_{6}]^{2 +} + 2 O H^{-} \to [C o (H_{2} O)_{4} (O H)_{2}] + 2 H_{2} O$

A blue-green precipitate of $[C o (H_{2} O)_{4} (O H)_{2}]$ forms.

3. $[C o (H_{2} O)_{6}]^{2 +}$ with Chloride Ions

When concentrated $H C l$ is added to $[C o (H_{2} O)_{6}]^{2 +}$ , the coordination number changes from 6 to 4:

$[C o (H_{2} O)_{6}]^{2 +} + 4 C l^{-} \to [C o C l_{4}]^{2 -} + 6 H_{2} O$

Complex	Formula	Colour	Geometry
Hexaaquacobalt(II)	$[C o (H_{2} O)_{6}]^{2 +}$	Pink	Octahedral
Tetrachlorocobaltate(II)	$[C o C l_{4}]^{2 -}$	Blue	Tetrahedral

Summary Table: Key Ligand Substitution Reactions

Metal Ion	Ligand Added	Product Complex	Colour Change
$C u^{2 +}$	$N H_{3}$ (excess)	$[C u (N H_{3})_{4} (H_{2} O)_{2}]^{2 +}$	Pale blue → Deep blue
$C u^{2 +}$	$O H^{-}$	$[C u (H_{2} O)_{4} (O H)_{2}]$	Pale blue → Pale blue precipitate
$C u^{2 +}$	$C l^{-}$ (conc.)	$[C u C l_{4}]^{2 -}$	Pale blue → Yellow
$C o^{2 +}$	$N H_{3}$ (excess)	$[C o (N H_{3})_{6}]^{2 +}$	Pink → Straw/yellow
$C o^{2 +}$	$O H^{-}$	$[C o (H_{2} O)_{4} (O H)_{2}]$	Pink → Blue-green precipitate
$C o^{2 +}$	$C l^{-}$ (conc.)	$[C o C l_{4}]^{2 -}$	Pink → Blue