Ligand Substitution Reactions (OCR A Level Chemistry A): Revision Note
Exam code: H432
Forming solutions
Ligand substitution (or ligand exchange) is when one ligand in a complex is replaced by another
Ligand substitution forms a new complex that is more stable than the original
The ligands in the original complex can be partially or entirely substituted by others
The substitution depends on the size of the ligands
For example, water ligands and ammonia ligands are a similar size
But, water ligands are smaller than chloride ion ligands
If the ligands are a similar size, there are no changes to the:
Coordination number
Geometry of the complex
If the ligands are a different size, there are changes to the:
Coordination number
Geometry of the complex
Substitution in copper(II) complexes
When a transition element ion is in solution, it can be assumed that it exists as a hexaaqua complex ion (i.e. it has six water ligands attached to it)
For example Cu2+(aq) is [Cu(H2O)6]2+(aq)
The [Cu(H2O)6]2+ (aq) complex ion is blue in colour
There are 3 substitution reactions to consider for copper(II) complexes:
Sodium hydroxide solution, NaOH (aq)
Ammonia solution, NH3 (aq)
Hydrochloric acid, HCl (aq)
NaOH substitution of copper(II) complexes
Dropwise addition of NaOH
Dropwise addition of sodium hydroxide solution produces a light blue precipitate
Partial ligand substitution occurs
Two water ligands are replaced by two hydroxide ligands:
[Cu(H2O)6]2+ (aq) + 2OH- → Cu(H2O)4(OH)2 (s) + 2H2O
pale blue solution → pale blue precipitate
Excess NaOH
There is no change with further addition of sodium hydroxide solution
NH3 substitution of copper(II) complexes
Dropwise addition of NH3
Dropwise addition of ammonia solution also produces a light blue precipitate
This is because the ammonia is in an aqueous solution
Again, this means that two water ligands are replaced by two hydroxide ligands:
[Cu(H2O)6]2+ (aq) + 2OH- → Cu(H2O)4(OH)2 (s) + 2H2O
pale blue solution → pale blue precipitate
Excess NH3
Upon addition of excess concentrated ammonia (NH3) solution, the pale blue precipitate dissolves to form a deep blue solution
Again, partial ligand substitution occurs
Two water ligands and two hydroxide ligands are replaced by four ammonia ligands:
Cu(H2O)4(OH)2 (s) + 4NH3 → [Cu(NH3)4(H2O)2]2+ (aq) + 2H2O + 2OH-
pale blue precipitate → dark blue solution
Ligand substitution summary
HCl substitution of copper(II) complexes
The water ligands in [Cu(H2O)6]2+ can also be substituted by chloride ligands, upon addition of concentrated hydrochloric acid (HCl)
Full ligand substitution occurs
All the water ligands are replaced by four chloride ion ligands:
[Cu(H2O)6]2+ (aq) + 4Cl- → [CuCl4]2- (aq) + 6H2O
blue solution → yellow solution
![Diagram showing colour change of copper solution from light blue, \([Cu(H_2O)_6]^{2+}\), to yellow, \([CuCl_4]^{2-}\), with an arrow indicating the reaction.](https://cdn.savemyexams.com/cdn-cgi/image/f=auto,width=3840/https://cdn.savemyexams.com/uploads/2025/06/22804_httpscdn-savemyexams-comuploads202110copperii-chloride-colour-change.png)
This ligand substiution causes:
A change in coordination number
This changes from 6 in [Cu(H2O)6]2+ (aq) to 4 in [CuCl4]2- (aq)
Since chloride ion ligands are larger than water ligands, only 4 will fit around the central metal ion
A change in geometry
This changes from octahedral in [Cu(H2O)6]2+ (aq) to tetrahedral in [CuCl4]2- (aq)
A colour change
The official colour change is from blue [Cu(H2O)6]2+ (aq) to yellow [CuCl4]2- (aq)
But, this is a reversible reaction
Therefore, some of the blue [Cu(H2O)6]2+ complex will still be present in the solution
So, the mixture of blue and yellow solutions in the reaction mixture may cause a green colour
Adding water to the solution will cause the chloride ligands to be displaced by the water molecules, and the [Cu(H2O)6]2+ (aq) ion and pale blue solution will return
Substitution in chromium(III) complexes
The [Cr(H2O)6]3+ (aq) complex ion is pale purple in colour
It is common for this to appear green, due to impurities
There are 2 substitution reactions to consider for chromium(III) complexes:
Sodium hydroxide solution, NaOH (aq)
Ammonia solution, NH3 (aq)
NaOH substitution of chromium(III) complexes
Dropwise addition of NaOH
Dropwise addition of sodium hydroxide solution produces a dark green precipitate
Partial ligand substitution occurs
Three water ligands are replaced by three hydroxide ligands:
[Cr(H2O)6]3+ (aq) + 3OH- → Cr(H2O)3(OH)3 (s) + 3H2O
pale purple solution → dark green precipitate
Excess NaOH
Upon addition of excess sodium hydroxide solution, the dark green precipitate dissolves to form a dark green solution
Full ligand substitution occurs
The 3 remaining water ligands are replaced by three hydroxide ligands:
Cr(H2O)3(OH)3 (s) + 6OH- → [Cr(OH)6]3- (aq) + 3H2O
dark green precipitate → dark green solution
NH3 substitution of chromium(III) complexes
Dropwise addition of NH3
Dropwise addition of ammonia solution produces a dark green precipitate
This is because the ammonia is in an aqueous solution
Partial ligand substitution occurs
Three water ligands are replaced by three hydroxide ligands:
[Cr(H2O)6]3+ (aq) + 3OH- → Cr(H2O)3(OH)3 (s) + 3H2O
pale purple solution → dark green precipitate
Excess NH3
Upon addition of excess concentrated ammonia (NH3) solution, the dark green precipitate dissolves to form a purple solution
Full ligand substitution occurs
All the water ligands are replaced by six ammonia ligands:
Cr(H2O)3(OH)3 (s) + 6NH3 → [Cr(NH3)6]3+ (aq) + 3H2O + 3OH-
dark green precipitate → purple solution
Examiner Tips and Tricks
For copper(II) complexes, you only need to know the ligand substitution reaction and accompanying colour change for the addition of excess ammonia
Be careful
If solid copper chloride (or any other metal) is dissolved in water, it forms the aqueous [Cu(H2O)6]2+ complex and not the chloride [CuCl4]2- complex!
Forming precipitates
When transition metal ions in aqueous solution react with aqueous sodium hydroxide and aqueous ammonia they form precipitates
However, some of these precipitates will dissolve in an excess of sodium hydroxide or ammonia to form complex ions in solution
Aqueous transition metal ions with sodium hydroxide
Transition metal ion | Metal-aqua ion | with OH- | with excess OH- |
|---|---|---|---|
Cu2+ | Pale blue solution | Pale blue precipitate Cu(H2O)4(OH)2 (s) | No change |
Fe2+ | Pale green solution | Dark green precipitate Fe(H2O)4(OH)2 (s) | No change |
Mn2+ | Pale pink solution | Pale brown precipitate Mn(H2O)4(OH)2 (s) | No change |
Cr3+ | Pale purple solution | Dark green precipitate Cr(H2O)3(OH)3 (s) | Dark green solution [Cr(OH)6]3- (aq) |
Fe3+ | Yellow solution | Orange-brown precipitate Fe(H2O)3(OH)3 (s) | No change |
Examples of ionic equations for the reactions in the table above
[Fe(H2O)6]2+ (aq) + 2OH- (aq) → [Fe(H2O)4(OH)2] (s) +2H2O (l)
[Cu(H2O)6]2+ (aq) + 2OH- (aq) → [Cu(H2O)4(OH)2] (s) +2H2O (l)
[Fe(H2O)6]3+ (aq) + 3OH- (aq) → [Fe(H2O)3(OH)3] (s) + 3H2O (l)
Aqueous transition metal ions with ammonia
Transition metal ion | Metal-aqua ion | with NH3 | with excess NH3 |
|---|---|---|---|
Cu2+ | Pale blue solution | Pale blue precipitate Cu(H2O)4(OH)2 (s) | Dark blue solution [Cu(NH3)4(H2O)2]2+ (aq) |
Fe2+ | Pale green solution | Dark green precipitate Fe(H2O)4(OH)2 (s) | No change |
Mn2+ | Pale pink solution | Pale brown precipitate Mn(H2O)4(OH)2 (s) | No change |
Cr3+ | Pale purple solution | Dark green precipitate Cr(H2O)3(OH)3 (s) | Purple solution [Cr(NH3)6]3+ (aq) |
Fe3+ | Yellow solution | Orange-brown precipitate Fe(H2O)3(OH)3 (s) | No change |
Examples of ionic equations for the reactions in the table above
[Fe(H2O)6]2+ (aq) + 2NH3 (aq) → [Fe(H2O)4(OH)2] (s) + 2NH4+ (aq)
[Cu(H2O)6]2+ (aq) + 2NH3 (aq) → [Cu(H2O)4(OH)2] (s) + 2NH4+ (aq)
[Fe(H2O)6]3+ (aq) + 3NH3 (aq) → [Fe(H2O)3(OH)3] (s) + 3NH4+ (aq)
[Cu(H2O)4(OH)2] (s) + 4NH3 (aq) → [Cu(H2O)2(NH3)4]2+ (aq) + 2H2O (l) + 2OH- (aq)
Solutions of metal aqua ions react as acids with aqueous ammonia, whilst some react further with excess ammonia
Initially, ammonia acts as a base to remove one H+ ion per ammonia molecule used
With excess ammonia, some metal ions undergo ligand substitution with NH3
Examiner Tips and Tricks
It is easiest to remember the formulas of the precipitates by remembering that the number of OH– ions substituted is the same as the value of the charge on the initial ion
All colours are aligned to the published OCR Colours of inorganic ions and complexes poster (opens in a new tab)
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