Substitution Reactions (Oxford AQA International A Level Chemistry)
Revision Note
Written by: Richard Boole
Reviewed by: Stewart Hird
Ligand Exchange
Ligand exchange, or substitution, is where one ligand in a complex is replaced by another
Ligand exchange forms a new complex that is more stable than the original one
During ligand exchange, ligands can undergo complete or incomplete substitution
This can sometimes lead to changes in:
The charge of the complex ion
The coordination number
The shape of the complex
Complete substitution of [Co(H2O)6]2+ (aq) by NH3
The [Co(H2O)6]2+(aq) complex ion is pink in colour
If ammonia solution is added to [Co(H2O)6]2+ (aq):
Incomplete substitution by hydroxide ions from the ammonia solution initially occurs forming a blue precipitate of Co(H2O)4(OH)2 (s)
Complete substitution by ammonia ligands forms a pale yellow / straw coloured solution of [Co(NH3)6 ]2+ (aq)
[Co(H2O)6]2+ (aq) + 6NH3 (aq) → [Co(NH3)6 ]2+ (aq) + 6H2O (l)
Complete ligand substitution of the water ligands by ammonia ligands has occurred
Since NH3 and H2O ligands are similar in size and uncharged, there is no change to:
The overall charge of the complex
The octahedral geometry of the complex
The co-ordination number of the complex
Examiner Tips and Tricks
If excess concentrated ammonia solution is added to [Co(H2O)6]2+, a brown solution will be formed
The ammonia ligands make the cobalt(II) ion so unstable that it readily gets oxidised in air to cobalt(III), [Co(NH3)6]3+ (aq)
Incomplete Ligand Substitution
There are examples where complete ligand substitution does not occur, examples include:
[Co(H2O)6]2+ (aq) with small amounts of dilute ammonia solution
[Cu(H2O)6]2+ (aq) with ammonia solution
[Co(H2O)6]2+ (aq), [Cu(H2O)6]2+ (aq) and [Fe(H2O)6]3+ (aq) with chloride ligands
Incomplete substitution of [Co(H2O)6]2+ (aq) by NH3
The [Co(H2O)6]2+(aq) complex is pink in colour
Initially, the dropwise addition of ammonia solution to [Co(H2O)6]2+(aq) forms a blue-green precipitate of Co(OH)2(H2O)4 (s)
The same reaction occurs with the addition of hydroxide ions
Further addition of hydroxide ions does not result in any further changes
Incomplete ligand substitution of two water ligands by two hydroxide (OH-) ligands has occurred
[Co(H2O)6]2+ (aq) + 2OH- (aq) → Co(OH)2(H2O)4 (s) + 2H2O (l)
The charge of the complex changes from +2 to 0 because two negative hydroxide ions have replaced two neutral water ligands
However, there is no change to:
The octahedral geometry of the complex
The co-ordination number of the complex
Incomplete substitution of [Cu(H2O)6]2+ (aq) by NH3
The [Cu(H2O)6]2+(aq) complex is blue in colour
Initially, the dropwise addition of ammonia solution to [Cu(H2O)6]2+(aq) forms a blue precipitate of Cu(OH)2(H2O)4 (s)
The same reaction occurs with the addition of hydroxide ions
Further addition of hydroxide ions does not result in any further changes
Incomplete ligand substitution of two water ligands by two hydroxide (OH-) ligands has occurred
[Cu(H2O)6]2+ (aq) + 2OH- (aq) → Cu(OH)2(H2O)4 (s) + 2H2O (l)
The charge of the complex changes from +2 to 0 because two negative hydroxide ions have replaced two neutral water ligands
However, there is no change to:
The octahedral geometry of the complex
The co-ordination number of the complex
Incomplete ligand exchange with copper(II)
Further addition of excess concentrated ammonia (NH3) solution, causes the pale blue Cu(OH)2(H2O)4 precipitate to dissolve to form a deep blue solution
Again, partial ligand substitution has occurred
[Cu(H2O)6]2+ (aq) + 4NH3 (aq) → [Cu(NH3)4(H2O)2]2+ (aq) + 2H2O (l) + 2OH– (aq)
There is no change to:
The overall charge of the complex
The octahedral geometry of the complex
The co-ordination number of the complex
Incomplete substitution by chloride ions
If the ligands undergoing substitution are of a different size, then:
The co-ordination number will change
The shape of the complex will change
Examples of different size ligands are water and the chloride ion
The chloride ion is larger than the water molecule, which means that less chloride ligands fit around the central metal ion
The chloride ion causes a change in the charge on the overall complex
The water ligands in [M(H2O)6]2+ (aq) and [M(H2O)6]3+ (aq) can also be substituted by chloride ligands, upon addition of concentrated hydrochloric acid (HCl)
[M(H2O)6]2+ (aq) + 4 Cl- (aq) → [MCl4]2- (aq) + 6H2O (l)
[M(H2O)6]3+ (aq) + 4 Cl- (aq) → [MCl4]- (aq) + 6H2O (l)
Chloride ions and hexaaqua cobalt(II)
The substitution of the water ligands causes the pink [Co(H2O)6]2+ solution to form a blue [CoCl4]2- (aq) solution
[Co(H2O)6]2+ (aq) + 4 Cl- (aq) → [CoCl4]2- (aq) + 6H2O (l)
The following changes occur:
The overall charge of the complex changes from +2 to -2
The shape of the complex changes from octahedral to tetrahedral
The co-ordination number of the complex changes from 6 to 4
Adding water to the solution:
Causes the chloride ligands to be displaced by the water ligands
So, the pink [Co(H2O)6]2+ (aq) solution will return
Chloride ions and hexaaqua copper(II)
The substitution of the water ligands causes the blue [Cu(H2O)6]2+ solution to form a yellow [CuCl4]2- (aq) solution
[Cu(H2O)6]2+ (aq) + 4 Cl- (aq) → [CuCl4]2- (aq) + 6H2O (l)
The following changes occur:
The overall charge of the complex changes from +2 to -2
The shape of the complex changes from octahedral to tetrahedral
The co-ordination number of the complex changes from 6 to 4
This is a reversible reaction, and some of the [Cu(H2O)6]2+ complex will still be present in the solution
The mixture of blue and yellow solutions in the reaction mixture will give it a green colour
Adding water to the solution:
Causes the chloride ligands to be displaced by the water ligands
So, the blue [Cu(H2O)6]2+ (aq) solution will return
Chloride ions and hexaaqua iron(III)
The substitution of the water ligands causes the yellow [Fe(H2O)6]3+ solution to form an orange [FeCl4]– (aq) solution
[Fe(H2O)6]3+ (aq) + 4 Cl- (aq) → [FeCl4]- (aq) + 6H2O (l)
The following changes occur:
The overall charge of the complex changes from +3 to -1
The shape of the complex changes from octahedral to tetrahedral
The co-ordination number of the complex changes from 6 to 4
Adding water to the solution:
Causes the chloride ligands to be displaced by the water ligands
So, the yellow [Fe(H2O)6]3+ (aq) solution will return
The Haem Complex
Haemoglobin is one of nature's complexes using a transition metal ion
The haem complex has:
An central metal iron(II) ion
A multidentate haem ligand
A square planar shape
Oxygen atoms form a dative covalent bond with the Fe(II) which enables oxygen molecules to be transported around the body in the blood
The haem complex structure
Oxygen molecules are not very good ligands and bond weakly to the iron(II)
The weak bonds allows them to break off easily and be transported into cells
The effect of carbon monoxide
Carbon monoxide is toxic because:
It is a better ligand than oxygen
So, it binds strongly and irreversibly to the iron(II)
This prevents oxygen from being carried to the cells
If oxygen attached to the haemoglobin (oxyhaemoglobin) is replaced by carbon monoxide (carboxyhaemoglobin), a darker red colour is produced in the haem complex
This is a sign of carbon monoxide poisoning
The Chelate Effect
The chelate effect is where the monodentate ligands in a complex are replaced by bidentate and multidentate ligands
It is an energetically favourable reaction, i.e. ΔGꝋ is negative
The driving force behind the reaction is entropy
The Gibbs equation reminds us of the link between enthalpy and entropy:
ΔGꝋ = ΔHreactionꝋ – TΔSsystemꝋ
Reactions in solution between aqueous ions usually come with relatively small enthalpy changes
However, the entropy changes are always positive in chelation
This is because chelation produces a net increase in the number of particles
A small enthalpy change and relative large positive entropy change generally ensures that the overall free energy change is negative
For example, when EDTA chelates with aqueous cobalt(II):
There are two reactants
There are seven products
This means that there is a large increase in entropy
[Co(H2O)6 ]2+ (aq) + EDTA4- (aq) → [CoEDTA]2- (aq) + 6H2O (l)
Examiner Tips and Tricks
Make sure you can explain the chelate effect in terms of the balance between entropy and enthalpy changes.
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