In this lesson, we will learn:
- The expected observations for common ligand exchange reactions and how to write their equations.
- To understand why coordination number changes in some ligand exchange reactions.
- To apply the idea of entropy in predicting reactions involving multidentate ligands.
- In the previous lesson Ligands and complex ions, we finished by looking at carbon monoxide – how, because of its stronger interactions, it can displace O2 in the haemoglobin in our blood. That’s what makes it toxic humans.
This is an example of a ligand exchange which is a very common type of reaction in transition metal chemistry. This lesson looks at some common ligand exchanges with several transition metal ions and the expected observations.
The most common ligand exchange reactions involve metal hexaaqua complexes reacting with common molecules such as ammonia and chloride (from concentrated HCl as the source).
These simple ligand exchange reactions are often reversible. With this said, recall Le Chatelier’s principle; if you add more of the reagents on one side of the reaction, the equilibrium will shift to the other side to rebalance conditions.
- For reactions with chloride ions:
- A good example is copper (II) hexaaqua. This blue solution will turn a green colour when concentrated hydrochloric acid is added.
[Cu(H2O)6]2+ + 4 Cl- [CuCl4]2- + 6 H2O
If you want the reactants back, add more water. The above equilibrium responds by shifting to the left, using the water up and producing more copper hexaaqua at the expense of the [CuCl4]2- and its green colour.
- Cobalt (II) hexaaqua is another example of this reversible ligand exchange with chloride. It is a pale pink solution which turns blue when concentrated hydrochloric acid is added.
[Co(H2O)6]2+ + 4 Cl- [CoCl4]2- + 6 H2O
Like with copper, this is a reversible process; add water and you will shift the equilibrium back to the reactants, which shifts the colour from blue back to pink.
As you can see in the product formulae, the chloride ligand exchange causes a change in coordination number – only four chloride ligands fit around a metal centre where six H2O molecules were able to before.
- Reactions with ammonia are more complicated than those with chloride because ammonia reacts with complex ions in two ways:
- In lower concentrations it will act as a base, removing H+ from H2O ligands and converting them into OH- ligands. This change from neutral to charged ligands means the complex ion will change overall charge. If it is neutral, it will precipitate out as an uncharged complex. The precipitate is easily noticed as it will cloud up the test tube.
- When ammonia is added in excess, a ligand exchange reaction occurs where either four or all six H2O/OH- ligands on the complex ion get replaced.
Because adding excessive amounts of ammonia starts out with it not being excessive amounts, the ligand exchange reaction precipitates first, then redissolves for the product:
- With copper hexaaqua as an example, a neutral copper (II) hydroxide precipitate forms with a small addition of ammonia:
[Cu(H2O)6]2+ (aq) + 2NH3 (aq)→
[Cu(H2O)4(OH)2] (s) + 2NH4+ (aq)
This neutral copper precipitate is also the product when strong bases like sodium hydroxide are added to copper hexaaqua.
When excess ammonia is added, the precipitate re-dissolves as ammonia replaces the OH- and H2O ligands – this is the ligand exchange. The overall change is:
[Cu(H2O)6]2+ (aq) + 4NH3 (aq)→
[Cu(NH3)4(H2O)2]2+ (aq) + 4H2O (l)
This changes the complex from a sky blue (hexaaqua) to a blue precipitate, then a dark blue colour (the ammonia ligand exchange product).
- The reaction with chromium (III) hexaaqua is slightly different to copper as the ligand exchange replaces all six water ligands in chromium, unlike the four in copper. Otherwise, the same two steps appear.
First, ammonia deprotonates three water ligands to give a precipitate, a neutral chromium (III) hydroxide complex.
In excess ammonia, the precipitate redissolves and a ligand exchange occurs where all six water/hydroxide ligands are replaced to give chromium (III) hexaammine:
[Cr(H2O)6]3+ (aq) + 6NH3 (aq) → [Cr(NH3)6]3+ (aq) + 6H2O (l)
This takes the chromium complex from being a blue-gray solution through a precipitate, which redissolves on excess ammonia for a violet solution which is the hexamine.
- The reaction with cobalt (II) hexaaqua is similar to chromium.
Chromium (II) hexaaqua is a pale pink solution which, in small amounts of ammonia, forms a neutral cobalt (II) precipitate.
[Co(H2O)6]2+ (aq) + 2NH3 (aq) → [Co(H2O)4(OH)2]
(s) + 2H2O (l)
In excess, all six water ligands are replaced by ammonia.
[Co(H2O)6]2+ (aq) + 6NH3 (aq) → [Co(NH3)6]2+ (aq) + 6H2O (l)
The cobalt (II) hexaammine product is a pale brown, straw colour.
- As said above, the size of the chloride ligand means ligand exchanges with it cause a decrease in coordination number. A metal centre can fit six water ligands around it, but only four chlorides.
This causes an increase in entropy of the surroundings, so entropy favours the products as more molecules are there than in the reactants.
Reactions with multidentate ligands are also driven by entropy, favouring the products. Substituting monodentate ligands for multidentate ones will increase the overall system. See the reaction with diethylamine, or ethylenediamine below:
Because only four molecules are present in the reactants but seven are produced, there is a large increase in entropy of the surroundings. We would say then, that the products are entropically favoured.
We can also measure the stability of transition metal complexes by their stability constant, which is really just an equilibrium constant for the products of a ligand exchange process. Complexes containing multidentate ligands are consistently higher Kstab than that of only monodentate ligands.
This is the chelate effect in action; complexes with multidentate ligands (chelates) are just more stable than monodentate ligand complexes.