# Ligand substitution reaction in coordination compounds

I know that the following substitution reaction can take place:

$$\ce{[Cu(H2O)4]^2+ (aq) + 4NH3 (aq) -> [Cu(NH3)4]^2+ (aq) + 4H2O (l)}$$

This is because ammonia is a stronger ligand than water.

What can we say about the product formed from the following reaction?

$$\ce{[Cu(NH3)4]^2+ (aq) + excess HNO3 (aq) ->}$$

Nitric acid is a strong oxidizing agent but $$\ce{Cu}$$ can not be further oxidised. Is there a possibility of forming $$\ce{[Cu(H2O)4]^2+ (aq)}$$?

• Consider the fact that nitric acid is able to protonate the ammonia molecules, resulting in a change in the position of equilibrium of the ammonia-ammonium equilibrium when it is added. Then consider the effect on the ligand exchange equilibrium. – Tan Yong Boon Dec 2 '19 at 4:54

The reaction you wrote down is wrong on two counts. The reactant is not a hypothetical tetraaqua complex and the product is not a hypothetical tetraammin complex. The correct reaction is as shown below:

$$\ce{[Cu(H2O)6]^2+ (aq) + 4 NH3 (aq) <=> [Cu(NH3)4(H2O)2]^2+ (aq) + 4 H2O (l)}\tag{1}$$

Note that I have used an equilibrium arrow here: the reaction can proceed in both directions depending on the concentration of ammonia in your sample (in reality, there is also the competing precipitation of $$\ce{Cu(OH)2}$$ at slightly basic pH levels but I will ignore that here).

As an equilibrium, this can be pushed either way. So let’s take a look at what nitric acid can do. It is both an acid and an oxidising agent – but as you noted, copper(II) is already very highly oxidised and it is generally not possible to sustain copper(III) or higher copper oxidation states in aquaeous solution (it does exist as an intermediate in various copper-catalysed organic reactions). On the other hand, $$\ce{HNO3}$$ can very much act as an acid in your solution as there is a basic compound able to receive protons (see spoiler after figuring out the reaction yourself).

$$\ce{HNO3 (aq) + NH3 (aq) -> NH4+ (aq) + NO3- (aq)}\tag{2}$$

Note that while $$(2)$$ is also a reversible reaction, the equilibrium is far to the product side so we can essentially treat it as completed. Obviously, this has an effect on the original reaction – and with a simple application of Le Chatelier’s principle the resulting answer should be obvious.

Yes, since ammonia will be protonated to give ammonium, equilibrium $$(1)$$ will be shifted back to the reactant side, regenerating hexaaquacopper(II) $$\ce{[Cu(H2O)6]^2+}$$.

• In equation (1), why ammonia is replacing exactly 4 water molecules? – Apurvium Apr 6 '20 at 16:41
• @Apurvium I’m afraid the best answer I can give is that ‘that’s just how nature works’. Analysing the complex formed in aquaeous solution shows that it contains four ammine ligands and two water ligands. In liquid ammonia, a hexaamminecopper(II) complex can be synthesised but that is unstable in water and will degrade to the tetraammine complex (if the ammonia concentration is sufficiently high) or all the way to the aqua complex. – Jan Apr 8 '20 at 1:28