# Aqueous copper and how it reacts with NH3

I have learnt that you can test whether an unknown metal is copper by doing a simple test tube reaction. The reaction involves adding aqueous $$\ce{NH3}$$ to the test tube which contains the copper.

The $$\ce{NH3}$$ as it is dissolved in water will form $$\ce{NH4+}$$ ions and $$\ce{OH-}$$ ions. The $$\ce{OH-}$$ ions react with the copper forming a precipitate $$\ce{([Cu(H2O)4(OH)2])}$$ but many other metals form a precipitate with $$\ce{NH3}$$ as well such as $$\ce{Fe^2+}$$, $$\ce{Fe^3+}$$ etc. So to make sure that it is copper present in the test tube you add an excess of $$\ce{NH3}$$ as this causes a ligand exchange reaction to occur producing $$\ce{[Cu(H2O)2(NH3)4]2+}$$.

Why doesn't this occur with the other metals? Why doesn't this occur for say $$\ce{Fe^2+}$$ to form $$\ce{[Fe(H2O)2(NH3)4]2+}$$ which would also be soluble?

• I think the formation of a blue precipitate is already an evidence that the metal is copper. Further addition of ammonia should be unnecessary. – mck Feb 8 at 2:45

## 1 Answer

[Fe(H2O)6]2+ is a high-spin d6 complex, while [Fe(H2O)2(NH3)4]2+ is a low-spin d6 complex. The ligand substitution reaction is not favourable in terms of spin, so aqueous Fe2+ does not react with excess ammonia.

However, both [Cu(H2O)6]2+ and [Cu(H2O)2(NH3)4]2+ are d9 complexes. Their spins are the same, and being a stronger field ligand, NH3 increases the crystal field stabilization energy, so ligand substitution is favourable.