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?

  • $\begingroup$ I think the formation of a blue precipitate is already an evidence that the metal is copper. Further addition of ammonia should be unnecessary. $\endgroup$ – mck Feb 8 '19 at 2:45

[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.

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