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 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.