$\ce{CuSO4}$ when reacts with $\ce{KCN}$ forms $\ce{CuCN}$ while it is insoluble in $\ce{H2O}$. It is soluble in excess of $\ce{KCN}$ due to the formation of the following complex:

a) $\ce{K2[Cu(CN)4]}$
b) $\ce{K3[Cu(CN)4]}$
c) $\ce{CuCN2}$
d) none

Answer given is b) but we know that $\ce{Cu^2+}$ is more stable hence anwser should be a), shouldn’t it?

  • 2
    $\begingroup$ Is the question title even appropriate? Where's the tautomerism? $\endgroup$
    – Varun
    Feb 8, 2016 at 16:21
  • $\begingroup$ Edited the title to be appropriate, titles should be short descriptions of problems. @Varun Edit your answer, you could have edited the title too. $\endgroup$
    – Mithoron
    Feb 8, 2016 at 19:58
  • $\begingroup$ True... just wanted OP's opinion. Will edit my answer shortly. Thanks for the response. $\endgroup$
    – Varun
    Feb 9, 2016 at 3:48
  • $\begingroup$ There are actually many complexing agents that stabilize copper(I) over copper(II). For another example see here. $\endgroup$ May 28, 2020 at 1:58

2 Answers 2


but we know that $\ce{Cu^2+}$ is more stable hence anwser should be a).

Ah, but do we? I hope you don’t ‘know’ that, because that ‘knowledge’ would defy years of chemical research.

The stability of an oxidation state, especially that of a transition metal, cannot be deduced easily or by a set of memorised rules. Many a time, a species will only be stable in one environment and unstable in another one. Take copper: In aquaeous solutions, copper(I) is usually unstable concerning oxidation to copper(II). However, copper(II) on the other hand is unstable in the presence of iodide and the two will react:

$$\ce{2Cu^2+ + 4I- -> I2 + 2CuI v}$$

So copper(I) is more stable in the presence of iodide while copper(II) is more stable in the presence of water. Notice a pattern? I don’t.

In your question specifically, the first step is a redox reaction between copper(II) and cyanide according to the following equation:

$$\ce{2 Cu^2+ + 4 CN- -> NCCN ^ + 2 CuCN v}$$

At least I notice a pattern here: Cyanide behaves much like a halide or the iodide introduced above.

By this reaction, we have established that copper(I) is more stable in the presence of cyanide than copper(II). So by adding more cyanide there is no way that we can reverse the reaction — not only because there is nothing that copper could reduce with its extra electron. So the redissolution and complex formation cannot be a redox reaction, it must be a simple complexation reaction which only allows for the correct solution to be:

$$\ce{CuCN (s) + 3 KCN (aq) <=> K3[Cu(CN)4] (aq)}$$


Copper (II) ions are generally more stable than Copper(I) ions in aqueous medium (however not in cases of iodides and bromides where it is reduced to copper (I)). But this does not hold good for the corresponding complexes. This is not an exception, there are many more like this, including that of cobalt.

Cu(II) is reduced when treated with cyanide to form tetra cyanide cuprate (I) complex. Simple cyano complexes of Cu(II) are unstable and are reduced to Cu(I). In fact, it is very difficult to even isolate them.

References: http://pubs.acs.org/doi/abs/10.1021/ic50132a014 (some stable cyanide complexes of Cu(II))


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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