Confusion about the oxidation state of copper in copper cyanide re-dissolved with excess cyanide
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:

$$\ce{2Cu^2+ + 4I- -&gt; I2 + 2CuI v}$$

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:

$$\ce{2 Cu^2+ + 4 CN- -&gt; NCCN ^ + 2 CuCN v}$$

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:

$$\ce{CuCN (s) + 3 KCN (aq) &lt;=&gt; K3[Cu(CN)4] (aq)}$$