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The d- and f-Block Elements
Compounds of Copper
1 question from 1 paper, each with its answer.
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Concepts tested
- Reduction of Cu(II) by cyanide
All Questions
2006 AIPMT-PRE 1 question
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Copper sulphate dissolves in excess of KCN to give :-Copper sulphate first reacts with KCN to give cupric cyanide: $2KCN + CuSO_4 \rightarrow K_2SO_4 + Cu(CN)_2$
Cupric cyanide is unstable and decomposes into cuprous cyanide and cyanogen: $2Cu(CN)_2 \rightarrow 2CuCN + (CN)_2$
In this step copper is reduced from +2 to +1.
Cuprous cyanide dissolves in excess of KCN to form a soluble complex: $CuCN + 3KCN \rightarrow K_3[Cu(CN)_4]$
The complex ion formed is $[Cu(CN)_4]^{3-}$, in which copper is in the +1 oxidation state ($x - 4 = -3$).
So copper sulphate dissolves in excess of KCN to give $[Cu(CN)_4]^{3-}$.