Looking for classes? Ksquare Career Institute, Bengaluru →
$CuSO_4$ when reacts with KCN forms CuCN, which is insoluble in water. It is soluble in excess of KCN, due to formation of the following complex : -
A
$K_2[Cu(CN)_4]$
B
$K_3[Cu(CN)_4]$
C
$CuCN_2$
D
$Cu[KCu(CN)_4]$
Detailed Solution
$Cu^{2+}$ first forms $Cu(CN)_2$, which decomposes to insoluble cuprous cyanide and cyanogen:
$2CuSO_4 + 4KCN \rightarrow 2CuCN + (CN)_2 + 2K_2SO_4$
With excess KCN, CuCN dissolves by forming a soluble complex of Cu(I):
$CuCN + 3KCN \rightarrow K_3[Cu(CN)_4]$
In $[Cu(CN)_4]^{3-}$, Cu is +1 and the charge is $+1 + 4(-1) = -3$, balanced by three $K^+$.
So the complex formed is $K_3[Cu(CN)_4]$.
$2CuSO_4 + 4KCN \rightarrow 2CuCN + (CN)_2 + 2K_2SO_4$
With excess KCN, CuCN dissolves by forming a soluble complex of Cu(I):
$CuCN + 3KCN \rightarrow K_3[Cu(CN)_4]$
In $[Cu(CN)_4]^{3-}$, Cu is +1 and the charge is $+1 + 4(-1) = -3$, balanced by three $K^+$.
So the complex formed is $K_3[Cu(CN)_4]$.