In quantitative analysis of second group in laboratory, H2S gas is passed in acidic medium for precipitation. When Cu2+ react with KCN, then for product, true statement is
A
K2[Cu(CN)4] more soluble
B
K2[Cd(CN)4] less stable
C
K2[Cu(CN)2] less stable
D
K2[Cd(CN)3] more stable
Explanation
K3[Cu(CN)2] = 3(+1)+x+2(-1) = 0
$$ \Rightarrow $$ x = -1
$$ \therefore $$ the oxidation no. of 'Cu' is -1 (-ve), so this complex is unstable so it is not formed.
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