In qualitative analysis of second group in laboratory, H2S gas is passed in acidic medium for precipitation. When Cu2+ and Cd2+ react with KCN, then for product, true statement is
- K2[Cu(CN)4] – more stable; K2[Cd(CN)4] – less stable
- K2[Cu(CN)4] – less stable; K2[Cd(CN)4] – more stable
- K3[Cu(CN)4] – more stable; K2[Cd(CN)4] – less stable
- K3[Cu(CN)4] – less stable; K3[Cd(CN)4] – more stable
Answer
(3) K3[Cu(CN)4] – more stable; K2[Cd(CN)4] – less stable
Solution
With excess KCN, Cu2+ is reduced to Cu+ (cyanogen is set free) and a cyano complex of copper(I) is formed:
2CuSO4 + 10KCN 2K3[Cu(CN)4] + 2K2SO4 + (CN)2
whereas Cd2+ forms a cyano complex without change in oxidation state:
CdSO4 + 4KCN K2[Cd(CN)4] + K2SO4
As per NCERT, the stability of a complex decides how far it gives free metal ions in solution (stability constant).
K3[Cu(CN)4] is very stable and gives too few Cu+ ions to precipitate a sulphide, so copper stays in solution. K2[Cd(CN)4] is less stable and gives enough Cd2+ ions, so H2S precipitates yellow CdS; this is how Cu2+ is separated from Cd2+.
∴ K3[Cu(CN)4] – more stable; K2[Cd(CN)4] – less stable.
Answer: (3)