Question

When an excess of aqueous KCN solution is added to an aqueous solution of copper sulphate, the complex $\left[Cu\right(CN{)}_4{]}^{2-}$ is formed. On passing $H_{2}S$ gas through this solution no precipitate of CuS is formed because:

• A. sulphide ions cannot replace $C{N}^{-}$ ions
• B. $\left[Cu\right(CN{)}_4{]}^{2-}$ does not give $C{u}^{2+}$ ion in the solution
• C. sulphide ions from $H_{2}S$ do not form complexes
• D. sulphide ions cannot replace sulphate ions from copper sulphate solution

Answer 1

Answer: (B)

$\left[Cu\right(CN{)}_4{]}^{2-}$ does not give $C{u}^{2+}$ ion in the solution

Copper sulphate reacts with excess potassium cyanide to form $\left[Cu\right(CN{)}_4{]}^{2-}$ ion.

$CuS{O}_4+4KCN\to \left[Cu\right(CN{)}_4{]}^{2-}+4K^{+}+S{O}_4^{2-}$

$\left[Cu\right(CN{)}_4{]}^{2-}$ is a complex ion. Hence, it does not dissociates into $C{u}^{2+}$ ions and $C{N}^{-}$ ions. In absence of $C{u}^{2+}$ ions, no precipitate of CuS is obtained.