Question

When excess of aqueous $KCN$ solution is added to an aqueous solution of copper sulphate, the complex $[Cu(CN{)}_4{]}^{2-}$ 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. $[Cu(CN{)}_4{]}^{2-}$ 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)

Copper sulphate reacts with excess potassium cyanide to form ${\left[Cu(CN{)}_4\right]}^{2-}$ ion.
$CuS{O}_4+4KCN\to {\left[Cu(CN{)}_4\right]}^{2-}+4K^{+}+S{O}_4^{2-}$
${\left[Cu(CN{)}_4\right]}^{2-}$ 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.