Question

When an aqueous solution of KCN is added to a solution containing $N{i}^{2+}$ ions, a precipitate forms, which redissolves upon the addition of more KCN solution. No precipitate forms when $H_{2}S$is bubbled into this solution. Write reactions describing what happens in this solution. [Hint:$C{N}_2$ is a Bronsted–Lowry base $\left(Kb<1025\right)$and a Lewis base.]

Short answer

When excess of KCN is added to aqueous solution of$N{i}^{2+}$ ions then precipitate gets form, the complex ${\left[Ni{\left(CN\right)}_4\right]}^{2-}$is formed. When $H_{2}S$is bubbled into this solution no precipitate of NiS is formed.

Step-by-step solution

Copper Sulphate

Copper sulphate reacts with excess potassium cyanide to form $\left[Cu\right(CN)4{]}^{2-}$ion.

$NiS{O}_4+4KCN\to [Ni{\left(CN\right)}_4{]}^{2-}+4K^{+}+SO{4}^{2-}$

Complex ion

$[Ni(CN{)}_4{]}^{2-}$is a complex ion. Hence, it does not $N{i}^{2+}$dissociate into ions and$C{N}^{-}$ ions. In absence of $N{i}^{2+}$ions, no precipitate of NiS is obtained.