A solution which is ${10}^{-3}\mathrm{M}$ each in ${\mathrm{Mn}}^{2+}$, ${\mathrm{Fe}}^{2+},{\mathrm{Zn}}^{2+}$ and ${\mathrm{Hg}}^{2+}$is treated with ${10}^{-16}\mathrm{M}$ sulphide ion. If ${\mathrm{K}}_{\mathrm{sp}}$of $\mathrm{MnS},\mathrm{FeS},\mathrm{ZnS}$ and $\mathrm{HgS}$ are ${10}^{-15},{10}^{-23},{10}^{-20}$ and ${10}^{-54}$ respectively. Which of the following will be precipitated first?

Ionic product in the solution $={10}^{-3}\times {10}^{-16}={10}^{-19}$. The metal sulphide having the lowest solubility will precipitate first provided the ionic product is higher than the ${\mathrm{K}}_{\mathrm{sp}}$. Here, all salts are of the same valence type. So, the sulphide having the lowest ${\mathrm{K}}_{\mathrm{sp}}$value will precipitate first provided ${\mathrm{K}}_{\mathrm{sp}}<{10}^{-19}$. HgS has the lowest ${\mathrm{K}}_{\mathrm{sp}}$ value $\left({10}^{-54}\right)$, so it will precipitate first.

Hence the correct option is (C) $\mathrm{HgS}$.