Coordination Compounds Ques 38
38. $\mathrm{HgCl}_2$ and $I_2$ both when dissolved in water containing $I^{-}$ ions, the pair of species formed is:
[2017]
(a) $\mathrm{HgI}_2, \mathrm{I}^{-}$
(b) $\mathrm{HgI}_4^{2-}, \mathrm{I}_3^{-}$
(c) $\mathrm{Hg}_2 \mathrm{I}_2, \mathrm{I}^{-}$
(d) $\mathrm{HgI}_2, \mathrm{I}_3^{-}$
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Answer:
Correct Answer: 38.(b)
Solution: (b) In a solution containing $\mathrm{HgCl}_2, \mathrm{I}_2$ and $\mathrm{I}^{-}$, both $\mathrm{HgCl}_2$ and $\mathrm{I}_2$ compete for $\mathrm{I}^{-}$.
Since formation constant of $\left[\mathrm{HgI}_4\right]^{2-}$ is very large $\left(1.9 \times 10^{30}\right)$ as compared with $\Gamma_3$
$ \left(K_f=700\right) \text {. } $
$\therefore \mathrm{I}^{-}$ will preferentially combine with $\mathrm{HgCl}_2$.
$\mathrm{HgCl}_2+2 \mathrm{I}^{-} \rightarrow \mathrm{HgI}_2 \downarrow+2 \mathrm{Cl}^{-}$ (Red ppt)
$\mathrm{HgI}_2+2 \mathrm{I}^{-} \rightarrow\left[\mathrm{HgI}_4\right]^{2-}$ (soluble)
BETA
