HgCl_2 and I_2 both when dissolved in water c | vedclass

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Question

(B) When $HgCl_2$ is dissolved in a solution containing $I^{-}$ ions,it first forms $HgI_2$ precipitate,which then reacts with excess $I^{-}$ to form

Vedclass · Accepted Answer

$HgI_4^{2-}, I_3^-$

Vedclass · Answer

(B) When $HgCl_2$ is dissolved in a solution containing $I^{-}$ ions,it first forms $HgI_2$ precipitate,which then reacts with excess $I^{-}$ to form the soluble complex $[HgI_4]^{2-}$.$HgCl_2 + 2I^{-} \longrightarrow HgI_2(s) + 2Cl^{-}$$HgI_2(s) + 2I^{-} \longrightarrow [HgI_4]^{2-}(aq)$When $I_2$ is dissolved in a solution containing $I^{-}$ ions,it forms the triiodide ion,which is water-soluble.$I_2 + I^{-} \longrightarrow I_3^{-}(aq)$Thus,the final species formed are $[HgI_4]^{2-}$ and $I_3^{-}$.

$HgCl_2$ and $I_2$ both when dissolved in water containing $I^{-}$ ions,the pair of species formed is

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