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HgCl $_2$ and $I_2$ both when dissolved in water containing $I^-$ ions form the pair of species.

medium

Coordination Compounds

2017

chemistry

HgI $_2, I^-$

HgI $_4^{2-}, I_3^-$

Explanation

To solve this problem, we need to understand the reactions of $HgCl_2$ and $I_2$ in the presence of $I^-$

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Hg $_2$ I $_2, I^-$

HgI $_2, I_3^-$

ions.

\newline

1. Reaction of

HgCl_2

with

I^-: \newline

•

HgCl_2

reacts with

I^-

ions to form

HgI_2

, which is a precipitate.

\newline

• The reaction is:

\newline HgCl_2 + 2I^- \rightarrow HgI_2 + 2Cl^- \newline

•

HgI_2

is insoluble in water, but in excess

I^-

ions, it forms a soluble complex:

\newline HgI_2 + 2I^- \rightarrow HgI_4^{2-} \newline

2. Reaction of

I_2

with

I^-: \newline

•

I_2

reacts with

I^-

ions to form

I_3^-

(triiodide ion).

\newline

• The reaction is:

\newline I_2 + I^- \rightarrow I_3^- \newline

Combining these reactions, the species formed in the solution are

HgI_4^{2-}

and

I_3^-. \newline

Therefore, the correct option is Option 2:

HgI_4^{2-}, I_3^-.