Addition of excess potassium iodide solution | vedclass

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Question

(A) When excess potassium iodide $(KI)$ is added to mercuric chloride $(HgCl_{2})$, the reaction is as follows:$HgCl_{2} + 4KI \longrightarrow K_{2}[H

Vedclass · Accepted Answer

tetrahedral $K_{2}[HgI_{4}]$

Vedclass · Answer

(A) When excess potassium iodide $(KI)$ is added to mercuric chloride $(HgCl_{2})$, the reaction is as follows:$HgCl_{2} + 4KI \longrightarrow K_{2}[HgI_{4}] + 2KCl$In the complex $[HgI_{4}]^{2-}$, the central metal ion is $Hg^{2+}$.The electronic configuration of $Hg$ $(Z=80)$ is $[Xe] 4f^{14} 5d^{10} 6s^{2}$.For $Hg^{2+}$, the configuration is $[Xe] 4f^{14} 5d^{10} 6s^{0}$.The four $I^-$ ligands donate electron pairs to the $6s$ and $6p$ orbitals of $Hg^{2+}$, resulting in $sp^{3}$ hybridization.This leads to a tetrahedral geometry for the complex $K_{2}[HgI_{4}]$.

Addition of excess potassium iodide solution to a solution of mercuric chloride gives the halide complex

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