In a solution containing $HgCl _{2}, I _{2}$ and $I ^{-}$, both $HgCl _{2}$ and $I _{2}$ compete for $I ^{-}$.
Since formation constant of $\left[ Hgl _{4}\right]^{2-}$ is $1.9 \times 10^{30}$ which is very large as compared with $I _{3}^{-}\left( K _{ f }=700\right)$
$\therefore I ^{-}$ will preferentially combine with $HgCl _{2}$.