In a solution containing $HgCl_2$, $I_2$ and $I^-$, both $HgCl_2$ and $I_2$ compete for $I^-$. Since the formation constant of $[HgI_4]^{2-}$ is $1.9\times10^{30}$, which is very large compared with $I_3^-$ ($K_f = 700$), $I^-$ will preferentially combine with $HgCl_2$. $HgCl_2 + 2I^- \rightarrow HgI_2\downarrow$ (red ppt) $+ 2Cl^-$ $HgI_2 + 2I^- \rightarrow [HgI_4]^{2-}$ (soluble) With excess $I^-$, $I_2$ also forms $I_3^-$, so the pair formed is $HgI_4^{2-}$ and $I_3^-$.