${\mathrm{HgCl}}_2$ and ${\mathrm{I}}_2$ both when dissolved in water containing ${\mathrm{I}}^{-}$ ions, the pair of species formed is

In a solution containing HgCl₂, I₂ and I⁻, both HgCl₂ and I₂ compete for I⁻.
Since formation constant of [Hgl4]²⁻ is 1.9×10³⁰, which is very large,
I⁻ will preferentially combine with HgCl₂.
${\mathrm{HgCl}}_2+2{\mathrm{I}}^{-}\to \underset{\text{Redppt}}{{\mathrm{Hgl}}_2+2{\mathrm{Cl}}^{-}}$

${\mathrm{Hgl}}_2+2\mathrm{l}\to \underset{\text{soluble}}{{\left[{\mathrm{Hgl}}_4\right]}^{2-}}$