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

When ${\text{HgCl}}_2$ is added to a solution containing iodide ions (${\text{I}}^{-}$), it first forms a red precipitate of ${\text{HgI}}_2$, which then dissolves in excess iodide to form the soluble complex $[{\text{HgI}}_4{]}^{2-}$.
$${\text{HgCl}}_2+4{\text{I}}^{-}\to [{\text{HgI}}_4{]}^{2-}+2{\text{Cl}}^{-}$$
When iodine (${\text{I}}_2$) dissolves in water containing ${\text{I}}^{-}$ ions, it forms the triiodide ion:
$${\text{I}}_2+{\text{I}}^{-}\to {\text{I}}_3^{-}$$
Thus, the pair of species formed is $[{\text{HgI}}_4{]}^{2-}$ and ${\text{I}}_3^{-}$.