To solve this problem, we need to understand the reactions of $HgCl_2$ and $I_2$ in the presence of $I^-$ ions.$\newline$1. Reaction of $HgCl_2$ with $I^-: \newline$• $HgCl_2$ reacts with $I^-$ ions to form $HgI_2$, which is a precipitate.$\newline$• The reaction is:$\newline HgCl_2 + 2I^- \rightarrow HgI_2 + 2Cl^- \newline$• $HgI_2$ is insoluble in water, but in excess $I^-$ ions, it forms a soluble complex:$\newline HgI_2 + 2I^- \rightarrow HgI_4^{2-} \newline$2. Reaction of $I_2$ with $I^-: \newline$• $I_2$ reacts with $I^-$ ions to form $I_3^-$ (triiodide ion).$\newline$• The reaction is:$\newline I_2 + I^- \rightarrow I_3^- \newline$Combining these reactions, the species formed in the solution are $HgI_4^{2-}$ and $I_3^-. \newline$Therefore, the correct option is Option 2: $HgI_4^{2-}, I_3^-.$