In the reaction X+ I _2+2 HCl SnCl_4+2 HI The correct option...

$\mathrm{SnCl}_2+$\mathrm{I}$_2+2 \mathrm{HCl} \longrightarrow \mathrm{SnCl}_4+2 \mathrm{HI}$ $\mathrm{SnCl}_2$ is an strong reducing agent because it is readily oxidised into $\mathrm{Sn}^{4+}$ ion. Since, Sn-atom in $\mathrm{SnCl}_2$ is $\mathrm{sp}^2$-hybridised. So, it is an angular molecule. On the other hand, it is used as a reagent in test of $\mathrm{Hg}^{2+}$ radical.