The solublity of ${\mathrm{I}}_2$ in $\text{ Kl }$ solution is more than  its solubility in pure water because

The rule of "like dissolves like" governs soluble substances. As a result, a non-polar molecule does not dissolve in a polar solvent but does so in a non-polar one, and vice versa. They cannot mix because ${\mathrm{I}}_2$ is non-polar and $H_{2}O$ is polar. ${\mathrm{I}}_2$ turns polar and becomes ${\mathrm{KI}}_3$- when$\text{ Kl }$is added. Additionally, the solubility of the ${\mathrm{KI}}_3$- anion in water increases as a result of ion-dipole interaction.

So, the required solution is option C) ${\mathrm{I}}_2$ forms soluble complex ${\mathrm{KI}}_3$ with$\text{ Kl }$