For the reaction: I - + C l O 3 - + H 2 S O 4 → C l - + H S O 4 - + I 2   The correct…

The balanced equation is,
$\mathrm{C}\mathrm{l}{\mathrm{O}}_3^{-}+6{\mathrm{I}}^{-}+6{\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4\to 3{\mathrm{I}}_2+\mathrm{C}{\mathrm{l}}^{-}+6\mathrm{H}\mathrm{S}{\mathrm{O}}_4^{-}+3{\mathrm{H}}_2\mathrm{O}$

Here, in this reaction,

${\mathrm{HSO}}_4^{-}$ is in one of the products, and by balancing the reaction, we get its stiochiometric coefficient as 6.

Oxidation number of iodine in reactant side is $-1$ and it is oxidised to  ${\mathrm{I}}_2 \mathrm{with} 0$ oxidation number in product side.