Consider the following list of reagents :

Acidified ${\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7$, alkaline ${\mathrm{KMnO}}_4,{\mathrm{CuSO}}_4,{\mathrm{H}}_2{\mathrm{O}}_2$, ${\mathrm{Cl}}_2,{\mathrm{O}}_3,{\mathrm{FeCl}}_3,{\mathrm{HNO}}_3$  and ${\mathrm{Na}}_2{\mathrm{S}}_2{\mathrm{O}}_3$ . The total number of reagents that can oxidise aqueous iodide to iodine is :

$6\mathrm{KI}+{\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7+7{\mathrm{H}}_2{\mathrm{SO}}_4\to 4{\mathrm{K}}_2{\mathrm{SO}}_4+{\mathrm{Cr}}_2{\left({\mathrm{SO}}_4\right)}_3$

$+3{\mathrm{I}}_2+7{\mathrm{H}}_2\mathrm{O}$

$2{\mathrm{CuSO}}_4+4\mathrm{KI}\to {\mathrm{Cu}}_2{\mathrm{I}}_2+2{\mathrm{K}}_2{\mathrm{SO}}_4+{\mathrm{I}}_2$

$\begin{array}{l}{\mathrm{H}}_2{\mathrm{O}}_2+2\mathrm{KI}\to {\mathrm{I}}_2+2\mathrm{KOH}\\ {\mathrm{Cl}}_2+2\mathrm{KI}\to 2\mathrm{KCl}+{\mathrm{I}}_2\\ {\mathrm{O}}_3+2\mathrm{KI}+{\mathrm{H}}_2\mathrm{O}\to 2\mathrm{KOH}+{\mathrm{I}}_2+{\mathrm{O}}_2\end{array}$

$\begin{array}{l}2{\mathrm{FeCl}}_3+2\mathrm{KI}\to 2{\mathrm{FeCl}}_2+2\mathrm{KCl}+{\mathrm{I}}_2\\ 4{\mathrm{HNO}}_3+2\mathrm{KI}\to 2{\mathrm{KNO}}_3+{\mathrm{NO}}_2+{\mathrm{I}}_2+2{\mathrm{H}}_2\mathrm{O}\end{array}$

${\mathrm{Na}}_2{\mathrm{S}}_2{\mathrm{O}}_3$ reacts with ${\mathrm{I}}_2$​ to form iodide

There are 7 such reactions in which oxidise aqueous iodide is converted into iodine and KMnO_4​ cannot.

Thus, the correct answer is 7.