Assuming complete ionization, same moles of which of the following compounds will require the least amount…

${\mathrm{M}\mathrm{n}\mathrm{O}}_4^{-}⟶\mathrm{M}{\mathrm{n}}^{2+}$ ; Change in oxidation no. = 5
In option,
i. $\mathrm{F}{\mathrm{e}}^{2+}\mathrm{ }⟶\mathrm{F}{\mathrm{e}}^{3+}$
Change in oxidation no. = 1
ii. $\mathrm{F}{\mathrm{e}}^{2+}\mathrm{ }⟶\mathrm{F}{\mathrm{e}}^{3+}$
Change in oxidation no. = 1
${\mathrm{S}\mathrm{O}}_3^{2-}⟶{\mathrm{S}\mathrm{O}}_4^{2-}$
Change in oxidation no. = 2
$=1+2=3$
iii. $\mathrm{F}{\mathrm{e}}^{2+}\mathrm{ }⟶\mathrm{F}{\mathrm{e}}^{3+}$
Change in oxidation no. = 1
${\mathrm{C}}_2{\mathrm{O}}_4^{2-}⟶2\mathrm{C}{\mathrm{O}}_2$
Change in oxidation no. = 2
$=1+2=3$
iv. $\mathrm{F}{\mathrm{e}}^{2+}\mathrm{ }⟶\mathrm{F}{\mathrm{e}}^{3+}$
Change in oxidation no. = 1
$2{\mathrm{N}\mathrm{O}}_2^{-}⟶2{\mathrm{N}\mathrm{O}}_3^{-}$
Change in oxidation no. = 4
$=1+4=5$