Assuming complete ionization, same moles of which of the following compounds will require the least amount of acidified ${\text{KMnO}}_4$ for complete oxidation?

${\text{KMnO}}_4$ acts as an oxidizing agent. The amount required depends on the total change in oxidation number ($n$ -factor) of the reducing agent.${\text{FeC}}_2{\text{O}}_4$: ${\text{Fe}}^{2+}\to {\text{Fe}}^{3+}$ ($1e^{-}$) and ${\text{C}}_2{\text{O}}_4^{2-}\to 2{\text{CO}}_2$ ($2e^{-}$). Total $n=3$.$\text{Fe}({\text{NO}}_2{)}_2$: ${\text{Fe}}^{2+}\to {\text{Fe}}^{3+}$ ($1e^{-}$) and $2{\text{NO}}_2^{-}\to 2{\text{NO}}_3^{-}$ ($4e^{-}$). Total $n=5$.${\text{FeSO}}_4$: Only ${\text{Fe}}^{2+}\to {\text{Fe}}^{3+}$ ($1e^{-}$). ${\text{SO}}_4^{2-}$ is not oxidized. Total $n=1$.${\text{FeSO}}_3$: ${\text{Fe}}^{2+}\to {\text{Fe}}^{3+}$ ($1e^{-}$) and ${\text{SO}}_3^{2-}\to {\text{SO}}_4^{2-}$ ($2e^{-}$). Total $n=3$.Since ${\text{FeSO}}_4$ has the lowest $n$ -factor ($1$), it requires the least amount of ${\text{KMnO}}_4$.