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

The amount of acidified $\text{KMnO}_4$ required for the complete oxidation of a given number of moles of a compound depends on the number of moles of electrons lost by the compound during oxidation (i.e., its n-factor). Let's evaluate the n-factor for $1$ mole of each compound:

• $\text{FeSO}_4$: $\text{Fe}^{2+}$ is oxidised to $\text{Fe}^{3+}$ (loss of $1e^-$). Sulphur is already in its highest oxidation state ($+6$) in $\text{SO}_4^{2-}$ and cannot be further oxidised. Total n-factor = $1$.
• $\text{FeSO}_3$: $\text{Fe}^{2+}$ is oxidised to $\text{Fe}^{3+}$ (loss of $1e^-$) and $\text{SO}_3^{2-}$ is oxidised to $\text{SO}_4^{2-}$ (loss of $2e^-$). Total n-factor = $1 + 2 = 3$.
• $\text{FeC}_2\text{O}_4$: $\text{Fe}^{2+}$ is oxidised to $\text{Fe}^{3+}$ (loss of $1e^-$) and $\text{C}_2\text{O}_4^{2-}$ is oxidised to $\text{CO}_2$ (loss of $2e^-$). Total n-factor = $1 + 2 = 3$.
• $\text{Fe}(\text{NO}_2)_2$: $\text{Fe}^{2+}$ is oxidised to $\text{Fe}^{3+}$ (loss of $1e^-$) and two $\text{NO}_2^{-}$ are oxidised to two $\text{NO}_3^{-}$ (loss of $2 \times 2e^- = 4e^-$). Total n-factor = $1 + 4 = 5$. Since $\text{FeSO}_4$ provides the minimum number of moles of electrons (lowest n-factor) per mole of the compound, it will require the least amount of $\text{KMnO}_4$ for complete oxidation.