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(D) $KMnO_4$ $(Mn^{7+})$ is reduced to $Mn^{2+}$,involving $5$ electrons per mole of $KMnO_4$.For the same number of moles of the compound,the one req

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$FeSO_4$

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(D) $KMnO_4$ $(Mn^{7+})$ is reduced to $Mn^{2+}$,involving $5$ electrons per mole of $KMnO_4$.For the same number of moles of the compound,the one requiring the least number of electrons for oxidation will require the least amount of $KMnO_4$.$(a)$ For $FeSO_3$:$Fe^{2+} \rightarrow Fe^{3+} + 1e^-$$SO_3^{2-} \rightarrow SO_4^{2-} + 2e^-$Total $e^- = 1 + 2 = 3$$(b)$ For $FeC_2O_4$:$Fe^{2+} \rightarrow Fe^{3+} + 1e^-$$C_2O_4^{2-} \rightarrow 2CO_2 + 2e^-$Total $e^- = 1 + 2 = 3$$(c)$ For $Fe(NO_2)_2$:$Fe^{2+} \rightarrow Fe^{3+} + 1e^-$$2NO_2^- \rightarrow 2NO_3^- + 4e^-$Total $e^- = 1 + 4 = 5$$(d)$ For $FeSO_4$:$Fe^{2+} \rightarrow Fe^{3+} + 1e^-$Total $e^- = 1$Since $FeSO_4$ involves the minimum number of electrons $(1)$,it requires the least amount of $KMnO_4$.

Assuming complete ionisation,same moles of which of the following compounds will require the least amount of acidified $KMnO_4$ for complete oxidation?

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