Question

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

• A. $Fe{C}_2{O}_4$
• B. $Fe(N{O}_2{)}_2$
• C. $FeS{O}_4$
• D. $FeS{O}_3$

Answer 1

The correct option is C $FeS{O}_4$

All of these compounds have the same positive ion $F{e}^{2+}$, which will get oxidised to $F{e}^{3+}$. So, the compound where the anion does not get oxidised by $KMn{O}_4$ would require the least amount to get oxidised completley.
$C_2{O}_4^{2-}\to C{O}_2$
$N{O}_2^{-}\to N{O}_3^{-}$
$S{O}_3^{2-}\to S{O}_4^{2-}$
Since $S{O}_4^{2-}$ has $S$ in the highest oxidation state already, the answer would be $FeS{O}_4$