Question

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

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

Answer 1

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

$KMn{O}_4\left(M{n}^{7+}\right)$ changes to $M{n}^{2+}$ i.e., number of electrons involved per mole of $KMn{O}_4$ is 5.
(a) For $FeS{O}_3$,
$F{e}^{2+}\to F{e}^{3+}$ (No. of $e^{-}$ s involved = 1)
$S{O}_3^{2-}\to S{O}_4^{2-}$ (No. of $e^{-}$ s involved = 2)
Total number of $e^{-}$ s involved = 1+2= 3
(b) For $Fe{C}_2{O}_4$,
$F{e}^{2+}\to F{e}^{3+}$ (No. of $e^{-}$ s involved =1)
$C_2{O}_3^{2-}\to 2C{O}_4^{2-}$ (No. of $e^{-}$s involved =2)
Total number of $e^{-}$ s involved = 1+2= 3
(c) For $Fe(N{O}_2{)}_2$,
$F{e}^{2+}\to F{e}^{3+}$ (No. of $e^{-}$ s involved =1)
$2N{O}_2\to 2N{O}_3$ (No. of $e^{-}$ s involved =4)
Total number of $e^{-}$ s involved = 1+4 =5
(d) For $FeS{O}_4$,
$F{e}^{2+}\to F{e}^{3+}$ (No. of $e^{-}$ s involved =1)
Total number of $e^{-}$ s involved =1
As $FeS{O}_4$ requires least number of electrons thus, it will require least amount of $KMn{O}_4$.