Question

Assuming complete ionisation, same mole 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)
$S{O}_3^{2-}\to S{O}_4^{2-}$ (N0. 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)
$C_2{O}_4^{2-}\to 2C{O}_2$ (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 =2)
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 $KMn{O}_4$.