Question

When $KMn{O}_4$ acts as an oxidizing agent and ultimately forms$Mn{O}_4^{-2}$ ,$Mn{O}_2$ ,$M{n}_2{O}_3$ and$M{n}^{+2}$ , then the number of electrons transferred ( per Mn atom) in each case respectively is:

• A. 4, 3, 1, 5
• B. 1, 5, 3, 7
• C. 1, 3, 4, 5
• D. 3, 5, 7, 1

Answer 1

The correct option is C

1, 3, 4, 5

Check the oxidation states of Mn.
$⟹$ No of electrons transferred will be change in oxidation state

$\stackrel{+7}{M}n{O}_4^{-}+e^{-}⟶\stackrel{+6}{M}n{O}_4^{-2}$

$\stackrel{+7}{M}n{O}_4^{-}+3e^{-}⟶\stackrel{+4}{M}n{O}_2$

$\stackrel{+7}{M}n{O}_4^{-}+4e^{-}⟶\stackrel{+3}{M}{n}_2{O}_3$;

$\stackrel{+7}{M}n{O}_4^{-}+5e^{-}⟶\stackrel{+2}{M}{n}^{2+}$