Question

The equivalent weight of $Mn{O}_2$ is equal to its molecular weight when it is converted to:

• A. $M{n}_2{O}_3$
• B. $MnO$
• C. $Mn{O}_4^{2-}$
• D. $Mn{O}_4^{-}$

Answer 1

The correct option is A $M{n}_2{O}_3$

$M{n}^{4+}{O}_2\to M{n}_2^{3+}{O}_3$

Oxidation state of Mn in $MN{O}_2$ is +4 and in $M{n}_2{O}_3$ is +3.

In the above reaction, change in the oxidation state of Mn is 1. Hence, the equivalent weight of $Mn{O}_2$ is the same as its molecular weight .
$(\text{Equivalent weight}=\frac{\text{Molecular weight}}{\text{Number of electrons transferred}}=\frac{Molecularweight}{1})$