Question

What will be the n-factor for $Ba{\left(Mn{O}_4\right)}_2$ in acidic medium ?(Where it behaves as an oxidant)

• A. $5$
• B. $10$
• C. $6$
• D. $3$

Answer 1

The correct option is B $10$

Solution:- (B) $10$

Since barium is a strong electropositive metal, thus it cannt be reduced.

In acidic medium, ${Mn{O}_4}^{-}$ gets reduced to ${Mn}^{+2}$.

Since $Ba{\left(Mn{O}_4\right)}_2$ contains two $Mn$ ion, thus the n-factor of $Ba{\left(Mn{O}_4\right)}_2$ is -

$n$-factor $=2\times 5=10$

Hence the n-factor of $Ba{\left(Mn{O}_4\right)}_2$ is $10$.