Question

n-factor of ${Ba\left({MnO}_4\right)}_2$ in acidic medium is :

• A. $2$
• B. $6$
• C. $10$
• D. none of the above

Answer 1

The correct option is C $10$

Consider following balance chemical equation (for dissociation) :
$Ba(MnO4{)}_2\to B{a}^{2+}+2Mn{O}_4^{-}$
Now, $2Mn{O}_4^{-}\to 2M{n}^{2+}$ + 10 electrons
As O.S. of Mn in $MnO{4}^{-}$ is +7, and in $M{n}^{2+}$ it is +2, means 1 mol of releases 5 electrons, hence, 2 mol releases 10 electrons.
As 1 mol $Ba(Mn{O}_4{)}_2$ contains 2 mol $Mn{O}_4^{-}$, hence, releases total 10 electrons.

n= 10