Question

What will be the n-factor for $Ba{(Mn{O}_4)}_2$ in acidic medium? (Where it behaves as oxidant)
A. 5
B. 10
C. 6
D. 3

Answer 1

Hint: For acids we can say that the n-factor is nothing but the number of hydrogen ions replaceable by 1 mole of acid in a chemical reaction. There is a formula to calculate the n-factor and it is as follows.
n-factor = (Change in oxidation number per atom) (Number of atoms present per molecule)

Complete Solution :
- In the question it is given that what will be the n-factor of $Ba{(Mn{O}_4)}_2$ in acidic medium.
- We have to find the n-factor for $Ba{(Mn{O}_4)}_2$ in acidic medium.

- The dissociation of $Ba{(Mn{O}_4)}_2$ in acidic medium is as follows.
$$Ba{(Mn{O}_4)}_2\to B{a}^{2+}+2\stackrel{+7}{Mn}{O}_4^{-}\to 2M{n}^{2+}$$
- In the above dissociation we can see clearly that the oxidation number of manganese is changed from +7 to +2 means the change in oxidation number per atom is 5.
- The number of atoms present per molecule of manganese is 2.
- Substitute the above known values in the below formula to get the n-factor of $Ba{(Mn{O}_4)}_2$ in acidic medium.
n-factor of $Ba{(Mn{O}_4)}_2$ = (Change in oxidation number per atom) (Number of atoms present per molecule)
= (5) (2)
= 10.
 - Therefore the n-factor for barium permanganate ($Ba{(Mn{O}_4)}_2$ ) in acidic medium is 10.
So, the correct answer is “Option B”.

Note: Means $Ba{(Mn{O}_4)}_2$ can replace 10 hydrogen ions from the acidic solution. The n-factor of barium permanganate ($Ba{(Mn{O}_4)}_2$ ) is calculated based on the capability to replace the hydrogen ions from the acidic solution.