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# What will be the n-factor for $Ba{{(Mn{{O}_{4}})}_{2}}$ in acidic medium? (Where it behaves as oxidant)A. 5B. 10C. 6D. 3

Last updated date: 06th Sep 2024
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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.
$Ba{{(Mn{{O}_{4}})}_{2}}\to B{{a}^{2+}}+2\overset{+7}{\mathop{Mn}}\,O_{4}^{-}\to 2M{{n}^{2+}}$
- 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.