What will be n-factor for $B a (M n O_{4})_{2}$ in acidic medium? (Where it behaves as oxidant)

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Solution

The correct option is B $10$
$^{+ 7}$
$B a (M n O_{4})_{2} ⟶ B a^{2 +} + 2 {M n O_{4}}^{-}$
$↓$
$2 M n^{2 +}$
$B a$ being most electro positive element does not get reduced to $B a$ , thus only ${M n O_{4}}^{-}$ gets reduced to $M n^{2 +}$ as it always does in acidic medium.

And since compound has $2$ atoms of $M n$ in it so, its' n factor is given by,
(Change in oxidation number per atom) $\times$ (Number of atoms per molecule or formula unit of the given reagent)
$= 5 \times 2 = 10$

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