$n -$ factor of $M n_{2} O_{7}$ in the change, $2 M n_{2} O_{7} ⟶ 4 M n O_{2} + 3 O_{2}$ , is :

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Solution

$2 M n_{2} O_{7} ⟶ 4 M n O_{2} + 3 O_{2}$
+7 +4
Change in oxidation state of Mn $= 7 - 4 = 3$ .
So, $n$ -factor of $M n_{2} O_{7} = 2 \times 3 = 6$ .

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