Find the n-factor in the following chemical changes.
$K M n O_{4} H_{2} O - --- \to M n^{4 +}$

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Solution

The correct option is B 3
Steps for finding valency factor/n-factor:
1. Find the oxidation state (O.S.) of metal in reactant side.
2. Find the oxidation state of metal in product side.
3. Take the magnitude of difference of two oxidation state $| O . S ._{P r o d u c t} - O . S ._{R e a c t a n t} | \times number of atom$ for the valency factor
$K M n O_{4} O H^{-} - --- \to M n^{4 +}$
O.S. of Mn in reactant side
$= + 1 + x + 4 \times (- 2) = 0$
$x = + 7$
O.S. of Mn in product side=+4
$valency factor = number of Mn atom \times | O . S ._{P r o d u c t} - O . S ._{R e a c t a n t} |$
$valency factor = 1 \times | + 4 - 7 |$
$valency factor = 3$

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