Find the valency factor of $B r O_{3}^{-}$ in the reaction $B r O_{3}^{-} + 6 H^{+} + 6 e^{-} + \to B r^{-} + 3 H_{2} O$

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Solution

The correct option is B 6
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
$B r O_{3}^{-} + 6 H^{+} + 6 e^{-} + \to B r^{-} + 3 H_{2} O$
O.S. of Br in reactant side
$= x - 6 = - 1$
$x = + 5$
O.S. of Br in product side=-1
$valency factor = number of Br atom \times | O . S ._{P r o d u c t} - O . S ._{R e a c t a n t} |$
$valency factor = 1 \times | - 1 - 5 |$
$valency factor = 6$

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