In the conversion of $B r_{2} t o B r O_{3}^{-}$ , the oxidation state of $B r$ changes from:

+1 to +5

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0 to -3

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+2 to +5

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0 to +5

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Solution

The correct option is C 0 to +5
Given: $B r_{2} \to B r {O_{3}}^{-}$
Oxidation state of an element in its standard state is zero.
Therefore, for $B r_{2}$ oxidation number = 0
In $B r {O_{3}}^{-}$ let the oxidation number of $B r$ be $x$ and as for $O$ it is -2
Therefore $x + 3 \times (- 2) = - 1$
$x = - 1 + 6 = + 5$
$0 B r_{2} \to + 5 B r {O_{3}}^{-}$
$∴$ In the conversion of $B r_{2} t o B r {O_{3}}^{-}$ , the oxidation state of $B r$ changes from 0 to +5.

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