In the conversion of $B r O_{3}^{-}$ to $B r O_{4}^{-}$ :

there is no change in oxidation number

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reduction takes place in basic solution

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reduction also takes place by

X e F_{2}

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equivalent mass of

B r O_{3}^{-}

is one-half of ionic mass

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Solution

The correct option is D equivalent mass of $B r O_{3}^{-}$ is one-half of ionic mass
A. During the conversion of $B r O_{3}^{-} t o B r O_{4}^{-}$ , the oxidation number of bromine changes from +5 to +7. Hence, the option A is incorrect.
B. and C. Since there is an increase in the oxidation number, the net reaction is the oxidation of bromine. Hence, the options B and C are incorrect.
D. The equivalent mass of $B r O_{3}^{-}$ is the ratio of the ionic mass to the change in the oxidation number. It is tone half of the ionic mass as the change in the oxidation number is 2. Thus, the option D is correct.

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Similar questions

B r O_{3}^{-}

B r O_{4}^{-}

X e F_{2}

X e F_{2} m_{1} + B r O_{3}^{-} m_{2} + H_{2} O ⟶ B r O_{4}^{-} + X e + 2 H F

m_{1}

m_{2}

X e F_{2}

B r O_{3}^{-}

X e F_{2}

B r O_{3}^{-}

X e F_{2} + B r O_{3}^{-} + H_{2} O ⟶ B r O_{4}^{-} + X e + 2 H F

There will be some change in mass when a chemical reaction takes place.

X e F_{2} + {B r O_{3}}^{-} + H_{2} O ⟶ X e + {B r O_{4}}^{-} + 2 H F

e q u i v a l e n t m a s s o f {B r O_{3}}^{-} = \frac{m o l a r m a s s}{. . . . . . . . . .}

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