In the conservation $B r_{2} ⟶ B R O_{3}^{-}$ , the oxidation state of bromine changes from

- 1 to - 1

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0 to - 1

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0 to + 5

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0 to - 5

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Solution

The correct option is C
0 to + 5

$0 B r_{2} ⟶ + 5 B r O_{3}^{-}$ , in this reaction oxidation state change from 0 to +5.

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Similar questions

B r_{2} \to B r O_{3}^{-}

B r_{2}

B r O_{3}^{-}

C l_{2} + H_{2} S ⟶ 2 H C l + S

When the physical state of $H_{2}$ $O_{2}$ changes from vapour state to solid state,Dihedral angle changes from

a : b = 1 : 6, a \neq 0, b \neq 0

a x^{2} + b x + 5 a = 0

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