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(D) In the reaction $H_2O + Br_2 	o HOBr + HBr$,the oxidation state of bromine changes as follows:$Br_2$ (elemental state) has an oxidation number of

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Both oxidised and reduced

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(D) In the reaction $H_2O + Br_2 	o HOBr + HBr$,the oxidation state of bromine changes as follows:$Br_2$ (elemental state) has an oxidation number of $0$.In $HOBr$,the oxidation number of $Br$ is $+1$.In $HBr$,the oxidation number of $Br$ is $-1$.Since the oxidation number of bromine increases from $0$ to $+1$ (oxidation) and decreases from $0$ to $-1$ (reduction) simultaneously,bromine undergoes both oxidation and reduction.Therefore,it is a disproportionation reaction,which is a type of redox reaction.

Which is the best description of the behaviour of bromine in the reaction given below $H_2O + Br_2 \to HOBr + HBr$?

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