d
(d) \({H_2}O + \mathop {B{r_2}}\limits_0 \to \,\mathop {HOBr}\limits_{ + 1} + \mathop {HBr}\limits_{ - 1} \) In the above reaction the oxidation number of \(B{r_2}\) increases from zero (in \(B{r_2}\)) to \(+1\) (in \(HOBr\)) and decrease from zero (\(B{r_2}\)) to \(-1\) (in \(HBr\)). Thus \(B{r_2}\) is oxidised as well as reduced & hence it is a redox reaction.