A) Oxidised only
B) Reduced only
C) Proton acceptor only
D) Both oxidised and reduced
Correct Answer: D
Solution :
\[{{H}_{2}}O+\underset{0}{\mathop{B{{r}_{2}}}}\,\xrightarrow{{}}\,\underset{+1}{\mathop{HOBr}}\,+\underset{-1}{\mathop{HBr}}\,\] 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.
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