A) Proton acceptor only
B) Both oxidised and reduced
C) Oxidised only
D) Reduced only
Correct Answer: B
Solution :
\[{{H}_{2}}o+\overset{0}{\mathop{B}}\,{{r}_{2}}\xrightarrow{{}}HO\overset{+1}{\mathop{Br}}\,+\overset{-1}{\mathop{HBr}}\,\] 0 +1 -1 In the above reaction, the oxidation number of \[B{{r}_{2}}\]increases from 0 (in\[B{{r}_{2}}\]) to + 1 (in HOBr) and decreases from 0 (in Br,) to -1 (in HBr). Then \[B{{r}_{2}}\]is oxidised as well as reduced and hence it is a redox reaction.
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