A) Proton acceptor only
B) Both oxidized and reduced
C) Oxidized only
D) Reduced only
Correct Answer: B
Solution :
[b] \[{{H}_{2}}O+\overset{0}{\mathop{B}}\,{{r}_{2}}\xrightarrow{{}}HO\overset{+1}{\mathop{Br}}\,+H\overset{-l}{\mathop{Br}}\,\] Thus here oxidation number of Br increases from 0 to +1 and also decreases from 0 to -1. Thus it is oxidised as well as reduced.
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