A) \[{{I}_{2}}\]
B) \[B{{r}_{2}}\]
C) \[C{{l}_{2}}\]
D) \[{{F}_{2}}\]
Correct Answer: D
Solution :
Fluorine has maximum, reduction electrode potential \[({{E}^{o}}_{F/{{F}^{-}}}=2.87\,V)\], hence, it is easily reduced inot \[{{F}^{-}}\]and consequently \[{{F}_{2}}\] is the best oxidizing agent.
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