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