A) \[{{F}_{2}}\]
B) \[{{I}_{2}}\]
C) \[C{{l}_{2}}\]
D) \[B{{r}_{2}}\]
Correct Answer: A
Solution :
Since all the halogens have strong tendency to accept electrons, they act as strong oxidising agent. \[{{F}_{2}}+2{{e}^{-}}\xrightarrow{{}}2{{F}^{-}};\,\,\,\,\,\,\,\,\,\,\,\,{{E}^{o}}=+2.87V\] \[C{{l}_{2}}+2{{e}^{-}}\xrightarrow{{}}2C{{l}^{-}};\,\,\,\,\,\,\,\,\,\,\,\,{{E}^{o}}=+1.36V\] \[B{{r}_{2}}+2{{e}^{-}}\xrightarrow{{}}2B{{r}^{-}};\,\,\,\,\,\,\,\,\,\,\,\,{{E}^{o}}=+1.09V\] \[{{I}_{2}}+2{{e}^{-}}\xrightarrow{{}}2{{I}^{-}};\,\,\,\,\,\,\,\,\,\,\,\,{{E}^{o}}=+0.54V\] The electrode potential of F; is the maximum while that of \[{{I}_{2}}\] is the minimum. So, \[{{F}_{2}}\] is the strongest oxidising agent while \[{{I}_{2}}\] is the weakest oxidising agent.
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