A) \[B{{r}_{2}}>{{F}_{2}}>C{{l}_{2}}>{{I}_{2}}\]
B) \[{{F}_{2}}>C{{l}_{2}}>B{{r}_{2}}>{{I}_{2}}\]
C) \[{{I}_{2}}>{{F}_{2}}>B{{r}_{2}}>C{{l}_{2}}\]
D) \[{{F}_{2}}>{{I}_{2}}>B{{r}_{2}}>C{{l}_{2}}\]
Correct Answer: B
Solution :
Bond energy of \[{{F}_{2}},C{{l}_{2}},B{{r}_{2}}\] and \[{{I}_{2}}\] are as: \[F-F\xrightarrow{{}}150.6\,mo{{l}^{-1}}(kJ)\] \[Cl-Cl\xrightarrow{{}}242.7\,mo{{l}^{-1}}(kJ)\] \[Br-Br\xrightarrow{{}}188.3\,mo{{l}^{-1}}(kJ)\] \[I-I\xrightarrow{{}}150.6\,mo{{l}^{-1}}(kJ)\] Bond energy of \[{{F}_{2}}\] is lower because of repulsion between non bonded electrons of small sized F atom. So the order is \[{{F}_{2}}<C{{l}_{2}}>B{{r}_{2}}>{{I}_{2}}\]
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