A) \[{{F}_{2}}<C{{l}_{2}}<B{{r}_{2}}<{{I}_{2}}\]
B) \[{{F}_{2}}>C{{l}_{2}}>B{{r}_{2}}>{{I}_{2}}\]
C) \[C{{l}_{2}}>B{{r}_{2}}>{{F}_{2}}>{{I}_{2}}\]
D) \[C{{l}_{2}}>{{F}_{2}}>B{{r}_{2}}>{{I}_{2}}\]
Correct Answer: C
Solution :
In general, as the bond length (or the size of atom) increases, bond dissociation energy decreases. But the bond energy of \[\text{F }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ F}\]bond is lesser as compared to \[\text{C}{{\text{l}}_{\text{2}}}\]and \[\text{B}{{\text{r}}_{2}}.\]This is because of greater inter electronic repulsions between the lone pairs of electrons of two fluorine atoms. Such a repulsion is not found in case of other halogens. Thus, the order of bond energy is| \[C{{l}_{2}}\] > \[B{{r}_{2}}\] > \[{{F}_{2}}\] > \[{{I}_{2}}\] | ||||
| \[(k-cal/mol)\] | 57 | 45.5 | 38 | 35.56 |
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