A) \[C{{l}_{2}}\]
B) \[B{{r}_{2}}\]
C) \[{{I}_{2}}\]
D) \[{{F}_{2}}\]
Correct Answer: D
Solution :
The bond dissociation energies of \[{{F}_{2}},\,C{{l}_{2}},\,\,B{{r}_{2}}\]la are 155, 244, 193 and 151 kJ \[mo{{l}^{-1}}\] respectively. It decreases from chlorine to iodine because size of the atom increases and hence, the larger atom are unable to exert a greater attraction on the electron pair that holds the molecule together. But the dissociation energy of fluorine is very low due to following two factors (i) The F?F bond in fluorine is weak due to large repulsion between the non-bonding electrons in the small fluorine molecule. (ii) No possibility of multiple bonding in fluorine due to non-availability of d-orbitals. But other halogens have possibility of forming multiple bonding by the overlap of filled p-orbital of one atom with an empty d-orbital of the other atom. Multiple bonds are always stronger than the single bond. Therefore, fluorine has the lowest bond dissociation energy and due to this the weakest bond will be F?F bond.
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