A) \[{{F}_{2}}>C{{l}_{2}}>B{{r}_{2}}>{{I}_{2}}:\] Oxidizing power
B) \[{{F}_{2}}>C{{l}_{2}}>B{{r}_{2}}>{{I}_{2}}:\] Electron gain enthalpy
C) \[{{F}_{2}}>C{{l}_{2}}>B{{r}_{2}}>{{I}_{2}}:\] Bond dissociation energy
D) \[{{F}_{2}}>C{{l}_{2}}>B{{r}_{2}}>{{I}_{2}}\]: Electronegativity
Correct Answer: C
Solution :
Key Idea: Generally bond dissociation energies decreases in a group. Bond dissociation energy also decreases with repulsion. \[\begin{matrix} \text{X}-\text{X}\,\text{Bond} & \text{F}-\text{F} & \text{Cl}-\text{Cl} & \text{Br}-\text{Br} & \text{I}-\text{I} \\ \text{Bond}\,\text{length}\,\text{({ }\!\!\mathrm{\AA}\!\!\text{ })} & \text{1}\text{.42} & \text{1}\text{.99} & \text{2}\text{.28} & \text{2}\text{.67} \\ \text{Bond}\,\text{dissociation} & \text{38} & \text{57} & \text{45}\text{.5} & \text{35}\text{.6} \\ \text{energy}\,\text{(kcal/mol)} & {} & {} & {} & {} \\ \end{matrix}\] In general the bond dissociation energy decreases as the bond length increases, but the bond dissociation energy of \[{{F}_{2}}\] is less than that of \[C{{l}_{2}}.\]It is due to greater inter electronic repulsions between the lone pair of electrons on the two bonded fluorine atoms. Hence, the order of bond dissociation energy is as: \[C{{l}_{2}}>B{{r}_{2}}>{{F}_{2}}>{{I}_{2}}\]
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