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Bond energy of $\mathrm{Cl}_2, \mathrm{Br}_2$ and $\mathrm{I}_2$ follow the order
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The correct answer is:
$\mathrm{Cl}_2>\mathrm{Br}_2>\mathrm{I}_2$
As the size of atom increases, $A-A$ bond length increases and hence, $A-A$ bond energy decreases. (where $A=\mathrm{Cl}, \mathrm{Br}, \mathrm{I})$
$\therefore$ The order of bond energy is
$\mathrm{Cl}_2>\mathrm{Br}_2>\mathrm{I}_2$
$\therefore$ The order of bond energy is
$\mathrm{Cl}_2>\mathrm{Br}_2>\mathrm{I}_2$
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