The bond dissociation energy $(E)$ and bond length $(R)$ of $\mathrm{O}_2, \mathrm{~N}_2$ and $\mathrm{F}_2$…
The bond dissociation energy $(E)$ and bond length $(R)$ of $\mathrm{O}_2, \mathrm{~N}_2$ and $\mathrm{F}_2$ follow the order as:
Solution
Greater the bond order, shorter is the bond length and thus, stronger is the bond.
Since, for the given molecules, bond order follows the trend
$\begin{array}{ccc}\mathrm{N}_2 & >\mathrm{O}_2 & >\mathrm{F}_2 \\ (\mathrm{~B} . \mathrm{O}=3) & (\mathrm{B} . \mathrm{O}=2) & (\mathrm{B} . \mathrm{O}=1)\end{array}$
Thus, the bond length decreases as $\mathrm{F}_2>\mathrm{O}_2>\mathrm{N}_2$ and bond dissociation energy follows the trend $\mathrm{N}_2>\mathrm{O}_2>\mathrm{F}_2$
*
