The bond dissociation energy (E) and bond length $(R)$ of $O_{2}, N_{2}, F_{2}$ follows the order as

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$(E) : O_{2} > N_{2} > F_{2} (R) : F_{2} > O_{2} > N_{2}$

$(E) : N_{2} > O_{2} > F_{2} {(R) : F}_{2} > O_{2} > N_{2}$

$(E) : N_{2} > O_{2} > F_{2} {(R) : N}_{2} > O_{2} > F_{2}$

$(E) : N_{2} > F_{2} > O_{2} (R) : N_{2} > O_{2} > F_{2}$

answer is C.

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Detailed Solution

$N_{2}$ has triple bond between $N$ atoms.
$O_{2}$ has double bond between O atoms.
$F_{2}$ has single bond between the atoms.
The bond energy will be higher, if the bonds between the atoms is high.
In $F_{2}$ , the atoms are closer and repulsions are more. So, bond energy of $F_{2}$ is less.
$∴$ Order of bond energies $= N_{2} > O_{2} > F_{2}$