Bond energy is maximum in

As bond order increases, bond energy increases.

$F_2:\sigma 1s^2\sigma *1s^2\sigma 2s^2\sigma *2s^2\sigma 2p_x^2\pi 2p_y^2\pi 2p_z^2\pi *2p_y^2\pi *2p_z^2 ;B.O=\frac{10-8}{2}=1$

$N_2 : \sigma 1s^2\sigma *1s^2\sigma 2s^2\sigma *2s^2\pi 2p_y^2\pi 2p_z^2\sigma 2p_x^2 ; B.O=\frac{10-4}{2}=3$

$O_2:\sigma 1s^2\sigma *1s^2\sigma 2s^2\sigma *2s^2\sigma 2p_x^2\pi 2p_y^2\pi 2p_z^2\pi *2p_y^1\pi *2p_z^1;B.O=\frac{10-6}{2}=2$

Bond energy of Br₂ is higher than that of F₂ due to small size and higher inter electronic repulsions in 'F'

Hence the order pof bond energy is, ${\mathrm{F}}_2<{\mathrm{Br}}_2<{\mathrm{O}}_2<{\mathrm{N}}_2$