$MO$ electronic configuration of $F_2$ molecule. $\sigma 1{s^2},\,{\sigma ^*}1{s^2},\,\sigma 2{s^2},\,{\sigma ^*}2{s^2},\,\sigma 2p_x^2,\,\pi 2p_y^2 \approx \,\pi 2p_z^2,\,{\pi ^*}2p_y^2 \approx {\pi ^*}2p_z^2$ Thus there are $10$ electrons in bonding orbitals and $8$ electrons in antibonding orbitals. Thus, reason is true and it is also the correct explanation of assertion because Bond order $=$ $\frac{{{N_b} – {N_a}}}{2} = \frac{{10 – 8}}{2} = 1$