The correct molecular orbital diagram for F _ 2 molecule in the...

The electronic configuration of fluorine atom is $1{s}^{2}2{s}^{2}2{p}^{5}$. The fluorine molecule has $18$ electrons. The $s - p$ mixing occurs when the $s$ and $p$ orbitals have similar energies. The energy difference between $2s\text{and}2p$ orbitals in $O, F, \text{and}Ne$ is greater than that in $Li, Be, B, C, \text{and}N$. Hence, the molecular orbital electronic configuration of ${F}_{2}$ is $\sigma\left(1{s}^{2}\right)\sigma * \left(1{s}^{2}\right)\sigma\left(2{s}^{2}\right)\sigma * \left(2{s}^{2}\right)\sigma\left(2{{p}_{z}}^{2}\right)\pi\left(2{{p}_{x}}^{2}\right) = \pi\left(2{{p}_{y}}^{2}\right)\pi * \left(2{{p}_{x}}^{2}\right) = \pi * \left(2{{p}_{y}}^{2}\right)\sigma * \left(2{p}_{z}\right)$