JEE Advanced 2023 Paper 2, Chemistry Section 1 Q1: Molecular Orbital Theory

$F_2$ has $2\times 9=18$ electrons. For $F_2$ (and $O_2$), $s$-$p$ mixing is small, so the bonding ${\sigma }_{2p_z}$ (here labelled $2\sigma$) lies below the bonding ${\pi }_{2p_x},{\pi }_{2p_y}$ (labelled $1\pi$).

The ground-state configuration is therefore

$(1\sigma {)}^2(1{\sigma }^{\ast }{)}^2(2\sigma {)}^2(1\pi {)}^4(1{\pi }^{\ast }{)}^4(2{\sigma }^{\ast }{)}^0$.

Bond order $=\frac{8-6}{2}=1$, consistent with the single bond in $F_2$.

Only diagram (C) shows this ordering with $2\sigma$ doubly occupied below the filled $1\pi$ and a fully filled $1{\pi }^{\ast }$ with $2{\sigma }^{\ast }$ empty.