JEE Advanced 2023 · Paper 2, Chemistry question 1

Electron count. $\ce{F2}$ has 14 valence electrons.

Level order. For $\ce{O2}$ and $\ce{F2}$ there is little 2s-2p mixing. So the $\sigma(2p_z)$ orbital (labelled $2\sigma$) lies below the two $\pi(2p)$ orbitals ($1\pi$). The order is
$$1\sigma<1\sigma^*<2\sigma<1\pi<1\pi^*<2\sigma^*.$$

Filling.
$$1\sigma^2\,1\sigma^{*2}\,2\sigma^2\,1\pi^4\,1\pi^{*4}\,2\sigma^{*0}$$
This gives bond order 1 and no unpaired electrons.

Matching the diagrams. Only diagram (C) has this order and filling.

• (A) puts $1\pi$ below $2\sigma$ (the $\ce{N2}$-type order).
• (B) and (D) put electrons in $2\sigma^*$ while leaving $1\pi^*$ half filled.