Correct Molecular Orbital Diagram for Fluorine Molecule in Ground State

To determine the correct molecular orbital (MO) diagram for the fluorine molecule ($\text{F}_2$) in its ground state, we analyze the atomic electronic configuration, $s\text{-}p$ mixing, and electron filling according to the Aufbau principle, Pauli exclusion principle, and Hund's rule.

1. Electronic Configuration of Fluorine Atom

A fluorine atom ($Z = 9$) has the ground-state electronic configuration: $\text{F}: 1s^2\, 2s^2\, 2p^5$ Thus, each fluorine atom contributes $7$ valence electrons, giving a total of $14$ valence electrons for the $\text{F}_2$ molecule.

2. Relative Energy Levels of Molecular Orbitals ($s\text{-}p$ Mixing)

For homonuclear diatomic molecules of period-2 elements heavier than nitrogen (e.g., $\text{O}_2$ and $\text{F}_2$), the energy difference between the $2s$ and $2p$ atomic orbitals is relatively large ($\Delta E(2s-2p) > 15\text{ eV}$). Consequently, $s\text{-}p$ mixing is negligible.

Without $s\text{-}p$ mixing, the bonding $\sigma$ molecular orbital formed from $2p_z$ orbitals ($2\sigma$) lies lower in energy than the bonding $\pi$ molecular orbitals ($1\pi$).

The relative energy sequence for the valence molecular orbitals in $\text{F}_2$ is: $\sigma_{2s} \;(1\sigma) < \sigma^*_{2s} \;(1\sigma^*) < \sigma_{2p_z} \;(2\sigma) < \pi_{2p_x} = \pi_{2p_y} \;(1\pi) < \pi^*_{2p_x} = \pi^*_{2p_y} \;(1\pi^*) < \sigma^*_{2p_z} \;(2\sigma^*)$

3. Filling of Valence Electrons

Filling the $14$ valence electrons into the molecular orbitals in order of increasing energy:

1. $1\sigma$ orbital receives $2$ electrons: $(1\sigma)^2$
2. $1\sigma^*$ orbital receives $2$ electrons: $(1\sigma^*)^2$
3. $2\sigma$ orbital receives $2$ electrons: $(2\sigma)^2$
4. $1\pi$ orbitals (doubly degenerate) receive $4$ electrons: $(1\pi)^4$
5. $1\pi^*$ orbitals (doubly degenerate) receive $4$ electrons: $(1\pi^*)^4$
6. $2\sigma^*$ orbital remains empty: $(2\sigma^*)^0$

The complete valence ground-state electron configuration of $\text{F}_2$ is: $(1\sigma)^2\, (1\sigma^*)^2\, (2\sigma)^2\, (1\pi)^4\, (1\pi^*)^4\, (2\sigma^*)^0$

4. Comparison with the Options

• Option (A) correctly places $2\sigma$ below $1\pi$ and fills $1\sigma$, $1\sigma^*$, $2\sigma$, $1\pi$, and $1\pi^*$ completely, leaving $2\sigma^*$ empty.
• Option (B) incorrectly places $2\sigma^*$ below $1\pi^*$ and misallocates electron pairs.
• Options (C) and (D) incorrectly show $1\pi$ lower in energy than $2\sigma$, which occurs only in molecules with significant $s\text{-}p$ mixing ($\text{B}_2, \text{C}_2, \text{N}_2$).

Thus, the correct molecular orbital diagram is given by Option (A).