The correct molecular orbital diagram for F₂ molecule in the ground state is

The correct answer is:

Step-by-Step Explanation:

1. Electronic Configuration of Fluorine Atom:

A fluorine atom (${\text{F}}_{}$) has an atomic number of 9. Its ground state electronic configuration is:

$\text{F: 1s}{2}^{}\text{2s}{2}^{}\text{2p}{5}^{}$

For the homonuclear diatomic fluorine molecule (${\text{F}}_2$), there are two fluorine atoms contributing a total of 18 electrons (or 14 valence electrons from the 2s and 2p levels).

2. Molecular Orbital Energy Level Ordering for ${\text{F}}_2$ (Total Electrons > 14):

For diatomic molecules with more than 14 electrons (such as ${\text{O}}_2$ and ${\text{F}}_2$), there is no significant s-p mixing. Consequently, the energy order of the molecular orbitals formed by valence shell atomic orbitals is as follows:

$1\sigma <1{\sigma }^{*}<2\sigma <1\pi <1{\pi }^{*}<2{\sigma }^{*}$

Here, the notation used in the diagrams labels the MOs formed from valence shell orbitals ($\text{2s}$ and $\text{2p}$) as:

• $1\sigma$ for ${\sigma }_{\text{2s}}$
• $1{\sigma }^{*}$ for ${\sigma }_{\text{2s}}^{*}$
• $2\sigma$ for ${\sigma }_{{\text{2p}}_{\text{z}}}$
• $1\pi$ for degenerate ${\pi }_{{\text{2p}}_{\text{x}}},{\pi }_{{\text{2p}}_{\text{y}}}$
• $1{\pi }^{*}$ for degenerate ${\pi }_{{\text{2p}}_{\text{x}}}^{*},{\pi }_{{\text{2p}}_{\text{y}}}^{*}$
• $2{\sigma }^{*}$ for ${\sigma }_{{\text{2p}}_{\text{z}}}^{*}$

3. Electron Filling in ${\text{F}}_2$:

Filling 14 valence electrons (4 from 2s and 10 from 2p orbitals) according to Aufbau principle, Pauli exclusion principle, and Hund's rule gives:

• $1\sigma$: 2 electrons (paired $\nleftrightarrow$)
• $1{\sigma }^{*}$: 2 electrons (paired $\nleftrightarrow$)
• $2\sigma$: 2 electrons (paired $\nleftrightarrow$)
• $1\pi$: 4 electrons (2 paired pairs in the two degenerate orbitals)
• $1{\pi }^{*}$: 4 electrons (2 paired pairs in the two degenerate antibonding orbitals)
• $2{\sigma }^{*}$: 0 electrons (empty)

4. Comparing with the Provided Options:

Let's analyze the energy order shown in the diagrams:

• In the correct diagram, the bonding orbital $2\sigma$ lies lower in energy than the degenerate $1\pi$ orbitals ($2\sigma <1\pi$).

• In the $1\pi$ level, there are 4 fully paired electrons.

• In the $1{\pi }^{*}$ level, there are 4 fully paired electrons.

• The highest energy orbital $2{\sigma }^{*}$ remains completely empty.

This exact energy level ordering and complete electron filling arrangement corresponds to the third image diagram: