1. Determine the total number of electrons in $\mathrm{F}_2$

Each fluorine atom has atomic number $9$, so its electronic configuration is: $1s^2\,2s^2\,2p^5$

Thus, for $\mathrm{F}_2$:

• Total electrons $= 9 + 9 = 18$
• Valence electrons $= 7 + 7 = 14$

For molecular orbital filling, we mainly focus on the valence MOs formed from $2s$ and $2p$ orbitals.

2. MO energy order for $\mathrm{F}_2$

For diatomic molecules from oxygen onward ($\mathrm{O}_2, \mathrm{F}_2, \mathrm{Ne}_2$), the correct order of molecular orbitals is: $\sigma(2s) < \sigma^*(2s) < \sigma(2p_z) < \pi(2p_x)=\pi(2p_y) < \pi^*(2p_x)=\pi^*(2p_y) < \sigma^*(2p_z)$

This is the key point.

For $\mathrm{F}_2$, the $\sigma(2p)$ orbital lies below the $\pi(2p)$ orbitals.

3. Fill the 14 valence electrons

Now fill the electrons in order:

$\sigma(2s)^2$ $\sigma^*(2s)^2$ $\sigma(2p_z)^2$ $\pi(2p_x)^2 = \pi(2p_y)^2$ $\pi^*(2p_x)^2 = \pi^*(2p_y)^2$

So the full valence MO configuration is: $\sigma(2s)^2\,\sigma^*(2s)^2\,\sigma(2p_z)^2\,[\pi(2p_x)]^2[\pi(2p_y)]^2[\pi^*(2p_x)]^2[\pi^*(2p_y)]^2$

The orbital $\sigma^*(2p_z)$ remains empty.

4. Bond order check

Bond order is: $\text{Bond order} = \frac{N_b - N_a}{2}$ where:

• $N_b =$ bonding electrons
• $N_a =$ antibonding electrons

Bonding electrons: $2(\sigma 2s) + 2(\sigma 2p_z) + 4(\pi 2p) = 8$

Antibonding electrons: $2(\sigma^* 2s) + 4(\pi^* 2p) = 6$

Therefore, $\text{Bond order} = \frac{8-6}{2} = 1$

This is correct for $\mathrm{F}_2$.

5. Magnetic nature

All electrons are paired, so $\mathrm{F}_2$ is diamagnetic. Thus the correct diagram must show:

• $\sigma(2p)$ below $\pi(2p)$
• both $\pi^*(2p)$ orbitals fully filled
• $\sigma^*(2p)$ empty
• all electrons paired

6. Match with the given options

Since the correct option must represent the above MO ordering and filling, the correct diagram is Option C.

7. Comparison with stored answer

Stored correct answer: C

My derived answer: C

They agree.