Animated Solution for Chemistry - Chemical Bonding and Molecular Structure: The correct molecular orbital diagram for ${\text{F}}_2$ molecule in the ground state is

The journey into Molecular Orbital (MO) Theory is one of the most fascinating transitions in chemistry. We move from simple Lewis dot structures to a quantum mechanical view where electrons flow across the entire molecule. Today, we are going to decode the molecular orbital diagram for the fluorine molecule, $F_2$.

I know these diagrams can look like a terrifying web of lines and arrows, but let's take a breath. Once you understand the underlying logic, it becomes as simple as filling seats in a theater.

The Electron Count

Before we draw a single line, we must know our inventory. How many electrons are we dealing with?

A single fluorine atom has an atomic number of 9, meaning it has 9 electrons. Therefore, the $F_2$ molecule has a total of $18$ electrons.

However, the inner core electrons (the $1s$ electrons) are held so tightly by the nucleus that they barely participate in bonding. To keep things clean, we focus entirely on the valence electrons. Each fluorine atom brings $7$ valence electrons to the table, giving us a total of $14$ valence electrons to distribute into our molecular orbitals.

Mixing the 2s Orbitals

Let's start from the ground up. When the two fluorine atoms approach each other, their $2s$ atomic orbitals interact.

According to MO theory, when two atomic orbitals combine, they form two molecular orbitals: one lower in energy (bonding) and one higher in energy (antibonding).

The $2s$ orbitals form the $1\sigma$ (bonding) and $1{\sigma }^{\ast }$ (antibonding) orbitals. Since we have two electrons from each $2s$ orbital, we have a total of $4$ electrons to place here. We fill the lower energy $1\sigma$ orbital with two electrons, and the remaining two go into the $1{\sigma }^{\ast }$ orbital.

So far, so good. The real magic happens in the $2p$ sublevel.

The s-p Mixing Trap

Here is where mistakes happen. This is a favorite concept for JEE!

For lighter elements like Boron ($B_2$), Carbon ($C_2$), and Nitrogen ($N_2$), the energy gap between the $2s$ and $2p$ orbitals is relatively small. This allows them to mix, which pushes the ${\sigma }_{2p}$ orbital higher in energy, placing it above the ${\pi }_{2p}$ orbitals.

But fluorine is highly electronegative. Its nucleus pulls the $2s$ electrons incredibly close, creating a massive energy gap between the $2s$ and $2p$ orbitals. Because of this large gap, s-p mixing is negligible in $O_2$ and $F_2$.

Without this mixing, the natural overlap of the orbitals dictates the energy. Head-on overlap (sigma) is stronger than sideways overlap (pi). Therefore, the $2\sigma$ bonding orbital drops lower in energy than the $1\pi$ bonding orbitals.

This is the critical constraint: in our diagram, the $2\sigma$ line must be below the $1\pi$ lines!

Filling the 2p Orbitals

We have $10$ valence electrons left to place in the $2p$ molecular orbitals. Let's fill them in order of increasing energy:

1. First, $2$ electrons go into the lowest available orbital, the $2\sigma$. 2. Next, $4$ electrons fill the degenerate $1\pi$ orbitals. 3. We still have $4$ electrons left. These go into the higher energy $1{\pi }^{\ast }$ antibonding orbitals, completely filling them.

If we look at the options provided in the question, we can instantly eliminate (A) and (B) because they incorrectly place the $1\pi$ orbitals below the $2\sigma$ orbital.

Between (C) and (D), we must check the electron count. Option (D) only has $2$ electrons in the $1{\pi }^{\ast }$ orbital, which corresponds to $O_2$ (with $12$ valence electrons). Option (C) correctly shows $4$ electrons in the $1{\pi }^{\ast }$ orbital, perfectly matching our $14$ valence electron count for $F_2$.

Always think beyond just finding the answer! What else does this diagram tell us?

We can calculate the bond order: This confirms that fluorine forms a single bond. Furthermore, since every single electron in the diagram is paired up, we can confidently state that $F_2$ is diamagnetic.

Mastering these diagrams gives you the power to predict the physical reality of molecules. Keep visualizing, and the math will follow!