## What is the molecular electron configuration of \(\mathrm{F}_{2} ?\)

Answer:

\(\left(\sigma_{2 s}\right)^{2}\left(\sigma_{2 s}^{*}\right)^{2}\left(\sigma_{2 p_{z}}\right)^{2}\left(\pi_{2 p_{x}}\right)^{2}\left(\pi_{2 p_{y}}\right)^{2}\left(\pi_{2 p_{x}}^{*}\right)^{2}\left(\pi_{2 p_{y}}^{*}\right)^{2}\)

Recall that there are orbital mixing effects for homonuclear diatomic molecules that decrease from left to right until \(\mathrm{N}_{2}\) (inclusive), which gives rise to a molecular orbital ordering of \(\left(\pi_{2 p_{x}}, \pi_{2 p_{y}}\right)\) and then \(\sigma_{2 p_{z}}\), upwards. See here if you don’t remember.

Notice how the \(\sigma_{g}(2 p)\), or the \(\sigma_{2 p_{z}}\) molecular orbital, dips down below the \(\pi\) molecular orbital energies after \(\mathrm{N}_{2}\).

Since \(\mathrm{F}_{2}\) is after \(\mathrm{N}_{2}\) in the second row of the periodic table (where these effects are not present), the orbital energy ordering is “normal”.

In general, the molecular orbital energies follow these rules:

The relative atomic orbital energy differences approximate the relative \(\sigma / \sigma\) orbital energy differences.

So \(\sigma_{2 s}\) molecular orbitals are significantly lower in energy than \(\sigma_{2 p_{z}}\) molecular orbitals because the corresponding \(2s\) atomic orbitals are significantly lower in energy than the \(2 p^{\prime} \mathrm{s}\).

\(\sigma\) molecular orbitals are singly-degenerate, and \(\pi\) molecular orbitals are doubly-degenerate.

For example, an \(n s / n s\) overlap for a homonuclear diatomic molecule gives rise to a partial MO diagram like this:

and an \(n p / n p\) overlap for \(\mathrm{O}_{2}\) and \(\mathrm{F}_{2}\) gives:

So, the full MO diagram is:

Thus, the valence electron configuration is:

\(\left(\sigma_{2 s}\right)^{2}\left(\sigma_{2 s}^{*}\right)^{2}\left(\sigma_{2 p_{z}}\right)^{2}\left(\pi_{2 p_{x}}\right)^{2}\left(\pi_{2 p_{y}}\right)^{2}\left(\pi_{2 p_{x}}^{*}\right)^{2}\left(\pi_{2 p_{y}}^{*}\right)^{2}\)