8.4 Molecular Orbital Theory

Molecular Orbital Theory

Molecular Orbital Theory (MO Theory) explains bonding by showing how atomic orbitals on separate atoms merge to form new orbitals that belong to the entire molecule. Unlike Lewis structures or VSEPR, MO Theory can predict magnetic properties and explain why some molecules (like $He_2$) don't exist at all.

Molecular Orbital Theory

Derivation of Molecular Orbitals

Molecular orbitals form through the Linear Combination of Atomic Orbitals (LCAO) method. The idea: when two atoms get close enough, their atomic orbitals (AOs) overlap and combine into new orbitals that spread across both nuclei.

Two AOs combine only if they have similar energy and compatible symmetry (s + s, p + p along the same axis). Every combination produces exactly two molecular orbitals:

• Bonding MOs result from constructive interference, where the wave functions add together. Electron density builds up between the nuclei, pulling them together. These orbitals are lower in energy than the original AOs. Examples: $\sigma_{1s}$, $\sigma_{2s}$, $\pi_{2p}$.
• Antibonding MOs result from destructive interference, where the wave functions cancel. A node (a region of zero electron density) forms between the nuclei, pushing them apart. These orbitals are higher in energy than the original AOs and are marked with an asterisk. Examples: $\sigma_{1s}^*$, $\sigma_{2s}^*$, $\pi_{2p}^*$.

The underlying math comes from solving the Schrödinger equation for the molecular system, but for this course, the key takeaway is: n atomic orbitals in → n molecular orbitals out, half bonding and half antibonding.

Bonding vs. Antibonding Orbitals

A quick comparison:

Electrons in antibonding orbitals don't just "cancel out" bonding electrons. They actively work against bond formation. That's why $He_2$ (bond order = 0) doesn't form a stable molecule even though helium atoms have electrons in the bonding $\sigma_{1s}$ orbital: the two electrons in $\sigma_{1s}^*$ completely offset the stabilization.

Bond Order Calculations

Bond order tells you the net number of bonds holding a molecule together:

$\text{Bond order} = \frac{1}{2}(\text{bonding electrons} - \text{antibonding electrons})$

Steps to calculate:

1. Fill in the MO diagram for your molecule (see next section).
2. Count all electrons in bonding MOs.
3. Count all electrons in antibonding MOs.
4. Plug into the formula.

What bond order tells you:

• Bond order = 1 → single bond. Example: $H_2$ has 2 bonding electrons, 0 antibonding → $\frac{1}{2}(2-0) = 1$.
• Bond order = 2 → double bond. Example: $O_2$ has 8 bonding electrons, 4 antibonding → $\frac{1}{2}(8-4) = 2$.
• Bond order = 3 → triple bond. Example: $N_2$ has 8 bonding electrons, 2 antibonding → $\frac{1}{2}(8-2) = 3$.
• Bond order = 0 → molecule doesn't form (e.g., $He_2$).

Higher bond order = stronger bond = shorter bond length.

Electron Configurations and Molecular Properties

Diatomic Molecule Configurations

Filling MO diagrams follows the same three rules you already know from atomic electron configurations:

1. Aufbau principle: Fill the lowest-energy MO first, then work upward.
2. Pauli exclusion principle: Each MO holds a maximum of 2 electrons with opposite spins.
3. Hund's rule: When you have degenerate orbitals (same energy, like the two $\pi_{2p}$ orbitals), place one electron in each before pairing any.

The standard MO energy ordering for $Li_2$ through $N_2$ is:

$\sigma_{1s} < \sigma_{1s}^* < \sigma_{2s} < \sigma_{2s}^* < \pi_{2p} < \sigma_{2p} < \pi_{2p}^* < \sigma_{2p}^*$

Notice that for these lighter molecules, the $\pi_{2p}$ orbitals are lower in energy than $\sigma_{2p}$. This is due to s-p mixing (interaction between the 2s and 2p orbitals).

For $O_2$ and $F_2$, the ordering switches so that $\sigma_{2p}$ drops below $\pi_{2p}$:

$\sigma_{1s} < \sigma_{1s}^* < \sigma_{2s} < \sigma_{2s}^* < \sigma_{2p} < \pi_{2p} < \pi_{2p}^* < \sigma_{2p}^*$

Example configurations:

• $H_2$ (2 electrons): $\sigma_{1s}^2$
• $N_2$ (10 electrons, using the $\pi$ before $\sigma$ ordering): $\sigma_{1s}^2\, \sigma_{1s}^{*2}\, \sigma_{2s}^2\, \sigma_{2s}^{*2}\, \pi_{2p}^4\, \sigma_{2p}^2$
• $O_2$ (12 electrons, using the $\sigma$ before $\pi$ ordering): $\sigma_{1s}^2\, \sigma_{1s}^{*2}\, \sigma_{2s}^2\, \sigma_{2s}^{*2}\, \sigma_{2p}^2\, \pi_{2p}^4\, \pi_{2p}^{*2}$

Molecular Stability Predictions

Bond order directly predicts relative stability. Among second-row diatomics: $N_2$ (bond order 3) > $O_2$ (bond order 2) > $F_2$ (bond order 1). This matches experimental bond energies: $N_2$ has one of the strongest bonds known (945 kJ/mol).

MO Theory also predicts magnetic behavior, which is something Lewis structures can't do:

• Paramagnetic molecules have unpaired electrons and are attracted into a magnetic field. $O_2$ is the classic example: its two $\pi_{2p}^*$ electrons occupy separate degenerate orbitals with parallel spins (by Hund's rule), leaving 2 unpaired electrons. This is why liquid oxygen is attracted to a magnet.
• Diamagnetic molecules have all electrons paired and are slightly repelled by a magnetic field. $N_2$ is diamagnetic because every one of its MOs is fully paired.

Molecular Symmetry and Orbital Interactions

• Orbital symmetry determines which AOs can combine. Only orbitals with the same symmetry along the bond axis interact (for example, a $p_z$ orbital can form a $\sigma$ bond with another $p_z$, but not with a $p_x$).
• s-p mixing occurs when the 2s and 2p energy levels are close enough to interact, which shifts the relative energies of $\sigma_{2p}$ and $\pi_{2p}$. This is why the MO ordering differs between $N_2$ and $O_2$.
• In molecules with extended $\pi$-systems, electrons can spread across multiple atoms (delocalization), which lowers overall energy and adds stability. You'll see this idea again when studying benzene and resonance structures.