MO Theory: Simplest Examples
Skills to Develop
- Construct MO diagrams for H 2 and He 2
- Define bond order in theory and calculation
MO Theory for H 2
Let's find the MO description of H 2 . We are going to use one electron on each H atom, each in a 1s orbital. When we bring the 2 atoms next to each other, we can make 2 MOs out of the 2 1s AOs. As always, we will do a + combination and a — combination. These combinations are illustrated in the figure below.
Notice that the + combination produces an MO with more electron density between the nuclei, because the waves interfere constructively. The — combination produces an MO with a node between the nuclei, and not much electron density there, because the waves interfere destructively. If you think about the Coulomb's law forces, and imagine putting 2 electrons in the + MO, they will usually be between the nuclei, and the attractions between the electrons and nuclei will hold the nuclei together, making the molecule. For this reason, the + MO is called a bonding MO . On the other hand, if we imagine putting 2 electrons in the — MO, they will usually be on the outside of the nuclei, so the repulsion between the nuclei will push them apart, and no molecule will form. For this reason, the — MO is called an anti-bonding MO .
You've seen before that more nodes means higher energy. (For instance, if you try to swing a jump rope so it has a standing wave with 2 nodes, that is much harder than making a standing wave with no nodes.) So it makes sense that the bonding MO (no nodes) is lower in energy than the anti-bonding MO (1 node). Thus, in H 2 , both electrons will go in the bonding MO, and the molecule is stable. In fact, the bonding orbital will be lower in energy than the AOs it was made from, because of the increased Coulomb attractions. The anti-bonding orbital will be higher in energy than the AOs because of the increased Coulomb repulsions. We can represent this with an MO diagram, shown in the figure.
MO Theory for He 2
Now let's think about He 2 . We still have a combination of 2 AOs, both 1s. The bonding and anti-bonding orbitals will look very similar. But now we have 4 electrons, so we will have to put 2 electrons in each MO. Because the molecule He 2 does not exist, we can conclude that the anti-bonding orbital increases in energy more than the bonding orbital decreases in energy, so that He 2 is higher energy than 2 He.
Bond Order in MO Theory
One great thing about MO theory is that it makes it really simple to think about partial bonds and weird molecules, like radicals. The table shows some data for a few examples.
| Molecule |
Bond Length
(Å) |
Bond energy
(kcal/mol) |
|---|---|---|
| He 2 | * | * |
| H 2 + | 1.06 | 61 |
| He 2 + | 1.08 | 55 |
| H 2 | 0.74 | 103 |
Now try drawing the MO diagram for each molecule. What do you notice? If we calculate
net bonding electrons = (number of bonding electrons) — (number of anti-bonding electrons)
we get:
| Molecule |
Bonding
Electrons |
Anti-Bonding
Electrons |
Net Bonding
Electrons |
Bond Length
(Å) |
Bond energy
(kcal/mol) |
|---|---|---|---|---|---|
| He 2 | 2 | 2 | 0 | * | * |
| H 2 + | 1 | 0 | 1 | 1.06 | 61 |
| He 2 + | 2 | 1 | 1 | 1.08 | 55 |
| H 2 | 2 | 0 | 2 | 0.74 | 103 |
The number of net bonding electrons predicts the length and strength of the bond! Since we normally think of a chemical bond as a 2-electron bond, we can define the bond order like this:
\[Bond\; Order = \frac{\left(bonding\; e^{-}\right) - \left(anti \mbox{-} bonding\; e^{-}\right)}{2}\]
If bond order = 1, there is a single bond (H 2 , for instance). If bond order = 0, we expect no bond. If bond order = 0.5, we have a 1-electron bond or a half bond. It's approximately half as strong as a 2-electron bond.