Let's find the MO description of H2. 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.
The formation of MOs from 1s orbitals, illustrated
with approximate 1D waves, showing Ψ vs radius. The small circles
show the positions of the nuclei. We add the 2 waves on the left to
get the total wave on the right.The formation of MOs from 1s orbitals, illustrated
with approximate 2D drawings. The small pink dots show the position of
the nuclei. Positive amplitude of
Ψ is shown with black, negative amplitude of Ψ is shown with
white, on a gray background. Where positive and negative amplitude
waves overlap, the sum approaches zero (gray
background).
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 jumprope 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 H2, 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 energy level diagram for H2. The vertical axis
represents energy and the thick bars represent orbitals. The outside
bars (labelled 1s) represent the H AOs. The middle bars represent the
bonding and antibonding MOs. Because the bonding MO is lower in
energy, both electrons (↿ and ⇂) go
there.
MO theory for He2
Now let's think about He2. 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 He2 does not
exist, we can conclude that the anti-bonding orbital increases in
energy more than the bonding orbital decreases in energy, so that
He2 is higher energy than 2He.
MO energy level diagram for
He2. Because He2 doesn't exist, we can conclude
that ΔEa + ΔEb > 0.
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 some examples.
Molecule
Bond Length (Å)
Bond energy
(kcal/mol)
He2
*
*
H2+
1.06
61
He2+
1.08
55
H2
0.74
103
Now try drawing the MO diagram for each molecule. What do you notice?
If we calculate
net bonding electrons = (number 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)
He2
2
2
0
*
*
H2+
1
0
1
1.06
61
He2+
2
1
1
1.08
55
H2
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 = [(bonding e—) — (anti-bonding
e—)]/2
If bond order = 1, there is a single bond (H2, 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.
