The properties of an electron can be described by a mathematical function called wave function (ψ). There is a radial and an angular part of every wave function. As the radial wave function changes sign from positive to negative, it crosses the x-axis and that point where ψ=0 is called a radial node. Angular nodes can be found by looking at the 3D diagram of the atomic orbitals. An angular node exists in a plane where the electron density is zero. For example,

This is an s orbital. It has no angular node as all the planes intersect the orbital.

These are p orbitals. They each have one angular node, which occurs on the plane which doesn’t touch the orbitals i.e. xy plane for 2pz, yz plane for 2px and xz plane for 2py.

These are d orbitals. They each have two angular nodes, again occurring on the plane that doesn’t touch the orbitals, i.e. yz and xy planes for 3dxz, xy and xz planes for 3dyz, xz and yz planes for 3dxy, planes bisecting the xz and yz planes for 3dx2-y2 and for 3dz2 are the two cones in the z axis facing in opposite direction with their vertices meeting at the origin, as shown in picture on the right

The total number of nodes (angular + radial) is equal to n-1, where n is the principal quantum number. Hence,
1s orbital has no angular no radial node
2s orbital has no angular node and one radial node
2p orbitals have one angular node and no radial node
3s orbital has no angular node and two radial nodes
3p orbitals have one angular node and one radial node
3d orbitals have two angular nodes and no radial node
and so on.

Therefore, by knowing the 3D shapes of the orbitals and therefore the number of angular nodes, we can find out the number of radial nodes the orbital has, which corresponds to the graphs below.

Molecular orbitals resulting from combining atomic orbitals can be constructed in phase (ψ+) and out-of-phase (ψ-). In phase addition occurs when the wavefunctions of the two atomic orbitals combined are of the same sign. For example, in the simplest case of 1s AO which is positive, adding two such AO will lead in a constructive interference. A bonding MO is thus formed. Similarly subtracting the AOs leads to a destructive interference forming an antibonding MO.

In the region between the two nuclei in in phase addition the value of the MO wavefunctions is greater than that of the AO wavefunctions. Thus it is more like likely to find an electron in the inter-nuclear region.   
The in-phase MO is lower in energy than the isolated AOs because:
1. When the MO is occupied, there is increased electron density between the nuclei. The attraction from both nuclei leads to a lowering of the potential energy of the system.
2.  An electron in in-phase MO is less constrained compared when in an AO (i.e. it is more delocalized) leading to a decrease in its kinetic energy.
The in-phase combination MO is lower in energy than the original AO and occupancy of this orbital gives rise to bonding, thus this orbital is called the bonding molecular orbital.

For the out of phase addition, the anti-bonding combination has a node – where the electron density is zero. This is shown by the crossing point on the x axis (horizontal line).
    
The out-of-phase MO is higher in energy than the isolated AOs because:
1. When the MO is occupied, there is much more electron density outside the inter-nuclear region than in between the nuclei pulling them apart from each other. The potential energy in this MO is therefore higher than in the separate species.
2. The out-of-phase MO contains a node. This means that the kinetic energy of the electron in this orbital is greater than when it is in either the AO or in-phase MO.
The out-of-phase combination MO leads to an increase of the energy with respect of the original AO, thus this orbital is known as the anti-bonding molecular orbital.

This can be summarized graphically using an energy level diagram:

As described in Part 2, the filling up of molecular orbitals follow the same principles and rules as filling up atomic orbitals. Now, we can use energy level diagrams to rationalise differences in the stability of molecules. Let us look at the molecules H­₂⁺, H₂, He₂⁺ and He₂

The molecular orbitals would look like:

The table below shows the electronic configurations, dissociation energies and bond lengths of the respective molecules.

MO theory helps us explain the difference in the dissociation energies of the molecules. H­₂⁺, H₂ and He₂⁺ are stable with respect to their dissociation into their constituents because they have more electrons in the bonding MO than the anti-bonding MO. He₂ however has two electrons in the bonding MO and two in the anti-bonding MO. The anti-bonding MO is raised more in energy than the bonding MO is lowered, so we can expect the two anti-bonding electrons to outweigh the two bonding ones. As a result, He₂ is unstable and will dissociate into He atoms.

We have discussed the concept of bond order in the previous post. The bond order can also be considered when determining whether a bond forms (i.e. whether the molecule is stable). For example,
H­₂⁺ has 1e in bonding, 0 in anti-bonding –> bond order = 1/2
H₂ has 2e in bonding, 0 in anti-bonding –> bond order = 1
He₂⁺ has 2e in bonding, 1 in anti-bonding –> bond order = 1/2
However He₂ has 2e in bonding, 2 in anti-bonding –> bond order = 0. Therefore it is unstable

The advantage of MO theory becomes more apparent when we think about π bonds, especially in those situations where two or more π bonds are able to interact with one another. Some of the delocalized molecular orbitals that result will be stabilised, while others will be destabilised.

Let us first consider the π bond in butadiene using MO theory.

From valence orbital theory we might expect that the C₂-C₃ bond in this molecule to be able to rotate freely since it is a sigma bond. Experimentally, however, it is found that there are significant barriers to rotation about this bond. It is also found that the entire molecule is planar.  The C₂-C₃ bond, while longer than the C₁-C₂ and C₃-C₄ double bonds, is also significantly shorter than a typical carbon-carbon single bond.

These observations can be attributed to the concept of delocalized π bonds. The four p orbitals are all parallel to one another, and thus there is π-overlap not just between C₁ and C₂ and C₃and C₄, but between C₂ and C₃ as well.  The 4 atomic orbitals have combined to form 4 π molecular orbitals.

In the π system, there are no degenerate MO i.e. all MO will be in different energy levels. Thus we will have 4 MOs, all with different energy levels.

The lowest energy molecular orbital, Ψ₁, has zero nodes, and is a bonding MO.
Slightly higher in energy is the Ψ₂ orbital. This orbital has one node between C₂ and C₃, but is still a bonding orbital due to the two constructive interactions between C₁-C₂ and C₃-C₄.
The two higher energy MOs are denoted Ψ₃* and Ψ₄*, and are anti-bonding. The energy of both of these MOs is higher than that of the AOs of which they are composed.
Ψ₃* has two nodes and one constructive interaction.
Ψ₄* has three nodes and zero constructive interactions.

Filling up the MOs with regards to Aufbau’s principle, we will get 2 pairs of electrons on the bonding MOs and no electron on the anti-bonding MOs. This gives us a bond order of ½(4-0) = 2. This thus corresponds to the number of π bonds that the molecule has!

Now, you might be wondering, how on earth are you supposed to come up with the MO diagram of the π system!? Fret not, there is a trick to this!

First, we draw out evenly spaced dot representing the positions of the atom (numbered 1 to 4).

Then we add two extra atomic positions – one dot to the left of 1 and one dot the right of 4 numbered 0 and 5. There are no atoms here but these are necessary for the graphical constructions of the desired MOs.

Then, draw a half sine wave is between the positions 0 and 5.

The contribution of each AO to the MO is given by the height of the sine wave at the position of that atom.

Similarly, all the other three MOs can be deduced this way. The MO above gives the MO with lowest energy, and it has zero nodes. With increasing energy the number of nodes increase, i.e. the next MO will have 1 node, and the one after that will have 2 nodes, and so on. The MOs for the π-system is shown below:

With this method, the MOs for a π-system composed by any number of p-orbitals can be predicted. 😀

Now, to determine whether a π MO is bonding, non-bonding or anti-bonding, we have to see whether there are more constructive or destructive interactions. Let us take a look at the above example.

For 1π, all interactions are constructive, hence MO is bonding.
For 2π, 2 interactions are constructive (between positions 1 & 2 and 3 & 4) while one is destructive (between positions 2 & 3). Hence MO is bonding.
For 3π, 1 interaction is constructive (between positions 2 & 3) while two are destructive (between positions 1 & 2 and 3 & 4). Hence MO is anti-bonding.
For 4π, all interactions are destructive, hence MO is anti-bonding.

Note that when there is an odd number of MOs, the MO with equal number of constructive and destructive interactions will be non-bonding.