The fundamental aspects of valence bond theory as outlined on the previous page are that superposition of wavefunctions leads to a low energy bonding wavefunction in which the probability of finding the electrons in the internuclear region has been substantially increased.

This discussion centred on the H2 molecule, which contains a single σ bond. To discuss more complicated molecules it is necessary to extend the theory slightly. For example, the molecule N2 contains only two atoms, but instead of a single σ bond has a triple bond joining the two atoms. The electron configuration of a nitrogen atom is 1s22s22px12py12pz1 where the z axis is, by convention, taken to be the molecular axis.

Each of the atoms that go to make up an N2 molecule therefore have a 2pz orbital pointing towards a like orbital on the other atom. These two orbitals can overlap to form a σ bond, as they have the correct cylindrical symmetry about the internuclear axis. The spatial wavefunction for this bond has the following form:

which has the same form as for the σ bond in the H2 molecule. However, in this case A and B refer to nitrogen 2p orbital wavefunctions on atoms A and B, rather than to hydrogen 1s orbitals.

The 2px and 2py orbitals do not have the correct cylindrical symmetry to overlap and form two more σ bonds. Instead, the orbitals overlap with their counterpart on the other atom to form two π bonds. π bonds arise from the side-on overlap of p orbitals, and have a nodal plane (which is the plane in which the atomic nuclei lie). Sausage-shaped regions of electron density lie above and below the plane:

There are two π bonds in N2, one formed from overlap and spin pairing between the px orbitals, and one formed likewise from the py orbitals.

Further extension of the theory is necessary to explain the bonding in carbon. The ground-state electronic configuration of a carbon atom is 1s22s22px12py1, which suggests that carbon should form two covalent bonds, using its two unpaired electrons. However, experience shows that carbon is almost invariably tetravalent (forms four bonds) in its stable compounds.

The explanation for this is that a carbon atom can promote one of its 2s electrons into its vacant 2pz orbital. Though energy is required for this, it is an overall energetically favourable process due to the extra energy recovered from forming four bonds rather than two.

The electronic configuration of carbon once it has performed this promotion is 1s22s12px12py12pz1, which still leaves a problem. It implies that of the four bonds formed by a carbon atom, three will involve a carbon 2p orbital and the other one will involve a carbon 2s orbital, so should show different properties (length, strength) to the other three. However, in the case of a molecule such as methane, CH, the four C-H bonds are entirely equivalent. The explanation of this lies in a process termed hybridisation. This is where the four valence orbitals ‘mix’ (or, more accurately, their wavefunctions interfere with each other) to produce four equivalent hybrid orbitals. These hybrid orbitals are called sp3 orbitals (as they are made from one s and three p orbitals), and are arranged so that they point towards the corners of a regular tetrahedron (this being the arrangement in which all the orbitals occupy an equivalent position and are separated by the maximum possible amount). Each hybrid orbital contains one electron, so can pair with another half-filled orbital to form a σ bond. The spatial arrangement of the hybrid orbitals explains the tetrahedral shape of molecules such as methane, and since all four hybrid orbitals are identical, this explains the equivalence of the four C-H bonds.

Finally we consider unsaturated molecules such as ethene. The carbons in this molecule have a double bond between them, composed of one σ and one π bond. This indicates that there must be one unhybridised p orbital in the valence shell of each carbon atom (to overlap and form the π bond). The remaining two 2p orbitals and the 2s orbital hybridise to produce three identical sp2 orbitals. The lowest energy arrangement of these three orbitals is when they are pointing to the corners of an equilateral triangle (this has the maximum separation between the orbitals). In this situation, the angle between two adjacent orbitals is 120º. The sp2 orbitals define a plane, and the unhybridised p orbital lies at 90º to this plane.

This explains well the shape of the ethene molecule. The σ bonds around each carbon are arranged at 120º to each other, as they are formed to the three sp2 orbitals. The hydrogen atoms all lie in the same plane, as in this conformation the unhybridised p orbitals on the two carbons lie parallel to one another, allowing maximum overlap and thus maximum strength of the π bond.

In molecules such as ethyne, which contains a triple bond, there are two unhybridised p orbitals on each carbon. The remaining p orbital and the s orbital hybridise to give two sp hybrid orbitals. The minimum energy arrangement for these is at 180º to each other. Thus the σ bonds in ethyne are at 180º to each other, resulting in the molecule’s linear shape.

Note throughout the important and general result that the number of hybrid orbitals produced is equal to the total number of orbitals that were mixed together to create them.

All the types of hybrid orbital described above have broadly similar shapes, which look something like this:

Bonds are usually formed with the major lobe, as the minor lobe is too small to overlap effectively with other orbitals.