The bonding we saw earlier was about the simplest example of bonding there is.  However, when we start to consider molecules more complicated than hydrogen (i.e. practically all the rest), we run into problems, unless we invoke the concept of hybridisation.

If we consider methane (CH₄) for a moment, various experiments, and perhaps also our intuition, tell us that all the C-H bonds are equivalent.  That is they all have the same length, and all the angles between the bonds are the same.  However, if we try to imagine this picture with three p and one s orbitals, it is very hard to see how these could result in a tetrahedral arrangement about carbon, as it would seem to lead to three mutually perpendicular bonds (resulting from the p orbitals, and one different, directionless bond (resulting from the s orbital).

The way chemists account for this is by the invocation of hybridisation (mixing) of orbitals.

One 2s	+ one 2px	+ one 2py	+ one 2pz	= four sp3 hybrids

The term sp³, is used to denote the fact that three p orbitals are mixed with one s.  The sp³ hybrid orbitals are arranged tetrahedrally, with the angles between them each being 109° 28'.  All the new hybrids are identical.

A single sp³ orbital looks like this:

This view can help us to se why hybridisation allows the carbon atom to have four equal bonds.  The sp³ orbital is asymmetrical, that is, one of the lobes is larger than than other.  This means that the large end can overlap much better with an atomic orbital from another atom, and form stronger bonds than either an s or p orbital can.

A similar process can be envisaged for ethane:

These two fragments represent CH₃ fragments, with their remaining sp³ orbitals shown pointing towards each other.  From here it is an easy step to imagine them overlapping, and the consequent formation of a C-C bond.