To get a picture of methane with four equivalent pairs of electrons we need to start with four equivalent AOs on C, which we don’t have. But we can get them if we combine the carbon 2s and 2p orbitals fi rst to make four new orbitals, each composed of one-quarter of the 2s orbital and three-quarters of one of the p orbitals. The new orbitals are called sp3 (that’s said s-p-three, not s-p-cubed) hybrid orbitals to show the proportions of the AOs in each. This process of mixing is called hybridization. The hybrid orbitals are mathematically equivalent to the 2s and 2p orbitals we started with, and they have the advantage that when we use them to make MOs the orbitals correspond to bonding pairs of electrons.

What do the four hybrid orbitals look like? Each sp3 orbital takes three-quarters of its char acter from a p orbital and one-quarter from an s orbital. It has a planar node through the nucleus like a p orbital but one lobe is larger than the other because of the extra contribution of the 2s orbital: the symmetry of the 2s orbital means that adding it to a 2p orbital will increase the size of the wavefunction in one lobe, but decrease it in the other.

The four sp³ orbitals point to the corners of a tetrahedron and we build up a molecule of methane by overlapping the large lobe of each sp3 orbital with the 1s orbital of a hydrogen atom, as shown in the margin. Each overlap forms an MO (2sp³ + 1s) and we can put two electrons in each to form a C–H σ bond. There will of course also be an antibonding MO, σ* (2sp³ – 1s) in each case, but these orbitals are empty. Overall, the electrons are spatially distributed exactly as they were in our previous model, but now we can think of them as being located in four bonds.

The great advantage of this method is that it can be used to build up structures of much larger molecules quickly and without having to imagine that the molecule is made up from isolated atoms. Take ethane, for example. Each carbon uses three sp3 AOs orientated towards the three hydrogen atoms, leaving one sp3 orbital on each carbon atom for the C–C bond. In the MO energy level diagram we now have both C–H bonding σ and antibonding σ* orbitals (made from combining sp³ orbitals on C with 1s orbitals on H) and also a C–C bonding σ and antibonding σ* orbital, made from two sp3 orbitals on C. The diagram below just shows the C–C bond.

For ethene (ethylene), the simplest alkene, we need a new set of hybrid orbitals. Ethene is a planar molecule with bond angles close to 120°. Our approach will be to hybridize all the orbitals needed for the C–H framework and see what is left over. In this case we need three equivalent bonds from each carbon atom (one to make a C–C bond and two to make C–H bonds). Therefore we need to combine the 2s orbital on each carbon atom with two p orbitals to make the three bonds. We could hybridize the 2s, 2py, and 2pz orbitals (that is, all the AOs in the plane) to form three equal sp² orbitals, leaving the 2pz orbital unchanged. These sp² hybrid orbitals will have one-third s character and only two-thirds p character.

The three sp² hybrid AOs on each carbon atom can overlap with three other orbitals (two hydrogen 1s AOs and one sp2 AO from the other carbon) to form three σ MOs. This leaves the two 2px orbitals, one on each carbon, which combine to form the π MO. The skeleton of the molecule has five σ bonds (one C–C and four C–H) in the plane and the central π bond is formed by two 2px orbitals above and below the plane.

This is the fi rst MO picture we have constructed with a C=C double bond, and it is worth taking the time to think about the energies of the orbitals involved. We’ll again ignore the C–H bonds, which involve two of the sp² orbitals of each C atom. Remember, we mixed two of the three 2p orbitals in with the 2s orbital to make 3 × sp² orbitals on each C atom, leaving behind one unhybridized 2p orbital. Now, fi rst we need to generate the σ and σ* orbitals by interacting one sp² orbital on each atom. Then we need to deal with the two p orbitals, one on each C, which interact side-on. The unhy-bridized p orbitals are a bit higher in energy than the sp² orbitals, but they interact less well (we discussed this on p. 93) so they give a π orbital and a π* orbital whose energies are in between the σ and σ* orbitals. Each C atom donates two electrons to these orbitals (the other two electrons are involved in the two bonds to H), so the overall picture looks like this. Two AOs give two MOs.

Ethyne (acetylene) has a C≡C triple bond. Each carbon bonds to only two other atoms to form a linear CH skeleton. Only the carbon 2s and 2px have the right symmetry to bond to the two atoms at once so we can hybridize these to form two sp hybrids on each carbon atom, leaving the 2p_y and 2p_z to form π MOs with the 2p orbitals on the other carbon atom. These sp hybrids have 50% each s and p character and form a linear carbon skeleton.

We could then form the MOs as shown below. Each sp hybrid AO overlaps with either a hydrogen 1s AO or with the sp orbital from the other carbon. The two sets of p orbitals com bine to give two mutually perpendicular π MOs.

Hydrocarbon skeletons are built up from tetrahedral (sp3), trigonal planar (sp2), or linear (sp) hybridized carbon atoms. Deciding what sort of hybridization any carbon atom has, and hence what sort of orbitals it will use to make bonds, is easy. All you have to do is count up the atoms bonded to each carbon atom. If there are two, that carbon atom is linear (sp hybridized), if there are three, that carbon atom is trigonal (sp² hybridized), and if there are four, that carbon atom is tetrahedral (sp³ hybridized). Since the remaining unhybridized p orbitals are used to make the π orbitals of double or triple bonds, you can also work out hybridization state just by counting up the number of π bonds at each carbon. Carbon atoms with no π bonds are tetrahedral (sp³ hybridized), those with one π bond are trigonal (sp² hybridized), and those with two π bonds are linear (sp hybridized). There’s a representative example on the left. This hydrocarbon (hex-5-en-2-yne) has two linear 6 H 5 4 H 3 hex-5-en-2-yne 2 CH3 1 sp carbon atoms (C2 and C3), two trigonal sp2 carbon atoms (C5 and C6), a tetrahedral sp3 CH2 group in the middle of the chain (C4), and a tetrahedral sp3 methyl group (C1) at the end of the chain. We had no need to look at any AOs to deduce this—we needed only to count the bonds.