We now need to look at ways of combining more than two atoms at a time. For some mole cules, such as H2S and PH3, which have all bond angles equal to 90°, the bonding should be straightforward—the 3p orbitals (which are at 90°) on the central atom simply overlap with the 1s orbitals of the hydrogen atoms. Now, you might imagine it would be similar for ammonia, NH3, since N lies above P in the periodic table. The trouble is, we know experimentally that the bond angles in ammonia, as in water and methane, are not 90°, but instead 104°, 107°, and 109°, respectively. All the covalent compounds of elements in the row Li to Ne raise this difficulty. How can we get 109° angles from orbitals arranged 90° apart? To see what has to happen, we’ll start with a molecule of methane enclosed in a cube. It is possible to do this since the opposite corners of a cube describe a perfect tetrahedron. The carbon atom is at the centre of the cube and the four hydrogen atoms are at four of the corners. Now let’s consider each of the carbon’s 2s and 2p AOs in turn. The carbon’s 2s orbital can overlap with all four hydrogen 1s orbitals at once with all the orbitals in the same phase.

Each of the 2p orbitals points to a pair of opposite faces of the cube. Once more all four hydrogen 1s orbitals can combine with each p orbital, provided the hydrogen AOs on the opposite faces of the cube are of opposite phases.

The three MOs generated in this way are degenerate, and this gives us four bonding orbitals. Along with four associated antibonding orbitals this gives us a total of eight MOs, which is correct since there were eight AOs (C gave us 2s and 3 × 2p, while 4 × H gave us 4 × 1s). Using this approach, it is possible to construct a complete MO picture of methane—and indeed for very much more complex molecules than methane. There is experimental evi dence too that these pictures are correct. But the problem is this: the four fi lled, bonding orbitals of methane are not all the same (one came from the interaction with the C 2s orbital and three from the C 2p orbitals). But we also know from experimental observations all four C–H bonds in methane are the same. Something seems to be wrong, but there is in fact no contradiction. The MO approach tells us that there is one MO of one kind and three of another but the electrons in them are shared out over all fi ve atoms. No one hydrogen atom has more or fewer electrons than any other— they are all equivalent. Techniques that tell us the structure of methane do not tell us where bonds are; they simply tell us where the atoms are located in space—we draw in bonds connecting atoms together. Certainly, the atoms form a regular tetrahedron but exactly where the electrons are is a different matter entirely. So, do we have to give up the idea that methane has four bonds, each made of two electrons, linking the C with an H? If we choose to, then for every reaction, even of the simplest molecules, we are going to need to calculate, by computer, a full set of MOs and all of their interactions. That would be using physics to do chemistry. It might be accurate but it would kill creativity and invention. So here is an alternative: we keep our tried and tested practical picture of molecules made from discrete bonds, each containing a pair of electrons, but we make it compatible with MO theory. To do this we need a concept known as hybridization.