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MO diagrams and bond angles

We normally draw the levels corresponding to the orbitals that are solutions of the Fock equation in MO diagrams. It is also useful to have a look at other orbitals, as this will allow us to say something about the structure of a molecule.

BeH₂ has a linear structure. We can explain this structure with sp-hybridization. An MO diagram including this hybridization will look as follows.

The column of the Be atom after hybridization has a degenerate level of sp-hybrids. The energy of this level is the average of the energies of the 2s and the 2p levels, because the hybrids consist of equal amounts of 2s and 2p orbitals. Each of these hybrids form a bonding and antibonding orbital with a 1s orbital of hydrogen. These orbitals are shown in the BeH₂ column.

The BeH₂ column in the MO diagram above does not show the levels one gets from solving the Fock equation for BeH₂. These levels can be obtained, however, from the ones in the diagram above using the rules of qualitative MO theory and the fact that the MO's from the diagram above will interact. As this interaction is larger for levels closer in energy, we can restrict ourselves to interactions within the degenerate levels. The 2p level does not change, of course, because these atomic orbitals do not participate in the bonding. The other orbitals will form bonding and antibonding orbitals. This gives us

BeH2: sp-hybrids and solution Fock equation

If we compare the energy (i.e., the sum of energies of the orbitals) of the left and the right situation we find that the energy on the right is a bit higher, because the bonding orbital is a bit less stabilized than the antibonding orbital is destabilized. However, if the overlap between the occupied orbitals on the left is small than the difference is small as well. We can therefore use the left situation to get an estimate for the energy of the molecule. This allows us to predict the structure of molecules as will now be shown.

Suppose we want to bend BeH₂. If we make the bond angle 120^° we have sp²-hybridization. This gives the following MO diagram.

The important difference between this MO diagram and the one with sp-hybridization is that the sp²-hybrids are higher in energy than the sp-hybrids, because they have more 2p character (i.e., the 2p orbitals contribute more). As a consequence the bonding orbitals in BeH₂ are also higher in energy, and the molecule is less stable. This is why BeH₂ is a linear molecule.

We can use the procedure above to explain why H₂O is not linear. Linear H₂O has the following MO diagram.

The difference with BeH₂ is that oxygen has more electrons which reside in the 2p orbitals. With sp³-hybridization we get

As for BeH₂ we see that using more 2p orbitals for the hybridization increases the energy if we only look at the bonding orbitals, but for H₂O there is a another effect, which turns out to be more important. We have four electrons that do not participate in the bonding. They are in sp³-hybrids. These orbitals are called lone pair orbitals. These sp³-hybrids are lower in energy than the 2p orbitals of linear H₂O, so forming these hybrids lowers the energy of the molecule. Which effect dominates is hard to say from the MO diagram. Moreover, there are other effects as well. For example, the sp³-hybrid has a different overlap with the 1s of hydrogen than the sp-hybrid. Quantum chemical calculations show that the lowering of the energy of the electrons that do not participate in bond is more important, and therefore H₂O is a bent molecule. This is the quantum chemical origin of the rules from VSEPR that tell you what the structure of a molecule is.

MO diagrams

Basics of MO diagrams.
MO diagrams of interacting orbitals.
MO diagrams and bond angles (this page).