Explaining energetics and bonds with Slater determinants.

The best Slater determinant (i.e., the one with the lowest energy) is obtained by solving the Fock equation and then using the Aufbau principle and Hund's rules to choose the orbitals to put into the Slater determinant. This procedure (often called MO theory) not only gives the best Slater determinant, but also other information on the energetics in the form of orbital energies. These can be interpreted as ionization energies or electron affinities using Koopmans theorem. The sum of the orbital energies of the orbitals in the Slater determinant is an approximation for the total energy of the molecule. The structure and stability of a molecule is completely determined by its energy, so MO theory in principle yields all information we need. However, we normally talk about molecules in terms of atoms and bonds between these atoms. So we also would like to know where the bonds are in MO theory.

As an example we take BeH₂. This is a linear molecule with Be forming a single bond with each H atom: H-Be-H. The molecule has six electrons. The 1s orbital of Be is very low in energy, doesn't interact with the other orbitals, and by itself forms an MO that gets two electrons. The other two occupied MO's describe the bonding. One is approximately of the form 2s+(1s_A+1s_B), where 2s is a Be orbital and 1s_A and 1s_B are the 1s orbitals on the H atoms. The other MO is approximately 2p_z+(1s_A-1s _B) with 2p_z on Be. (The z axis is parallel to the molecular axis.) Note that the 2p_z orbital participates in the bonding which is vacant in the Be atom. The reason is shown in the following MO diagrams.

Plots of these two MO's (see the next figure) show that neither of them corresponds to a Be-H bond.

It is very important to realize that all electronic properties of a molecule depend on its electronic wave function which we have written as a Slater determinant. The orbitals are only of secondary importance: we use them to construct the Slater determinant. Moreover, we know from linear algebra that there are operations on a matrix that do not change its determinant. In particular if we have two matrices A and B that are related via A = UB, then we have det(A) = det(U)det(B). If U is an orthogonal matrix, then det(A) and det(B) can at most differ a factor -1. If we write A = UB in matrix elements we get

Amn = å k  UmkBkn.

From this we see that the rows of A are all linear combinations of the rows of B. In Slater determinants these rows are the occupied orbitals. So if we apply an orthogonal transformation to the occupied orbitals, we change the Slater determinant (i.e., the wave function) by at most a factor -1. As this is physically irrelevant, the wave function effectively remains the same. So there is an infinite number of sets of occupied orbitals that all yield the same Slater determinant. The solutions of the Fock equation form just one of these sets.

For BeH₂ we had the orbitals 2s+(1s_A+1s_B) and 2p_z+(1s_A-1s_B). The orthogonal transformation that we will use is

1 Ö2 æ ç è 1 1 1 -1 ö ÷ ø .

This form is chosen because in the resulting new orbitals either 1s_A or 1s_B is removed. Applying the transformation yields

[2s+(1s_A+1s_B)]/Ö2 +[2p_z+(1s_A-1s_B)]/Ö2

= (2s+2p_z)/Ö2+Ö2.1s_A

[2s+(1s_A+1s_B)]/Ö2 -[2p_z+(1s_A-1s_B)]/Ö2

= (2s-2p_z)/Ö2+Ö2.1s_B.

These orbitals are each constructed from atomic orbitals on just two atoms. This means that they are localized between Be and one H atom (see the next figure), and each of them corresponds to a Be-H bond.

The transformation on the solutions of the Fock equation above is not only possible for BeH₂, but also for many other molecules can we localize the MO's so that they correspond to bonds. An exception is formed by aromatic molecules. These molecules contain one or more p systems that consist of MO's that cannot be localized. This gives them their specific character.