Chem 327

Multielectron Wavefunctions and Bond Formation

The word ‘orbital’ is a synonym for one-electron wavefunction.

A molecular orbital is usually expressed as a linear combination of atomic orbitals located on different atom, say for instance hydrogen atom A and hydrogen atom B of the H2 molecule.

an example of such a molecular orbital is:

If I want to explicitly include the label of the one electron in this ‘orbital’ then I can:

Now, if I want to write a two-electron wavefunction for H2, then I simply write something like this where I’ve been sure to include an antisymmetric spin wavefunction to ensure in conjunction with the symmetric spatial part that the overall wavefunction antisymmetric with respect to interchange of electron labels 1 and 2. Because there is only one spin function associated with this spatial function (and because these spins are paired) this is called a singlet.

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If I need to represent an excited state, I can do so in a couple of different ways. First, I have a singlet excited state resulting from the distribution of one electron in a  molecular orbital and a second in a * MO. This singlet corresponds to a symmetric spatial part and an antisymmetric spin part.

If instead I had wanted to use an antisymmetric spatial part I could have used a symmetric spin wavefunction for which there are three alternatives and is therefore called a triplet

Problem

You have an expression for a ground state wavefunction for the H2 molecule above. Write down the spatial part of this wavefunction (don’t worry about the spin part) use the definition of the molecular orbital for each electron (1) and (2) and FOIL it out. If you do this properly, you will get four term (expressed in terms of products of 1sA or 1sB, etc.) in the expansion of this ground state wavefunction. Can you find a meaning to each of these terms.

The valence bond wavefunction for H2 (as opposed to the molecular orbital wavefunction that you have already seen) is simply another, different way of expression a molecular wavefunction. The VB wavefunction for H2 is as follows

Compare this (VB) wavefunction to the MO wavefunction for H2, what are their similarities and differences?

Figure 3. Names for and interpretations of commonly found integrals in quantum mechanical descriptions of bonding.

Potential energy surface for H2+

Here we will examine the potential energy surface of the hydrogen molecule ion, H2+. This is a one-electron system so we can actually write down a relatively simple expression for the expectation value for total energy of the molecule. Still, you can unambiguously see that there is a bond between the two H and the point of this illustration is to examine the origin of the molecular (ionic in this case) potential energy surface. That is, to examine how each of the energy expressions that show up in the total energy (expectation value) expression gives rise to the overall character of the PES.

The wavefunction

Again, because this is a one-electron problem, the form of the wavefunction is particularly simple. We say that the electron occupies a  bonding molecular orbital (defined above). We can multiply this spatial orbital by either one of two spin functions for one electron ( or )- so this state is a doublet.

Where the normalization constant N is given below.

Prove that the one-electron (approximate) wavefunction above is normalized.

Energy Expression

The expression we need to evaluate in order to generate the potential energy surface is constructed just like it is for any expectation value evaluation. The operator for total energy, the Hamiltonian operator H, is given below. The denominator of this expression is simply the square of the wavefunction which we know is normalized to unity so we don’t really need to worry about it here.

The Hamiltonian

The Hamiltonian operator is relatively simple as well, notice that it has only four terms. The first is simply the kinetic energy operator, the second and third terms are operators that represent the energy of attraction between the negatively charged electron with each of the two (equivalent) positively charged nuclei. The final term simply represents the mutual repulsion of the two positively charged nuclei

Evaluating the energy expression

Notice that in the next equation I have simplified the energy formula already. First, I have dropped the spin part of the wavefunction. I can do this because there are no terms in the Hamiltonian operator that depend directly on spin, so the spin functions do not contribute to the energy and therefore can be ignored.

Editors Note: You probably recall from our discussions in class that given the same spatial components (orbitals) of molecular wavefunctions that triplets tend to be lower in energy than singlets: doesn’t this argue against my position that spin doesn’t matter. Well, actually no, spin still does not directly affect energy. However, there is an indirect effect. If you could write out an energy expression for a two (or more) electron molecule or atom, you would find that the number an types of terms you write out would be different for singlet and triplet wavefunctions. Among other things, the energy expectation value expression for the triplet wavefunction has more of the mysterious quantum mechanical “exchange” integrals which lower the energy.

FOILing this expression out we get

Now, we make use of the fact that because nuclei A and B are identical the following two relationships between integrals hold:

Using this these facts we can simplify the energy expression to get

Where in the second expression above I have simply inserted the explicit form of the Hamiltonian operator

I can rearrange these terms any way I like, after all addition is commutative. I choose this way and you’ll see that I do this for a reason.

The reason I have chosen this way to arrange the terms in this way is that it leads straightforwardly to

Notice that what is different is the first term on each line. How can I do this? Well the sum of terms is simply the Hamiltonian operator for the hydrogen atom and because the atomic contribution to the molecular orbital is 1sA, the eigenvalue E1s is the energy of the hydrogen atomic 1s orbital. (atom B is the same as atom A so the same argument holds).

Because E1s is simply a number, I can write.

Furthermore, because the atomic wavefunction ,1sA, is normalized the integral of its square is just 1. In addition, RAB is independent of electron coordinate ( ie the electronic wavefunction) so we can pull it out of the integral in the third term on each line. Similarly, we use the definition of overlap S to simplify further

I can now collect terms and rearrange to get

Problem

Write a similar energy expectation value expression for the H2+ ion in its excited electronic state, *(1). What kind of PES does this give rise to? Step one is to find the normalization constant N, it is different here than it was for (1)- but not too much different.

Figure 3. Potential energy surface for the H2+ ion.