INTRODUCTION
This exercise follows directly from the experiment on atomic orbitals that you have already done. The text that follows is not intended to be complete, but only a summary of material that should be covered in the lectures.
In studying atomic structure, you learned that a Schrödinger equation could be written for the hydrogen atom, and that this equation could be exactly solved. The previous exercise illustrated some of the solutions of this equation: the wave fuctions ψ (psi). Rather crude representations - the sort of diagram you might sketch yourselves under "exam conditions" - of the atomic orbitals called 1s and 2p are shown in Figure 1.
Figure 1
Remember that the sign (+ or -) shown on the figure has no physical significance, and simply indicates the sign of the numerical value of ψ in this region of space. Nevertheless, these signs are of importance.
It is a simple matter to write down the Scrödinger equation for a molecule. However, the resultant equation cannot be solved exactly. Various approximate methods are used to circumvent this difficulty. The one we will illustrated in this dry lab is called the molecular orbital (MO) approach. The objective is to answer the following questions:
The first objective is to derive molecular orbitals into which we can place all the electrons in the molecule in much the same way as we do for atoms. (Using the Aufbau Principle.) To obtain these molecular orbitals, we use a method called the linear combination of atomic orbitals (LCAO). As the name implies, the molecular orbitals are made by adding or subtracting the atomic orbitals. The simplest example is obtained by considering the hydrogen molecule, H2, which we write as HA-HB, using the subscripts A and B to label the two atoms. LCAO theory states that one of the molecular orbitals can be written as:
The constants c and d are weighting constants indicating the relative amounts of each atomic orbital that will be used. For H2, where the two hydogen atoms are equivalent, c = d, and we can replace them by a single "normalizing" constant N:
(If the probability of finding the electron in such an orbital somewhere in all space is to be equal to 1, N = 1/√2.)So far we have done nothing difficult, but simply assumed that equation 1 represents our MO. Use the Applet
to have a look at what this orbital looks like. You can adjust the distance between the nuclei to see what happens as the atomic orbital ovelap more and more. Figure 2 gives a rather crude picture of what you should see.
Figure 2
There is another linear combination that we should have considered according to LCAO theory:
Use the Applet
to have a look at what this combination looks like. Figure 3 shows a crude representation of what you should see. This combination is referred to as antibonding. Whenever we combine two atomic orbitals in a way which produces a change in the sign of ψ between the two component atomic orbitals, anti-bonding results.
Figure 3
Figure 4
The probability plot for ψ22 is also shown in Figure 4. What would be its equation? (The equivalent of equation 3) What do you notice about the electron density between the nuclei as compared to two individual hydrogen atoms simply placed side by side?Further insight into the bonding of HA and HB can be obtained by considering the energies of the electrons in ψ1 and ψ2 compared to their energies in the non-interacting atoms. This can be done by plugging the LCAO wave functions for the molecule back into the appropriate Scrödinger equation (just as it can be done for the individual atoms using the atomic wave functions). The results are shown in Figure 5, where ψ1 and ψ2 are sometimes renamed s(1s) and s*(1s), respectively, to indicate the type of molecular orbital and their parentage.
Figure 5
Question 2.
Using Figure 5, comment on the observations about the stability of the diatomic species listed above. (Would you expect all the negative molecule-ions to be unstable? Are there other species not listed which might be observed? Your answer should make reference to the electronic configuration and bond orders in these species
SECOND-ROW HOMONUCLEAR DIATOMICS
Second row atoms have 2s and 2p orbitals available for use in bonding. As in the case of two atoms with 1s orbitals interacting, two atoms with 2s orbitals interacting lead to two molecular orbitals called s(2s) and s*(2s).
The symbol s is used when the molecular orbital has no nodal plane which contains both nuclei. For the bonding combination, there is only one region of high electon density between the two nuclei. If there is a single nodal plane containing both nuclei, the orbital is of type p. In this case, the bonding combination will have two regions of high electron density separated by the node. There are rare cases in certain transition metal compounds, where two nodes per molecular orbital contain both nuclei. These are designated d orbitals.
In valence bond theory terms, a single bond would have only a s symmetry bond. A double bond consists of a s and a p bond, a triple bond would have a s and two p bonds, and the esoteric quadruple bond has one s, two p and one d combination. In a multiple bond, the various orbitals co-exist in the same region of space between the nuclei. Do not mistake the two regions of overlap of a p-bonding orbital for a double bond!
Use the Applet
to display the molecular orbitals derived from the 2s atomic orbitals on two atoms.
Question 3.
Prepare sketches of the s(2s) and s*(2s) orbitals similar to Figures 2 and 3. Do not attempt to copy all shading; just show all the nodes and the phase (sign) of ψ. Do they differ at all from figures 2 and 3?
Next use the computer program to display the overlap of two 2px orbitals. (The x axis is taken as the internuclear axis by the computer program.)
Question 4.
Prepare crude sketches of the resulting orbitals. Which combination, the sum or the difference, corresponds to the bonding combination this time? What would be the names for these orbitals?
Now use the Applet
to examine the molecular orbitals that result from the linear combinations of the 2py orbitals. (Note that the results would be the same for the combinations of 2pz orbitals, which, if you have time you can check by varying the value of z.)
Question 5.
Once again prepare simplified sketches of what you see. Give the proper labels for the two combinations, including the * to indicate which is the antibonding orbital.
At this point you should have seen all the all the molecular orbitals formed by pairs of 2s and 2p orbitals. We can construct an energy level diagram to illustrate the relative energies of all these molecular orbitals and the atomic orbitals from which they are derived (Figure 6).
Figure 6
- What is the bond order in Li2 and O2?
- What homonuclear molecule-cation and molecule-anion species should have the same bond order as O2? Consider cases between X22− and X22+ where X = Li to Ne.
- Predict which two second-row elements are unlikely to give a diatomic molecule. Does simple Lewis bonding theory agree with your predictions?
Figure 6 does not quite tell the whole story. It can be refined a little by considering some other types of combination. Use the computer program to examine the linear combination of a 2s orbital on one atom with a 2px orbital on the other.
Question 7.
Again sketch the results. There is no special way to label these combinations, but say if they are of type s or p, and whether they are bonding or antibonding.
Use the computer to form a combination of a 2s orbital on one atom with a 2py orbital on the other.
Question 8.
Sketch the resulting orbital. Can you classify this orbital as s or p and bonding or antibonding? Explain. Go back to the beginning of this section if you are not sure.
Your answers to questions 7 and 8 should have convinced you that an LCAO of the type illustrated in question 7 is a reasonable combination, but that the other combination, described in question 8, is not useful. (Such a combination is called non-bonding.)
The results of including the additional overlap of 2s with 2p orbitals is shown in Figure 7. The main changes are shown in red.
Figure 7
Figure 8
What is the difference for B2?
Question 10.
Go back and briefly answer, in the context of the homonuclear diatomic molecules, the four questions which were posed at the beginning of this lab.
Feel free to play with the Applet
to examine some of the other combinations of atomic orbitals that are possible. Because the projection direction is down the z-axis certain combinations cannot be displayed.