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Walsh Diagrams

Walsh diagrams are very useful in making quick predictions about the geometries of small molecules. Introduced in 1953 [5], these are ranked among the great achievements of molecular orbital theory. Walsh constructed his diagrams by plotting ``orbital binding energies'' against bond angles. The basic idea is that the total energy is the sum of all the orbital binding energies; therefore, by considering the stabilization or destabilization of all the orbitals by a change in the angle, one can predict (roughly) the equilibrium bond angle for a given state of a molecule.

Today, we usually use orbital energies (also called orbital eigenvalues) in Walsh diagrams. The truth is that nobody really knows exactly what Walsh's ``orbital binding energies'' were! However, whatever Walsh plotted, his diagrams are very helpful. In this lab, you will construct your own Walsh diagram for a particular molecule using the Hartree-Fock orbital energies.

Here we will briefly discuss how Walsh diagrams are interpreted. Usually, core orbitals (1s for B, C, N, O, F, Ne) are not included in Walsh diagrams, since they are very low in energy and not changed much by bond angle bending. Only valence orbitals are included. Note that some of the valence orbitals are usually unoccupied. Figure 1 presents the Walsh diagram for AH2 molecules. Let us consider a simple molecule, H2O. The electronic configuration of H2O is 1a12 2a12 1b22 3a12 1b1, as you could determine by hand (if you knew how) or by running a program like Q-Chem.

Figure: Walsh diagram for XH2 molecules from Herzberg [6].

From the Walsh diagram, we can see that if we doubly occupied the 2a1, 1b2, 3a1, and 1b1 orbitals (remember, the core orbital doesn't enter into the diagram), the lowest energy would come at roughly 100$^{\rm o}$. Although the 2a1 and 1b2 orbitals go up in energy as the molecule is bent from 180$^{\rm o}$ towards 90$^{\rm o}$, their energies do not increase nearly as rapidly as the energy of the 3a1 orbital decreases. Hence, the molecule is stabilized by bending. Note that the 1b1 orbital energy is basically flat and doesn't care whether the molecule is linear or bent. As you can see from the diagram, the energies of 1b2 and 2a1 start to rise more rapidly just before 90$^{\rm o}$, which is why the equilibrium geometry is a little larger, around 100$^{\rm o}$. Although one could get out a ruler to determine more exactly the minimum energy angle, there would be little point, since the Walsh diagram is capable only of approximate results (and can even make totally wrong predictions in some cases). In this case, the diagram works well, since the equilibrium bond angle for water is 104.52$^{\rm o}$.

What if an electron was promoted from the 1b1 orbital to the 4a1orbital? This would give us a 3B1 or 1B1 state of H2O, depending on whether the unpaired electrons were singlet or triplet coupled. What would the geometry of this excited electronic state be? From the Walsh diagram, we can see that the 1b1 orbital doesn't care if the molecule is linear or bent. Therefore, it has no influence on the equilibrium geometry. However, we now have one electron in the 4a1 orbital. Since there is only one electron in this orbital, instead of two, the influence of this orbital will only be half as much as the doubly occupied orbitals. We can see that the 4a1 orbital prefers linear geometries, but only very slightly. Therefore, we predict the equilibrium geometry for this excited state to be very similar to the ground state, but slightly closer to linear (if the ground state is 104.52$^{\rm o}$, maybe the excited state is 107$^{\rm o}$?). I have not yet been able to find an experimental value for the bond angle of the 3B1 or 1B1 states of H2O.

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C. David Sherrill