href="#fb3_img_img_c4835c5f-d14a-5491-8d12-ac15dcb859ef.png" alt="upper H prime Subscript AA Baseline equals upper H prime Subscript BB"/> and . With this, Eq. (2.6) becomes
(2.7)
This equation system has a nontrivial solution when the determinant of the coefficient matrix vanishes. This gives two possible solutions
(2.8)
With these solutions, we can calculate the relation between the coefficients and from Eq. (2.7), and we find that for the “” solution and for the “” solution (see Problem 3 and note that ).
The energy levels and wave functions from this calculation are illustrated in Figure 2.2a. For a large distance between the nuclei, the overlap integral and the matrix element vanish and the only solution is . This makes sense as the second proton is far away from , so is approximately the ground state energy of atomic hydrogen. A corresponding solution does, of course, exist for the electron located on the other proton. For a short distance , on the other hand, this energy level splits up into the bonding and antibonding molecular states . The single electron will occupy the lower level and this will result in the energy gain of covalent bonding. The wave function corresponding to this is the one for which , which increases the electron density between the nuclei. The wave function corresponding to the antibonding upper energy level is constructed by choosing , and it shows the characteristics of an antibonding state with a node between the two nuclei.
Figure 2.2 (a) Formation of bonding and antibonding energy levels in the
ion. The electronic energy level of an isolated H atom splits into the levels
according to
Eq. (2.8). The radial wave function of an isolated H atom is shown at the left, and the bonding and antibonding wave functions along the molecular axis of the
ion are shown at the right (for proper normalization see Problem 3). (b) Bonding and antibonding energy levels
as a function of internuclear distance
. Zero energy corresponds to the ground state energy of a free hydrogen atom.
Figure 2.2b shows how the energy levels vary as a function of the distance between the nuclei. The antibonding energy increases monotonically as the nuclei approach each other, but the bonding energy has a minimum of eV at approximately twice the Bohr radius . With the single electron placed into the bonding state , the total energy gain resulting from formation of the covalent bond is thus 1.77 eV. It is tempting to extend this model to the two electrons in the molecule by placing the second electron in the same energy level. This electron would then also lower its energy by 1.77 eV, amounting to a total energy gain of 3.54 eV for the formation of the hydrogen molecule. Handling the additional electron in this way is too simplistic, as we shall see below, but the energy gain thus computed is at least similar to the experimental value of eV. It is also clear that no additional electrons can be placed in the energy level, as this would violate the Pauli principle. A third electron would need to occupy the level, destabilizing the molecule as compared to
