0.8 Chemical bonding and molecular energy levels  (Page 3/7)

Recall that the electron orbitals in the hydrogen atom are described by a set of quantum numbers. One of these quantum numbers is related to the symmetry or shape of the atomic orbital and is generally depicted by a letter. Recall that an “s” orbital is spherical in shape, and a “p” orbital has two lobes aligned along one axis. Similarly, the molecular orbitals for the $H_2^{+}$ molecular ion are described by a set of numbers which give the symmetry (or shape) of the orbital. For our purposes, we need only one of these descriptors, based on the symmetry of the orbital along the bond: if the molecular orbital has the symmetry of a cylinder, we refer to it as a “ $\sigma$ orbital.” The orbital in Fig. 2c satisfies this condition.

We conclude that chemical bonding results from an electron in a molecular orbital which has substantial probability for the electron to be between two nuclei. However, this example illustrates chemical bonding with a single electron. Our rules of valence indicate that bonding typically occurs with a pair of electrons, rather than a single electron. Furthermore, this model of bonding does not tell us how to handle molecules with many electrons (say, $F_2$ ) where most of the electrons do not participate in the bonding at all.

Observation 2: bonding and non-bonding in diatomic molecules

We now consider molecules with more than one electron. These are illustrated most easily by diatomic molecules (molecules with only two atoms) formed by like atoms, beginning with the hydrogen molecule, $H_2$ . The most direct experimental observation of a chemical bond is the amount of energy required to break it. This is called the bond energy, or somewhat less precisely, the bond strength. Experimentally, it is observed that the bond energy of the hydrogen molecule $H_2$ is 458 kJ/mol. By contrast, the bond energy of the $H_2^{+}$ molecular ion is 269 kJ/mol. Therefore, the bond in $H_2$ is stronger than the bond in $H_2^{+}$ . Thus, the pair of shared electrons in $H_2$ generates a stronger attractive force than does the single electron in $H_2^{+}$ .

Before deducing an explanation of this in terms of electron orbitals, we first recall the valence shell electron pair description of the bonding in $H_2$ . Each hydrogen atom has a single electron. By sharing these two electrons, each hydrogen atom can fill its valence shell, attaining the electron configuration of helium.

How does this translate into the electron orbital picture of electron sharing that we have just described for the $H_2^{+}$ molecular ion? There are two ways to deduce the answer to this question, and, since they are both useful, we will work through them both. The first way is to imagine that we form an $H_2$ molecule by starting with an $H_2^{+}$ molecular ion and adding an electron to it. As a simple approximation, we might imagine that the first electron’s probability distribution (its orbital) is not affected by the addition of the second electron. The second electron must have a probability distribution describing its location in the molecule as well. We recall that, in atoms, it is possible to put two electrons into a single electron orbital, provided that the two electrons have opposite values of the spin quantum number, ms. Therefore, we expect this to be true for molecules as well, and we place the added second electron in $H_2$ into the same $\sigma$ orbital as the first. This results in two electrons in the region between the two nuclei, thus adding to the force of attraction of the two nuclei into the bond. This explains our observation that the bond energy of $H_2$ is almost (although not quite) twice the bond energy of $H_2^{+}$ .