0.7 Chemical bonding and molecular energy levels  (Page 6/7)

As a third example, we consider fluorine, $F_2$ . In this case, we find that the ionization energy of molecular fluorine is 1515 kJ/mol, which is smaller than the ionization energy of a fluorine atom, 1681 kJ/mol. This seems inconsistent with the bonding orbital concept we have developed above, which states that the electrons in the bond have a lower energy than in the separated atoms. If the electron being ionized has a higher energy in $F_2$ than in F, why is $F_2$ a stable molecule? Apparently, we need a more complete description of the molecular orbital concept of chemical bonding.

To proceed further, we compare bond energies in several molecules. Recall that the bond energy (or bond strength) is the energy required to separate the bonded atoms. We observe that the bond energy of $N_2$ is 956 kJ/mol. This is very much larger than the bond energy of $H_2$ , 458 kJ/mol, and of $F_2$ , which is 160 kJ/mol. We can account for the unusually strong bond in nitrogen using both our valence shell electron pair sharing model and our electron orbital descriptions. A nitrogen atom has three unpaired electrons in its valence shell, because the three 2p electrons distribute themselves over the three 2p orbitals, each oriented along a different axis. Each of these unpaired electrons is available for sharing with a second nitrogen atom. The result, from valence shell electron pair sharing concepts, is that three pairs of electrons are shared between two nitrogen atoms, and we call the bond in $N_2$ a “triple bond.” It is somewhat intuitive that the triple bond in $N_2$ should be much stronger than the single bond in $H_2$ or in $F_2$ .

Now consider the molecular orbital description of bonding in $N_2$ . Each of the three 2p atomic orbitals in each nitrogen atom must overlap to form a bonding molecular orbital, if we are to accommodate three electron pairs. Each 2p orbital is oriented along a single axis. One 2p orbital from each atom is oriented in the direction of the other atom, that is, along the bond axis. When these two atomic orbitals overlap, they form a molecular orbital which has the symmetry of a cylinder and which is therefore a $\sigma$ orbital. Of course, they also form a ${\sigma }^{\mathrm{*}}$ orbital. The two electrons are then paired in the bonding orbital.

The other two 2p orbitals on each nitrogen atom are perpendicular to the bond axis. The constructive overlap between these orbitals from different atoms must therefore result in a molecular orbital somewhat different that what we have discussed before. As shown in Fig. 4, the molecular orbital which results now does not have the symmetry of a cylinder, and in fact, looks something more like a cylinder cut into two pieces. This we call a $\pi$ orbital. There are two such $\pi$ orbitals since there are two sets of p orbitals perpendicular to the bond axis. Figure 4 also shows that an anti-bonding orbital is formed from the destructive overlap of 2p orbitals, and this is called a ${\pi }^{\mathrm{*}}$ orbital. There are also two ${\pi }^{\mathrm{*}}$ orbitals formed from destructive overlap of 2p orbitals. In $N_2$ , the three shared electron pairs are thus in a single $\sigma$ orbital and in two $\pi$ orbitals. Each of these orbitals is a bonding orbital, therefore all six electrons have their energy lowered in comparison to the separated atoms.