Absolutely Small - Michael D. Fayer [87]
In the series F2, O2, and N2, we have a single bond, double bond, and triple bond. The electron sharing gives each atom the Ne configuration. The next element to the left of nitrogen in the Periodic Table is carbon. One might assume that carbon would make a quadruple bond to form C2 and get to the Ne configurations. C2 doesn’t exist as a stable molecule. You can see why by looking at Figure 13.9, the MO diagram for N2, and removing the two electrons with the highest energy, the bonding MO. This would be the C2 electron configuration. However, it would have a double bond from the four electrons in the two π bonding MOs, not a quadruple bond. These two bonds would mean that the carbons in C2 would only have gained two extra electrons through sharing, not the four extra electrons each carbon needs to obtain the Ne configuration. Carbon needs to make four bonds to obtain the Ne configuration by forming molecules such as CH4. It can’t make four bonds by forming the C2 molecule, and C2 doesn’t exist.
FIGURE 13.9. The MO energy level diagram for nitrogen, N2. There is one extra pair of σ bonding electrons and two extra pairs of p bonding electrons. N2has a triple bond.
F2 has a single bond, O2 has a double bond, and N2 has a triple bond. Table 13.1 illustrates how the bond order strongly affects the properties of the bond. As the bond order increases, the bond length gets shorter and the bond energy increases. The bond energy is the energy that needs to be put into the molecule to break the bond. Breaking the bond means separating the atoms to a distance so far apart that they no longer feel each other. As will be discussed in the next chapter, carbon can make single, double, and triple bonds to another carbon atom while at the same time forming bonds to other atoms, such has hydrogen. Before we can discuss molecules that are larger than diatomics, we need to go beyond homonuclear diatomic molecules and examine heteronuclear diatomics to see how molecular orbitals are formed from nonidentical atoms.
HETERONUCLEAR DIATOMICS
In homonuclear diatomics, the MOs are formed from atomic orbitals with identical energies. In a heteronuclear diatomic, for example hydrogen fluoride (HF), the two atoms are different. Because the atoms are different, the atomic orbitals’ energies of one atom will not match those of the other. For HF, the hydrogen atom has a single electron in the 1s orbital. F has nine electrons in the 1s, 2s, and 2p orbitals. Both F2 and H2 have single bonds. Looking at Figure 13.6, the single bond in F2 isaσbond due to the bonding MO, . This bonding MO was formed from two 2pz atomic orbitals, one on each F atom. H2 has a single σ bond due to the bonding MO formed from two 1s orbitals (see Figure 12.7). To make HF, the question is, which orbital on F will combine with the 1s orbital on H to give MOs involved in bonding? Quantum theory calculations show that states (atomic orbitals) that are close in energy can combine to produce electron sharing MOs. Atomic orbitals with significantly different energies form MOs that are essentially equal to the individual atomic orbitals, and they do not contribute to bonding.
TABLE 13.1. The Effect of Bond Order on Bond Properties.
The hydrogen 1s orbital has an energy of—2.2 × 10-18 J. (Recall that the negative sign means that the electron is bound.) The energy of the fluorine 1s orbital (measured in the F2 molecule) is —1.1 × 10-16 J. So the 1s orbital of F is about 50 times lower in energy than the 1s orbital of H. This is a tremendous difference in energy, and the H 1s will not form a MO with the F 1s. In contrast, an F 2p orbital has an energy of—2.8 × 10-18 J. This energy is about 25% below the H 1s energy, which is close in energy. So an F 2p and the H 1s orbital energies are similar enough that they will form true MOs.
There are three F 2p orbitals, 2pz, 2px, and 2py. To decide which of these will interact with the H 1s, we need to define how we bring the atoms together. Let’s