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

This is depicted in Fig. 5 in what is called a “molecular orbital energy diagram.” Each pair of atomic orbitals, one from each atom, is overlapped to form a bonding and an anti-bonding orbital. The three 2p orbitals from each atom form one $\sigma$ and ${\sigma }^{\mathrm{*}}$ pair and two $\pi$ and ${\pi }^{\mathrm{*}}$ pairs. The lowering of the energies of the electrons in the $\sigma$ and $\pi$ orbitals is apparent. The ten n=2 electrons from the nitrogen atoms are then placed pairwise, in order of increasing energy, into these molecular orbitals. Note that, in agreement with the Pauli Exclusion Principle, each pair in a single orbital consists of one spin up and one spin down electron.

Recall now that we began the discussion of bonding in $N_2$ because of the curious result that the ionization energy of an electron in $F_2$ is less than that of an electron in an F atom. By comparing the molecular orbital energy level diagrams for $N_2$ and $F_2$ we are now prepared to answer this puzzle. There are five p electrons in each fluorine atom. These ten electrons must be distributed over the molecular orbitals whose energies are shown in Fig. 6. (Note that the ordering of the bonding 2p orbitals differ between $N_2$ and $F_2$ .) We place two electrons in the $\sigma$ orbital, four more in the two ${\pi }^{}$ orbitals, and four more in the two ${\pi }^{\mathrm{*}}$ orbitals. Overall, there are six electrons in bonding orbitals and four in anti-bonding orbitals. Since $F_2$ is a stable molecule, we must conclude that the lowering of energy for the electrons in the bonding orbitals is greater than the raising of energy for the electrons in the antibonding orbitals. Overall, this distribution of electrons is, net, equivalent to having two electrons paired in a single bonding orbital.

This also explains why the ionization energy of $F_2$ is less than that of an F atom. The electron with the highest energy requires the least energy to remove from the molecule or atom. The molecular orbital energy diagram in Fig. 6 clearly shows that the highest energy electrons in $F_2$ are in anti-bonding orbitals. Therefore, one of these electrons is easier to remove than an electron in an atomic 2p orbital, because the energy of an anti-bonding orbital is higher than that of the atomic orbitals. (Recall that this is why an anti-bonding orbital is, indeed, anti-bonding.) Therefore, the ionization energy of molecular fluorine is less than that of atomic fluorine. This clearly demonstrates the physical reality and importance of the anti-bonding orbitals.

A particularly interesting case is the oxygen molecule, $O_2$ . In completing the molecular orbital energy level diagram for oxygen, we discover that we must decide whether to pair the last two electrons in the same ${\mathrm{2p\pi }}^{\mathrm{*}}$ orbital, or whether they should be separated into different ${\mathrm{2p\pi }}^{\mathrm{*}}$ orbitals. To determine which, we note that oxygen molecules are paramagnetic, meaning that they are strongly attracted to a magnetic field. To account for this paramagnetism, we recall that electron spin is a magnetic property. In most molecules, all electrons are paired, so for each “spin up” electron there is a “spin down” electron and their magnetic fields cancel out. When all electrons are paired, the molecule is diamagnetic meaning that it responds only weakly to a magnetic field.

If the electrons are not paired, they can adopt the same spin in the presence of a magnetic field. This accounts for the attraction of the paramagnetic molecule to the magnetic field. Therefore, for a molecule to be paramagnetic, it must have unpaired electrons. The correct molecular orbital energy level diagram for an $O_2$ molecule is shown in Fig. 7.

In comparing these three diatomic molecules, we recall that $N_2$ has the strongest bond, followed by $O_2$ and $F_2$ . We have previously accounted for this comparison with Lewis structures, showing that $N_2$ is a triple bond, $O_2$ is a double bond, and $F_2$ is a single bond. The molecular orbital energy level diagrams in Figs. 5 to 7 cast a new light on this analysis. Note that, in each case, the number of bonding electrons in these molecules is eight. The difference in bonding is entirely due to the number of antibonding electrons: 2 for $N_2$ , 4 for $O_2$ , and six for $F_2$ . Thus, the strength of a bond must be related to the relative numbers of bonding and antibonding electrons in the molecule. Therefore, we now define the bond order as

$\text{Bond}\text{Order}=\frac{1}{2}\left(\text{\# bonding electrons}-\text{\# antibonding electrons}\right)$

Note that, defined this way, the bond order for $N_2$ is 3, for $O_2$ is 2, and for $F_2$ is 1, which agrees with our conclusions from Lewis structures. We conclude that we can predict the relative strengths of bonds by comparing bond orders.

Review and discussion questions

• Why does an electron shared by two nuclei have a lower potential energy than an electron on a single atom? Why does an electron shared by two nuclei have a lower kinetic energy than an electron on a single atom? How does this sharing result in a stable molecule? How can this affect be measured experimentally?
• Explain why the bond in an $H_2$ molecule is almost twice as strong as the bond in the $H_2^{+}$ ion. Explain why the $H_2$ bond is less than twice as strong as the $H_2^{+}$ bond.
• ${\mathrm{Be}}_2$ is not a stable molecule. What information can we determine from this observation about the energies of molecular orbitals?
• Less energy is required to remove an electron from an $F_2$ molecule than to remove an electron from an F atom. Therefore, the energy of that electron is higher in the molecule than in the atom. Explain why, nevertheless, $F_2$ is a stable molecule, i.e., the energy of an $F_2$ molecule is less than the energy of two F atoms.
• Why do the orbitals of an atom "hybridize" when forming a bond?
• Calculate the bond orders of the following molecules and predict which molecule in each pair has the stronger bond:
  • $C_2$ or $C_2^{+}$
  • $B_2$ or $B_2^{+}$
  • $F_2$ or $F_2^{-}$
  • $O_2$ or $O_2^{+}$
• Which of the following diatomic molecules are paramagnetic: CO, ${\mathrm{Cl}}_2$ , NO, $N_2$ ?
• $B_2$ is observed to be paramagnetic. Using this information, draw an appropriate molecular orbital energy level diagram for $B_2$ .