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Chemical Bonding and Molecular
Energy Levels
John S. Hutchinson
This work is produced by The Connexions Project and licensed under the Creative Commons Attribution License †
Abstract A development of the quantum mechanical concepts of bonding using valence bond and molecular orbital descriptions to account of bond strength and molecular ionization energies.
1 Foundation
Our basis for understanding chemical bonding and the structures of molecules is the electron orbital descrip- tion of the structure and valence of atoms, as provided by quantum mechanics. We assume an understanding of the periodicity of the elements based on the nuclear structure of the atom and our deductions concerning valence based on electron orbitals.
2 Goals
Our model of valence describes a chemical bond as resulting from the sharing of a pair of electrons in the valence shell of the bonded atoms. This sharing allows each atom to complete an octet of electrons in its valence shell, at least in the sense that we count the shared electrons as belonging to both atoms. However, it is not clear that this electron counting picture has any basis in physical reality. What is meant, more precisely, by the sharing of the electron pair in a bond, and why does this result in the bonding of two atoms together? Indeed, what does it mean to say that two atoms are bound together? Furthermore, what is the signi cance of sharing a pair of electrons? Why aren't chemical bonds formed by sharing one or three electrons, for example? We seek to understand how the details of chemical bonding are related to the properties of the molecules formed, particularly in terms of the strengths of the bonds formed.
3 Observation 1: Bonding with a Single Electron
We began our analysis of the energies and motions of the electrons in atoms by observing the properties of the simplest atom, hydrogen, with a single electron. Similarly, to understand the energies and motions of electrons which lead to chemical bonding, we begin our observations with the simplest particle with a chemical bond, which is the H+ 2 molecular ion. Each hydrogen nucleus has a charge of +1. An H 2 + molecular ion therefore has a single electron. It seems inconsistent with our notions of valence that a single electron, rather than an electron pair, can generate a chemical bond. However, these concepts have been based on
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observations on molecules, not molecular ions like H+ 2. And it is indeed found that H 2 + is a stable bound molecular ion. What forces and motions hold the two hydrogen nuclei close together in the H 2 + ion? It is worth keeping in mind that the two nuclei must repel one another, since they are both positively charged. In the absence of the electron, the two nuclei would accelerate away from one another, rather than remaining in close proximity. What is the role of the electron? Clearly, the electron is attracted to both nuclei at the same time, and, in turn, each nucleus is attracted to the electron. The eect of this is illustrated in Fig. 1. In Fig. 1a, the electron is outside of the two nuclei. In this position, the electron is primarily attracted to the nucleus on the left, to which it is closer. More importantly, the nucleus on the right feels a greater repulsion from the other nucleus than attraction to the electron, which is farther away. As a result, the nucleus on the right experiences a strong force driving it away from the hydrogen atom on the left. This arrangement does not generate chemical bonding, therefore. By contrast, in Fig. 1b, the electron is between the two nuclei. In this position, the electron is roughly equally attracted to the two nuclei, and very importantly, each nucleus feels an attractive force to the electron which is greater than the repulsive force generated by the other nucleus. Focusing on the electron's energy, the proximity of the two nuclei provides it a doubly attractive environment with a very low potential energy. If we tried to pull one of the nuclei away, this would raise the potential energy of the electron, since it would lose attraction to that nucleus. Hence, to pull one nucleus away requires us to add energy to the molecular ion. This is what is meant by a chemical bond: the energy of the electrons is lower when the atoms are in close proximity than when the atoms are far part. This holds the nuclei close together, since we must do work (add energy) to take the nuclei apart.
On second thought, though, this description must be inaccurate. We have learned our study of Energy Levels in Atoms that an electron must obey the uncertainty principle and that, as a consequence, the electron does not have a de nite position, between the nuclei or otherwise. We can only hope to specify a probability for observing an electron in a particular location. This probability is, from quantum mechanics, provided by the wave function. What does this probability distribution look like for the H 2 + molecular ion? To answer this question, we begin by experimenting with a distribution that we know: the 1s electron orbital in a hydrogen atom. This we recall has the symmetry of a sphere, with equal probability in all directions away from the nucleus. To create an H 2 + molecular ion from a hydrogen atom, we must add a bare second hydrogen nucleus (an H+ion). Imagine bringing this nucleus closer to the hydrogen atom from a very great distance (see Fig. 2a). As the H+ion approaches the neutral atom, both the hydrogen atom's nucleus and electron respond to the electric potential generated by the positive charge. The electron is attracted and the hydrogen atom nucleus is repelled. As a result, the distribution of probability for the electron about the nucleus must become distorted, so that the electron has a greater probability of being near the H+^ ion and the nucleus has a greater probability of being farther from the ion. This distortion, illustrated in Fig. 2b, is called polarization: the hydrogen atom has become like a dipole, with greater negative charge to one side and greater positive charge to the other.
by starting with two separated hydrogen atoms. Each of these atoms has a single electron in a 1s orbital. As the two atoms approach one another, each electron orbital is polarized in the direction of the other atom. Once the atoms are close enough together, these two orbitals become superimposed. Now we must recall that these orbitals describe the wave-like motion of the electron, so that, when these two wave functions overlap, they must interfere, either constructively or destructively. In Fig. 3, we see the consequences of constructive and destructive interference. We can deduce that, in H 2 the electron orbitals from the atoms must constructively interfere, because that would increase the electron probability in the region between the nuclei, resulting in bonding as before. Therefore, the σ molecular orbital describing the two electrons in H 2 can be understood as resulting from the constructive overlap of two atomic 1s electron orbitals. We now add to our observations of diatomic molecules by noting that, of the diatomic molecules formed from like atoms of the rst ten elements, H 2 , Li 2 , B 2 , C 2 , N 2 , O 2 , and F 2 are stable molecules with chemical bonds, whereas He 2 , Be 2 , and N e 2 are not bound. In examining the electron con gurations of the atoms of these elements, we discover a correspondence with which diatomic molecules are bound and which ones are not. H, Li, B, N, andF all have odd numbers of electrons, so that at least one electron in each atom is unpaired. By contrast, He, Be, and Ne all have even numbers of electrons, none of which are unpaired. The other atoms, C and O both have an even number of electrons. However, as deduced in our understanding of the electron con gurations in atoms, electrons will, when possible, distribute themselves into dierent orbitals of the same energy so as to reduce the eect of their mutual repulsion. Thus, in C and O, there are three 2p orbitals into which 2 and 4 electrons are placed, respectively. Therefore, in both atoms, there are two unpaired electrons. We conclude that bonds will form between atoms if and only if there are unpaired electrons in these atoms. In H 2 , the unpaired electrons from the separated atoms become paired in a molecular orbital formed from the overlap of the 1s atomic electron orbitals. In the case of a hydrogen atom, then, there are of course no paired electrons in the atom to worry about. In all other atoms, there certainly are paired electrons, regardless of whether there are or are not unpaired electrons. For example, in a lithium atom, there are two paired electrons in a 1s orbital and an unpaired electron in the 2s orbital. To form Li 2 , the unpaired electron from each atom can be placed into a molecular orbital formed from the overlap of the 2s atomic electron orbitals. However, what becomes of the two electrons paired in the 1s orbital in a Li atom during the bonding of Li 2? To answer this question, we examine He 2 , in which each atom begins with only the two 1s electrons. As we bring the two He atoms together from a large distance, these 1s orbitals should become polarized, as in the hydrogen atom. When the polarized 1s orbitals overlap, constructive interference will again result in a σ molecular orbital, just as in H 2. Yet, we observe that He 2 is not a stable bound molecule. The problem which prevents bonding for He 2 arises from the Pauli Exclusion Principle: only two of the four electrons in He 2 can be placed into this σ bonding molecular orbital. The other two must go into a dierent orbital with a dierent probability distribution. To deduce the form of this new orbital, we recall that the bonding orbital discussed so far arises from the constructive interference of the atomic orbitals, as shown in Fig. 3. We could, instead, have assumed destructive interference of these orbitals. Destructive interference of two waves eliminates amplitude in the region of overlap of the waves, also shown in Fig. 3. In the case of the atomic orbitals, this means that the molecular orbital formed from destructive interference decreases probability for the electron to be between in the nuclei. Therefore, it increases probability for the electron to be outside the nuclei, as in Fig. 1a. As discussed there, this arrangement for the electron does not result in bonding; instead, the nuclei repel each other and the atoms are forced apart. This orbital is thus called an anti-bonding orbital. This orbital also has the symmetry of a cylinder along the bond axis, so it is also a σorbital; to indicate that it is an anti-bonding orbital, we designate it with an asterisk, σ∗
Figure 3
In He 2 , both the bonding and the anti-bonding orbitals must be used in order to accommodate four electrons. The two electrons in the bonding orbital lower the energy of the molecule, but the two electrons in the anti-bonding orbital raise it. Since two He atoms will not bind together, then the net eect must be that the anti-bonding orbital more than osets the bonding orbital. We have now deduced an explanation for why the paired electrons in an atom do not contribute to bonding. Both bonding and anti-bonding orbitals are always formed when two atomic orbitals overlap.
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 σ orbital. Of course, they also form a σ∗orbital. The two electrons are then paired in the bonding orbital.
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Figure 5
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 ve p electrons in each uorine 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 dier between N 2 and F 2 .) We place two electrons in the σ orbital, four more in the two π orbitals, and four more in the two π∗^ 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.
Figure 6
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 uorine is less than
Figure 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 dierence 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 de ne the bond order as BondOrder = 12 (# bonding electrons − # antibonding electrons) Note that, de ned 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.
6 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 aect 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.
- 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: a. C 2 or C 2 + b. B 2 or B+ 2 c. F 2 or F 2 − d. O 2 or O+ 2
- Which of the following diatomic molecules are paramagnetic: CO, 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.
