Valence Bond Theory: A Localized Bonding Approach

In Chapter 8 "Ionic versus Covalent Bonding", you learned that as two hydrogen atoms approach each other from an infinite distance, the energy of the system reaches a minimum. This region of minimum energy in the energy diagram corresponds to the formation of a covalent bond between the two atoms at an H–H distance of 74 pm (Figure 8.9 "A Plot of Potential Energy versus Internuclear Distance for the Interaction between Two Gaseous Hydrogen Atoms"). According to quantum mechanics, bonds form between atoms because their atomic orbitals overlap, with each region of overlap accommodating a maximum of two electrons with opposite spin, in accordance with the Pauli principle. In this case, a bond forms between the two hydrogen atoms when the singly occupied 1s atomic orbital of one hydrogen atom overlaps with the singly occupied 1s atomic orbital of a second hydrogen atom. Electron density between the nuclei is increased because of this orbital overlap and results in a localized electron-pair bond (Figure 9.10 "Overlap of Two Singly Occupied Hydrogen 1").

Although Lewis and VSEPR structures also contain localized electron-pair bonds, neither description uses an atomic orbital approach to predict the stability of the bond. Doing so forms the basis for a description of chemical bonding known as valence bond theoryA localized bonding model that assumes that the strength of a covalent bond is proportional to the amount of overlap between atomic orbitals and that an atom can use different combinations of atomic orbitals (hybrids) to maximize the overlap between bonded atoms., which is built on two assumptions:

1. The strength of a covalent bond is proportional to the amount of overlap between atomic orbitals; that is, the greater the overlap, the more stable the bond.
2. An atom can use different combinations of atomic orbitals to maximize the overlap of orbitals used by bonded atoms.

Figure 9.11 "Three Different Ways to Form an Electron-Pair Bond" shows an electron-pair bond formed by the overlap of two ns atomic orbitals, two np atomic orbitals, and an ns and an np orbital where n = 2. Maximum overlap occurs between orbitals with the same spatial orientation and similar energies.

Let’s examine the bonds in BeH₂, for example. According to the VSEPR model, BeH₂ is a linear compound with four valence electrons and two Be–H bonds. Its bonding can also be described using an atomic orbital approach. Beryllium has a 1s²2s² electron configuration, and each H atom has a 1s¹ electron configuration. Because the Be atom has a filled 2s subshell, however, it has no singly occupied orbitals available to overlap with the singly occupied 1s orbitals on the H atoms. If a singly occupied 1s orbital on hydrogen were to overlap with a filled 2s orbital on beryllium, the resulting bonding orbital would contain three electrons, but the maximum allowed by quantum mechanics is two. How then is beryllium able to bond to two hydrogen atoms? One way would be to add enough energy to excite one of its 2s electrons into an empty 2p orbital and reverse its spin, in a process called promotionThe excitation of an electron from a filled $n{s}^2$ atomic orbital to an empty $np$ or $(n-1)d$ valence orbital.:

In this excited state, the Be atom would have two singly occupied atomic orbitals (the 2s and one of the 2p orbitals), each of which could overlap with a singly occupied 1s orbital of an H atom to form an electron-pair bond. Although this would produce BeH₂, the two Be–H bonds would not be equivalent: the 1s orbital of one hydrogen atom would overlap with a Be 2s orbital, and the 1s orbital of the other hydrogen atom would overlap with an orbital of a different energy, a Be 2p orbital. Experimental evidence indicates, however, that the two Be–H bonds have identical energies. To resolve this discrepancy and explain how molecules such as BeH₂ form, scientists developed the concept of hybridization.

Hybridization of s and p Orbitals

The localized bonding approach uses a process called hybridizationA process in which two or more atomic orbitals that are similar in energy but not equivalent are combined mathematically to produce sets of equivalent orbitals that are properly oriented to form bonds., in which atomic orbitals that are similar in energy but not equivalent are combined mathematically to produce sets of equivalent orbitals that are properly oriented to form bonds. These new combinations are called hybrid atomic orbitalsNew atomic orbitals formed from the process of hybridization. because they are produced by combining (hybridizing) two or more atomic orbitals from the same atom.

In BeH₂, we can generate two equivalent orbitals by combining the 2s orbital of beryllium and any one of the three degenerate 2p orbitals. By taking the sum and the difference of Be 2s and 2p_z atomic orbitals, for example, we produce two new orbitals with major and minor lobes oriented along the z-axes, as shown in Figure 9.12 "The Formation of ".Because the difference A − B can also be written as A + (−B), in Figure 9.12 "The Formation of " and subsequent figures we have reversed the phase(s) of the orbital being subtracted, which is the same as multiplying it by −1 and adding. This gives us Equation 9.1, where the value $\frac{1}{\sqrt{2}}$ is needed mathematically to indicate that the 2s and 2p orbitals contribute equally to each hybrid orbital.

The nucleus resides just inside the minor lobe of each orbital. In this case, the new orbitals are called sp hybrids because they are formed from one s and one p orbital. The two new orbitals are equivalent in energy, and their energy is between the energy values associated with pure s and p orbitals, as illustrated in this diagram:

Each singly occupied sp hybrid orbitalThe two equivalent hybrid orbitals that result when one $ns$ orbital and one $np$ orbital are combined (hybridized). The two $sp$ hybrid orbitals are oriented at 180° from each other. They are equivalent in energy, and their energy is between the energy values associated with pure $s$ and pure $p$ orbitals. can now form an electron-pair bond with the singly occupied 1s atomic orbital of one of the H atoms. As shown in Figure 9.13 "Explanation of the Bonding in BeH", each sp orbital on Be has the correct orientation for the major lobes to overlap with the 1s atomic orbital of an H atom. The formation of two energetically equivalent Be–H bonds produces a linear BeH₂ molecule. Thus valence bond theory does what neither the Lewis electron structure nor the VSEPR model is able to do; it explains why the bonds in BeH₂ are equivalent in energy and why BeH₂ has a linear geometry.

Because both promotion and hybridization require an input of energy, the formation of a set of singly occupied hybrid atomic orbitals is energetically uphill. The overall process of forming a compound with hybrid orbitals will be energetically favorable only if the amount of energy released by the formation of covalent bonds is greater than the amount of energy used to form the hybrid orbitals (Figure 9.14 "A Hypothetical Stepwise Process for the Formation of BeH"). As we will see, some compounds are highly unstable or do not exist because the amount of energy required to form hybrid orbitals is greater than the amount of energy that would be released by the formation of additional bonds.

The concept of hybridization also explains why boron, with a 2s²2p¹ valence electron configuration, forms three bonds with fluorine to produce BF₃, as predicted by the Lewis and VSEPR approaches. With only a single unpaired electron in its ground state, boron should form only a single covalent bond. By the promotion of one of its 2s electrons to an unoccupied 2p orbital, however, followed by the hybridization of the three singly occupied orbitals (the 2s and two 2p orbitals), boron acquires a set of three equivalent hybrid orbitals with one electron each, as shown here:

The hybrid orbitals are degenerate and are oriented at 120° angles to each other (Figure 9.15 "Formation of "). Because the hybrid atomic orbitals are formed from one s and two p orbitals, boron is said to be sp² hybridized (pronounced “s-p-two” or “s-p-squared”). The singly occupied sp² hybrid atomic orbitalsThe three equivalent hybrid orbitals that result when one $ns$ orbital and two $np$ orbitals are combined (hybridized). The three $s{p}^2$ hybrid orbitals are oriented in a plane at 120° from each other. They are equivalent in energy, and their energy is between the energy values associated with pure $s$ and pure $p$ orbitals. can overlap with the singly occupied orbitals on each of the three F atoms to form a trigonal planar structure with three energetically equivalent B–F bonds.

Looking at the 2s²2p² valence electron configuration of carbon, we might expect carbon to use its two unpaired 2p electrons to form compounds with only two covalent bonds. We know, however, that carbon typically forms compounds with four covalent bonds. We can explain this apparent discrepancy by the hybridization of the 2s orbital and the three 2p orbitals on carbon to give a set of four degenerate sp³ (“s-p-three” or “s-p-cubed”) hybrid orbitals, each with a single electron:

The large lobes of the hybridized orbitals are oriented toward the vertices of a tetrahedron, with 109.5° angles between them (Figure 9.16 "Formation of "). Like all the hybridized orbitals discussed earlier, the sp³ hybrid atomic orbitalsThe four equivalent hybrid orbitals that result when one $ns$ orbital and three $np$ orbitals are combined (hybridized). The four $s{p}^3$ hybrid orbitals point at the vertices of a tetrahedron, so they are oriented at 109.5° from each other. They are equivalent in energy, and their energy is between the energy values associated with pure $s$ and pure $p$ orbitals. are predicted to be equal in energy.

In addition to explaining why some elements form more bonds than would be expected based on their valence electron configurations, and why the bonds formed are equal in energy, valence bond theory explains why these compounds are so stable: the amount of energy released increases with the number of bonds formed. In the case of carbon, for example, much more energy is released in the formation of four bonds than two, so compounds of carbon with four bonds tend to be more stable than those with only two. Carbon does form compounds with only two covalent bonds (such as CH₂ or CF₂), but these species are highly reactive, unstable intermediates that form in only certain chemical reactions.

The bonding in molecules such as NH₃ or H₂O, which have lone pairs on the central atom, can also be described in terms of hybrid atomic orbitals. In NH₃, for example, N, with a 2s²2p³ valence electron configuration, can hybridize its 2s and 2p orbitals to produce four sp³ hybrid orbitals. Placing five valence electrons in the four hybrid orbitals, we obtain three that are singly occupied and one with a pair of electrons:

The three singly occupied sp³ lobes can form bonds with three H atoms, while the fourth orbital accommodates the lone pair of electrons. Similarly, H₂O has an sp³ hybridized oxygen atom that uses two singly occupied sp³ lobes to bond to two H atoms, and two to accommodate the two lone pairs predicted by the VSEPR model. Such descriptions explain the approximately tetrahedral distribution of electron pairs on the central atom in NH₃ and H₂O. Unfortunately, however, recent experimental evidence indicates that in CH₄ and NH₃, the hybridized orbitals are not entirely equivalent in energy, making this bonding model an active area of research.

Hybridization Using d Orbitals

Hybridization is not restricted to the ns and np atomic orbitals. The bonding in compounds with central atoms in the period 3 and below can also be described using hybrid atomic orbitals. In these cases, the central atom can use its valence (n − 1)d orbitals as well as its ns and np orbitals to form hybrid atomic orbitals, which allows it to accommodate five or more bonded atoms (as in PF₅ and SF₆). Using the ns orbital, all three np orbitals, and one (n − 1)d orbital gives a set of five sp³d hybrid orbitalsThe five hybrid orbitals that result when one $ns\text{,}$ three $np$ and one $(n-1)d$ orbitals are combined (hybridized). that point toward the vertices of a trigonal bipyramid (part (a) in Figure 9.17 "Hybrid Orbitals Involving "). In this case, the five hybrid orbitals are not all equivalent: three form a triangular array oriented at 120° angles, and the other two are oriented at 90° to the first three and at 180° to each other.

Similarly, the combination of the ns orbital, all three np orbitals, and two nd orbitals gives a set of six equivalent sp³d² hybrid orbitalsThe six equivalent hybrid orbitals that result when one $ns$, three $np$, and two $(n-1)d$ orbitals are combined (hybridized). oriented toward the vertices of an octahedron (part (b) in Figure 9.17 "Hybrid Orbitals Involving "). In the VSEPR model, PF₅ and SF₆ are predicted to be trigonal bipyramidal and octahedral, respectively, which agrees with a valence bond description in which sp³d or sp³d² hybrid orbitals are used for bonding.

Hybridization using d orbitals allows chemists to explain the structures and properties of many molecules and ions. Like most such models, however, it is not universally accepted. Nonetheless, it does explain a fundamental difference between the chemistry of the elements in the period 2 (C, N, and O) and those in period 3 and below (such as Si, P, and S).

Period 2 elements do not form compounds in which the central atom is covalently bonded to five or more atoms, although such compounds are common for the heavier elements. Thus whereas carbon and silicon both form tetrafluorides (CF₄ and SiF₄), only SiF₄ reacts with F⁻ to give a stable hexafluoro dianion, SiF₆²⁻. Because there are no 2d atomic orbitals, the formation of octahedral CF₆²⁻ would require hybrid orbitals created from 2s, 2p, and 3d atomic orbitals. The 3d orbitals of carbon are so high in energy that the amount of energy needed to form a set of sp³d² hybrid orbitals cannot be equaled by the energy released in the formation of two additional C–F bonds. These additional bonds are expected to be weak because the carbon atom (and other atoms in period 2) is so small that it cannot accommodate five or six F atoms at normal C–F bond lengths due to repulsions between electrons on adjacent fluorine atoms. Perhaps not surprisingly, then, species such as CF₆²⁻ have never been prepared.