# 4.6: Hybridization using d Orbitals

## 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 orbitals that point toward the vertices of a trigonal bipyramid (part (a) in Figure 4.6.7). 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 orbitals oriented toward the vertices of an octahedron (part (b) in Figure 4.6.7). 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. Figure 4.6.7 Hybrid Orbitals Involving d Orbitals. The formation of a set of (a) five sp3d hybrid orbitals and (b) six sp3d2 hybrid orbitals from ns, np, and nd atomic orbitals where n = 4.

Example 4.6.2

What is the hybridization of the central atom in each species? Describe the bonding in each species.

1. XeF4
2. SO42−
3. SF4

Given: three chemical species

Asked for: hybridization of the central atom

Strategy:

1. Determine the geometry of the molecule using the strategy in Example 1. From the valence electron configuration of the central atom and the number of electron pairs, determine the hybridization.
2. Place the total number of electrons around the central atom in the hybrid orbitals and describe the bonding.

Solution:

1. A Using the VSEPR model, we find that Xe in XeF4 forms four bonds and has two lone pairs, so its structure is square planar and it has six electron pairs. The six electron pairs form an octahedral arrangement, so the Xe must be sp3d2 hybridized. B With 12 electrons around Xe, four of the six sp3d2 hybrid orbitals form Xe–F bonds, and two are occupied by lone pairs of electrons.
2. A The S in the SO42− ion has four electron pairs and has four bonded atoms, so the structure is tetrahedral. The sulfur must be sp3 hybridized to generate four S–O bonds. B Filling the sp3 hybrid orbitals with eight electrons from four bonds produces four filled sp3 hybrid orbitals.
3. A The S atom in SF4 contains five electron pairs and four bonded atoms. The molecule has a seesaw structure with one lone pair: To accommodate five electron pairs, the sulfur atom must be sp3d hybridized. B Filling these orbitals with 10 electrons gives four sp3d hybrid orbitals forming S–F bonds and one with a lone pair of electrons.

Exercise 4.6.2

What is the hybridization of the central atom in each species? Describe the bonding.

1. PCl4+
2. BrF3
3. SiF62−
sp3 with four P–Cl bonds
sp3d with three Br–F bonds and two lone pairs
sp3d2 with six Si–F bonds

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.

Example 4.6.3: $$OF_4$$

What is the hybridization of the oxygen atom in OF4? Is OF4 likely to exist?

Given: chemical compound

Strategy:

1. Predict the geometry of OF4 using the VSEPR model.
2. From the number of electron pairs around O in OF4, predict the hybridization of O. Compare the number of hybrid orbitals with the number of electron pairs to decide whether the molecule is likely to exist.

Solution:

A The VSEPR model predicts that OF4 will have five electron pairs, resulting in a trigonal bipyramidal geometry with four bonding pairs and one lone pair. B To accommodate five electron pairs, the O atom would have to be sp3d hybridized. The only d orbital available for forming a set of sp3d hybrid orbitals is a 3d orbital, which is much higher in energy than the 2s and 2p valence orbitals of oxygen. As a result, the OF4 molecule is unlikely to exist. In fact, it has not been detected.

Exercise 4.6.3

What is the hybridization of the boron atom in $$BF_6^{3−}$$? Is this ion likely to exist?