10.7: Hybridization of Atomic Orbitals II

sp³d and sp³d ² Hybridization

To describe the five bonding orbitals in a trigonal bipyramidal arrangement, we must use five of the valence shell atomic orbitals (the s orbital, the three p orbitals, and one of the d orbitals), which gives five sp³d hybrid orbitals. With an octahedral arrangement of six hybrid orbitals, we must use six valence shell atomic orbitals (the s orbital, the three p orbitals, and two of the d orbitals in its valence shell), which gives six sp³d ² hybrid orbitals. These hybridizations are only possible for atoms that have d orbitals in their valence subshells (that is, not those in the first or second period).

In a molecule of phosphorus pentachloride, PCl₅, there are five P–Cl bonds (thus five pairs of valence electrons around the phosphorus atom) directed toward the corners of a trigonal bipyramid. We use the 3s orbital, the three 3p orbitals, and one of the 3d orbitals to form the set of five sp³d hybrid orbitals that are involved in the P–Cl bonds. Other atoms that exhibit sp³d hybridization include the sulfur atom in SF₄ and the chlorine atoms in ClF₃ and in ClF₄⁺.

The sulfur atom in sulfur hexafluoride, SF₆, exhibits sp³d ² hybridization. A molecule of sulfur hexafluoride has six bonding pairs of electrons connecting six fluorine atoms to a single sulfur atom. There are no lone pairs of electrons on the central atom. To bond six fluorine atoms, the 3s orbital, the three 3p orbitals, and two of the 3d orbitals form six equivalent sp³d ² hybrid orbitals, each directed toward a different corner of an octahedron. Other atoms that exhibit sp³d ² hybridization include the phosphorus atom in PCl₆⁻, the iodine atom in the interhalogens IF₆⁺, IF₅, ICl₄⁻, IF₄⁻, and the xenon atom in XeF₄.

This text has been adapted from Openstax, Chemistry 2e, Section 8.2: Hybrid Atomic Orbitals.

The trigonal bipyramidal, octahedral, and other molecular shapes can be explained by assuming the participation of 3d orbitals in the process of hybridization.

The phosphorus pentachloride molecule has a trigonal bipyramidal shape, and it contains 5 valence electrons. Phosphorus uses the 3s orbital, the three 3p orbitals, and one of the 3d orbitals to form five sp³d hybrid orbitals that are involved in the phosphorus–chlorine bonds.

Sulfur hexafluoride has an octahedral structure and it contains 6 valence electrons. The 3s orbital, the three 3p orbitals, and two of the 3d orbitals on sulfur form six equivalent sp³d² hybrid orbitals. These six sp³d² orbitals form an octahedral structure around sulfur and participate in the formation of sulfur–fluorine bonds.

The concept of hybridization also provides an explanation for the formation of multiple bonds. The side-on overlap of two p orbitals gives rise to a π bond. 

However, a π-bond can only be formed in double and triple bonds when a σ bond already exists between two atoms. Because the π bond exists on opposite sides of the internuclear axis, π bonds are unable to rotate around this axis.

In the ethene molecule, both carbons exhibit sp² hybridization. The mixing of one s orbital and two p orbitals of a carbon atom produce three identical sp² hybrid orbitals, and one p orbital remains unhybridized.

The carbon–carbon σ bond is formed by the overlap of two sp² hybrid orbitals, one on each carbon atom.

The two carbon–hydrogen σ bonds on each carbon are formed by the overlap of two sp² hybrid orbitals with the 1s orbitals on the hydrogen atom. Thus, five σ bonds are formed in the ethene molecule. 

The unhybridized 2p orbitals on the carbons overlap sideways with each other to produce a π bond. All six atoms lie in the same plane, and therefore 2p orbitals can overlap effectively.

Thus, the double bond in ethene consists of one σ and one π bond.

The triple bonds and the linear geometry of ethyne can be explained using sp hybridization. The 2s and 2p orbitals of both carbon atoms undergo hybridization to produce two sp orbitals each, and two p orbitals remain unhybridized. 

One of the sp-orbitals forms a σ bond with the other carbon atom, whereas the remaining sp orbital forms a σ bond with a hydrogen atom. The two unhybridized 2p orbitals are perpendicular and intersect at the principal axis of the sp hybrid orbitals. 

These 2p orbitals overlap sideways with the 2p orbitals of the other carbon atom resulting in the formation of two π bonds. Therefore, the triple bond in ethyne consists of one σ bond and two π bonds between the two carbon atoms.