3.3 Hybridization

Learning Objectives

After Chapter 3.3, you will be able to:

• Recall the percentage of s character present in a given hybridization level, such as sp²
• Describe the relationship between electron density and resonance structures
• Identify the hybridization of an atom within a complex molecule:

  

              

Carbon has the electron configuration 1s²2s²2p² and therefore needs four electrons to complete its octet (2s²2p⁶). A typical molecule formed by carbon is methane, CH₄. Experimentation shows that the four σ bonds in methane are equivalent. This may seem inconsistent with what we know about the asymmetrical distribution of carbon’s valence electrons: two electrons in the 2s-orbital, one in the p_x-orbital, one in the p_y-orbital, and none in the p_z-orbital. This apparent discrepancy is accounted for by the theory of orbital hybridization.

sp³

Hybrid orbitals are formed by mixing different types of orbitals. Just as with molecular orbitals, we can use advanced mathematics to merge three p-orbitals and one s-orbital. The result? As shown in Figure 3.4, this forms four identical sp³ orbitals with new, hybridized shapes.

Figure 3.4. sp³-Hybridized Orbitals An atom with these orbitals has tetrahedral geometry, and there are no unhybridized p-orbitals to form π bonds.

All four of these orbitals point toward the vertices of a tetrahedron to minimize repulsion, which explains why carbon prefers tetrahedral geometry. The hybridization is accomplished by promoting one of the 2s electrons into the 2p_z-orbital, as shown in Figure 3.5. This produces four valence orbitals, each with one electron, which can be mathematically mixed to model the hybrid orbitals.

Figure 3.5. Hybridization of Carbon Orbitals

Key Concept

Hybridization is a way of making all of the bonds to a central atom equivalent to each other. The sp³ orbitals are the reason for the tetrahedral shape that is a hallmark of carbon-containing compounds.

The MCAT sometimes tests how much “s character” a certain hybrid orbital has. To answer such questions, we simply need to determine what type of hybridization exists and use the name to solve the problem. For example, in sp³ orbitals, we have one s- and three p-orbitals, so the bond has 25% s character and 75% p character.

sp²

Although carbon is most often bonded with sp³ hybridization, there are two other possibilities. When one s-orbital is mixed with two p-orbitals, three sp²-hybridized orbitals are formed, as shown in Figure 3.6. These orbitals have 33% s character and 67% p character.

This is the hybridization seen in alkenes. The third p-orbital of each carbon is left unhybridized. This is the orbital that participates in the π bond. The three sp² orbitals are oriented 120° apart, which allows for maximum separation. We know that the unhybridized p-orbital is involved in the π component of the double bond, but what about the hybrid orbitals? In ethene, two of the sp² hybridized orbitals will participate in C–H bonds, and the other hybrid orbital will line up with the π bond and form the σ component of the C=C double bond.

Figure 3.6. sp²-Hybridized Orbitals A molecule with these orbitals has trigonal planar geometry, and the one unhybridized p-orbital can be used to form a π bond.

sp

To form a triple bond, we need two of the p-orbitals to form π bonds, and the third p-orbital will combine with the s-orbital to form two sp-orbitals, as shown in Figure 3.7. These orbitals have 50% s character and 50% p character. These orbitals are oriented 180° apart, which explains the linear structure of molecules containing sp-hybridized carbons. The two π bonds can be between the carbon and one other atom (forming a triple bond, like ethyne), or between the carbon and two different atoms (forming two double bonds in a row, like carbon dioxide). In both cases, the molecule is linear about the sp-hybridized carbon.

Figure 3.7. sp-Hybridized Orbitals A molecule with these orbitals has linear geometry, and the two unhybridized p-orbitals can be used to form π bonds.

Resonance

Resonance delocalization of electrons occurs in molecules that have conjugated bonds. Conjugation requires alternating single and multiple bonds because this pattern aligns a number of unhybridized p-orbitals down the backbone of the molecule. π electrons can then delocalize through this p-orbital system, adding stability to the molecule. Resonance structures are drawn as the various transient forms the molecule takes, as shown in Figure 3.8.

Figure 3.8. Resonance Forms of Carbonate These forms have equal stability and therefore contribute equally to the true electron density of the molecule.

However, these forms aren’t in any sort of equilibrium—the electron density is distributed throughout, making the true form a hybrid of the resonance structures, as shown with ozone in Figure 3.9.

Figure 3.9. Structure of Ozone The true electron density of ozone is somewhere between the two resonance forms, creating 1.5 bonds between each oxygen, and leaving each oxygen with a charge.

If the stability of the various resonance forms differs, then the true electron density will favor the most stable form. Particular resonance structures can be favored because they lack formal charges or form full octets on highly electronegative atoms, like oxygen and nitrogen. Stabilization of positive and negative charges through induction and aromaticity can also favor certain resonance structures.

MCAT Concept Check 3.3

Before you move on, assess your understanding of the material with these questions.

1. What is the s character of sp-, sp²-, and sp³-hybridized orbitals?

   • sp:  

   • sp²:

   • sp³:

2. What are resonance structures? How does the true electron density of a compound relate to its resonance structures?