5.2 Hybridization of atomic orbitals

Hybridization of Atomic Orbitals

Hybridization explains how atoms mix their standard atomic orbitals to form new, equivalent hybrid orbitals better suited for bonding. Without this concept, we can't account for the observed geometries of most molecules. For example, carbon's ground-state electron configuration ($1s^2 2s^2 2p^2$) suggests it should form only two bonds, yet methane ($CH_4$) has four identical bonds arranged tetrahedrally. Hybridization resolves this discrepancy.

By identifying an atom's hybridization state, you can predict molecular geometry, bond angles, and the types of bonds (sigma vs. pi) a molecule contains.

Hybridization in Covalent Bonding

Concept and Role of Hybridization

Hybridization is the mathematical mixing of atomic orbitals on the same atom to produce new hybrid orbitals with intermediate shapes and energies. The number of hybrid orbitals produced always equals the number of atomic orbitals mixed.

• Hybrid orbitals form by combining s and p orbitals (and sometimes d orbitals) within the same principal energy level
  • Example: sp hybridization combines one s orbital and one p orbital to produce two equivalent sp hybrid orbitals
• The resulting hybrid orbitals point in specific directions, which maximizes overlap with orbitals on neighboring atoms. Greater overlap means stronger, more stable sigma ($\sigma$) bonds.
• The type of hybridization an atom undergoes depends on the number of electron domains (bonding pairs and lone pairs) surrounding it

Why Hybridization Matters for Covalent Bonding

Standard atomic orbitals (s, p, d) don't explain the geometries we actually observe. Pure p orbitals are oriented at 90° to each other, yet molecules like $CH_4$ have bond angles of 109.5°. Hybridization accounts for this.

• Hybrid orbitals have shapes and orientations that maximize overlap between bonding atoms, producing stronger bonds than unhybridized orbitals would
• Hybridization also explains multiple bonds: after an atom forms its hybrid orbitals, any leftover unhybridized p orbitals are available for pi ($\pi$) bonding
• Understanding hybridization lets you predict structure, stability, and reactivity from a Lewis structure alone

Hybridization State of Atoms

Determining Hybridization State

The hybridization state of a central atom is set by the number of electron domains around it. An electron domain is any region of electron density: a single bond, a double bond, a triple bond, or a lone pair. (A double bond counts as one electron domain, not two.)

To find the hybridization of any central atom:

1. Draw the Lewis structure of the molecule
2. Count the total electron domains around the central atom (bonds + lone pairs, where each multiple bond counts as one domain)
3. Match the count to the table above

Exceptions and Special Cases

• Multiple bonds: A double or triple bond counts as a single electron domain for determining hybridization. In $C_2H_4$ (ethene), each carbon has three electron domains (two C–H bonds and one C=C bond), so each carbon is sp² hybridized. The second bond in the double bond is a $\pi$ bond from unhybridized p orbitals, not from hybrid orbitals.
• Expanded octets: Atoms in the third period and below can use d orbitals in hybridization. For example, $SF_6$ has six electron domains around sulfur and undergoes sp³d² hybridization, producing an octahedral geometry. Similarly, $PCl_5$ is sp³d hybridized with a trigonal bipyramidal geometry (five electron domains).

Formation of Hybrid Orbitals

Types of Hybrid Orbitals and Their Geometries

Each type of hybridization produces a characteristic number of orbitals with a specific spatial arrangement:

• sp hybridization: One s + one p orbital → two sp hybrid orbitals oriented 180° apart (linear). Two p orbitals remain unhybridized.
  • Example: In $CO_2$, carbon is sp hybridized. The two unhybridized p orbitals form $\pi$ bonds with each oxygen.
• sp² hybridization: One s + two p orbitals → three sp² hybrid orbitals oriented 120° apart (trigonal planar). One p orbital remains unhybridized.
  • Example: In $BF_3$, boron is sp² hybridized with all three hybrid orbitals forming $\sigma$ bonds to fluorine.
• sp³ hybridization: One s + three p orbitals → four sp³ hybrid orbitals oriented 109.5° apart (tetrahedral). No p orbitals remain unhybridized.
  • Example: In $NH_3$, nitrogen is sp³ hybridized. Three hybrid orbitals form bonds to hydrogen; the fourth holds a lone pair.

Unhybridized Orbitals and Multiple Bonds

After hybridization, any leftover p orbitals don't disappear. They sit perpendicular to the hybrid orbital plane and can overlap sideways with p orbitals on adjacent atoms to form pi ($\pi$) bonds.

• A single bond = one $\sigma$ bond (head-on overlap of hybrid orbitals)
• A double bond = one $\sigma$ bond + one $\pi$ bond
• A triple bond = one $\sigma$ bond + two $\pi$ bonds

For example, in $N_2$, each nitrogen is sp hybridized. One sp orbital on each nitrogen overlaps head-on to form the $\sigma$ bond. Each nitrogen has two unhybridized p orbitals, and these overlap sideways to form two $\pi$ bonds, giving the triple bond ($N \equiv N$).

In $C_2H_4$ (ethene), each carbon is sp² hybridized. The remaining unhybridized p orbital on each carbon overlaps sideways to form one $\pi$ bond above and below the molecular plane.

Hybridization Theory for Molecular Structure

Predicting Molecular Structure Using Hybridization

Here's a systematic approach:

1. Draw the Lewis structure of the molecule, showing all bonds and lone pairs
2. Count electron domains around the central atom (single bonds, double bonds, triple bonds, and lone pairs each count as one domain)
3. Assign the hybridization based on the electron domain count (2 → sp, 3 → sp², 4 → sp³)
4. Determine the electron domain geometry from the hybridization (linear, trigonal planar, or tetrahedral)
5. Identify the molecular shape by considering only the positions of atoms (not lone pairs)
6. Assign bond types: hybrid orbitals form $\sigma$ bonds; unhybridized p orbitals form $\pi$ bonds

Bond Angles and Molecular Shape

The ideal bond angles come directly from the hybridization:

• sp → 180°
• sp² → 120°
• sp³ → 109.5°

However, lone pairs compress bond angles because they occupy more space than bonding pairs. Lone pair electron density is held closer to the nucleus and spreads out more, squeezing bonding pairs together.

• In $NH_3$, the H–N–H bond angle is about 107°, slightly less than the ideal 109.5°, because of one lone pair
• In $H_2O$, the H–O–H bond angle is about 104.5°, compressed further because oxygen has two lone pairs

This distinction between electron domain geometry (includes lone pairs) and molecular shape (atoms only) is critical. The hybridization tells you the electron domain geometry; the molecular shape depends on how many of those domains are lone pairs versus bonds.