Structure 2.2.16 — Hybridization and Molecular Geometry

Hybridization

Hybridization is the process by which atomic orbitals (like s and p) mix to form new, equivalent hybrid orbitals.

These hybrid orbitals are used to form covalent bonds that match the observed molecular geometries.

Hybrid orbitals explain:

• The geometry of molecules (VSEPR-based)
• The shape and angle between bonds
• The equivalence of bond lengths in some structures (e.g., \( \text{CH}_4 \))

Hybridization Types (IBDP Syllabus)

1. \( \text{sp} \) Hybridization:

• Orbitals involved: One s orbital + one p orbital = 2 hybrid orbitals.
• Number of electron domains: 2
• Geometry: Linear
• Bond angle: 180°
• Unhybridized p orbitals: Two remain unhybridized, which can form π bonds.
• Commonly observed in: Atoms with triple bonds or two double bonds.

2. \( \text{sp}^2 \) Hybridization:

• Orbitals involved: One s orbital + two p orbitals = 3 hybrid orbitals.
• Number of electron domains: 3
• Geometry: Trigonal planar
• Bond angle: 120°
• Unhybridized p orbital: One unhybridized p orbital can form a π bond.
• Commonly observed in: Atoms with a double bond and two other electron regions.

3. \( \text{sp}^3 \) Hybridization:

• Orbitals involved: One s orbital + three p orbitals = 4 hybrid orbitals.
• Number of electron domains: 4
• Geometry: Tetrahedral
• Bond angle: 109.5°
• All orbitals are hybridized: No unhybridized p orbitals remain.
• Commonly observed in: Atoms with only single bonds (σ bonds), like carbon in methane or nitrogen in ammonia.

Hybridization and Electron Domains

Each hybrid orbital corresponds to one electron domain around the central atom. The number of electron domains in the VSEPR model helps predict the hybridization:

• 2 domains → \( \text{sp} \)
• 3 domains → \( \text{sp}^2 \)
• 4 domains → \( \text{sp}^3 \)

Examples: Organic Compounds

• \( \text{CH}_4 \): Carbon forms 4 σ bonds (tetrahedral). → \( \text{sp}^3 \)
• \( \text{C}_2\text{H}_4 \) (ethene): Each C forms 3 σ bonds, 1 π bond → \( \text{sp}^2 \)
• \( \text{C}_2\text{H}_2 \) (ethyne): Each C forms 2 σ bonds, 2 π bonds → \( \text{sp} \)

Examples: Inorganic Compounds

• \( \text{BeCl}_2 \): Linear shape → \( \text{sp} \) hybridization (2 bonding pairs)
• \( \text{BF}_3 \): Trigonal planar → \( \text{sp}^2 \) hybridization
• \( \text{NH}_3 \): Trigonal pyramidal → \( \text{sp}^3 \) hybridization
• \( \text{H}_2\text{O} \): Bent → \( \text{sp}^3 \) (2 bonding + 2 lone pairs)

 Step-by-Step: How to Determine Hybridization

1. Draw the Lewis structure.
2. Count the number of electron domains around the central atom (bonds + lone pairs).
3. Assign the hybridization:
   • 2 domains → \( \text{sp} \)
   • 3 domains → \( \text{sp}^2 \)
   • 4 domains → \( \text{sp}^3 \)
4. Predict geometry based on electron domain geometry.

Relationship Between Hybridization, Bonding, and Geometry

 Notes on π Bonds and Hybridization

• σ bonds are formed using hybrid orbitals
• π bonds come from unhybridized p orbitals
• π bonds appear in double/triple bonds and restrict rotation