Pattern: Steric number determines hybridization directly!

Types of Hybridization

sp Hybridization

Orbitals mixed: 1 s + 1 p = 2 sp hybrid orbitals

Electron geometry: Linear

Bond angle: 180°

Remaining p orbitals: 2 unhybridized p orbitals (can form π bonds)

Example: CO₂, BeCl₂, HCN, C₂H₂ (acetylene)

Carbon in acetylene (C₂H₂):

$H-C≡C-H$

• Each C is sp hybridized
• 2 sp orbitals form σ bonds (1 to H, 1 to other C)
• 2 unhybridized p orbitals form 2 π bonds (triple bond = 1 σ + 2 π)

Geometry: Linear (all atoms in line)

sp² Hybridization

Orbitals mixed: 1 s + 2 p = 3 sp² hybrid orbitals

Electron geometry: Trigonal planar

Bond angle: 120°

Remaining p orbitals: 1 unhybridized p orbital (can form π bond)

Example: C₂H₄ (ethylene), BF₃, formaldehyde (H₂CO)

Carbon in ethylene (C₂H₄):

$H2C=CH2$

• Each C is sp² hybridized
• 3 sp² orbitals form σ bonds (2 to H, 1 to other C)
• 1 unhybridized p orbital forms 1 π bond (double bond = 1 σ + 1 π)

Geometry: Trigonal planar around each C (all atoms in same plane)

sp³ Hybridization

Orbitals mixed: 1 s + 3 p = 4 sp³ hybrid orbitals

Electron geometry: Tetrahedral

Bond angle: 109.5°

Remaining p orbitals: None (all p orbitals used)

Example: CH₄, H₂O, NH₃, CCl₄

Carbon in methane (CH₄):

• C is sp³ hybridized
• 4 sp³ orbitals form 4 σ bonds to H atoms
• No π bonds (only single bonds)

Geometry: Tetrahedral

Note on lone pairs:

• H₂O: O is sp³ hybridized (2 bonds + 2 lone pairs = 4 electron domains)
• NH₃: N is sp³ hybridized (3 bonds + 1 lone pair = 4 electron domains)
• Lone pairs occupy hybrid orbitals!

sp³d Hybridization

Orbitals mixed: 1 s + 3 p + 1 d = 5 sp³d hybrid orbitals

Electron geometry: Trigonal bipyramidal

Bond angles: 90°, 120°

Example: PCl₅, SF₄

Only in Period 3+ (need d orbitals)

Phosphorus in PCl₅:

• P is sp³d hybridized
• 5 sp³d orbitals form 5 σ bonds to Cl atoms
• Trigonal bipyramidal geometry

sp³d² Hybridization

Orbitals mixed: 1 s + 3 p + 2 d = 6 sp³d² hybrid orbitals

Electron geometry: Octahedral

Bond angles: 90°

Example: SF₆, XeF₄

Only in Period 3+ (need d orbitals)

Sulfur in SF₆:

• S is sp³d² hybridized
• 6 sp³d² orbitals form 6 σ bonds to F atoms
• Octahedral geometry

Xenon in XeF₄:

• Xe is sp³d² hybridized
• 4 sp³d² orbitals form σ bonds to F atoms
• 2 sp³d² orbitals contain lone pairs
• Square planar molecular geometry

Sigma (σ) and Pi (π) Bonds

Sigma (σ) Bonds

Definition: Bond formed by head-on overlap of orbitals along the internuclear axis

Characteristics:

• Strongest type of covalent bond
• Electron density concentrated between nuclei
• Allows rotation around bond axis
• Can form from: s-s, s-p, p-p (head-on), or hybrid-hybrid overlap

In every bond:

• Single bond: 1 σ bond
• Double bond: 1 σ bond (+ 1 π)
• Triple bond: 1 σ bond (+ 2 π)

There is ALWAYS one (and only one) σ bond between any two bonded atoms

Pi (π) Bonds

Definition: Bond formed by side-by-side overlap of parallel p orbitals

Characteristics:

• Weaker than σ bonds
• Electron density above and below internuclear axis
• Restricts rotation (causes rigidity)
• Can only form from unhybridized p orbitals (or d orbitals)

Occurrence:

• Single bond: 0 π bonds
• Double bond: 1 π bond
• Triple bond: 2 π bonds

π bonds only exist in multiple bonds!

Visualizing σ and π Bonds

Single bond (C-C):

• 1 σ bond (sp³-sp³ overlap)
• Total: 1 σ, 0 π

Double bond (C=C):

• 1 σ bond (sp²-sp² overlap)
• 1 π bond (p-p side overlap)
• Total: 1 σ, 1 π

Triple bond (C≡C):

• 1 σ bond (sp-sp overlap)
• 2 π bonds (two sets of p-p overlaps)
• Total: 1 σ, 2 π

Determining Hybridization - Step by Step

Step 1: Draw Lewis structure

Step 2: Identify central atom

Step 3: Count electron domains (steric number)

• Bonds (each counts as 1, regardless of single/double/triple)
• Lone pairs (each counts as 1)

Step 4: Match steric number to hybridization

Step 5: Count σ and π bonds

• Count all bonds
• Each bond has 1 σ bond
• Double bond: +1 π bond
• Triple bond: +2 π bonds

Examples of Hybridization

Example 1: Methane (CH₄)

Lewis structure: C with 4 single bonds to H

Step 1: Central atom = C

Step 2: Count domains

• 4 C-H bonds = 4 domains
• 0 lone pairs
• SN = 4

Step 3: Hybridization

• SN = 4 → sp³

Step 4: Bonds

• 4 single bonds = 4 σ bonds
• No π bonds
• Total: 4 σ, 0 π

Example 2: Ethylene (C₂H₄)

Lewis structure: H₂C=CH₂

For each C:

Step 1: Count domains

• 2 C-H bonds = 2 domains
• 1 C=C bond = 1 domain (double bond counts as 1!)
• 0 lone pairs
• SN = 3

Step 2: Hybridization

• SN = 3 → sp²

Step 3: Bonds (entire molecule)

• 4 C-H bonds = 4 σ bonds
• 1 C=C double bond = 1 σ + 1 π
• Total: 5 σ, 1 π

Example 3: Acetylene (C₂H₂)

Lewis structure: H-C≡C-H

For each C:

Step 1: Count domains

• 1 C-H bond = 1 domain
• 1 C≡C bond = 1 domain (triple bond counts as 1!)
• 0 lone pairs
• SN = 2

Step 2: Hybridization

• SN = 2 → sp

Step 3: Bonds (entire molecule)

• 2 C-H bonds = 2 σ bonds
• 1 C≡C triple bond = 1 σ + 2 π
• Total: 3 σ, 2 π

Example 4: Water (H₂O)

Lewis structure: H-O-H with 2 lone pairs on O

For O:

Step 1: Count domains

• 2 O-H bonds = 2 domains
• 2 lone pairs = 2 domains
• SN = 4

Step 2: Hybridization

• SN = 4 → sp³

Important: Lone pairs occupy hybrid orbitals!

Step 3: Bonds

• 2 O-H bonds = 2 σ bonds
• Total: 2 σ, 0 π

Hybridization in Complex Molecules

For molecules with multiple central atoms, analyze each atom separately.

Example: Formaldehyde (H₂CO)

$H2C=O$

For C:

• 2 C-H bonds + 1 C=O bond = 3 domains
• SN = 3 → sp² hybridization

For O:

• 1 C=O bond + 2 lone pairs = 3 domains
• SN = 3 → sp² hybridization

Bonds:

• 2 C-H: 2 σ bonds
• 1 C=O: 1 σ + 1 π
• Total: 3 σ, 1 π

Properties Related to π Bonds

Rotation

Single bonds (only σ): Free rotation possible

• σ bond allows rotation around bond axis
• Examples: C-C in ethane

Double bonds (σ + π): No rotation

• π bond restricts rotation (would break π overlap)
• Creates geometric isomers (cis/trans)
• Examples: C=C in ethylene

Triple bonds (σ + 2π): No rotation

• Even more restricted than double bonds
• Linear geometry maintained

Bond Strength and Length

More π bonds → Stronger, shorter bonds

C-C bonds:

Pattern:

• More bonds → shorter distance
• More bonds → higher energy (stronger)

Common Mistakes to Avoid

1. Counting multiple bonds as multiple domains: C=O is ONE domain, not two
2. Forgetting lone pairs in hybridization: Lone pairs occupy hybrid orbitals
3. Confusing σ and π count: Every bond has exactly 1 σ; π bonds only in multiple bonds
4. Wrong hybrid orbital count: sp³ makes 4 orbitals, not 3
5. Assuming Period 2 can expand octet: C, N, O, F cannot use d orbitals (no sp³d or sp³d²)

Summary Table: Complete Reference

Applications

1. Predicting molecular properties: Hybridization explains observed geometries
2. Understanding reactivity: π bonds are more reactive than σ bonds
3. Isomerism: Restricted rotation around double bonds creates geometric isomers
4. Conjugation: Alternating single/double bonds create extended π systems
5. Spectroscopy: π→π* transitions explain UV-visible absorption

N has 3 N-H bonds and 1 lone pair

Step 2: Identify central atom

Central atom: N (nitrogen)

Step 3: Count electron domains on N

• 3 N-H single bonds = 3 domains
• 1 lone pair = 1 domain
• Total: 4 electron domains

Step 4: Determine steric number

$\text{SN} = 4$

Step 5: Determine hybridization

SN = 4 → sp³ hybridization

Important: The lone pair occupies one of the sp³ hybrid orbitals!

Step 6: Count sigma (σ) bonds

• 3 N-H bonds = 3 σ bonds
• Each single bond = 1 σ bond

Step 7: Count pi (π) bonds

• No double or triple bonds
• 0 π bonds

Answer:

Hybridization of N: sp³

Bonds:

• σ bonds: 3
• π bonds: 0

Explanation:

Nitrogen uses sp³ hybridization because it has 4 electron domains (3 bonds + 1 lone pair). The 4 sp³ hybrid orbitals are arranged tetrahedrally:

• 3 sp³ orbitals form σ bonds with H atoms (overlap with H 1s orbitals)
• 1 sp³ orbital contains the lone pair

The molecular geometry is trigonal pyramidal (not tetrahedral) because we only "see" the 3 H atoms in the molecular shape, though the electron geometry is tetrahedral.

Bond angle: ~107° (less than ideal 109.5° due to lone pair repulsion)

Key takeaway: Hybridization is determined by TOTAL electron domains (including lone pairs), not just bonding domains.