Hybridization: The mixing of atomic orbitals to form new hybrid orbitals of equal energy that can form bonds

Why hybridization occurs:

• Pure atomic orbitals (s, p, d) don't always explain observed molecular geometries
• Hybrid orbitals better describe bonding in molecules
• Provides framework for understanding molecular shapes

Key concept: Number and type of hybrid orbitals = number and type of electron domains (from VSEPR)

Relationship: Hybridization and Steric Number

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

Or more clearly: C has 2 single bonds to H and 1 double bond to O O has 2 lone pairs

Part (a): Determine hybridization

For Carbon (C):

Count electron domains:

• 2 C-H single bonds = 2 domains
• 1 C=O double bond = 1 domain (multiple bonds count as 1!)
• 0 lone pairs
• Total: 3 electron domains

Steric number: SN = 3

Hybridization: SN = 3 → sp² hybridization

For Oxygen (O):

Count electron domains:

• 1 C=O double bond = 1 domain
• 2 lone pairs = 2 domains
• Total: 3 electron domains

Steric number: SN = 3

Hybridization: SN = 3 → sp² hybridization

Part (b): Count σ and π bonds

Sigma bonds:

• C-H bond (left): 1 σ (sp²-s overlap)
• C-H bond (right): 1 σ (sp²-s overlap)
• C=O bond: 1 σ (sp²-sp² overlap)
• Total σ bonds: 3

Pi bonds:

• C=O double bond: 1 π (p-p side overlap)
• Total π bonds: 1

Verification:

• Total bonds: 2 single + 1 double = 4 bonds
• 4 bonds = 3 σ + 1 π ✓

Part (c): Explain planarity

Why is H₂CO planar?

1. Carbon hybridization: C is sp² hybridized

   • sp² creates trigonal planar electron geometry
   • All 3 sp² orbitals lie in the same plane
   • Bond angles ~120°

2. Pi bond requirement: The C=O double bond requires:

   • 1 σ bond (sp² hybrid overlap)
   • 1 π bond (unhybridized p orbital overlap)

3. P orbital orientation:

   • Each sp² hybridized atom (C and O) has 1 unhybridized p orbital
   • These p orbitals must be parallel to overlap and form π bond
   • P orbitals are perpendicular to the plane of sp² orbitals
   • This forces all atoms to lie in the same plane

4. Restricted rotation:

   • π bond prevents rotation around C=O
   • If molecule tried to twist, p orbitals wouldn't overlap
   • Planarity is maintained

Answer:

(a) Hybridization:

• Carbon: sp²
• Oxygen: sp²

(b) Bonds:

• Sigma (σ) bonds: 3
• Pi (π) bonds: 1

(c) Planarity explanation:

The molecule is planar because both C and O are sp² hybridized. sp² hybridization creates a trigonal planar arrangement where all atoms lie in the same plane. Additionally, the π bond in the C=O double bond requires parallel p orbitals, which forces the molecule to remain flat. Any deviation from planarity would disrupt the p orbital overlap needed for the π bond.

Molecular geometry:

• Around C: Trigonal planar (3 groups)
• Around O: Bent (1 bond + 2 lone pairs, but sp² geometry)
• All 4 atoms (H, C, H, O) in same plane

Bond angles:

• H-C-H: ~120°
• H-C=O: ~120°
• Total around C: ~360° (planar)

Each C has:

• 1 single bond to H
• 1 triple bond to other C

Part (a): Determine hybridization of each C

For left carbon (C₁):

Count electron domains:

• 1 C-H single bond = 1 domain
• 1 C≡C triple bond = 1 domain (triple bond counts as 1!)
• 0 lone pairs
• Total: 2 electron domains

Steric number: SN = 2

Hybridization: SN = 2 → sp hybridization

For right carbon (C₂):

Same analysis as C₁:

• SN = 2 → sp hybridization

Orbital composition for sp hybridization:

• Mix 1 s + 1 p orbital → 2 sp hybrid orbitals
• Remaining: 2 unhybridized p orbitals (py and pz)

Part (b): Count σ and π bonds

Sigma bonds:

• H-C bond (left): 1 σ
• C≡C bond: 1 σ (sp-sp overlap)
• C-H bond (right): 1 σ
• Total σ bonds: 3

Pi bonds:

• C≡C triple bond: 2 π bonds (two sets of p-p overlaps)
• Total π bonds: 2

Total bonds: 3 σ + 2 π = 5 bonds

• 2 single bonds (H-C, C-H) = 2 bonds
• 1 triple bond (C≡C) = 3 bonds
• Total: 5 bonds ✓

Part (c): Describe orbital overlap for each bond

Left H-C bond:

• Type: σ bond
• Orbitals: H (1s) overlaps with C (sp hybrid)
• Overlap: Head-on (along internuclear axis)

C≡C bond (triple bond):

σ bond (1):

• Orbitals: C (sp hybrid) overlaps with C (sp hybrid)
• Overlap: Head-on (sp-sp)
• Orientation: Along C-C axis

π bond (1):

• Orbitals: C (unhybridized py) overlaps with C (unhybridized py)
• Overlap: Side-by-side (parallel p orbitals)
• Orientation: Above and below C-C axis

π bond (2):

• Orbitals: C (unhybridized pz) overlaps with C (unhybridized pz)
• Overlap: Side-by-side (parallel p orbitals)
• Orientation: Front and back of C-C axis

Right C-H bond:

• Type: σ bond
• Orbitals: C (sp hybrid) overlaps with H (1s)
• Overlap: Head-on

Visualization of p orbitals:

• py orbitals: create π bond in vertical plane
• pz orbitals: create π bond in horizontal plane
• Together: cylindrical electron cloud around C-C axis

Part (d): Explain geometry and collinearity

Why are all atoms collinear (in a straight line)?

1. sp Hybridization geometry:

• sp hybridization → linear electron geometry
• 2 sp hybrid orbitals point in opposite directions (180° apart)
• This minimizes repulsion between 2 electron domains

2. Orbital arrangement:

Left carbon:

• One sp orbital points toward H (forms H-C σ bond)
• One sp orbital points toward other C (forms C-C σ bond)
• 180° angle between H and C

Right carbon:

• One sp orbital points toward other C (forms C-C σ bond)
• One sp orbital points toward H (forms C-H σ bond)
• 180° angle between C and H

3. Pi bond requirement:

• Two sets of p orbitals must be perpendicular to each other
• Both sets must be perpendicular to the sp orbital axis
• This arrangement requires linear geometry
• Any bending would destroy p orbital overlap

4. Maximum separation:

• SN = 2 means 2 electron domains
• Maximum separation = 180° (linear)
• Minimizes electron-electron repulsion

Answer:

(a) Hybridization:

• Both carbon atoms: sp

(b) Bonds:

• Sigma (σ) bonds: 3 (2 C-H + 1 C-C)
• Pi (π) bonds: 2 (both in C≡C)

(c) Orbital overlaps:

• H-C bonds: H(1s)-C(sp) σ bonds
• C≡C bond:
  • 1 σ bond from C(sp)-C(sp) overlap
  • 2 π bonds from C(py)-C(py) and C(pz)-C(pz) overlaps

(d) Geometry:

All atoms are collinear (linear) because:

1. sp hybridization creates 2 hybrid orbitals at 180°
2. Steric number = 2 requires linear geometry (maximum separation)
3. π bonds require unhybridized p orbitals to be perpendicular to internuclear axis
4. This forces H-C-C-H into straight line

Bond angle: H-C-C = C-C-H = 180°

Overall molecular geometry: Linear (all 4 atoms in straight line)

Key insight: The triple bond (1 σ + 2 π) creates a very strong, rigid, linear structure. The two π bonds form a cylindrical electron cloud around the C-C σ bond, preventing any rotation or bending.