Hybridization in Organic Chemistry

Master sp, sp2, and sp3 hybridization, bond angles, geometry, and their role in organic molecules for JEE Chemistry.

Introduction

Hybridization is the key to understanding the 3D shapes of organic molecules! It explains why methane is tetrahedral, ethene is planar, and ethyne is linear. Master this concept, and organic chemistry becomes 10x easier!

Drug Design in Real Life
In pharmaceutical companies (depicted in Painkiller 2024), chemists design drug molecules with EXACT 3D shapes! Whether a drug fits into a receptor depends on sp³ vs sp² hybridization. The difference between a life-saving medicine and an inactive compound can be just one hybridization change - tetrahedral vs planar changes everything about how molecules interact!

What is Hybridization?

Definition

Hybridization is the concept of mixing atomic orbitals to form new hybrid orbitals of equal energy, which are better suited for bonding.

$$\boxed{\text{Atomic orbitals} \xrightarrow{\text{Mix}} \text{Hybrid orbitals} \xrightarrow{\text{Overlap}} \text{Bonds}}$$

Why Hybridization?

Problem: Ground state carbon has 2 unpaired electrons, should form only 2 bonds!

Carbon ground state: 1s² 2s² 2p²

2s: [↑↓]
2p: [↑] [↑] [ ]

But carbon forms 4 bonds in CH₄!

Solution: Hybridization

  1. Excitation: One 2s electron promoted to 2p

    2s: [↑]
    2p: [↑] [↑] [↑]
    
  2. Hybridization: 2s and 2p orbitals mix to form hybrid orbitals

  3. Result: 4 equivalent hybrid orbitals for bonding!


Types of Hybridization

Summary Table

HybridizationOrbitals MixedHybrid OrbitalsGeometryBond AngleExamples
sp³1s + 3p4 sp³Tetrahedral109.5°CH₄, C₂H₆, Diamond
sp²1s + 2p3 sp² + 1pTrigonal planar120°C₂H₄, C₆H₆, Graphite
sp1s + 1p2 sp + 2pLinear180°C₂H₂, CO₂, HCN

sp³ Hybridization

Formation

Mixing: 1 s orbital + 3 p orbitals → 4 sp³ hybrid orbitals

graph LR
    A[2s: 1 orbital] --> C[4 sp³ orbitals]
    B[2p: 3 orbitals] --> C
    C --> D[Tetrahedral arrangement]

Energy diagram:

Before:           After hybridization:
2p: [↑] [↑] [↑]   sp³: [↑] [↑] [↑] [↑]
2s: [↑]               (equal energy)

Characteristics

Geometry: Tetrahedral

Bond Angle: 109.5° (109° 28')

% s-character: 25% (1/4)

% p-character: 75% (3/4)

Shape of sp³ orbital:

  • Large lobe pointing in one direction
  • Small lobe in opposite direction
  • Better overlap → stronger bonds

Examples

1. Methane (CH₄)

Structure:

  • C is sp³ hybridized
  • 4 sp³ orbitals overlap with H 1s orbitals
  • All C-H bonds equivalent
  • Perfect tetrahedral: All angles = 109.5°
$$\boxed{\text{sp³ C} + 4 \text{ H(1s)} \rightarrow 4 \text{ C-H } \sigma\text{-bonds}}$$

2. Ethane (C₂H₆)

Structure:

  • Both carbons sp³ hybridized
  • C-C bond: sp³-sp³ overlap (σ-bond)
  • C-H bonds: sp³-1s overlap (σ-bonds)
  • Free rotation around C-C single bond

3. Ammonia (NH₃)

Structure:

  • N is sp³ hybridized
  • 3 sp³ orbitals form N-H bonds
  • 1 sp³ orbital contains lone pair
  • Pyramidal shape (not tetrahedral!)
  • Bond angle: 107° (less than 109.5° due to lone pair repulsion)

4. Water (H₂O)

Structure:

  • O is sp³ hybridized
  • 2 sp³ orbitals form O-H bonds
  • 2 sp³ orbitals contain lone pairs
  • Bent shape
  • Bond angle: 104.5° (lone pair-lone pair > lone pair-bond pair repulsion)
JEE Alert

Bond angle trend due to lone pairs:

$$\boxed{CH_4 (109.5°) > NH_3 (107°) > H_2O (104.5°)}$$

Rule: More lone pairs → smaller bond angle (lone pairs repel more strongly)

5. Diamond

Structure:

  • Each C is sp³ hybridized
  • 3D network of C-C single bonds
  • Tetrahedral geometry throughout
  • Extremely hard (strong C-C σ-bonds in 3D)

sp² Hybridization

Formation

Mixing: 1 s orbital + 2 p orbitals → 3 sp² hybrid orbitals + 1 unhybridized p orbital

graph LR
    A[2s: 1 orbital] --> C[3 sp² orbitals]
    B[2p: 2 orbitals] --> C
    D[2p: 1 orbital] --> E[Unhybridized p]
    C --> F[Trigonal planar]
    E --> G[π-bonding]

Energy diagram:

Before:           After hybridization:
2p: [↑] [↑] [↑]   sp²: [↑] [↑] [↑]  (in plane)
2s: [↑]           p:   [↑]          (perpendicular)

Characteristics

Geometry: Trigonal planar (all atoms in one plane)

Bond Angle: 120°

% s-character: 33.3% (1/3)

% p-character: 66.7% (2/3)

Unhybridized p orbital: Perpendicular to the plane, used for π-bonding

Examples

1. Ethene (C₂H₄)

Structure:

  • Both C atoms sp² hybridized
  • σ-bonds:
    • C-C: sp²-sp² overlap
    • C-H: sp²-1s overlap (4 bonds)
  • π-bond:
    • Sideways overlap of unhybridized p orbitals
    • Above and below the plane

Double bond = 1 σ + 1 π

Important features:

  • All 6 atoms in same plane
  • No free rotation around C=C (π-bond restricts rotation)
  • Bond angle: ~120° (slightly less due to π-electron repulsion)
$$\boxed{C=C: \text{ sp²-sp² overlap (σ)} + \text{ p-p overlap (π)}}$$

2. Benzene (C₆H₆)

Structure:

  • All 6 C atoms sp² hybridized
  • σ-bonds:
    • C-C: sp²-sp² overlap (6 bonds forming hexagon)
    • C-H: sp²-1s overlap (6 bonds)
  • π-bonds:
    • 6 unhybridized p orbitals (one per C)
    • Delocalized over the entire ring
    • Forms π-electron cloud above and below the ring

Features:

  • Perfectly planar molecule
  • All C-C bonds equal length (1.39 Å)
  • All bond angles = 120°
  • Aromatic stability due to delocalized π-electrons

3. Graphite

Structure:

  • All C atoms sp² hybridized
  • Forms hexagonal layers
  • Within layer: Strong C-C σ-bonds (sp²-sp²)
  • Between layers: Weak van der Waals forces
  • Delocalized π-electrons move freely (electrical conductivity!)

Properties from sp²:

  • Soft (layers slide easily)
  • Good conductor (delocalized electrons)
  • Black color (π-electron transitions)

4. Carbonyl Group (C=O)

Structure:

  • C is sp² hybridized
  • O is sp² hybridized (+ 2 lone pairs in sp²)
  • σ-bond: sp²(C) - sp²(O)
  • π-bond: p(C) - p(O)
  • Planar geometry around C=O
Memory Trick

“sp² = 2 Doubly bonded”

  • sp² hybridization is present when there’s a DOUBLE bond
  • Geometry is PLANAR (all in 2D plane)
  • Bond angle is 120° (three bonds spread out)

Examples: C=C, C=O, C=N, benzene


sp Hybridization

Formation

Mixing: 1 s orbital + 1 p orbital → 2 sp hybrid orbitals + 2 unhybridized p orbitals

graph LR
    A[2s: 1 orbital] --> C[2 sp orbitals]
    B[2p: 1 orbital] --> C
    D[2p: 2 orbitals] --> E[2 unhybridized p]
    C --> F[Linear]
    E --> G[2 π-bonds]

Energy diagram:

Before:           After hybridization:
2p: [↑] [↑] [↑]   sp: [↑] [↑]       (linear)
2s: [↑]           p:  [↑] [↑]       (perpendicular)

Characteristics

Geometry: Linear (180° apart)

Bond Angle: 180°

% s-character: 50% (1/2)

% p-character: 50% (1/2)

Unhybridized p orbitals: 2 perpendicular p orbitals for π-bonding

Examples

1. Ethyne/Acetylene (C₂H₂)

Structure:

  • Both C atoms sp hybridized
  • σ-bonds:
    • C-C: sp-sp overlap
    • C-H: sp-1s overlap (2 bonds)
  • π-bonds:
    • Two p-p overlaps (perpendicular to each other)
    • Form cylindrical π-electron cloud around C-C axis

Triple bond = 1 σ + 2 π

Important features:

  • Linear molecule (H-C-C-H all in one line)
  • Bond angle: 180°
  • No free rotation (two π-bonds lock the molecule)
  • Very strong C≡C bond
$$\boxed{C \equiv C: \text{ sp-sp (σ)} + 2 \times \text{ p-p (π)}}$$

2. Carbon Dioxide (CO₂)

Structure:

  • C is sp hybridized
  • Each O is sp² hybridized
  • σ-bonds: sp(C) - sp²(O) on each side
  • π-bonds: 2 π-bonds (one with each O)
  • Linear molecule: O=C=O (180°)

3. Hydrogen Cyanide (HCN)

Structure:

  • C is sp hybridized
  • σ-bonds:
    • C-H: sp-1s
    • C-N: sp-sp
  • π-bonds: 2 π-bonds (C≡N triple bond)
  • Linear: H-C≡N (180°)

4. Alkynes (General)

R-C≡C-R’

  • Both triple-bonded carbons: sp
  • Linear geometry at triple bond
  • Bond angle: 180°
JEE Pattern

Bond length and strength trend:

$$\boxed{C \equiv C < C=C < C-C \text{ (length)}}$$ $$\boxed{C \equiv C > C=C > C-C \text{ (strength)}}$$

Reason:

  • sp has 50% s-character (most s → shortest, strongest)
  • sp² has 33% s-character (intermediate)
  • sp³ has 25% s-character (least s → longest, weakest)

More s-character → Shorter, stronger bonds

Interactive Demo: Visualize Orbital Shapes

Explore sp, sp², and sp³ hybrid orbitals in 3D space.


Comparison of Hybridizations

Properties Comparison

Propertysp³sp²sp
Orbitals mixed1s + 3p1s + 2p1s + 1p
Hybrid orbitals432
Unhybridized p012
GeometryTetrahedralTrigonal planarLinear
Bond angle109.5°120°180°
s-character25%33.3%50%
p-character75%66.7%50%
Bond typeOnly σσ + πσ + 2π
RotationFreeRestrictedNo rotation
ExampleC₂H₆C₂H₄C₂H₂

Bond Properties

s-character effects:

  1. Bond length: More s → shorter bond

    $$sp < sp² < sp³$$
  2. Bond strength: More s → stronger bond

    $$sp > sp² > sp³$$
  3. Acidity: More s → more acidic H

    $$HC \equiv CH > H_2C=CH_2 > H_3C-CH_3$$

    (pKa: 25 > 44 > 50)

  4. Electronegativity: More s → more electronegative C

    $$sp > sp² > sp³$$
Common JEE Mistake

Mistake: Assuming all carbon atoms have the same hybridization in a molecule.

Correct: Different carbons can have different hybridizations!

Example: CH₂=CH-CH₃ (propene)

  • C1 (in =CH₂): sp²
  • C2 (in CH=): sp²
  • C3 (in -CH₃): sp³

Always identify each carbon’s bonding separately!


Determining Hybridization

Method 1: Counting Regions of Electron Density

Steric Number = σ-bonds + Lone pairs

Steric NumberHybridizationGeometry
2spLinear
3sp²Trigonal planar
4sp³Tetrahedral

Examples:

  1. CH₄: C has 4 σ-bonds, 0 lone pairs → Steric # = 4 → sp³

  2. NH₃: N has 3 σ-bonds, 1 lone pair → Steric # = 4 → sp³

  3. C₂H₄: Each C has 3 σ-bonds (2 C-H, 1 C-C), 0 lone pairs → sp²

  4. CO₂: C has 2 σ-bonds (2 C-O), 0 lone pairs → sp

Method 2: Counting Bonds

Bond TypeHybridization
4 single bonds (4σ)sp³
1 double + 2 single (3σ + 1π)sp²
1 triple + 1 single (2σ + 2π)sp
2 double bonds (2σ + 2π)sp

Key: Count σ-bonds only for hybridization!


Resonance and Hybridization

Resonance Structures

When resonance is present, hybridization is determined by the resonance hybrid, not individual structures.

Example: Carboxylate ion (CH₃COO⁻)

Both oxygens are equivalent due to resonance:

    O⁻        O
    ‖    ↔    ‖
R—C         R—C
    |         |
    O         O⁻

Hybridization:

  • C (carbonyl): sp²
  • Both O atoms: sp² (not one sp² and one sp³!)
  • Resonance averages the bonding

Benzene Revisited

All 6 carbons are sp² due to resonance:

  • Delocalized π-system
  • All C-C bonds equivalent (1.39 Å)
  • Cannot assign double/single bonds

Applications and Examples

Example 1: Allene (H₂C=C=CH₂)

Structure:

  • C1 and C3: sp² (in double bonds, planar)
  • C2: sp (forms two double bonds, linear)

Geometry:

  • C2 is linear
  • C1 and C3 planes are perpendicular to each other!

Example 2: Vitamin C (Ascorbic Acid)

Multiple hybridizations:

  • Ring carbons: Mostly sp³
  • Carbonyl C: sp²
  • Enol C: sp²

Example 3: Formaldehyde (H₂C=O)

Structure:

  • C: sp² (3 σ-bonds: 2 C-H, 1 C-O)
  • O: sp² (1 σ-bond + 2 lone pairs in sp²)
  • Planar molecule
  • H-C-H angle ≈ 120°

Practice Problems

Level 1: Basic Identification

  1. Identify the hybridization of each carbon:

    • a) CH₃-CH₃
    • b) CH₂=CH₂
    • c) HC≡CH
    • d) CH₃-CH=CH₂
  2. What is the bond angle in:

    • a) Methane (CH₄)
    • b) Ethene (C₂H₄)
    • c) Acetylene (C₂H₂)
  3. How many σ and π bonds in:

    • a) C₂H₆
    • b) C₂H₄
    • c) C₂H₂

Level 2: Application

  1. Explain why:

    • C-H bond in acetylene is more acidic than in ethane
    • Benzene is planar
    • There’s no free rotation around C=C double bond
  2. Determine hybridization of each atom:

    • a) Formaldehyde (H₂CO)
    • b) Acetone (CH₃COCH₃)
    • c) Carbon dioxide (CO₂)
  3. Arrange in order of increasing C-C bond length: CH₃-CH₃, CH₂=CH₂, HC≡CH

  4. Which molecule is linear?

    • a) H₂O
    • b) NH₃
    • c) CO₂
    • d) CH₄

Level 3: JEE Advanced

  1. In the molecule CH₃-C≡C-CHO, how many carbon atoms are sp hybridized?

    • (a) 1
    • (b) 2
    • (c) 3
    • (d) 4
  2. The H-C-H bond angle in ethene is closest to:

    • (a) 109.5°
    • (b) 120°
    • (c) 117°
    • (d) 180°
  3. Which statement is INCORRECT about sp² hybridization?

    • (a) Bond angle is 120°
    • (b) Geometry is trigonal planar
    • (c) One p orbital remains unhybridized
    • (d) All four orbitals are used in σ-bonding
  4. Assertion (A): sp hybridized carbon forms stronger C-H bonds than sp³. Reason (R): sp has 50% s-character while sp³ has 25% s-character.

    • (a) Both A and R true, R explains A
    • (b) Both true, R doesn’t explain A
    • (c) A true, R false
    • (d) Both false
  5. In benzene, each C-C bond length is 1.39 Å. This is:

    • (a) Equal to C-C single bond
    • (b) Equal to C=C double bond
    • (c) Between single and double bond
    • (d) Equal to C≡C triple bond
Quick Check
Test yourself: Draw the Lewis structure of CH₃-CH=CH-C≡N and identify the hybridization of each carbon. Can you predict all bond angles?

Memory Tricks

“SLIP” for Hybridization

  • Single bonds only → sp³
  • Linear geometry → sp
  • In-plane (planar) → sp²
  • Pi bonds: 0(sp³), 1(sp²), 2(sp)

Bond Angles Quick Memory

“109, 120, 180 - All in Line!”

  • sp³: 109.5° (tetrahedral)
  • sp²: 120° (planar)
  • sp: 180° (linear)

s-character Rule

“More s, More Short, More Strong!”

  • More s-character → Shorter bonds
  • More s-character → Stronger bonds
  • More s-character → More acidic H
  • sp (50%) > sp² (33%) > sp³ (25%)

Counting Method

“Count sigma, ignore pi!”

  • 4 σ-bonds → sp³
  • 3 σ-bonds → sp²
  • 2 σ-bonds → sp

Within Organic Principles

Chemical Bonding Foundation

Atomic Structure

Hydrocarbon Applications

  • Alkanes - sp³ hybridization examples
  • Alkenes - sp² hybridization and addition
  • Alkynes - sp hybridization, triple bonds
  • Benzene - sp² and aromaticity

Functional Group Applications

Cross-Subject Connections


Isomerism Types →