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Hybridization, Bonding &
Molecular Geometry

Discover how atoms mix orbitals to form new hybrid shapes, create sigma and pi bonds, and produce the 3D molecular geometries that define chemistry.

Organic Chemistry 1

What Is Hybridization?

When atoms form covalent bonds, their atomic orbitals (s, p, and sometimes d) blend together to create new orbitals of equal energy and shape. These are called hybrid orbitals. Hybridization explains why molecules adopt specific 3D geometries that pure atomic orbitals cannot.

Atomic Orbitals: The Building Blocks

s Orbital
Spherical shape. One per energy level. Holds 2 electrons.
p Orbitals
Dumbbell shape. Three per level (\(p_x, p_y, p_z\)), oriented at \(90^\circ\). Each holds \(2e^-\).
Hybrid Orbitals
New orbitals formed by mixing s and p. Equal energy, directed toward bonding partners.

Why Does Hybridization Matter?

Pure atomic orbitals predict bond angles of \(90^\circ\) (from p orbitals). But real molecules show angles of \(109.5^\circ\), \(120^\circ\), and \(180^\circ\). Hybridization explains this discrepancy and determines:

1
Molecular geometry: The 3D shape of the molecule
2
Bond angles: The precise angles between bonded atoms
3
Reactivity: How molecules interact in organic reactions

The Three Types of Hybridization

\(sp^3\)
 Tetrahedral
\(109.5^\circ\)
 4 regions
\(sp^2\)
 Trigonal Planar
\(120^\circ\)
 3 regions
\(sp\)
 Linear
\(180^\circ\)
 2 regions

Sigma vs Pi Bonds

Sigma (\(\sigma\)) Bond
 Head-to-head overlap Along the bond axis Stronger, free rotation
Pi (\(\pi\)) Bond
 Side-by-side overlap Above and below axis Weaker, restricts rotation

The Bond Counting Rule

Single Bond
\(1\sigma\) bond
Formed by hybrid orbitals
Double Bond
\(1\sigma + 1\pi\)
One hybrid + one unhybridized \(p\)
Triple Bond
\(1\sigma + 2\pi\)
One hybrid + two unhybridized \(p\)

Key Distinction: \(sp^3\) vs \(sp^2\) vs \(sp\)

Property \(sp^3\) \(sp^2\) \(sp\)
Orbitals Mixed \(1s + 3p\) \(1s + 2p\) \(1s + 1p\)
Hybrid Orbitals 4 3 2
Unhybridized p 0 1 2
Geometry Tetrahedral Trigonal Planar Linear
Bond Angle \(109.5^\circ\) \(120^\circ\) \(180^\circ\)
Example \(\text{CH}_4\) \(\text{CH}_2\text{O}\) \(\text{CO}_2\)

How to Determine Hybridization

1
Draw the Lewis structure of the molecule, showing all bonds and lone pairs.
2
Pick the atom you want to hybridize (usually the central atom).
3
Count electron regions: each bond (single, double, or triple counts as one region) plus each lone pair.
4
Match the count: 4 regions = \(sp^3\), 3 regions = \(sp^2\), 2 regions = \(sp\).
5
Assign geometry based on hybridization and check for lone pairs (which modify shape).

Example 1: Methane (\(\text{CH}_4\))

C H H H H Tetrahedral geometry (dashed = behind plane)

Example 2: Formaldehyde (\(\text{CH}_2\text{O}\))

C O
\(\sigma\)
\(\pi\)
 H H Trigonal planar around carbon

Example 3: Acetylene (\(\text{C}_2\text{H}_2\))

H C C H
\(\sigma\)
\(\pi\)
\(\pi\)
\(180^\circ\) (Linear)
 Each carbon: sp hybridized

Lone Pairs Change the Shape

Lone pairs occupy hybrid orbitals but are invisible in molecular geometry. They repel bonding pairs more strongly, compressing bond angles below the ideal values.

\(\text{NH}_3\)
\(sp^3\) hybridized, 3 bonds + 1 lone pair. Shape: trigonal pyramidal. Angle: \(\approx 107^\circ\)
\(\text{H}_2\text{O}\)
\(sp^3\) hybridized, 2 bonds + 2 lone pairs. Shape: bent. Angle: \(\approx 104.5^\circ\)

Walkthrough: Acetic Acid (\(\text{CH}_3\text{COOH}\))

\(\text{C}_1\)
\(sp^3\)
 tetrahedral H H H
\(\sigma\)
\(\text{C}_2\)
\(sp^2\)
 trig. planar O O
\(\sigma\)
 H
\(sp^3\)

Common Mistakes to Avoid

✕ Counting each bond in a double bond as two regions
A double bond is one electron region, not two. Count bonds to atoms, not bond order.
✕ Forgetting lone pairs when counting regions
Lone pairs occupy hybrid orbitals. \(\text{NH}_3\) has 4 regions (3 bonds + 1 LP), making it \(sp^3\), not \(sp^2\).
✕ Confusing electron geometry with molecular shape
\(\text{H}_2\text{O}\) has tetrahedral electron geometry but bent molecular shape. Report the shape you see.

Summary Table

Regions Hybridization Geometry Angle \(\pi\) Bonds
4 \(sp^3\) Tetrahedral \(109.5^\circ\) 0
3 \(sp^2\) Trigonal Planar \(120^\circ\) 1
2 \(sp\) Linear \(180^\circ\) 2

Question 1 of 5

What is the hybridization of the carbon atom in \(\text{CO}_2\)?

Question 2 of 5

How many sigma (\(\sigma\)) and pi (\(\pi\)) bonds are in a molecule of ethylene (\(\text{C}_2\text{H}_4\))?

Question 3 of 5

The nitrogen atom in \(\text{NH}_3\) is \(sp^3\) hybridized, yet its molecular shape is not tetrahedral. Why?

Question 4 of 5

In acetylene (\(\text{C}_2\text{H}_2\)), how many unhybridized \(p\) orbitals does each carbon atom have?

Question 5 of 5

Which molecule has a bond angle of approximately \(120^\circ\)?