🧬 Hybridization and Sigma/Pi Bonds

Part 1 of 7 — Introduction to Hybridization

Atomic orbitals (s, p, d) describe where electrons are likely to be found in isolated atoms. But when atoms form covalent bonds, something remarkable happens — the orbitals mix to create new, equivalent orbitals that are better suited for bonding.

This mixing process is called hybridization.

Why Do We Need Hybridization?

Consider carbon in methane (CH₄):

• Carbon's electron configuration is $1s^2\,2s^2\,2p^2$
• The $2s$ orbital and three $2p$ orbitals have different shapes and energies
• Yet all four C–H bonds in CH₄ are identical in length and strength

How can different orbitals produce identical bonds? The answer: the orbitals hybridize into four equivalent sp³ hybrid orbitals before bonding occurs.

The Hybridization Process

Hybridization is the mathematical combination of atomic orbitals on the same atom to form new hybrid orbitals.

Key Principles

1. Number of hybrid orbitals = number of atomic orbitals mixed

   • Mix 1 s + 3 p → get 4 sp³ hybrid orbitals
   • Mix 1 s + 2 p → get 3 sp² hybrid orbitals
   • Mix 1 s + 1 p → get 2 sp hybrid orbitals

2. Hybrid orbitals are equivalent — they have the same shape and energy

3. Hybrid orbitals are oriented to minimize repulsion — just like VSEPR predicts

4. Unhybridized orbitals remain unchanged — they can form pi bonds (more on this later)

Energy Perspective

The energy of hybrid orbitals is an average of the contributing atomic orbitals:

$E_{\text{sp}^3} = \frac{E_s + 3E_p}{4}$

This costs some energy (promoting an electron from s to p), but the energy is more than recovered by forming stronger bonds.

sp³ Hybridization — Tetrahedral

When one s orbital mixes with three p orbitals, four equivalent sp³ hybrid orbitals form.

Properties of sp³ Orbitals

How sp³ Works in Methane (CH₄)

1. Carbon starts: $1s^2\,2s^2\,2p^2$ (only 2 unpaired electrons)
2. One $2s$ electron is promoted to the empty $2p$ orbital
3. The one $2s$ and three $2p$ orbitals hybridize → four $sp^3$ orbitals
4. Each $sp^3$ orbital overlaps with a hydrogen $1s$ orbital → 4 equivalent C–H bonds
5. The four $sp^3$ orbitals point toward the corners of a tetrahedron (109.5° apart)

Other Examples of sp³ Hybridization

• NH₃: N is sp³ (3 bonds + 1 lone pair = 4 electron domains)
• H₂O: O is sp³ (2 bonds + 2 lone pairs = 4 electron domains)
• CCl₄: C is sp³ (4 bonds + 0 lone pairs)

Rule: Any atom with 4 electron domains is sp³ hybridized.

Test your understanding of hybridization fundamentals.

Shape of Hybrid Orbitals

Each hybrid orbital has a distinctive shape: one large lobe pointing in the bonding direction and one small lobe on the opposite side.

sp³ vs. Unhybridized Orbitals

Why Hybrids Bond Better

The large lobe of an sp³ orbital extends further from the nucleus than either an s or p orbital alone. This produces:

• Greater overlap with the bonding partner
• Stronger bonds
• More directed electron density

Key Takeaway

Hybridization is the atom's way of optimizing orbital geometry for bonding. The "cost" of promoting an electron is more than repaid by the stronger, more directional bonds that hybrid orbitals form.

Determine the number of electron domains around the central atom and confirm sp³ hybridization.

Select the correct answer for each statement about hybridization.

Summary — Introduction to Hybridization

Key Ideas

1. Hybridization = mixing atomic orbitals on the same atom to create new, equivalent hybrid orbitals
2. The number of hybrid orbitals formed equals the number of atomic orbitals mixed
3. sp³ hybridization: 1 s + 3 p → 4 equivalent orbitals, tetrahedral geometry, 109.5° angles
4. Lone pairs also occupy hybrid orbitals
5. Hybrid orbitals form stronger, more directional bonds than unhybridized orbitals

The Pattern

Coming up in Part 2: sp² and sp hybridization — what happens when not all p orbitals are used.

🔺 sp² and sp Hybridization

Part 2 of 7 — Trigonal Planar and Linear Geometries

In Part 1 we saw that four electron domains lead to sp³ hybridization. But what happens when an atom has only three or two electron domains?

The atom uses fewer p orbitals in hybridization, leaving unhybridized p orbitals available. These leftover p orbitals play a crucial role — they form pi (π) bonds, which we'll explore in depth in Part 3.

For now, let's focus on the geometry and properties of sp² and sp hybrid orbitals.

sp² Hybridization

When one s orbital mixes with two p orbitals, three equivalent sp² hybrid orbitals form.

Properties of sp² Orbitals

The Unhybridized p Orbital

This is critical: the one p orbital that is not used in hybridization remains perpendicular to the plane of the three sp² orbitals. It is available for pi bonding.

Example: BF₃ (Boron Trifluoride)

1. Boron has 3 valence electrons and forms 3 bonds to fluorine
2. 3 electron domains → sp² hybridization
3. The three sp² orbitals point to the corners of an equilateral triangle
4. All F–B–F angles = 120°
5. The molecule is perfectly flat (planar)

Example: Formaldehyde (H₂C=O)

1. Carbon has 3 electron domains (2 single bonds to H + 1 double bond to O)
2. Carbon is sp² hybridized
3. The three sp² orbitals form sigma bonds in a trigonal planar arrangement
4. The unhybridized p orbital on carbon overlaps with a p orbital on oxygen → pi bond

Test your understanding of sp² hybridization.

sp Hybridization

When one s orbital mixes with one p orbital, two equivalent sp hybrid orbitals form.

Properties of sp Orbitals

Two Unhybridized p Orbitals

An sp-hybridized atom has two leftover p orbitals, both perpendicular to each other and to the axis of the sp hybrids. These can form two pi bonds, enabling triple bonds.

Example: CO₂ (Carbon Dioxide)

1. Carbon has 2 double bonds to oxygen → 2 electron domains
2. Carbon is sp hybridized
3. The two sp orbitals point in opposite directions (180°)
4. Each unhybridized p orbital overlaps with a p orbital on an oxygen atom → 2 pi bonds
5. Result: O=C=O is perfectly linear

Example: C₂H₂ (Acetylene)

1. Each carbon has 1 C–H bond + 1 C≡C triple bond = 2 electron domains
2. Each carbon is sp hybridized
3. The molecule is linear: H–C≡C–H (180° angles)
4. The triple bond consists of 1 sigma bond + 2 pi bonds

Determine the hybridization and geometry for each molecule.

The Complete Hybridization Table

The Quick Rule

$\text{Hybridization} = \text{based on number of electron domains (steric number)}$

Important Note

Remember that a double bond counts as 1 electron domain and a triple bond counts as 1 electron domain. Only the number of groups matters, not the bond order.

Select the correct hybridization for the central atom in each molecule.

Verify your mastery of sp² and sp hybridization.

⚡ Sigma and Pi Bonds

Part 3 of 7 — Two Kinds of Covalent Bonds

Not all covalent bonds are the same. There are two fundamentally different types based on how the orbitals overlap:

• Sigma (σ) bonds — formed by head-on overlap
• Pi (π) bonds — formed by lateral (side-by-side) overlap

Understanding the difference between sigma and pi bonds is essential for explaining molecular structure, rotation, rigidity, and reactivity.

Sigma Bonds — Head-On Overlap

A sigma bond forms when two orbitals overlap end-to-end (head-on), with electron density concentrated along the bond axis (the line connecting the two nuclei).

Types of Sigma Bond Overlap

Key Properties of Sigma Bonds

1. Every single bond is a sigma bond
2. Sigma bonds allow free rotation around the bond axis (the cylindrical symmetry of electron density means rotation doesn't break the overlap)
3. Sigma bonds are stronger than pi bonds (greater orbital overlap)
4. Every bonding pair shares exactly one sigma bond — it's always the first bond formed between two atoms

Pi Bonds — Lateral Overlap

A pi bond forms when two unhybridized p orbitals overlap side-by-side (laterally), with electron density concentrated above and below (or in front and behind) the bond axis.

Key Properties of Pi Bonds

1. Pi bonds form from unhybridized p orbitals only
2. Electron density is in two lobes — one above and one below the internuclear axis
3. Pi bonds prevent rotation around the bond axis (rotating would break the side-by-side overlap)
4. Pi bonds are weaker than sigma bonds (less overlap)
5. Pi bonds are always the second and third bonds between two atoms

Why Pi Bonds Prevent Rotation

Imagine two p orbitals overlapping side-by-side. If you rotate one atom 90°, the orbitals no longer overlap — the pi bond breaks. This is why:

• Ethane (C–C single bond, no π): free rotation
• Ethene (C=C double bond, 1 π): no free rotation → rigid and planar
• Acetylene (C≡C triple bond, 2 π): no free rotation → rigid and linear

Distinguish between sigma and pi bonds.

The Counting Rules

The Simple Rules

$\text{Single bond} = 1\sigma + 0\pi$ $\text{Double bond} = 1\sigma + 1\pi$ $\text{Triple bond} = 1\sigma + 2\pi$

Counting Strategy

1. Draw or visualize the Lewis structure
2. Count each bond:
   • Every line in a Lewis structure between two atoms = 1 sigma bond
   • For double bonds: the second line = 1 pi bond
   • For triple bonds: the second and third lines = 2 pi bonds

Worked Example: Ethene (C₂H₄)

Structure: H₂C=CH₂

• 4 C–H single bonds → 4σ
• 1 C=C double bond → 1σ + 1π
• Total: 5σ + 1π

Worked Example: Acetylene (C₂H₂)

Structure: HC≡CH

• 2 C–H single bonds → 2σ
• 1 C≡C triple bond → 1σ + 2π
• Total: 3σ + 2π

Worked Example: HCN (Hydrogen Cyanide)

Structure: H–C≡N

• 1 H–C single bond → 1σ
• 1 C≡N triple bond → 1σ + 2π
• Total: 2σ + 2π

Count the total number of sigma and pi bonds in each molecule.

Select the correct description for each scenario.

Verify your understanding of sigma and pi bond concepts.

🔍 Hybridization from Molecular Structure

Part 4 of 7 — Using Steric Number to Assign Hybridization

Now that you know the three main hybridization types (sp, sp², sp³), it's time to develop a systematic method for determining the hybridization of any atom in any molecule.

The key is the steric number — the total count of electron domains around the atom of interest.

Step-by-Step: Finding Hybridization

Step 1: Draw the Lewis Structure

This gives you bond types and lone pairs.

Step 2: Count Electron Domains

For the atom in question, count:

• Each single bond = 1 domain
• Each double bond = 1 domain
• Each triple bond = 1 domain
• Each lone pair = 1 domain

$\text{Steric Number} = \text{bonded atoms} + \text{lone pairs}$

Step 3: Match Steric Number to Hybridization

Critical Reminder

Lone pairs count as electron domains! An atom with 2 bonds and 2 lone pairs has steric number 4 → sp³ (not sp).

Worked Examples

Example 1: Water (H₂O)

• O has 2 bonds + 2 lone pairs = 4 electron domains
• Steric number = 4 → sp³
• Electron geometry: tetrahedral
• Molecular geometry: bent (because 2 lone pairs are "invisible")
• Bond angle: ~104.5° (less than 109.5° due to lone pair compression)

Example 2: Ozone (O₃)

• Central O has 1 single bond + 1 double bond + 1 lone pair = 3 electron domains
• Steric number = 3 → sp²
• Electron geometry: trigonal planar
• Molecular geometry: bent
• Bond angle: ~117° (lone pair compresses slightly from 120°)

Example 3: Carbon in Cyanide Ion (CN⁻)

• Carbon has 1 triple bond to nitrogen + 1 lone pair = 2 electron domains
• Steric number = 2 → sp
• Geometry: linear

Example 4: Nitrogen in Ammonia (NH₃)

• N has 3 bonds + 1 lone pair = 4 electron domains
• Steric number = 4 → sp³
• Even though NH₃ has a trigonal pyramidal molecular shape, the nitrogen is still sp³ hybridized

Determine the hybridization of the indicated atom.

Calculate the steric number and identify the hybridization.

Common Mistakes to Avoid

Mistake 1: Confusing Molecular Geometry with Hybridization

The molecular geometry (shape you "see") may differ from what the hybridization implies:

• H₂O is bent but oxygen is sp³
• NH₃ is trigonal pyramidal but nitrogen is sp³

Hybridization depends on the total number of electron domains (steric number), not just the bonding arrangement.

Mistake 2: Counting Double/Triple Bonds as Multiple Domains

• C=O is 1 electron domain (not 2)
• C≡N is 1 electron domain (not 3)

Mistake 3: Forgetting Lone Pairs

Always check for lone pairs on the atom of interest. They occupy hybrid orbitals just like bonding pairs.

Mistake 4: Over-Applying Hybridization

Hybridization applies to atoms that form covalent bonds. Terminal atoms (like H or F) don't need hybridization analysis — focus on central atoms or atoms bonded to multiple groups.

Select the correct answer for each scenario.

Apply the steric number method to determine hybridization.

🔗 Multiple Bonds and Hybridization

Part 5 of 7 — Double Bonds, Triple Bonds, and Orbital Pictures

Now let's bring together hybridization and sigma/pi bond counting to analyze molecules with multiple bonds in detail. We'll examine how the orbital picture explains the geometry and bond composition of key molecules.

Double Bonds = 1 Sigma + 1 Pi

Every double bond consists of exactly one sigma bond and one pi bond.

Orbital Picture of Ethene (C₂H₄)

Each carbon in ethene:

• Has 3 electron domains (2 C–H + 1 C=C) → sp² hybridized
• Uses 3 sp² orbitals for sigma bonds
• Has 1 unhybridized p orbital perpendicular to the molecular plane

The C=C double bond:

• The sigma bond forms from sp²–sp² head-on overlap
• The pi bond forms from p–p lateral overlap (unhybridized p orbitals)

Bond count for C₂H₄:

• 4 C–H bonds → 4σ
• 1 C=C bond → 1σ + 1π
• Total: 5σ + 1π = 6 bonds total

Why Ethene Is Planar

The pi bond requires the unhybridized p orbitals to be parallel (side-by-side). This forces all 6 atoms into the same plane. Rotation around the C=C bond would break the pi bond — this is why double bonds are rigid.

Triple Bonds = 1 Sigma + 2 Pi

Every triple bond consists of one sigma bond and two pi bonds.

Orbital Picture of Acetylene (C₂H₂)

Each carbon in acetylene:

• Has 2 electron domains (1 C–H + 1 C≡C) → sp hybridized
• Uses 2 sp orbitals for sigma bonds (pointing 180° apart)
• Has 2 unhybridized p orbitals, perpendicular to each other

The C≡C triple bond:

• 1 sigma bond: sp–sp head-on overlap
• 1 pi bond: p–p lateral overlap in one plane
• 1 pi bond: p–p lateral overlap in the perpendicular plane

Bond count for C₂H₂:

• 2 C–H bonds → 2σ
• 1 C≡C bond → 1σ + 2π
• Total: 3σ + 2π = 5 bonds total

Orbital Picture of HCN

Carbon in HCN:

• Has 2 electron domains (1 C–H + 1 C≡N) → sp hybridized
• The C≡N triple bond: 1σ + 2π

Bond count for HCN:

• 1 C–H bond → 1σ
• 1 C≡N bond → 1σ + 2π
• Total: 2σ + 2π

Nitrogen in HCN:

• Has 2 electron domains (1 triple bond + 1 lone pair) → sp hybridized
• The lone pair occupies an sp orbital

Analyze the bonds in molecules with double and triple bonds.

Bond Strength and Length Trends

Adding pi bonds to a sigma bond makes the overall bond stronger and shorter:

Key Observations

1. A double bond is not twice as strong as a single bond

   • C=C (614) < 2 × C–C (696)
   • This is because the pi bond (~266 kJ/mol) is weaker than the sigma bond (~348 kJ/mol)

2. Higher bond order → shorter bond length

   • More electron density between nuclei pulls them closer

3. The pi bond energy can be estimated: $E_{\pi} \approx E_{\text{double}} - E_{\text{single}} = 614 - 348 = 266 \text{ kJ/mol}$

Count the sigma and pi bonds in each molecule.

Select the correct answer for each scenario.

Test your complete understanding of multiple bonds and hybridization.

🛠️ Problem-Solving Workshop

Part 6 of 7 — Mixed Practice on Hybridization and Sigma/Pi Bonds

Time to put everything together. In this workshop you'll practice:

• Determining hybridization from Lewis structures
• Counting sigma and pi bonds
• Connecting hybridization to molecular geometry
• Identifying bond angles

Work through each problem carefully — these are the types of questions that appear on the AP exam.

Problem-Solving Strategy

For Any Molecule:

Step 1: Draw the Lewis structure (show all bonds and lone pairs).

Step 2: For each atom of interest, count electron domains: $\text{Steric Number} = \text{bonded atoms} + \text{lone pairs}$

Step 3: Assign hybridization:

Step 4: Count bonds:

• Single bond = 1σ
• Double bond = 1σ + 1π
• Triple bond = 1σ + 2π

Step 5: Total σ = total number of bonds between atom pairs. Total π = total number of extra bonds beyond the first in each pair.

Analyze the formaldehyde molecule (H₂C=O). Carbon is the central atom with 2 C–H single bonds and 1 C=O double bond.

Analyze CO₂ (O=C=O). Answer questions about its hybridization and bonds.

Analyze these more complex molecules. For CH₃CHO (acetaldehyde), the structure is H₃C–CH=O.

Select the correct hybridization for the central or indicated atom.

Connect hybridization to molecular geometry.

Workshop Takeaways

Quick Reference Card

Common AP Exam Patterns

1. "Determine the hybridization of atom X" → count electron domains
2. "How many sigma/pi bonds?" → use the counting rules
3. "Explain why molecule X is planar/nonplanar" → connect to hybridization and pi bonds
4. "Compare bond lengths/strengths" → higher bond order = shorter and stronger

Coming up in Part 7: Connecting hybridization to VSEPR and polarity for AP-level synthesis.

🎓 Synthesis & AP Review

Part 7 of 7 — Connecting Hybridization to VSEPR, Polarity, and the AP Exam

In this final part, we tie hybridization and sigma/pi bonds into the broader picture of molecular structure. The AP Chemistry exam expects you to move fluently between Lewis structures, VSEPR, hybridization, and polarity — so let's practice that integration.

From Lewis Structure to Full Analysis

For any molecule on the AP exam, you should be able to perform this complete analysis:

The Full Workflow

1. Draw the Lewis structure → bonds, lone pairs, formal charges
2. Count electron domains → steric number
3. Assign hybridization → sp, sp², or sp³
4. Determine electron geometry → tetrahedral, trigonal planar, or linear
5. Determine molecular geometry → remove lone pairs from the picture
6. Predict bond angles → ideal angles modified by lone pair effects
7. Count σ and π bonds → single=1σ, double=1σ+1π, triple=1σ+2π
8. Assess polarity → symmetry of geometry + bond dipoles

Worked Example: Sulfur Dioxide (SO₂)

How Hybridization Connects to Polarity

The Link

Hybridization → Geometry → Symmetry → Polarity

When Lone Pairs Are Present

Lone pairs break symmetry:

AP Exam Connection

A common AP question: "Explain why CO₂ is nonpolar but SO₂ is polar."

Answer: Both have polar bonds (electronegativity difference between atoms). CO₂ is sp hybridized → linear → symmetric → dipoles cancel → nonpolar. SO₂ is sp² with a lone pair → bent → asymmetric → dipoles don't cancel → polar.

These questions mimic the style and difficulty of AP Chemistry exam questions.

Perform a complete analysis of the given molecules.

Select the correct answer that connects hybridization, geometry, and polarity.

These questions represent the highest level of AP Chemistry integration.

🏆 Congratulations — Hybridization and Sigma/Pi Bonds Complete!

The Big Picture

You now understand the orbital-level explanation for molecular bonding:

AP Exam Tips

1. Always start with the Lewis structure — hybridization flows from electron domains
2. Don't confuse electron geometry with molecular geometry — lone pairs affect shape but still count for hybridization
3. Bond order determines properties — higher order = shorter, stronger, less reactive bonds
4. Connect hybridization to polarity — geometry determines if dipoles cancel
5. Practice sigma/pi counting — it's a guaranteed easy point if you know the rules

The Connection Chain

$\text{Lewis Structure} \rightarrow \text{Electron Domains} \rightarrow \text{Hybridization} \rightarrow \text{Geometry} \rightarrow \text{Polarity}$

Master this chain and you'll ace the bonding questions on the AP exam! 🎯