Hybridization and Sigma/Pi Bonds

Understand orbital hybridization (sp, sp², sp³, sp³d, sp³d²), distinguish between sigma and pi bonds, and relate hybridization to molecular geometry.

Hybridization and Sigma/Pi Bonds

What is Hybridization?

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

| Steric Number | Hybridization | Electron Geometry | Bond Angle | |---------------|---------------|-------------------|------------| | 2 | sp | Linear | 180° | | 3 | sp² | Trigonal planar | 120° | | 4 | sp³ | Tetrahedral | 109.5° | | 5 | sp³d | Trigonal bipyramidal | 90°, 120° | | 6 | sp³d² | Octahedral | 90° |

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):

\ceH−C≡C−H\ce{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):

\ceH2C=CH2\ce{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

| SN | Hybridization | |----|---------------| | 2 | sp | | 3 | sp² | | 4 | sp³ | | 5 | sp³d | | 6 | sp³d² |

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)

\ceH2C=O\ce{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:

| Bond | Length (pm) | Energy (kJ/mol) | σ bonds | π bonds | |------|-------------|-----------------|---------|---------| | C-C | 154 | 348 | 1 | 0 | | C=C | 134 | 614 | 1 | 1 | | C≡C | 120 | 839 | 1 | 2 |

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

| Molecule | Central Atom | SN | Hybridization | Geometry | σ bonds | π bonds | |----------|--------------|-----|---------------|----------|---------|---------| | CO₂ | C | 2 | sp | Linear | 2 | 2 | | BF₃ | B | 3 | sp² | Trig planar | 3 | 0 | | CH₄ | C | 4 | sp³ | Tetrahedral | 4 | 0 | | H₂O | O | 4 | sp³ | Bent | 2 | 0 | | PCl₅ | P | 5 | sp³d | Trig bipyramidal | 5 | 0 | | SF₆ | S | 6 | sp³d² | Octahedral | 6 | 0 | | C₂H₄ | C | 3 | sp² | Trig planar | 5 | 1 | | C₂H₂ | C | 2 | sp | Linear | 3 | 2 | | NH₃ | N | 4 | sp³ | Trig pyramidal | 3 | 0 |

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

📚 Practice Problems

1Problem 1easy

❓ Question:

Determine the hybridization of the central atom in ammonia (NH₃) and count the number of sigma and pi bonds.

💡 Show Solution

Solution:

Given: NH₃ (ammonia) Find: Hybridization of N, count σ and π bonds

Step 1: Draw Lewis structure

\ceH−N−H\ce{H - N - H} \ce∣\ce{ | } \ceH\ce{ H }

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

SN=4\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.

2Problem 2medium

❓ Question:

For the molecule formaldehyde (H₂CO), determine: (a) the hybridization of both C and O, (b) the total number of sigma and pi bonds, and (c) explain why the molecule is planar.

💡 Show Solution

Solution:

Given: H₂CO (formaldehyde) Find: (a) Hybridization of C and O, (b) σ and π bond counts, (c) why planar

Step 1: Draw Lewis structure

\ceH\ce{ H } \ce∣\ce{ | } \ceH−C=O\ce{H - C = O}

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)

3Problem 3hard

❓ Question:

Consider the molecule acetylene (C₂H₂) with structure H-C≡C-H. (a) Determine the hybridization of each carbon atom. (b) Count the total number of sigma and pi bonds. (c) Describe the orbital overlap for each bond. (d) Explain the geometry and why all atoms are collinear.

💡 Show Solution

Solution:

Given: C₂H₂ (acetylene), structure H-C≡C-H Find: (a) Hybridization, (b) σ and π counts, (c) orbital overlaps, (d) geometry explanation

Step 1: Draw Lewis structure

\ceH−C≡C−H\ce{H - C ≡ C - H}

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.

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