VSEPR and Hybridization

Learning Objective 2.7.A Based on the relationship between Lewis diagrams, VSEPR theory, bond orders, and bond polarities: a. Explain structural properties of molecules b. Explain electron properties of molecules

Quick Notes

VSEPR theory predicts molecular geometry by minimizing electron pair repulsion.
Electron pairs (bonding and lone pairs) arrange themselves as far apart as possible.
Common geometries: linear, trigonal planar, tetrahedral, trigonal pyramidal, bent, trigonal bipyramidal, seesaw, T-shaped, octahedral, square pyramidal, square planar.
Bond angles depend on geometry and lone pair repulsion.
Bond order affects bond length and bond energy: higher bond order = shorter, stronger bond.
Hybridization is the process by which atomic orbitals combine to form new hybrid orbitals for bonding and is linked to geometry:
- sp hybridized → 180° bond angle
- sp² hybridized→ 120° bond angles
- sp³ hybridized→ 109.5° bond angles
Sigma (σ) bonds = single bonds, formed by direct overlap of orbitals - stronger.
Pi (π) bonds = found in double/triple bonds, formed by sideways overlap of p-shaped orbitals, weaker, restricted rotation.
Molecular polarity depends on bond polarity and shape.

Full Notes

To predict the shape and bonding properties of molecules, we combine information from Lewis structures, VSEPR theory, bond order, and electronegativity. These models help explain molecular geometry, bonding patterns, and electron behavior in both molecules and ions.

VSEPR Theory (Valence Shell Electron Pair Repulsion)

The shape of a molecule is determined by repulsions between regions of electron density (electron domains) around the central atom. These regions include:

Bonding pairs of electrons (in covalent bonds)
Lone pairs of non-bonded electrons

Key principles:

Electron pairs repel each other and arrange themselves as far apart as possible in 3D space.
Lone pairs repel more strongly than bonding pairs because they are closer to the nucleus and occupy more space.
These repulsions lead to predictable molecular geometries.

Matt’s exam tip

Double and triple bonds count as one region of electron density in VSEPR, but they exert slightly stronger repulsion than single bonds, which may reduce adjacent bond angles slightly.

How Lone Pairs Affect Bond Angles

Lone pairs occupy more space than bonding pairs, causing greater repulsion. This pushes bonding pairs closer together and reduces bond angles – the more lone pairs present, the greater the compression.

AP Chemistry overview of how lone pairs increase repulsion and compress bond angles in common VSEPR shapes

Common Molecular Shapes and Bond Angles

Linear (180° Bond Angle)

AP Chemistry linear molecular shape with 180° bond angle

2 bonding pairs, no lone pairs → Equal repulsion forces keep the bonds in a straight line.

Examples:CO₂ (O=C=O), BeCl₂.

Trigonal Planar (120° Bond Angle)

AP Chemistry trigonal planar molecular shape with 120° bond angles

3 bonding pairs, no lone pairs → Electrons spread out evenly in a flat triangle.

Examples:BF₃, NO₃⁻.

Tetrahedral (109.5° Bond Angle)

AP Chemistry tetrahedral molecular shape with 109.5° bond angles

4 bonding pairs, no lone pairs → Electrons arrange in a 3D tetrahedral shape.

Examples:CH₄, NH₄⁺.

Trigonal Pyramidal (107° Bond Angle)

AP Chemistry trigonal pyramidal molecular shape with 107° bond angle due to lone pair

3 bonding pairs, 1 lone pair → Lone pair repulsion reduces bond angle from 109.5° to 107°.

Examples:NH₃, PCl₃.

Non-linear (104.5° Bond Angle)

AP Chemistry bent (non-linear) molecular shape such as water with 104.5° bond angle

2 bonding pairs, 2 lone pairs → Extra lone pair repulsion reduces bond angle further to 104.5°.

Examples:H₂O, OF₂.

Trigonal Bipyramidal (90° & 120° Bond Angles)

AP Chemistry trigonal bipyramidal molecular shape with 90° and 120° bond angles

5 bonding pairs, no lone pairs → Atoms arrange in a 3D two-layer shape.

Example:PCl₅.

Octahedral (90° Bond Angle)

AP Chemistry octahedral molecular shape with 90° bond angles

6 bonding pairs, no lone pairs → Electrons spread out in a symmetrical 3D shape.

Example:SF₆.

Square Planar (90° Bond Angle)

AP Chemistry square planar molecular shape with 90° bond angles

4 bonding pairs, 2 lone pairs → Electrons spread out in a flat, planar arrangement.

Example:[Ni(CN)₄]²⁻.

Application in Ions

The same rules apply for polyatomic ions:
NH₄⁺ → Tetrahedral (109.5°).
NO₃⁻ → Trigonal Planar (120°).
OH⁻ → Bent (~104.5°).

Recap - Bond Order, Bond Length, and Bond Strength

Bond order refers to the number of electron pairs shared between two atoms:

Single bond = 1 shared pair → bond order = 1
Double bond = 2 shared pairs → bond order = 2
Triple bond = 3 shared pairs → bond order = 3

As bond order increases:

Bond length decreases
Bond strength (bond energy) increases

Why?
More shared electrons lead to greater electrostatic attraction between the negatively charged electrons and both nuclei.

AP Chemistry bond order vs bond length and bond energy comparison

This stronger attraction brings the nuclei closer together (shorter bond) and requires more energy to break (stronger bond).

Molecular Polarity

A molecule can be polar or non-polar, depending on whether it contains polar bonds and the molecules symmetry. Remember a bond can be polar if the two bonding atoms have different electronegativities (see electronegativity).

1. Non-Polar molecules have no permanent dipole
If polar bonds are arranged symmetrically, dipoles cancel out → Non-polar molecule.

AP Chemistry non‑polar molecule example CO₂ where bond dipoles cancel

Example CO₂ (O=C=O)
Each C=O bond is polar, but the molecule is linear, so dipoles cancel.
No overall dipole = Non-polar molecule.

Example CCl₄ (Tetrachloromethane)
Each C–Cl bond is polar, but tetrahedral shape means dipoles cancel.
CCl₄ is non-polar despite having polar bonds.

2. Polar molecules have a permanent dipole
If dipoles do not cancel due to asymmetry, the molecule is polar.

AP Chemistry polar molecule examples where dipoles do not cancel, including water and chloroform

Example H₂O
O–H bonds are polar and form a bent shape (104.5°). Dipoles do not cancel → Water is polar.

Example CHCl₃ (Chloroform)
The C–H and C–Cl bonds have different polarities. Dipoles do not cancel → CHCl₃ is polar.

Hybridization of Atomic Orbitals

Hybridization is the process by which atomic orbitals (s, p) combine to form new hybrid orbitals for bonding. These hybrid orbitals have equal energy and are oriented in specific geometries to minimise electron repulsion. It explains shapes predicted by VSEPR theory and observed experimentally.

Determining Type of Hybridization

To determine hybridization:

Count the number of electron domains (regions of electron density) around the atom.
This includes single, double, and triple bonds, as well as lone pairs.
Match this number to the hybridization.

4 regions = sp³
3 regions = sp²
2 regions = sp

AP Chemistry carbon hybridisation showing sp, sp², and sp³ orbital sets and geometries

Hybridisation can be used to explain why atoms bond in the way they do (for example, why carbon atoms form four covalent bonds).

AP Chemistry hybridisation and bonding patterns explaining why atoms form specific numbers of bonds

Note: You are not required to understand d-orbital hybridization for the AP Exam.

Sigma (σ) and Pi (π) Bonds

σ (sigma) bonds:
Covalent bonds formed by direct (end-to-end) overlap of orbitals along the bond axis. Present in all covalent bonds.

AP Chemistry sigma bond formed by direct end-to-end orbital overlap

π (pi) bond:
Covalent bonds formed by sideways overlap of adjacent p orbitals above and below the bonding axis. Only found in double and triple bonds.

AP Chemistry pi bond formed by sideways overlap of p orbitals above and below the bond axis

Note the orbital overlap is stronger in sigma than pi bonds, which is reflected in sigma bonds having greater bond energy than pi bonds. Pi bonds restrict rotation around the bond axis. This explains the existence of geometric (cis-trans) isomerism in alkenes.

Determining Number of Sigma and Pi Bonds

Each single bond is 1 σ
Each double bond = 1 σ + 1 π
Each triple bond = 1 σ + 2 π

Examples AP Chemistry worked examples counting sigma and pi bonds in ethene, carbon dioxide, and ethyne C₂H₄ (ethene)
Structure: H–C=C–H
5 σ bonds (1 C–C and 4 C–H)
1 π bonds (from C=C double bond)

CO₂
O=C=O
2 double bonds → 2 σ + 2 π
C₂H₂ (ethyne)

Structure: H–C≡C–H
3 σ bonds (1 C–C and 2 C–H)
2 π bonds (from C≡C triple bond)

Matt’s exam tip

Be careful to distinguish electron geometry (based on all regions) from molecular shape (only bonded atoms). Always count total electron regions and match the hybridization accordingly.

Summary

VSEPR theory and hybridization help us predict the structure and bonding in molecules. Molecular shape depends on the number of electron regions around the central atom.
Hybridization gives insight into the types of orbitals involved in bonding.
Bond polarity, bond order, and molecular geometry together determine a molecule’s polarity and reactivity.
The presence of pi bonds also explains properties like restricted rotation and multiple bond strength.
Key points to remember:
- VSEPR: electron pairs repel → shape predicts geometry
- Lone pairs affect bond angles and shape
- Hybridization: sp (2 regions), sp² (3), sp³ (4)
- Sigma (σ) = all single bonds and the first bond in multiple bonds
- Pi (π) = the second and third bonds in double/triple bonds