Introduction

Lewis structures provide insight into properties such as molecular geometry, bond order, bond length, and dipole moments. The Valence Shell Electron Pair Repulsion (VSEPR) model predicts the shapes of molecules by considering how electron pairs arrange themselves to minimize repulsion between them. This prediction is based on the Coulombic repulsion that occurs between negatively charged electrons.

What do I need to know about VSEPR?

For the AP Exam, it’s essential to commit the VSEPR tables to memory. These charts summarize everything needed to predict molecular shapes, and with enough practice, any VSEPR-based problem essentially becomes guaranteed points.

To determine molecular geometry, follow these steps in order:

Here’s what each column in the table represents:

VSEPR

Family 5: Five Electron Domains

When the central atom contains five regions of electron density, the electrons organize themselves into a trigonal bipyramidal arrangement to reduce repulsive forces. In this geometry, there are two unique types of positions:

$\begin{array}{ccccc}\text{General Formula}& \text{Molecular Shape}& \text{Description}& \text{Bond Angles}& \text{Example}\\ {\mathrm{MX}}_5& \text{Trigonal bipyramidal}& \text{All 5 positions are occupied by atoms}& 90^{\circ },120^{\circ },180^{\circ }& {\mathrm{PF}}_5,{\mathrm{PCl}}_5\\ {\mathrm{MX}}_4\mathrm{E}& \text{Seesaw}& \text{1 lone pair in equatorial position}& <90^{\circ },<120^{\circ },\sim 180^{\circ }& {\mathrm{SF}}_4\\ {\mathrm{MX}}_3{\mathrm{E}}_2& \text{T-shaped}& \text{2 lone pairs in equatorial positions}& \sim 90^{\circ },<180^{\circ }& {\mathrm{ClF}}_3,{\mathrm{BrF}}_3\\ {\mathrm{MX}}_2{\mathrm{E}}_3& \text{Linear}& \text{3 lone pairs in equatorial positions}& 180^{\circ }& {\mathrm{XeF}}_2,{\mathrm{I}}_3^{-}\end{array}$

Lone pairs tend to occupy the equatorial positions in a trigonal bipyramidal geometry because these locations provide greater spatial separation, which reduces electron–electron repulsion with the other domains.

Family 6: Six Electron Domains

When a central atom has six regions of electron density, the resulting electron domain geometry is octahedral. In this arrangement, all positions are identical, with each bond angle measuring 90°.

$\begin{array}{ccccc}\text{General Formula}& \text{Molecular Shape}& \text{Description}& \text{Bond Angles}& \text{Example}\\ {\mathrm{MX}}_6& \text{Octahedral}& \text{All 6 positions occupied by atoms}& 90^{\circ }& {\mathrm{SF}}_6,{\mathrm{PCl}}_6^{-}\\ {\mathrm{MX}}_5\mathrm{E}& \text{Square pyramidal}& \text{1 lone pair creates pyramid with square base}& <90^{\circ }& {\mathrm{BrF}}_5,{\mathrm{IF}}_5\\ {\mathrm{MX}}_4{\mathrm{E}}_2& \text{Square planar}& \text{2 lone pairs opposite each other}& 90^{\circ }& {\mathrm{XeF}}_4,{\mathrm{ICl}}_4^{-}\end{array}$

Bonding

Sigma and Pi Bonds

Covalent bonds are formed through the overlap of atomic orbitals, and the way these orbitals overlap determines both the type and the strength of the bond.

Sigma (σ) bonds arise from the direct, head-on overlap of orbitals along the internuclear axis—the line that connects the nuclei of the bonded atoms. This overlap can occur between:

Because this head-on overlap creates a concentrated electron density directly between the nuclei, sigma bonds are the strongest kind of covalent bond.

Pi ($\pi$) bonds form differently. They result from the sideways overlap of parallel p orbitals above and below the internuclear axis. This overlap produces two distinct regions of electron density, one above and one below the bond axis. Since the sideways overlap is less effective than the direct overlap in sigma bonds, pi bonds are weaker. This distinction explains several important bonding features:

For quick reference:

As the number of pi bonds increases, bond energy rises, bond length decreases, and rotational freedom becomes more restricted.

This lack of free rotation around double bonds has an important consequence: it allows the existence of geometric (cis-trans) isomers. These are structural isomers with the same molecular formula and connectivity but different spatial arrangements of atoms due to the fixed geometry of the double bond.

For example, in 2-butene (${\text{CH}}_3$${\text{CH=CHCH}}_3$):

Even though both isomers share the same molecular formula, their physical properties—such as boiling point, melting point, and dipole moment—differ significantly. This difference arises entirely from the rigidity imposed by the pi bond within the double bond.

Hybridization

Hybridization describes how atomic orbitals combine to create new, hybridized orbitals that affect both bonding behavior and molecular geometry. This concept builds on valence bond theory, helping to explain molecular structures that otherwise wouldn’t make sense.

There are five main types of hybridization—${\text{sp}}^3$, ${\text{sp}}^2$, $\text{sp}$, ${\text{sp}}^3\text{d}$, and ${\text{sp}}^3{\text{d}}^2$—though you’ll primarily be tested on the first three. In each case, orbitals mix so that electrons can be distributed across subshells in a way that lowers the overall energy of the atom.

Hybridization also accounts for molecules like methane (${\text{CH}}_4$). Without hybrid orbitals, the arrangement of carbon’s valence electrons would not allow for the formation of four equivalent sigma bonds. The ${\text{sp}}^3$ hybridization explains how all four $\text{C–H}$ bonds in methane are identical in strength and orientation.

It’s important to note that lone pairs take up more space than bonding pairs, which pushes the atoms slightly closer together and reduces the bond angles from their ideal values. By connecting hybridization to geometry and bond angles, you can more quickly predict and analyze the three-dimensional structures of molecules.

Dipole Moments

A molecule exhibits a dipole moment when its electron density is unevenly distributed, resulting in partial positive and negative regions. To determine whether a molecule has a dipole:

For instance, ${\text{CO}}_2$ contains two polar $\text{C=O}$ bonds, but because the molecule is linear, the dipoles oppose each other and cancel, so the molecule is nonpolar. By contrast, ${\text{H}}_2\text{O}$ has two polar $\text{O–H}$ bonds arranged in a bent shape, meaning the dipoles reinforce rather than cancel, giving the molecule an overall dipole moment.

Bond Length and Atomic Radius

Bond length depends not only on bond order (single, double, triple) but also on the size of the atoms involved. Larger atoms have valence electrons farther from the nucleus, which results in longer bonds. For example, when comparing similar bonds:

This pattern aligns with the periodic trend in atomic radius: atoms increase in size as you move down a group, causing longer bonds. Therefore, when evaluating bond lengths, it’s important to consider both the bond order (more shared electrons shorten the bond) and the atomic radii of the atoms involved.