VSEPR: Electron-Domain Geometry
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You should be comfortable drawing Lewis structures and identifying the number of bonding pairs and lone pairs around a central atom. Familiarity with basic vector addition and the concept of electrostatic repulsion is helpful.
Why this matters
Molecular geometry dictates reactivity, polarity, and biological recognition.
Knowing the exact three-dimensional arrangement of atoms is not merely an academic exercise; it is the foundation of chemical behaviour. The shape of a molecule determines whether it can fit into an enzyme active site, whether it possesses a permanent dipole moment that affects boiling point, and how it interacts with other molecules in solution. While quantum mechanics provides the rigorous description of electron density, the Valence Shell Electron Pair Repulsion (VSEPR) model offers a remarkably accurate, intuitive, and computationally simple method for predicting these shapes based solely on the electron count around the central atom.
The Core Idea: Electron Domains
Imagine the central atom of a molecule as a hub, with electron pairs radiating out from it like spokes on a wheel. VSEPR theory posits that these electron pairs—whether they are involved in bonding with another atom or sitting as lone pairs on the central atom—act as regions of negative charge. Because like charges repel, these regions will arrange themselves in three-dimensional space to be as far apart from each other as possible. This minimisation of repulsion energy dictates the geometry. Crucially, VSEPR treats a double or triple bond as a single "domain" or "region" of electron density, even though it contains more electrons. The key variable is the steric number, which is simply the count of these distinct regions around the central atom. By counting the sigma bonds and lone pairs, we can map the molecule to one of several standard geometric arrangements, such as linear, trigonal planar, tetrahedral, or octahedral.
Steric Number and Idealised Geometries
The steric number (\(SN\)) is the sum of the number of \(\sigma\) bonds and the number of lone pairs on the central atom. This number directly corresponds to the electron-domain geometry. The idealised angles are derived from the maximum separation of points on a sphere or plane.
For \(SN=4\), the tetrahedral arrangement is the unique minimum-energy configuration for four equivalent domains. The angle of \(109.5^\circ\) is significantly larger than the \(90^\circ\) of a square planar arrangement, reducing repulsion. For \(SN=5\), the trigonal bipyramidal geometry places three domains in an equatorial plane and two in axial positions. This is preferred over a square pyramid because it minimises the number of \(90^\circ\) interactions.
Lone Pair Repulsion and Molecular Geometry
While electron-domain geometry considers all regions, molecular geometry describes only the positions of the atoms. Lone pairs occupy more space than bonding pairs because they are held by only one nucleus and are thus more diffuse. This leads to a hierarchy of repulsion strengths: lone pair–lone pair > lone pair–bonding pair > bonding pair–bonding pair.
Consequently, the presence of lone pairs compresses the bond angles between atoms. For example, in ammonia (\(\mathrm{NH_3}\)), the steric number is 4 (three bonds, one lone pair). The electron-domain geometry is tetrahedral, but the molecular geometry is trigonal pyramidal. The lone pair pushes the N–H bonds closer together, reducing the H–N–H angle from the ideal \(109.5^\circ\) to approximately \(107^\circ\). In water (\(\mathrm{H_2O}\)), with two lone pairs, the repulsion is stronger, compressing the H–O–H angle further to approximately \(104.5^\circ\).
Multiple Bonds and Domain Counting
A common point of confusion is how multiple bonds are treated. In VSEPR, a double bond or triple bond counts as a single electron domain. This is because the electron density is concentrated in the same region between the two nuclei, and the repulsive effect on neighbouring domains is similar to that of a single bond, albeit slightly stronger due to higher electron density.
Consider carbon dioxide (\(\mathrm{CO_2}\)). The central carbon forms two double bonds. The steric number is 2 (two domains, zero lone pairs). The geometry is linear with a bond angle of \(180^\circ\). If we incorrectly counted the double bonds as two domains each, we would predict a tetrahedral geometry, which is wrong. Similarly, in formaldehyde (\(\mathrm{H_2C=O}\)), the carbon has three domains (two C–H single bonds and one C=O double bond). The steric number is 3, resulting in a trigonal planar geometry with bond angles close to \(120^\circ\).
The result
VSEPR and molecular shape
The three-dimensional shape a molecule adopts is set, to good approximation, by the number of electron domains (bonding regions plus lone pairs, treating a multiple bond as one domain) around the central atom, which arrange themselves to maximise their mutual angular separation. The steric number \(SN=(\text{number of }\sigma\text{ bonds})+(\text{number of lone pairs on the central atom})\) determines an idealised electron-domain geometry; removing the lone-pair positions from that geometry (while they still occupy space and distort the remaining angles) gives the observed molecular geometry.
Open the full result →A frequent misconception is that the molecular geometry is always the same as the electron-domain geometry. This is only true when there are no lone pairs on the central atom. For molecules like \(\mathrm{SF_4}\) (steric number 5, one lone pair), the electron-domain geometry is trigonal bipyramidal, but the lone pair occupies an equatorial position to minimise \(90^\circ\) repulsions, resulting in a "seesaw" molecular geometry. The bond angles are distorted: the axial F–S–F angle is less than \(180^\circ\) and the equatorial F–S–F angle is less than \(120^\circ\). Always distinguish between the electron-domain arrangement and the atomic arrangement. C
Worked example
Problem. Determine the steric number, electron-domain geometry, and molecular geometry of the sulfite ion, \(\mathrm{SO_3^{2-}}\). Assume the sulfur atom is the central atom.
- Calculate the number of valence electrons: Sulfur (6) + 3 \(\times\) Oxygen (6) + 2 (charge) = 26 electrons.
- Draw the Lewis structure: Sulfur is central, bonded to three oxygens. To satisfy the octet rule for oxygen and minimise formal charge, one S=O double bond and two S–O single bonds (with negative formal charges on O) are formed, or resonance structures with partial double bond character. Regardless of resonance, there are three bonding regions around sulfur.
- Determine lone pairs on sulfur: with three S–O bonds and three lone pairs on each oxygen, \(26 - 6 - 18 = 2\) electrons remain, i.e. one lone pair on sulfur. (Drawing an S=O resonance form changes the formal charges but not the count of domains.)
- Calculate steric number: \(SN = 3 \text{ (bonds)} + 1 \text{ (lone pair)} = 4\).
- Identify geometries: \(SN=4\) corresponds to a tetrahedral electron-domain geometry. Removing the lone pair position leaves a trigonal pyramidal molecular geometry.
Units check. Steric number is a dimensionless count; geometry is a qualitative descriptor.
Three quick tests
- What is the steric number of the central atom in \(\mathrm{PCl_5}\)?
Answer
5 (five bonding pairs, zero lone pairs). - Why is the bond angle in \(\mathrm{H_2O}\) smaller than in \(\mathrm{CH_4}\)?
Answer
\(\mathrm{H_2O}\) has two lone pairs which repel bonding pairs more strongly than bonding pairs repel each other, compressing the angle. - Does a double bond count as one or two domains in VSEPR?
Answer
One domain.
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Every lecture and every result in this unit, with prerequisites and threads.
Lecture notes drafted with a local model and checked line by line by the editor; every number in the worked example was recomputed.