VSEPR theory

Draw a Lewis structure of $\text{H}_2\text{O}$ and it tells you which atoms are bonded, but not what the molecule looks like in space — is it straight, bent, pyramidal? VSEPR (Valence Shell Electron Pair Repulsion) theory answers that with one idea: every electron pair around a central atom, bonding or lone, carries negative charge, so the pairs push each other as far apart as geometry allows. Count the pairs, and the shape falls out almost automatically. The payoff is large for such a simple rule. VSEPR predicts why $\text{CH}_4$ is a perfect tetrahedron while $\text{NH}_3$ , with the same four electron pairs, is pyramidal, and why $\text{H}_2\text{O}$ 's bond angle is squeezed even further to about $104.5°$ . It also explains the two-length bonding pattern of $\text{PCl}_5$ and the perfectly symmetric geometry of $\text{SF}_6$ . This article builds the repulsion order, the standard geometries, and walks through the syllabus's named molecules step by step.

The core insight

The electron pairs around a central atom — bonding and lone — arrange themselves to minimise mutual repulsion, so counting bonding regions and lone pairs on a Lewis structure directly predicts a molecule's electron-pair geometry and, once lone pairs are set aside, its actual molecular shape.

The core idea

Why the pairs push apart. Every electron pair, whether it is shared between two nuclei (a bond pair) or belongs to just the central atom (a lone pair), is a region of concentrated negative charge. Like charges repel, so the set of pairs around a central atom settles into whichever arrangement keeps them as far from each other as possible — this is the entire content of VSEPR theory, and everything else follows from it.

Counting the domains. Only the number of electron-pair "domains" matters, not how many electrons a bond contains: a single, double, or triple bond between the central atom and one neighbour still counts as one domain. The total domain count — bonding regions plus lone pairs — fixes the electron-pair geometry:

SN = n_{\text{bond}} + n_{\text{lone}}

where: $SN$ is the steric number (total electron-pair domains around the central atom), $n_{\text{bond}}$ is the number of bonding regions (each multiple bond still counts as one), and $n_{\text{lone}}$ is the number of lone pairs on the central atom. The steric number is looked up in the standard table below to name the electron-pair geometry and its ideal bond angle:

Steric number	Electron-pair geometry	Ideal bond angle
2	Linear	$180°$
3	Trigonal planar	$120°$
4	Tetrahedral	$109.5°$
5	Trigonal bipyramidal	$90°/120°$
6	Octahedral	$90°$

Geometry versus shape. When every domain is a bond pair (no lone pairs), the electron-pair geometry is the molecular shape — as in $\text{BeCl}_2$ (linear, $SN = 2$ ), $\text{BF}_3$ (trigonal planar, $SN = 3$ ) and $\text{CH}_4$ (tetrahedral, $SN = 4$ ). Once lone pairs appear, they still occupy one of the $SN$ positions and still repel, but they are invisible when naming the shape, since "molecular shape" describes only atom positions. So $\text{NH}_3$ and $\text{H}_2\text{O}$ , both $SN = 4$ like $\text{CH}_4$ , get different, smaller names — pyramidal and bent — because one or two of their four domains are lone pairs, not atoms.

The repulsion order. Because a lone pair is held by only one nucleus, it spreads out more and occupies more angular space than a bond pair, which is pulled taut between two nuclei. This gives a fixed order of repulsion strength, strongest first:

\text{lone pair–lone pair} > \text{lone pair–bond pair} > \text{bond pair–bond pair}

The consequence is bond-angle compression: every lone pair on a central atom pushes the surrounding bond pairs slightly closer together than the "ideal" geometric angle predicts, and the more lone pairs present, the greater the squeeze.

Figure

Side by side ball and stick diagrams of ammonia pyramidal with bond angle 107 degrees and water bent with bond angle 104.5 degrees compared against the ideal tetrahedral angle of 109.5 degrees

Why it works

VSEPR is not a bonding theory in the sense of explaining why atoms bond — it only predicts shape, purely from electrostatics, which is why it needs no orbital calculations. Picture the central atom's electron pairs as points of charge free to slide over a sphere around it; the arrangement that minimises total pairwise repulsion is the one where the points spread as evenly as possible, and that is exactly the set of standard geometries in the table — two points to opposite ends of a line, three into a flat triangle, four into a tetrahedron, and so on. The electron-pair geometry therefore depends only on the count of domains, not on what they are made of chemically. Lone pairs distort the observed atomic angles, even though the underlying electron-pair geometry is unchanged, because a lone pair's electron cloud is fatter and closer to the nucleus (held by one nucleus only, with nothing pulling it outward), so it repels its neighbours more fiercely than a bond pair does. The bond pairs get pushed together in response, shrinking the angle between atoms below the ideal value even though the domain positions themselves stay as symmetric as electrostatics can make them. This one picture — pairs repel, lone pairs repel hardest — explains both the standard geometries and every deviation from them.

Worked example

Predict and compare the molecular shapes of $\text{NH}_3$ , $\text{H}_2\text{O}$ , $\text{PCl}_5$ , and $\text{SF}_6$ , explaining the bond-angle trend.

Step 1. Find $SN$ for $\text{NH}_3$ . Nitrogen forms 3 N–H bonds and carries 1 lone pair, so $SN = 3 + 1 = 4$ , giving a tetrahedral electron-pair geometry (ideal $109.5°$ ). Removing the one lone pair to name the atomic arrangement leaves a trigonal pyramidal shape, and because lone pair–bond pair repulsion exceeds bond pair–bond pair repulsion, the H–N–H angle compresses from $109.5°$ to about $107°$ .

Step 2. Find $SN$ for $\text{H}_2\text{O}$ . Oxygen forms 2 O–H bonds and carries 2 lone pairs, so $SN = 2 + 2 = 4$ — the same steric number as $\text{NH}_3$ , hence the same tetrahedral electron-pair geometry. Removing both lone pairs leaves a bent (angular) shape. With two lone pairs now compressing the structure instead of one, the H–O–H angle shrinks further, to about $104.5°$ — smaller than $\text{NH}_3$ 's $107°$ , confirming that more lone pairs cause greater compression even at the same steric number.

Step 3. Find $SN$ for $\text{PCl}_5$ . Phosphorus forms 5 P–Cl bonds and carries 0 lone pairs, so $SN = 5 + 0 = 5$ , giving a trigonal bipyramidal electron-pair geometry that is also the molecular shape (no lone pairs to remove). Three Cl atoms sit in equatorial positions ( $120°$ apart, in one plane through P) and two sit in axial positions ( $90°$ to each equatorial Cl, $180°$ to each other) — the two distinct positions are why $\text{PCl}_5$ has two slightly different P–Cl bond lengths, axial bonds being marginally longer because axial atoms feel more neighbouring repulsion.

Step 4. Find $SN$ for $\text{SF}_6$ . Sulphur forms 6 S–F bonds and carries 0 lone pairs, so $SN = 6 + 0 = 6$ , giving an octahedral electron-pair geometry and shape, with all six S–F bonds equivalent, each at $90°$ to four neighbours and $180°$ to the one opposite. Answer: $\text{NH}_3$ is trigonal pyramidal ( $107°$ ), $\text{H}_2\text{O}$ is bent ( $104.5°$ ), $\text{PCl}_5$ is trigonal bipyramidal (axial/equatorial $90°/120°$ ), and $\text{SF}_6$ is a regular octahedron ( $90°$ ) — the bond-angle sequence $109.5° \rightarrow 107° \rightarrow 104.5°$ tracks the rising lone-pair count on an $SN = 4$ centre.

Competitive stretch

$\text{PCl}_5$ and $\text{SF}_6$ share a feature the first four molecules do not: an expanded octet, 10 and 12 electrons respectively around the central atom. This is possible only because P and S are period 3 elements with valence shells large enough to hold more than 8 electrons, unlike the period 2 elements (Be, B, C, N, O) above, whose central atoms never exceed $SN = 4$ . The period of the central atom is a fast check on whether $SN > 4$ is even chemically allowed before counting domains.

Common traps

Confusing electron-pair geometry with molecular shape. They coincide only when the central atom has no lone pairs; whenever lone pairs are present, the atomic shape (bent, pyramidal) differs in name from the underlying domain arrangement (tetrahedral).
Miscounting multiple bonds. A double or triple bond still counts as one domain in $SN$ , not two or three — in $\text{CO}_2$ , carbon has $SN = 2$ (two double bonds, no lone pairs), giving a linear shape, not four domains.
Treating all positions in a trigonal bipyramid as equal. Axial and equatorial positions are genuinely different, with different bond angles ( $90°$ and $120°$ ) and slightly different bond lengths — do not assume a uniform angle as in the tetrahedral or octahedral cases.
Reversing the repulsion order. Getting lone pair–lone pair, lone pair–bond pair, and bond pair–bond pair out of order leads to predicting the wrong direction of bond-angle compression.

Ask the tutor: Ask the AI tutor to walk through predicting the shape of a molecule not covered here, such as $\text{ClF}_3$ or $\text{XeF}_4$ .

Why does a lone pair repel more strongly than a bond pair if both are just electron pairs?

A bond pair is shared between two nuclei, both of which pull it inward and hold it in a fairly tight, well-defined region between the atoms. A lone pair belongs to only one nucleus, so nothing pulls it toward a second centre — it spreads out and occupies a broader angular region close to the central atom. That extra spatial spread is what makes its electrostatic push on neighbouring pairs stronger, which is why the repulsion order places lone pair interactions above bond pair interactions.

Does VSEPR ever fail to predict the correct shape?

For the syllabus examples here it works reliably, but VSEPR is a simple electrostatic model, not a full quantum treatment, so it can struggle with unusual cases such as some transition-metal complexes or molecules where other effects (like d-orbital involvement or crystal-field effects) compete with simple pair repulsion. For the standard main-group molecules and ions in this unit — $\text{BeCl}_2$ through $\text{SF}_6$ — VSEPR gives the correct experimentally observed shape every time.

How do I handle a central atom with a multiple bond when counting for VSEPR?

Count the multiple bond as a single domain, exactly like a single bond — what matters for VSEPR is the number of separate regions of electron density around the central atom, not how many electron pairs sit inside one of those regions. So a central atom with two single bonds and one double bond has a steric number of 3, not 4, and you read its geometry off the $SN = 3$ row of the table (trigonal planar) just as you would for three single bonds.