VSEPR Theory

Why is $\text{CH}_4$ a perfect tetrahedron while $\text{NH}_3$ , with almost the same central atom and the same four electron groups, folds into a pyramid with a noticeably smaller bond angle? The answer is not memorised shape charts — it is one counting rule. VSEPR theory (Valence Shell Electron Pair Repulsion theory) says that electron pairs around a central atom, whether they are shared in a bond or sitting alone as a lone pair, behave like mutually repelling balloons tied to a common knot: they push apart as far as geometry allows. Count the electron groups, place them to maximise their separation, then read off where the atoms actually sit, and the molecular shape falls out — no memorisation needed. This article turns that counting rule into a formula you can apply to any central atom, shows why lone pairs squeeze bond angles below the ideal geometric value, and works through $\text{NH}_3$ and $\text{H}_2\text{O}$ side by side so you can see exactly how one extra lone pair changes the answer.

The core insight

A molecule's shape follows from counting the total electron groups (bonding pairs plus lone pairs) around the central atom, arranging them to minimise repulsion, and then reading the geometry from atom positions alone, with each lone pair compressing the surrounding bond angles because lone pair–bond pair repulsion exceeds bond pair–bond pair repulsion.

The core idea

Electron groups repel, not atoms. VSEPR treats every region of electron density around the central atom — a single bond, a double or triple bond (still counted as one group), or a lone pair — as a point of negative charge that repels every other such point. The stable arrangement is whichever one keeps these groups as far apart as possible on a sphere around the central atom.

Two different geometries matter. The electron-pair geometry (also called the electron-domain geometry) is determined by all the groups, including lone pairs. The molecular shape is determined only by where the atoms sit — lone pairs are invisible to shape, though very much felt in the angles. These two coincide only when there are zero lone pairs on the central atom, which is why $\text{CH}_4$ , $\text{PCl}_5$ , and $\text{SF}_6$ have textbook-exact bond angles while $\text{NH}_3$ and $\text{H}_2\text{O}$ do not.

Repulsion strength is not uniform. A lone pair is held by only one nucleus, so it spreads into a fatter angular cloud than a bonding pair, which is pulled taut between two nuclei. This gives the repulsion ordering:

\text{lone pair–lone pair} \;>\; \text{lone pair–bond pair} \;>\; \text{bond pair–bond pair.} \qquad (1)

where: each term denotes the strength of electrostatic repulsion between two electron groups of the stated type sharing a central atom. This ordering is the entire reason bond angles shrink whenever a lone pair is present — nothing else in the model is needed to explain it. The count of electron groups (the steric number) sets the parent geometry directly:

Figure

Steric number two through six mapped to their electron pair geometries linear trigonal planar tetrahedral trigonal bipyramidal and octahedral with bond angles labelled

Understanding VSEPR theory

VSEPR predictions rest on a genuine physical mechanism, not a lookup table, so the reasoning below stands in for a numerical derivation: given any central atom and its valence electron count, you can predict a molecule's shape and its bond angle distortion from first principles, using nothing beyond electron counting and the repulsion ordering in equation (1).

Step 1. Count the central atom's valence electrons, $V$ , from its group number in the periodic table — the same count you already use to draw Lewis structures.

Step 2. Count $n$ , the number of atoms directly bonded to the central atom by a single bond each (a double or triple bond still counts as one bonded neighbour and one electron group, per VSEPR postulate 3). Each such bond uses one electron from the central atom's valence shell.

Step 3. The electrons not used in bonding form lone pairs on the central atom. For simple single-bonded, neutral central atoms this gives the lone-pair count directly:

L = \dfrac{V - n}{2}. \qquad (2)

where: $L$ is the number of lone pairs on the central atom, $V$ is its valence electron count, and $n$ is the number of atoms singly bonded to it. (The general Lewis-structure count, needed for multiple bonds or ions, simply replaces $n$ with the electrons actually used in bonding and adjusts $V$ for charge — but for the syllabus examples here, equation (2) is exact.)

Step 4. The steric number, $\text{SN} = n + L$ , is the total number of electron groups, and it is $\text{SN}$ — not $n$ alone — that fixes the electron-pair geometry from the table implicit in the figure above (2 → linear, 3 → trigonal planar, 4 → tetrahedral, 5 → trigonal bipyramidal, 6 → octahedral).

Step 5. Because lone pair–bond pair repulsion exceeds bond pair–bond pair repulsion (equation 1), every lone pair squeezes the remaining bond angles below the ideal value for that $\text{SN}$ . Each additional lone pair compresses the angle further, which is exactly the trend seen going from $\text{CH}_4$ (109.5°, $L=0$ ) to $\text{NH}_3$ (~107°, $L=1$ ) to $\text{H}_2\text{O}$ (~104.5°, $L=2$ ), all at $\text{SN}=4$ .

Worked example

Use equation (2) to find the number of lone pairs on the central atom in $\text{NH}_3$ and $\text{H}_2\text{O}$ , then predict each molecular shape and rank their H–X–H bond angles.

Step 1 (NH₃). Nitrogen has $V = 5$ valence electrons and bonds to $n = 3$ hydrogen atoms, so $L = (5-3)/2 = 1$ lone pair.

Step 2 (NH₃). Steric number $\text{SN} = n + L = 3 + 1 = 4$ , so the electron-pair geometry is tetrahedral; with one corner occupied by a lone pair, the molecular shape (atom positions only) is trigonal pyramidal, and the H–N–H angle is compressed from 109.5° to the observed ~107°.

Step 3 (H₂O). Oxygen has $V = 6$ valence electrons and bonds to $n = 2$ hydrogen atoms, so $L = (6-2)/2 = 2$ lone pairs.

Step 4 (H₂O). Steric number $\text{SN} = n + L = 2 + 2 = 4$ , again tetrahedral electron-pair geometry; with two corners occupied by lone pairs, the molecular shape is bent (angular), and the H–O–H angle compresses further, to the observed ~104.5°.

Step 5 (ranking). Since two lone pairs push harder on the remaining bonds than one, the bond angle order follows the lone-pair count directly: $\text{CH}_4\,(109.5°,\ L=0) > \text{NH}_3\,(107°,\ L=1) > \text{H}_2\text{O}\,(104.5°,\ L=2)$ .

The H–N–H angle in $\text{NH}_3$ is about 107° and the H–O–H angle in $\text{H}_2\text{O}$ is about 104.5°, both compressed from the ideal 109.5° tetrahedral angle, with $\text{H}_2\text{O}$ compressed more because it carries two lone pairs against $\text{NH}_3$ 's one.

Competitive stretch

Apply equation (2) to $\text{PCl}_5$ ( $V=5$ , $n=5$ ): $L = (5-5)/2 = 0$ , so $\text{SN}=5$ with zero lone pairs, giving an exact trigonal bipyramidal shape with two distinct angles, 90° (axial–equatorial) and 120° (equatorial–equatorial) — no compression, because there is no lone pair to cause any.

Common traps

Mixing up electron-pair geometry with molecular shape — they are identical only when $L = 0$ ; whenever a lone pair is present, name the shape from atom positions alone, not from the parent geometry.
Forgetting to include lone pairs when counting $\text{SN}$ — this is the single most common cause of a wrong shape prediction, since $n$ alone (bonded atoms) always undercounts the true steric number when $L > 0$ .
Counting a double or triple bond as two or three separate electron groups — VSEPR counts a multiple bond as exactly one group (postulate 3), even though it repels somewhat more strongly than a single bond.
Applying bond-angle compression to a molecule with no lone pairs — $\text{CH}_4$ , $\text{PCl}_5$ , and $\text{SF}_6$ all have $L=0$ , so their bond angles equal the ideal geometric value exactly; equation (2) confirms this by giving $L=0$ for each.

Ask the tutor: Want to run equation (2) on a molecule not covered here, like $\text{BF}_3$ or a polyatomic ion, and check your predicted shape?

Why does a double bond still count as only one electron group in VSEPR

VSEPR cares about regions of electron density around the central atom, not the number of electron pairs in each region. A double bond keeps both its bonding pairs confined to the same direction, between the same two nuclei, so it occupies one angular position exactly like a single bond does — it simply repels somewhat more strongly because it is more electron-dense. Splitting it into two separate groups would place two "phantom" repulsion centres in the same location, which does not match the observed geometry.

Does equation (2), L = (V minus n)/2, work for every molecule, including ions and expanded octets

It works directly for any neutral central atom that forms $n$ single bonds and no multiple bonds, which covers every example in this syllabus, including $\text{SF}_6$ and $\text{PCl}_5$ where the central atom simply uses more of its valence electrons in bonding (period 3 elements can access $d$ orbitals to hold more than eight electrons). For ions you would adjust $V$ by adding an electron per negative charge or subtracting one per positive charge before applying the formula; for molecules with multiple bonds you instead count electron groups directly from the Lewis structure rather than using equation (2), since a double or triple bond still only removes one group's worth of "slots," not two or three.

If lone pairs and bonding pairs are both just electron pairs, why don't they repel identically

A bonding pair is pulled taut between two positively charged nuclei, which confines its electron cloud into a narrower angular region. A lone pair is attracted by only one nucleus, so it is free to spread into a fatter, more diffuse cloud that occupies more angular space near the central atom. That extra spatial spread is exactly why lone pair interactions are more repulsive than bonding pair interactions of the same count — it is a geometric consequence of having one anchor instead of two, not a difference in charge.