Molecular geometry and VSEPR: the guide

End of September 2026, Year 3 at the liceo scientifico (the Italian science-focused high school): if your teacher follows the order of the textbook, the class has just finished (or is finishing) electron configuration and the periodic table, and chemical bonds arrive in the next few weeks, between the end of October and November. This post arrives a little early on purpose: molecular geometry is the chapter that follows straight after the type of bond, and the test that includes it typically falls between the end of November and the beginning of December, so eight to ten weeks from now. If you get there in a few weeks, the method below is the same; if your teacher has already introduced Lewis, you are in the right place for the next piece.

On the atom, the periodic table and the type of bond (ionic, covalent, metallic) I have already written in Chemistry in Year 3 in the human-sciences track: atoms and bonds: that post stops at the type of bond. Here we start from there and go further: how the atoms of a molecule are arranged in space, and why that arrangement decides whether the molecule is polar or not.

Lewis structures: the drawing before the shape

Before talking about geometry you need to know how to draw the molecule with Lewis structures, which show the valence electrons as dots or pairs around the symbol of the atom.

Step 1 — count the total valence electrons. Add up the valence electrons of all the atoms in the molecule (the group number in the periodic table, for main-group elements). For ${\text{NH}}_3$: nitrogen has 5 valence electrons, each hydrogen has 1, total $5+3\times 1=8$.

Step 2 — draw the skeleton. The central atom is almost always the least electronegative one (excluding hydrogens, which are always on the outside, never in the centre: they have only one electron to share). Connect each outer atom to the centre with a single bond, that is, a shared pair of electrons.

Step 3 — complete the octet of the outer atoms, then that of the central atom, using the remaining pairs of electrons as lone pairs (unshared pairs, which stay on one atom only).

Step 4 — if the central atom lacks electrons for the octet, turn one or more lone pairs of an outer atom into a double or triple bond shared with the centre.

Main exceptions, in one line: hydrogen stops at 2 electrons (the “duet”), boron can stop at 6 (incomplete octet), and elements of the third period or later (sulfur, phosphorus, chlorine) can exceed the octet, reaching 10 or 12 electrons.

VSEPR theory: the shape depends on the pairs around the centre

VSEPR theory (Valence Shell Electron Pair Repulsion) starts from a simple idea: the pairs of electrons around the central atom — both bonding and lone pairs — repel each other electrically, and arrange themselves in space in the way that keeps them as far from one another as possible. The geometry of the molecule is the direct consequence of this arrangement.

The notation ${\text{AX}}_n{\text{E}}_m$ is used: $\text{A}$ is the central atom, $\text{X}$ each atom bonded to it ($n$ times), $\text{E}$ each lone pair on the centre ($m$ times). What counts is the number of groups of electrons around the centre, not the number of single electrons: a double or triple bond counts as a single group, exactly like a single bond, because it still occupies only one direction in space.

The main geometries

The point that causes most confusion: ${\text{NH}}_3$ and ${\text{H}}_2\text{O}$ both have 4 electron groups around the centre (like methane), so the basic arrangement is tetrahedral in all three cases. What changes is how many of those groups are visible bonds (which define the shape of the molecule) and how many are lone pairs (invisible in the shape, but just as bulky). A lone pair takes up more space than a bond — it is not shared with another nucleus, so it stays closer to the central atom and “pushes” harder — and that is why it slightly compresses the angles between the bonds: from 109.5° of the perfect tetrahedron to about 107° in ${\text{NH}}_3$ (one lone pair) and to about 104.5° in ${\text{H}}_2\text{O}$ (two lone pairs, which compress even more).

Electronegativity, bond polarity and molecular polarity

A covalent bond is polar when the two atoms have different electronegativity: the bonding electrons are not shared uniformly, but stay closer to the more electronegative atom. The difference in electronegativity $\Delta EN$ between the two atoms indicates how polar the single bond is — the same scale already seen for telling ionic and covalent bonds apart in Chemistry in Year 3 in the human-sciences track.

The point that the test almost always asks you to distinguish is this: bond polarity and molecular polarity are two different things. A molecule can have polar bonds and be, overall, non-polar — if the geometry is symmetric and the dipole moments of the single bonds cancel vectorially.

${\text{CO}}_2$ is non-polar, even though it has two strongly polar $\text{C}=\text{O}$ bonds (oxygen is more electronegative than carbon). The molecule is linear (${\text{AX}}_2$ geometry): the two dipole moments, equal in magnitude and opposite in direction, cancel exactly.

${\text{H}}_2\text{O}$ is polar, even though it has $\text{O}-\text{H}$ bonds less polar than those of ${\text{CO}}_2$ in terms of $\Delta EN$. The geometry is bent (~104.5°, not 180°): the two dipole moments are not opposite, and their vector sum does not cancel. The result is a net dipole, with oxygen on the negative side and the two hydrogens on the positive side.

The general rule: a molecule is non-polar only if both conditions are true — identical bonds (same type of outer atom, same polarity) and a symmetric geometry that allows the dipole moments to cancel. If just one of the two is missing (a different outer atom, or an asymmetric geometry because of lone pairs) the molecule becomes polar.

Intermolecular forces and consequences for boiling points

The polarity of the molecule decides with what force molecules attract one another, and this can be seen directly in the boiling point: the stronger the attraction between molecules, the more energy (temperature) is needed to separate them and make them pass to the gas state.

• London (dispersion) forces: present in all molecules, even non-polar ones, due to temporary instantaneous dipoles. They are the weakest, but grow with the size and number of electrons of the molecule.
• Dipole-dipole forces: between polar molecules, the positive end of one molecule attracts the negative end of another. Stronger than London forces for the same molecular mass.
• Hydrogen bond: a special, and particularly strong, case of dipole-dipole attraction: it forms when a hydrogen atom is bonded directly to fluorine, oxygen or nitrogen (the smallest and most electronegative atoms), and that hydrogen is attracted by the lone pair of an electronegative atom of a neighbouring molecule.

Three worked examples

1. ${\text{NH}}_3$ — Lewis, VSEPR, polarity

Valence electrons: nitrogen 5, three hydrogens $3\times 1$, total 8, that is 4 pairs. Skeleton: N in the centre, three H bonded with single bonds (3 shared pairs, 6 electrons). One pair remains: it goes on the nitrogen as a lone pair, which completes its octet ($3\times 2$ bonding electrons $+2$ of the lone pair $=8$).

VSEPR notation: ${\text{AX}}_3\text{E}$ — 3 bonds, 1 lone pair, so 4 groups in total around the nitrogen. Geometry: trigonal pyramidal, bond angle ~107° (compressed compared with the 109.5° of the tetrahedron because of the bulk of the lone pair).

Polarity: the three N–H bonds are polar (nitrogen is more electronegative than hydrogen), and the pyramidal geometry is asymmetric — the three dipole moments do not cancel, and the lone pair itself contributes a component of its own. ${\text{NH}}_3$ is polar.

2. ${\text{CO}}_2$ — why it is non-polar despite the polar bonds

Valence electrons: carbon 4, two oxygens $2\times 6$, total 16, that is 8 pairs. Skeleton: C in the centre, one oxygen on each side. With single bonds there would be too few electrons to complete the carbon’s octet: two double bonds form, $\text{O}=\text{C}=\text{O}$, one on each side (each oxygen gives up a lone pair to form the second bond with the carbon). Carbon thus reaches 4 bonds in total (two double), the oxygens complete the octet with two remaining lone pairs each.

VSEPR notation: ${\text{AX}}_2$ — 2 groups around the carbon (each double bond counts as a single group, not two). Geometry: linear, angle 180°.

Polarity: the two $\text{C}=\text{O}$ bonds are polar, but since the molecule is linear and the two oxygens are identical, the two dipole moments are equal in magnitude and opposite in direction: they cancel vectorially. ${\text{CO}}_2$ is non-polar, even though it has polar bonds — it is the example that the test almost always asks you to argue in words, not only to draw.

An honest note: our simulations for high school cover maths and physics, not yet chemistry. If in Year 3 you are also keeping physics and maths going, there you find a mock test per topic: the correct answer to every question is free, the guided solution unlocks with the first lesson.

3. ${\text{SO}}_2$ — a case with a lone pair on the centre

Valence electrons: sulfur 6, two oxygens $2\times 6$, total 18, that is 9 pairs. Skeleton: S in the centre, one oxygen on each side. With one double bond $\text{S}=\text{O}$ and one single bond $\text{S}-\text{O}$ the octets of both oxygens are completed; on the sulfur one lone pair remains (sulfur, from the third period, also tolerates more than 8 electrons, but here 8 are enough: two from the single bond, four from the double, two from the lone pair).

VSEPR notation: ${\text{AX}}_2\text{E}$ — 2 bonds (counting the double as one group) plus 1 lone pair on the sulfur, so 3 groups in total. With 3 groups the basic arrangement is trigonal planar, but with a lone pair the geometry of the molecule (defined only by the visible atoms) is bent, with a bond angle slightly below the theoretical 120° because of the repulsion of the lone pair.

Polarity: bent, asymmetric geometry, polar $\text{S}=\text{O}$ and $\text{S}-\text{O}$ bonds (oxygen is more electronegative than sulfur): the dipole moments do not cancel. ${\text{SO}}_2$ is polar.

Comparing boiling points: ${\text{H}}_2\text{O}$ versus ${\text{H}}_2\text{S}$

${\text{H}}_2\text{O}$ boils at 100 °C, ${\text{H}}_2\text{S}$ (same bent geometry, same group as oxygen in the periodic table) boils at about $-60$ °C: almost 160 degrees of difference, even though ${\text{H}}_2\text{S}$ has a greater molecular mass (34 u against 18 u), which would usually favour stronger London forces and so a higher boiling point, not a lower one.

The difference is explained by the hydrogen bond: in ${\text{H}}_2\text{O}$ the hydrogen is bonded to oxygen, one of the atoms that form hydrogen bonds; in ${\text{H}}_2\text{S}$ the hydrogen is bonded to sulfur, which is too large and not electronegative enough to give a significant hydrogen bond. Water molecules therefore stay “stuck” to one another much more strongly than those of ${\text{H}}_2\text{S}$, and much more energy is needed to separate them and make them boil. It is the example to keep in mind every time a boiling point seems “wrong” relative to molecular mass: first of all you have to check whether a hydrogen bond is present.

Typical mistakes

• Confusing the number of atoms with the number of electron groups. A double or triple bond counts as a single group in VSEPR notation, not as two or three. Whoever counts electrons rather than groups systematically gets the geometry of ${\text{CO}}_2$ and the like wrong.
• Forgetting the lone pairs in the count of groups. The “visible” geometry (the one with a name: linear, bent, pyramidal…) depends only on the atoms, but the number of groups that decides the angles also includes the lone pairs on the centre.
• Confusing polar bond and polar molecule. They are two different questions: the first concerns a single bond ($\Delta EN$), the second the whole molecule (bonds plus geometry). ${\text{CO}}_2$ is the example to cite always to show that they can have opposite answers.
• Ignoring the hydrogen bond when comparing boiling points. A lighter molecule with hydrogen bonding can boil at a much higher temperature than a heavier one without: molecular mass alone is not enough to predict the boiling point.

How I work on this topic in lessons

With Year 3 students I always start from the hand-drawn Lewis structure, step by step, before naming any geometry: the shape of the molecule is a logical consequence of the Lewis scheme, not a new list of names to learn by heart in parallel. How I work on chemistry and biology you will find on the science page. To count the valence electrons that go into the scheme you need electron configuration: Electron configuration and quantum numbers: the method.

If the chapter on the type of bond (ionic, covalent, metallic) is not yet solid, start from Chemistry in Year 3 in the human-sciences track: atoms and bonds. If instead you need to revise the mole and stoichiometric calculations, which in many classes are done before bonds or in parallel, you will find the method in The mole and stoichiometry: the method in Year 3, and to give the correct name to the compounds you draw with Lewis there is Chemical nomenclature: traditional and IUPAC without mixing them up.

If you want to set up geometry and polarity well before the test, or already have a marked test to go over together, start with a half-hour free introductory call to work out where to concentrate your revision.