Last updated: 16 August 2026
A displayed formula tells us which atoms are connected, but it does not tell us where those atoms actually sit in three-dimensional space. CO₂ and H₂O both have two atoms bonded to a central atom, yet CO₂ is straight while H₂O is bent. CH₄, meanwhile, is not a flat cross even though it is often drawn that way on paper.
The missing idea is that the electrons around the central atom occupy space. Because electron regions repel one another, they arrange themselves in particular three-dimensional patterns. Once you learn to identify those regions and compare their repulsions, molecular shapes become something you can work out rather than simply memorise.
Diagnostic Quiz
Try these five questions before reading the chapter. They check the electron-pair ideas that you will use to predict molecular shapes and bond angles.
- 5 quick questions
- Tests the ideas behind VSEPR
- Every answer option is explained
What do molecular shape and bond angle mean?
The shape of a molecule describes the three-dimensional arrangement of its atoms. A formula drawn on a flat page is therefore not necessarily a picture of the molecule's actual shape.
Methane, CH₄, is a good example. Its displayed formula may be drawn with four C–H bonds forming a flat cross, but the real molecule is three-dimensional. The four hydrogen atoms point towards the corners of a tetrahedron around the carbon atom.
A bond angle is the angle between two bonds meeting at the same atom. For example:
- O–C–O in CO₂ is 180°;
- F–B–F in BF₃ is 120°;
- H–C–H in CH₄ is approximately 109.5°; and
- H–O–H in H₂O is approximately 104.5°.
The question behind every shape
Ask: How many regions of electron density surround the central atom, and how can they arrange themselves to minimise repulsion?
VSEPR theory: why electron regions determine shape
VSEPR stands for Valence Shell Electron Pair Repulsion. The model begins with a simple electrostatic idea: electrons are negatively charged, so regions containing electrons repel one another.
Around a central atom, the different regions of electron density therefore arrange themselves so that the overall repulsion is as small as possible. In symmetrical cases, this corresponds to placing the regions as far apart as the geometry allows.
The VSEPR principle
Regions of electron density around a central atom repel one another and adopt an arrangement that minimises the repulsion between them.
What counts as one electron region?
When deciding the basic shape around a central atom:
- a single bond counts as one bonding region;
- a double bond counts as one bonding region;
- a triple bond counts as one bonding region; and
- a lone pair on the central atom counts as one non-bonding region.
This distinction is especially important for CO₂. Each C=O bond contains two shared pairs of electrons, but both pairs are directed towards the same oxygen atom. Each C=O double bond therefore forms one region of electron density around carbon.
Shared pairs are not the same as VSEPR regions
Carbon in CO₂ participates in four shared electron pairs altogether, but there are only two directions of bonding electron density around the carbon atom. The molecule is therefore linear rather than tetrahedral.
Why lone pairs change bond angles
Not all electron regions repel neighbouring regions equally strongly.
A bonding pair is attracted to the nuclei of two atoms. Its electron density is therefore concentrated substantially in the region between those nuclei.
A lone pair is not shared with another atom. It is associated mainly with the central atom and occupies more space around that central atom. It therefore exerts a stronger repulsion on neighbouring electron regions.
Order of electron-pair repulsion
lone pair–lone pair > lone pair–bonding pair > bonding pair–bonding pair
This order explains one of the most important trends in molecular shape:
CH₄, 109.5° > NH₃, 107° > H₂O, 104.5°
All three species have four electron regions around their central atoms. The difference is the number of lone pairs. As lone pairs replace bonding pairs, the stronger lone-pair repulsions squeeze the remaining bonds closer together.
Do not memorise a “subtract 2.5°” rule
The values 109.5°, 107° and 104.5° are useful examples, but the decrease is not a universal arithmetic rule. VSEPR explains the trend qualitatively through differences in electron-pair repulsion. Other molecules with the same general shape need not have exactly the same bond angle.
How to predict the shape of a simple molecule
Step 1: Identify the central atom
Find the atom around which the other atoms are arranged. For the syllabus examples:
- carbon is central in CO₂ and CH₄;
- boron is central in BF₃;
- nitrogen is central in NH₃;
- oxygen is central in H₂O; and
- sulfur is central in SF₆.
Step 2: Determine the bonding and lone pairs
Use the electronic structure, dot-and-cross diagram or Lewis structure to determine:
- how many atoms are bonded to the central atom; and
- how many lone pairs remain on the central atom.
Do not guess the number of lone pairs simply from the molecular formula.
Step 3: Count the electron regions
Add the bonding regions and lone pairs around the central atom. Remember that a multiple bond counts as one bonding region for the basic VSEPR arrangement.
Step 4: Arrange the electron regions to minimise repulsion
The arrangements required for the syllabus examples are:
- 2 regions → linear;
- 3 regions → trigonal planar;
- 4 regions → tetrahedral; and
- 6 regions → octahedral.
Step 5: Name the molecular shape from the positions of the atoms
Lone pairs affect the positions of the bonded atoms, but the molecular shape itself is named using the positions of the atoms.
This distinction matters for NH₃ and H₂O. Both have four electron regions and therefore a tetrahedral electron-region arrangement, but neither molecule has a tetrahedral molecular shape because some of those regions are lone pairs.
Step 6: State and explain the bond angle
Give the expected angle and, where lone pairs are present, explain why it is compressed relative to the ideal arrangement.
A strong examination explanation
State the number of bonding regions and lone pairs around the central atom, explain that these regions repel one another and arrange themselves to minimise repulsion, identify the molecular shape, and then state the bond angle. If lone pairs are present, explain their stronger repulsion.
CO₂: linear shape and a 180° bond angle
Carbon dioxide has the structure:
O=C=O
The central carbon atom forms two double bonds and has no lone pairs.
Although each double bond contains two shared electron pairs, each C=O bond counts as one region of electron density. Carbon therefore has:
- 2 bonding regions; and
- 0 lone pairs.
Two regions minimise their mutual repulsion by pointing directly away from one another.
CO₂
- Electron regions around carbon: 2
- Bonding regions: 2
- Lone pairs on carbon: 0
- Shape: linear
- O–C–O bond angle: 180°
A species with the same pattern of two bonding regions and no lone pairs around its central atom will also have a linear basic geometry.
BF₃: trigonal planar shape and 120° bond angles
In BF₃, boron forms three B–F bonds and has no lone pairs on the central atom.
There are therefore three regions of electron density around boron. To minimise repulsion, the three bonding regions spread out evenly in one plane.
The fluorine atoms occupy the corners of an equilateral triangle around boron.
BF₃
- Electron regions around boron: 3
- Bonding regions: 3
- Lone pairs on boron: 0
- Shape: trigonal planar
- F–B–F bond angles: 120°
Because all three bonding regions are equivalent, the three B–F bonds are arranged symmetrically around the central boron atom.
The tetrahedral family: CH₄, NH₃ and H₂O
CH₄, NH₃ and H₂O provide the clearest demonstration of why you must distinguish between an electron-region arrangement and a molecular shape.
Each has four electron regions around its central atom, so in all three cases those regions are based on a tetrahedral arrangement. What changes is how many of those four regions contain bonds and how many contain lone pairs.
CH₄: tetrahedral
Carbon has four outer-shell electrons and forms four C–H bonds in methane. There are four bonding regions and no lone pairs on the carbon atom.
If the four bonds were arranged in a flat square, some would be only 90° apart. The bonds can get farther away from one another by pointing into three-dimensional space towards the corners of a tetrahedron.
CH₄
- Electron regions around carbon: 4
- Bonding regions: 4
- Lone pairs on carbon: 0
- Electron-region arrangement: tetrahedral
- Molecular shape: tetrahedral
- H–C–H bond angle: 109.5°
NH₃: trigonal pyramidal
Nitrogen has five outer-shell electrons. In NH₃, three of these electrons are used to form three N–H bonds, leaving one lone pair on nitrogen.
Nitrogen therefore has:
- 3 bonding regions; and
- 1 lone pair.
These four electron regions adopt a tetrahedral arrangement. However, when the molecular shape is named, only the positions of the atoms are considered.
The three hydrogen atoms form a triangular base, with nitrogen above it, giving a trigonal pyramidal molecule.
The lone pair repels the bonding pairs more strongly than the bonding pairs repel one another. The N–H bonds are therefore pushed slightly closer together.
NH₃
- Electron regions around nitrogen: 4
- Bonding regions: 3
- Lone pairs on nitrogen: 1
- Electron-region arrangement: tetrahedral
- Molecular shape: trigonal pyramidal
- H–N–H bond angle: approximately 107°
H₂O: bent or V-shaped
Oxygen has six outer-shell electrons. In H₂O, oxygen forms two O–H bonds and retains two lone pairs.
Oxygen therefore has:
- 2 bonding regions; and
- 2 lone pairs.
Again, the four electron regions are based on a tetrahedral arrangement. But when the lone pairs are omitted from the molecular shape, only the two hydrogen atoms remain visible around oxygen.
The molecule is therefore bent, also described as V-shaped.
H₂O has two lone pairs rather than one. The stronger lone pair–lone pair and lone pair–bonding pair repulsions compress the O–H bonds even more strongly than in NH₃.
H₂O
- Electron regions around oxygen: 4
- Bonding regions: 2
- Lone pairs on oxygen: 2
- Electron-region arrangement: tetrahedral
- Molecular shape: bent (V-shaped)
- H–O–H bond angle: approximately 104.5°
Why 109.5° becomes 107° and then 104.5°
CH₄ has four bonding regions and no lone pairs, so its H–C–H angles are 109.5°.
NH₃ replaces one bonding region with a lone pair. The stronger lone pair–bonding pair repulsion compresses the H–N–H angle to about 107°.
H₂O has two lone pairs. The still greater effect of the lone-pair repulsions compresses the H–O–H angle further to about 104.5°.
SF₆: octahedral shape and 90° bond angles
In SF₆, sulfur is surrounded by six fluorine atoms. There are six S–F bonding regions and no lone pairs on the sulfur atom.
The six bonding regions minimise repulsion by adopting an octahedral arrangement.
Four fluorine atoms lie in a square plane around sulfur. A fifth fluorine lies above this plane and a sixth lies below it.
SF₆
- Electron regions around sulfur: 6
- Bonding regions: 6
- Lone pairs on sulfur: 0
- Shape: octahedral
- Angle between adjacent S–F bonds: 90°
- Angle between opposite S–F bonds: 180°
Why is it called octahedral if there are six fluorine atoms?
The name comes from the geometry of an octahedron, which has eight triangular faces. An octahedral molecule has six positions around the central atom: four in one plane and two perpendicular to that plane.
Summary of shapes and bond angles
| Molecule | Bonding regions | Lone pairs on central atom | Electron-region arrangement | Molecular shape | Bond angle |
|---|---|---|---|---|---|
| CO₂ | 2 | 0 | Linear | Linear | 180° |
| BF₃ | 3 | 0 | Trigonal planar | Trigonal planar | 120° |
| CH₄ | 4 | 0 | Tetrahedral | Tetrahedral | 109.5° |
| NH₃ | 3 | 1 | Tetrahedral | Trigonal pyramidal | ≈107° |
| H₂O | 2 | 2 | Tetrahedral | Bent (V-shaped) | ≈104.5° |
| SF₆ | 6 | 0 | Octahedral | Octahedral | 90° and 180° |
Using VSEPR to predict analogous molecules
Examination questions do not have to use exactly the six molecules above. Once you understand the electron-region pattern, you can apply the same reasoning to an unfamiliar but analogous species.
For example, suppose you are asked to predict the shape of PCl₃.
Phosphorus has five outer-shell electrons. It forms three P–Cl bonds and retains one lone pair. Around phosphorus there are therefore:
- 3 bonding regions;
- 1 lone pair; and
- 4 electron regions altogether.
The electron regions are arranged tetrahedrally, but the positions of the three chlorine atoms give a trigonal pyramidal molecular shape.
Analogous shape does not mean identical bond angle
PCl₃ and NH₃ both contain three bonding regions and one lone pair on their central atoms, so both are trigonal pyramidal. However, you should not automatically assign the exact NH₃ bond angle of 107° to every trigonal pyramidal molecule. The precise angle also depends on the atoms and the distribution of electron density.
What if the central atom forms a multiple bond?
For the basic shape, a double or triple bond is still counted as one region of electron density.
However, a multiple bond generally contains more electron density than a single bond and may repel neighbouring electron regions somewhat more strongly. This can cause deviations from ideal bond angles in less symmetrical molecules.
VSEPR should therefore be treated as a model that predicts the broad arrangement extremely well, rather than a machine that calculates every exact bond angle.
Common mistakes and why they are wrong
Mistake 1: Predicting the shape directly from the displayed formula
A displayed formula is normally two-dimensional. You must determine the electron regions around the central atom before deciding the three-dimensional shape.
Mistake 2: Counting a double bond as two separate VSEPR regions
A double bond contains two shared pairs but both are directed towards the same bonded atom. It therefore counts as one bonding region when predicting the basic shape.
Mistake 3: Ignoring lone pairs on the central atom
Lone pairs occupy space and repel other electron regions. Ignoring them would incorrectly predict NH₃ as trigonal planar and H₂O as linear.
Mistake 4: Calling NH₃ tetrahedral
NH₃ has a tetrahedral electron-region arrangement, but one of those four regions is a lone pair. The arrangement of the atoms is trigonal pyramidal.
Mistake 5: Calling H₂O linear because it has two O–H bonds
Oxygen also has two lone pairs. The four electron regions are arranged tetrahedrally, producing a bent molecule with an H–O–H angle of about 104.5°.
Mistake 6: Saying that lone pairs attract bonding pairs
Electron regions are negatively charged and repel one another. Lone pairs repel nearby bonding pairs more strongly, pushing the bonds closer together.
Mistake 7: Assuming every tetrahedral-based molecule has a 109.5° bond angle
109.5° is the ideal angle when four equivalent bonding regions are arranged tetrahedrally, as in CH₄. Lone pairs or unequal bonding regions can alter the measured angle.
Mistake 8: Using the “subtract 2.5° per lone pair” idea as a general law
The CH₄, NH₃ and H₂O values happen to give a useful progression, but there is no universal 2.5° subtraction rule. The real reason for the decrease is the different strength of the electron-pair repulsions.
Mistake 9: Saying that SF₆ has only 90° bond angles
Adjacent S–F bonds are 90° apart, but bonds pointing in opposite directions form angles of 180°.
Molecular-shape checklist
Find the electron regions
- Identify the central atom.
- Determine its bonding regions.
- Determine its lone pairs.
- Count a multiple bond as one region.
Choose the arrangement
- 2 regions → linear.
- 3 regions → trigonal planar.
- 4 regions → tetrahedral.
- 6 regions → octahedral.
Account for lone pairs
- Lone pairs repel more strongly.
- NH₃ is trigonal pyramidal.
- H₂O is bent or V-shaped.
- Lone pairs compress nearby bond angles.
Finish the exam answer
- State bonding regions and lone pairs.
- Explain that electron regions repel.
- Name the molecular shape precisely.
- State and explain the bond angle.
Mastery Quiz
Now apply VSEPR reasoning to familiar and unfamiliar molecules. These questions focus on explaining shapes and comparing bond angles rather than simply recalling names.
- 10 application-focused questions
- Includes unfamiliar molecular examples
- Every answer option is explained
Frequently Asked Questions
Questions students often ask about molecular shapes
These questions deal with the small distinctions that make VSEPR explanations much more precise.
What is the difference between electron-region arrangement and molecular shape?
The electron-region arrangement includes both bonding regions and lone pairs around the central atom. Molecular shape describes only the arrangement of the atoms. NH₃ therefore has four electron regions arranged tetrahedrally, but because one region is a lone pair, the molecular shape formed by the atoms is trigonal pyramidal.
Why are lone pairs not included when naming the molecular shape?
Molecular shape describes the arrangement of atomic nuclei. Lone pairs affect where those nuclei are positioned, but a lone pair is not an atom. It is therefore included when working out the geometry but omitted when the final molecular shape is named.
Why does a double bond count as only one region of electron density?
Both shared pairs in the double bond lie in the same general direction from the central atom towards the same bonded atom. For basic VSEPR shape prediction, the whole multiple bond is therefore treated as one region of electron density.
Does a double bond repel exactly like a single bond?
Not necessarily. A multiple bond generally contains more electron density than a single bond and may repel neighbouring regions somewhat more strongly. However, both are still counted as one region when determining the basic electron-region arrangement.
Why isn't H₂O linear if oxygen forms only two bonds?
Oxygen also has two lone pairs. Together with the two O–H bonding regions, this gives four electron regions around oxygen. They adopt a tetrahedral arrangement, and the two visible O–H bonds therefore form a bent molecule rather than a straight line.
Why is NH₃ not trigonal planar?
Nitrogen has three N–H bonding regions and one lone pair. The lone pair is the fourth electron region, so the regions are based on a tetrahedral rather than trigonal planar arrangement. Considering only the atoms gives a trigonal pyramidal molecular shape.
Why do lone pairs reduce bond angles?
A lone pair occupies more space around the central atom than a bonding pair and therefore repels neighbouring electron regions more strongly. It pushes nearby bonding regions closer together, decreasing the angle between their bonds.
Should I memorise 2.5° less for every lone pair?
No. The progression from 109.5° in CH₄ to about 107° in NH₃ and 104.5° in H₂O is useful to know, but it is not a universal subtraction rule. The important explanation is that lone pairs repel more strongly than bonding pairs. Exact angles depend on the particular molecule.
Are all molecules with three bonding regions trigonal planar?
No. BF₃ has three bonding regions and no lone pairs, so it is trigonal planar. NH₃ also has three bonding regions, but nitrogen has an additional lone pair. Its four electron regions are arranged tetrahedrally, producing a trigonal pyramidal molecular shape.
Do analogous molecules always have exactly the same bond angle?
No. Molecules with the same numbers of bonding regions and lone pairs normally have the same broad VSEPR shape, but their exact bond angles can differ because the atoms and bonding electron distributions are different.
What should I write when an exam question asks me to explain a molecular shape?
State the number of bonding regions and lone pairs around the central atom. Explain that the electron regions repel one another and arrange themselves to minimise repulsion. Then give the molecular shape and bond angle. If lone pairs are present, explain that their stronger repulsion compresses the angle between the bonding regions.
Molecular shapes become much easier once you stop treating linear, trigonal planar, tetrahedral, trigonal pyramidal, bent and octahedral as unrelated names. Begin with the electronic structure, count the bonding regions and lone pairs around the central atom, and let electron-pair repulsion lead you to the geometry. The same reasoning then explains not only what shape a molecule has, but also why its bond angles have the values they do.