Chemical Bonding and Molecular Structure · Lesson 6 of 12
Predicting Molecular Shape with VSEPR
“Use electron-domain repulsion to predict three-dimensional shapes and explain how lone pairs alter bond angles.”
• Apply the main postulates of valence-shell electron-pair repulsion theory. • Distinguish electron-domain arrangement from molecular shape. • Predict shapes with two to six electron domains, including lone-pair cases. • Explain angle compression and equatorial lone-pair placement using relative repulsions.
A flat drawing is not the molecular shape
A Lewis drawing identifies bonds and lone pairs, but the positions on the page do not determine the real three-dimensional arrangement. Four lines drawn around carbon could make a cross on paper, even though methane is tetrahedral. We need a model that explains how the electron regions arrange in space.
Valence-shell electron-pair repulsion theory predicts arrangements around a central atom by placing electron domains as far apart as their geometry allows.
An electron domain is a region of electron density around the central atom: a lone pair or a bond to a neighbouring atom. A double or triple bond counts as one domain when predicting the basic arrangement, because its shared electrons lie in the same general direction. Its repulsion can differ from that of a single bond, so this counting step is a starting model rather than an exact angle calculation.
Repulsion establishes the basic arrangement
Electron-rich regions repel one another. With two domains, opposite directions give the greatest separation. With three, an arrangement in one plane at 120° spreads the domains evenly. Four domains gain more separation by pointing into three dimensions towards tetrahedral corners than by remaining in a flat square.
| Total domains | Electron-domain arrangement | Ideal angles | Example without central lone pairs |
|---|---|---|---|
| 2 | Linear | 180° | BeCl₂, HgCl₂ |
| 3 | Trigonal planar | 120° | BF₃ |
| 4 | Tetrahedral | About 109.5° | CH₄, ammonium |
| 5 | Trigonal bipyramidal | 90°, 120°, 180° | PCl₅ |
| 6 | Octahedral | 90°, 180° | SF₆ |
In a trigonal bipyramid, three equatorial domains lie in one plane and two axial domains lie above and below it. These positions are not equivalent: an axial domain makes three 90° contacts with equatorial domains. In an octahedron, four directions can be viewed as a square plane with one direction above and one below.
Problem
Predict the arrangement around boron in BF₃.
- 1.The Lewis structure has three B–F bonding domains and no lone pair on boron.
- 2.Three domains arrange in a trigonal plane to maximise their separation. Adjacent ideal angles are 120°.
- 3.Since every domain corresponds to a bonded atom, the electron-domain arrangement and molecular shape are both trigonal planar.
Lone pairs change the visible shape
An electron-domain arrangement includes lone pairs, but molecular shape names describe the positions of the atoms. We therefore first count all domains, then name the arrangement of the bonded atoms. Removing the lone-pair markers from a diagram does not mean the electron pairs vanish; they still influence the geometry.
A bonding pair is shared between nuclei, whereas a lone pair is concentrated around one atom. A lone pair therefore tends to occupy more angular space and exert a stronger repulsion on neighbouring domains. The usual order is lone pair–lone pair greater than lone pair–bond pair greater than bond pair–bond pair.
Methane has four bonds and no central lone pairs, so its molecular shape is tetrahedral. Ammonia has three bonds and one lone pair: the electron-domain arrangement is tetrahedral but the shape is trigonal pyramidal. Water has two bonds and two lone pairs: its shape is bent.
Problem
Explain the approximate bond-angle sequence CH₄, NH₃, H₂O.
- 1.All three species have four electron domains around the central atom in the basic VSEPR treatment.
- 2.CH₄ has only bonding domains and an angle near 109.5°. NH₃ has a lone pair, whose stronger repulsion pushes the N–H bonds closer, giving about 107°.
- 3.H₂O has two lone pairs. Their stronger repulsions compress the H–O–H angle further, to about 104.5°.
- 4.The sequence is explained qualitatively by changing domain types. VSEPR does not obtain the exact measured angles merely by subtracting a fixed amount for each lone pair.
For SO₂, the two S–O bond directions and a central lone pair give three domains. The electron arrangement is approximately trigonal planar, but the shape is bent, with an angle slightly below 120°. Treat the S–O multiple-bond region as one directional domain on each side when counting.
Lone pairs in five- and six-domain arrangements
With five domains, deciding where the lone pairs go is part of the reasoning. In SF₄, placing the lone pair axially would give three 90° lone-pair–bond-pair interactions. An equatorial placement gives only two. The equatorial choice therefore reduces the particularly strong close-angle interactions.
Problem
Use five-domain arrangements to explain SF₄ and ClF₃.
- 1.SF₄ has four bonding domains and one lone pair, giving five domains in total. Put the lone pair equatorially.
- 2.The remaining four atoms form a see-saw shape, also called a distorted tetrahedral arrangement in some descriptions.
- 3.ClF₃ has three bonds and two lone pairs. Both lone pairs occupy equatorial sites, leaving two axial fluorines and one equatorial fluorine.
- 4.The three fluorine positions form a T-shaped molecule. Comparing possible placements is more informative than memorising the shape name alone.
Problem
Predict BrF₅ and XeF₄ from their six-domain arrangements.
- 1.BrF₅ has five bonding domains and one lone pair around bromine. The six-domain arrangement is octahedral.
- 2.One octahedral direction contains the lone pair. The five remaining atomic positions form a square pyramid.
- 3.XeF₄ has four bonding domains and two lone pairs. Place the two lone pairs opposite one another in the octahedral arrangement to minimise their mutual repulsion.
- 4.The four fluorines then lie in a square plane. The electron arrangement remains octahedral, while the molecular shape is square planar.
| Domain notation | Central bonds / lone pairs | Molecular shape | Example |
|---|---|---|---|
| AB₂E | 2 / 1 | Bent | SO₂ or O₃ |
| AB₃E | 3 / 1 | Trigonal pyramidal | NH₃ |
| AB₂E₂ | 2 / 2 | Bent | H₂O |
| AB₄E | 4 / 1 | See-saw | SF₄ |
| AB₃E₂ | 3 / 2 | T-shaped | ClF₃ |
| AB₅E | 5 / 1 | Square pyramidal | BrF₅ |
| AB₄E₂ | 4 / 2 | Square planar | XeF₄ |
In this notation A is the central atom, B represents a bonded neighbouring atom, and E represents a lone pair on A. The subscript counts how many are present; an omitted subscript means one. For example AB₂E₂ has four domains in total, not two. Multiple bonds do not increase the number of B directions.
A reliable prediction sequence
Start from a valid Lewis structure. Count the bonded directions and central lone pairs, choose the arrangement for the total domains, place lone pairs to reduce repulsion, and then name the arrangement of atoms. Finally, explain any expected angle distortion. Where resonance occurs, equivalent contributors must lead to the same essential domain arrangement.
Calling water “tetrahedral” without qualification confuses four electron domains with the positions of its three atoms. Its electron-domain arrangement is tetrahedral; its molecular shape is bent. Also avoid reading a perspective drawing as a literal flat molecular structure.
VSEPR is especially useful for many p-block compounds. It predicts shapes qualitatively, but is not a complete quantum-mechanical explanation and does not calculate every observed angle or bond energy. Its strength is a clear structural prediction from electron-domain repulsion.
Check your understanding
For each question, count all central domains first and name the molecular shape only after identifying lone pairs. Use “bonding domain” consistently when a multiple bond is present.
Quiz
What is the molecular shape of a species with two bonding domains and two central lone pairs?
How many domains does a central atom with two double bonds and no lone pairs have?
Which repulsion is usually strongest?
Why does the lone pair in SF₄ favour an equatorial position?
Which pair correctly matches electron arrangement and molecular shape for XeF₄?
Practice Problems
- Predict the shapes of BeCl₂, BCl₃, SiCl₄, and AsF₅, showing your central-domain count.
- Use electron-domain arrangements to explain why NH₃ and H₂O have different molecular shapes and angles.
- Draw projected trigonal-bipyramidal arrangements for SF₄ and ClF₃, marking lone pairs and identifying axial and equatorial sites.
- Predict BrF₅ and XeF₄ and explain why the lone pairs in XeF₄ are placed opposite one another.
- Explain why a double bond counts as one domain when selecting a basic electron arrangement.
- Use H₂S and PH₃ to identify bent and pyramidal shapes, while explaining why ideal tetrahedral angles should not be assumed to be exact measured values.
- A student draws methane as a flat square. Explain why a tetrahedral arrangement separates its four equivalent bonding domains more effectively.
Key Takeaways
• VSEPR arranges electron domains to reduce their repulsions. • Count a multiple bond as one domain when choosing the basic arrangement. • Molecular shape names refer to atoms; electron-domain arrangements also include lone pairs. • Stronger lone-pair repulsions help explain angle compression and site preference. • Two to six domains generate the linear, planar, tetrahedral, trigonal-bipyramidal, and octahedral families.