Shapes of molecules and ions

CIE A-Level Chemistry· Unit 3: Chemical Bonding· 25 min read

1. VSEPR Theory Fundamentals★★☆☆☆⏱ 5 min

📘 Definition

VSEPR Theory

VSEPR

Valence Shell Electron Pair Repulsion theory states that electron domains (groups of electrons) around a central atom arrange themselves to minimize repulsion between them, producing a predictable 3D shape.

Example:

A central atom with 2 electron domains will arrange them 180° apart.

Electron domains include both bonding pairs (any single, double or triple bond counts as one domain) and lone pairs of electrons. Repulsion strength follows a clear order: lone pair-lone pair > lone pair-bonding pair > bonding pair-bonding pair. This order explains deviations from ideal bond angles.

📐 Worked Example

Count the total number of electron domains around the central atom in $N H_{3}$

1
Identify the central nitrogen atom, which has 5 valence electrons. Each N-H bond uses 1 valence electron from N, so 3 electrons are used in bonding.
2
Remaining valence electrons on N: $5 - 3 = 2$ electrons, which form one lone pair.
3
Total electron domains = 3 bonding domains + 1 lone pair = 4 total domains.

Exam tip:

Remember that multiple bonds (double/triple) only count as one electron domain, even though they contain more electrons.

2. Common Shapes (2-4 Electron Domains)★★☆☆☆⏱ 8 min

For any number of electron domains, there is an ideal electron domain geometry, while molecular shape describes only the positions of the bonded atoms (ignoring lone pairs). 2 to 4 electron domains are the most frequently tested shapes in CIE exams.

📘 Definition

Molecular Geometry

The 3D arrangement of only bonded atoms around the central atom; lone pairs are not included in the shape name or description.

Total electron domains	Lone pairs	Electron geometry	Molecular shape	Ideal bond angle
2	0	Linear	Linear	180°
3	0	Trigonal planar	Trigonal planar	120°
3	1	Trigonal planar	Bent (V-shaped)	<120°
4	0	Tetrahedral	Tetrahedral	109.5°
4	1	Tetrahedral	Trigonal pyramidal	<109.5°
4	2	Tetrahedral	Bent (V-shaped)	<<109.5°

📐 Worked Example

Predict the shape and approximate bond angle of the sulfite ion $S O_{3}^{2 -}$

1
Central S atom has 6 valence electrons. Add 2 electrons to account for the 2- negative charge, giving 8 total valence electrons on S.
2
There are 3 S-O bonds, each uses 2 electrons total, leaving 2 unbonded electrons = 1 lone pair.
3
Total electron domains = 3 bonding domains + 1 lone pair = 4.
4
Electron domain geometry is tetrahedral; molecular shape (ignoring the lone pair) is trigonal pyramidal.
5
Lone pair-bonding pair repulsion pushes bonds closer, so the bond angle is ~107°, less than the ideal 109.5°.

3. Shapes (5-6 Electron Domains)★★★☆☆⏱ 7 min

Central atoms can have expanded octets (more than 8 valence electrons) to give 5 or 6 total electron domains, with their own characteristic geometries. Lone pairs always occupy positions that minimize repulsion.

Total electron domains	Lone pairs	Electron geometry	Molecular shape
5	0	Trigonal bipyramidal	Trigonal bipyramidal
5	1	Trigonal bipyramidal	See-saw
5	2	Trigonal bipyramidal	T-shaped
5	3	Trigonal bipyramidal	Linear
6	0	Octahedral	Octahedral
6	1	Octahedral	Square pyramidal
6	2	Octahedral	Square planar

📐 Worked Example

Predict the shape of xenon tetrafluoride $X e F_{4}$

1
Central Xe atom has 8 valence electrons. 4 electrons are used in 4 Xe-F bonds.
2
Remaining electrons = $8 - 4 = 4$ electrons = 2 lone pairs.
3
Total electron domains = 4 bonding + 2 lone pairs = 6.
4
Lone pairs occupy opposite axial positions to minimize 90° lone pair-lone pair repulsion.
5
Molecular shape (only atoms) is square planar, with 90° F-Xe-F bond angles.

Exam tip:

For 6 electron domains with 2 lone pairs, lone pairs are always opposite each other, never adjacent, to reduce repulsion.

4. Shapes of Polyatomic Ions★★★☆☆⏱ 5 min

The same VSEPR rules apply to polyatomic ions, the only adjustment is adding or subtracting electrons to account for the ion's overall charge. Negative charges add electrons to the central atom, while positive charges remove electrons, which changes the number of lone pairs.

📐 Worked Example

Predict the shape of the ammonium ion $N H_{4}^{+}$

1
Central N atom has 5 valence electrons. Subtract 1 electron to account for the +1 charge, giving 4 total valence electrons.
2
All 4 electrons form N-H bonding pairs, so there are 0 lone pairs.
3
Total electron domains = 4 bonding domains, 0 lone pairs = 4 total domains.
4
Molecular shape is tetrahedral, with an ideal bond angle of 109.5°.

5. Common Pitfalls

Wrong move:

Counting double or triple bonds as multiple electron domains

Why:

VSEPR counts any bond between two atoms as a single domain, regardless of bond order

Correct move:

Count every distinct bonded atom to the central atom as one bonding domain

Wrong move:

Forgetting to adjust the electron count for the charge of an ion

Why:

Charge changes the number of valence electrons on the central atom, which changes the number of lone pairs

Correct move:

Add 1 electron per negative charge, subtract 1 electron per positive charge when counting valence electrons

Wrong move:

Calling electron domain geometry the same as molecular shape when lone pairs are present

Why:

Examiners specifically ask for molecular shape (the shape of atoms only) in most questions

Correct move:

Always state molecular shape unless explicitly asked for electron domain geometry

Wrong move:

Stating the ideal 109.5° bond angle for water

Why:

Two lone pairs on oxygen cause greater repulsion than bonding pairs, pushing bonds closer together

Correct move:

State bond angle as ~104.5°, or less than 109.5° if an exact value is not required

Wrong move:

Placing two lone pairs adjacent in $X e F_{4}$

Why:

Adjacent lone pairs have high repulsion at 90°, which is less stable than opposite placement at 180°

Correct move:

Place two lone pairs opposite each other for 6 electron domains with 2 lone pairs

6. Quick Reference Cheatsheet

Total electron domains	0 lone pairs	1 lone pair	2 lone pairs	3 lone pairs
2	Linear (180°)
3	Trigonal planar (120°)	Bent (<120°)
4	Tetrahedral (109.5°)	Trigonal pyramidal (<109.5°)	Bent (<<109.5°)
5	Trigonal bipyramidal	See-saw	T-shaped	Linear
6	Octahedral	Square pyramidal	Square planar

7. Frequently Asked

Do I need to memorize all bond angles for the exam?

Yes, CIE examiners expect you to recall approximate bond angles for all common geometries from 2 to 6 electron domains, including deviations caused by lone pairs.

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2022 · 12
Identify shape of sulfate ion
2023 · 21
Explain bond angle in water
2021 · 13
Predict shape of XeF4

Going deeper

simulationPhET Molecule Shapes SimulationInteractive visualization of molecular shapes

What's Next

Mastering VSEPR prediction of molecular and ionic shapes is a high-yield foundational skill for CIE A-Level Chemistry, required for understanding molecular polarity, intermolecular forces, and stereochemistry in organic reactions. Exam questions regularly test your ability to draw shapes, state bond angles, and explain deviations from ideal angles, so consistent practice of this topic is key. Next, you will connect this knowledge to molecular polarity, which depends directly on 3D shape and bond dipole symmetry, and use these shape concepts in later topics like intermolecular forces and organic stereochemistry.