Predict analogous molecular shapes and bond angles
Transfer the VSEPR electron-region method from named examples to analogous molecules and ions.
Section 3 · Chemical bonding
AS Level; Cambridge International Chemistry 9701, 2025–2027 syllabus. Use the supplied Pauling values when given, show particle and bond reasoning, and distinguish intramolecular bonds from intermolecular forces. Written practice is self-checked
Before you begin
- Read element symbols and simple molecular formulae; use the Periodic Table.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What should be counted around the central atom in VSEPR?
What you will learn
- Predict shapes and bond angles for species analogous to the syllabus examples
- Distinguish a bonding region from a lone pair
Use the same region-counting method for new species
For an unfamiliar molecule or ion analogous to the syllabus examples, draw a Lewis structure, count regions around the central atom, and arrange them to reduce repulsion. A multiple bond counts as one region for this VSEPR step.
Include lone pairs in the electron-region count, then ignore them when naming the positions of atoms. For ions, include the charge when counting valence electrons and use brackets or a charge label when representing the species.
Bond angles depend on which regions are present
Species with the same number and type of electron regions are analogous to a named geometry. Lone pairs can compress bond angles. Compare with the relevant syllabus example and state whether the angle is the ideal geometry or a compressed value.
When the question gives a specific molecule or ion beyond the memorised examples, explain the prediction from its electron regions instead of guessing by formula appearance.
Pause and explain
How many VSEPR regions does a double bond occupy?
Worked example
Predict the shape of NH₄⁺ using VSEPR.
Show the worked solution
- Count valence electrons including the positive charge: N has four H bonds and no remaining lone pair.
- There are four bonding regions around nitrogen.
- Four regions give a tetrahedral arrangement and shape.
Answer NH₄⁺ is tetrahedral, analogous to CH₄, with an ideal angle near 109.5°.
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Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Predict the shape of NH₄⁺ and explain how its charge affects the electron count.
Give me a hint
Count valence electrons for the whole ion before assigning lone pairs.
Compare my reasoning
- The positive charge means one fewer valence electron than neutral NH₃ plus an added H atom would have.
- Four N–H bonds leave no lone pair on nitrogen.
- Four regions form a tetrahedral shape.
NH₄⁺ is tetrahedral. Counting the positive charge leaves four N–H bonding regions and no lone pair on nitrogen.
Look for these in your work
- Charge included
- Lone-pair count correct
- Shape derived
A central atom has three bonded atoms and one lone pair. Predict the molecular shape and compare it with the ideal four-region arrangement.
Give me a hint
Name atom positions, not the lone pair, in the molecular shape.
Compare my reasoning
- There are four electron regions total.
- The electron regions are approximately tetrahedral.
- Excluding the lone pair leaves a pyramidal molecular shape with a compressed angle.
The molecular shape is pyramidal. Four electron regions are approximately tetrahedral, but the lone pair compresses the bonded-atom angle.
Look for these in your work
- Four regions counted
- Pyramidal molecular shape stated
- Angle compression explained
A species has two bonding regions and two lone pairs around its central atom. Predict its shape and compare it with the water example.
Give me a hint
Use the water electron-region pattern.
Compare my reasoning
- Four regions arrange approximately tetrahedrally.
- Two positions are occupied by lone pairs.
- The two bonded atoms form a non-linear molecule or ion with a compressed angle.
It is non-linear, analogous to H₂O: two lone pairs compress the angle between the two bonded atoms.
Look for these in your work
- Region arrangement identified
- Molecular shape stated
- Analogy and lone-pair effect linked
Common mistakes
- Forgetting to include ionic charge in electron counting
- Counting each line in a multiple bond as a separate VSEPR region
- Including lone pairs as atoms in the molecular shape name
What can you explain now?
What should be counted around the central atom in VSEPR?
Compare with the explanation
Bonding regions and lone pairs
Both occupy electron regions and affect repulsion.
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Make it stick
What extra counting step is needed when the unfamiliar species is an ion?
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.