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v0.1.1436
NotesChemistry HLTopic 2.2Molecular shapes (VSEPR)
Back to Chemistry HL Topics
2.2.22 min read

Molecular shapes (VSEPR)

IB Chemistry • Unit 2

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Contents

  • Why molecules have shapes
  • Counting domains → the shape
  • Lone pairs squeeze the angle
  • Exam-style question
The big idea (VSEPR): Valence Shell Electron Pair Repulsion theory says: the electron domains around a central atom repel each other and spread out as far apart as possible.

An electron domain is any group of electrons: a single, double or triple bond each counts as one domain, and so does each lone pair. The arrangement that keeps them furthest apart fixes the shape and the bond angle.

4 bonding pairs, 0 lone pairs → tetrahedral, 109.5°.

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Count domains, not bonds: A double or triple bond is still one domain (one direction). So count the number of atoms bonded to the centre plus the number of lone pairs — that total decides the geometry.

Count the bonding domains and lone pairs on the central atom, then read off the shape and angle:

Bonding domainsLone pairsShapeBond angleExample
20Linear180°CO2, HCN
30Trigonal planar120°BF3
21Bent~118°SO2
40Tetrahedral109.5°CH4
31Trigonal pyramidal107°NH3
22Bent104.5°H2O
The two-step method: (1) Draw the Lewis structure → count bonding atoms + lone pairs on the centre. (2) Match that count to the table to name the shape and state the angle.

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A lone pair repels more strongly than a bonding pair, so it pushes the bonds closer together — shrinking the bond angle. Compare three molecules that all have four domains around the central atom:

3 bonding pairs + 1 lone pair → trigonal pyramidal, 107°.

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2 bonding pairs + 2 lone pairs → bent, 104.5°.

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The 109.5 → 107 → 104.5 pattern: All three have four electron domains, but the angle falls as lone pairs increase:

- CH_{4} — 0 lone pairs → 109.5° - NH_{3} — 1 lone pair → 107° - H_{2}O — 2 lone pairs → 104.5°

Each lone pair squeezes the angle by roughly 2–3°.
How this is tested: 'State and explain the shape / bond angle of …' is a routine Paper 2 question, and Paper 1A likes to ask you to order molecules by bond angle.

For full marks: give the shape name, the bond angle, and the reason (number of bonding domains and lone pairs, and lone-pair repulsion if relevant).

IB-style question — hydrogen cyanide (a)

(a) State and explain the molecular shape of hydrogen cyanide, HCN. [2]

How to score the marks

  1. From the Lewis structure (H–C≡N), the central carbon is bonded to 2 atoms and has 0 lone pairs.
  2. Two electron domains repel to 180° apart.
  3. So HCN is linear with a bond angle of 180°. (The triple bond counts as one domain.)

Final answer

Linear, 180° — carbon has two bonding domains and no lone pairs.

2 bonding domains around C (the triple bond counts as one) → linear, 180°.

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IB-style question — sulfur dioxide (b)

(b) Deduce the electron-domain geometry and the molecular shape of sulfur dioxide, SO2. [2]

How to score the marks

  1. From the Lewis structure, sulfur has 2 bonding domains (to the two O atoms) and 1 lone pair → 3 electron domains in total.
  2. Three domains give a trigonal planar electron-domain geometry.
  3. But one domain is a lone pair, so the molecular shape is bent, with an angle of about 119° (slightly less than 120° because the lone pair repels more).

Final answer

Electron-domain geometry trigonal planar; molecular shape bent (~119°).

2 bonding domains + 1 lone pair on S → bent, ~119°.

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The molecules BF3 and NF3 both have a central atom bonded to three other atoms.

why BF3 is trigonal planar (120°) but NF3 is trigonal pyramidal (~102°).
[2 marks]

Related Chemistry HL Topics

Continue learning with these related topics from the same unit:

2.1.1Formation of ions and ionic bonding
2.1.2Formulas and names of ionic compounds
2.1.3Ionic lattices and their properties
2.2.1Covalent bonding and Lewis structures
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