3.5 Shapes of Molecules

11. Shapes of Molecules — A Level Chemistry
🎯 Learning Outcomes

Candidates should be able to:

  1. State and explain the shapes of, and bond angles in, molecules by using VSEPR theory, including CO₂ (linear, 180°), BF₃ (trigonal planar, 120°), CH₄ (tetrahedral, 109.5°), NH₃ (pyramidal, 107°), H₂O (non-linear, 104.5°), PF₅ (trigonal bipyramidal, 120° and 90°) and SF₆ (octahedral, 90°).
  2. Predict the shapes of, and bond angles in, molecules and ions analogous to those listed above.
1 VSEPR Theory — The Basics ▾
VSEPR Theory Valence Shell Electron Pair Repulsion theory. Because all electrons carry the same (negative) charge, pairs of electrons around a central atom repel each other. This repulsion pushes the electron pairs as far apart as possible, and it is this arrangement of electron pairs that determines the shape of the molecule.

The valence shell electrons are the electrons in the main outer shell of an atom. The shape and bond angles of a covalently bonded molecule depend on:

  • the number of pairs of electrons around the central atom, and
  • whether those pairs are lone pairs or bonding pairs.
Key Rules
  • Pairs of electrons repel each other because they carry the same charge.
  • A lone pair repels other pairs more strongly than a bonding pair does.
  • Repulsion from a double or triple bond is treated the same as a single bond for working out shape — except that repulsion between two double bonds is greater than between two single bonds.
  • The most stable shape is the one that minimises the forces of repulsion between all the electron pairs.
⚠ Order of Repulsion Lone pair–lone pair (most repulsion) > lone pair–bond pair > bond pair–bond pair (least repulsion).
O H H Greatest repulsion between lone pairs Intermediate repulsion (lone pair – bond pair) H–O–H angle is 104.5° Least repulsion between bonding pairs
In water, the two lone pairs on oxygen repel each other more strongly than they repel the bonding pairs, which in turn repel each other the least — compressing the H–O–H angle to 104.5°
2 Shapes With No Lone Pairs ▾

When every electron pair around the central atom is a bonding pair, the pairs spread out as evenly as possible, giving these three classic shapes:

O C O 180° Carbon dioxide, CO₂ — linear
2 bonding regions, 0 lone pairs: linear, bond angle 180°
B F F F 120° Boron trifluoride, BF₃ — trigonal planar
3 bonding pairs, 0 lone pairs: trigonal planar, bond angle 120°
C H H H H 109.5° Methane, CH₄ — tetrahedral
4 bonding pairs, 0 lone pairs: tetrahedral, bond angle 109.5°
3 Shapes With Lone Pairs ▾

Ammonia and water both start from the same tetrahedral arrangement of four electron pairs as methane — but one or two of those pairs are lone pairs instead of bonding pairs, which changes the observed shape and squeezes the bond angle.

N H H H 107° Ammonia, NH₃ — pyramidal (1 lone pair)
3 bonding pairs + 1 lone pair: pyramidal, bond angle 107° (compressed from 109.5° by the extra repulsion from the lone pair)
O H H 104.5° Water, H₂O — non-linear (bent) (2 lone pairs)
2 bonding pairs + 2 lone pairs: non-linear (bent/V-shaped), bond angle 104.5° (compressed further still by two lone pairs)
Why the Angle Shrinks Going from CH₄ (109.5°) → NH₃ (107°) → H₂O (104.5°), each extra lone pair adds more repulsion that pushes the remaining bonding pairs closer together, reducing the bond angle by roughly 2–3° each time.
4 Expanded Octet Shapes ▾

Elements in Period 3 and beyond can accommodate more than 8 electrons around the central atom, giving rise to two further shapes with no lone pairs.

P F F F F F 120° 90° Phosphorus pentafluoride, PF₅ trigonal bipyramidal
5 bonding pairs, 0 lone pairs: trigonal bipyramidal — 120° between the 3 equatorial atoms, 90° between an axial and an equatorial atom
S F F F F F F 90° Sulfur hexafluoride, SF₆ — octahedral
6 bonding pairs, 0 lone pairs: octahedral, bond angle 90° throughout
5 Predicting Shapes of Related Molecules & Ions ▾

The same seven reference shapes let you predict the shape of a molecule or ion you haven’t seen before — just work out how many electron pairs surround the central atom, and how many of those are lone pairs.

Method
  1. Draw the dot-and-cross structure (or count valence electrons) to find the total number of electron pairs around the central atom.
  2. Identify how many of those pairs are bonding pairs and how many are lone pairs.
  3. Match this combination to the reference shape with the same number of bonding pairs and lone pairs.

Worked Examples

SpeciesBonding / Lone PairsShape (like…)
NH₄⁺4 bonding, 0 loneTetrahedral, 109.5° (like CH₄)
H₃O⁺3 bonding, 1 lonePyramidal, ~107° (like NH₃)
SO₃3 bonding, 0 loneTrigonal planar, 120° (like BF₃)
PCl₅5 bonding, 0 loneTrigonal bipyramidal (like PF₅)
[SiF₆]²⁻6 bonding, 0 loneOctahedral, 90° (like SF₆)
📝 Exam Tips & Quick Summary
✔ Do
  • Count all electron pairs around the central atom first, then classify each as bonding or lone.
  • Remember lone pairs repel more strongly, and compress bond angles when present.
  • Treat a multiple bond the same as a single bond when counting regions of electron density.
  • State both the shape name and the bond angle — both are usually needed for full marks.
✘ Don’t
  • Don’t forget to include lone pairs on the central atom when working out the shape.
  • Don’t say NH₃ and CH₄ have the same shape — the extra lone pair makes NH₃ pyramidal, not tetrahedral.
  • Don’t give a single “90°” for PF₅ — trigonal bipyramidal has two different bond angles (120° and 90°).
  • Don’t confuse the shape of the electron pairs with the shape of the molecule — they’re the same only when there are no lone pairs.
Quick Recap — The 7 Reference Shapes
  1. CO₂: linear, 180° (2 bonding, 0 lone)
  2. BF₃: trigonal planar, 120° (3 bonding, 0 lone)
  3. CH₄: tetrahedral, 109.5° (4 bonding, 0 lone)
  4. NH₃: pyramidal, 107° (3 bonding, 1 lone)
  5. H₂O: non-linear/bent, 104.5° (2 bonding, 2 lone)
  6. PF₅: trigonal bipyramidal, 120° and 90° (5 bonding, 0 lone)
  7. SF₆: octahedral, 90° (6 bonding, 0 lone)

Similar Posts

Leave a Reply

Your email address will not be published. Required fields are marked *