Shapes of molecules and intermolecular forces

What you'll learn

The shape of a molecule and the forces between molecules determine many of a substance's physical properties — its boiling point, its solubility, its viscosity and whether it is a gas, a liquid or a solid at room temperature. This guide covers valence shell electron pair repulsion (VSEPR) theory for predicting the shapes and bond angles of molecules and ions, the effect of lone pairs, how shape determines whether a molecule is polar, and the three main types of intermolecular force: London (dispersion) forces, permanent dipole–dipole forces and hydrogen bonding. It also covers the unusual properties of water that arise from hydrogen bonding, and introduces the kinetic theory and the behaviour of ideal gases, since intermolecular forces explain why real gases deviate from ideal behaviour.

CAPE questions regularly ask you to draw and name shapes with bond angles, to explain differences in boiling points, and to explain the anomalous properties of water. Diagrams should be clear, and explanations should identify the type and strength of intermolecular forces involved.

Key terms and definitions

VSEPR theory — Electron pairs around a central atom repel each other and arrange themselves as far apart as possible.

Bonding pair — A pair of electrons shared in a covalent bond.

Lone pair — A pair of outer electrons not involved in bonding.

Polar molecule — A molecule with a net dipole moment because its bond dipoles do not cancel.

London (dispersion) forces — Weak attractions caused by temporary, induced dipoles in all molecules.

Permanent dipole–dipole forces — Attractions between the δ+ end of one polar molecule and the δ− end of another.

Hydrogen bond — A strong intermolecular attraction between a hydrogen atom bonded to N, O or F and a lone pair on another N, O or F atom.

Ideal gas — A theoretical gas whose particles have no volume and no intermolecular forces.

Boiling point — The temperature at which the vapour pressure of a liquid equals the external pressure.

Core concepts

VSEPR: predicting shapes

1. Count the electron pairs (bonding and lone) around the central atom. A double or triple bond counts as one "region" of electron density.
2. Arrange the regions as far apart as possible.
3. Name the shape from the positions of the atoms (not lone pairs).

Repulsion strength: lone pair–lone pair > lone pair–bonding pair > bonding pair–bonding pair. Lone pairs are closer to the central atom and repel more strongly, reducing bond angles by about 2.5° per lone pair.

Shape and polarity

A molecule is polar if it has polar bonds and those bonds are arranged so their dipoles do not cancel.

• CO₂ has polar C=O bonds but is linear, so the dipoles cancel and the molecule is non-polar.
• H₂O has polar O–H bonds and is bent, so the dipoles do not cancel and the molecule is polar.
• CCl₄ is tetrahedral with identical polar bonds, so it is non-polar; CHCl₃ is polar because the bonds are not identical.

London (dispersion) forces

At any instant, electrons in a molecule may be unevenly distributed, creating a temporary dipole. This induces a dipole in a neighbouring molecule, and the two attract. London forces exist between all molecules.

They increase with:

• number of electrons (larger molecules) — so boiling points rise down Group VII from F₂ to I₂;
• surface area / shape — straight-chain alkanes have higher boiling points than their branched isomers because they can pack closer and have more points of contact.

Permanent dipole–dipole forces

Polar molecules attract each other through their permanent dipoles, in addition to London forces. Propanone (polar) has a higher boiling point than butane (non-polar), even though they have similar numbers of electrons.

Hydrogen bonding

Hydrogen bonding occurs when hydrogen is bonded to a highly electronegative, small atom — nitrogen, oxygen or fluorine. The hydrogen becomes strongly δ+ and is attracted to a lone pair on an N, O or F atom of another molecule. Hydrogen bonds are the strongest intermolecular forces (though much weaker than covalent bonds).

Evidence: the hydrides of Groups V, VI and VII show increasing boiling points down each group because of increasing London forces — except NH₃, H₂O and HF, which have unexpectedly high boiling points because of hydrogen bonding.

The anomalous properties of water

Hydrogen bonding gives water unusual properties that are vital for life — and for Caribbean ecosystems:

• High boiling point and melting point for such a small molecule.
• High specific heat capacity — water absorbs a lot of energy for a small temperature rise, so the Caribbean Sea moderates island climates.
• High surface tension.
• Ice is less dense than liquid water — in ice, hydrogen bonds hold molecules in an open hexagonal lattice, so ice floats. This insulates water beneath it in cold climates.
• Good solvent for ionic and polar substances, because water molecules can surround ions and form hydrogen bonds with polar molecules.

Kinetic theory and gases

The kinetic theory states that gas particles move rapidly and randomly, colliding elastically with each other and the container walls. An ideal gas assumes particles have negligible volume and no intermolecular forces, and obeys pV = nRT, where p is pressure (Pa), V is volume (m³), n is amount (mol), R = 8.31 J K⁻¹ mol⁻¹ and T is temperature (K).

Real gases deviate from ideal behaviour at high pressure (particle volume becomes significant) and low temperature (intermolecular forces become significant). Gases with weak intermolecular forces, such as helium, behave most ideally.

Liquids and solids

Increasing temperature gives particles more kinetic energy to overcome intermolecular forces, so solids melt and liquids boil. The stronger the intermolecular forces, the more energy is needed and the higher the melting and boiling points.

Worked examples

Example 1: Predicting shape

Predict the shape and bond angle of PH₃.

Phosphorus has 5 outer electrons; three form bonds with H, leaving one lone pair. Four electron regions with one lone pair → trigonal pyramidal, about 107° (in practice slightly smaller for PH₃).

Example 2: Comparing boiling points

Explain why ethanol (C₂H₅OH) has a much higher boiling point than propane (C₃H₈), although they have similar Mᵣ. (3 marks)

Propane molecules are held together only by London forces (1). Ethanol molecules have an O–H group and can form hydrogen bonds with each other (1). Hydrogen bonds are much stronger than London forces, so more energy is needed to separate ethanol molecules (1).

Example 3: Using pV = nRT

Calculate the volume occupied by 0.500 mol of gas at 300 K and 100 kPa.

V = nRT ÷ p = (0.500 × 8.31 × 300) ÷ 100 000 = 0.0125 m³ = 12.5 dm³.

Example 4: Polarity

Explain why BF₃ is non-polar but NF₃ is polar.

BF₃ is trigonal planar with three identical B–F dipoles at 120°, which cancel. NF₃ has a lone pair, giving a trigonal pyramidal shape, so the N–F dipoles do not cancel and the molecule has a net dipole.

Example 5: Shapes of ions and molecules with expanded octets

Predict the shapes of (a) SF₆, (b) PCl₅, (c) NH₄⁺ and (d) SO₂.

(a) S has six bonding pairs, no lone pairs → octahedral, 90°. (b) P has five bonding pairs, no lone pairs → trigonal bipyramidal, 90° and 120°. (c) N has four bonding pairs (one is dative), no lone pairs → tetrahedral, 109.5°. (d) S has two double-bond regions and one lone pair (three regions) → bent, about 119°, slightly less than 120° because of the lone pair.

Example 6: Comparing H₂O and H₂S

Explain why water (Mᵣ 18) is a liquid at room temperature while hydrogen sulfide (Mᵣ 34) is a gas.

Hydrogen sulfide has more electrons and therefore stronger London forces, but sulfur is not electronegative enough for hydrogen bonding. Water molecules form hydrogen bonds, because the O–H bond is highly polar and oxygen has lone pairs. Hydrogen bonds are much stronger than the London and dipole–dipole forces between H₂S molecules, so more energy is needed to separate water molecules.

Example 7: Branching and boiling point

Pentane boils at 36 °C and 2,2-dimethylpropane boils at 10 °C, although both are C₅H₁₂. Explain.

Both experience only London forces and have the same number of electrons. Pentane is a long, straight chain with a larger surface area of contact between molecules, so its London forces are stronger. 2,2-dimethylpropane is nearly spherical, with fewer points of contact, so its intermolecular forces are weaker and it boils at a lower temperature.

Example 8: When gases are not ideal

Explain why ammonia deviates more from ideal gas behaviour than helium.

Ammonia molecules are polar and form hydrogen bonds, so the attractions between them are significant, especially at low temperatures and high pressures. Helium atoms are very small and have only very weak London forces, so they come closest to the ideal assumptions of negligible particle volume and no intermolecular forces.

Intermolecular forces at a glance

Common mistakes and how to avoid them

• Naming the shape by electron arrangement instead of atoms. Water's electrons are tetrahedral, but its shape is bent.
• Forgetting lone pairs reduce bond angles.
• Saying hydrogen bonds form with any hydrogen. Hydrogen must be bonded to N, O or F.
• Confusing intermolecular forces with covalent bonds. Boiling breaks intermolecular forces, not covalent bonds.
• Using °C or dm³ in pV = nRT. Convert to K and m³.

Self-test questions

1. Predict the shape and bond angle of the H₃O⁺ ion. Answer: Three bonding pairs and one lone pair → trigonal pyramidal, about 107°.
2. Is CH₂Cl₂ polar or non-polar? Explain. Answer: Polar: it is tetrahedral, but the C–Cl and C–H bond dipoles are different and do not cancel.
3. Arrange HF, HCl, HBr and HI in order of increasing boiling point. Answer: HCl < HBr < HI < HF. Boiling points rise from HCl to HI with increasing London forces, but HF is highest because of hydrogen bonding.
4. Convert 2.00 dm³ and 27 °C into SI units for use in pV = nRT. Answer: 2.00 × 10⁻³ m³ and 300 K.

How this links to your Internal Assessment

An SBA might investigate how chain length affects the boiling point or rate of evaporation of alcohols, or compare the viscosity of liquids with different intermolecular forces. Good investigations control temperature, use the same volume of liquid and repeat measurements. In your analysis, explain trends using London forces and hydrogen bonding, and evaluate sources of error such as heat loss or inaccurate temperature readings.

Exam technique for Shapes of molecules and intermolecular forces

• Draw shapes with wedges and dashes for 3D, show lone pairs, and label bond angles.
• For boiling point comparisons, identify all the intermolecular forces present and say which is strongest.
• When explaining hydrogen bonding, draw a diagram showing the δ+ hydrogen, the lone pair and the dotted hydrogen bond in a straight line.
• In gas calculations, show unit conversions clearly.

Quick revision summary

• VSEPR: electron pairs repel to maximum separation; lone pairs repel more than bonding pairs.
• Linear 180°, trigonal planar 120°, tetrahedral 109.5°, pyramidal 107°, bent 104.5°, trigonal bipyramidal 90°/120°, octahedral 90°.
• Polarity needs polar bonds that do not cancel by symmetry.
• London forces (all molecules; increase with electrons and surface area) < permanent dipole–dipole < hydrogen bonds (H bonded to N, O, F).
• Water's hydrogen bonding explains its high boiling point, high heat capacity, surface tension and less-dense ice.
• Ideal gases obey pV = nRT; real gases deviate at high pressure and low temperature.