CHEMISTRY · LESSON 05 OF 12
Intermolecular forces, phases and solubility
Forces between molecules — dispersion, dipole–dipole, hydrogen bonding and ion–dipole — decide boiling points, phase changes and which substances mix.
What this lesson explains
A covalent bond holds atoms together inside a molecule. Intermolecular attractions act between molecules. This distinction explains why a liquid can boil without changing each molecule’s chemical formula.
These attractions explain the properties chemical engineers use daily: boiling points for distillation, why water dissolves salts, and why oil and water separate.
Before you begin
Polarity from bonding
A molecule is polar when its bond dipoles do not cancel (see Chemical bonding). Electronegativity differences create bond dipoles; molecular shape decides whether they add or cancel.
Covalent bond versus attraction between molecules
Breaking a covalent bond changes the substance; overcoming an intermolecular attraction (boiling, dissolving) does not.
The idea, made visible
London dispersion forces arise from fluctuating electron distributions and induced dipoles. They occur between all atoms and molecules, including non-polar ones. Greater polarisability and more effective contact between molecules can make these attractions stronger. Molar mass is a useful comparison within some related families, but it is not a universal rule.
Dipole–dipole attractions occur between molecules with permanent dipoles. All such molecules also have dispersion forces. Shape matters: CO₂ contains polar C=O bonds, but its bond dipoles cancel, so an isolated CO₂ molecule has no permanent net dipole. Do not assign a molecule’s polarity from one bond alone.
In the usual introductory model, hydrogen bonding needs a donor H covalently bonded to N, O or F and a suitable lone pair on an acceptor, commonly N, O or F. Water can both donate and accept. An ether can accept a hydrogen bond from water but cannot donate one because its hydrogens are bonded to carbon. Hydrogen bonds between molecules are not the covalent O–H bonds inside those molecules.
Ion–dipole attractions occur between an ion and a polar molecule. Water’s partially negative oxygen points towards a cation; its partially positive hydrogens point towards an anion. This helps stabilise dissolved ions. Dissolution still depends on the balance of separating the original particles, forming new interactions and the entropy change. Hydration alone does not guarantee that every salt is soluble.
For comparable molecular substances, stronger net intermolecular attractions often mean higher boiling temperature, lower equilibrium vapour pressure at the same temperature, and larger energy required for vaporisation. Comparisons need the same external pressure for boiling points. Molecular size, shape and the set of interactions all matter; there is no fixed ranking that works for every pair of substances.
A liquid boils when its equilibrium vapour pressure equals the external pressure. Lower external pressure allows boiling at a lower temperature. Evaporation can occur below the boiling temperature. In vaporising molecular water, molecules become widely separated but remain H₂O; making H₂ and O₂ would be a chemical reaction with a different energy requirement.
| Substance or mixture | Particle type | Important attractions |
|---|---|---|
| CH₄, pure | Non-polar molecules | Dispersion |
| HCl, pure | Polar molecules | Dispersion and dipole–dipole |
| H₂O, pure | Polar molecules with O–H bonds | Dispersion, dipole–dipole and hydrogen bonding |
| Na⁺ in water | Ion surrounded by polar water | Ion–dipole hydration |
| NaCl crystal | Ionic lattice | Ion–ion electrostatic interactions; not a molecular liquid |
The formulas and what they mean
| Symbol | Meaning | Unit |
|---|---|---|
| δ+ and δ− | partial atomic charges in a polar bond | not integer ionic charges |
| ΔHvap | molar enthalpy of vaporisation at stated conditions | kJ/mol |
| n | amount vaporised | mol |
| q | heat absorbed for the stated change | kJ |
| Pvap | equilibrium vapour pressure at a given temperature | pressure |
Heat for vaporisation
q = nΔHvap
Conditions and limits: For vaporising the stated amount at the specified phase-change temperature and pressure. This excludes heating the liquid to that temperature and heating the vapour afterwards.
Boiling condition
Pvap(Tboil) = Pexternal
Conditions and limits: Equilibrium boiling condition; the boiling temperature depends on external pressure.
Separate heating from a phase change
qtotal = mcpΔT + nΔHvap
Conditions and limits: For a simple path that first warms a liquid to its boiling point, then fully vaporises it; approximate cₚ as constant over the interval. Additional stages need additional terms.
A first worked example
How to compare intermolecular forces
- List the particles present (molecules or ions) and decide whether each molecule is polar.
- Every molecule has dispersion forces; they grow with the number of electrons and with larger, more elongated molecules.
- Add dipole–dipole forces for polar molecules, hydrogen bonding when H is bonded to N, O or F and an N/O/F lone pair is available, ion–dipole forces for ions in polar solvents.
- Compare like with like (similar molar mass) before concluding; state the prediction as a tendency.
Compare two molecules with the same molecular formula
Problem. Ethanol CH₃CH₂OH and dimethyl ether CH₃OCH₃ both have formula C₂H₆O. Explain why ethanol is expected to have the higher boiling point at the same pressure.
Both have similar molar mass and both have dispersion forces. Both molecules are polar.
A fair comparison includes the interactions shared by the substances.
Ethanol has an O–H donor and oxygen lone pairs, so neighbouring ethanol molecules can hydrogen-bond.
The molecular structure supplies both a donor and an acceptor.
Dimethyl ether has oxygen lone pairs but no O–H bond, so pure ether cannot form the same donor–acceptor network.
An acceptor alone does not supply a hydrogen-bond donor.
Result: Ethanol is expected to boil at the higher temperature because its molecules have additional hydrogen-bonding interactions.
What it means: Dimethyl ether can still accept hydrogen bonds from water. The statement about the pure substance does not prohibit interactions in mixtures.
A different case
Explain solubility without an absolute rule
Problem. Explain why water generally mixes more favourably with ethanol than with a non-polar hydrocarbon such as hexane.
Water molecules form an extensive hydrogen-bond network.
Separating water molecules changes those interactions.
Ethanol’s OH group can form hydrogen bonds with water.
New solute–solvent interactions help compensate for the interactions disrupted on mixing.
Hexane cannot form hydrogen bonds with water, although dispersion forces still act between them.
The full energetic and entropic balance is less favourable for extensive mixing.
Result: Polarity and hydrogen-bond compatibility help explain the difference; “like dissolves like” is a guide, not a complete law.
What it means: Actual solubility depends on temperature and the free-energy balance; do not predict exact solubility from one interaction label.
More worked cases
Each case below uses a different skill. Every step and result is shown.
Vaporise a stated amount of liquid water
Problem. At its boiling temperature under the stated conditions, 0.250 mol of liquid water vaporises. Take ΔHvap = 40.7 kJ/mol. Find the heat absorbed.
The water is already at the phase-change temperature, so q = nΔHvap.
No preliminary warming is included in this problem.
q = 0.250 mol × 40.7 kJ/mol = 10.175 kJ.
Moles cancel, leaving energy.
Report 10.2 kJ absorbed to three significant figures.
Vaporisation requires energy input.
Result: q = +10.2 kJ for the water.
What it means: H₂O molecules remain H₂O in the vapour. For condensation of the same amount under the same conditions, q has the opposite sign.
Changing the pressure changes the boiling temperature
Problem. A liquid boils in a vessel. Explain what happens to its boiling temperature if the external pressure is lowered.
At boiling, the liquid’s vapour pressure equals the external pressure.
This is the condition to compare.
A lower external pressure can be matched at a lower temperature on the same vapour-pressure curve.
Equilibrium vapour pressure increases with temperature along the liquid–vapour curve.
Result: The boiling temperature falls when external pressure is reduced.
What it means: This explains vacuum distillation. It does not mean that the substance’s covalent bonds have become weaker.
Common misunderstandings
Misunderstanding: Boiling breaks each molecule into atoms.
Correct idea: Boiling a molecular liquid separates molecules; it normally preserves the covalent bonds within them.
Misunderstanding: Every substance containing H can hydrogen-bond to itself.
Correct idea: Check for an appropriate donor and acceptor; a C–H bond does not supply the usual introductory donor.
Misunderstanding: Dispersion acts only in non-polar molecules.
Correct idea: Dispersion is present in all atoms and molecules, alongside any other interactions.
Misunderstanding: A stronger named force always guarantees a higher boiling point.
Correct idea: Compare the whole molecules: size, polarisability, shape and all available interactions.
Keep in mind
- q = nΔHvapHeat for vaporisation
- Pvap(Tboil) = PexternalBoiling condition
- qtotal = mcpΔT + nΔHvapSeparate heating from a phase change
Scope of this lesson
- Qualitative comparisons only; vapour-pressure equations (Clausius–Clapeyron) and phase diagrams are not developed here. Check your outline.
Next: Writing formulas and naming compounds. Write formulas from ion charges and name ionic and molecular compounds so that the formula, the name and the charges always agree.
Further reading: OpenStax Chemistry 2e — Intermolecular forces · OpenStax Chemistry 2e — Phase transitions. Sources and credits.