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Intermolecular forces are the forces of attraction that act between molecules. They are fundamentally different from the covalent bonds that hold atoms together within molecules. Intermolecular forces are much weaker than covalent bonds, but they have enormous influence on physical properties such as boiling points, melting points, and solubility.
There are three types of intermolecular force, in order of increasing strength: London dispersion forces, permanent dipole–dipole interactions, and hydrogen bonding.
London dispersion forces (also called van der Waals forces, dispersion forces, or temporary dipole–induced dipole forces) are the weakest intermolecular force, but they are present in all molecules and are the only intermolecular force in non-polar molecules.
At any given instant, the electrons in an atom or molecule may be unevenly distributed, creating a temporary (instantaneous) dipole — a brief, random imbalance of charge. This temporary dipole induces a dipole in a neighbouring molecule by attracting or repelling its electrons. The result is a fleeting attraction between the two molecules.
These temporary dipoles are constantly forming, disappearing, and re-forming. Although each individual interaction is extremely short-lived, the cumulative effect across billions of molecules produces a significant attractive force.
The strength of London forces depends on the number of electrons in the molecule (which correlates with molecular mass and surface area):
More electrons → stronger London forces. More electrons means a larger electron cloud that is more easily distorted (more polarisable), creating larger temporary dipoles. This is why boiling points increase down a group of elements.
Greater surface area → stronger London forces. A long-chain molecule has more surface area for interactions than a compact, branched molecule with the same number of electrons. This is why pentane (straight chain, bp 36°C) has a higher boiling point than 2,2-dimethylpropane (compact sphere, bp 10°C), despite both having the formula C₅H₁₂.
| Noble gas | Number of electrons | Boiling point / °C |
|---|---|---|
| He | 2 | −269 |
| Ne | 10 | −246 |
| Ar | 18 | −186 |
| Kr | 36 | −152 |
| Xe | 54 | −108 |
The steady increase in boiling point directly tracks the increasing number of electrons and hence the increasing strength of London dispersion forces.
Permanent dipole–dipole interactions occur between polar molecules — molecules that have a permanent dipole moment due to differences in electronegativity and asymmetric geometry.
The δ+ end of one polar molecule is attracted to the δ− end of a neighbouring polar molecule. Unlike London forces, which are random and fleeting, these attractions have a fixed orientation — the molecules preferentially align so that opposite charges are near each other.
Permanent dipole–dipole interactions are generally stronger than London forces for molecules of similar size. However, London forces still operate between polar molecules in addition to the dipole–dipole forces.
For example, propanone (CH₃COCH₃, polar, bp 56°C) has a higher boiling point than butane (C₄H₁₀, non-polar, bp −1°C), even though butane has a slightly larger molecular mass. The permanent dipole in propanone creates additional attractive forces beyond the London forces that both molecules experience.
Students sometimes think that polar molecules only have dipole–dipole interactions. In reality, polar molecules have both London forces and dipole–dipole forces. The total intermolecular attraction is the sum of all forces present.
Hydrogen bonding is the strongest type of intermolecular force and has a dramatic effect on physical properties. Despite its name, a hydrogen bond is not a covalent bond — it is an intermolecular attraction, albeit an unusually strong one.
Three conditions must all be met:
The hydrogen bond forms between the δ+ hydrogen on one molecule and the lone pair on the δ− N, O, or F atom of another molecule.
These three elements are small and highly electronegative. When bonded to hydrogen, they create an exceptionally strong δ+ charge on the hydrogen atom. The hydrogen atom is also very small (just a proton with no inner electrons), so the lone pair on a neighbouring N, O, or F atom can get very close to it. This combination of strong polarity and close approach creates the unusually strong hydrogen bond.
Chlorine is also electronegative (3.0) but is too large — the lone pairs on chlorine are too far from the hydrogen to create a true hydrogen bond.
Hydrogen bonds are typically about 10% as strong as a covalent bond. A typical O–H···O hydrogen bond has an energy of about 20 kJ mol⁻¹, compared to around 460 kJ mol⁻¹ for a covalent O–H bond. However, they are significantly stronger than London forces or ordinary dipole–dipole interactions for molecules of similar size.
| Force | Typical strength / kJ mol⁻¹ | Present in | Origin |
|---|---|---|---|
| London (dispersion) | 1–10 | All molecules | Temporary dipole–induced dipole |
| Permanent dipole–dipole | 5–25 | Polar molecules only | δ+···δ− attraction |
| Hydrogen bonding | 10–40 | H bonded to N, O, or F | δ+ H···lone pair on N/O/F |
Boiling point is the temperature at which molecules have enough kinetic energy to overcome all intermolecular forces and escape into the gas phase. Stronger intermolecular forces mean higher boiling points.
| Group 4 hydride | Mr | bp / °C | Group 6 hydride | Mr | bp / °C |
|---|---|---|---|---|---|
| CH₄ | 16 | −162 | H₂O | 18 | 100 |
| SiH₄ | 32 | −112 | H₂S | 34 | −60 |
| GeH₄ | 77 | −88 | H₂Se | 81 | −41 |
| SnH₄ | 123 | −52 | H₂Te | 130 | −2 |
Group 4 hydrides are all non-polar with only London forces — boiling points increase steadily with molecular mass.
Group 6 hydrides show the same trend for H₂S, H₂Se, and H₂Te. But H₂O is dramatically out of line: it boils at 100°C instead of the predicted ~−80°C. The ~180°C anomaly is entirely due to hydrogen bonding.
The same anomaly is seen for NH₃ in the Group 5 hydrides and HF in the Group 7 hydrides — the first member has an anomalously high boiling point due to hydrogen bonding.
Structural isomers have the same molecular formula but different shapes, which affects London forces:
| Isomer of C₅H₁₂ | Shape | Boiling point / °C |
|---|---|---|
| Pentane | Straight chain | 36 |
| 2-Methylbutane | One branch | 28 |
| 2,2-Dimethylpropane | Compact/spherical | 10 |
All three have identical molecular formulae, so the same number of electrons and the same types of intermolecular force (London only). The difference is surface area: pentane has the most surface area for contact, giving the strongest cumulative London forces and the highest boiling point. 2,2-Dimethylpropane is the most compact, with the least surface area.
To identify the intermolecular forces in a substance:
Then consider the relative strengths to explain physical properties. For large non-polar molecules, London forces can be stronger than the hydrogen bonds in small molecules. Context matters — always consider the size and type of molecules being compared.
Common exam mistake: When comparing boiling points, students sometimes only mention the type of intermolecular force without considering molecular size. Iodine (I₂, non-polar, London forces only) has a higher melting point (114°C) than water (0°C) despite water having hydrogen bonding. This is because I₂ has 106 electrons, generating very strong London forces that outweigh the hydrogen bonds in the much smaller water molecule. Always consider both the type and the size factor.
Edexcel 9CH0 Topic 2.6 — intermolecular forces is the home address of this material, restricted here to London (dispersion / induced-dipole–induced-dipole) forces and permanent dipole–dipole forces. Hydrogen bonding, although formally part of the same sub-topic, is treated in the next lesson because its mechanism, geometry and biological consequences merit separate treatment. The cross-paper footprint is unusually wide. Paper 1, Topic 4 (Inorganic Chemistry — Periodicity) uses IMFs to explain why the boiling points of the Period 3 elements rise across S, P, then collapse at S₈ → P₄ → Cl₂ → Ar; the molecular solids depend wholly on London forces, while the giant covalent and metallic structures depend on directional bonding. Paper 2, Topic 6 (Organic — alkanes and alkenes) invokes London forces to explain the steady ~20–30 °C rise in boiling point per added CH₂ across a homologous series. Paper 2, Topic 7 (Alcohols, haloalkanes) demands a comparison between alkanes, haloalkanes and alcohols of similar Mr, where permanent dipoles, London forces and (in the next lesson) hydrogen bonding all compete. Paper 3 (synoptic / practical) routinely uses the rule "like dissolves like" — a compact summary of IMF matching between solute and solvent — to predict miscibility, recrystallisation choices and chromatographic behaviour. (refer to the official Pearson Edexcel specification document for exact wording)
Question (7 marks): The noble gases have the following boiling points: helium −269 °C, neon −246 °C, argon −186 °C, krypton −153 °C, xenon −108 °C. The halogens fluorine (F₂) and iodine (I₂) are both non-polar diatomic molecules, yet F₂ boils at −188 °C while I₂ boils at 184 °C.
(a) Explain the trend in boiling points down the noble-gas group. (4) (b) Explain why I₂ has a much higher boiling point than F₂. (3)
Solution with mark scheme:
(a) Step 1 — identify the only IMF acting. Noble gases exist as monatomic species with no permanent dipole; the only intermolecular force present is the London (dispersion) force, arising from temporary instantaneous dipoles caused by random fluctuation of electron density which induce dipoles in neighbouring atoms.
B1 — explicit statement that London forces are the only IMF acting on noble gas atoms.
Step 2 — link descent of group to electron count. Down the group from He (2 electrons) to Xe (54 electrons), the number of electrons increases, so the electron cloud is larger and more polarisable.
M1 — connecting the trend to increasing number of electrons / increasing polarisability.
Step 3 — explain consequence on London force strength. A more polarisable cloud allows larger temporary dipoles to form, which in turn induce larger dipoles in neighbouring atoms. The instantaneous dipole–induced dipole attractions are therefore stronger, requiring more energy to overcome during boiling.
M1 — strength of London forces increases with polarisability / size of electron cloud.
Step 4 — link to boiling point. A higher boiling point reflects greater energy input needed to separate atoms in the liquid; therefore the boiling point rises monotonically from He through Xe.
A1 — final link to "more energy required to overcome IMFs, therefore higher boiling point".
(b) Step 1 — identify forces. Both F₂ and I₂ are non-polar diatomic molecules (the bond is between identical atoms, so there is no permanent dipole). The only IMF in either liquid is the London force.
B1 — both species are non-polar; only London forces act.
Step 2 — count electrons. F₂ has 18 electrons; I₂ has 106 electrons. The much larger electron cloud of I₂ is far more polarisable, so the temporary instantaneous dipoles are much larger and the induced-dipole attractions correspondingly stronger.
M1 — explicit comparison of electron count and polarisability.
Step 3 — link to boiling point. Stronger London forces in I₂ require considerably more energy to overcome; this is why I₂ is a solid at room temperature and boils at 184 °C, whereas F₂ is a gas boiling at −188 °C.
A1 — clear final link from polarisability → London force strength → boiling point.
Total: 7 marks (B2 M3 A2 split as shown). Note how the answer never appeals to "stronger bonds" — covalent bonds within the molecule are not broken when boiling.
Question (6 marks): Pentane (C₅H₁₂) is a straight-chain alkane that boils at 36 °C. Its structural isomer 2,2-dimethylpropane (also C₅H₁₂) boils at 9.5 °C. The two compounds have identical molecular formulae and the same number of electrons.
(a) Identify the type of intermolecular force responsible for holding either liquid together and justify your choice. (2) (b) Explain, in terms of molecular shape, why pentane has a higher boiling point than 2,2-dimethylpropane. (4)
Mark scheme decomposition by AO:
(a)
(b)
Total: 6 marks (B2 M2 A2; AO1 ≈ 2, AO2 ≈ 3, AO3 ≈ 1). The branching question is an Edexcel favourite because it cleanly separates candidates who memorise "Mr → BP" from those who reason about molecular geometry.
Connects to:
Topic 2.5 — Electronegativity and bond polarity: permanent dipole–dipole forces only exist in molecules where polar bonds fail to cancel by symmetry. CO₂ has two polar C=O bonds but is linear, so the bond dipoles cancel and the molecule is non-polar (London forces only). H₂O has two polar O–H bonds in a bent geometry, giving a net dipole. Vector reasoning learned in Topic 2.5 underpins every classification of "polar vs non-polar" required here.
Topic 6 — Alkanes (Paper 2): the homologous series CH₄, C₂H₆, C₃H₈, C₄H₁₀, … shows boiling points rising by approximately 20–30 °C per added CH₂ unit. Each extra CH₂ adds 8 electrons and a longer surface, both of which strengthen London forces. Beyond C₁₇H₃₆ the alkanes are waxy solids at room temperature, held together entirely by London forces — a vivid demonstration that "weak" forces summed over many contacts can become very strong indeed.
Topic 7 — Alcohols (Paper 2): short alcohols (methanol, ethanol, propan-1-ol) are fully miscible with water; long alcohols (heptan-1-ol upward) are essentially insoluble. The competition is between hydrogen bonding (treated in the next lesson) which favours dissolving in water, and the long hydrocarbon tail whose London-force interactions favour partitioning into other hydrocarbons. The same logic explains why detergents have a polar head and a non-polar tail.
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