Intermolecular Forces: Dipoles and Van der Waals Revision Notes | A Level AQA Chemistry

1. Why Molecules Attract Each Other

Covalent molecules have no overall ionic charge — so why do they stick together at all? The answer lies in the uneven distribution of electrons, both permanent and temporary, that creates electrical attractions between neighbouring molecules.

These attractions are called intermolecular forces — forces that act between molecules rather than within them. They are far weaker than covalent bonds, but they govern boiling points, melting points, viscosity, and solubility. Understanding them is central to explaining why, for example, $\text{HCl}$ boils at $-85\,^{\circ}\text{C}$ while $\text{H}_2\text{O}$ boils at $100\,^{\circ}\text{C}$.

In this lesson, we cover two types of intermolecular force: permanent dipole–dipole forces and induced dipole–dipole forces (also called van der Waals forces, London dispersion forces, or dispersion forces). Both arise from electrostatic attraction, but their origins are different. Note that hydrogen bonding — a third, distinct type of intermolecular force required at A Level — is covered in a separate lesson.

2. Permanent Dipole–Dipole Forces

Start by picturing a $\text{HCl}$ molecule. Chlorine is considerably more electronegative than hydrogen, so the shared electron pair is pulled towards the chlorine end. This creates a permanent dipole: the chlorine end carries a partial negative charge ($\delta-$) and the hydrogen end carries a partial positive charge ($\delta+$).

When many $\text{HCl}$ molecules are close together, they align so that the $\delta+$ hydrogen of one molecule is attracted to the $\delta-$ chlorine of its neighbour. This alignment is not perfect — thermal motion constantly disrupts it — but on average there is a net attractive force between the molecules.

For a permanent dipole to exist, two conditions must both be met: the molecule must contain polar bonds (bonds between atoms of different electronegativity), and the molecular geometry must be such that the individual bond dipoles do not cancel. A linear $\text{CO}_2$ molecule has two polar $\text{C=O}$ bonds, but they point in exactly opposite directions and cancel — so $\text{CO}_2$ has no permanent dipole. Contrast this with a bent $\text{SO}_2$ molecule, where the bond dipoles do not cancel and a net dipole exists.

3. Induced Dipole–Dipole Forces (Van der Waals / London Dispersion Forces)

Now consider a non-polar molecule such as $\text{Cl}_2$. It has no permanent dipole, yet it exists as a liquid at room temperature under modest pressure and boils at $-34\,^{\circ}\text{C}$ — well above what you would expect if there were no intermolecular forces at all. Something is holding the molecules together.

The key is that electrons are never perfectly evenly distributed. At any given instant, the electron cloud of a molecule may be slightly denser on one side than the other, creating a temporary (instantaneous) dipole. This temporary dipole distorts the electron cloud of a neighbouring molecule, inducing a dipole in it. The two temporary dipoles then attract each other. This is the origin of induced dipole–dipole forces, also called London dispersion forces or van der Waals forces.

These forces are present in every substance — polar or non-polar, ionic or covalent. In polar molecules, both permanent dipole–dipole forces and induced dipole–dipole forces operate simultaneously. In non-polar molecules, induced dipole–dipole forces are the only intermolecular force present.

The strength of induced dipole–dipole forces depends on how easily the electron cloud can be distorted — a property called polarisability. Larger molecules with more electrons have more diffuse, more easily distorted electron clouds, so they experience stronger induced dipole–dipole forces. This is why boiling points of the noble gases increase from helium to xenon, and why larger alkanes have higher boiling points than smaller ones.

The trend in the chart above is entirely due to increasing numbers of electrons as you descend Group 0. More electrons means greater polarisability, stronger induced dipole–dipole forces, and therefore more energy needed to separate the molecules — hence a higher boiling point.

4. Comparing the Two Forces: Strength and Molecular Factors

A common question asks you to compare permanent dipole–dipole forces and induced dipole–dipole forces. The key point is that neither is universally stronger than the other — it depends on the specific molecules being compared.

For small, highly polar molecules, permanent dipole–dipole forces can make a significant contribution to the total intermolecular attraction. However, as molecules become larger and contain more electrons, the induced dipole–dipole forces grow rapidly and can easily dominate over permanent dipole–dipole forces. This is why large non-polar molecules such as iodine ($\text{I}_2$) or long-chain alkanes have higher boiling points than small polar molecules such as $\text{HCl}$.

Molecular shape also matters for induced dipole–dipole forces. A long, straight-chain molecule such as pentane has a larger surface area of contact between neighbouring molecules than a compact, branched isomer such as 2,2-dimethylpropane (neopentane). Greater surface contact means more points of induced dipole interaction, so pentane has a higher boiling point than its branched isomer despite having the same molecular formula and the same number of electrons.

5. Applying These Forces to Familiar and Unfamiliar Molecules

AQA examiners frequently present unfamiliar molecules and ask you to identify and explain the intermolecular forces present. The approach is always the same: first determine whether the molecule is polar (check for polar bonds and non-cancelling geometry), then consider the size and electron count to assess the strength of induced dipole–dipole forces.

Worked example: Identifying and Explaining Intermolecular Forces in an Unfamiliar Molecule

Phosphine, $\text{PH}_3$, is a gas at room temperature with a boiling point of $-87.7\,^{\circ}\text{C}$. Identify the intermolecular forces present in liquid $\text{PH}_3$ and explain their origin.

1. P has electronegativity ≈ 2.1; H has electronegativity ≈ 2.2. The difference is very small (≈ 0.1). — The P–H bonds are very weakly polar. The electronegativity difference is so small that the bonds are considered essentially non-polar for most purposes. Compare this with N–H in ammonia, where the difference is much larger (N ≈ 3.0, H ≈ 2.2).
2. PH₃ is trigonal pyramidal (like NH₃), with one lone pair on P. Even if the P–H bonds were polar, the geometry would give a small net dipole. — Because the P–H bonds are so weakly polar, the net permanent dipole of PH₃ is very small — much smaller than that of NH₃. Permanent dipole–dipole forces in PH₃ are therefore negligible.
3. PH₃ has 18 electrons (P: 15, 3×H: 3). All molecules experience induced dipole–dipole forces. — The electron cloud of PH₃ can become temporarily uneven, creating an instantaneous dipole that induces a dipole in a neighbouring molecule. These induced dipole–dipole (van der Waals) forces are the dominant intermolecular force in liquid PH₃.
4. Dominant force: induced dipole–dipole (van der Waals/London dispersion) forces. — The low boiling point of −87.7 °C is consistent with weak intermolecular forces. PH₃ has more electrons than HCl (18 vs 18 — similar), but lacks the significant permanent dipole of HCl, so its boiling point is comparable to other molecules of similar electron count and weak polarity.

Answer: Liquid PH₃ experiences induced dipole–dipole (van der Waals/London dispersion) forces as the dominant intermolecular force. The P–H bonds are so weakly polar that permanent dipole–dipole forces are negligible. The induced dipole–dipole forces arise from instantaneous uneven electron distribution in the PH₃ molecule creating a temporary dipole, which induces a dipole in an adjacent molecule.

Now consider a molecule where both types of force operate. Propanone ($\text{CH}_3\text{COCH}_3$) contains a polar $\text{C=O}$ bond and a non-symmetrical structure, giving it a significant permanent dipole. It also has 32 electrons, giving it reasonably strong induced dipole–dipole forces. Both permanent dipole–dipole forces and induced dipole–dipole forces act between propanone molecules in the liquid state.

The carbonyl group ($\text{C=O}$) in propanone is highly polar because oxygen is considerably more electronegative than carbon. This creates a significant $\delta-$ on oxygen and $\delta+$ on carbon, and the asymmetric molecular geometry means the dipoles do not cancel. The result is a permanent dipole across the molecule, enabling permanent dipole–dipole interactions between neighbouring propanone molecules in addition to the ever-present induced dipole–dipole forces.

6. Exam Technique and Common Mistakes

Intermolecular forces questions appear across all three A Level papers and carry marks at every level — from one-mark state questions to six-mark extended explanations. The marks are very accessible once you know the precise language examiners expect.

Bringing everything together: when you encounter any molecule — familiar or unfamiliar — the systematic approach is to ask three questions. Does it have polar bonds? Do those bond dipoles cancel in the molecular geometry? How many electrons does it have? The answers tell you which forces are present and give you the language to explain physical properties with precision.