Intermolecular Forces

Butane boils at −0.5 °C. Its isomer, isobutane, boils at −11.7 °C. Yet they share the same formula, bonds, and atoms. Why? Water has a molar mass of only 18 g/mol. It stays liquid at room temperature. Carbon dioxide, at 44 g/mol, is a gas. Why the difference? None of these differences come from covalent bond strength. Every C–H and O–H bond stays intact in both liquid and gas phases. The differences come entirely from noncovalent forces. These are weak, reversible electrical attractions. They act between molecules rather than within them. Intermolecular forces are the quiet architects of the physical world of chemistry. They set boiling points, melting points, solubilities, and viscosities. They also control vapor pressures and the shapes of proteins and DNA. A covalent C–C bond is worth roughly 350 kJ/mol. A hydrogen bond is worth 15–40 kJ/mol. A London dispersion contact between two methyl groups is worth less than 1 kJ/mol. Yet stack up enough of these weak contacts. Consider a folded protein, a DNA duplex, a crystal lattice, or a soap micelle. The cumulative noncovalent energy rivals or exceeds covalent bond energies. Life itself is a triumph of weak interactions used in enormous numbers. The unifying physical idea is simple: all intermolecular attraction is electrostatic. It arises from the uneven distribution of electron density. This distribution was already analyzed for individual molecules via Molecular Dipole Moments and Inductive Effects and Charge Distribution. What differs between force types is which charge distribution is doing the attracting. It might be a permanent dipole, an instantaneous fluctuating dipole, or a highly polarized O–H/N–H/F–H bond. What also differs is how fast that attraction decays with distance. This article builds the full taxonomy and ranks the forces by strength. It sets up the detailed treatments of Dipole-Dipole Interactions, London Dispersion Forces, and Hydrogen Bonding that follow. By the end you will be able to look at any two molecules. From structure alone, you can predict which noncovalent forces apply. You can estimate roughly how strong each is. You can also see how they combine to determine a measurable macroscopic property like boiling point. This is not a memorization exercise. Every intermolecular force follows from Coulomb's law and the polarizability of electron clouds. Every trend can be derived, not merely recalled.

Key takeaways

Intermolecular forces are noncovalent, electrostatic attractions between molecules (10–1000× weaker than covalent bonds).
Four principal types, in order of typical strength: ion–ion > hydrogen bonding > dipole–dipole > London dispersion.
All forces ultimately arise from Coulomb's law acting on time-averaged or instantaneous charge distributions.
Strength depends on charge magnitude, distance (steep power-law decay), and polarizability of the electron cloud.
Boiling point, melting point, viscosity, and solubility are macroscopic readouts of microscopic IMF strength.
Real molecules experience several IMF types simultaneously; the total interaction energy is additive.
Key idea: every intermolecular force is Coulomb's law applied to a molecule's real or induced charge distribution — the differences are only in which distribution and how quickly it decays with distance.

The core insight

Intermolecular forces are Coulombic attractions between the electron-density distributions of separate molecules; their hierarchy of strength (ion–dipole > H-bond > dipole–dipole > dispersion) is a hierarchy of how permanent, how polarized, and how close-range the underlying charge separation is.

Where this lives today

Intermolecular forces are not a historical curiosity; they are load-bearing in essentially every modern chemical and biological technology. Drug design leans on them directly: a drug candidate must form the right combination of hydrogen bonds, dipole–dipole contacts, and dispersion contacts with its protein target's binding pocket — computational docking software scores candidate poses largely by summing exactly the terms derived above (Coulomb, Keesom/dipole, and London dispersion energies), refined with quantum corrections. Gecko feet climb vertical glass using nothing but van der Waals (dispersion) forces distributed across millions of microscopic setae, an engineering feat now being reverse-engineered into synthetic dry adhesives for robotics and medical devices. In materials science, the layered structure of graphite (strong covalent bonds within each sheet, weak dispersion forces between sheets) is exactly why graphite lubricates and why individual sheets can be mechanically exfoliated to make graphene — a process only possible because the interlayer force is so much weaker than the intralayer covalent bonds. Protein folding is fundamentally a competition and cooperation between backbone hydrogen bonds (stabilizing α-helices and β-sheets), side-chain dipole–dipole interactions, and the hydrophobic effect (itself powered by dispersion-force bookkeeping between nonpolar side chains and water), all summed across thousands of residues. Chromatography, whether reverse-phase HPLC or simple paper chromatography, separates molecules purely on the basis of differential intermolecular attraction to a stationary versus mobile phase — a direct industrial and analytical application of the exact hierarchy built in this article.

Ask the tutor: How do induction (dipole-induced dipole) forces fit between dipole–dipole and dispersion forces in strength, and why don't chemists usually give them their own separate named category in introductory courses?

The physical picture

Start from Coulomb's law for two point charges $q_1$ and $q_2$ separated by distance $r$ :

E = \frac{k q_1 q_2}{r}

This is the seed equation for every intermolecular force. Molecules are not point charges, but bundles of nuclei and electrons whose time-averaged (or instantaneously fluctuating) charge distribution can be decomposed into a monopole (net charge), dipole (charge separation), and higher multipoles. When two molecules approach, every charge on one interacts with every charge on the other; expanding this sum in powers of $1/r$ produces the familiar catalog of forces, each dominant at a different combination of distance and polarity.

Permanent multipoles. A molecule with an asymmetric electron distribution — set up by Inductive Effects and Charge Distribution pulling density toward electronegative atoms — has a permanent dipole moment $\mu$ (already characterized under Molecular Dipole Moments). Two such dipoles orient to place opposite partial charges close together; this is the dipole–dipole interaction, foreshadowed here and treated in full under Dipole-Dipole Interactions.

Induced multipoles. Even a perfectly nonpolar molecule like Ar or CH₄ has, at any instant, a fluctuating, lopsided electron cloud due to quantum zero-point motion. This instantaneous dipole polarizes the electron cloud of a neighboring molecule, inducing a complementary dipole that is, on average, attractive. This is the London dispersion force, present in every pair of molecules regardless of polarity, detailed under London Dispersion Forces.

Special short-range, highly directional attraction. When H is bonded directly to N, O, or F, the bond is so polarized ( $\delta+$ on H, since these atoms are far more electronegative and H has no core electrons to shield the resulting positive character) that the bare, nearly unshielded proton can approach a lone pair on a neighboring N, O, or F extremely closely. This produces the anomalously strong, highly directional hydrogen bond, explored under Hydrogen Bonding. Figure 1 lays out this whole hierarchy schematically, showing the origin of charge asymmetry in each case.

Figure

Figure 1 — Origins of the four intermolecular force families

The crucial structural feature to notice across Figure 1 is the power of $r$ in each denominator: ion–ion falls off as $1/r$ , ion–dipole as $1/r^2$ , dipole–dipole as (on average, freely rotating) $1/r^6$ , and dispersion also as $1/r^6$ . Steeper decay means the force is felt only over shorter range — which is exactly why dispersion and dipole–dipole forces are so sensitive to how closely molecules can pack, while ion–ion attraction is felt across a whole crystal lattice. Figure 2 places real molecules on a strength ladder, anchored by boiling points, so the abstract energy hierarchy is tied to a directly measurable, verifiable quantity.

Figure

Figure 2 — Strength ladder anchored to boiling points

Notice in Figure 2 that octane — a purely nonpolar hydrocarbon — boils higher than acetone, a polar molecule with genuine dipole–dipole attraction. This apparent inversion is not a contradiction; it demonstrates that dispersion forces, while individually weak per contact, scale with surface area of contact, and a long alkane chain accumulates many more such contacts than a small polar molecule can offer through its single dipole. This is the first hint of a general rule developed fully under London Dispersion Forces: total interaction energy is a sum over all pairwise contacts, and the number of contacts often matters more than the strength of any single one.

The core ideas

The unifying mathematical structure is that every intermolecular energy is a term in the multipole expansion of the Coulomb interaction between two charge distributions, evaluated after appropriate thermal/orientational averaging. The general multipole energy for two point charges is

E_{\text{ion-ion}} = \frac{q_1 q_2}{4\pi\varepsilon_0 r}

where:

$q_1, q_2$ = the two point charges (C)
$\varepsilon_0$ = permittivity of free space, $8.854\times10^{-12}\ \text{C}^2\text{N}^{-1}\text{m}^{-2}$
$r$ = center-to-center separation (m)

For an ion interacting with a permanent dipole (freely rotating case, thermally averaged), the leading term becomes

E_{\text{ion-dipole}} = -\frac{q^2 \mu^2}{6(4\pi\varepsilon_0)^2 k_B T\, r^4}

where:

$q$ = ion charge
$\mu$ = dipole moment of the neutral partner
$k_B T$ = thermal energy, since free rotation averages the orientation-dependent $1/r^2$ term into an effective $1/r^4$ decay
the negative sign shows the averaged interaction is still net-attractive because the dipole preferentially samples attractive orientations

For two freely rotating permanent dipoles (the Keesom/orientation term, foreshadowing Dipole-Dipole Interactions):

E_{\text{dipole-dipole}} = -\frac{2\mu_1^2\mu_2^2}{3(4\pi\varepsilon_0)^2 k_B T\, r^6}

where:

$\mu_1, \mu_2$ = permanent dipole moments of the two molecules
the $1/r^6$ and $1/T$ dependence both arise from thermal averaging over relative orientations, exactly analogous to the ion-dipole case one power of $r$ further out

And for the universal dispersion term (foreshadowing London Dispersion Forces), London's approximate formula is

E_{\text{dispersion}} = -\frac{3}{4}\frac{\alpha_1\alpha_2}{r^6}\cdot\frac{I_1 I_2}{I_1+I_2}

where:

$\alpha_1,\alpha_2$ = polarizabilities of the two interacting electron clouds
$I_1, I_2$ = first ionization energies, used as proxies for the characteristic electronic excitation energy
again $1/r^6$ , but this time arising from second-order perturbation theory rather than thermal orientational averaging

A reading of these four equations together: every one is Coulomb's law diluted by successive layers of averaging or induction. Ion–ion needs no averaging (both charges are fixed) and decays slowest, $1/r$ . Ion–dipole needs one orientational average, decays as $1/r^4$ . Dipole–dipole needs two orientational averages, decays as $1/r^6$ . Dispersion needs no static charge at all — only quantum fluctuation and induced polarization — but ends up at the same $1/r^6$ decay as dipole–dipole because it is also a second-order (two-molecule, two-fluctuation) effect.

Derivation AFrom Coulomb's law to the

1/r^6

dipole–dipole law

Goal. Show, with explicit curved-arrow-style charge bookkeeping (here, algebraic charge terms rather than mechanism arrows, since this is an energetics derivation), why two freely rotating permanent dipoles interact with an energy that decays as $1/r^6$ , not $1/r^3$ as the fixed-orientation formula alone would suggest.

Step 1 — Fixed-orientation dipole–dipole energy. Model each dipole as two point charges $+\delta$ and $-\delta$ separated by a small distance $d$ , with $\mu = \delta d$ . For two dipoles held in a fixed relative orientation at center-to-center distance $r \gg d$ , expanding the four pairwise Coulomb terms (unlike-unlike attractive, unlike-unlike attractive, like-like repulsive ×2) in powers of $d/r$ gives the classical point-dipole result:

E(\theta_1,\theta_2,\phi) = \frac{\mu_1\mu_2}{4\pi\varepsilon_0 r^3}\Big[\sin\theta_1\sin\theta_2\cos\phi - 2\cos\theta_1\cos\theta_2\Big]

This fixed-geometry energy scales as $1/r^3$ — much longer-ranged than what is observed for real, freely tumbling molecules in a fluid.

Step 2 — Molecules in a fluid are not fixed; they rotate. In a gas or liquid, molecules tumble on a timescale fast compared to how long any given orientation persists, sampling all $(\theta_1,\theta_2,\phi)$ combinations. Averaging $E$ directly over all orientations with equal (Boltzmann, $T\to\infty$ ) weight gives exactly zero — attractive and repulsive orientations cancel perfectly to first order. This is the key subtlety: a naive average would (wrongly) predict no net dipole–dipole force at all.

Step 3 — Boltzmann-weight the orientations, not just average them uniformly. The correct treatment recognizes that attractive orientations (low $E$ ) are thermally more populated than repulsive ones (high $E$ ), per the Boltzmann factor $e^{-E/k_BT}$ . Expanding this exponential for $|E| \ll k_BT$ (valid except at very low temperature or very short range):

e^{-E/k_BT} \approx 1 - \frac{E}{k_BT} + \frac{E^2}{2k_B^2T^2} - \cdots

Averaging this expansion over all orientations: the $-E/k_BT$ term still averages to zero (by the same symmetry as Step 2), but the $E^2/2k_B^2T^2$ term does not average to zero, because $E^2 \geq 0$ always. This quadratic term survives and dominates the orientational average.

Step 4 — Extract the surviving term. Since $E \propto \mu_1\mu_2/r^3$ , we have $E^2 \propto \mu_1^2\mu_2^2/r^6$ . Carrying through the full orientational integral (a standard but lengthy solid-angle average) gives the coefficient $2/3$ , reproducing exactly the formula stated above:

E_{\text{dipole-dipole}} = -\frac{2}{3}\frac{\mu_1^2\mu_2^2}{(4\pi\varepsilon_0)^2 k_BT\,r^6}

The minus sign shows the surviving, thermally-averaged interaction is net attractive (low-energy antiparallel-ish alignments are visited more often than high-energy ones), and the extra $1/r^3$ relative to Step 1 comes entirely from squaring the fixed-orientation energy in Step 3 — an entropic/statistical effect, not a new physical force.

Limiting case check. As $T\to\infty$ , $E_{\text{dipole-dipole}}\to 0$ : infinite thermal energy randomizes orientations completely and washes out any net attraction, exactly as physical intuition demands (a very hot gas behaves as if dipoles were absent). As $T\to 0$ , the approximation $|E|\ll k_BT$ breaks down and molecules instead lock into the fixed-orientation $1/r^3$ regime of Step 1 — this is why the $1/r^6$ law is specifically a fluid-phase, finite-temperature result.

Go deeper · Derivation B · 2 minWhy dispersion forces exist even with zero permanent dipole: a perturbation argument

Derivation BWhy dispersion forces exist even with zero permanent dipole: a perturbation argument

Goal. Show why two nonpolar atoms (He, Ar) or nonpolar bonds (C–H, C–C) attract at all, given that their time-averaged dipole moment is exactly zero.

Step 1 — The instantaneous picture. At any single instant, the electrons in an atom are not smeared uniformly; quantum mechanically, the electron density fluctuates, and at any snapshot in time the atom has some nonzero instantaneous dipole moment $\mu_{\text{inst}}(t)$ , even though $\langle \mu_{\text{inst}}(t)\rangle_{\text{time}} = 0$ . This is analogous to how a coin flipped many times averages to 50% heads, yet any single flip is not "half a head."

Step 2 — The instantaneous dipole polarizes the neighbor. This fluctuating dipole on atom 1 creates a small, fluctuating electric field at the location of neighboring atom 2. Atom 2's electron cloud, being polarizable (characterized by polarizability $\alpha_2$ ), distorts in response — its electron density shifts to place a compensating induced dipole moment $\mu_2^{\text{ind}} = \alpha_2 E_1$ oriented so as to be attractive toward atom 1's instantaneous dipole (any induced dipole in a field is, by definition of induction, oriented to lower the electrostatic energy — this is the polarization response, formally identical to how Inductive Effects and Charge Distribution describes electron density shifting toward electron-withdrawing centers).

Step 3 — This attraction is not random-signed, unlike a static dipole–dipole fluctuation. Crucially, because $\mu_2^{\text{ind}}$ is induced by, and therefore correlated with, $\mu_1^{\text{inst}}$ , the two dipoles are never anti-correlated — the induced dipole always "follows" the inducing one in the attractive sense. This is fundamentally different from two independent, uncorrelated permanent dipoles (Step 2 of the previous derivation), where correlation only survives at second order via Boltzmann weighting. Here, the correlation is built in at first order because one dipole is literally caused by the other.

Step 4 — Second-order perturbation theory quantifies the energy. Treating the instantaneous dipole–induced dipole interaction as a perturbation on the unperturbed atomic Hamiltonians, second-order perturbation theory gives an energy lowering

\Delta E^{(2)} = -\sum_{n\neq 0}\frac{|\langle 0|\hat{H}'|n\rangle|^2}{E_n - E_0}

where $\hat H'$ is the instantaneous dipole–dipole coupling operator between the two atoms, scaling as $1/r^3$ ; because the matrix element enters squared, $\Delta E^{(2)} \propto 1/r^6$ , and because it is a second-order (energy-lowering, denominator-positive) correction, the sign is always negative — always attractive, for any pair of atoms, regardless of what they are.

Step 5 — London's closed-form approximation. Replacing the sum over excited states by an effective single "average excitation energy" (approximated by the ionization energy $I$ ) gives London's compact formula already stated:

E_{\text{dispersion}} = -\frac{3}{4}\frac{\alpha_1\alpha_2}{r^6}\cdot\frac{I_1I_2}{I_1+I_2}

Limiting case check. Setting $\alpha_1$ or $\alpha_2 \to 0$ (a hypothetical rigid, unpolarizable electron cloud) makes $E_{\text{dispersion}}\to 0$ : an atom that cannot distort its electron density in response to a field cannot support an induced dipole, and dispersion vanishes entirely — consistent with the physical requirement that dispersion is fundamentally a polarizability effect, not a charge effect.

Worked examples

Example 1 — Foundational: ranking three small molecules by boiling point from structure alone. Consider CH₄ (methane), CH₃F (fluoromethane), and CH₃OH (methanol). Molar masses are 16, 34, and 32 g/mol respectively — nearly identical for the latter two, so mass-driven dispersion differences cannot explain their boiling points. Step 1: identify available IMFs. Methane has only C–H bonds (electronegativity difference C 2.55, H 2.20 — negligible polarity), so the only force is London dispersion. Fluoromethane has a strong C–F dipole (F electronegativity 3.98) giving it dispersion plus dipole–dipole attraction, but no H directly bonded to F, N, or O, so no hydrogen bonding. Methanol has an O–H bond, so it has dispersion, dipole–dipole, and hydrogen bonding (both as donor via O–H and acceptor via O lone pairs). Step 2: rank by cumulative force count and strength. Methane (dispersion only, weak, small molecule) < fluoromethane (dispersion + dipole–dipole) < methanol (dispersion + dipole–dipole + hydrogen bonding, the strongest per-molecule contact). Step 3: compare to real data.

T_{\text{bp}}(\text{CH}_4) = -161.5\ ^\circ\text{C}, \quad T_{\text{bp}}(\text{CH}_3\text{F}) = -78.4\ ^\circ\text{C}, \quad T_{\text{bp}}(\text{CH}_3\text{OH}) = 64.7\ ^\circ\text{C}

The ranking $-161.5 < -78.4 < 64.7$ matches the force-count prediction exactly, and the gap between fluoromethane and methanol (83 °C) is far larger than the gap between methane and fluoromethane (83 °C too, coincidentally similar in this case but for different underlying reasons — added dipole moment in one step, added hydrogen bonding network in the next), underscoring that hydrogen bonding contributes disproportionately once an N–H/O–H/F–H bond is present.

Example 2 — Applied: predicting relative water solubility of two isomeric alcohols. Compare 1-butanol (CH₃CH₂CH₂CH₂OH) and tert-butanol ((CH₃)₃COH), both C₄H₁₀O, both capable of hydrogen bonding to water via their –OH groups. Step 1: identify the competing effect. Both molecules can form comparably strong hydrogen bonds to water (similar O–H bond polarity, similar steric accessibility of the lone pairs in this particular comparison), so hydrogen-bonding capacity alone does not distinguish them. The distinguishing feature is the hydrocarbon portion, which must be accommodated by disrupting water's own hydrogen-bond network (the "hydrophobic effect"). Step 2: reason about dispersion contact area. 1-Butanol's chain is linear, exposing a larger effective nonpolar surface area to solvent per molecule in an extended conformation; tert-butanol is more compact and spherical (branched at the carbinol carbon), presenting less surface area for a given number of carbons, which is energetically cheaper to accommodate in water because it disrupts fewer water–water hydrogen bonds. Step 3: compare to real solubility data. Water solubility of tert-butanol is essentially unlimited (miscible in all proportions), while 1-butanol's solubility is limited to about 7.7 g per 100 g water at 20 °C. This matches the surface-area reasoning: compact, branched hydrocarbons disturb the solvent's hydrogen-bond network less than extended, linear ones of identical molecular formula. Step 4: sanity-check against the limiting case. As chain length increases without branching (propanol → butanol → pentanol → hexanol, all unbranched), water solubility falls monotonically (fully miscible → 7.7 g/100 g → 2.7 g/100 g → 0.6 g/100 g), confirming that increasing exposed nonpolar surface area — not the presence of the –OH group, which is constant across the series — is the controlling variable.

Example 3 — Exam/competition level: quantitative comparison of dipole–dipole vs. dispersion contributions using the derived formulas. Estimate the ratio of the dipole–dipole interaction energy to the dispersion interaction energy for two acetone molecules (CH₃COCH₃) at a face-to-face contact distance of $r = 0.45$ nm and $T = 298$ K, using:

\mu(\text{acetone}) = 2.88\ \text{D} = 9.61\times10^{-30}\ \text{C}\,\cdot\,\text{m}

\alpha(\text{acetone}) \approx 6.4\times10^{-30}\ \text{m}^3\ (\text{in CGS-polarizability-volume units, converted via } \alpha_{SI} = 4\pi\varepsilon_0\alpha_{CGS})

I(\text{acetone}) \approx 9.7\ \text{eV} = 1.554\times10^{-18}\ \text{J}

Step 1: compute the dipole–dipole (Keesom) energy.

E_{\text{dipole-dipole}} = -\frac{2\mu^4}{3(4\pi\varepsilon_0)^2 k_BT\,r^6}

(using $\mu_1=\mu_2=\mu$ ). Plug in $\mu = 9.61\times10^{-30}$ C·m, $4\pi\varepsilon_0 = 1.113\times10^{-10}\ \text{C}^2\text{N}^{-1}\text{m}^{-2}$ , $k_BT = 1.381\times10^{-23}\times298 = 4.116\times10^{-21}$ J, and $r^6 = (0.45\times10^{-9})^6 = 8.30\times10^{-57}\ \text{m}^6$ .

\mu^4 = (9.61\times10^{-30})^4 = 8.53\times10^{-118}\ \text{C}^4\text{m}^4

(4\pi\varepsilon_0)^2 = 1.239\times10^{-20}

E_{\text{dipole-dipole}} = -\frac{2\times 8.53\times10^{-118}}{3\times1.239\times10^{-20}\times4.116\times10^{-21}\times8.30\times10^{-57}}

= -\frac{1.706\times10^{-117}}{1.264\times10^{-96}} = -1.35\times10^{-21}\ \text{J per pair}

Converting to a molar quantity: $-1.35\times10^{-21}\times6.022\times10^{23} = -0.81\ \text{kJ/mol}$ . Step 2: compute the dispersion (London) energy.

E_{\text{dispersion}} = -\frac{3}{4}\frac{\alpha^2}{r^6}\cdot\frac{I^2}{2I} = -\frac{3}{8}\frac{\alpha^2 I}{r^6}

(setting $I_1=I_2=I$ simplifies $I_1I_2/(I_1+I_2) = I/2$ ). With $\alpha = 6.4\times10^{-30}\ \text{m}^3$ converted to SI polarizability $\alpha_{SI}=4\pi\varepsilon_0\times6.4\times10^{-30} = 7.12\times10^{-40}\ \text{C}^2\text{m}^2\text{J}^{-1}$ :

\alpha_{SI}^2 = 5.07\times10^{-79}

E_{\text{dispersion}} = -\frac{3}{8}\times\frac{5.07\times10^{-79}\times1.554\times10^{-18}}{8.30\times10^{-57}}

= -\frac{3}{8}\times\frac{7.88\times10^{-97}}{8.30\times10^{-57}} = -\frac{3}{8}\times9.49\times10^{-41} = -3.56\times10^{-41}\ \text{J per pair}

This is a nonsensically small number, signaling that the naive Gaussian/CGS-to-SI polarizability conversion needs the standard practical shortcut: chemists usually quote $C_6$ dispersion coefficients directly (in J·m⁶) rather than reconstructing them from $\alpha$ and $I$ via unit-fragile formulas. Using acetone's tabulated $C_6 \approx 6.0\times10^{-78}\ \text{J}\,\cdot\,\text{m}^6$ (a standard literature value) in $E = -C_6/r^6$ :

E_{\text{dispersion}} = -\frac{6.0\times10^{-78}}{8.30\times10^{-57}} = -7.23\times10^{-22}\ \text{J per pair}

which corresponds to $-7.23\times10^{-22}\times6.022\times10^{23} = -0.44$ kJ/mol. Step 3: take the ratio.

\frac{E_{\text{dipole-dipole}}}{E_{\text{dispersion}}} = \frac{-0.81}{-0.44} \approx 1.8

Interpretation and limiting-case check. For a small, quite polar carbonyl compound like acetone, dipole–dipole and dispersion contributions are genuinely comparable in magnitude (same order of magnitude, dipole–dipole roughly twice as large at contact distance) — neither dominates outright. As molecular size increases at fixed dipole moment (e.g., moving to a long-chain ketone with the same C=O but a much larger alkyl framework), $\alpha$ grows roughly linearly with the number of polarizable atoms while $\mu$ stays fixed, so $E_{\text{dispersion}}$ grows while $E_{\text{dipole-dipole}}$ does not — this is exactly the mechanism behind the methane/fluoromethane/methanol size-independent comparison in Example 1 breaking down for larger polar molecules, and it is why dispersion eventually dominates boiling-point trends within any homologous series regardless of polarity.

Limiting cases, scope & common traps

Situation	What actually happens	Why
Two nonpolar atoms, $T\to 0$	Dispersion attraction still present, unlike dipole–dipole	Dispersion arises from quantum fluctuation, not thermal orientational averaging — it has no $1/T$ dependence and survives at absolute zero
Dipole–dipole at very short $r$ (comparable to molecular size)	Simple point-dipole $1/r^6$ law breaks down	The multipole expansion assumed $r\gg d$ (charge separation); at contact, higher multipoles and exchange repulsion dominate
Very polar small molecule vs. very large nonpolar molecule	The large nonpolar molecule can have the higher boiling point	Dispersion scales with polarizability (roughly with number of electrons / surface area), which can outgrow a fixed dipole moment
N–H, O–H, F–H bonds	Anomalously strong, highly directional attraction beyond generic dipole–dipole prediction	Small size of H means very short donor–acceptor distances; treated separately as Hydrogen Bonding rather than folded into generic dipole–dipole scaling
Ions in solution	Ion–dipole and ion-induced-dipole forces dominate over dipole–dipole and dispersion	$q$ in the Coulomb numerator is a full elementary charge, vastly larger than any partial charge $\delta$ in a neutral dipole
Branched vs. linear isomers	Branched isomers usually have lower boiling points	Reduced surface-area contact reduces total dispersion, even though total electron count (hence total polarizability) is identical

The chemist behind it

Johannes Diderik van der Waals is the figure most closely associated with recognizing that real gases deviate from ideal behavior because molecules attract one another and occupy finite volume — the physical insight underlying the entire modern taxonomy of intermolecular forces, still collectively called "van der Waals forces" in casual usage even though the detailed mechanisms (dispersion, induction, orientation) were only worked out decades after his 1873 doctoral thesis by Debye, Keesom, and London.

Origins · 1 minJohannes Diderik van der Waals and how this was discovered

Johannes Diderik van der Waals

1837–1923 · Leiden, correcting the ideal gas law

Van der Waals was a Dutch physicist, largely self-taught in advanced mathematics as a young man while working as a schoolteacher, who earned his doctorate in 1873 at the University of Leiden with a thesis titled Over de Continuïteit van den Gas- en Vloeistoftoestand ("On the Continuity of the Gas and Liquid State"). His equation of state, \left(P + \frac{an^2}{V^2}\right)(V - nb) = nRT corrected the ideal gas law with two empirical terms: $b$ accounting for the finite volume occupied by real molecules (excluded volume, a repulsive effect), and $a$ accounting for the mutual attraction between molecules that reduces the effective pressure they exert on container walls. This attractive term $a$ was the first quantitative, testable acknowledgment that molecules pull on one another even when uncharged and un-bonded — precisely the phenomenon later dissected into dipole–dipole, induction, and dispersion contributions. He received the 1910 Nobel Prize in Physics for this work.

History

- 1873 — van der Waals proposes the corrected equation of state, introducing the attractive pressure term

a/V^2

- 1912 — Willem Hendrik Keesom derives the thermally-averaged orientation (dipole–dipole) interaction between freely rotating permanent dipoles.

- 1920 — Peter Debye extends the treatment to include induction forces, where a permanent dipole polarizes a neighboring nonpolar molecule.

- 1930 — Fritz London provides the quantum-mechanical, second-order perturbation-theory derivation of the universal dispersion force, closing the gap left by Keesom and Debye for nonpolar species.

- 1936–present — Linus Pauling and successors systematize hydrogen bonding as a distinct, exceptionally strong subclass of dipole-type interaction central to protein and nucleic acid structure.

Are intermolecular forces "weak" in an absolute sense, or only relative to covalent bonds?

Only relative to covalent bonds. A single hydrogen bond (15–40 kJ/mol) is roughly 10–20× weaker than a typical C–C or C–H bond (~350–410 kJ/mol), and a single dispersion contact between two methyl groups can be under 1 kJ/mol. But these numbers are per-contact; a folded protein or a DNA duplex involves thousands of simultaneous hydrogen bonds and dispersion contacts, whose summed stabilization energy can rival or exceed a covalent bond's strength. This is why breaking a protein's fold (denaturation) can happen with mild heating, while breaking its peptide backbone requires much harsher chemical treatment — the individual forces are weak, but their sum is not.

Do all molecules experience London dispersion forces, even highly polar ones like water?

Yes. Dispersion arises from the polarizability of any electron cloud, regardless of whether the molecule also has a permanent dipole. Water experiences dispersion forces alongside its dominant hydrogen bonding and dipole–dipole contributions; dispersion is simply a smaller fraction of water's total intermolecular cohesion because hydrogen bonding is so strong in comparison. There is no molecule, polar or nonpolar, charged or neutral, that lacks a dispersion contribution — it is the one universal intermolecular force.

How can I quickly predict, without doing any calculation, which of two molecules has the higher boiling point?

First check for hydrogen bonding (an H directly bonded to N, O, or F) — its presence usually dominates unless molecular sizes differ drastically. If both or neither molecule can hydrogen bond, compare dipole moments for a secondary dipole–dipole contribution. Finally, and often decisively for molecules of significantly different size, compare molecular surface area and total number of electrons, since dispersion forces scale with both and frequently outweigh polarity differences in larger molecules. Always compare molecules of similar size first; size differences can overwhelm every other consideration.

Why does the dipole–dipole interaction energy depend on temperature, while the dispersion interaction energy does not?

The dipole–dipole (Keesom) formula's $1/T$ dependence comes specifically from Boltzmann-averaging over molecular tumbling, i.e., orientations of permanent dipoles that are constantly randomized by thermal motion — hotter systems randomize orientation faster, weakening the net averaged attraction. Dispersion forces arise from correlated electronic charge fluctuations, a quantum mechanical effect operating on electronic timescales far faster than molecular rotation or translation; it is present even at absolute zero and has no explicit thermal-averaging step in its derivation, so it carries no $1/T$ factor.

Is it correct to call all of these forces collectively "van der Waals forces"?

Usage varies. Strictly, "van der Waals forces" often refers only to the three additive terms from classical electrostatics acting between neutral molecules: the Keesom (dipole–dipole orientation), Debye (dipole-induced dipole), and London (dispersion) terms — explicitly excluding hydrogen bonding and ion-based interactions, which are usually treated as distinct, stronger categories. Some introductory treatments use "van der Waals forces" loosely to mean "all noncovalent intermolecular forces," which is imprecise; the rigorous convention restricts the term to the three purely electrostatic/polarization terms just listed.

Why does a hydrogen bond get its own category instead of being classified as an unusually strong dipole–dipole interaction?

It is defensible to view a hydrogen bond as an extreme case of dipole–dipole attraction, but the effect is anomalously large and highly directional in a way that generic dipole–dipole scaling badly underestimates — a fact confirmed both experimentally (O–H⋯O bond strengths of 20+ kJ/mol vastly exceed what the Keesom formula predicts for typical O–H bond dipoles at van der Waals contact distances) and computationally (hydrogen bonds show measurable covalent/charge-transfer character from lone-pair donation into the σ* orbital of the X–H bond). Because of this extra covalent-like component and its strict geometric requirements (near-linear X–H⋯Y arrangement), chemists treat it as its own category rather than folding it into generic dipole–dipole discussion.

Does molecular shape matter for intermolecular forces, or only molecular composition?

Shape matters enormously, especially for dispersion forces, which depend on the total area of close molecular contact rather than composition alone. Linear molecules can pack side-by-side with many simultaneous close contacts; branched or spherical molecules of identical formula present less total contact area and therefore weaker net dispersion attraction, as shown directly by comparing boiling points of n-pentane (36 °C) and neopentane (9.5 °C), both C₅H₁₂. Shape is just as fundamental a variable as which functional groups are present.

Intermolecular Forces

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Worked examples

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