van der Waals forces — London dispersion, dipole-dipole, dipole-induced dipole: Worked examples

Before we start, one reminder of the three tools, in plain words:

Formula The three van der Waals energies (magnitude ∝)

$U_{London} \propto - \frac{α _{1} α _{2}}{r ^{6}} \frac{I _{1} I _{2}}{I _{1} + I _{2}}, U_{dd} \propto - \frac{μ _{1}^{2} μ _{2}^{2}}{k _{B} T r ^{6}}, U_{di} \propto - \frac{α μ ^{2}}{r ^{6}}$

$μ$ (mu) = dipole moment: how lopsided the charge is (bigger = more polar). See Dipole moment and molecular polarity.
$α$ (alpha) = polarisability: how squishy/deformable the electron cloud is (bigger, heavier atoms = squishier). See Polarisability and Fajans' rules.
$I_{1}, I_{2}$ = the ionisation energies of the two atoms/molecules: how tightly each holds its outermost electrons. They appear in the London formula because a cloud that lets go of electrons more easily (smaller $I$ ) flickers into instantaneous dipoles more readily, giving stronger dispersion.
$k_{B}$ = the Boltzmann constant. It is nature's fixed "exchange rate" that turns a temperature (in kelvin) into an energy: $k_{B} = 1.38 \times 1 0^{- 23} J K^{- 1}$ . The subscript $B$ honours Ludwig Boltzmann. So the product $k_{B} T$ has units of joules and measures the typical thermal jiggle energy a molecule carries at temperature $T$ .
$r$ = distance between the molecules; $T$ = absolute temperature (kelvin). The minus sign means the energy drops (gets more negative) as molecules approach — that is what "attraction" means. Lower energy = happier = stuck together.

The scenario matrix

Every van der Waals question is really asking "which of these three arrows is longest, and why?" Here is the full set of case-classes this topic can throw at you:

Cell	Case class	Degenerate/limit knob	Which example
A	Both non-polar ( $μ_{1} = μ_{2} = 0$ ) → London only	$μ \to 0$	Ex 1
B	One polar, one non-polar → London + dipole–induced dipole	mixed	Ex 2
C	Both polar → London + dipole–dipole	full case	Ex 3
D	Same electrons, different polarity (isolate the dipole term)	control $α$ , vary $μ$	Ex 4
E	Rising electrons fights falling dipole (which wins?)	$α$ ↑ while $μ$ ↓	Ex 5
F	Same formula, different shape (surface area effect)	vary contact area	Ex 6
G	Distance limits: die-off ( $r \to \infty$ ) and touching ( $r \to 0$ )	$r \to \infty$ , $r \to 0$	Ex 7
H	Temperature limit: which term feels heat? ( $T \to \infty$ )	$T \to \infty$	Ex 8
I	Real-world word problem (solubility / sticking)	applied	Ex 9
J	Exam twist: trap where "bigger $μ$ ⇒ higher BP" fails	misdirection	Ex 10

We now hit every cell.

Cell A — both non-polar (London only)

Worked example Ex 1 — Boiling points of the halogens Cl₂, Br₂, I₂

All three are non-polar diatomics ( $μ = 0$ ). Rank their boiling points and explain the mechanism.

Forecast: Which one boils highest? Guess before reading — think "which cloud is squishiest?"

Identify the only force. Since $μ = 0$ for a symmetric homonuclear molecule, both dipole–dipole and dipole–induced dipole vanish. Only London dispersion survives. Why this step? You must kill the dead terms first — a zero dipole makes two of three formulas exactly $0$ , so the whole ranking rests on $α$ .
Rank polarisability by electron count. Electrons per molecule: Cl₂ $= 34$ , Br₂ $= 70$ , I₂ $= 106$ . More electrons in a bigger cloud ⇒ larger $α$ . Why this step? $U_{London} \propto α^{2}$ , so the electron count is the single knob that decides the order.
Translate to boiling point. Larger $α$ ⇒ more negative (stronger) $U$ ⇒ more thermal energy needed to separate ⇒ higher BP. So I₂ > Br₂ > Cl₂. Why this step? Boiling means overcoming intermolecular attraction; see Boiling point and intermolecular forces.

Verify: Real BPs: Cl₂ $238 K$ , Br₂ $332 K$ , I₂ $457 K$ — order matches, and each jump ( $94$ K, $125$ K) grows with the electron jump. ✓

Cell B — one polar, one non-polar (dipole–induced dipole)

Worked example Ex 2 — O₂ dissolving in water

Non-polar O₂ ( $μ = 0$ ) dissolves slightly in polar water ( $μ_{H_{2} O} = 1.85 D$ ). Name the force and write its scaling.

Forecast: With one partner having $μ = 0$ , is there any attraction at all?

Check dipole–dipole. Needs two permanent dipoles; O₂ has $μ = 0$ , so $U_{dd} \propto μ_{1}^{2} μ_{2}^{2} = 0$ . Dead. Why this step? Ruling out dd shows the attraction is not the "obvious" one.
Water's field induces a dipole in O₂. Water's dipole makes a field $E \propto μ / r^{3}$ ; it squishes O₂'s cloud into $μ_{ind} = α_{O_{2}} E$ . Why this step? This is exactly the dipole–induced dipole (Debye) mechanism — one factor of $E$ to make the dipole.
Interaction energy. $U_{di} \propto - μ_{ind} E \propto - α_{O_{2}} μ_{H_{2} O}^{2} / r^{6}$ . Why this step? Second factor of $E$ to interact; two factors of $1/ r^{3}$ give the $1/ r^{6}$ tail.

Verify: Units — $[α]$ (C·m²·V⁻¹) $\times [μ^{2}]$ (C²·m²) $/ r^{6}$ collapses to energy after the $ε_{0}$ constants; sign is negative (attractive), consistent with O₂ dissolving rather than being expelled. ✓

Cell C — both polar (London + dipole–dipole)

Worked example Ex 3 — HCl vs F₂, nearly equal electrons

HCl and F₂ both have $\approx 18$ electrons. HCl boils at $188 K$ , F₂ at $85 K$ . Why the gap?

Forecast: With equal electron count the London parts are similar — so what breaks the tie?

London terms nearly cancel out of the comparison. Equal electrons ⇒ similar $α$ ⇒ similar $U_{London}$ for both. Why this step? Matching the controlled variable ( $α$ ) isolates the effect we care about.
Add the dipole term (and simplify it). F₂ is symmetric ⇒ $μ = 0$ ⇒ no dd. HCl interacts with another HCl, so both partners are the same molecule: $μ_{1} = μ_{2} = μ$ . The general Keesom form $U_{dd} \propto - μ_{1}^{2} μ_{2}^{2}$ then collapses to $- μ^{2} \cdot μ^{2} = - μ^{4}$ . With $μ_{HCl} = 1.08 D \neq = 0$ this bonus term is non-zero. Why this step? Only HCl gets a bonus attraction; showing the $μ_{1} = μ_{2} = μ$ reduction makes clear why we can write a single $μ^{4}$ .
Conclude. Extra attraction ⇒ HCl needs more heat ⇒ higher BP. $188 > 85$ . ✓ Why this step? Converts the energy comparison back into the measured quantity.

Verify: Ratio $188/85 \approx 2.2$ — a large gap consistent with an added force channel, not just a small tweak. ✓

Cell D — same electrons, vary only polarity

Worked example Ex 4 — Isolating the pure dipole effect: N₂ vs CO

N₂ and CO are isoelectronic (both 14 electrons, near-identical size, so nearly equal $α$ ). N₂ has $μ = 0$ ; CO has a tiny $μ = 0.12 D$ . BPs: N₂ $77 K$ , CO $82 K$ .

Forecast: With $α$ held fixed, will the tiny CO dipole matter at all?

Freeze the London term. Equal electron count ⇒ equal $α$ ⇒ equal $U_{London}$ . This is a controlled experiment. Why this step? To see the dipole effect alone, everything else must be equal.
Only CO adds dd (identical partners ⇒ $μ^{4}$ ). Two CO molecules meeting each other have $μ_{1} = μ_{2} = μ_{CO}$ , so $U_{dd} \propto - μ_{1}^{2} μ_{2}^{2} = - μ^{4}$ . Because $μ_{CO} = 0.12 D$ is tiny, $μ^{4}$ is very tiny. Why this step? Stating $μ_{1} = μ_{2} = μ$ justifies the single-symbol $μ^{4}$ ; the fourth power warns you a small dipole gives an almost invisible boost.
Predict a small rise. CO should boil slightly higher than N₂. Observed: $82 > 77$ , a mere $5 K$ . ✓ Why this step? Confirms dd is real but weak when $μ$ is small.

Verify: Gap is only $5/77 \approx 6.5%$ — matching the " $μ^{4}$ of a tiny $μ$ " prediction, versus HCl's much larger gap in Ex 3 where $μ$ was $9 \times$ bigger. ✓

Cell E — electrons rise while dipole falls

Worked example Ex 5 — The classic fight: HCl, HBr, HI

Down the series the dipole moment falls ( $1.08 \to 0.82 \to 0.44 D$ ) but the electron count rises ( $18 \to 36 \to 54$ ). BPs: $188 \to 206 \to 238 K$ . Who wins?

Forecast: Falling $μ$ pushes BP down; rising $α$ pushes it up. Which arrow is longer?

Write the two opposing trends. For identical partners $U_{dd} \propto μ^{4}$ decreases; $U_{London} \propto α^{2}$ increases. Why this step? Naming the tug-of-war stops you from blindly using "bigger $μ$ ⇒ higher BP".
Estimate the dipole loss. $(μ_{HI} / μ_{HCl})^{4} = (0.44/1.08)^{4} \approx 0.027$ — the dd term collapses to $\sim 3%$ . Why this step? Quantifying shows dd is nearly wiped out down the group.
Estimate the London gain quantitatively. Polarisability rises roughly in step with electron count; a good rule of thumb is $α \propto N_{e}$ (electrons). From HCl to HI, $N_{e}$ goes $18 \to 54$ , a $3 \times$ rise, so $α$ rises $\sim 3 \times$ and $U_{London} \propto α^{2}$ rises $\sim 3^{2} = 9 \times$ . Why this step? A $9 \times$ gain in London utterly swamps the $0.027 \times$ collapse in dd — now the "which wins" is numeric, not a hand-wave.
Conclude. BP rises despite falling polarity: London dominates. $188 < 206 < 238$ . ✓ Why this step? Delivers the exam lesson: for heavier molecules London usually wins.

Verify: $(0.44/1.08)^{4} \approx 0.0275$ (dd nearly gone) while $(54/18)^{2} = 9$ (London up ninefold) — the ninefold gain dominates, and BP indeed climbs. ✓

Cell F — same formula, different shape

Look at the two shapes below — same atoms, same electron count, but different contact area. This figure is the whole argument of Ex 6, so read it before the steps.

Figure — van der Waals forces — London dispersion, dipole-dipole, dipole-induced dipole

Figure (Ex 6) — shape controls contact area. Left: two n-pentane chains lying side by side (blue and green), with red bars marking the many points where their surfaces touch. Right: two neopentane spheres, touching at just one point (single red bar). Same atoms, but the linear molecule presents far more surface to snuggle against, so it feels more London attraction. (Alt-text: two side-by-side zig-zag chains joined by many short red contact bars on the left; two circles touching at a single red bar on the right.)

Worked example Ex 6 — n-pentane vs neopentane (both C₅H₁₂)

Same molecular formula, same electron count, same $α$ -potential. Yet BPs differ: n-pentane $309 K$ , neopentane $283 K$ . Why?

Forecast: If electrons are equal, how can shape change London strength?

Recall London acts over touching surface. Instantaneous dipoles must be close; a bigger contact patch means more of the cloud interacts. Why this step? London strength depends not just on how many electrons, but how much of the surface can snuggle up.
Compare shapes (look at the figure above). n-pentane is a long chain — the red bars on the left panel show many side-on contact points; neopentane is a compact sphere — the right panel shows a single red contact bar. Why this step? The picture makes "surface area" concrete: count the red bars — more bars mean more London contact.
Conclude. More contact ⇒ stronger total London ⇒ higher BP for n-pentane. $309 > 283$ . ✓ Why this step? Turns geometry into the measured trend.

Verify: Gap $309 - 283 = 26 K$ , a real but moderate shift — exactly what a shape-only change (not an electron change) should produce. ✓

Cell G — the distance limits (far AND touching)

The plot below is the punchline of Ex 7: it draws the vdW energy against separation and marks what happens both when you double $r$ and when you push $r$ toward zero. Read the curve first, then the steps.

Figure (Ex 7) — how fast attraction dies, and why it can't win at zero. Blue curve: the van der Waals attraction

- 1/ r^{6}

; orange curve: a bare point charge

- 1/ r

, both against separation

r

. The two red dots on the blue curve show that moving from

r

2 r

drops the attraction to

U (r) /64

. The purple dashed curve is the short-range repulsion

+ 1/ r^{12}

that dominates as

r \to 0

, and the green curve is their sum (the Lennard-Jones potential) with its minimum-energy "resting" separation marked. (Alt-text: attraction curve flattening quickly to the right; a steep positive repulsion wall rising to the left; their sum forms a well with a marked minimum.)

Worked example Ex 7 — Both distance limits: pulling apart AND pushing together

Two molecules attract with London energy $U (r) \propto - 1/ r^{6}$ . (a) Pull them from $r$ to $2 r$ : by what factor does attraction weaken, versus a bare charge $1/ r$ ? (b) What happens as $r \to 0$ (molecules almost touching)?

Forecast: For (a) most people say "half" — is that right? For (b), does the $- 1/ r^{6}$ attraction just keep growing forever as they touch?

(a) Use the $r^{- 6}$ tail. $\frac{U ( 2 r )}{U ( r )} = \frac{( 1/ ( 2 r ) ) ^{6}}{( 1/ r ) ^{6}} = \frac{1}{2 ^{6}} = \frac{1}{64}$ . Why this step? The whole van der Waals family shares this $r^{- 6}$ decay; that steep power is the point. On the figure this is the drop between the two red dots on the blue curve.
(a) Contrast with Coulomb $1/ r$ . A point charge would only weaken to $1/2$ (the gently-falling orange curve). See Coulomb's law. Why this step? Shows why vdW is "short-range": it dies $32 \times$ faster than a bare charge over the same move.
(a) Limit $r \to \infty$ . $U \to 0$ extremely fast — far-apart molecules feel essentially nothing (blue curve hugs the axis). Why this step? Covers the large-distance degenerate case: no long-range vdW glue.
(b) Limit $r \to 0$ . The naive $- 1/ r^{6}$ would blow up to $- \infty$ (infinite attraction), which is unphysical — atoms cannot overlap. In reality, electron clouds repel strongly at short range: a $+ A / r^{12}$ term (Pauli repulsion, the purple curve) grows even faster than $- 1/ r^{6}$ and dominates. So the true energy (green curve, the Lennard-Jones potential $U = A / r^{12} - C_{6} / r^{6}$ ) turns upward and shoots to $+ \infty$ as $r \to 0$ . Why this step? Without this the reader would wrongly think molecules collapse together; the $r^{- 12}$ wall is what gives matter its finite size and sets the equilibrium spacing.
(b) The balance point. Between the far-off zero and the near-in wall lies a minimum — the green dot — where net force is zero. That is the molecules' comfortable resting separation. Why this step? Explains why solids and liquids have a fixed inter-particle distance rather than collapsing; see States of matter and condensation.

Verify: $2^{6} = 64$ , so $1/64 \approx 0.0156$ — a $98.4%$ collapse from a mere doubling; and at small $r$ , since $12 > 6$ , the $1/ r^{12}$ term outruns $1/ r^{6}$ so $U \to + \infty$ (repulsive), not $- \infty$ . ✓

Cell H — the temperature limit

Worked example Ex 8 — Which force feels heat?

Of the three vdW energies, which one weakens as temperature $T$ rises, and which two do not?

Forecast: Guess which formula has a $T$ in it before checking.

Scan for $T$ . Only $U_{dd} \propto - \frac{μ _{1}^{2} μ _{2}^{2}}{k _{B} T r ^{6}}$ contains $T$ in the denominator (recall $k_{B} T$ is the thermal jiggle energy). Why this step? You solve this by reading the formulas, not guessing.
Explain the $1/ T$ . Dipole–dipole relies on molecules aligning + to −. Heat tumbles them, randomising orientation, so the net alignment bias shrinks as $1/ T$ . Why this step? Connects the algebra to the physical picture of tumbling molecules.
The other two are $T$ -independent. London and dipole–induced dipole create fresh aligned dipoles each instant, so tumbling can't spoil them. Why this step? Covers the limiting case $T \to \infty$ : dd $\to 0$ , but London and Debye survive, so they are the forces that keep e.g. noble gases condensable.

Verify: As $T \to \infty$ , $U_{dd} \to 0$ while $U_{London}, U_{di}$ stay finite — so at very high $T$ only the temperature-free channels remain. ✓ (sign check: $U_{dd} < 0$ for all finite $T > 0$ , always attractive.)

Cell I — real-world word problem

Worked example Ex 9 — Why does a gecko stick to glass?

A gecko's foot has millions of tiny hairs (setae). No glue, no suction. Explain the adhesion using vdW, and why many small contacts beat one big one.

Forecast: A single vdW contact is $\sim 10 pN$ — laughably weak. How does a gecko hold its whole body?

Identify the force. Gecko keratin and glass are both non-polar-ish; the universal force present between any two surfaces is London dispersion. Why this step? Kills the "must be glue/suction" misconception — London works even with no bonds, just like the condensation of noble gases.
Use surface-area logic (from Ex 6). London strength scales with real contact area. Splitting a foot into millions of nanohairs lets the surface conform to microscopic bumps, multiplying true contact. Why this step? Reuses the shape/area principle in a new setting.
Add up. Take $1, 000, 000$ setae, each holding $\sim 10 pN = 1 0^{- 11} N$ . Total $= 1 0^{6} \times 1 0^{- 11} N = 1 0^{- 5} N$ per patch; scaled across the whole foot's $\sim 1 0^{9}$ finer spatulae the total reaches tens of newtons. Why this step? Shows a weak force made strong by number — the real insight, and why vdW is not "negligible" in bulk.

Verify: $1 0^{6} \times 1 0^{- 11} N = 1 0^{- 5} N$ ; scaling to $\sim 1 0^{9}$ spatulae gives order tens of newtons — far above a $\sim 0.5 N$ gecko weight. The order of magnitude works. ✓

Cell J — the exam trap

Worked example Ex 10 — "Bigger dipole ⇒ higher BP?" — spot the trap

Statement on an exam: "Since HI is less polar than HCl, HI must boil at a lower temperature." True or false, and why is it a trap?

Forecast: Sounds airtight. Is it?

Name the assumed rule. The claim uses only $U_{dd} \propto μ^{4}$ , ignoring London entirely. Why this step? Exposing the hidden assumption is how you defuse the trap.
Recall Ex 5's numbers. From HCl to HI, dd falls to $\sim 0.027 \times$ but London rises $\sim 9 \times$ (electron count triples, $α^{2}$ up ninefold). London dominates. Why this step? Brings in the quantified term the trap conveniently forgot.
Verdict. False. HI boils higher ( $238 K$ ) than HCl ( $188 K$ ) despite lower $μ$ . Why this step? States the corrected conclusion cleanly.

Verify: $238 > 188$ contradicts the claim's prediction of "lower", confirming it is false. ✓

Recall

Recall Which cell does each fact live in?

A zero-dipole molecule can still attract — which force? ::: London dispersion (Cell A/I) Doubling $r$ weakens vdW by what factor? ::: $2^{6} = 64$ (Cell G) What stops molecules collapsing as $r \to 0$ ? ::: The steep $+ 1/ r^{12}$ Pauli repulsion outruns the $- 1/ r^{6}$ attraction (Cell G) Which of the three energies contains temperature $T$ ? ::: Dipole–dipole (Keesom), $\propto 1/ T$ (Cell H) What does $k_{B}$ mean and what are its units? ::: The Boltzmann constant, $1.38 \times 1 0^{- 23} J K^{- 1}$ ; $k_{B} T$ is thermal energy Why can HI boil higher than HCl despite lower $μ$ ? ::: More electrons ⇒ larger $α$ (about $3 \times$ ) ⇒ London up $\sim 9 \times$ , which dominates (Cells E, J) n-pentane vs neopentane: what property differs? ::: Shape/contact surface area, not electron count (Cell F)