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

What decides this: London dispersion is always present. Dipole–dipole is added only if the molecule has a permanent dipole. Dipole–induced dipole appears only when a polar and a non-polar partner mix — not between two identical molecules.

• (a) Argon atoms are non-polar and identical → London dispersion only.
• (b) ${\text{Cl}}_2$ is non-polar (two identical atoms, symmetric bond) → London dispersion only.
• (c) $\text{HCl}$ has a permanent dipole (Cl pulls electrons) → ==London dispersion and dipole–dipole==.

False. Even non-polar molecules have flickering instantaneous dipoles, so London dispersion always acts. That is exactly why non-polar gases such as $\text{Ar}$ or ${\text{CH}}_4$ can still liquefy — see States of matter and condensation.

What controls it: only London dispersion acts (all are single atoms). London strength grows with electron count → polarisability $\alpha$ → the coefficient $C_6$ in $U=-C_6/r^6$. Electron count: $\text{He}(2)<\text{Ne}(10)<\text{Ar}(18)<\text{Kr}(36)<\text{Xe}(54)$. Boiling point order: $\text{He}<\text{Ne}<\text{Ar}<\text{Kr}<\text{Xe}$. Why boiling point tracks this: boiling means giving molecules enough thermal energy to escape their neighbours' attraction, so stronger vdW ⇒ higher BP (see Boiling point and intermolecular forces).

Same electron count ⇒ same intrinsic polarisability, so the tie-breaker is shape / contact area.

• $n$-pentane is a long, extended chain: neighbouring molecules can lie alongside each other over a large surface, giving many close London contacts.
• Neopentane is nearly spherical and compact: less surface touches a neighbour, fewer contacts. More effective contact ⇒ stronger total dispersion ⇒ higher BP for $n$-pentane. Order confirmed.

Water has a permanent dipole ${\mu }_{{\text{H}}_2\text{O}}$; ${\text{O}}_2$ is non-polar but polarisable (${\alpha }_{{\text{O}}_2}$). Water's field induces a dipole in ${\text{O}}_2$ → dipole–induced dipole (Debye) attraction: $U_{\text{di}}\propto -\frac{{\alpha }_{{\text{O}}_2}{\mu }_{{\text{H}}_2\text{O}}^2}{r^6}.$

Equal electron count ⇒ their London contributions are comparable. The difference is the extra channel:

• ${\text{F}}_2$ is non-polar → London only.
• $\text{HCl}$ has a permanent dipole → London plus dipole–dipole (Keesom). That additional attraction raises HCl's boiling point. So even with matched electrons, the polar molecule wins here.

Two competing terms:

• Dipole–dipole $U_{\text{dd}}\propto -{\mu }^2$ — decreasing down the series.
• London $U_{\text{London}}\propto -{\alpha }^2$ — rising sharply because electron count jumps ($\text{HCl}18\to \text{HBr}36\to \text{HI}54$). The London rise is much larger than the dipole–dipole fall, so total attraction increases → BP increases. London dominates for heavy molecules.

Two factors of the field appear, one to make the induced dipole and one to interact with it.

• Induce: ${\mu }_{\text{ind}}=\alpha E$, and $E\propto \mu /r^3$, so ${\mu }_{\text{ind}}\propto \mu /r^3$.
• Interact: $U\propto -{\mu }_{\text{ind}}E\propto -\left(\frac{\mu }{r^3}\right)\left(\frac{\mu }{r^3}\right)=-\frac{{\mu }^2}{r^6}$. $U_{\text{di}}\propto -\frac{\alpha {\mu }^2}{r^6}.$ Two $1/r^3$ fields multiply → $1/r^6$. (The Keesom orientation-average and London's quantum result reach the same $1/r^6$ tail by their own routes.)

$U=-C_6/r^6$, so scaling $r\to 2r_0$ multiplies energy by $(1/2{)}^6=1/64$. $U(2r_0)=\frac{U_0}{64}=\frac{-8.0}{64}=-0.125\text{kJ/mol}.$ Interpretation: doubling the distance kills the attraction to about $1.6\%$ of its value — this steep drop is why vdW forces are short-range.

At $r=1\text{nm}$, $r^6=1$, so:

• (a) $U=-(a+b)=-(90+30)=-120\text{kJ/mol}$.
• (b) London fraction $=\frac{a}{a+b}=\frac{90}{120}=0.75=75\%$. Reading it: London supplies three-quarters of the attraction here — the quantitative face of "London dominates."

All three are tetrahedral and symmetric → net dipole $=0$ → London only for every one. So rank purely by electron count / polarisability.

• ${\text{CH}}_4$: $6+4(1)=10$ electrons.
• ${\text{CF}}_4$: $6+4(9)=42$ electrons.
• ${\text{CCl}}_4$: $6+4(17)=74$ electrons. More electrons → larger $\alpha$ → stronger dispersion → higher BP. Order: ${\text{CH}}_4<{\text{CF}}_4<{\text{CCl}}_4$. (Real BPs: $112,145,350\text{K}$ — consistent.)

Two mechanisms compete along the group ${\text{H}}_2\text{S},{\text{H}}_2\text{Se},{\text{H}}_2\text{Te}$ vs the odd-one-out ${\text{H}}_2\text{O}$.

1. From ${\text{H}}_2\text{S}$ downward: O is absent, so only vdW acts. Electron count rises (${\text{H}}_2\text{S}\to {\text{H}}_2\text{Se}$), London grows, BP rises — hence ${\text{H}}_2\text{Se}>{\text{H}}_2\text{S}$. Normal vdW trend.
2. ${\text{H}}_2\text{O}$ breaks the pattern because O is small and very electronegative, so water forms hydrogen bonds — a specially strong, directional dipole–dipole interaction restricted to H–O/N/F. This extra energy jumps its BP far above what its small electron count would predict. Conclusion: ordinary members obey the electron-count (London) trend; water's anomaly is hydrogen bonding, not ordinary van der Waals force. Keep the two categories separate.

Procedure:

flowchart TD
  A["Same electron count question"] -->|"yes"| B["compare polarity"]
  A -->|"no"| C["more electrons usually wins via London"]
  B -->|"one is polar"| D["polar one adds dipole-dipole so it wins"]
  B -->|"both non-polar"| E["compare shape and surface area"]

Apply to ${\text{Br}}_2$ vs $\text{ICl}$: electron counts are equal ($70$ each), so London contributions are comparable. Tie-break on polarity:

• ${\text{Br}}_2$ — two identical atoms → non-polar → London only.
• $\text{ICl}$ — different atoms (I vs Cl) → permanent dipole → London + dipole–dipole. The extra dipole–dipole term makes $\text{ICl}$ boil higher. Prediction: $\text{ICl}>{\text{Br}}_2$ — matches the data.

$r^6=(0.40{)}^6=0.004096{\text{nm}}^6$.

• London: $U_{\text{L}}=-\frac{1.2\times 10^{-3}}{0.004096}=-0.2930\text{kJ/mol}$.
• Keesom: $U_{\text{K}}=-\frac{4.0\times 10^{-4}}{0.004096}=-0.09766\text{kJ/mol}$.
• (a) Total $U=-0.3906\text{kJ/mol}$.
• (b) London share $=\frac{1.2\times 10^{-3}}{1.2\times 10^{-3}+4.0\times 10^{-4}}=\frac{1.2}{1.6}=0.75=75\%$. The distance factor cancels in the ratio, so London supplies $75\%$ regardless of $r$ — the fraction depends only on the coefficients, the magnitude on distance.

The three vdW energies all fall off as which power of distance? ::: $1/r^6$ Which vdW force acts even in a single-element non-polar substance? ::: London dispersion When two molecules have equal electron counts, what breaks the boiling-point tie? ::: the presence of a permanent dipole (dipole–dipole)