Question bank — van der Waals forces — London dispersion, dipole-dipole, dipole-induced dipole

London dispersion ::: instantaneous dipole ↔ induced dipole; exists in ALL matter; scales with polarisability $\alpha$ (electron count / cloud size). Dipole–dipole (Keesom) ::: permanent dipole ↔ permanent dipole; needs a real, permanent $\mu$; weakened by thermal tumbling. Dipole–induced dipole (Debye) ::: permanent dipole ↔ induced dipole; needs one polar partner and one polarisable partner.

"Noble gases have no van der Waals forces because they have no bonds." ::: False — London dispersion needs no bonds and no permanent charge; flickering electron clouds make it exist in every atom, which is exactly why He, Ne, Ar all condense. "A molecule with zero permanent dipole moment can still attract another identical molecule." ::: True — instantaneous dipoles arise from momentary electron lopsidedness, so non-polar $C{H}_4$, $C{l}_2$ and Ar all attract via London forces. "van der Waals forces are a type of chemical bond." ::: False — they involve no electron sharing or transfer and are ~10–1000× weaker; they act between molecules (intermolecular), not within them. "Dipole–dipole forces are always stronger than London forces." ::: False — for larger, electron-rich molecules London usually dominates; e.g. HI (less polar) boils higher than HCl because its bigger cloud gives far stronger dispersion. "All three van der Waals mechanisms have the same $1/r^6$ distance dependence in their attractive energy." ::: True — Keesom, Debye and London all end up as $U\propto -1/r^6$, which is why the combined tail is written $-C_6/r^6$. "Hydrogen bonding is completely separate from van der Waals forces and unrelated in origin." ::: Half-false — it is electrostatically a special, strong dipole–dipole interaction (H bonded to N/O/F), but by convention we treat it separately because it is much stronger and directional. "Raising temperature weakens all van der Waals forces equally." ::: False — only orientation-dependent ones (dipole–dipole) carry an explicit $1/k_{B}T$ factor from tumbling; London and Debye have no $1/T$ in their leading term. "Dipole–induced dipole forces can be repulsive depending on orientation." ::: False — the induced dipole always lines up favourably with the inducing field, so this force is always attractive (unlike raw permanent-dipole pairs which can momentarily repel).

"F₂ boils higher than HCl because F₂ has more electrons." ::: Wrong on both counts — they have similar electron counts (~18), and HCl boils higher (188 K vs 85 K) because HCl adds permanent dipole–dipole attraction on top of London. "Branching increases boiling point because it packs molecules closer." ::: Wrong — branching makes a molecule more compact/spherical, reducing surface contact area, so London forces decrease; neopentane boils lower than linear n-pentane. "London forces scale with dipole moment $\mu$." ::: Wrong — London forces scale with polarisability $\alpha$ (electron count and cloud size), not with any permanent dipole; a molecule can have $\mu =0$ and still have strong London forces. "The dipole field falls off as $1/r^2$, just like a point charge." ::: Wrong — a dipole is a near-cancelling $+q$/$-q$ pair, so from afar you see only the small difference; its field falls faster, as $1/r^3$. "Because O₂ is non-polar, it cannot interact with water at all." ::: Wrong — water's permanent dipole induces a dipole in O₂ (Debye force, $U\propto {\alpha }_{O_2}{\mu }_{H_{2}O}^2/r^6$), which is why O₂ has some solubility in water. "Van der Waals energies are around 100–1000 kJ/mol." ::: Wrong — that range is covalent/ionic bonds; van der Waals energies are only ~0.1–10 kJ/mol, which is why they are easily overcome and why vdW solids melt at low temperature. "Since dipole–dipole energy for one fixed pair goes as $1/r^3$, that is the value we use for gases." ::: Wrong — in a tumbling gas we Boltzmann-average over orientations, and the tiny net bias turns it into the Keesom $1/r^6$ dependence with a $1/k_{B}T$ factor.

Why does boiling point rise He → Ne → Ar → Kr → Xe? ::: Down the group electron count rises, so polarisability $\alpha$ and thus $C_6$ rise, giving stronger London dispersion and requiring more thermal energy to separate the atoms. Why do all three van der Waals forces end up as $1/r^6$ despite different mechanisms? ::: Each attraction depends on a dipole field ($\propto 1/r^3$) acting through another dipole field ($\propto 1/r^3$), or on a $1/r^3$ field averaged/squared — two factors of $1/r^3$ multiply to $1/r^6$. Why does thermal motion appear ($1/k_{B}T$) only in the dipole–dipole term? ::: Permanent dipoles are free to tumble, so heat scrambles their orientations and weakens the net alignment; induced dipoles are created by the field itself and always align, so tumbling does not undo them. Why is the Debye (dipole–induced dipole) force always attractive but the raw permanent dipole–dipole pair not always so? ::: The induced dipole is generated by the neighbour's field and so points the favourable way by construction; two independent permanent dipoles can be locked head-to-head and momentarily repel before averaging. Why can a bigger (more polarisable) molecule beat a more polar molecule in boiling point? ::: Polarisability scales with electron count and grows fast down a group, so its London contribution can overtake the shrinking dipole–dipole contribution — the classic HI > HBr > HCl reversal. Why do we even need Coulomb's law if molecules are neutral overall? ::: Because Coulomb's law between the separated $+$ and $-$ regions (dipoles), not the zero net charge, is what survives; van der Waals forces are Coulomb attraction between charge-separated regions.

Two identical non-polar atoms with $\mu =0$ each — what force remains? ::: London dispersion only; there is no permanent or purely induced-from-permanent contribution, yet instantaneous dipoles still give net attraction. A perfectly non-polarisable atom ($\alpha =0$): what happens to Debye and London? ::: Both vanish — with $\alpha =0$ no dipole can be induced, so $U_{di}\propto \alpha {\mu }^2=0$ and $U_{London}\propto {\alpha }^2=0$; only a hypothetically rigid cloud would show this. As $r\to \infty$, how fast does van der Waals attraction vanish compared with an ion pair? ::: Much faster — vdW dies as $1/r^6$ while ionic Coulomb energy dies only as $1/r$, so vdW is genuinely short-range. At very small $r$ (clouds overlapping), why doesn't $-C_6/r^6$ keep growing forever? ::: A steep repulsive term (e.g. $+A/r^{12}$ from Pauli exclusion of overlapping clouds) dominates and pushes back, giving the Lennard-Jones potential its energy minimum instead of infinite attraction. One polar molecule and one non-polar molecule together — which of the three forces are present? ::: London (always) plus Debye (polar induces a dipole in the non-polar partner); no dipole–dipole, since only one partner has a permanent dipole. Between two permanent dipoles, do we ever get London forces too, or only dipole–dipole? ::: Both are present — every real molecule has a fluctuating cloud, so London always coexists with dipole–dipole; the total is their sum. A gas at extremely high temperature: what happens to the dipole–dipole contribution? ::: It shrinks toward zero because its $1/k_{B}T$ factor makes orientations nearly random, leaving London and Debye (which lack the $1/T$ factor) as the surviving attractions.