Predicting Boiling Points ranked boiling points with a simple ladder: dispersion < dipole–dipole < hydrogen bonding. That ladder works when the molecules are about the same size. When they are not, size can beat polarity, and you need to know what makes dispersion forces strong.
Key idea
Dispersion forces act between all molecules and grow with the number of electrons (bigger, more polarizable electron clouds) and with the contact area between molecules (long, flat shapes beat compact, round ones). In large molecules dispersion often outweighs dipole–dipole forces: $\mathrm{CCl_4}$ (nonpolar, 74 electrons) boils at $77^\circ\text{C}$, while $\mathrm{CH_3F}$ (polar, 18 electrons) boils at $-78^\circ\text{C}$.
Why it works
A dispersion force comes from a fleeting, lopsided moment in one molecule's electron cloud that induces a matching one in its neighbor, so the two attract for an instant. A large cloud, with many electrons far from the nuclei, is easily distorted (polarizable), so its temporary dipoles are big. Each molecule also feels these attractions at every point where it touches a neighbor, so more surface in contact means more total attraction. A permanent dipole adds a steady attraction on top, but for a small molecule that bonus is modest. Hydrogen bonds (H on N, O or F) are the strongest of the three per contact, which is why water still beats molecules several times its size.
| Substance | Electrons | Polar? | Boiling point |
|---|---|---|---|
| $\mathrm{HCl}$ | 18 | yes | −85 °C |
| $\mathrm{HBr}$ | 36 | yes (less) | −66 °C |
| $\mathrm{HI}$ | 54 | yes (least) | −35 °C |
| $\mathrm{HF}$ | 10 | yes, H-bonds | 20 °C |
| n-pentane | 42 | no | 36 °C |
| neopentane | 42 | no | 10 °C |
Worked example: Size against polarity
Which has the higher boiling point, $\mathrm{CCl_4}$ or $\mathrm{CH_3F}$? Explain.
- $\mathrm{CH_3F}$ is polar, so it has dipole–dipole forces; $\mathrm{CCl_4}$ is nonpolar (its bond dipoles cancel).
- But $\mathrm{CCl_4}$ has 74 electrons to $\mathrm{CH_3F}$'s 18, a much larger and more polarizable cloud.
- Its dispersion forces are far stronger than all of $\mathrm{CH_3F}$'s attractions combined.
Answer: $\mathrm{CCl_4}$ ($77^\circ\text{C}$ against $-78^\circ\text{C}$).
Worked example: Same formula, different shape
n-Pentane and neopentane (2,2-dimethylpropane) are both $\mathrm{C_5H_{12}}$. Which boils higher?
- Same electrons, both nonpolar: only dispersion forces, so the difference must be shape.
- n-Pentane is a long chain that can lie alongside its neighbors; neopentane is nearly a ball that touches neighbors only at a few points.
Answer: n-Pentane ($36^\circ\text{C}$ against $10^\circ\text{C}$).
Worked example: Ranking a series
Rank $\mathrm{CH_4}$, $\mathrm{C_3H_8}$ and $\mathrm{C_8H_{18}}$ by boiling point.
- All nonpolar hydrocarbons: only dispersion forces.
- More carbons, more electrons and a longer chain: stronger dispersion.
Answer: $\mathrm{CH_4}$ ($-162$) < $\mathrm{C_3H_8}$ ($-42$) < $\mathrm{C_8H_{18}}$ ($126^\circ\text{C}$).
Common mistake
Thinking dispersion forces are always weaker than dipole–dipole forces, and so predicting that the polar $\mathrm{CH_3F}$ boils above $\mathrm{CCl_4}$. The ladder "dispersion < dipole < H-bond" only compares molecules of similar size. Check: compare electron counts first. If one molecule has several times as many electrons, expect it to win unless the other has hydrogen bonds.