London dispersion forces (LDFs) are the weakest form of van der Waals forces, arising from temporary, instantaneous dipoles caused by the random fluctuation of electron density in any atom or molecule. Although individually very weak (0.1–40 kJ/mol), they are the only intermolecular force present in nonpolar molecules and noble gases, and become significant in large, polarisable molecules. LDFs increase with molecular size, surface area, and number of electrons, explaining why larger alkanes have higher boiling points than smaller ones.
E_disp ≈ -(3/2) * (α₁α₂/r⁶) * (I₁I₂/(I₁+I₂))
LaTeX: E_{\text{disp}} \approx -\frac{3}{2}\frac{\alpha_1 \alpha_2}{r^6}\frac{I_1 I_2}{I_1+I_2}
| Symbol | Meaning | Unit |
|---|---|---|
| α₁, α₂ | Polarisabilities of the two molecules | C²·s²/kg |
| r | Intermolecular distance | m |
| I₁, I₂ | Ionisation energies of the two molecules | J |
| E_disp | Dispersion interaction energy | J |
Problem
Methane (CH₄, MW = 16 g/mol) has a boiling point of –161 °C, while octane (C₈H₁₈, MW = 114 g/mol) has a boiling point of 126 °C. Both are nonpolar. Explain the difference in boiling points using London dispersion forces.
Solution
Step 1: Both CH₄ and C₈H₁₈ are nonpolar, so only London dispersion forces act between molecules. Step 2: Octane has 114/16 ≈ 7× more electrons (58 vs 10) than methane. Step 3: More electrons → larger, more polarisable electron cloud → stronger instantaneous dipoles → stronger London forces. Step 4: Additionally, octane has a larger molecular surface area, allowing more contact points between molecules. Step 5: Stronger intermolecular forces → more energy needed to separate molecules → higher boiling point.
Answer
Octane's boiling point (126 °C) is 287 °C higher than methane's (–161 °C) because it has a much larger, more polarisable electron cloud and greater surface area, resulting in far stronger London dispersion forces.
| Alkane | Formula | Molecular Weight (g/mol) | Boiling Point (°C) |
|---|---|---|---|
| Methane | CH₄ | 16 | –161 |
| Ethane | C₂H₆ | 30 | –89 |
| Propane | C₃H₈ | 44 | –42 |
| Butane | C₄H₁₀ | 58 | –1 |
| Pentane | C₅H₁₂ | 72 | 36 |
| Hexane | C₆H₁₄ | 86 | 69 |
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Van der Waals forces are a collective term for weak, short-range intermolecular attractions that arise from transient or permanent electric dipoles, including London dispersion forces, dipole-dipole interactions, and dipole-induced dipole interactions. Named after Dutch physicist Johannes Diderik van der Waals, these forces explain deviations of real gases from ideal behaviour and govern properties such as boiling points, surface tension, and the adhesion of geckos to surfaces. Their strength scales with molecular size and polarisability, making them significant for large molecules and noble gases.
Dipole-dipole interactions are intermolecular forces that occur between polar molecules, where the partially positive end (δ+) of one molecule is attracted to the partially negative end (δ–) of a neighbouring molecule. These forces are stronger than London dispersion forces but weaker than hydrogen bonds, typically ranging from 1–20 kJ/mol. They are responsible for the elevated boiling points of polar molecules such as HCl, SO₂, and acetone compared to nonpolar molecules of similar molecular weight.
Electronegativity is a measure of the tendency of an atom to attract a shared pair of electrons toward itself within a covalent bond, expressed on a dimensionless scale. The most widely used scale is the Pauling scale, where fluorine is assigned the highest value of 3.98, making it the most electronegative element, and caesium the lowest at 0.79. Electronegativity determines bond polarity, the character of chemical bonds (ionic vs. covalent), and influences molecular properties such as reactivity, acid strength, and solubility.
Named after the German-American physicist Fritz London (1900–1954), who first explained the quantum mechanical origin of these forces in 1930. London showed that temporary fluctuations in electron distribution create instantaneous dipoles that induce dipoles in neighbouring molecules.