Intermolecular Forces: Van der Waals Radius and London Dispersion (Transcript Notes)

Van der Waals radius and atom size

• The Van der Waals radius for an atom is one half the distance between atoms at the potential energy minimum. In equation form: $R<em>{ ext{vdW}} = frac{1}{2} d</em>{ ext{min}}$
• In the figure discussed, the black arrows show the internuclear distance, and the red arrows show the Van der Waals radii.
• Example comparison: helium vs xenon. Xenon has 54 electrons; helium has 2 electrons. More electrons mean a larger electron cloud, which contributes to a larger atomic size.

Electron cloud, polarizability, and size trends

• Electrons are attracted to the nucleus; the farther the electrons are from the nucleus, the weaker the attractive force they feel.
• When electrons are farther from the nucleus, the electron cloud is easier to distort, i.e., more polarizable.
• Xenon has electrons farther from the nucleus than helium, making its electron cloud more polarizable and thus more easily distorted ("floppier").

London dispersion forces

• London dispersion forces arise because instantaneous dipoles occur due to fluctuations in the electron distribution.
• They are stronger for bigger atoms or molecules because bigger atoms have more electrons and are more polarizable.
• A larger electron cloud leads to a larger separation of charge during fluctuations, enhancing dispersion forces.
• London dispersion forces are also stronger for objects with more surface area, because there are more contact regions to interact.
• Conceptual note: London dispersion forces are one type of intermolecular force (see next section).

Intermolecular vs intramolecular forces

• Intermolecular forces: forces between molecules (you must have two or more particles to observe an intermolecular force).
• Prefix inter- means between; related mnemonic: inter means between, as in intermolecular interactions.
• Contrast with intra- in words like intramural sports, which occur within one campus or set of walls (within a single group).

Practice question 1

• Question prompt: Compared to helium, the London dispersion forces between xenon atoms are expected to be stronger than those between helium atoms.
• Answer: Yes — the London dispersion forces between xenon atoms are stronger.
• Rationale: Xenon has a larger electron cloud (more electrons) than helium, making it more polarizable. Increased polarizability allows for larger instantaneous dipoles and greater fluctuations in charge distribution, which strengthens London dispersion forces.
• General takeaway: London dispersion forces tend to be stronger for larger atoms or molecules due to more electrons and greater polarizability.

Practice question 2: Melting/boiling points and strength of interactions

• Core concept: To melt (solid to liquid) or boil (liquid to gas), you must break intermolecular interactions.
• The amount of energy (heat) required to break those interactions depends on their strength.
• Prediction: Xenon should have a higher melting and/or boiling point than helium because xenon has stronger intermolecular interactions.
• Given values: Helium melting point $T<em>m( ext{He}) ext{ around } 1~ ext{K}$; Xenon melting point $T</em>m( ext{Xe}) ext{ around } 161~ ext{K}$.
• Conclusion: The prediction is correct; xenon’s stronger dispersion forces lead to a much higher melting point relative to helium.

Key numerical references and definitions

• Electron counts used for illustration: Xenon has $54$ electrons; Helium has $2$ electrons.
• Van der Waals radius definition: $R<em>{ ext{vdW}} = frac{1}{2} d</em>{ ext{min}}$ where $d_{ ext{min}}$ is the internuclear distance at the potential energy minimum.
• Melting point comparison values cited: $T<em>m( ext{He}) \,\approx\, 1\ \text{K}$; $T</em>m( ext{Xe}) \,\approx\, 161\ \text{K}$.

Connections to broader concepts and implications

• Larger electron clouds correlate with greater polarizability, which enhances dispersion forces and affects macroscopic properties like phase transitions (melting/boiling points).
• Trends in van der Waals radii as atoms get larger will influence predictions of condensation, gas behavior, and interactions in condensed phases.
• Practical relevance: understanding these forces helps in predicting liquefaction of gases, cryogenics, material design, and interpreting real-world phenomena such as vapor pressures and boiling behaviors.

Summary of core ideas

• Van der Waals radius is a geometric quantity linked to the distance at the potential energy minimum: $R<em>{ ext{vdW}} = frac{1}{2} d</em>{ ext{min}}$.
• Larger atoms (more electrons) have bigger, more polarizable electron clouds, leading to stronger London dispersion forces.
• London dispersion forces strengthen with increased size and surface area, impacting melting/boiling points and overall intermolecular interactions.
• Intermolecular forces require two or more particles; intra- refers to forces within a single particle or structure.
• In the xenon-helium comparisons, xenon’s larger size yields stronger dispersion forces and a much higher melting point compared to helium (approximately 161 K vs 1 K in the given notes).