Concept:London dispersion forces are the only intermolecular forces present in non-polar diatomic halogens. The strength of these forces depends directly on atomic size and polarizability.
Formula:$$ \text{LDF Strength} \propto \text{Polarizability} \propto \text{Electron Cloud Size} $$
Solution:- As you move down Group 17 (Fluorine, Chlorine, Bromine, Iodine), the atoms acquire more electron shells, dramatically increasing in size.
- A larger electron cloud is held less tightly by the nucleus, making it easier to distort into temporary dipoles (high polarizability).
- Fluorine (\( \text{F}_2 \)) is at the very top of the group. It is the smallest molecule with the fewest electrons.
- Because its electrons are held tightly to the nucleus, it has very low polarizability, resulting in the weakest London dispersion forces. (This is why it is a highly volatile gas at room temperature, while Iodine is a solid).
Why other options are incorrect:Chlorine is larger, Bromine is a liquid, and Iodine is a solid, proving that their dispersion forces are increasingly stronger.
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