I've read in a textbook that in the fluorine atom, for example, high ionization energy reflects the fact that the electrons are tightly held. Thus compounds containing many fluorine atoms tend to have low polarizabilities, and so the London dispersion interaction is weak. This accounts for the fact that many nonmetal fluorides, such as $\ce{SF6}$, $\ce{PF5}$, and $\ce{IF7}$ occur as gases at room temperature. Even $\ce{UF6}$, despite its high relative molar mass of $\pu{352 g/mol}$, sublimes at only $\pu{57 ^\circ C}$.

I don't really understand why polarizability accounts for the compounds thermodynamic properties. Can someone explain this? Thanks a lot!

  • 3
    $\begingroup$ Re. formatting: the $\ce{...}$ and $\pu{...}$ macros are pretty useful for chemical formulae and quantities with units respectively (these should be typeset in upright font $\ce{SF6}$, not italic $SF_6$). Section 3 of chemistry.meta.stackexchange.com/questions/86 has more info on how to use these. Also, you explained it yourself, no? It is because dispersion forces are weaker. $\endgroup$ – orthocresol Feb 5 at 0:34

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