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My textbook states that polonium has a lower boiling point than tellurium because it has weaker intermolecular forces of attractions (van der Waals forces).

Why are van der Waals forces of attraction less in polonium as compared to tellurium?

Polonium's atomic size is greater than tellurium, and I believe van der Waals forces are directly proportional to the surface area. Is this the reason?

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  • $\begingroup$ Strongly related: chemistry.stackexchange.com/questions/33815/… $\endgroup$ – Nilay Ghosh Feb 20 '18 at 12:49
  • $\begingroup$ I was told by my friend that, "Due to inert pair effect valence electrons in Po will be less availably and hence van der Walls forces will be less".... Seems dubious, please someone confirm me if this is right/wrong. $\endgroup$ – Zenix Mar 4 '20 at 5:30
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    $\begingroup$ @Zenix It's wrong. Also it's not reasonable to think that that crude rules of thumb would be helpful here. $\endgroup$ – Mithoron Mar 4 '20 at 16:54
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It's mainly about relativistic effects. Relativistic affects mainly affect heavy nuclei, such as polonium etc, so makes their properties slightly different to the trend.[ref] The outer electron configuration of Polonium could differ so much from its lighter family member tellurium due to the electron constriction, causing an outer electron to fall into a lower shell, that the intramolecular forces are weaker. Thus a lower boiling point.

It explains a lot of things that do not seem realistic with very heavy nuclei. Such as the "golden" colour of Gold. It should be silvery, but due to relativistic effects it is a "golden" colour. It's also like Copernicium behaving more like a noble metal, unlike the other metals of group 12, even being gaseous in room temperature.


Reference: Wikipedia article on relativistic quantum chemistry

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