I am confused about what seems to me two very different explanations for the relativistic contraction of orbitals for heavy elements.

My teacher told me that since 1s electrons in heavy elements travel very fast their momenta is much larger than predicted by classical physics. Hence the 1s orbital contracts lead to the contraction of all ns orbitals so that their overlap continues to be zero. D and f orbitals on the other hand do not contract. Because of their symmetry, they do not need to contract for overlap with the 1s orbital to remain 0. They in fact expand because of increased shielding from s electrons.

My book however says the reason for the contraction is that since ns orbitals are non-zero at the nucleus they experience the effects of nuclear charge more and hence relativistic contraction is most significant in them (it as if implies that the ns orbitals are all traveling at relativistic speeds). Which one is correct? Or are both of them oversimplifications?

  • 1
    $\begingroup$ I would rather close this then put a bounty... $\endgroup$
    – Mithoron
    Oct 29, 2020 at 13:54
  • $\begingroup$ Both effects are happening. For orbitals to contract two conditions must be met, 1) the electrons in the orbital must travel at relativistic speeds and 2) the orbitals must have a non-zero probability density at the nucleus (s, p). See this earlier answer for more details. $\endgroup$
    – ron
    Oct 29, 2020 at 15:10
  • $\begingroup$ s orbitals always overlap. More generally, there is no requirement for zero orbital overlapping and they generally do overlap each other. $\endgroup$
    – Poutnik
    Oct 30, 2020 at 9:24


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