Why do relativistic effects that cause contraction of the $s$ and $p$ orbitals impact the $s$ and $o$ orbitals further away from the nucleus?

Question

I was reading quite a well-known paper titled 'Relativistic Effects in Structural Chemistry' by PEKKA PYYKKO.

In the paper, he describes how the electrons occupying $s$ and $p$ orbitals move at relativistic speeds and so orbit closer to the nucleus due to relativistic mass increase. He then states that due to orthogonality this causes the s orbitals in the higher principal quantum number $n$ orbitals to also contract. However, he states that for higher s shells because of 'interacting relativistic and shell structure effects, their contraction can in fact be even larger' than the 1s or lower s orbitals.

What I do not understand is what exactly the author means by interacting relativistic and shell structure effects, what are these affects specifically.

Answer 1

It is implicitly explained on the second page of the paper. Interacting relativistic effects should be intended as the spin-orbit effects, while by shell structure effects Pyykko means the effect due to the combination of the contraction of the inner $s$ and $p$ shells and the need for $d$ and $f$ shells to be orthogonal to the core states. The two effects result in the more effective screening of the nuclear charge for those states and then in an expansion of those orbitals.

In passing, I would note that the way Pyykko described the traditional way of introducing relativistic effects is nowadays outdated for a physicist. Speaking in terms of velocity of the electrons is not consistent with quantum mechanics and should be intended as a conventional way to speak about the average kinetic energy associated with one electronic state. Furthermore, the relativistic mass could be eliminated, in terms of the relativistic energy.