It is known that flerovium, with a [core]7s²7p² configuration, is predicted to be noble-gas-like; this is due to the "7p²" part of flerovium actually being (7p(1/2))², i.e. mathematically having an azimuthal number of 1/2 instead of 1, that splits well below the other four "7p" orbitals that become (7p(3/2))⁴.

However, I do not understand why the 7p(1/2) orbital would make flerovium noble-gas-like; the spherical harmonic corresponding to n=7 and l=1/2 is just mathematically not spherically symmetric.

Further complicating the matter is that oganesson, with a [core]7s²(7p(1/2))²(7p(3/2))⁴ configuration, should have a spherically symmetric "combination of the six 7p subshells" like the lighter group 18 elements(cf. Unsöld's theorem), but trying to plug in the mathematical 7p(1/2) and 7p(3/2) functions into Unsöld fails to give a spherically symmetric function as well.

Is there a way to mathematically "fix" this, or is this purely an artifact of relativistic effects in period 7?

  • 1
    $\begingroup$ Isn’t “theoretical evidence” an oxymoron? $\endgroup$
    – Andrew
    Oct 21, 2021 at 11:58
  • 1
    $\begingroup$ Do we have information about the electronic structure of Flerovium ? I know its nuclei have been found. But their half-life is so short that they may disappear before having the chance of meeting enough lwctrons to become neutral atoms. Does anybody know ? $\endgroup$
    – Maurice
    Oct 21, 2021 at 13:57
  • $\begingroup$ @Andrew It is and it is not. It is, if it would try to relate to facts about the observed reality. It is not if it is about to confirm/reject if some statement or theoretical structure is or is not part of particular theoretical model. It is e.g. possible to find evidence that classical electrostatics says $F=\frac{q_1 \cdot q_2 }{ 4 \pi \epsilon r^2}$ $\endgroup$
    – Poutnik
    Oct 21, 2021 at 13:58
  • $\begingroup$ @Maurice I'm sure five seconds of half-life time are enough to reach an electronic ground state, but you can never catch such a particle. You only know you had one after detecting it's decay. $\endgroup$
    – Karl
    Oct 21, 2021 at 16:37

1 Answer 1


In this answer the relativistic spin-orbit coupled probability density distributions as a function of angle are tabulated below for angular momentum states we encounter in known atoms:

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The $p_{1/2}$ angular dependence is constant and thus already spherically symmetric with just the one orbital. For $p_{3/2}$ it is slightly more complicated. Of the two orbitals in this spinor one will have an azimuthal component of $3/2$ corresponding to the probability density function $(3/2)\sin^2\theta$ given in the table; the other will have an azimuthal component of $1/2$ and a probability density function $(1/2)(3\cos^2\theta+1)$. These two functions, applied to one $p_{3/2}$ orbital apiece, then add up to a constant $2$ as shown in right column of the table.

Thus, both spinors separately have spherical symmetry in their electronic probability distributions when fully occupied, leading to spherical symmetry in both the flerovium and oganesson valence-electron distributions.


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