I'm learning about orbital quantum numbers. While checking several elements on the periodic table I noticed that there aren't any atoms that have electrons in the g or h subshells. Why is this?


3 Answers 3


Well, the first available sub-shell for "g" electrons would be 5g (i.e., 1s, 2p, 3d, 4f, so 5g). Based on current trends, we'd need row 8 of the periodic table. We just haven't found those elements yet.

Let's look at it this way.

  • After row 1 (1s electrons only), there are 2 rows of the s and p-block elements before "d" opens up with $\ce{3d^1 4s^2}$
  • After two rows of s- p- and d-block elements, the "f" block opens up with cerium: $\ce{4f^1 5d^1 6s^2}$

So we'd need two more rows of the f-block (i.e., the lanthanides and actinides) and then complete the row.

Right now, we have found element 118, so if we can synthesize a few more, we can open up the "g" block.

Now, there are elements that have g electrons in excited states, so g and h-orbitals are relevant to some chemistry. But the ground-state elements haven't been discovered yet.

  • $\begingroup$ Could you elaborate on elements having g electrons in excited states? Any references? Thanks! $\endgroup$
    – leongz
    May 15, 2019 at 6:33

According to Umemoto and Saito,[1] starting with element 126, elements would possess $\mathrm{5g}$ electrons. The calculated ground-state electronic configurations for elements 126–131 are:

  • element 126: $\ce{[Og] 8s^2 8p^1 6f^4 5g^1}$
  • element 127: $\ce{[Og] 8s^2 8p^2 6f^3 5g^2}$
  • element 128: $\ce{[Og] 8s^2 8p^2 6f^3 5g^3}$
  • element 129: $\ce{[Og] 8s^2 8p^2 6f^3 5g^4}$
  • element 130: $\ce{[Og] 8s^2 8p^2 6f^3 5g^5}$
  • element 131: $\ce{[Og] 8s^2 8p^2 6f^3 5g^6}$

[1] Umemoto, K.; Saito, S. Electronic Configurations of Superheavy Elements. J. Phys. Soc. Jpn. 1996, 65 (10), 3175–3179. DOI: 10.1143/JPSJ.65.3175.

  • $\begingroup$ Why filling of g-orbitals do not start with element- 121 as $\ce{[Og] 8s^2 5g^1}$ according to order of energy of orbitals $\endgroup$
    – Apurvium
    Nov 19, 2022 at 10:44
  • 1
    $\begingroup$ @Apurvium due to relativist effects there is large splitting between the 8p1/2 and 8p3/2 levels and the 8p1/2 is lower energy than 5g, 6f and 7d. Calculations before 1970 suggested that 7d would fill at 121, but later calculations at higher levels of theory showed 8p aip.scitation.org/doi/10.1063/1.1674338 You could ask another question about this for more information. Maybe someone else can explain better. $\endgroup$
    – DavePhD
    Nov 19, 2022 at 13:42

Although, through a massive amount of energy, you could excite the electrons into g and h level orbitals no elements have electrons in those orbitals at the ground state. Think about how difficult it would be to have an electron in such large and excited state. You would need more protons than any of the elements we currently know about.

In short, it takes to much energy for any known elements to hold an electron in a nonexcited state in the g or h orbital.

  • $\begingroup$ This is incorrect. The reason there are no 5g atoms in the ground state is due to the instability of large nuclei. Any element with enough protons to charge-balance 5g electrons would fall to bits due to nucleons being able to overcome the strong force and leave the nucleus by quantum tunnelling. It would not require energy to "have an electron" in a 5g orbital since its energy would still be lower than a free electron (though not by much). $\endgroup$ Jan 8, 2022 at 12:58

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