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Some of the places, the electronic configuration of darmstadtium ($\ce{Ds}$) is given as $\mathrm{6d^9 7s^1}$ while at some other places, it is $\mathrm{6d^8 7s^2}$. Which among the two is correct?

darmstadtium (encyclopedia Britanica

darmstadtium (PubChem)

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    $\begingroup$ Electron configurations are merely fictional or just a book keeping work. Don't worry too much about it. Find me a chemist who can tell you how to experimentally determine the electron configuration of sodium. Forget about Ds. $\endgroup$
    – ACR
    Commented Jul 12, 2020 at 15:54
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    $\begingroup$ They should not be asking such questions whose answers they do not know themselves. As I said, ask them how did someone experimentally determine the electron configuration of a simple element such as Na (Z=11)? I bet nobody can answer because I have been searching for this answer for decades without any decent response. $\endgroup$
    – ACR
    Commented Jul 12, 2020 at 17:32
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    $\begingroup$ Anyway, what is the use of such a question ? Up to now, the number of Ds atoms produced in the world is increasing with the time, but it is still much lower than 100. And they all live less than 10 seconds, often less than 1 millisecond... $\endgroup$
    – Maurice
    Commented Jul 12, 2020 at 18:13
  • $\begingroup$ Sir could you please elaborate we cannot determine the electron configuration? $\endgroup$ Commented Jul 13, 2020 at 5:13
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    $\begingroup$ From practical point of view: electronic configuration of an atom strictly speaking referring to the isolated, single atom. If there are more than one low energy state, chances you can observe the ones that are preferred by the given environment. Ie when the answer is difficult, the answer generally doesn't matter. $\endgroup$
    – Greg
    Commented May 30, 2023 at 10:49

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Wikipedia suggests the electronic configuration of Ds as $\ce{[Rn] 5f^{14} 6d^8 7s^2}$ while some others suggests $\ce{[Rn] 5f^{14} 6d^9 7s^1}$.

Relativistic stabilization of 7s orbital in darmstadtium explains its predicted configuration to be $\ce{[Rn] 5f^{14} 6d^8 7s^2}$. It contracts the $l = 0$ subshell i.e. the s subshell which in this case contains two electrons. According to this reference:

The relativistic stabilization of the 7s orbital is 2.6 eV

The effect of the ns orbital contraction reaches its maximum in the 6th period with Au (17.3%) and in the 7th period with element 112 (31%)

The shift of the maximum to element 112 in the 7th period in contrast to gold in the 6th period is due to the fact that in both elements 111 and 112 the ground state electronic configuration is $\ce{d^\ce{q}s^2}$

which means the relativistic effects on 7s subshell are quite significant in the element adjacent to element 111 which is darmstadtium (element 110).

These factors makes darmstadtium to have a configuration of $\ce{[Rn] 5f^{14} 6d^8 7s^2}$ instead of $\ce{[Rn] 5f^{14} 6d^9 7s^1}$.

Reference: The Chemistry of the Actinide and Transactinide Elements, $\ce3^{rd}$ edition, by Lester R. Morss, Norman M. Edenstein, Jean Fuger & Joseph J. Katz

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    $\begingroup$ Are there other sources, especially computational articles, apart from just Wikipedia? The energy levels between s/f/d are probably not very large (the gaps get smaller with larger $n$) and I wonder if there are other factors (particularly relativistic effects) that are important in determining the electronic configuration. $\endgroup$ Commented May 29, 2023 at 11:42
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    $\begingroup$ I dont think we can trust wikipedia for things books or university lectures dont have much about.Though ofcourse it is better than nothing.... $\endgroup$
    – Volpina
    Commented May 29, 2023 at 12:01
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    $\begingroup$ @Proscionexium what I am saying is that we havent studied much elements on the 7th row of the periodic table and predictions rely mostly on the trends of lighter elements which we dont know if they apply to these elements. $\endgroup$
    – Volpina
    Commented May 29, 2023 at 12:04
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    $\begingroup$ @Proscionexium I really believe its a GOOD idea not to continue the discussion in chat and leave the comments in plain sight because the problem here has its roots at the scientific procedure. $\endgroup$
    – Volpina
    Commented May 29, 2023 at 12:10
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    $\begingroup$ @Volpina For sure, experimental evidence would be the best. But in the absence of that, you can still do theoretical calculations (and that's what I was hoping might exist out there). It's not right to say that we can only make an educated guess. Wikipedia is perfectly fine as a place to look up infromation but it is not a source in and of itself — you must always look for the sources which it cites. $\endgroup$ Commented May 30, 2023 at 21:41

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