Is element 111 considered to act as an eka-aurum? Being under the same column of group 11, which usually provides enough evidence for correlation of properties for an eka-element, would this element have the similar properties such as conductivity or inertness with gold?

With the understanding of the correlation between group 11 being minimal to me, is anyone able to predict perhaps the colour, ductility, or other physical properties of this element with reason?


1 Answer 1


As far as I understand, there has been very limited experimental investigation of the properties of roentgenium (Z=111), if at all, simply due to the fact that so few atoms have been produced and they decay so quickly. It is possible to do some chemistry with even single atoms in seconds, but as you might expect it's an extremely arduous endeavour and is riddled with massive error bars. For now, just the synthesis of the element seems to dominate experimental roentgenium research.

There has, however, been ample theoretical investigation of roentgenium's chemical properties, both by extrapolation of behaviour for group 11 elements and by ab initio computational methods, at varying levels of approximation. One difficulty is that roentgenium displays strong relativistic effects, which make accurate calculations harder. Of course, atomic properties are easier to predict than molecular properties, which are themselves much easier to predict than condensed phase properties. So while estimates for ionization energy for atoms may be relatively accurate, physical properties of the pure materials such as ductility or melting point are to some extent little more than informed guesses.

Here is a list of some predicted basic properties of the period 7 $d$-block elements, including roentgenium (source, p. 1691):

That source expands upon some of the general properties of roentgenium. Its electronic configuration is predicted to be $\mathrm{[Rn](5f)^{14} (6d)^9 (7s)^2}$, which is dissimilar to the configuration for gold $\mathrm{[Xe](4f)^{14} (5d)^{10} (6s)^1}$. This is because relativistic effects cause a strong stabilization of $s$ orbitals (now more accurately referred to as $s_{1/2}$ due to spin-orbit coupling) by pulling them closer to the nucleus, while simultaneously destabilising the $d$ orbitals (as well as splitting the $d$ subshell into two groups of degenerate orbitals, two $d_{3/2}$ orbitals and three $d_{5/2}$ orbitals) by pushing them away. This means complete population of the $s$ subshell becomes preferable as relativistic effects get stronger. For example, see the effect here (ibid, p. 1667) in the triad Nb/Ta/Db in group 5.

Roentgenium is expected to be a noble metal, like gold, and in fact from its predicted standard reduction potential, is more noble than gold ($\mathrm{Au^{3+}(aq) + 3\ e^{-} \longrightarrow Au^0(s), \ \ \Delta E^0=+1.52\ V}$, for comparison). That said, once the roentgenium atom is ionized to Rg(III), it can reach higher oxidation states more easily due to less stable filled $6d_{5/2}$ orbitals, and so Rg(V) compounds are expected to be more stable. Interestingly, though in the 6th period transition metals gold (group 11) shows the local "relativistic maximum" with respect to stabilization of the $6s$ subshell, in the 7th period the maximum stabilization of the $7s$ subshell shift to copernicium (group 12) rather than roentgenium. This means roentgenium is expected to be less noble than copernicium.

With respect to physical properties, not much is predicted about roentgenium. It will likely be a very dense metal, with good electrical conductivity. The question about its colour is an interesting one. Wikipedia mentions it is expected to be silvery, even though roentgenium displays stronger relativistic effects, which is likely the cause of the colour of gold. This may be related to the switch in electronic ground state between gold and roentgenium. Unfortunately the source for the claim on Wikipedia is no longer accessible.

  • $\begingroup$ All versions of articles on Wikipedia are available via their history (unless deleted by admins, in such case only they can see them). $\endgroup$
    – Mithoron
    Commented May 16, 2020 at 16:00
  • $\begingroup$ @Mithoron It seems I made a small mistake. The reference for that sentence in Wikipedia is currently the same one from 2015. While the URL appears to have suffered from link rot, its DOI did and still does link to the freely accessible article. $\endgroup$ Commented May 16, 2020 at 23:30

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