Is seven a hard theoretical limit (there can’t be an eighth period) or is it there just because we haven’t discovered/synthesized anything beyond that?
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2$\begingroup$ It's an interesting and complex subject: en.wikipedia.org/wiki/Extended_periodic_table $\endgroup$– theoristCommented Mar 7, 2021 at 5:18
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1$\begingroup$ The stability of heavy nuclei with too many protons has little to do with periodicity of patterns in electron behaviour. $\endgroup$– PoutnikCommented Mar 7, 2021 at 7:10
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5$\begingroup$ I don't have any trouble theorizing the 27th period. Synthesizing those elements is a "bit" more complicated. $\endgroup$– MaxWCommented Mar 7, 2021 at 11:17
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1$\begingroup$ chemistry.stackexchange.com/questions/2786/… chemistry.stackexchange.com/questions/43144/… $\endgroup$– MithoronCommented Mar 7, 2021 at 14:32
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4 Answers
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There is no theoretical limit to the number of rows, but...
- On one hand, as Tom Lehrer states, there may be many others but they haven't been discovard.
- On the other hand, after all s, p, d, f orbitals are filled in the seventh period, there might be a new row, or there might be an extension of the seventh row, as a new type of orbital (g? Anyone know it there's a name?). Each orbital has double an odd number (double because the Pauli Exclusion Principle allows pairs of electrons of opposite spins), so the next orbital would have 18 electrons. Is it possible? Inner electrons would be moving at very high speeds, and they'd also shield outer electrons from attraction by the positive nucleus.
- On the gripping hand, transuranics may last a few seconds... or a few attoseconds, so getting spectra might be problematic.
answered Mar 7, 2021 at 5:09
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$\begingroup$ The S, P, D and F orbitals aren't all filled. Each of the outer shells has more S, P, D, F and G orbitals with room for more electrons. It's just that the lower number elements we already know don't use those outer orbitals. $\endgroup$– erebusCommented Mar 7, 2021 at 13:34
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1$\begingroup$ This is more of a few random comments then an answer :/ And you ask about thing you should know before you start writing! $\endgroup$– MithoronCommented Mar 7, 2021 at 21:23
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$\begingroup$ @Mithoron, agreed, this is not intended as an answer, but as hints to the questioner where to look for info. $\endgroup$ Commented Mar 8, 2021 at 18:26
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1$\begingroup$ Meh, finding duplicates is way more useful. $\endgroup$– MithoronCommented Mar 8, 2021 at 19:17
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On the other hand, it will be difficult to synthetize elements heavier than Oganesson ($\pu{Z = 118}$) for lack of neutrons. When the number of protons increases through the periodic table, the ratio neutrons/protons is also increasing. In oxygen-$16$, this ratio $\ce{n/p = 8/8 = 1.0}$. In iron-$56$, this ratio is $\ce{30/26 = 1.15}$. In uranium-$238$, it is $\ce{146/92 = 1.59}$. For heavier atoms, this ratio keeps increasing.
One of the problems to be solved for creating the last elements (tenessine, oganesson) was to find a not too heavy nucleus having enough neutrons to be used as missile sent on a target already rich in neutrons. Calcium-$48$ is a rare isotope of calcium, but a good choice with $20$ protons and a ratio neutron/proton = $28/20 = 1.4$. This value is the highest possible value among the $30$ first elements. When $\ce{Ca-48}$ is sent on a target like uranium-$238$ ($92$ protons), no nucleus can be made with a number of protons greater than $92 + 20 = 112$, which is copernicium Cn. And this collision will produce a nucleus with a maximum of $28 + 146 = 174$ neutrons, which gives the isotope $\ce{Cn-286}$. This nucleus is already rather instable, as its half-life is shorter than $1$ minute. If a lighter isotope of calcium had been used as missile, there would not be enough neutrons to make any isotope of copernicium. Of course lighter isotopes of copernicium have also been found.
It is extremely difficult to produce a nucleus with more protons than Copernicium-$252$. For making heavier atoms, it is necessary to hit a nucleus having more protons than uranium. But these targets are more and more radioactive. For getting the last atom, oganesson ($\ce{Z = 118}$), it was necessary to send missiles made of $\ce{Ca-48}$ nucleus on a target made of Californium-$252$ ($\ce{98}$ protons and $154$ neutrons), whose half-life is $\pu{2.2 y}$. Used as a target hit by $\ce{Ca-48}$ nucleus, it may produce a nucleus with a maximum of $98 + 20 = 118$ protons, and $28 + 154 = 182$ neutrons, which is oganesson-$300$. This is a maximum formula. For some reason, some neutrons have been ejected in the collisions, and real oganesson nuclei have been obtained with less than $182$ neutrons.
Now the yield of this synthesis is wretched. A total of only $14$ atoms of oganesson has been produced in the first publication, and they last less than one second. And to obtain this result, it was necessary to bombard the target of californium with an intense bundle of calcium-$48$ non-stop for three months and registering non-stop all particles emitted by this bombardment. And those particles were mostly made of smaller objects than oganesson. $14$ atoms have been obtained in three months, during which the target was quickly decaying.
To produce heavier atoms than oganesson, one should find a missile heavier than $\ce{Ca-48}$ with a higher ratio neutron/proton, like $\ce{Se-82}$, where $\ce{n/p = 1.41}$, or like $\ce{Sn-124}$, with a ratio $\ce{n/p = 1.48}$. One should also use a target heavier than Californium. But heavier atoms have shorter and shorter half-lives.
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2$\begingroup$ Probably a very naive question, but could one send two streams of missiles at the target? Or are the chances of three atoms combining into one essentially zero? $\endgroup$ Commented Mar 7, 2021 at 15:54
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2$\begingroup$ Good question. I am not a specialist of this domain. But the probability that one missile hits the target and produces a new nucleus is extremely low, I doubt the yield would be better in a triple collision. In order to produce 14 nuclei of oganesson, they had to bombard many millions calcium ions per second, and wait three months. About one nucleus par week ! $\endgroup$– MauriceCommented Mar 7, 2021 at 16:41
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One other possibility is that if we were to advance into an eighth period with the creation of new elements for which we can characterize chemical properties (we have begun to do so with hassium, atomic number 108 and copernicium, 112), we may find that the periodic recurrences we have come to know from previous periods break down. We may have the elements but not a good placement for them in the Periodic Table as it is currently rendered.
Two factors might be considered, one theoretical and one "experimental" (meaning "experiments" with computer calculations). On the theoretical side, the eighth period would include elements with atomic numbers equal or greater than 137, this number being the reciprocal of the fine-structure constant. The idea that such elements are impossible has become passe, but the theories that enable those elements also imply that relativistic effects will become a predominant factor in the electronic structure and properties, much more so than we currently see in Periods 6 and 7. Such relativistic effects were not accessible when Mendeleev first conceived the Periodic Table and not prominent across known elements even after Einstein's theories came. We should therefore expect the relativistic effects to scramble the periodic expectations based on lighter elements.
More concretely, computer calculations indicate that the electron-shell structure on which the Periodic Table is based is already breaking down in known superheavy elements. According to Ref. [1] and as reported in Chemistry World, oganesson (atomic number 118) is approaching a non-shell, Fermi-gas electron structure, driven by those same relativistic effects described in the previous paragraph. Such a breakdown in the shell structure would imply a breakdown in the periodic evolution of chemical properties based on such a shell structure. The Periodic Table may end in a fog.
Reference
1. Paul Jerabek, Bastian Schuetrumpf, Peter Schwerdtfeger, and Witold Nazarewicz, "Electron and Nucleon Localization Functions of Oganesson: Approaching the Thomas-Fermi Limit", Phys. Rev. Lett. 120, 053001 – Published 31 January 2018.
answered Mar 7, 2021 at 15:50
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1$\begingroup$ Oh good I don't have to answer. I have slightly different information expecting the breakdown at 120 because the energy difference after putting in the relativistic corrections says that 120 should be 1s8 1p8 rather than 2s8 but the difference is guessing whose theory is better. $\endgroup$– JoshuaCommented Mar 7, 2021 at 17:53
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$\begingroup$ We want to see your take on this breakdown @joshua! $\endgroup$ Commented Mar 7, 2021 at 20:58
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1$\begingroup$ I don't have anything much to add. I show a breakdown at 120 because the orbital energies are wrong, another breakdown at 137 because electron orbital velocity is too high and another breakdown at 173 because the electric charge energy is enough to create electron-positron pairs. The result of these is relativistic corrections are no longer remotely negligable, which means it can't be periodic anymore. $\endgroup$– JoshuaCommented Mar 7, 2021 at 21:04
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The periodic system is based on chemical properties and electronic structure.
Even the seventh period does not have a relevant chemistry. The lifetime of the respective isotopes is so short an atom hardly has time to relax into its electronic ground state before the nucleus decays. Everything we could "know" about the electronics of the eigth period would be based on conjecture, speculation. Unscientific, because not only practically, but principally unfalsifiable.
Unless someone goes and finds stable isotopes out there ... ;)
