26
$\begingroup$

Why is argon a noble gas given that the 3d subshell is still empty? More generally, why is it that the filling of a p sub-shell makes an element noble rather than s, d, or f sub-shells, or completed n-levels?

Let's start with the prior that in a hydrogen atom the principle quantum number determines the energy - so maybe full shells correspond to noble gases.

Now, I understand that in helium the n=1 level is full, so it is a noble gas. Then, beryllium has a full 2s subshell, but empty 2p, so it is a metal. Neon has the n=2 level full, and is a noble gas. Magnesium has a full 3s subshell, but not a full n=3 level and is a metal - so far so good. But, oops, argon has full 3s and 3p subshells but empty 3d, yet it is a noble gas.

At this point we look at the aufbau principle and say "ah, your prior is no good, the 4s is lower energy than 3d, and this is the first time it has kicked in - so there you go". But, as we continue down the periodic table there are more noble gasses and they happen only with the filling of a p subshell. Why is krypton noble, but not zinc? Even palladium cleanly fills n=4 up through 4d (with only 4f empty) but it is still not a noble gas. Ytterbium fills an f subshell for the first time and it isn't noble. What makes the filling of a p subshell so specially inert compared to s, d, and f subshells?

The aufbau principle does not explain this. Draw the classic "diagonal aufbau chart" and circle where the noble gasses are. It doesn't add any insight. Noble gasses only happen when you fill a p subshell. Why?

$\endgroup$
  • 4
    $\begingroup$ Something for chemistry.stackexchange.com? $\endgroup$ – Avantgarde Dec 31 '18 at 19:51
  • 5
    $\begingroup$ And - no! - not chemistry stack exchange! - this topic should lie within the bounds of quantum physics, IMHO $\endgroup$ – Paul Young Dec 31 '18 at 20:03
  • $\begingroup$ I've removed some partial answers and (very interesting) side-discussions from the comments. $\endgroup$ – rob May 6 at 2:09
3
+50
$\begingroup$

Aufbau arises mainly from the fact that the screened potential is not the same as the $1/r$ potential of hydrogen. In the screened potential, some levels are pushed up in energy and some are pushed down. Basically the high-l orbitals have less probability concentrated near the nucleus, where the screened potential is the most negative, so they're pushed up in energy relative to the low-l orbitals.

In addition to this, you get effects that are the way they are not because of any semiclassical argument because of the way the Schrodinger equation works. You're basically diagonalizing a big matrix, and there is no guarantee that the results will come out to be easy to explain. Sometimes such results can be at least partially described by heuristics, such as a tendency for energy levels to "repel" each other, although it's not actually a physical repulsion. For this "level repulsion" to happen, the levels have to have the same angular momentum. (It's actually pretty surprising that this kind of level repulsion leaves the $n$ shells degenerate in hydrogen. You would think it would split them up.)

Noble gases exist when there is a large gap between the single-particle energy levels. These gaps do not occur for the screened potential at the same particle numbers where they would occur for hydrogen.

Here are some energy levels of states in noble gases, calculated using the Hartree-Fock method, from Johnson, Lectures on Atomic Physics, p. 86:

table of Hartree-Fock energy levels for noble gases

The Hartree-Fock method calculates the energies self-consistently for a particular atom, so the energies are different for the different atoms. It would be convenient to have the calculation for something like potassium, so we could see the gap between the last filled state in argon and the final electron for potassium, but he doesn't actually give those. However, you can clearly see the relevant shell gap in the calculation for krypton. There is a gap of 4.5 Ha between 3p and 3d. This is considerably larger than the gap between 3s and 3p (2.5 Ha) and between 3d and 4s (2.7 Ha).

So - what is special about p sub shells such that when they fill there is a larger gap up to the next level than when s,d, or f sub shells fill?

There is nothing special about p shells. These calculations are based on diagonalizing a big matrix. There is no particular reason to expect the spacing between subshells to follow any simple rules.

Beryllium isn’t a noble gas.

If you look at the levels for argon, you can see that the gap between 2s and 2p is less than 3 Ha, while the gap between 2p and 3s is about 8 Ha. The reason for the sizes of the gaps is ultimately just that that's what you get when you diagonalize the matrix. However, I don't find the result particularly surprising, since the unscreened potential has 2s degenerate with 2p, and a big gap between 2p and 3s.

$\endgroup$
  • 5
    $\begingroup$ I like this answer a lot, and I get the "big matrix" Hartree-Fock approach, but I'm not sure I buy "there's nothing special about p-shells." All of the nonmetals (save H and the noble gases) have partially-filled p-shells, and filling a p-shell seems to produce a noble gas. It seems there is something about a partially-filled p-shell that supports more complex chemistry than a partially-filled s- or d-shell --- though I wonder whether someone who does lanthanide/actinide chemistry would say the same thing about f-shells, or whether the "shell" approximation gets squashier in such heavy nuclei. $\endgroup$ – rob May 6 at 2:23
  • $\begingroup$ I think @rob makes a good point. It is just barely possible to separate rare earth’s from each other, and transition metals vary meaningfully ... but those p filling elements sure seem to vary a lot .. it feels there is ... something more? $\endgroup$ – Paul Young May 6 at 2:32
  • $\begingroup$ @Ben - I can imagine 2 reasons why only filled p shells are noble. The 1st is that forming a bond requires "reaching out" to another atom in one direction. But s-levels aren't directional and higher L orbitals don't go as far from the nucleus, so p is the "sweet spot". Landau claims something like this. The 2nd is that multi-electron atom energy gaps "just so happen" to be large after a p-level is filled - your point. How confident are you that the 1st point doesn't matter? How confident are you that there is no intuition for why the H-F gaps are so big after p-levels fill? $\endgroup$ – Paul Young May 6 at 14:56
  • $\begingroup$ @Ben - the units in the table are Hartrees $\endgroup$ – Paul Young May 6 at 15:36
  • $\begingroup$ @PaulYoung: I think your point about "reaching out" would be more relevant if we had two conflicting criteria for labeling something as a noble gas: high ionization and inability to easily form chemical bonds. But in fact these criteria agree. If there is some way of making this "reaching out" idea rigorous, I would be happy to hear about it. $\endgroup$ – Ben Crowell May 7 at 0:18
2
$\begingroup$

The definition of a noble gas has changed over the years as our understanding of the atom improved. The first relevent definition is that a noble gas is one that has a completed octet. In other words the outer most shell is supposed to have $8$ electrons (with the obvious exception of helium where the last shell can have only $2$ electrons). This definition was found purely experimentally. Later on as quantum mechanics developed this definition changed. Now the definition is that a noble gas has a completed $p$-orbital (with the exception of helium again). Hence one answer to your question is that it is defined that way. But this answer is really unsatisfactory.

For a satisfactory answer one must ask what makes a noble gas noble? As we all know noble gases are not supposed to react, i.e. they should not accept, donate or share electrons easily. This means that the electron affinity and ionisation energy of noble gases should be very high. I am not going to mention the exact numbers but they can be checked here. The electron affinity and ionisation energy values are higher for a filled $p$-orbital than a filled $d$-orbital. This can be shown by a simple example of Argon and Zinc. To add an electron to Zinc is easier than Argon because in Zinc the screening will be provided significantly only by the $4s$ orbital. But in argon the screening will be provided by the $3p$ orbital, which is significantly higher. Hence Argon will be more stable than Zinc. Zinc is not a noble element since it has very different properties like lower screening effect, EA, IE, etc when compared to other noble gases. But this does not mean that it does not form bonds. Only after Quantum Mechanics we were able to explain the existence of compounds like $XeF_4$. These compounds were a major drawback of the previous theory.

Hope this helps.

$\endgroup$
  • $\begingroup$ @MS - all this looks fine to me, but it seems to say "filled p shells have all these various properties which make elements noble" without addressing why p shells, but not s, d or f have these properties. $\endgroup$ – Paul Young May 6 at 15:17
  • $\begingroup$ @PaulYoung I have mentioned why adding and removing electrons to (from) d-orbital is much easier than p-orbital. f-orbital has a similar reasoning. For a filled s-orbital an electron can be added to p orbital without much repulsive force. You are right in saying that all element with last orbital filled will be stable. But it will not be as stable as noble gases due to the mentioned reasons. $\endgroup$ – Manvendra Somvanshi May 7 at 4:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.