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Why isn't Beryllium a noble gas?

  1. Jun 25, 2010 #1
    Hello All,

    I'm a physics major and trying to figure out how "full" electronic shells are determined. It really reduces to the question why isn't Beryllium a noble gas?

    If an angular momentum subshells are what matters then the shells are:
    [tex] 1S^2 2S^2 [/tex].

    Clearly the [tex]2S[/tex] subshell is full, yet this isn't considered a noble gas.


    Thanks a lot for any help.
  2. jcsd
  3. Jun 25, 2010 #2


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    It's not the subshells that matter. The second electron shell consists of both the 2s and 2p orbitals. When both of those are full you have the noble gas Neon.

    i.e., the important thing is that all states of the same quantum number [itex]n[/itex] are filled. In Beryllium the n=2, l = 1 states are not filled.
  4. Jun 25, 2010 #3
    Thanks for the reply!

    However, if the quantum number n is all that matters, how can Argon be a noble gas??


    but 2*1^2 + 2*2^2 + 2*3^2 = 28.

    What I mean is, based on n, the order of noble gasses should be

    What is called a "shell"? It's evidently not angular momentum, and not energy...
  5. Jun 25, 2010 #4


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    Ok, I guess it's slightly more complicated than I led on/I made a mistake. It's when the outer shell of valence electrons is full that matters. The order in which the orbitals are filled is not strictly in order of increasing quantum number n. For instance, the d orbitals for a given n tend to get filled after the n+1 s orbitals. I guess it ends up that when the p orbitals (the outer shell) are fully filled you get a noble gas (for Ne to Rn - after that relativistic effects apparently screw things up so a hypothetical element that appears under Rn on the periodic table isn't expected to be a noble gas).

    See this wikipedia page for more information: http://en.wikipedia.org/wiki/Electron_configuration

    The filling of orbitals can get tricky (e.g., Co and Cu seem like they have their electron configurations mixed up based on the order of the periodic table)
  6. Jun 25, 2010 #5
    Thanks again for the replies. That seems like a pretty simple rule.

    Is there any kind of physical intuition behind this? Naively, one would expect that when s subshells (spherically symmetric) are filled that would fill a 'shell'.

    Why of all the subshells is p the significant one?
  7. Jun 25, 2010 #6


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    I would say that's because energy levels of s & p are usually not very different.
  8. Jun 25, 2010 #7


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    It is energy, it's just not all of it.

    Every filled sub-shell is spherically symmetric, and every filled subshell is relatively stable. (Leading to why copper is 4s1 3d10 rather than 4s2 3d9.
  9. Jun 25, 2010 #8
    That's a throwback to the original question.

    If all subshells are spherically symmetric and stable, what makes p a special subshell, that when it gets filled the element is inert?
  10. Jun 25, 2010 #9


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    Not p, p & s.
  11. Jun 25, 2010 #10


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    Because a filled p-shell for a given n is so much lower in energy than the s-shell with n+1, whereas d and f shells for a given n end up in-between the s and p energies for n+1. Hence the structure of the periodic table. Put another way, they're ordered by n + l, and for two sub-shells with the same n + l, the one with the lower l comes first. This is called Madelung's rule.

    A perhaps better way is to invoke MO theory, drawing an MO diagram will show the noble gases won't form stable compounds. And it will show that beryllium (for instance) should form compounds. This is of the essence here, because as soon as another atom comes into play, it changes everything about the relative ordering of the orbitals.

    But - very notably in this context - it also predicts that Beryllium should not form a stable compound with itself. It was not until the 50's or 60's or so that it was shown that Beryllium does in fact form a stable (if reactive) dimer. And it's not some weakly-bound van-der-Waals interaction either, even if it's not a proper covalent bond. In fact, the Beryllium dimer bond can't really be characterized, because it cannot be described in terms of any of the simplified models chemists use. Or even using explicit quantum mechanics using the more approximate methods. It's quite an exceptional case.
  12. Jun 25, 2010 #11

    So if I understand correctly, there's a big jump in energy between a full p shell and the next s shell.

    Madelung's rule applies to ordering the subshells in terms of their energy for increasing atomic number (and increasing number of electrons). But the question here is more about proving that given a fixed number of electrons, is the material inert? (i.e. which subshell is the most important to fill to qualify as a noble gas).

    An s subshell is significantly higher in energy than it's preceding p subshell. Therefore it might be possible to gain a lot of energy by shedding these electrons (being doubly ionized). In contrast dropping from a p shell would require six fold ionization.

    Am I getting warmer?
  13. Jun 28, 2010 #12


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    Yes, I suppose that's the best I can come up with. Although the s shell is not necessarily always lower either. (Ar- has its last electron in the p shell, but K, with the same number of electrons, has it in the s shell, which illustrates well how tricky this stuff can be)

    But filling a shell of a higher n is obviously quite a bit higher in energy, which one can tell from the fact that it almost never happens. O2- is everywhere, but F2- simply doesn't exist, even though the latter should be at least as stable - if you ignore the quantum mechanics involved. There's really just Xenon hexafluoride and a handful of other such exotic and unstable compounds where this occurs.

    Yes. Well remember here that for isolated atoms in vacuum, the neutral species is always lower in energy. The "stability" of noble gas configurations is not in itself large enough to offset the energy required for ionization. So it's only once other atoms/molecules come into play that this effect becomes noticable. So for instance, when you put an alkali metal in water (a highly dielectric medium), you stabilize the charge and hence lower the ionization energy. Only then does it become energetically beneficial for an alkali metal to drop its outermost electron (or for a halogen to gain one) and go to the noble gas configuration.

    One has to watch out here with thinking too much in pure electrostatic terms. Electrons don't behave at all either like classical point charges or like a semi-classical charge-distribution 'cloud'. The relative stabilities here are governed to a large extent by purely quantum-mechanical effects, like Pauli repulsion/exchange energy. A filled orbital is not only more stable due to spherical symmetry, but also due to the fact that it minimizes the exchange energy. But this is not the case for an s orbital, which has no exchange energy since it only contains one electron of each spin. So this is the reason why a filled shell is more stable, but a filled s-shell is not so much more stable. And so you have Al3+ but not Al+. (Easy theoretical rationalization: Take a peek at the Hartree-Fock equations and note the negative sign in front of the exchange operator)

    You could make an argument also in terms of Pauling electronegativities and such, but I think that would be cheating, since those are based on ionization potentials/electron affinities to begin with.
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