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To: Stosh; DustyMoment
The energy levels that the electrons can possess are a function of their attraction to the nucleus, which is largely a function of the nuclear charge (which equals the number of protons).

Only true in hydrogenic atoms. The energy levels of multiple electron atoms are dominated by the Exclusion Principle.

As long as you don’t change the number of protons, you don’t change the [nuclear] identity of the element

True in terms of the physics, but not the chemistry. An He+ ion does not behave like Helium for chemical purposes, (though Z is the same) and it certainly doesn't behave like Hydrogen (though Ne is the same.)

An iron nucleus [nucleus, yes!] identifies it as iron whether it has all its electrons (as in metallic iron) or whether it’s missing three of its electrons (as it the form of iron found in iron oxide / rust).

Again, too oversimplified to be true. There is significant overlap of the state vectors of the oxygen and iron in rust, and the delocalized electrons are in molecular orbitals; they are not ions, and they are not atomic orbitals, either. If oxidized iron identified as iron when three of its electrons are significantly delocalized into molecular orbitals, there would be no reason for rust to form (and it would not form.) It is iron in terms of the nuclear chemistry, but is no longer iron in terms of atomic (ordinary) chemistry.

Losing 20 out of 22 electrons is a pretty big perturbation

It's not a perturbation from that perspective, and a perturbation approach wouldn't be done that way: the perturbation would treat the atom as a hydrogenic atom with Z=22, and one electron. The second electron is then added as a perturbation. Just one additional electron is still a very large effect.

I know you will think I am picking nits; however, in terms of what's reported here, the researchers are essentially examining the atomic (not nuclear) properties of certain atoms. For the purposes of these studies, what they're saying is: we expect the emission spectra to look like an "overcharged" hydrogen or helium nucleus. The energy levels of those are very well known (the former -- hydrogen -- is, in fact, exactly calculable, with any Z you want, and is an exercize for junior level nuclear physics or P-chem courses at undergraduate level. With a hydrogenic wavefunction with Z=22, the Schroedinger equation doesn't apply very well. Still, the values with a full relativistic treatment with QED are calculable to a very high degree of precision.)

So ... quite surprising. I bet it turns out to be an experimental artifact and is not new science.

16 posted on 12/03/2012 4:58:19 PM PST by FredZarguna (Shut 'er down Clancy. She's pumpin' mud.)
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To: FredZarguna

No problem with the “nits” - they were well-picked. At what level one presents this stuff depends a lot on what audience you’re shooting at. Given the nature of the original question, I’m not sure delocalized MO’s and the wave equation would be the best place to start - but I do appreciate your well-presented elaborations.


19 posted on 12/03/2012 6:25:20 PM PST by Stosh
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To: FredZarguna
It's not a perturbation from that perspective, and a perturbation approach wouldn't be done that way: the perturbation would treat the atom as a hydrogenic atom with Z=22, and one electron. The second electron is then added as a perturbation. Just one additional electron is still a very large effect.

I thought perturbation theory for electronic structure was supplanted by configuration interaction; and that people generally liked to use Density Functional Theory rather than actually solving for the wavefunction explicitly.

Or am I just having a senior moment?

Cheers!

26 posted on 12/03/2012 7:06:27 PM PST by grey_whiskers (The opinions are solely those of the author and are subject to change without notice.)
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