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Generally, noble gases are very unreactive. Why?

I understand that it has to do with electron shielding prospective electrons that could join the atom or leave and the energies associated with both actions, but why are 5 electrons in the p orbital (to take Fluorine as an example) so much worse in shielding the nuclear charge than 6 (Neon)?

What IS electron shielding, anyway? I understand it in the classical lie-to-children way, but not satisfactorily.

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marked as duplicate by Freddy, Klaus-Dieter Warzecha, J. LS, Burak Ulgut, ron Apr 14 '15 at 13:06

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It's all about the energy levels of electrons. Electrons exist in different energy levels and each energy level can only be occupied by a specific number of electrons (google quantum theory).

Example: Neon has an electron configuration of $1s^2 2s^2 1p^6$, which simply means that it has a full $1s$ orbital (the maximum is 2 electrons as denoted by the superscript), and has a full valence shell which is the $2s$ and $1p$ orbitals (which is the valence shell for molecules in row 2).

Now for neon to lose an electron (to bond) it would have to take an electron out of a full orbital ($1p$), which is what makes molecules reactive (a non-full valence shell), and would just take the electron right back.

Think of Ne$^+$ as more electronegative (wanting electrons) than elemental Fluorine, so it will pull an electron from almost anything and go back to its elemental form.

If this answer is not satisfactory then I can explain further, without diving head first into quantum theory.

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  • $\begingroup$ You're right, it wasn't quite satisfactory, maybe I didn't state it obviously enough in the question, but I understood most of that already. As useful as that explanation is, it contains some tautologies and because of that doesn't explain much of WHY (simply restates the meaning of stability). So if possible, could you wander to the "real" explanation? $\endgroup$ – Meow Oct 3 '12 at 21:16

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