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Why adding just one electron changes tremendously the ionization potential from any of the nobel atoms ?

If it is the screening why adding a second electron increases the ionization potential ?

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Related: – John Rennie Sep 26 '12 at 7:30
up vote 1 down vote accepted

It's because of shell filling--- when you add electrons they go into discrete orbits similar to those of the H-atom, and at higher Z, all these discrete levels bind tighter to the nucleus. There is also screening, so that the outer shells see a reduced charge, but this is in combination with shell filling. The shells are at a certain distance from the nucleus. When you fill a shell, you have the highest ionization energy, because the outermost electrons are all seeing a high nuclear charge, less screening from the core, and are in a lower level. When you add one more electron to a filled shell, it is loosely bound since it sees an effective core charge of 1 unit. There is nothing qualitatively mysterious in the ionization energy graph after these screening and shell filling are understood.

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Alkaline has s shell filled and huge difference from the nobel atoms in ionization potential. Nobel atoms have p shell filled has is that more stable? From left to right the atom size is decreasing so adding more protons and electron is reducing the size of the atom ? And the other thing that puzzles me is that it is reversed when you look at the second ionization potential and higher. – user12445 Sep 26 '12 at 16:33
@user12445: Alkaline does not have S-shell filled, the column after that has S-shell filled (the second column from the left), there are two spin states for an electron. Filling the "S" shell only adds two units to "Z", while filling a "P" shell adds 6 units to Z, and this is proportionally more nuclear charge at lower Z. Think of adding electrons in a shell as increasing "Z" for all the electrons in the shell without increasing the distance of any of the electrons in this shell. The "D" shell doesn't follow this rule because it gets filled later than higher n S,P shells, same for F... – Ron Maimon Sep 26 '12 at 17:39
This is because these orbits are circular (higher l at fixed n is a more circular classical orbit), so they are further from the nucleus on average, and see a more highly screened charge, so when the 1S 2S 2P 3S 3P is filled, 3D is higher energy than 4S. – Ron Maimon Sep 26 '12 at 17:45

Pauli exclusion principle prevents an extra electron from entering a closed shell, it works as an effective repulsive force. So the extra electron fill an extra (higher) energy level.

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