Why Do Ionization Energies Vary with Odd and Even Electron Counts?

AI Thread Summary
Ionization energy varies with electron count due to the stability of electron configurations, particularly in relation to filled and half-filled orbitals. Noble gases exhibit the highest ionization energies because of their stable filled orbitals. When adding electrons, an odd number can lead to decreased ionization energy due to increased repulsion in unpaired electrons, while even numbers often result in increased stability and higher ionization energy. The behavior of ionization energy can also be explained through the filling of S and P orbitals, where filled orbitals confer greater stability. Understanding these principles is crucial for grasping the quantum mechanics underlying ionization energy trends.
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This question is not a homework question. I am having a little trouble with the ionization
energy graph. I understand that the noble gasses are the most stable
and hence require the most energy to be ionised, but why is it that after you
add even number of electrons the ionization energy increases but when it
is odd numba of electrons the ionization energy decreases. I understand that as you add shells the sheilding effect increases, but why is it that when you add an odd number of electrons the ionization energy decreases??

Is it because when electrons are in pairs they are more stable?? What is the nuclear physics or qunatum physics explanation behind this


Thanks in Advance
 
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The most stable configurations tend to be either filled or half-filled orbitals. I'm not sure where you are getting that even numbers of electrons are necessarily more stable than odd numbers.

For example, nitrogen (7 electrons) has a higher IE than oxygen (8 electrons). This is due to the stability of a half-filled 2P shell for nitrogen.

For example, see the table here:
http://dl.clackamas.cc.or.us/ch104-07/ionization_energy.htm
 
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But u can see in that figure, once u add an extra electron after the shell is filled it automatically drops off right, once u add the second the ionization energy increases, once you add the third it drops off again... i obviously don't understand what's going on here so by no means am i saying I am right about what i said, i just want to know how it all works.
 
Okay, I think I can give a more helpful explanation now.

Have you learned about S orbitals and P orbitals? The important thing to know is that S orbitals need only 2 electrons to be filled, while P orbitals need 6 electrons.

So, looking at what happens after neon, for example: neon has a filled P orbital, so the next element (Na) has a drop in ionization energy. Two elements past neon is Mg, which now has a filled S orbital -- because S orbitals need only 2 electrons to be filled. Because of it's filled orbital, Mg has a higher I.E. than the elements next to it.

The next element after Mg is Al, at which point we are adding electrons to P orbitals, and it takes 6 more electrons to fill that. These filled orbitals are filled 6 elements after Mg, or at Ar. So Ar has a high I.E. again. Also, note that when the P orbitals are half-filled with 3 electrons, at P (Z=15 on the graph), there is a slight rise in I.E. there. Half-filled orbitals are also stable, though not as stable as filled orbitals.

Hope you're able to follow all this on the graph I linked to before.
 
But is there a quantum physics explanation for this. As this is for a physics course, and they don't really go into how many electrons fill a shell etc, i understand what ur talking about coz I've done chem b4 but this is for a quantum mechanics course.
 
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