Exploring E-e Repulsion: Stability of Half and Full P Sub-Shells Explained

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I am fully aware that a half filled p sub-shell is stable due to no repulsion as electrons first occupy the orbitals singly before pairing up according to Hunds rule, why is that however a fully occupied p sub-shell with 6 electrons is also considered stable? Where as a filled s sub-shell with 2 electrons is said to undergo e-e repulsion, hence the I.E value of He is less than that of Hydrogen's I.E by a factor of 2 (90% to be exact).

In short: Why is a full p sub-shell with 6 electrons not said to undergo e-e repulsion (considered stable), whereas a full s sub-shell with 2 electrons is said to undergo e-e repulsion (unstable).
 
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The ionization energy of Helium is a little less than twice that of atomic hydrogen, i.e. helium is more inert than atomic hydrogen.
See the table of values here.
 
Doesn't it have to do with e-e repulsion?
 
htneT said:
Doesn't it have to do with e-e repulsion?
Does what have to do with electron-electron repulsion? You said that helium has a lower ionization energy than helium, which is not the case. Similarly you are talking about p sub-shells but only referring to hydrogen and helium, which only have half and fully filled 1s sub-shells respectively.
 
Vagn said:
Does what have to do with electron-electron repulsion? You said that helium has a lower ionization energy than helium, which is not the case. Similarly you are talking about p sub-shells but only referring to hydrogen and helium, which only have half and fully filled 1s sub-shells respectively.
My main question is why is a full p sub-shell considered stable, whereas a full s sub-shell considered unstable and its electrons undergo e-e repulsion?
 
htneT said:
Where as a filled s sub-shell with 2 electrons is said to undergo e-e repulsion
Each electron in a multi-electron atom will of course experience e-e repulsion regardless of which subshell the electron under consideration occupies.
htneT said:
a full s sub-shell considered unstable
What is the measure of stability/instability you are using, I.E.? Ionization energy is not only determined by the number of electrons present in the atom, but also by the nuclear charge.