Ionisation energies of paired vs unpaired electrons.

[Robsta]

Homework Statement

Why is it harder to rip off a paired electron than an unpaired electron? I'm trying to work out why the ionisation energy is lower for chlorine than for argon.

Homework Equations

Configuration for argon: $$1s^22s^22p^63s^23p^6 $$
Configuration for chlorine $$1s^22s^22p^63s^23p^5 $$
I know that argon has three pairs of electrons in p orbitals with opposite spins. Chlorine has two pairs and a lone electron. Why do electrons get more strongly bound when they have a spin pair? I know that electrons with opposite spins can get closer to each other because they're not anticorrelated, but I'm not sure why they'd have a lower energy than an electron without a partner. Can anybody explain?

The Attempt at a Solution

 

[blue_leaf77]

This is because electrons in the same subshell (in your example, subshell 3p) have similar wavefunctions, i.e, their radial distances from the nucleus do not vary significantly from one another. When the proton number increases, one electron must be added in order to make the atom neutral, but as long as the added electron still occupies the same subshell as the last electron of the previous atom, the additional screening provided by the new electron is rather negligible compared to the increased binding force of the nucleus due to the addition of one proton. Therefore, the ionization energy increases. If you add one more electron to an already filled subshell, this electron will occupy a new subshell which is at some distance away from the last subshell. So the screening of the electrons in the lower subshells will be effective enough to counter the nuclear binding force, and thus the ionization energy decreases as you move to the next row in the periodic table.