Which electrons get removed to form "Ar"^+ and "Ar"^(2+)?

1 Answer
Truong-Son N.
May 26, 2016

Argon ("Ar") has the atomic number 18 (it's right by chlorine and neon). By default, its electron (script) configuration is written as 1s^2 2s^2 2p^6 3s^2 color(blue)(3p^6).

Any electron removed during an ionization is from the orbital that is currently highest in energy.

You can see in Appendix B.9 here that the energy of the 3p orbital is higher by "13.42 eV" ("1294.83 kJ/mol").

So, the 3p electrons will get removed:

"Ar"^(+): 1s^2 2s^2 2p^6 3s^2 color(blue)(3p^5)

"Ar"^(2+): 1s^2 2s^2 2p^6 3s^2 color(blue)(3p^4)

As a general observation, the energies for orbitals of the same principal quantum number n are usually:

E_"nd" > E_"np" > E_"ns"

Related questions
See all questions in Electron Configuration
Impact of this question
9245 views around the world
You can reuse this answer
Creative Commons License