Question

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

Answer 1

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"`