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

May 26, 2016

Argon ($\text{Ar}$) has the atomic number $18$ (it's right by chlorine and neon). By default, its electron (script) configuration is written as $1 {s}^{2} 2 {s}^{2} 2 {p}^{6} 3 {s}^{2} \textcolor{b l u e}{3 {p}^{6}}$.

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 $3 p$ orbital is higher by $\text{13.42 eV}$ ($\text{1294.83 kJ/mol}$).

So, the $3 p$ electrons will get removed:

${\text{Ar}}^{+}$: $1 {s}^{2} 2 {s}^{2} 2 {p}^{6} 3 {s}^{2} \textcolor{b l u e}{3 {p}^{5}}$

${\text{Ar}}^{2 +}$: $1 {s}^{2} 2 {s}^{2} 2 {p}^{6} 3 {s}^{2} \textcolor{b l u e}{3 {p}^{4}}$

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

${E}_{\text{nd" > E_"np" > E_"ns}}$