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

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

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Answer 1 · 2016-05-26T22:46:54

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

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 $3 p$ $3p$ orbital is higher by $13.42 eV$ $"13.42 eV"$ ( $1294.83 kJ/mol$ $"1294.83 kJ/mol"$ ).

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

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

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

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

$E_{nd} > E_{np} > E_{ns}$ $E_"nd" > E_"np" > E_"ns"$

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

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