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"#

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