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"#
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Answer 2

The electrons that get removed to form "Ar"^+ are the outermost electrons in the argon atom's electron configuration, specifically from the 3p orbital. To form "Ar"^(2+), additional electrons are removed, typically from the 3s orbital, resulting in the removal of two electrons in total.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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