Electron Configuration for Ions

[Andonios Karas 4H]

Scandium = [Ar] 3d1 4s2
Scandium 2+ = [Ar] 3d1

My question regarding these electron configurations is why does the electron remain in the 3d shell for the ion instead of moving the the 4s shell.

For potassium keeps its one valence electron in the 4s shell because the electron is at a lower energy level in a partially filled 4s shell rather than a partially filled 3d shell.

Why does the 3d shell have lower energy in the ion but not in the atom, for they have the same number of electrons?

[Elisa Bass 4L]

The 3d orbital is closer to the nucleus than the 4s orbital, making the electrons harder to remove once they have been put in there. This is why the 3d is written before the 4s in the atoms that have a 3d shell and why electrons in 3d do not move to the 4s orbital when it is vacant. Therefore, in the case of scandium, when ionized the electrons are lost from the higher energy shell, the 4s.

In the case of the potassium atom, it does not have enough electrons to fill a 3d shell and will probably not gain enough electrons to do so. The 3d shell is only used if the atom has enough electrons to fill the 4s FIRST but potassium ions rarely if ever do because they tend to form cations rather than anions.

[Kobe_Wright]

Because potassium in this instance does not ever fill it's d shell because it doesnt have the atomic configuration to be in that orbital while for scandium it has the atomic configuration(protons+neutrons+electrons) that has it fill the 3d orbital which is of higher energy than the 4s in the ion, and atom, so when Scandium loses two electrons it loses them from its lowest energy orbital 4s, because pulling the one electron from 3d and one from 4s you would have only one electron in the 4s orbital instead of the 3d orbital which would put it in an un-optimum state.