Which is the electron configuration for $\ce{Cr^2+}$
(A) $\mathrm{[Ar]4s^23d^4}$
(B) $\mathrm{[Ar]4s^23d^2}$
(C) $\mathrm{[Ar]4s^03d^4}$
(D) $\mathrm{[Ar]4s^43d^2}$
I referred to the periodic table and got the orbital notation for $\ce{Cr}$ as $$\mathrm{[Ar]4s^23d^4}\tag1$$ So $\ce{Cr^2+}$ would be $\mathrm{[Ar]4s^23d^2}$, or (B). But the teacher counted that wrong and said it was actually (C).
Question: Shouldn't the answer be (B)? Why does $\ce{Cr}$ lose the $\mathrm{s}$ shell first?
Chromium’s standard electron configuration as given by the aufbau principle is actually more of an exception than you may think initially. See for example this question to catch a small glimpse of the truth behind the simplifications.
The aufbau principle tells you to fill in the order $\ce{1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, }\dots$ That order, however, is only valid for neutral atoms (and not always as the linked question implies). As soon as you add charge to an atom, the relative orbital energies shift. The $\ce{3d}$ orbitals, belonging to the M shell, are stabilised more than the $\ce{4s}$ orbital which belongs to the N shell. This is a general rule: charges typically stabilise orbitals belonging to lower shells more than those in higher shells.
Thus, while in neutral chromium the $\ce{4s}$ and the $\ce{3d}$ shell are very close in energy, this balance completely tips towards $\ce{3d}$ (which then has a lower energy) upon oxidation. Therefore, one should always assume an empty $\ce{4s}$ orbital when discussing the electronic configuration of charged d-block elements.
