What is the electron configuration of \(\mathrm{Co}^{3+}\)?
Answer 1:
The electron configuration of \(\mathrm{Co}^{3+} \text { is }[\mathrm{Ar}] 4 s 3 d^{5} \text {. }\).
\(Co\) is in Period 4 of the Periodic Table, and \(Ar\) is the preceding noble gas.
Cobalt is also in Group 9, so it must have 9 valence electrons.
The valence shell configuration is therefore \(4 s^{2} 3 d^{7}\) and the core notation is \(\text { Co: }[\mathrm{Ar}] 4 s^{2} 3 d^{7}\)
When a transition metal forms an ion, the \(s\) electrons are removed before the \(d\) electrons.
We would predict the electron configuration of \(\mathrm{Co}^{3+}\) to be
\(\mathbf{C o}^{\mathbf{3}+}:[\mathrm{Ar}] 3 d^{6}\).
The \(4s\) and \(3d\) sublevels are nearly identical in energy, so the ion can become more stable by moving one of the \(3d\) electrons to the \(4s\) level.
Then, both the \(4s\) and \(3d\) levels are half-filled, and the ion gains a little more stability.
The electron configuration becomes:
\(\mathbf{C o}^{3+}:[\mathrm{Ar}] 4 s 3 d^{5}\)