# Why does Co(I) have a 3d8 configuration?

Why is the electron configuration of $$\ce{Co^+}$$ $$[\ce{Ar}](\mathrm{3d})^8$$?

Since neutral $$\ce{Co}$$ itself has a $$[\ce{Ar}](\mathrm{4s})^2(\mathrm{3d})^7$$ configuration, wouldn't the ionised electron be lost from the $$\mathrm{4s}$$ orbital, leading to an electron configuration of $$[\ce{Ar}](\mathrm{4s})^1(\mathrm{3d})^7$$?

For lighter elements, the shells fill in order. Starting at the transition metals, an outer s orbital may fill before an inner d orbital, so the electron configuration of unioninzed cobalt is written $$\ce{[Ar]}4\mathrm s^1\,3\mathrm d^7$$, rather than $$\ce{[Ar]}3\mathrm d^7\,4\mathrm s^1$$.
There is a video diagramming the electron configuration of $$\ce{Co}$$, $$\ce{Co^{2+}}$$ and $$\ce{Co^{3+}}$$, thought it does not explain the reasoning, nor does it cover the less common $$\ce{Co^{+}}$$ ion, produced by photoionization or as found in some esoteric metal-organic compounds, or in the theoretical $$\ce{CoCl}$$.
That said, the situation becomes murky for transition elements, and downright turbid for lanthanides and actinides, where f orbitals are added. For example the outermost shells of $$\ce{La, Ce and Pr}$$ are $$\ce5\mathrm d^1\,$$, then $$\ce4\mathrm f^1\,5\mathrm d^1$$, and then $$\ce4\mathrm f3$$. What happened to the Pr d electron? My understanding is that energy levels are quite close for the larger outer shells, and it is not intuitive which orbital fills first.
• I have always been under the impression that for transition metals, 3d < 4s in energy. The $3d^8$ configuration of Co(I) ion is thus the norm, and it's the $4s^2 3d^{n-2)}$ configuration of neutral atoms that is the exception. Also, what exactly is the energetic drawback of having an incomplete orbital? Exchange energy only really plays a crucial role in the d5 and d10 case; it's always there for d6/d7/d8/d9, of course, but isn't large enough to affect the electron configuration. – orthocresol May 2 at 1:37
• What is it meant by 'has no incomplete orbital'. Are there not 2 half-filled orbitals in a $\ce{d^8}$ configuration? – M.L May 2 at 3:03