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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 $[\ce{Ar}](\mathrm{4s})^1(\mathrm{3d})^7$?

orthocresol
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jukam
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1 Answers1

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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.

DrMoishe Pippik
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  • 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 02 '21 at 01:37
  • I'm probably wrong about the 3d/4s thing, though, sorry. cf. references in https://chemistry.stackexchange.com/a/33310/16683. I think I should probably do some research before commenting on that. – orthocresol May 02 '21 at 01:39
  • 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 02 '21 at 03:03
  • Thanks, @M.L, my mind must have been wandering. Max numbers are, of course, s2, p6, and d10. The answer will be corrected. – DrMoishe Pippik May 02 '21 at 19:31
  • So there is now way once can predict the configuration of transition metals, lanthanides, actinides ions? – jukam May 03 '21 at 14:17
  • So there is no way to predict the configuration for lanthanides, actinides or transition metal ions? I thought 4s was prefered to 3d and 3d9 or 3d1p prefferes to 3d9 e.g. Ni, Pd? – jukam May 03 '21 at 14:20
  • There are rules, but calculation can be very difficult. Consider the effort to calculate wave functions for H https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Map%3A_Physical_Chemistry_for_the_Biosciences_(Chang)/11%3A_Quantum_Mechanics_and_Atomic_Structure/11.10%3A_The_Schr%C3%B6dinger_Wave_Equation_for_the_Hydrogen_Atom – DrMoishe Pippik May 03 '21 at 16:24