The electron configuration of the ground state of $\ce{Fe}$ is $\mathrm{1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^6}$. What are the configurations of its first and third excited states? The first excited states seems to be $\mathrm{[Ar]4s^1 3d^7}$, but isn't the angular momentum not conserved because $l=0$ for $\mathrm{s}$ and $l=2$ for $\mathrm{d}$?

  • $\begingroup$ I am somewhat puzzled at the “isn't the angular momentum not conserved” part of the question… Among excited states, not all will have the same principal quantum numbers (such as angular momentum). $\endgroup$
    – F'x
    Commented Aug 12, 2016 at 7:29

1 Answer 1


The electronic configurations of the excited states of neutral iron are listed on this NIST Database page:

  • 3d64s2 — a 5D — ground state
  • 3d7(4F)4s1 — a 5F — 6928.268 cm–1
  • 3d7(4F)4s1 — a 3F — 11976.238 cm–1
  • 3d7(4P)4s1 — a 5P — 17550.180 cm–1
  • 3d64s2 — a 3P — 18378.185 cm–1
  • 3d6(5D)4s14p1(3P) — z 7D0 — 19350.892 cm–1

The reference give for this data is: G. Nave, S. Johansson, R. C. M. Learner, A. P. Thorne, and J. W. Brault, Astrophys. J. Suppl. Ser. 94, 221 (1994).

So, while in general predicting electronic configurations for excited states is a tricky business, you are right: the first excited electronic configuration is to “promote” one 4s electron into the 3d orbitals. And the second excited state configuration (regardless of the different spectroscopic terms and sublevels) is to promote the 4s electron into 4p.


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