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Depending on the textbook there are two different electronic configurations stated for cerium. On the one hand

$$[\ce{Xe}]\mathrm{4f^15d^16s^2}\quad$$

and on the other hand

$$[\ce{Xe}]\mathrm{4f^26s^2}$$

Are both configurations possible? If not, why so?

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  • $\begingroup$ There is no need to waste your time in memorizing the differences. Yes, you will hear and read "stories" of this electron configuration and that electron configurations without saying why so. I have yet to find a spectroscopist (asked so many leading ones) who can tell me how these electron configurations were obtained from either experiment or by theory. I bet your teacher will have no clue either. So the question you should ask is how would we experimentally distinguish your first configuration from the other. $\endgroup$ – M. Farooq Feb 5 at 0:29
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The configuration of cerium is Ce -> 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6 6s2 4f1 5d1 https://en.wikipedia.org/wiki/Cerium

The filling of electrons follows Aufbau principle nicely till Z<20, however, introduction of d and f orbitals and resulting poor shielding affects predicted energy order. Lanthanoids have their atomic configuration with a variable occupancy of 4f energy level and a common 6s2 level.The energy level of the 5d sub-shell increases than that of the 4f sub-shell moving across period(57–71). As a result of the low energy of 4f sub-shell electrons occupy these sub-shells before occupying the 5d sub-shell. As we progress from atomic number 57 to 71, an electron is added in 4f energy level with increase in atomic number with the exception of some metals where the electron occupies a 5d energy level.

In case of Cerium, energy of 4f and 5d subshells are quite similar and thus explains the electronic configuration stated above.

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