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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?
The filling of electrons follows Aufbau principle. However, introduction of $\mathrm{d}$ and $\mathrm{f}$ orbitals and resulting poor shielding affects predicted energy order, and it consists last 2 or three shells incompletely filled. The energy associated with this subshells is nearly same, and one of the subshell of outer shell is bigger in size than subshell of inner shell. So, to perform minimum repulsion between electrons, one or two electrons jump to a bigger subshell (with similar energy) so this exceptions occur.
In case of cerium, energy of $\mathrm{4f}$ and $\mathrm{5d}$ subshells are quite similar and $\mathrm{5d}$ subshell is bigger than $\mathrm{4f}$ subshell. So 1 electron in $\mathrm{4f}$ jump to $\mathrm{5d}$ subshell.
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.
