So, I am studying electronic configuration but the elements of the series of the lanthanides confused me. The electronic configuration of $\ce{La}$ is $\mathrm{6s^2\,5d^1}$ and that of cerium is $\mathrm{6s^2\,5d^1\,4f^1}$.

So, why does the extra electron that was added make the $\mathrm{5d}$ orbital have a higher energy?


1 Answer 1


Due to two contradictory factors, one is the effective nuclear charge which is greater in $\mathrm{5d}$ making it of lesser energy and the second one is a nodal factor. Nodal factor depicts localization of electrons, which indicates of higher energy. Hence to balance these two effects out, for Ce the second electron goes to $\mathrm{4f}$ and its electronic configuration stands as $[\ce{Xe}]~\mathrm{(4f)^1 (5d)^1 (6s)^2}.$


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