Q.Give the general electronic configuration of the lanthanoids and explain why cerium's configuration is often written as an exception to the general pattern.

Across the fourteen lanthanoid elements, cerium ($Z=58$) to lutetium ($Z=71$), electrons are progressively added to the 4f subshell, built on the xenon core and the $6s^2$ outer shell already present in the preceding element lanthanum, giving the general configuration

For most lanthanoids, this simplifies to $[\text{Xe}]4f^n6s^2$, with no electron in $5d$. Cerium, however, is a well-known exception: rather than the configuration $[\text{Xe}]4f^26s^2$ that a simple extrapolation from lanthanum ($[\text{Xe}]5d^16s^2$, no 4f electron) would suggest, cerium's actual ground-state configuration is usually written $[\text{Xe}]4f^15d^16s^2$ -- one electron in 4f and one retained in 5d, rather than both of the "new" electrons entering 4f.

The reason is that the 4f and 5d orbitals are extremely close in energy at this point in the periodic table -- close enough that the precise ground-state configuration can depend on subtle balances of electron-electron repulsion and orbital penetration that are not always intuitive from a simple Aufbau-filling argument. This is directly analogous to (though a distinct phenomenon from) the chromium/copper $4s$-to-$3d$ promotions seen in the first-row transition series: in both cases, the two subshells competing for the "extra" electron(s) are close enough in energy that the atom's actual ground state does not follow the naive orbital-filling order. …

State the general [Xe]4f^(1-14)5d^(0-1)6s^2 pattern, then single out cerium's 4f^1 5d^1 configuration as the exception and explain it via the close 4f/5d e …

Writing cerium as simply $4f^26s^2$ without noting it is commonly cited as an exception; confusing this 4f/5d closeness with the completely separate d-block Cr/Cu 4s-t …