The outer electronic configuration of cerium (Z = 58) is

The expected outer electronic configuration of ${ }_{58}\mathrm{Ce}$ is $\left(4\mathrm{f}{)}^1\right(5\mathrm{d}{)}^1(6\mathrm{s}{)}^2$. But due to very similar energies of 4f and 5d orbitals in all the lanthanides excepting Gd $\left(4{\mathrm{f}}^{7}5{\mathrm{d}}^{1}6{\mathrm{s}}^2\right)$ and ${ }_{71}\mathrm{Lu}\left(4{\mathrm{f}}^{14},5{\mathrm{d}}^1,6{\mathrm{s}}^2\right),(5\mathrm{d}{)}^1$electron is shifted to (4f) orbital and hence 5d orbital remains unoccupied. Thus, the observed configuration of ${ }_{58}\mathrm{Ce}$ is $\left(4\mathrm{f}{)}^2\right(5\mathrm{d}{)}^0(6\mathrm{s}{)}^2$.