Chemistry - f-block Question with Solution | TestHub

The atomic number of Cerium (Ce) is $58$.

The electronic configuration of Ce is $[Xe]4f^{1}5d^{1}6s^2$.

In $+3$ oxidation state, the electronic configuration of $C{e}^{3+}$ is $[Xe]4f^1$.

By losing one more electron, it forms $C{e}^{4+}$ ion.

The electronic configuration of $C{e}^{4+}$ is $[Xe]4f^0$, which is a highly stable noble gas configuration.

Thus, Cerium shows $+4$ oxidation state because after losing one more electron from $+3$ state, it acquires $4f^0$ electronic configuration.

Answer: After losing one more electron, it acquires $4f^0$ electronic configuration.