Although $(+3)$ oxidation state is the characteristic oxidation state of lanthanoids,cerium shows $(+4)$ oxidation state also. Why?
(N/A) Cerium ($Ce$,atomic number $58$) has the electronic configuration $[Xe] 4f^1 5d^1 6s^2$.
By losing four electrons,it attains the stable noble gas configuration of Xenon $([Xe])$.
This configuration is highly stable because it corresponds to an empty $4f$ orbital,which is energetically favorable.
Therefore,$Ce^{4+}$ is a stable ion,allowing cerium to exhibit the $(+4)$ oxidation state.
By losing four electrons,it attains the stable noble gas configuration of Xenon $([Xe])$.
This configuration is highly stable because it corresponds to an empty $4f$ orbital,which is energetically favorable.
Therefore,$Ce^{4+}$ is a stable ion,allowing cerium to exhibit the $(+4)$ oxidation state.
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