Class 12 NEET Chemistry D And F Block Elements Questions

Concept:Cerium shows +4 oxidation state because it attains a stable empty $4f$ subshell ($4f^0$) after losing four electrons.Chemical Equation / Formula:$\mathrm{Ce}$ (Z = 58): $[\mathrm{Xe}]\,4f^1\ 5d^1\ 6s^2$$\mathrm{Ce}^{3+}$: $[\mathrm{Xe}]\,4f^1$$\mathrm{Ce}^{4+}$: $[\mathrm{Xe}]\,4f^0$ (stable empty $4f$ orbital)Explanation:The most common oxidation state of lanthanoids is +3.However, cerium can lose one more electron to form $\mathrm{Ce}^{4+}$.Losing this extra electron gives cerium a completely empty $4f$ subshell ($4f^0$).An empty $4f$ subshell is electronically stable due to its symmetric nature.Hence, cerium exhibits the +4 oxidation state.Shortcut:In lanthanoids, empty ($4f^0$), half-filled ($4f^7$), and fully filled ($4f^{14}$) configurations are exceptionally stable.Cerium achieves $4f^0$, so it readily forms $\mathrm{Ce}^{4+}$.Answer:Option D: After losing one more electron, it acquires $4f^0$ electronic configuration.