Class 12 NEET Chemistry D And F Block Elements Questions

Concept:Cerium shows the +4 oxidation state because, after losing four electrons, it attains the stable noble gas configuration of xenon, i.e. a completely empty $4f$ subshell ($4f^0$).Electronic Configurations:$\mathrm{Ce}$ $(Z = 58)$: $[\mathrm{Xe}]\,4f^1\,5d^1\,6s^2$$\mathrm{Ce}^{3+}$: $[\mathrm{Xe}]\,4f^1$$\mathrm{Ce}^{4+}$: $[\mathrm{Xe}]\,4f^0$ (stable empty $4f$ subshell, i.e. noble gas configuration)Explanation:The most common oxidation state of lanthanoids is +3.Cerium, however, can lose one more electron from $\mathrm{Ce}^{3+}$ to give $\mathrm{Ce}^{4+}$.This extra electron removal empties the $4f$ subshell completely, giving the stable $[\mathrm{Xe}]$ configuration.Hence cerium readily exhibits the +4 oxidation state.Why the other options are wrong:Option A is incorrect: $\mathrm{Ce}^{4+}$ is not $4f^{14}$; the fully filled $4f^{14}$ configuration is shown by species such as $\mathrm{Yb}^{2+}$ or $\mathrm{Lu}^{3+}$, not by cerium.Option B is incorrect: the nearest noble gas to cerium is xenon ($Z = 54$), not radon ($Z = 86$).Option C is incorrect: the atomic number of cerium is 58, whereas 61 is the atomic number of promethium.Shortcut:In lanthanoids, the empty ($4f^0$), half-filled ($4f^7$) and fully filled ($4f^{14}$) configurations are exceptionally stable.Cerium achieves $4f^0$ on forming $\mathrm{Ce}^{4+}$, so it readily shows the +4 oxidation state.Answer:Option D: After losing one more electron, it acquires $4f^0$ electronic configuration.