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(C) The electronic configuration of Cerium ($Ce$,atomic number $58$) is $[Xe] 4f^1 5d^1 6s^2$. When $Ce$ loses four electrons to form $Ce^{4+}$,it los

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$Ce$

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(C) The electronic configuration of Cerium ($Ce$,atomic number $58$) is $[Xe] 4f^1 5d^1 6s^2$. When $Ce$ loses four electrons to form $Ce^{4+}$,it loses the two $6s$,one $5d$,and one $4f$ electron. The resulting configuration of $Ce^{4+}$ is $[Xe] 4f^0$,which is a stable noble gas configuration. Therefore,$Ce$ is the lanthanoid that exhibits a $+4$ oxidation state with an $f^0$ configuration.

Which lanthanoid from the following may exhibit $+4$ oxidation state with $f^{0}$ configuration?

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