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(C) The electronic configuration of $Ce$ $(Z=58)$ is $[Xe] 4f^1 5d^1 6s^2$.When $Ce$ forms $Ce^{4+}$,it loses four electrons to achieve the stable nob

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Due to empty $f$-orbitals

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(C) The electronic configuration of $Ce$ $(Z=58)$ is $[Xe] 4f^1 5d^1 6s^2$.When $Ce$ forms $Ce^{4+}$,it loses four electrons to achieve the stable noble gas configuration of $Xe$ $([Xe] 4f^0 5d^0 6s^0)$.In the $+4$ oxidation state,all $4f$,$5d$,and $6s$ orbitals are empty,resulting in a stable configuration similar to the noble gas $Xe$.

Which of the following is responsible for the stability of $Ce^{4+}$?

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