Cerium can show an oxidation state of +4 because

The correct answer is Option (3) → of its tendency to attain noble gas configuration of xenon

The common oxidation state of all the lanthanoids is +3. A few of them also show +2 or +4 when the ion formed has an extra-stable f subshell: empty (f⁰), half-filled (f⁷) or completely filled (f¹⁴).

• Configuration: Cerium (Z = 58) is \( \ce{Ce}:\quad [\ce{Xe}]\,4f^1\,5d^1\,6s^2 \).

• Electrons removed: it loses its four outer electrons (two 6s, one 5d and one 4f) to form \( \ce{Ce^4+} \): \( [\ce{Xe}]\,4f^0 \).

With the 4f subshell empty, \( \ce{Ce^4+} \) has the same electron arrangement as the noble gas xenon. This stable arrangement favours the +4 state.

Why the other options are incorrect

Option (1) it resembles alkali metals — Incorrect. Alkali metals have one valence electron (ns¹) and show only the +1 state. A likeness to them could never explain a +4 state.

Option (2) it has very low value of ionization energy — Incorrect.

• Removing a fourth electron from an ion that already carries a +3 charge needs a large amount of energy.

• Cerium reaches +4 because the ion formed is stable (empty 4f), not because its ionisation energies are low; that is also why most lanthanoids stop at +3.

Option (4) of its tendency to attain the half-filled f⁷ configuration — Incorrect. \( \ce{Ce^4+} \) is 4f⁰, an empty f subshell, not a half-filled one. The f⁷ case belongs to ions such as \( \ce{Eu^2+} \):

• Configuration: europium (Z = 63) is [Xe]4f⁷6s².

• Result: losing the two 6s electrons gives 4f⁷.

NCERT (2026-27 edition):
4.5.3 Oxidation States – “In the lanthanoids, La(II) and Ln(III) compounds are predominant species. However, occasionally +2 and +4 ions in solution or in solid compounds are also obtained. This irregularity (as in ionisation enthalpies) arises mainly from the extra stability of empty, half-filled or filled f subshell. Thus, the formation of Ce^IV is favoured by its noble gas configuration, but it is a strong oxidant reverting to the common +3 state.”