Choose the symbol of a lanthanoid metal which is well known to exhibi…

The correct answer is Option (3) → Ce

The lanthanoids are the 14 elements after lanthanum, cerium (Z = 58) to lutetium (Z = 71), in which the 4f orbitals are being filled. Their common oxidation state is +3. A +2 or +4 state appears only where it gives an empty, half-filled or completely filled 4f subshell, which is extra stable.

Cerium (Z = 58) is [Xe]4f¹5d¹6s². Losing all four outer electrons gives \( \ce{Ce^4+} \): [Xe]4f⁰, an empty 4f subshell with the noble gas configuration. So cerium is the lanthanoid well known for the +4 state. Ce(IV) is a strong oxidising agent (E° for \( \ce{Ce^4+/Ce^3+} \) = +1.74 V), because it tends to return to the common +3 state.

Why the other options are incorrect

• Option (1) La — Incorrect. Lanthanum (Z = 57) is [Xe]5d¹6s². Losing three electrons already gives the noble gas configuration [Xe], so it shows only the +3 state; it has no fourth outer electron to lose.

• Option (2) Th — Incorrect. Thorium (Z = 90) does show +4, its main oxidation state, but it is an actinoid (5f series), not a lanthanoid.

• Option (4) Cm — Incorrect. Curium (Z = 96) is also an actinoid, not a lanthanoid; its common oxidation state is +3.

Note: The NCERT book prints La(II) in the first line of the passage below; it should read La(III).

NCERT (2026-27 edition):
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. The E^o value for Ce⁴⁺/ Ce³⁺ is + 1.74 V which suggests that it can oxidise water. However, the reaction rate is very slow and hence Ce(IV) is a good analytical reagent.”