Cerium (Z = 58) is an important member of the lanthanoids. Which of t…

The correct answer is Option (3) → The +4 oxidation state of cerium is not known in solution

• Configuration: Ce (Z = 58) is [Xe]4f¹5d¹6s².

• Ce³⁺: losing the two 6s electrons and the 5d electron gives [Xe]4f¹, the common +3 state of the lanthanoids.

• Ce⁴⁺: losing the 4f electron as well gives [Xe]4f⁰, a noble gas configuration, which favours the +4 state.

Ce(IV) does exist in solution. It is a strong oxidant (E° for Ce⁴⁺/Ce³⁺ = +1.74 V) that tends to go back to the more stable +3 state, but it reacts with water so slowly that its solutions are used as an analytical reagent.

Option (1) The common oxidation states of cerium are +3 and +4 — Correct about cerium. Besides the usual +3 state, cerium is the lanthanoid well known for the +4 state.

Option (2) The +3 oxidation state of cerium is more stable than +4 oxidation state — Correct about cerium. Ce(IV) is a strong oxidant reverting to the common +3 state, so +3 is the more stable one.

Option (3) The +4 oxidation state of cerium is not known in solution — Incorrect about cerium. +4 ions occur in solution as well as in solids; Ce(IV) is a good analytical reagent.

Option (4) Cerium(IV) acts as an oxidising agent — Correct about cerium. Its high E° of +1.74 V means Ce⁴⁺ readily accepts an electron and is reduced to Ce³⁺.

NCERT (2026-27 edition):
4.5.3 Oxidation States – “… 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.”