The characteristic oxidation state of Lanthanoids is +3 but Ce shows +4 oxidation state because .............
(A) it has variable ionization enthalpy
(B) it has a tendency to attain noble gas configuration
(C) it has a tendency to attain \( f^0 \) configuration
(D) it resembles \( Pb^{4+} \)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (B) and (C) only
The lanthanoids mostly show +3, but +2 and +4 ions also appear where they give an empty, half-filled or completely filled 4f subshell, which is extra stable.
Ce (Z = 58) is \( \mathrm{[Xe]\,4f^{1}\,5d^{1}\,6s^{2}} \). Losing all four outer electrons gives \( \ce{Ce^4+} \): \( \mathrm{[Xe]\,4f^{0}} \). This is the configuration of the noble gas xenon with an empty f subshell, so the +4 state is favoured. \( \ce{Ce^4+} \) is still a strong oxidant (\( E^\circ = +1.74 \) V) and tends to return to +3.
(A) it has variable ionization enthalpy — Incorrect. This does not explain the +4 state; it is the stable configuration reached that does.
(B) it has a tendency to attain noble gas configuration — Correct. \( \ce{Ce^4+} \) has the xenon configuration.
(C) it has a tendency to attain \(f^0\) configuration — Correct. All its 4f electrons are lost, leaving \( \mathrm{4f^{0}} \).
(D) it resembles \(Pb^{4+}\) — Incorrect. Lead is a p-block element; its oxidation states have nothing to do with the 4f configuration of cerium.
NCERT (2026-27 edition):
Oxidation States
