Derive the electronic configuration of \( Ce^{3+} \) ion by using Hund's rule.
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \([Xe] 4f^1\)
To find the configuration of a lanthanoid ion, write the configuration of the neutral atom from its atomic number and remove electrons in the order 6s first, then 5d, then 4f. Hund’s rule then places the remaining f electrons singly in the 4f orbitals (here there is only one).
Cerium (Z = 58):
\( \displaystyle \text{Ce}: \mathrm{[Xe]\,4f^{1}\,5d^{1}\,6s^{2}} \)
Forming \( \ce{Ce^3+} \) removes the two 6s electrons and the 5d electron:
\( \displaystyle \ce{Ce^3+}: \mathrm{[Xe]\,4f^{1}} \)
So the configuration of \( \ce{Ce^3+} \) is [Xe] 4f1, with one unpaired electron. (Losing that 4f electron as well gives \( \ce{Ce^4+} \), [Xe], a stable noble gas configuration.)
Why the other options are incorrect
Option (1) \([Xe] 4f^1 5d^1 6s^2\) — Incorrect. This is the configuration of the neutral cerium atom; no electrons have been removed.
Option (3) \([Rn] 5f^2 6d^1 7s^2\) — Incorrect. A radon core belongs to the actinoids; this is the configuration of protactinium (Z = 91), not of a cerium ion.
Option (4) \([Rn] 5f^1\) — Incorrect. This also has an actinoid (radon) core; it is the configuration of \( \ce{Th^3+} \), the actinoid counterpart of \( \ce{Ce^3+} \).
NCERT (2026-27 edition):
Electronic Configurations – …
