The electronic configuration of the last three orbitals in cerium (At. No-58) in ground state is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(4f^15d^16s^2\)
Note: NTA dropped this question in the official Answer Key. However, to provide additional question practice to the students, the question has been slightly modified so that the question has one clear answer.
Cerium (Z = 58) is the first lanthanoid, one place after lanthanum ([Xe]5d16s2). Its electronic configuration is
Ce: [Xe] 4f1 5d1 6s2
The 4f and 5d levels are so close in energy that cerium keeps one electron in 5d; from praseodymium onwards the 5d electron moves into 4f.
NCERT (2026-27 edition):
Electronic Configurations – …
