The atomic number of Lanthanum is 57. Its electronic configuration will be..........
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \([Xe]5d^16s^2\)
Lanthanum (Z = 57) comes right after barium, \( \mathrm{[Xe]\,6s^{2}} \), so it has three electrons beyond the xenon core. Two fill the 6s orbital. The third enters the 5d orbital, not 4f, because in lanthanum the 4f orbitals have not yet dropped below 5d in energy:
\( \displaystyle \text{La}\ (Z = 57): \mathrm{[Xe]\,5d^{1}\,6s^{2}} \)
The 4f filling starts only from cerium, \( \mathrm{[Xe]\,4f^{1}\,5d^{1}\,6s^{2}} \).
Why the other options are incorrect
Option (2) \([Xe]4f^15d^2\) — Incorrect. It leaves the 6s orbital empty, which fills before 4f and 5d.
Option (3) \([Xe]4f^3\) — Incorrect. It puts all three outer electrons in 4f; La has no 4f electron, and 6s is filled first.
Option (4) \([Xe]4f^15d^16s^1\) — Incorrect. La has no 4f electron, and both of its 6s electrons are present; the 4f filling starts only from cerium.
NCERT (2026-27 edition):
Electronic Configurations – …
