What is the correct electronic configuration of Palladium (Pd, Atomic…

The correct answer is Option (3) → [Kr] 4d¹⁰, 5s⁰

By the aufbau order, transition elements are expected to have the outer configuration (n−1)d^x ns². But the (n−1)d and ns orbitals are very close in energy, so an electron sometimes moves from ns to (n−1)d when this gives the extra stability of a half-filled (d⁵) or completely filled (d¹⁰) d subshell. Common examples:

• Cr (Z = 24): 3d⁵4s¹   and   Mo (Z = 42): 4d⁵5s¹

• Cu (Z = 29): 3d¹⁰4s¹   and   Ag (Z = 47): 4d¹⁰5s¹

• Pd (Z = 46): 4d¹⁰5s⁰, where both 5s electrons move into 4d

Palladium has ten electrons outside the krypton core. Instead of 4d⁸5s², all ten occupy the 4d subshell, which becomes completely filled. So the configuration of Pd is [Kr]4d¹⁰5s⁰, and it is the only element of the d-block with no electron in its outermost s orbital.

Why the other options are incorrect

• Option (1) [Kr] 4d⁸, 5s² — Incorrect. This is the expected aufbau configuration (like nickel, 3d⁸4s²), but it is not the observed one for Pd.

• Option (2) [Kr] 4d⁹, 5s¹ — Incorrect. Neither subshell is full; this pattern (5d⁹6s¹) belongs to platinum.

• Option (4) [Kr] 4d⁹, 5s² — Incorrect. This has eleven electrons outside the krypton core, but palladium (Z = 46) has only ten.

NCERT (2026-27 edition):
Electronic Configurations of the d-Block Elements

Note: The same/similar question has been asked in different years by NTA as per the details given below: