Correct electronic configuration of Cu and Cr are ____.

The correct answer is Option (2) → \(3d^{10} 4s^1\) and \(3d^5 4s^1\) respectively

By the aufbau order, the 4s orbital fills before 3d, so one would expect every first-row transition element to end in \( 4s^2 \). Chromium and copper are the two exceptions. The 3d and 4s energies are very close, and a half-filled (\( d^5 \)) or completely filled (\( d^{10} \)) d subshell is extra stable because of its symmetry and larger exchange energy. So one 4s electron moves into the 3d subshell.

• Cr (Z = 24): expected \( \mathrm{[Ar]\,3d^{4}\,4s^{2}} \); actual \( \mathrm{[Ar]\,3d^{5}\,4s^{1}} \) (half-filled 3d).

• Cu (Z = 29): expected \( \mathrm{[Ar]\,3d^{9}\,4s^{2}} \); actual \( \mathrm{[Ar]\,3d^{10}\,4s^{1}} \) (completely filled 3d).

So, for Cu and Cr respectively, the configurations are \( 3d^{10}4s^1 \) and \( 3d^5 4s^1 \).

NCERT (2026-27 edition):
Electronic Configurations of the d-Block Elements – “… Furthermore, half and completely filled sets of orbitals are relatively more stable. A consequence of this factor is reflected in the electronic configurations of Cr and Cu in the 3d series. For example, consider the case of Cr, which has 3d⁵ 4s¹ configuration instead of 3d⁴4s²; the energy gap between the two sets (3d and 4s) of orbitals is small enough to prevent electron entering the 3d orbitals. Similarly in case of Cu, the configuration is 3d¹⁰4s¹ and not 3d⁹4s².”