Find the number of orbitals that contain at least one electron (singly or doubly occupied) in a ground-state molybdenum atom ( $Z = 42$ ).

Numerical answer type, tolerance 0.1.

Answer: 24

Mo is

[Kr] 4 d^{5} 5 s^{1}

(exception).

[Kr]

occupies

1 s, 2 s, 2 p (3), 3 s, 3 p (3), 3 d (5), 4 s, 4 p (3)

= 1 + 1 + 3 + 1 + 3 + 5 + 1 + 3 = 18

orbitals.

4 d^{5}

uses 5 orbitals and

5 s^{1}

uses 1. Total

= 18 + 5 + 1 = 24