The correct set of four quantum numbers for the valence electron of rubidium atom (Z=37) is :

Solution

Chemistry ma’am · 3 steps

Rubidium's last electron sits alone in 5s, so read off n = 5, l = 0, and m has no choice but 0.

1. Valence orbital. The configuration of Rb is [Kr] 5s¹. Krypton accounts for 36 electrons, so the 37th electron goes into the 5s orbital.
2. Read the numbers. For 5s, the principal quantum number is n = 5 and an s subshell has l = 0. The magnetic quantum number runs from -l to +l, so with l = 0 the only value is m = 0.

   Option	n, l, m	Problem
   (a)	5, 1, 1	l = 1 is a 5p orbital, not 5s
   (b)	5, 0, 0	none: this is 5s
   (c)	5, 0, 1	m = 1 is impossible when l = 0
   (d)	5, 1, 0	l = 1 is a 5p orbital, not 5s

3. So. The single 5s electron has spin +1/2 by convention, giving 5, 0, 0, +1/2, which is option (b).

The idea to keep. Write the configuration first, find the subshell of the last electron, then read n and l from it. Always check that |m| ≤ l before trusting a set.

Answer from the NTA final answer key. Solution by PrepWiser.