Answer:
Option A
Explanation:
Given, atomic number of Rb, Z=37
Thus, its elecronic configuration is [Kr]5s1. Since the last electron or valence electron enter in 5s subshell.
So, the quantum numbers are n=5, l=0, (for s orbital) m=0
$(\because m=+1 to -1),s=+\frac{1}{2}or -\frac{1}{2}$