Question

Correct set of four quantum numbers for valence electron of rubidium ($Z = 37$) is

• A. 5, 0, 0, $ +\frac{1}{2} $
• B. 5,1, 0, $ +\frac{1}{2} $
• C. 5, 1, 1, $ +\frac{1}{2} $
• D. 6, 0, 0, $ +\frac{1}{2} $

Answer 1

Answer: (A)

Electronic configuration of $Rb (Z=37)$ is
$1 s^{2}, 2 s^{2}, 2 p^{6}, 3 s^{2}, 3 p^{6}, 3 d^{10}, 4 s^{2}, 4 p^{6}, 5 s^{1}$
So, for the valence shell electron $\left(5 s^{1}\right)$
$n=5, l=0, m=0, s=+\frac{1}{2}$