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)

$5,0,0, + \frac{1}{2}$

Answer 2

Electronic configuration of $Rb_{(37)}$ is

$1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{10}4s^{2}4p^{6}5s^{1}$

So for the valence shell electron $(5s^{1})$

$n = 5,l = 0,m = 0,s = + \frac{1}{2}$