Question

Correct set of four quantum numbers for the valence (outermost) 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

The correct option is A $5,0,0,+\frac{1}{2}$

The four quantum numbers are determined by looking at the valence shell configuration:
The valence shell configuration of rubidium (Rb) is $\left[Kr\right]5s^1$

$\therefore$ The valence falls in the 5s orbital.

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

Hence, the correct option is $A$