Question

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

• A. $n=5,l=0,m=0,s=+\frac{1}{2}$
• B. $n=5,l=1,m=0,s=+\frac{1}{2}$
• C. $n=5,l=1,m=1,s=+\frac{1}{2}$
• D. $n=6,l=0,m=0,s=+\frac{1}{2}$

Answer 1

The correct option is A $n=5,l=0,m=0,s=+\frac{1}{2}$

The electron configuration for Rubidium is:
$R{b}_{37}$: $\left[Kr\right]5s^1$
So, the valence electron is in $5^{th}$ shell and $s$-subshell i.e $n=5,l=0,m=0ands=+\frac{1}{2}or-\frac{1}{2}$.