Question

The correct set of four quantum numbers for the outermost electrons 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 $5,0,0,+\frac{1}{2}$

Electronic configuration of $Rb=\left[Kr\right]5s^1$

For outermost electron of Rb, the configuration is $5s^1$.

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

Therefore, option A is correct.