Question

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

• A. $5,0,0,+1/2$
• B. $5,1,0,+1/2$
• C. $5,1,1,+1/2$
• D. $5,0,1,+1/2$

Answer 1

The correct option is A $5,0,0,+1/2$

The electronic configuration of rubidium is $\left[Kr\right]5s^1$

Thus, there is one electron in last or in 5th shell

For one electron value of $l,m$ must be 0 and $m_s$ may be $-1/2$or $+1/2$

So, the correct option is $A$