Question

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

Electronic configuration of rubidium
$1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{10}4s^{2}4p^{6}5s^1$
Therefore, valence shell of rubidium has:
Principle quantum number $\left(n\right)=5$
Azimuthal quantum number $\left(l\right)=0$
Magnetic quantum number $\left(m_l\right)=0$
Spin quantum number can be $=\pm \frac{1}{2}$