Find the electronic configuration of rubidium / $(Z=37)$

Rubidium has 37 electrons.

Its ground-state electronic configuration is: $1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,4s^2\,3d^{10}\,4p^6\,5s^1$

So, the valence electron is in the $5s$ orbital.

Assign quantum numbers for a $5s$ electron

For the valence electron in $5s^1$:

• Principal quantum number: $n=5$
• Azimuthal quantum number for an $s$-orbital: $l=0$
• Magnetic quantum number: $m_l=0$ because for $l=0$, only possible value is $0$.
• Spin quantum number: $m_s=+\frac{1}{2}$ or $-\frac{1}{2}$

Since the options give $+\frac{1}{2}$, that is acceptable.

Compare with options

• A: $(5,1,1,+\tfrac{1}{2})$ → corresponds to a $5p$ electron, not $5s$.
• B: $(5,0,0,+\tfrac{1}{2})$ → correct for $5s^1$.
• C: $(5,0,1,+\tfrac{1}{2})$ → impossible, because if $l=0$, then $m_l$ must be $0$.
• D: $(5,1,0,+\tfrac{1}{2})$ → corresponds to a $5p$ electron, not $5s$.

Final answer

The correct set of quantum numbers is: $n=5,\; l=0,\; m_l=0,\; m_s=+\frac{1}{2}$

Hence, the correct option is B.