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

The correct set of four quantum numbers for valence electron of Rubidium (Z = 37) is :  $5,0,0,+\frac{1}{2}$.

Rubidium's electronic structure is 

$1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{10}4s^{2}4p^{6}5s^1$
In light of this, the rubidium valence shell has:
Quantum number (n) in principle equals 5.
The azimuthal quantum number (l) equals 0.
$\left(m_1\right)$ = 0 is the magnetic quantum number.
The spin quantum number could be = $\pm \frac{1}{2}$

Hence, option A is correct.