Comment
Share
Q)

# The correct set of four quantum number for the valence electron of rubidium atom $(Z=37)$ is :

$\begin{array}{1 1} (a)\;5,1,1,+1/2 \\ (b)\;6,0,0,+1/2 \\ (c)\;5,0,0,+1/2 \\ (d)\;5,1,0,+1\end{array}$

$_{37}Rb=_{36} [Kr]5s^1$
Its valance electron is $5s^1$
So, $n= 5,l=0$
$m=0$
$s= +\large\frac{1}{2}$