The correct set of four quantum numbers for the valence electrons of rubidium atom ( Z = 37) is                                                                

Given, atomic number of Rb, Z = 37 Thus, its electronic configuration is [Kr]  $5s^1$. Since, the last electron or valence electron enter in 5s – sub shell. 

So, the quantum numbers are n = 5 l = 0 (for s – orbitals)

$m=0(\because m=+to-l),s=+1/2or-1/2$