Question
The correct set of four quantum numbers for the valence electrons of rubidium atom $$( Z = 37 )$$ is :
A.
$$5,0,0, + \frac{1}{2}\,$$
B.
$$5,1,0, + \frac{1}{2}$$
C.
$$5,1,1, + \frac{1}{2}$$
D.
$$5,0,1, + \frac{1}{2}$$
Answer :
$$5,0,0, + \frac{1}{2}\,$$
Solution :
The electronic configuration of Rubidium $$(Rb=37)$$ is
$$1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}3{d^{10}}4{s^2}4{p^6}5{s^1}$$
Since last electron enters in $$5s$$ orbital
$${\text{Hence}}\,\,n = 5,\,\,l = 0,\,\,m = 0,\,\,s = \pm \frac{1}{2}$$