Correct set of four quantum numbers for the valence (outermost) electron of rubidium `(Z = 37)` is

Options

A. `5, 1, 1, +(1)/(2)`
B. `6,0,0,+(1)/(2)`
C. `5,0,0,+(1)/(2)`
D. `5,1,0,+(1)/(2)`

Answer

C

Explanation

(c) `._(37)Rb=._(36)[Kr]5s^(1)`
Its valence electron is `5s^(1)`.
n=5
l=0 (For s-orbital)
m=0 (As m = -l to +l)
`s=+(1)/(2)`