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

Options

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

Answer

A

Explanation

`Rb = 1s^(2), 2s^(2), 2p^(6), 3s^(2), 3p^(6), 3d^(10), 4s^(2), 4p^(6), 5s^(1)`
Last electron enters in the 5s orbital
`n=5, 1=0, m=0, s =+1//2`