The correct set of four quantum numbers for the valence elections of rubidium atom (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

B

Explanation

`Rb(z = 37)`
`[Kt]_(36)5s^(1)`
Find the last electron
`n = 5,l = 0,m = 0,s= 1//2`