The correct set of four quantum numbers for valence electrons 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. `5,0,1,+(1)/(2)`

Answer

D

Explanation

The electronic configuration of Rubidum (Rb = 37) is `1s^(2) 2s^(2)2p^(2)3s^(3)3p^(6) 3d^(10) 4s^(2) 4p^(6)5s^(1)`
Since last electron enters in 5s orbital .
Hence n = 5 , l = 0 , m = 0 , `s = pm (1)/(2)`