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 1

Answer: (A)

E.C of Rubidium is $( Z =37)=1 s ^{2} \,2 s ^{2}\, 2 p ^{6}\, 3 s ^{2} \,3 p ^{6}\, 4 s ^{2} \,3 d ^{10} \,4 p ^{6} \,5 s ^{1}$
The valence electron is in $5 \,s$ orbital.
Its 4 quantum numbers are- $n =5, l =0, m =0, s =+\frac{1}{2}$