Question

The correct set of four quantum numbers for the valence electrons of rubidium atom $(Z=37)$ is:

• A. $5,1,1,+\frac{1}{2}$
• B. $5,0,1,+\frac{1}{2}$
• C. $5,0,0,+\frac{1}{2}$
• D. $5,1,0,+\frac{1}{2}$

Answer 1

Answer: (C)

$5,0,0,+\frac{1}{2}$

E.C of Rubidium is (Z=37) = $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^1$

The valence electron is in 5s orbital.

Its 4 quantum numbers are- $n=5,l=0,m=0,s=+\frac{1}{2}$

Therefore, the correct option is $C$