Q. The correct set of four quantum number for the valence electron of rubidium ($Z$=37) is
A) $n = 5,\ l = 0,\ m = 0,\ s = + 1/2$
B) $n = 5,\ l = 1,\ m = 1,\ s = + 1/2$
C) $n = 5,\ l = 1,m = 1,s = + 1/2$
D) $n = 6,l = 0,m = 0,s = + 1/2$
Explanation
(a)
$Rb -$Atomic number is 37,
So configuration is
$1s^{2},2s^{2},2p^{6},3s^{2},3p^{6},4s^{2},3d^{10},4p^{6},5s^{1}$
∴ Last electron (valence electron) is $5s^{1}$
$\therefore n = 5\ \ \ \ \ \ \ \$(∵ Electron enters 5 energy level)
$\ \ \ \ \ l = 0$ (∵It is $s$-subshell)
$\ \ \ m = 0$
$\ \ \ \ \ s = \pm 1/2$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.