Structure of Atom - Result Question 38
####42. The correct set of four quantum numbers for the valence electron of rubidium atom $(Z=37)$ is
[2012]
(a) $5,1,1+1 / 2$
(b) $6,0,0,+1 / 2$
(c) $5,0,0,+1 / 2$
(d) $5,1,0,+1 / 2$
Show Answer
Solution:
- (c) Electronic configuration of $Rb=[Kr] 5 s^{1}$
Set of quantum numbers, $n=5$
$\ell=0, \therefore s$-orbital
$m=0, s=+1 / 2$
The other possible answer is 5, 0, 0, - $1 / 2$ In and orbital any electron can take the value of $\pm 1 / 2$ and the other elctron of the same orbital, takes the value with opposite sign.
