Atomic Structure Question 206
Question: Correct set of four quantum numbers for valence electron of rubidium (Z = 37) is
Options:
A) $ 5,,0,,0,,+\frac{1}{2} $
B) $ 5,,1,,0,,+\frac{1}{2} $
C) $ 5,,1,,1,,+\frac{1}{2} $
D) $ 6,,0,,0,,+,\frac{1}{2} $
Show Answer
Answer:
Correct Answer: A
Solution:
Electronic configuration of $ R{b_{(37)}} $ is $ 1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{10}4s^{2}4p^{6}5s^{1} $ So for the valence shell electron $ (5s^{1}) $
$ n=5,,l=0,,m=0,,s=+\frac{1}{2} $
