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} $



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