Atomic Structure Result Question 15-1
15. The correct set of four quantum numbers for the valence electrons of rubidium atom $(Z=37)$ is
(2013 Main)
(a) $5,0,0,+\dfrac{1}{2}$
(b) $5,1,0,+\dfrac{1}{2}$
(c) $5,1,1,+\dfrac{1}{2}$
(d) $5,0,1,+\dfrac{1}{2}$
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Answer:
Correct Answer: 15. ( a )
Solution:
- Given, atomic number of $\mathrm{Rb}, Z=37$
Thus, its electronic configuration is $[\mathrm{Kr}] 5 s^1$. Since, the last electron or valence electron enter in $5 s$ subshell.
So, the quantum numbers are $n=5, l=0$, (for $s$-orbital) $m=0$
$(\because m=+l$ to $-l), s=+1 / 2$ or $-1 / 2$.
BETA
