MCQ
Correct set of four quantum numbers for valence electron of rubidium $ (Z = 37)$  is
  • $5,\,0,\,0,\, + \frac{1}{2}$
  • B
    $5,\,1,\,0,\, + \frac{1}{2}$
  • C
    $5,\,1,\,1,\, + \frac{1}{2}$
  • D
    $6,\,0,\,0,\, + \,\frac{1}{2}$

Answer

Correct option: A.
$5,\,0,\,0,\, + \frac{1}{2}$
a
(a) Electronic configuration of $R{b_{(37)}}$ is

$1{s^2}\,2{s^2}\,2{p^6}\,3{s^2}\,3{p^6}\,3{d^{10}}\,4{s^2}\,4{p^6}\,5{s^1}$

So for the valence shell electron $(5{s^1})$

$n = 5,\,l = 0,\,m = 0,\,s = + \frac{1}{2}$

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