The correct set of four quantum numbers for the valence electron of rubidium atom $(Z=37)$ is

The correct set of four quantum numbers for the valence electron of rubidium atom $(Z=37)$ is
  1. $5,1,1,+\frac{1}{2}$
  2. $6,0,0,+\frac{1}{2}$
  3. $5,0,0,+\frac{1}{2}$
  4. $5,1,0,+\frac{1}{2}$

Solution

${ }_{37} \mathrm{Rb}={ }_{36}[\mathrm{Kr}] 5 s^1$
Its valence electron is $5 s^1$.
So,
$\begin{aligned}
& n =5 \\
& l =0 \text { (For } s \text { orbital) } \\
& m =0 (\text { As } m=-l \text { to }+l) \\
& s =+\frac{1}{2}
\end{aligned}$

Asked in: NEET 2012 (Screening)

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