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Correct set of four quantum numbers for the valence electron of rubidium (Z = 37) is
Rubidium is $$[\text{Kr}]5s^1$$, so its valence electron is in the $$5s$$ orbital. That gives $$n = 5$$, $$l = 0$$, $$m = 0$$ and $$s = +\frac{1}{2}$$.
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