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

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$5, 0, 0, + 1 ⁄ 2$

$5, 1, 0, + 1 ⁄ 2$

$5, 1, 1, + 1 ⁄ 2$

$6, 0, 0, + 1 ⁄ 2$

answer is A.

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Detailed Solution

$Rb (z = 37)$

$[Kr]_{36} 5 s^{1}$

For the last electron $n = 5, l = 0, m = 0, s = 1 / 2$

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