The possible set of quantum numbers $(n, l, m, s)$ of the last electron of unipositive potassium ion :

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$3, 2, 0, \frac{1}{2}$

$4, 0, 0, - \frac{1}{2}$

$4, 1, 0, - \frac{1}{2}$

None of these

answer is D.

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Detailed Solution

From the given options their is no possible set of quantum numbers $(n, l, m, s)$ of the last electron of unipositive potassium ion.

The electronic configuration of potassium ion is,

$_{19} K + = 1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6}$

It is observed that the last $e^{-}$ goes into 3p subshell

The value of the quantum number are, $n = 3; l = 1; m = - 1, 0, 1; s = + \frac{1}{2}, - \frac{1}{2}$ .

Hence, the option D is correct.

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