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The set of quantum number for 19^th electrons of chromium (Z = 24) is

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

$4, 1, 1, + \frac{1}{2}$

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

$3, 2, - 2, + \frac{1}{2}$

answer is A.

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Detailed Solution

$(Z = 24) {= 1s}^{2} {,2s}^{2} {,2p}^{6} {,3s}^{2} {,3p}^{6} {,4s}^{1} {,3d}^{5}$ For 19^th electron $(i . e . 4 s^{1})$

$n = 4, l = 0, s = + \frac{1}{2}$

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