Which of the following correctly represents the set of quantum numbers $(n, l, m, s)$ for the $21^{s t}$ electron in Chromium?

4, 0, 0, + 1 / 2

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3, 2, 0, + 1 / 2

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3, 2, - 1, + 1 / 2

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3, 2, + 1, + 1 / 2

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Solution

The correct option is C $3, 2, - 1, + 1 / 2$
Cr - Electronic Configuration - $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{1} 3 d^{5}$

$21^{s t}$ electron is present in 3d orbital.

So, Its set of quantum numbers are
n = 3
l = 2
m = -1
s = +1/2

Option C is correct

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Similar questions

Which one of the following represents a wrong set of quantum numbers for an electron in an atom? (arranged as n, l, m, and s)

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

n = 3, l = 1, m = 1, s = - \frac{1}{2}

n = 3, l = 2, m = 0, s = + \frac{1}{2}

n = 3, l = 0, m = 0, s = - \frac{1}{2}

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

n = 3, l = 1, m = 1, s = - \frac{1}{2}

n = 3, l = 2, m = 0, s = + \frac{1}{2}

n = 3, l = 0, m = 0, s = - \frac{1}{2}

P = 3, 2, - 2, + \frac{1}{2}

Q = 3, 0, 0, + \frac{1}{2}

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