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Quantum Numbers

Which of the ...

Question

Which of the following pairs of electrons is excluded from an atom?

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

and

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

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

and

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

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

and

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

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

and

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

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Solution

The correct option is C $n = 2, l = 0, m = 0, s = + \frac{1}{2}$ and $n = 2, l = 0, m = 0, s = + \frac{1}{2}$
According to Pauli's exclusion principle, no two electrons in an atom can have the same set of values of the four quantum numbers.

Therefore, the excluded pair:

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

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Similar questions

Which set of quantum numbers is possible for the last electron of Mg⁺ ion?

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}

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

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

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

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

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