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Multi-Electron Configurations and Orbital Diagrams

Consider the ...

Question

Consider the following sets of quantum numbers. Which of the below setting is not permissible arrangement of electrons in an atom?

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

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n = 5, l = 3, m = 0, s = + \frac{1}{2}

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

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

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Solution

The correct option is C $n = 3, l = 2, m = - 3, s = + \frac{1}{2}$
For any value of $n$ the value of $m$ can lie only between $- l$ to $+ l$ . Hence, for $n = 3, m = - 3$ is not possible.

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Similar questions

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}

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}

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

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