The maximum number of electrons that can have principal quantum number, $n = 3$ , and spin quantum number, $m_{s} = - 1 / 2$ , is:

7

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8

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9

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10

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Solution

The correct option is B $9$
For principal quantum number $(n = 3)$
Number of orbitals $= n^{2} = 9$
So, number of electrons with $m_{s} = \frac{1}{2}$ will be 9

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n = 3

m_{s} = - \frac{1}{2}

m_{s} = - \frac{1}{2}

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m_{s} = - \frac{1}{2}

m_{s} = + \frac{1}{2}

2

1

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