In an atom, the total number of electrons having quantum numbers $n = 4, | m_{l} | = 1$ and $m_{s} = - \frac{1}{2}$ is :

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Solution

The correct option is A 6
Given,
$| m_{l} | = 1$ means $m_{l}$ can be +1 and -1.
Since n = 4, the possible values of ` $l$ ' are 0,1,2 and 3.
For, $l$ =0 the orbitals of required $m_{l}$ value are not possible.

For the rest of the sub shells we will get a total of 6 orbitals having $m_{l}$ values +1 and -1.

So 6 orbitals can accomodate maximum of 12 electrons according to Pauli's exclusion principle. Since we are considering only $+ \frac{1}{2}$ spin, thus we will get 6 electrons only.

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n = 4, | m_{l} | = 1

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

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n = 4, | m_{l} | = 1

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