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(D) For $n=4$,the possible values of $l$ are $0, 1, 2, 3$. The condition $|m_{l}|=1$ implies $m_{l} = +1$ or $m_{l} = -1$. For each $l$ value,we check

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$6$

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(D) For $n=4$,the possible values of $l$ are $0, 1, 2, 3$. The condition $|m_{l}|=1$ implies $m_{l} = +1$ or $m_{l} = -1$. For each $l$ value,we check if $m_{l} = \pm 1$ is possible: - $l=0$ ($s$-orbital): $m_{l}=0$ (Not possible) - $l=1$ ($p$-orbital): $m_{l}=-1, 0, +1$ (Possible: $m_{l}=-1, +1$) - $l=2$ ($d$-orbital): $m_{l}=-2, -1, 0, +1, +2$ (Possible: $m_{l}=-1, +1$) - $l=3$ ($f$-orbital): $m_{l}=-3, -2, -1, 0, +1, +2, +3$ (Possible: $m_{l}=-1, +1$) Total orbitals with $|m_{l}|=1$ are $2$ $(p)$ + $2$ $(d)$ + $2$ $(f)$ = $6$ orbitals. Each orbital can hold one electron with $m_{s}=-\frac{1}{2}$. Therefore,the total number of electrons is $6 	imes 1 = 6$.

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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