Answer
a) $8$
b) $6$
c) $2$
d) $1$
Work Step by Step
a) If $n=2$, possible values for angular quantum number are $1)$ $l=0$ and $2)$ $l=1$. If $l=0$, the only possible value for magnetic quantum number is $m_{l}=0$. We have two possibilities for the spin quantum number: $m_{s}=+\frac{1}{2}$ and $m_{s}=-\frac{1}{2}$. Hence, there are $2$ electrons with this combination of quantum numbers. If $l=1$, possible values for magnetic quantum number are $-1$, $0$ and $1$, and for every of these options, there are two possible values for the spin quantum number. Therefore, there are $3\times2=6$ electrons with this combination of quantum numbers. In total, that is $2+6=8$ electrons.
b) If $n=3$ and $l=1$, there are three possible values for magnetic quantum number: $-1$, $0$ and $1$, and for every of these options there are two possible spin quantum numbers ($+\frac{1}{2}$ and $-\frac{1}{2}$). In total, there are $3\times2=6$ electrons which have the required values of principal and angular quantum number.
c) If $n=3$, $l=1$ and $m_{l}=0$, there are 2 possible values for the spin quantum number ($+\frac{1}{2}$ and $-\frac{1}{2}$). Therefore, only $2$ electrons can have this combination of quantum numbers.
d) Since all of the quantum numbers given are correct (in terms of allowed range), there can be only one electron with such combination.
