Answer
a) $3$
b) $3$
c) $1$
d) $5$
e) $5$
f) $7$
g) $16$
h) $1$
Work Step by Step
a) For $3p$ orbitals, we have $n=3$, $l=1$ and $m_{l}$ can be $-1$, $0$ or $1$, therefore there are $3$ $3p$ orbitals ($3p_{x}$, $3p_{y}$, $3p_{z}$).
b) For $4p$ orbitals, we have $n=4$, $l=1$ and $m_{l}$ can be $-1$, $0$ or $1$, therefore there are $3$ $4p$ orbitals ($4p_{x}$, $4p_{y}$, $4p_{z}$).
c) There is only one $4p_{x}$ orbital, since the principal, angular and magnetic quantum numbers are unambiguously determined.
d) For $6d$ orbitals, we have $n=6$, $l=2$ and $m_{l}$ can be $-2$, $-1$, $0$, $1$ or $2$, therefore there are $5$ $6d$ orbitals.
e) For $5d$ orbitals, we have $n=5$, $l=2$ and $m_{l}$ can be $-2$, $-1$, $0$, $1$ or $2$, therefore there are $5$ $5d$ orbitals.
f) For $5f$ orbitals, we have $n=5$, $l=3$ and $m_{l}$ can be $-3$, $-2$, $-1$, $0$, $1$, $2$ or $3$, therefore there are $7$ $5f$ orbitals.
g) If $n=5$, we have the following orbitals: $5s$ ($1$ orbital), $5p$ ($3$ orbitals), $5d$ ($5$ orbitals) and $5f$ ($7$ orbitals). In total, that is $16$ orbitals.
h) For $7s$ orbital, we have $n=7$, $l=0$ and $m_{l}=0$. Therefore, there is only one $7s$ orbital.
