Answer
(a) and (d) represent impossible combinations of $n$ and $l$.
Work Step by Step
*NOTES TO REMEMBER:
In an orbital designation,
- the number represents the value of $n$.
- the letter represents the value of $l$, according to the following rule:
$l=0$, the letter used is $s$.
$l=1$, the letter used is $p$.
$l=2$, the letter used is $d$.
$l=3$, the letter used is $f$.
- The value of $l$ cannot exceed or even be equal with the value of $n$
(a) Orbital designation: $1p$. Therefore, $n=1$ and $l=1$.
Since in this case, $l$ is equal with $n$, so this is an impossible combination.
(b) Orbital designation: $4s$. Which means, $n=4$ and $l=0$.
The value of $l$ in this case is smaller than the value of $n$. Therefore, this is a possible combination.
(c) Orbital designation: $5f$. Which means $n=5$ and $l=3$.
The value of $l$ in this case is smaller than the value of $n$. Therefore, this is a possible combination.
(d) Orbital designation: $2d$. Which means $n=2$ and $l=2$.
The value of $l$ in this case is equal with the value of $n$. So this is an impossible combination.
