# Which of the following is NOT a valid quantum number set for an orbital?

## Is $A$ wrong? A) $\left(n , l , {m}_{l}\right) = \left(3 , - 3 , 2\right)$ B) $\left(n , l , {m}_{l}\right) = \left(4 , 3 , 0\right)$ C) $\left(n , l , {m}_{l}\right) = \left(2 , 1 , 1\right)$ D) $\left(n , l , {m}_{l}\right) = \left(3 , 2 , 0\right)$

##### 1 Answer
Jun 19, 2017

Recall the following relevant rules:

1. The value of the angular momentum quantum number $l$ can at most be $n - 1$, one less than the principal quantum number $n$. Also, $l = 0 , 1 , 2 , . . .$, so $l \ge 0$.
2. ${m}_{l}$ can only be in the set $\left\{- l , - l + 1 , . . . , 0 , . . . , l - 1 , l\right\}$. That is, the magnitude of any given ${m}_{l}$ can at most be $l$.

In fact, you are correct that it is $A$ that does NOT occur. $n = 3$ specifies all atomic orbitals at the third energy level ($3 s , 3 p , 3 d$), while $l = 3$, an $f$ orbital, is only possible at $n = 4$ or above (for example, $4 f , 5 f , . . .$ orbitals). Furthermore, $l \ge 0$, so the value of $l = - 3$ isn't even valid.

$B$, $C$, and $D$ are all valid configurations, which are not the incorrect one you are looking to eliminate.

• $B$ specifies a $4 f$ orbital of ${m}_{l} = 0$, namely the $4 {f}_{{z}^{3}}$. • $C$ specifies a $2 p$ orbital of ${m}_{l} = + 1$, such as the $2 {p}_{x}$. • $D$ specifies a $3 d$ orbital of ${m}_{l} = 0$, namely the $3 {d}_{{z}^{2}}$. 