# Which of the following sets of 3 quantum numbers is/are possible?

## a) $2 , 1 , 1$ b) $3 , 2 , - 2$ c) $2 , 0 , 0$ d) $2 , 0 , - 1$ e) $1 , 1 , - 1$

May 15, 2017

This is a situation where you'll just have to remember the rules... The following rules are relevant:

• The principal quantum number $n$ is always one more than the maximum angular momentum $l$. That is, ${l}_{\max} = n - 1$.
• The set of valid ${m}_{l}$ must be in the range $\left\{0 , \pm 1 , \pm 2 , . . . , \pm l\right\}$. Thus, if for example, $l = 2$, ${m}_{l}$ cannot be $- 3$ because $| - 3 | > 2$.

These are all that are necessary to determine the impossible combinations. I will identify the possible quantum number combinations, and I will leave you to determine why the remaining combinations are incorrect.

Possible:

• a): $\left(n , l , {m}_{l}\right) = \left(2 , 1 , 1\right)$. This designates one of the $2 p$ orbitals.
• b) $\left(n , l , {m}_{l}\right) = \left(3 , 2 , - 2\right)$. This designates one of the $3 d$ orbitals.
• c) $\left(n , l , {m}_{l}\right) = \left(2 , 0 , 0\right)$. This designates the one $2 s$ orbital.

Impossible: the rest, because they violate the above rules. Which one violates which rule? What is a valid correction to the combination?

Here is one example; d) has $l = 0$, but ${m}_{l}$ is not in the set $\left\{0\right\}$ since $| - 1 | > 0$. Therefore, a) is impossible. One can fix this by letting ${m}_{l} = 0$, since $0$ is in the set $\left\{0\right\}$. If we have $\left(n , l , {m}_{l}\right) = \left(1 , 0 , 0\right)$, we have a $1 s$ orbital.