# How many sets of quantum numbers specify all the 2p orbitals?

Jul 30, 2017

Well, there are three sets... The $2 p$ orbitals are triply-degenerate, i.e. $2 l + 1 = 3$, and as such, three sets of quantum numbers specify such orbitals.

$\left(n , l , {m}_{l}\right) = \left(2 , 1 , \left\{- 1 , 0 , + 1\right\}\right)$

Do you spot three here?

Recall:

• The principal quantum number $n$ specifies what energy level one is in: $n = 1 , 2 , 3 , . . .$
• The angular momentum quantum number $l$ describes the shape of the orbital.

$l = 0 , 1 , 2 , 3 , 4 , 5 , 6 , 7 , . . . , n - 1 \leftrightarrow s , p , d , f , g , h , i , k , . . .$.

If $l = 0$, the orbital is sharp (spherical), an $s$ orbital.
If $l = 1$, the orbital is principal (polarized), an $p$ orbital.
If $l = 2$, the orbital is diffuse, an $d$ orbital.
If $l = 3$, the orbital is "fundamental", an $f$ orbital.

• The magnetic quantum number ${m}_{l}$ corresponds to each orbital "orientation", and describes a separate orbital orthogonal to the rest. ${m}_{l} = \left\{- l , - l + 1 , . . . , 0 , . . . , l - 1 , l\right\}$.

These are all three quantum numbers required to describe an orbital. If you want to describe orbital occupations, you'll need to specify the electron spin, ${m}_{s}$.