# What characteristic is given by the angular momentum quantum number?

Nov 7, 2016

The angular momentum quantum number $l$ describes the shape of an orbital. It corresponds to each orbital type, i.e. $\left(0 , 1 , 2 , 3 , . . . , n - 1\right) = \left(s , p , d , f , g , h , . . .\right)$.

Note that:

• ${l}_{\max} = n - 1$
• $l = 0 , 1 , 2 , 3 , . . . , n - 1$
• $2 l + 1$ is the number of orbitals in a given $l$ subshell.
• $l$ is the number of angular nodes in an orbital.

So two examples: $1 s$ orbital:

• $\boldsymbol{n = 1}$ for the principal quantum number (read it off the orbital name).

Thus, ${l}_{\max} = \boldsymbol{l = 0}$. This also means there is no angular node in the $1 s$ orbital.

• $2 l + 1 = 2 \left(0\right) + 1 = \boldsymbol{1}$. Thus, there is only one $1 s$ orbital.

That is also determined by realizing that for the magnetic quantum number, ${m}_{l} = \left\{0\right\}$ due to $l = 0$. i.e. the only $1 s$ orbital there is has $l = 0$.

$2 s , 2 p$ orbital(s):

• $\boldsymbol{n = 2}$ for the principal quantum number (read it off the orbital name).

Thus, $\boldsymbol{l = 0 , 1}$, and ${l}_{\max} = 1$. This also means there is no angular node in the $2 s$ orbital, but $\boldsymbol{1}$ angular node in each $2 p$ orbital.

• For the $2 s$ orbital(s), $2 l + 1 = 2 \left(0\right) + 1 = \boldsymbol{1}$. Thus, there is only one $2 s$ orbital.

That is also determined by realizing that for the magnetic quantum number, ${m}_{l} = \left\{0\right\}$ due to $l = 0$. i.e. the only $1 s$ orbital there is has $l = 0$.

• For the $2 p$ orbital(s), $2 l + 1 = 2 \left(1\right) + 1 = \boldsymbol{3}$. Thus, there are three $2 p$ orbitals.

That is also determined by realizing that for the magnetic quantum number, ${m}_{l} = \left\{- 1 , 0 + 1\right\}$ due to $l = 1$. i.e. each $2 p$ orbital uniquely corresponds to either $l = - 1$, $0$, or $+ 1$. As there are three ${m}_{l}$ values, there are three $2 p$ orbitals.