# What is the maximum number of f orbitals in any single energy level in an atom?

Jun 1, 2017

Recall that the angular momentum quantum number $l$ gives the shape of an orbital. It corresponds as:

$l \text{ "" ""type}$
$0 \text{ "" } s$
$1 \text{ "" } p$
$2 \text{ "" } d$
$3 \text{ "" } f$
$\vdots \text{ } \vdots$

The number of orbitals of a given shape is given by the degeneracy, $2 l + 1$. So, there are

$2 l + 1 = 2 \left(3\right) + 1 = \textcolor{b l u e}{\boldsymbol{7}}$

$\left(n - 2\right) f$ orbitals for a given principal quantum number $n$. This is also represented in the number of magnetic quantum numbers ${m}_{l}$.

${m}_{l} = \left\{- l , - l + 1 , . . . , 0 , . . . , l - 1 , l\right\}$

So in the case of $\left(n - 2\right) f$ orbitals,

${m}_{l} = \left\{- 3 , - 2 , - 1 , 0 , + 1 , + 2 , + 3\right\}$

and there are $\textcolor{b l u e}{\boldsymbol{7}}$ such ${m}_{l}$ values, each corresponding to one of the $\left(n - 2\right) f$ orbitals.