# Question #66a8c

Jan 28, 2017

$\text{9 orbitals}$

#### Explanation:

The principal quantum number, $n$, corresponds to the energy shell that holds an electron in an atom. Each energy shell contains a specific number of subshells, which in turn contain a specific number of orbitals.

The number of shells is given by the angular momentum quantum number, $l$, which can take the following values

$l = \left\{0 , 1 , \ldots , n - 1\right\}$

In your case, the third energy shell contains $3$ subshells, since

$l = \left\{0 , 1 , 2\right\}$

The number of orbitals is given by the magnetic quantum number, ${m}_{l}$, which can take the following values

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

• $l = 0 \implies {m}_{l} = 0$
• $l = 1 \implies {m}_{l} = \left\{- 1 , 0 , 1\right\}$
• $l = 2 \implies {m}_{l} = \left\{- 2 , - 1 , 0 , 1 , 2\right\}$

This means that the third energy level holds $9$ orbitals, each given by a specific combination of $l$ and of ${m}_{l}$.

Keep in mind that the number of orbitals that can be found on a given energy level $n$ is equal to

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{\text{no. of orbitals} = {n}^{2}}}}$

For $n = 3$, you will once again find

$\text{no. of orbitals} = {3}^{2} = 9$