# Question #a7fdc

Feb 7, 2018

${m}_{l} = 0$

#### Explanation:

Start by writing the electron configuration of a neutral sodium atom.

$\text{Na: } 1 {s}^{2} 2 {s}^{2} 2 {p}^{6} 3 {s}^{1}$

The last electron in a neutral sodium atom is added to the $3 s$ subshell, which, as you know, contains a single orbital, the $3 s$ orbital.

Now, the magnetic quantum number, ${m}_{l}$, tells you the orientation of the orbital in which an electron is located inside an atom.

In the case of an $s$ orbital, the magnetic quantum number can only take one value.

${m}_{l} = 0$

This is the case because the value of the magnetic quantum number depends on the value of the angular momentum quantum number, $l$.

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

The angular momentum quantum number tells you the shape of the orbital(s), i.e. the energy subshell in which the electron is located.

An $s$ subshell is denoted by

$l = 0$

which implies that the magnetic quantum number can only take one possible value here.

$l = 0 \implies {m}_{l} = 0$

This tells you that an $s$ subshell contains a single orbital, the $s$ orbital.