# How would you determine the quantum number, ml, for an element?

Feb 19, 2016

${m}_{l}$ is the magnetic quantum number, corresponding to the projection of the angular momentum of an orbital, i.e. its orientation in space.

As the symbol suggests, it has to do with $l$, the angular momentum quantum number. $l$ describes the shape of the orbital. Let's look at various values of $l$ and their corresponding ${m}_{l}$.

• $l = 0 \to {m}_{l} = 0$, orbital = $s$
• $l = 1 \to {m}_{l} = - 1 , 0 , + 1$, orbital = $p$
• $l = 2 \to {m}_{l} = - 2 , - 1 , 0 , + 1 , + 2$, orbital = $d$
• $l = 3 \to {m}_{l} = - 3 , - 2 , - 1 , 0 , + 1 , + 2 , + 3$, orbital = $f$

and so on.

The general pattern is that we have:

${m}_{l} = - l , - l + 1 , - l + 2 , . . . , 0 , + 1 , + 2 , . . . , + l - 2 , + l - 1 , + l$

or

$\textcolor{b l u e}{{m}_{l} = 0 , \pm 1 , \pm 2 , . . . , \pm l}$

In short, we have $2 l + 1$ values of ${m}_{l}$ for a particular $l$ for a particular orbital.

If, let's say, we chose boron ($Z = 5$), it has access to the valence orbitals $2 s$ and $2 p$, but it also has the $1 s$ technically as a core orbital.

$1 s$:

$\left(n , l , \textcolor{b l u e}{{m}_{l}}\right) = \left(1 , 0 , \textcolor{b l u e}{0}\right)$

Hence, there is only one $1 s$ orbital.

$2 s$:

$\left(n , l , \textcolor{b l u e}{{m}_{l}}\right) = \left(2 , 0 , \textcolor{b l u e}{0}\right)$

So, there is only one $2 s$ orbital.

$2 p$:

$\left(n , l , \textcolor{b l u e}{{m}_{l}}\right) = \left(2 , 1 , \left[\textcolor{b l u e}{- 1 , 0 , + 1}\right]\right)$

Therefore, there are only three $2 p$ orbitals ($2 {p}_{x}$, $2 {p}_{y}$, and $2 {p}_{z}$).

For its valence orbitals, since it has one $2 s$ and three $2 p$ orbitals, it can have up to $2 \times 1 + 3 \times 2 = 8$ valence electrons. Thus, it is not expected to exceed $8$ valence electrons in its molecular structures.