# What are m and s for a 3p electron?

Jan 6, 2015

Could you be referring to the quantum numbers ${m}_{l}$ and ${m}_{s}$, corresponding maybe to the electrons in a 3p-orbital?

Let's assume you are talking about chlorine, which has an electron configuration

"Cl": ["Ne"]3s^(2)3p^5

and you want to determine the magnetic quantum number, ${m}_{l}$, and the spin quantum number, ${m}_{s}$, for the unpaired electron sitting in a 3p-orbital.

Since you are dealing with a 3p-orbital, your principal and angular momentum quantum numbers will be $n = 3$ and $l = 1$.

The magnetic quantum number can only take integer values ranging from $- l$ to $+ l$, so you have three possible values for ${m}_{l}$:

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

each corresponding to one of the three 3p-orbitals $3 {p}_{x}$, $3 {p}_{y}$, and $3 {p}_{z}$. Let's ssume that

"Cl": ["Ne"]3s^(2)3p_x^(2)3p_y^(2)3p_z^(1)

In this case, the unpaired electron can be found in the $3 {p}_{z}$ orbital, which corresponds to ${m}_{l} = + 1$.

Since the spin quantum number can only be $- \frac{1}{2}$ or $+ \frac{1}{2}$, and since no electrons are already present to influence the spin, let's assume your electron has a positive spin, $+ \frac{1}{2}$.

So, the quantum numbers that describe chlorine' unpaired electron are:

$n = 3$
$l = 1$
${m}_{l} = + 1$
${m}_{s} = + \frac{1}{2}$