# Question #9b3cf

##### 1 Answer
Sep 14, 2016

Here's what I got.

#### Explanation:

The first thing to do here is to write the electron configuration of a neutral aluminium atom.

Aluminium is located in period 3, group 13 of the Periodic Table of Elements and has a total of $13$ electrons surrounding its nucleus, as given by its atomic number.

The electron configuration of a neutral aluminium atom looks like this

$\text{Al: } 1 {s}^{2} 2 {s}^{2} 2 {p}^{6} 3 {s}^{2} \textcolor{b l u e}{3} {p}^{1}$

As you can see, the last electron present in an aluminium atom is located in a $3 p$ orbital.

Now, we need to find the values of the four quantum numbers used to describe the position and spin of an electron inside an atom. In your case, the principal quantum number, $n$, which gives the energy level on which the electron is located, is equal to $\textcolor{b l u e}{3}$.

The angular momentum quantum number, $l$, which gives you the subshell in which the electron is located, is equal to $1$, since

• $l = 0 \to$ designates the s subshell
• $l = 1 \to$ designates the p subshell
• $l = 2 \to$ designates the d subshell

and so on. The magnetic quantum number, ${m}_{l}$, which gives you the orbital which holds the electron, can take one of three possible values here

• ${m}_{l} = - 1 \to$ the $3 {p}_{x}$ orbital
• ${m}_{l} = \textcolor{w h i t e}{-} 0 \to$ the $3 {p}_{z}$ orbital
• ${m}_{l} = \textcolor{w h i t e}{-} 1 \to$ the $3 {p}_{y}$ orbital

Because in the case of an aluminium atom the p subshell contains a single electron, you can pretty much pick any of these three values for ${m}_{l}$.

Let's say that we have ${m}_{l} = 0$ for an electron located in the $3 {p}_{z}$ orbital.

Finally, the spin quantum number, ${m}_{s}$, which tells you the spin of the electron, can be $+ \frac{1}{2}$ for an electron that has spin-up and $- \frac{1}{2}$ for an electron that has spin-down.

Since your electron is alone in the ${p}_{z}$ orbital, you can pick either value for ${m}_{s}$. Let's say that we have ${m}_{s} = + \frac{1}{2}$.

Therefore, you can say that a valid set of quantum numbers that describe the last electron added to an aluminium atom could be

$n = 3 , l = 1 , {m}_{l} = 0 , {m}_{s} = + \frac{1}{2}$

This describes an electron located on the third energy level, in the 3p-subshell, in the $3 {p}_{z}$ orbital, that has spin-up

You could also have, for example

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

This describes an electron located on the third energy level, in the 3p-subshell, in the $3 {p}_{x}$ orbital, that has spin-up

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

This describes an electron located on the third energy level, in the 3p-subshell, in the $3 {p}_{y}$ orbital, that has spin-up

Here is a video with more explanation of quantum numbers.