# What is a valid set of quantum numbers for the outermost electron of an aluminum atom?

## I've scoured Google for a possible explanation, and I think this is along the right path...? n=3, l=1, ml=1, ms= +1/2 However, I'm not sure I completely understand the explanation... Could someone please explain and check my answer? Thanks!

Oct 31, 2016

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

#### Explanation:

You are very close :), just the ${m}_{l}$ or what I simply call $m$ is incorrect. The variable $m$ has allowed values from $- l \le m \le + l$ and it represents the shell that the last electron of Aluminum is located. If you were to draw the electron energy diagram for it, you find that it is in the first box, or shell, thus since $m$ can in this case be -1, 0, 1, it is -1.

Hope this helps,