# Question #01fe2

Dec 10, 2016

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

#### Explanation:

I assume that you're talking about the quantum numbers.

So, if you're not familiar with the concepts behind quantum numbers, I'd recommend you take a look at this video first. If you're good, let's go...

So to quickly recap here's what each quantum number describes:

$n$ = energy level.
$l$ = subshell; ranges from 0 up to $n - 1$
${m}_{l}$ = orbital; ranges from $- l$ to $l$
${m}_{s}$ = tells the spin of the electron; is either $+ \frac{1}{2}$ or $- \frac{1}{2}$

Now, your question asks what are the possible values for $l$, ${m}_{l}$, and ${m}_{s}$ for $n = 4$. So, let's just work through this from top to bottom.

Firstly, if $n = 4$, let's consider what $l$ could be. We know that $l$ ranges between 0 to $n - 1$. Therefore, we can say that:

$l = 0 , 1 , 2 , 3$.

Side note: this essentially means that the n=4 energy level can hold electrons in the s, p, d and f subshells

Okay, so now that we know all possible values of $l$, let's look at ${m}_{l}$. We know that ${m}_{l}$ ranges between $- l$ and $l$. But which of the above $l$'s do we use? Well, consider this: if we just looked at the highest value of $l$ present, it would take care of all the lower ones. See what I mean below:

${m}_{l} = - 3 , - 2 , - 1 , 0 , 1 , 2 , 3$

See how values like $- 2$ and $1$ are taken care of?

So, that's really it. ${m}_{s}$ will only be either $+ \frac{1}{2}$ or $- \frac{1}{2}$ regardless of the other 3 quantum numbers.

Hope that helped :)