# What are the quantum numbers for atoms?

##### 1 Answer
Nov 29, 2016

There is actually a lot of memorization involved here. I would recommend you try flash cards, and write down these relations:

• The total number of orbitals in a subshell is equal to $2 l + 1$.
• The total number of orbitals in one quantum level is ${n}^{2}$.
• The maximum number of electrons per orbital is $2$.
• The total number of radial nodes (spherical nodal shells) is $n - l - 1$.
• The angular momentum (azimuthal) quantum number $l$ is the total number of angular nodes (nodal planes). Another way to say it is that the number of nodal planes is $l$.

Along with those overarching rules that arise from the definitions of the quantum numbers, you should know what the quantum numbers directly tell you.

• The principal quantum number $n$ tells you what quantum level you are on.

$n = 1 , 2 , 3 , . . .$

So, at $n = 4$, you are looking at an element from the fourth row on the periodic table.

• The angular momentum (azimuthal) quantum number $l$ tells you what the shape of the orbital is.

$l = 0 , 1 , 2 , . . . , n - 1$, and each $l$ corresponds to an orbital shape.

$\left(0 , 1 , 2 , 3 , 4 , . . .\right) \leftrightarrow \left(s , p , d , f , g , . . .\right)$.

Thus, $l$ CANNOT be equal to $n$. That is why the $1 p$, $2 d$, $3 f$, $4 g$, [...] orbitals do not exist.

• The magnetic quantum number ${m}_{l}$ represents each actual orbital in the subshell. So, ${m}_{l} = \left\{- l , - l + 1 , . . . , 0 , . . . , l - 1 , l\right\}$.

Thus, $| {m}_{l} | \le l$, and ${m}_{l}$ cannot be greater in magnitude than $l$. For example, a $p$ subshell, with $l = 1$, would have three orbitals. That is because their ${m}_{l} = \left\{- 1 , 0 , + 1\right\}$, so there are three orbitals, corresponding to three total ${m}_{l}$ values.

• The spin quantum number ${m}_{s}$ is simple; it is only $\pm \text{1/2}$ for electrons, no exceptions.

PRACTICE PROBLEM

What is the set of quantum numbers corresponding to a single, spin-up electron in a $4 {p}_{x}$ orbital? Assume the ${p}_{z}$ has ${m}_{l} = 0$ and that the ${p}_{y}$ has ${m}_{l} = - 1$.

ANSWER:

• $\textcolor{w h i t e}{n = 4}$$\textcolor{w h i t e}{\text{, because the}}$ $\textcolor{w h i t e}{4}$ $\textcolor{w h i t e}{\text{in front tells you what}}$ $\textcolor{w h i t e}{n}$ $\textcolor{w h i t e}{\text{is.}}$
• $\textcolor{w h i t e}{\text{l = 1}}$$\textcolor{w h i t e}{\text{, because}}$ $\textcolor{w h i t e}{l = 1}$ $\textcolor{w h i t e}{\text{corresponds to a}}$ $\textcolor{w h i t e}{p}$ $\textcolor{w h i t e}{\text{subshell.}}$
• $\textcolor{w h i t e}{{m}_{l} = + 1}$$\textcolor{w h i t e}{\text{, because we've accounted for}}$ $\textcolor{w h i t e}{{m}_{l} = 0 , - 1}$ $\textcolor{w h i t e}{\text{already, and}}$ $\textcolor{w h i t e}{{m}_{l} = \left\{- 1 , 0 , + 1\right\}}$ $\textcolor{w h i t e}{\text{since}}$ $\textcolor{w h i t e}{l = 1}$ color(white)("tells you the range of" $\textcolor{w h i t e}{{m}_{l}}$ $\textcolor{w h i t e}{\text{values allowed.}}$
• $\textcolor{w h i t e}{{m}_{s} = + \text{1/2}}$$\textcolor{w h i t e}{\text{, because a spin-up electron has a}}$
$\textcolor{w h i t e}{\text{positive spin quantum number value.}}$

(highlight to see)