# How do you write the three quantum numbers that define an orbital?

## Can someone explain to me how to write the three specific quantum numbers that define and orbital, such as 1s? Thank you!!

Oct 7, 2017

Here's how you can do that.

#### Explanation:

For starters, you know that you have The three quantum numbers that you can use to describe a specific orbital are

1) the principal quantum number, $n$

The principal quantum number tells you the energy shell in which the orbital is located.

2) the angular momentum quantum number, $l$

The angular momentum quantum number tells you the energy subshell in which the orbital is located, i.e. the type of orbital that you're dealing with: $s$, $p$, $d$, and so on.

It's important to remember that the angular momentum quantum number can take the following values

• $l = 0 \to$ the $s$ subshell
• $l = 1 \to$ the $p$ subshell
• $l = 2 \to$ the $d$ subshell
• $l = 3 \to$ the $f$ subshell
$\vdots$

3) the magnetic quantum number, ${m}_{l}$

The magnetic quantum number tells you the orientation of the orbital, i.e. it distinguishes a specific orbital from the other orbitals present in the same subshell.

So, let's say that you want to write the three quantum numbers that can describe the $1 s$ orbital.

The coefficient added in front of the orbital tells you the energy shell in which the orbital is located, i.e. the value of the principal quantum number.

$\textcolor{b l u e}{1} s \to n = \textcolor{b l u e}{1}$

The type of the orbital tells you the energy subshell in which it's located. In this example, you're dealing with an $s$ orbital, so you can conclude that it must be located in the $s$ subshell.

$l = 0$

Finally, the $s$ subshell can hold a single orbital because ${m}_{l}$ can take a single value for $l = 0$.

$l = 0 \implies {m}_{l} = 0$

This means that you have

$n = 1 , l = 0 , {m}_{l} = 0$

For another example, let's say that you want to write the three quantum numbers that describe the $\textcolor{b l u e}{3} {p}_{z}$ orbital.

Right from the start, you know that you have

$n = \textcolor{b l u e}{3}$

Since you're dealing with a $p$ orbital, you can say that the orbital will be located in a $p$ subshell.

$l = 1$

Finally, the subscript added to the name of the orbital tells you its orientation.

For the $p$ subshell, you have three possible values for ${m}_{l}$

${m}_{l} = \left\{- 1 , 0 , + 1\right\}$

which means that you have three orbitals that occupy this subshell. By convention, the ${p}_{z}$ orbital is described by

${m}_{l} = 0$

This means that you have

$n = 3 , l = 1 , {m}_{l} = 0$

This orbital is located on the third energy level, in the $3 p$ subshell, i.e. in the $p$ subshell present on the third energy level, and is oriented along the $z$ axis.

By comparison, you can have

$n = 2 , l = 1 , {m}_{l} = 0$

This orbital is located on the second energy level, in the $2 p$ subshell, i.e. in the $p$ subshell present on the second energy level, and is oriented along the $z$ axis. 