# Question f473f

Apr 5, 2016

Here's what I got.

#### Explanation:

As you know, four quantum numbers are used to describe the position and spin of an electron in an atom.

A good starting point here will be chlorine's electron configuration, which looks like this - I'll use the noble gas shorthand notation

"Cl: " ["Ne"] 3s^2 color(red)(3)p^5#

As you can see, chlorine's outermost electrons are located in its 3p-subshell. Four of these five electrons are located in the $3 {p}_{z}$ and $3 {p}_{y}$ orbitals, and one electron is located in the ${p}_{z}$ orbital.

Let's try to write a quantum number set for the electron located in the ${p}_{z}$ orbital.

The principal quantum number, $n$, tells you the energy level on which the electron is located. In chlorine's case, this electron is located on the third energy level, $n = \textcolor{red}{3}$.

The angular momentum quantum number, which gives you the subshell in which the electron is located, takes values that range from $0$ to $n - 1$. More specifically, you have

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

In your case, the electron is located in the p-subshell, you know that you have $l = 1$.

The magnetic quantum number, ${m}_{l}$, gives you the exact orbital in which you can find the electron. For a p-subshell, the magnetic quantum number can be

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

Since we';re looking for the magnetic quantum number of an electron located in the $3 {p}_{z}$ orbital, you can say that you have ${m}_{l} = 1$.

Finally, the spin quantum number, which gives you the spin of the electron, can only have two possible values

${m}_{s} = = \frac{1}{2} \text{ }$ or $\text{ } {m}_{s} = - \frac{1}{2}$

Since the electron occupies the orbital by itself, you can attribute to it a spin magnetic number of ${m}_{s} = + \frac{1}{2}$.

This means that the quantum number set that describes the position and spin of the electron located in chlorine's $3 {p}_{z}$ orbital will be

$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{n = 3 , l = 1 , {m}_{l} = 1 , {m}_{s} = + \frac{1}{2}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$