# What is a set of four quantum numbers that could represent the last electron added (using the Aufbau principle) to the Cl atom?

##### 1 Answer
Jul 3, 2016

Here's what I got.

#### Explanation:

Your starting point here will be the electron configuration of a neutral chlorine atom.

Chlorine is located in period 3, group 17 of the periodic table and has an atomic number equal to $17$. This tells you that the electron configuration of a chlorine atom must account for a total of $17$ electrons that surround the nucleus of the atom.

The electron configuration of a chlorine atom looks like this

$\text{Cl: } 1 {s}^{2} 2 {s}^{2} 2 {p}^{6} 3 {s}^{2} 3 {p}^{5}$

Now, the last subshell to be filled with electrons, which is also the highest in energy, is the $3 p$ subshell.

As you can see from the electron configuration, this subshell contains a total of $5$ electrons. These electrons are distributed in $3$ orbitals labelled $3 {p}_{x}$, $3 {p}_{y}$, and $3 {p}_{z}$.

As you know, we can use a set of four quantum numbers to describe the location and spin of an electron in an atom

Let's start with the principal quantum number, $n$. Since this last electron is added to the third energy level, you will have

$n = 3 \to$ the third energy level

The angular momentum quantum number, $l$, describes the subshell in which the electron is located. In this case, the last electron is added to the $3 p$ subshell, so you will have

$l = 1 \to$ the p-subshell

Now, this is where things can get a little tricky. According to Hund's Rule, every orbital in a given subshell must be occupied with $1$ electron before a second electron is added to any of these orbitals.

You know that the $3 p$ subshell contains a total fo $5$ electrons. In this case, each of the three $3 p$ orbitals will first be occupied with a spin-up electron. This will account for $3$ of the $5$ electrons.

After this happens, the second-to-last electron will occupy the $3 {p}_{x}$ orbital, this time having spin-down.

Finally, the last electron to be added will be placed in the $3 {p}_{z}$ orbital, once again having spin-down. Here's a diagram showing the electron configuration of chlorine, with the last electron added highlighted

So, the magnetic quantum number, ${m}_{l}$, tells you the specific orbital in which the electron is located. By convention, you have

• ${m}_{l} = - 1 \to$ the $3 {p}_{x}$ orbital
• ${m}_{l} = \textcolor{w h i t e}{-} 0 \to$ the $3 {p}_{z}$ orbital
• ${m}_{l} = + 1 \to$ the $3 {p}_{y}$ orbital

In this case, you would have

${m}_{l} = 0 \to$ the $3 {p}_{z}$ orbital

Finally, the spin quantum number, ${m}_{s}$, tells you the spin of the electron. In this case, you have

${m}_{s} = - \frac{1}{2} \to$ a spin-down electron

Therefore, a possible quantum number set for the last electron added to a chlorine atom is

$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{n = 3 , l = 1 , {m}_{l} = 0 , {m}_{s} = - \frac{1}{2}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$