Question

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

Answer 1

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

`"Cl: " 1s^2 2s^2 2p^6 3s^2 3p^5`

Now, the last subshell to be filled with electrons, which is also the highest in energy, is the `3p` 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 `3p_x`, `3p_y`, and `3p_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

![figures.boundless.com]()

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 ->` 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 `3p` subshell, so you will have

`l=1 ->` 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 `3p` subshell contains a total fo `5` electrons. In this case, each of the three `3p` 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 `3p_x` orbital, this time having spin-down.

Finally, the last electron to be added will be placed in the `3p_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 ->` the `3p_x` orbital
• `m_l = color(white)(-)0 ->` the `3p_z` orbital
• `m_l = +1 ->` the `3p_y` orbital

In this case, you would have

`m_l = 0 ->` the `3p_z` orbital

Finally, the spin quantum number, `m_s`, tells you the spin of the electron. In this case, you have

`m_s = -1/2 ->` a spin-down electron

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

`color(green)(|bar(ul(color(white)(a/a)color(black)(n=3, l=1, m_l = 0, m_s = -1/2)color(white)(a/a)|)))`