Respuesta :

sharmaenderp
An element has four quantum numbers associated with it. We have

1. principal quantum number ([tex]n[/tex])
2. azimuthal quantum number ([tex]l[/tex])
3. magnetic quantum number ([tex]m_{l}[/tex])
4. spin quantum number ([tex]m_{s}[/tex])

To look for each quantum  number of Chlorine (Cl), we mus first recall its electronic configuration: [tex]1s^{2}2s^{2}2p^{6}3s^{2}3p^{5}[/tex].

1. For [tex]n[/tex], we need to look for its highest energy level or by simply checking the period that the element belongs to. So for this case, it's [tex]n = 3[/tex].

2. For [tex]l[/tex], we need to look for the sublevel where the last electron was assigned. The sublevels, p, d, f are assigned to a value of 0, 1, 2, and 3, respectively. Since the last electron was assigned in p, then [tex] l = 1 [/tex].

3. And for [tex]m_{l}[/tex], the number of orbitals present in Cl is [tex]2l+1 = 3[/tex]. The possible magnetic quantum numbers for [tex]p[/tex] are -1, 0, +1. Since there 5 electrons we count starting at -1. So, we have [tex]m_{l}=+1[tex].

4.  Lastly, for [tex]m_{s}[/tex], we need to determine the spin. Since we have 5 electrons, all the arrows upward are filled in. That means the fifth electron has a spin of +1/2. 

Answers: [tex] n =3, l = 1, m_{l} = 0, m_{s} =+1/2 [/tex]


aksnkj

Answer:

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

[tex]n=3, l=1, m_{l} =0, m_{s} = \frac{-1}{2}[/tex]

Step-by-step 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:  [tex]1s^{2} 2s^{2}2p^{6} 3s^{2}3p^{5}[/tex]

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

As you can see from the electron configuration, this subshell contains a total of  5 electrons. These electrons are distributed in  3  orbitals labelled [tex]3p_{x}, 3p_{y}[/tex] and [tex]3p_{z}[/tex].

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  →  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 ⇒ 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  [tex]3p_{x}[/tex]   orbital, this time having spin-down.

Finally, the last electron to be added will be placed in the [tex]3p_{z}[/tex]   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

[tex]m_{l}[/tex]  =  − 1  ⇒  the  [tex]3p_{x}[/tex]  orbital

[tex]m_{l}[/tex] =  − 0  ⇒  the  [tex]3p_{z}[/tex]  orbital

[tex]m_{l}[/tex]  =  +  1  ⇒ the   [tex]3p_{y}[/tex]  orbital

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

[tex]n=3, l=1, m_{l} =0, m_{s} = \frac{-1}{2}[/tex]

For more information:

https://brainly.com/question/11181071?referrer=searchResults

Ver imagen aksnkj
Ver imagen aksnkj

Otras preguntas