Question #f473f

1 Answer
Apr 5, 2016

Answer:

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.

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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 #3p_z# and #3p_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 = color(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 -># the s-subshell
  • #l=1 -># the p-subshell
  • #l=2 -># 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 -># the #p_x# orbital
  • #m_l = color(white)(-)0 -># the #p_y# orbital
  • #m_l = color(white)(-)1 -># the #p_z# orbital

Since we';re looking for the magnetic quantum number of an electron located in the #3p_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 = =1/2" "# or #" "m_s = -1/2#

Since the electron occupies the orbital by itself, you can attribute to it a spin magnetic number of #m_s = +1/2#.

This means that the quantum number set that describes the position and spin of the electron located in chlorine's #3p_z# orbital will be

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