Provide one set of acceptable quantum numbers for each of the valence electrons in sulfur?

1 Answer
Oct 14, 2017

I assume you mean sulfur atom. Well, sulfur atom has an electron configuration of:

#1s^2 2s^2 2p^6 3s^2 3p^4#

or in short,

#[Ne] 3s^2 3p^4#

or diagrammatically,

#underbrace(ul(uarr darr)" "ul(uarr color(white)(darr))" "ul(uarr color(white)(darr)))_(3p)#

#ul(uarr darr)#
#""" "^(3s)#

Sulfur can be seen to have six distinct electron quantum states:

  1. Spin-up in #3s#
  2. Spin-down in #3s#
  3. Spin-up in #3p_x#
  4. Spin-down in #3p_x#
  5. Spin-up in #3p_y#
  6. Spin-up in #3p_z#

The principal quantum number is #n = 3#, for #3s#, #3p#, and #3d# orbitals.

The angular momentum quantum number is #l = 0# for #3s# orbitals and #l = 1# for #3p# orbitals.

The magnetic quantum number is #m_l = {0}# for #s# orbitals and #m_l = {-1,0,+1}# for the set of #{p_x, p_z, p_y}# orbitals.

The spin quantum number is #m_s = pm1/2# for spin up/down electrons, respectively.

And so, we have the following quantum number sets...

  1. #(n,l,m_l,m_s) = (3,0,0,+1/2)#
  2. #(n,l,m_l,m_s) = (3,0,0,-1/2)#
  3. #(n,l,m_l,m_s) = (3,1,-1,+1/2)#
  4. #(n,l,m_l,m_s) = (3,1,-1,-1/2)#
  5. #(n,l,m_l,m_s) = (3,1,+1,+1/2)#
  6. #(n,l,m_l,m_s) = (3,1,0,+1/2)#