What are the quantum numbers for the forty third electron in Au?

The following assumes you already understand what it means to be assigned the principal quantum number #n#, angular momentum quantum number #l#, magnetic quantum number #m_l#, and spin quantum number #m_s#.

The basic numerical properties are:

• #n = 1, 2, 3, . . . #, #" "n > 0#
• #l = 0, 1, 2, 3, . . . , n-1#, #" "0 <= l <= n-1#
• #m_l = {-l, -l+1, . . . , 0, . . . , l-1, l+1}#, #" "-l <= m_l <= l#
• #m_s = pm1/2#

Well, a krypton core has #36# electrons, whose electron configuration is

#[Kr] = 1s^2 2s^2 2p^6 3s^2 3p^6 3d^10 4s^2 4p^6#,

so the #43#rd electron is somewhere either in #n = 5# for the #s# orbitals, or #n = 4# for the #d# orbitals. We don't know yet.

For any atomic orbital subshell, there are #2l+1# orbitals of the same energy, the same as the number of #bb(m_l)# values for a given #l#.

• Recall that #s# orbitals have #l = 0#, so #2(0) + 1 = 1 xx 5s# orbital, and so we can fill it with two electrons. Any atomic orbital holds at most #2# electrons, and there is only one #5s# orbital.

• Thus, we are #overbrace(43 - 36)^("5s + some 4d electrons") - overbrace(2)^("5s electrons") = 5# electrons into the #4d# orbitals.

• Recall that #d# orbitals have #l = 2#, so #2(2) + 1 = 5 xx 4d# orbitals, and having #5# electrons in them half-fills these #4d# orbitals.

As a result, it is conventionally chosen spin-up and can be assigned any ONE of these five quantum number sets:

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

These correspond as follows. From left to right we have these values of #m_l#:

#ul(uarr color(white)(darr))" "ul(uarr color(white)(darr))" "ul(uarr color(white)(darr))" "ul(uarr color(white)(darr))" "ul(uarr color(white)(darr))#
#underbrace(-2" "" "-1" "" "0" "" "+1" "color(white)(.)+2)_(4d)#

You would pick exactly ONE (and only one) of these as the ONE electron that is the "43rd" electron. However, you will not be able to distinguish which one is which, so you have one try and five options.