How many electrons have #n = 3#, #m_l = pm1#, and #m_s = +1/2#?

1 Answer
Oct 31, 2016

For #n = 3#, we are in the 3rd energy level.

  • In the #n = 3# level, we have the values #color(green)(l = 0, 1)#, and #color(green)(2)# available, as #bb(l_max = n - 1)# and #n - 1 = 2#.

Therefore, we have the #bb(3s)#, #bb(3p)#, and #bb(3d)# orbitals available (#l = 0 harr s#, #l = 1 harr p#, #l = 2 harr d#).

  • For #m_l = pm1#, we have exactly two specified orbitals (each orbital corresponds to one #m_l#) for a given #bbl#. Now we examine each #l# and see what restriction #m_l = pm1# places on our set of orbitals.

For #color(green)(l = 0)#, #m_l = pm1# is not valid because it falls outside the range of #l# (we see that #pm1# is not in the "range" of #0#). So, disregard this for the #3s# orbitals, because specifying #m_l = pm1# has disallowed the #3s# orbital, whose #m_l = bb(0)#.

For #color(green)(l = 1)#, #m_l = pm1# is valid, since we would have allowed #m_l = {-1,0,+1}#. So we include two of the #3p# orbitals.

For #color(green)(l = 2)#, #m_l = pm1# is also valid, since we would have allowed #m_l = {-2,-1,0,+1,+2}#. So we include two of the #3d# orbitals.

  • For #m_s = +"1/2"#, it just says that we specify a single electron with spin "up" of magnitude #"1/2"#. (Ordinarily #m_s# could have been #pm"1/2"#.)

As a result, our set of orbitals and electron(s) includes:

  • Two #3p# orbitals (one corresponds to #m_l = -1#, the other #m_l = +1#)
  • Two #3d# orbitals (one corresponds to #m_l = -1#, the other #m_l = +1#)
  • One spin-up electron in each (#m_s = +"1/2"#).

Therefore, we have an (isolated, theoretical) configuration of #3p^2 3d^2#, or four electrons.