How do you find the total number of electrons allowed in the #n = 3# energy level?

Well, it would be based on the four atomic quantum numbers.

[There is an equation, but I would rather not promote simple memorization unless the reasons are truly something beyond the scope of the course.]

• #n = 1, 2, 3, . . . # is the principal quantum number, describing the size of each orbital and their energy levels.
• #l = 0, 1, 2, . . . , n-1# is the angular momentum quantum number, describing the shape of each orbital and their energy sublevels. These values correspond to the #s, p, d, f, . . . # orbitals.
• #m_l = {-l, -l+1, . . . , l-1, l}# is the magnetic quantum number, corresponding to each unique orbital in the sublevel defined by #l#. Its range is limited by the value of #l#, and there are #2l+1# such values of #m_l# for a given #l#.
• #m_s = pm1/2# is the spin quantum number for an electron.

There also exists the Pauli Exclusion Principle.

1. To know which energy level you are in, you need #n#.
2. To know which sublevel of that energy level you are in, you need #l#.
3. To know which orbital in that sublevel within that energy level you are in, you need #m_l#.
4. To know which electron you are in that orbital in that sublevel within that energy level you are in, you need #m_s#.

That Principle states, no two electrons can be identical AND in the same orbital. Thus, two electrons in the same orbital must differ in their #m_s#.

• The maximum #l# is #n-1#.
• For a given #l#, there exist #2l+1# orbitals for that value of #l#.
• There are only two #m_s# values possible, so only two electrons can be in one orbital.

Therefore, if you are given #n = 3#, then the maximum #l# is #2#. So, you have six sets of quantum numbers:

• #n = 3#, #l = 0#, #m_l = {0}#, #m_s = pm1/2#

#-># #3s# electrons

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

#-># #3p# electrons

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

#-># #3d# electrons

Each set of quantum numbers corresponds to an allowed electron in a particular orbital in a particular sublevel in a particular energy level.

Therefore, there are

#overbrace([2(0) + 1])^("s orbitals") + overbrace([2(1) + 1])^"p orbitals" + overbrace([2(2) + 1])^"d orbitals" = 9# atomic orbitals

in #n = 3# and

#"2 electrons"/cancel"orbital" xx 9 cancel"orbitals" = bb18# electrons

that can be in such orbitals on #n = 3#. After the fact, if you notice, #3^2 = 9#, so there are #n^2# orbitals for a given #n#, and #2n^2# electrons that can go into those orbitals in total.