It can be very confusing to think of it this way.
Potassium has access to its 1s, 2s, 2p, 3s, 3p, and 4s orbitals. Its electron configuration is color(green)(1s^2 2s^2 2p^6 3s^2 3p^6 4s^1).
Hence, your "KLMN" configuration would be:
color(white)([(color(black)(ul("e"^(-)"'s/subshell"))), (color(black)(2)), (color(black)(2,6)), (color(black)(2,6)), (color(black)(1))] = [(color(black)(ul("Total e"^(-)"'s"))), (color(black)(2)),(color(black)(8)),(color(black)(8)), (color(black)(1))] = [(color(black)(ul(l))), (color(black)(0)),(color(black)(0,1)),(color(black)(0,1)), (color(black)(0))] = [(color(black)(ul(n))), (color(black)(1)),(color(black)(2)),(color(black)(3)), (color(black)(4))]).
i.e. You should have written 2,8,8,1.
Thus, in your words, the 19th electron goes into the "4th shell". It does have access to n = {1,2,3,4}, being on the 4th period of the periodic table.
The "KLMN" configuration numbers arise from the number of m_l values possible for all values of l possible at a given n (recall these are quantum numbers), combined with the maximum of two electrons of opposite spin (m_s = pm1/2) in the same orbital.
Recall that l = 0, 1, 2, . . . , n-1.
- For n = 1, we have l = 0 available.
- For n = 2, we have l = 0 or 1.
- For n = 3, we have l = 0, 1, or 2.
- For n = 4, we have l = 0, 1, 2, or 3.
For each of these cases, we thus have...
1S ORBITAL
For n = 1 and l = 0, we have m_l = {0} and m_s = pm1/2. Thus, for the 1s orbital, 2 electrons can fill the set of subshells.
2S AND 2P ORBITALS
For n = 2 and l = 0,1, we have m_l = {0} and m_l = {0, pm1}, and m_s = pm1/2. Thus, for the 2s and 2p orbitals combined, 2 + 6 = 8 electrons can fill the set of subshells. (Hence, we have the octet rule.)
3S, 3P, AND 3D ORBITALS
For n = 3 and l = 0,1,2, we have m_l = {0}, m_l = {0, pm1}, and m_l = {0, pm1, pm2}, as well as m_s = pm1/2. Thus, for the 3s, 3p, and 3d orbitals combined, 2 + 6 + 10 = 18 electrons can fill the set of subshells.
4S, 4P, 4D, AND 4F ORBITALS
For n = 4 and l = 0,1,2,3, we have m_l = {0}, m_l = {0, pm1}, m_l = {0, pm1, pm2}, and m_l = {0, pm1, pm2, pm3}, as well as m_s = pm1/2. Thus, for the 4s, 4p, 4d, and 4f orbitals combined, 2 + 6 + 10 + 14 = 32 electrons can fill the set of subshells.
