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.
