How many total orbitals in shell n=4? What is the relationship between the total number of shell and the quantum number n for that shell?

1 Answer
Truong-Son N.
Jan 26, 2018

#n^2# orbitals in each energy level, and #n# subshells in each energy level.

I assume you kind of recognize quantum numbers...

  • #n# is the principal quantum number, the energy level. #n = 1, 2, 3, . . . #
  • #l# is the angular momentum quantum number, corresponding to the shape of the orbitals of that kind. #l = 0, 1, 2, 3, . . . , n-1#. That is, #l_max = n-1#.
  • #m_l# is the magnetic quantum number, corresponding to each orbital of that shape. #m_l = {-l, -l+1, . . . , 0, . . . , l-1, l+1}#. That is, #|m_l| <= l#.
  • #m_s# is the spin quantum number for electrons. #m_s = pm1/2#.

For #n = 4#, the maximum #l# is therefore #4-1 = 3#. Of course, there is more than one value of #l# for one value of #n#.

That means:

#bbul(n = 4)#

#l = 0#:
#m_l = {0}#

#l = 1#:
#m_l = {-1, 0, +1}#

#l = 2#
#m_l = {-2, -1, 0, +1, +2}#

#l = 3 -= l_max#:
#m_l = {-3, -2, -1, 0, +1, +2, +3}#

and each #m_l# value corresponds to one orbital. We have #bbul4# subshells in this case; #s,p,d,f# #harr# #0,1,2,3# for the value of #l#.

We have an odd number of orbitals per subshell (#2l+1#), and so:

#overbrace(2(0) + 1)^(s) + overbrace(2(1) + 1)^(p) + overbrace(2(2) + 1)^(d) + overbrace(2(3) + 1)^(f)#

#= 1 + 3 + 5 + 7#

#= bbul16# orbitals in the #bb(n = ul4)# energy level.

  • If you repeat the process for #n = 3#, you would find #l_max = 2# and there are #bbul9# orbitals in #n = bbul3#.

#bbul(n = 3)#

#l = 0#:
#m_l = {0}#

#l = 1#:
#m_l = {-1, 0, +1}#

#l = 2 -= l_max#
#m_l = {-2, -1, 0, +1, +2}#

and each #m_l# value corresponds to one orbital. We have #bbul3# subshells in this case; #s,p,d# #harr# #0,1,2# for the value of #l#.

  • If you repeat the process for #n = 2#, you would find #l_max = 1# and there are #bbul4# orbitals in #n = bbul2#.

#bbul(n = 2)#

#l = 0#:
#m_l = {0}#

#l = 1 -= l_max#:
#m_l = {-1, 0, +1}#

and each #m_l# value corresponds to one orbital. We have #bbul2# subshells in this case; #s,p# #harr# #0,1# for the value of #l#.

  • If you repeat the process for #n = 1#, you would find #l_max = 0# and there is #bbul1# orbital in #n = bbul1#.

#bbul(n = 1)#

#l = 0 -= l_max#:
#m_l = {0}#

and each #m_l# value corresponds to one orbital. We have #bbul1# subshell in this case; #s# #harr# #0# for the value of #l#.

Thus, we have #bb(n^2)# orbitals in one energy level, and #bbn# subshells in one energy level.

Related questions
Impact of this question
58493 views around the world
You can reuse this answer
Creative Commons License