Circle all of the allowed sets of quantum numbers for an electron in a hydrogen atom?

n=6, l=3, ml =-2, ms=+1/2
n=2, l=1, ml =0, ms=-1/2
n=2, l=2, ml =1, ms=+1/2
n=3, l=0, ml =1/2, ms=-1/2
n=3, l=1, ml =0, ms=0
n=0, l=0, ml =0, ms=-1/2

1 Answer
May 11, 2016

Recall the quantum numbers:

The three main quantum numbers describe the energy level, shape, and projection of the orbitals onto the xyz axes. Bonus: there is a fourth which describes the spin of the electron(s) in the orbital.

  • #n# is the principle quantum number which describes the energy level. #n >= 1# and is in the set of integers. That is, #n = 1,2, . . . , N#, for some finite #N# (only one of those numbers at a time).
  • #l# is the orbital angular momentum quantum number which describes the shape of the orbital. #l >= 0# and is an integer. #l = 0, 1, 2, ..., n-1# for #s, p, d, ...# orbitals, respectively (only one of those numbers at a time).
  • #m# (more specifically, #m_l#) is the magnetic quantum number, which describes the projection of the orbital in the #0,pm1, pm2, . . . , pml# directions.

It takes on every possible value of #l# and #0# in integer steps (not just the value of #l#, but every integer from #-l# to #0# to #+l#).

That is, #m_l = {0,pm1,pm2, . . . , pml}#. Note that #m_l# tells you how many subshells there are for a given orbital type (#s,p,d, . . .#).

  • #m_s# (see why we need to say #m_l#?) is the electron spin quantum number, which describes the spin of the electron. It is only #pm1/2#, independent of the other quantum numbers.

(1)

With #n = 6# and #l = 3#, we are looking at a #6f# orbital, since #l = 3# corresponds to an #f# orbital.

For #l = 3#, we really have #m_l = {0,pm1,pm2,pm3}#, so #m_l = -2# is allowed. Lastly, #m_s = +1/2# is allowed.

This is really saying that there is a spin-up electron in the #6f# orbital with an orbital angular momentum of #-2# (the second orbital from the left, for a #-3,-2,-1,0,+1,+2,+3# order).

(2)

With #n = 2# and #l = 1#, we are looking at a #2p# orbital, since #l = 1# corresponds to an #p# orbital.

For #l = 1#, we really have #m_l = {0,pm1}#, so #m_l = 0# is allowed. Lastly, #m_s = -1/2# is allowed.

This is really saying that there is a spin-down electron in the #2p# orbital with an orbital angular momentum of #0# (the middle orbital for a #-1,0,+1# order).

(3)

With #n = 2# and #l = 2#, we are looking at a so-called "#2d#" orbital, which doesn't exist.

For #l = 0,1,2, . . . , n-1#, we walk through #0 harr s, 1 harr p, 2 harr d# and find that #l = 2 harr d#.

For #d# orbitals, we note that #n >= 3# for #d# orbitals, but #n = 2# is impossible if #l = 2#.

Thus, this is an impossible configuration.

(4)

With #n = 3# and #l = 0#, we are looking at a #3s# orbital.

For #l = 0#, #m_l = {0}#. So, #m_l = 1/2# is impossible.

Thus, this is an impossible configuration.

(5)

With #n = 3# and #l = 1#, we are looking at a #3p# orbital.

For #l = 1#, #m_l = {0,pm1}#. So, #m_l = 0# is allowed. However, only #m_s = pm1/2# is allowed. So, #m_s = 0# is forbidden.

Thus, this is an impossible configuration.

(6)

With #n = 0#, consider that #n = 1,2,3, . . . , N#, for some finite #N#. But of course, #0 < 1#. It's outside the range of possible values of #n#.

Thus, this is an impossible orbital.