How many sets of quantum numbers specify all the #2p# orbitals?

1 Answer
Jul 30, 2017

Well, there are three sets... The #2p# orbitals are triply-degenerate, i.e. #2l + 1 = 3#, and as such, three sets of quantum numbers specify such orbitals.

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

Do you spot three here?


Recall:

  • The principal quantum number #n# specifies what energy level one is in: #n = 1, 2, 3, . . . #
  • The angular momentum quantum number #l# describes the shape of the orbital.

#l = 0, 1, 2, 3, 4, 5, 6, 7, . . . , n-1 harr s, p, d, f, g, h, i, k, . . . #.

If #l = 0#, the orbital is sharp (spherical), an #s# orbital.
If #l = 1#, the orbital is principal (polarized), an #p# orbital.
If #l = 2#, the orbital is diffuse, an #d# orbital.
If #l = 3#, the orbital is "fundamental", an #f# orbital.

  • The magnetic quantum number #m_l# corresponds to each orbital "orientation", and describes a separate orbital orthogonal to the rest. #m_l = {-l, -l+1, . . . , 0, . . . , l-1, l}#.

These are all three quantum numbers required to describe an orbital. If you want to describe orbital occupations, you'll need to specify the electron spin, #m_s#.