What quantum numbers refer to a 4d orbital?

1 Answer
Jun 11, 2015

The four quantum numbers of interest are #n# (principal quantum number), #l# (angular momentum), #m_l# (magnetic), and #m_s# (spin).

A generic #4d_(z^2)# orbital has #n = 4# and #l = 2#. #n = 4# specifies the energy level, and #l# specifies the orbital's shape. #s -> l = 0, p -> l = 1#, etc. Thus, its #m_l# varies as #0, pm1, pm2#, and the orbital has projections above the plane and below the plane.

http://www.udbquim.frba.utn.edu.ar/

Depending on how full the orbital is, #m_s# varies. If it happens to be a #4d^1# configuration, for example, then one of five orbitals are filled (#d_(x^2-y^2), d_(z^2), d_(xy), d_(xz), d_(yz)#) with one electron. In that case, the electron is, by default, spin #pm1/2#. Thus, #m_s = pm1/2#.

In this case, it would give a term symbol of #""^(2)D_("1/2")#, #""^(2)D_("3/2")#, and #""^(2)D_("5/2")#. The notation is:

#""^(2S+1) L_("J")#

where #J = L+S#.

(The most stable one would be the #""^(2)D_("1/2")# state, according to Hund's rules for less-than-half-filled orbitals with the same #S# and the same #L#.)

Here, the spin multiplicity is #2S+1 = 2("1/2")+1 = 2#, and the total angular momentum #J = L + S = |m_l| + |m_s|#
# = 0 + "1/2", 1 + "1/2", and 2 + "1/2" = "1/2", "3/2", and "5/2"#.

(#2 - "1/2" = 1 + "1/2", and 1 - "1/2" = 0 + "1/2"#, which are duplicates, while by the selection rules, #DeltaL = 0, pm1, DeltaS = 0, # and #DeltaJ =0, pm1# )