# What quantum numbers refer to a 4d orbital?

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 $4 {d}_{{z}^{2}}$ orbital has $n = 4$ and $l = 2$. $n = 4$ specifies the energy level, and $l$ specifies the orbital's shape. $s \to l = 0 , p \to l = 1$, etc. Thus, its ${m}_{l}$ varies as $0 , \pm 1 , \pm 2$, and the orbital has projections above the plane and below the plane. Depending on how full the orbital is, ${m}_{s}$ varies. If it happens to be a $4 {d}^{1}$ configuration, for example, then one of five orbitals are filled (${d}_{{x}^{2} - {y}^{2}} , {d}_{{z}^{2}} , {d}_{x y} , {d}_{x z} , {d}_{y z}$) with one electron. In that case, the electron is, by default, spin $\pm \frac{1}{2}$. Thus, ${m}_{s} = \pm \frac{1}{2}$.

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 $2 S + 1 = 2 \left(\text{1/2}\right) + 1 = 2$, and the total angular momentum $J = L + S = | {m}_{l} | + | {m}_{s} |$
$= 0 + \text{1/2", 1 + "1/2", and 2 + "1/2" = "1/2", "3/2", and "5/2}$.

($2 - \text{1/2" = 1 + "1/2", and 1 - "1/2" = 0 + "1/2}$, which are duplicates, while by the selection rules, $\Delta L = 0 , \pm 1 , \Delta S = 0 ,$ and $\Delta J = 0 , \pm 1$ )