Question

What quantum numbers refer to a 4d orbital?

Answer 1

Answer
The four quantum numbers of interest are n (principal quantum number), l (angular momentum) $m_1$ (magnetic , and $m_s$ (spin )
A generic $4d_z^2$ orbital has $n=4$ and $l=2.n=4$ specifies the energy level ,and $l$ specifices the orbitl's shape . $s\to l=0,p\to l=1$ etc, .Thus , its $m_1$ varies as $0,\pm 1,\pm 2$ and the orbital has projection above the plane and below the plane .
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 ( $(d{x}^2-y^2,d_z^2{,}_{dx},d_{xz},d_{yz})$) 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 $02S+1=2\left(1/2\right)+1=2$ ,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 $5/5$
$(2--1/2=1+1+1/2$ , and $1-1/2=0+1/2$ , which are duplicates , while by the selection rules , $\Delta L=0,\pm 1,\Delta S=0,$ and $\Delta J=0,\pm 1$