# How many 3d_(z^2) orbitals have n = 3 and l = 2?

The $3 {d}_{{z}^{2}}$ is the specific orbital with:
• $\boldsymbol{n = 3}$, because the $3$ in front says that. Thus, it is in the 3rd quantum level.
• $\boldsymbol{l = 2}$, which corresponds to a $d$ orbital, since the values of $l$ correspond as $\left(0 , 1 , 2 , 3 , 4 , . . .\right) \leftrightarrow \left(s , p , d , f , g , . . .\right)$.
• ${m}_{l}$ for this orbital is one of the values in the following set: $\left\{- 2 , - 1 , 0 , + 1 , + 2\right\}$. It does not matter which one, but it is only one of them, since you have specified only one $d$ orbital. That is, ${m}_{l}$ is known, but is yours to choose, since all the $3 d$ orbitals in an atom are the same energy (degenerate).
• ${m}_{s}$ requires more information to specify, but it can either be $\pm \text{1/2}$ in general. It depends on how many electrons were in that orbital to begin with, whether it's spin-up or spin-down for the electron of choice.