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

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Answer 1 · 2016-11-09T06:26:05

You already specified an orbital, so... one...

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The $3 d_{z^{2}}^{2}$ $3d_(z^2)$ is the specific orbital with:

$n = 3$ $bb(n = 3)$ , because the $3$ $3$ in front says that. Thus, it is in the 3rd quantum level.
$l = 2$ $bb(l = 2)$ , which corresponds to a $d$ $d$ orbital, since the values of $l$ $l$ correspond as $(0, 1, 2, 3, 4, . . .) \leftrightarrow (s, p, d, f, g, . . .)$ $(0,1,2,3,4, . . . ) harr (s,p,d,f,g, . . . )$ .
$m_{l}$ $m_l$ for this orbital is one of the values in the following set: ${- 2, - 1, 0, + 1, + 2}$ ${-2,-1,0,+1,+2}$ . It does not matter which one, but it is only one of them, since you have specified only one $d$ $d$ orbital. That is, $m_{l}$ $m_l$ is known, but is yours to choose, since all the $3 d$ $3d$ orbitals in an atom are the same energy (degenerate).
$m_{s}$ $m_s$ requires more information to specify, but it can either be $\pm 1/2$ $pm"1/2"$ 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.

How many 3d_(z^2)3dz23d_(z^2) orbitals have n = 3n=3n = 3 and l = 2l=2l = 2?