3

Why are the $d$-orbitals so named?

The naming of $p$-orbitals is fine. I can easily name any p orbital according to the axis along which it is oriented. But what about $d$- orbitals? Why are there subscripts such as $x^2-y^2$ and $z^2$ ? Are we supposed to read it like "x squared- y squared"?

Also why does the $d_{z^2}$ orbital has a ring like structure around it?

Is there any order in why the degenerate d orbitals are filled? Like for $p$-orbitals we have an order of filling: $p_x$ then $p_y$ then $p_z$.(Not too sure of this.My friend told me about this.)

My friend also told me that for d-orbitals, $d_z$ is conventionally taken as the principal orbital.(Again not too sure what he meant by that).

Although I wouldn't want to get into knowing about f orbitals,but information on that might help someone looking for answers on that.

I wonder why this hasn't been asked on CSE before.

NB: I have visited this link , but couldn't understand much.

Thank You.

Community
  • 1
Karan Singh
  • 3,795
  • 9
  • 37
  • 50
  • 2
    That answer should have told you everything you needed to know already, so strictly speaking this is a duplicate. The whole point is, the $p_x$, $p_y$ and $p_z$ orbitals are so named because the corresponding wavefunctions are of a form $xf$, $yf$ and $zf$ respectively where $f$ is a function of the the coordinates $r, \theta, \phi$. The same thing can be said of the $d$ orbitals. – orthocresol Oct 12 '15 at 14:44
  • Well there is much more to the question than that.And also, I didn't understand much from that as stated in the question.I really don't have much knowledge on wavefunctions as well. :( – Karan Singh Oct 12 '15 at 14:47
  • 2
    Exactly the same thing can be said of the $d$ orbitals. They are of the form $xyg$, $yzg$, ... where $g$ is also a function of $r, \theta, \phi$. Same goes for the $f$ orbitals. There is no correct order for filling. I could call the x-axis the y-axis, and I could call the y-axis the x-axis. That would make the $p_y$ orbital the $p_x$ orbital and vice versa. But obviously the molecule does not know which axis I have called which. Lastly, I have absolutely no idea what a principal orbital is, and I don't think that such a thing even exists. – orthocresol Oct 12 '15 at 14:47
  • The idea about wavefunctions is that each orbital is not a real object, it is a mathematical function obtained by solving the Schrodinger equation. The diagrams that you see are just graphs of these functions, just like how you can draw a graph for $f(x) = x^2$. It just so happens that the graph of $d_{z^2}$ has a ring. Of course before you are introduced to quantum mechanics, the way you view an orbital is like a container, but that is not really the case. Ultimately, there is no good way to explain the reason behind the naming of the orbitals without quantum mechanics. – orthocresol Oct 12 '15 at 14:50
  • Well that definitely helped.Thank you.You should go ahead and post this as an answer. – Karan Singh Oct 12 '15 at 14:55
  • 2
    Technically, not even $\ce{p}x$, $\ce{p}_y$ and $\ce{p}_z$ are accurate names, they should be called $\ce{p}{-1}$, $\ce{p}{+1}$ and $\ce{p}{0}$. The first two are, in fact, complex orbitals (i.e. have a contribution with $i$ in it) and only their linear combinations gives us $\ce{p}_x$ and $\ce{p}_y$. – Jan Oct 12 '15 at 15:27
  • Of course, and so are the d orbitals, they are $\mathrm{d}{-2}$, $\mathrm{d}{-1}$, $\mathrm{d}{0} (= \mathrm{d}{z^2})$, $\mathrm{d}{+1}$ and $\mathrm{d}{+2}$. On top of that, I went back to check and it turns out that $f = f(r)$ and $g = g(r)$, i.e. they are independent of $\theta$ and $\phi$. And the $\mathrm{d}_{z^2}$ orbital is of the form $(3z^2-r^2)g(r)$, but apparently we just say $z^2$ because it is easier to say. – orthocresol Oct 12 '15 at 15:50

0 Answers0