We were given the following question in chemistry class (under the topic “Atomic Structure”):

Which orbital does the following wavefunction represent? $$φ(r) = \frac{1}{81(6π)^{1/2}}\left(\frac{Z}{a}\right)^{7/2}r^2\exp\left(\frac{-Zr}{3a}\right)(3\cos^2θ - 1),$$

where $Z$ is the atomic mass, $a$ is the Bohr radius, and $r$ is the distance from the origin.

The given answer is a $3\mathrm d_{z^2}$ orbital.

This question was to be solved without having needed to plot diagrams, and the method was to apparently equate the power of $r$ to the azimuthal quantum number value and then use it to find the principal quantum number value. I felt a bit lost on how to do it here, and couldn't find this method anywhere else.

It would be great if you could help me on how to approach this question (on how we could specify that the orbital was a $3\mathrm d$ orbital and more specifically a $3\mathrm d_{z^2}$) since I feel I lack the understanding of how to relate the wavefunction and the orbitals, and would also be great if you could perhaps link any useful resources that I could refer to, to better my understanding of these concepts relating to wavefunctions.


2 Answers 2


An easier way is to search for nodes, i.e. regions where the wavefunction is zero.

Clearly, the $r^2$ term means that $\psi = 0$ at $r = 0$ (i.e. the nucleus). Also, the $\exp(-kr)$ term means that $\psi \to 0$ as $r \to +\infty$. Neither of those are particularly useful on their own; however, the absence of any other values of $r$ which yield $\psi = 0$ means that there are zero radial nodes.

This already narrows it down a lot: it's going to essentially either be a 1s, 2p, or 3d orbital. (Yes, in theory it could be 4f, 5g, ... but when was the last time you saw an illustration of one of those, let alone their mathematical form?)

Now, the angular part is:

$$3\cos^2\theta - 1$$

This part is zero when $\theta = \arccos(1/\sqrt{3})$, i.e., $\theta = 54.7^\circ$ or $125.3^\circ$. Assuming you know your spherical coordinates, then you can infer from this that the nodes are not planes (as you might be used to thinking) but rather cones*, which is a very special feature of the $\mathrm{d}_{z^2}$ orbital. See my answer here for an illustration: https://chemistry.stackexchange.com/a/59145/16683.

All in all, this lets us conclude that it's the $3\mathrm{d}_{z^2}$ orbital.

A useful further exercise is to inspect the mathematical form of the $\mathbf{4}\mathrm{d}_{z^2}$ orbital. It will look something like this:

$$\psi \propto r^2(c-r)(3\cos^2\theta - 1)$$

The angular form is the same, which allows us to tell that this is a $\mathrm{d}_{z^2}$ orbital. But crucially there is an extra term of $c - r$ (for some $c$), which implies a radial node at $r = c$, and hence a larger principal quantum number.

* If there's a technical / mathematical term for this, please let me know.


The general form of the radial wave function for hydrogen like atom is, $$ R(r) = - \sqrt{ \Bigl(\frac{2Z}{na_o}\Bigr)^3 \frac{(n-l-1)!}{2n(n+l)!}}e^{-\frac{Zr}{na_o}} \Bigl(\frac{2Zr}{na_o}\Bigr)^l L$$ where, $L$ is generalised Laguerre polynomial of degree $(n-l-1)$.
Here are some steps followed by me to identify the orbital which corresponds to the given wave function :

  1. Let us first concentrate on the radial wave function. In this step we are interested to find the principal quantum number ($n$) of the orbital. Observe the exponential part of the radial wave function. It is given by $ e^{-\frac{Zr}{na_o}}$. So clearly the value of n is $3$.
  2. Observe the power of $r$. Radius (r) is always raised to the power of $l$ i.e., $r^l$. So the value of $l$ is $2$.
  3. Now observe the angular part of the wave function. If it doesn't contain any imaginary part then it is corresponds to the orbital with zero magnetic quantum number ($m$).
    Therefore, your orbital has quantum numbers $n=3$, $l=2$, and $m=0$ which corresponds to $3d_{z^2}$ orbital.

$1.$ Generally orbitals with $m=0$ corresponds to the orbitals oriented along $Z-$axis.
$2.$ If the angular part of wave function contains imaginary part then exponential part of it is given by $e^{im\phi}$ where $i= \sqrt{-1}$. So from exponential part you can easily find the magnetic quantum number also.
And finally you can also deduce the values of $n$ or $l$ from constant part or from the power of atomic number ($Z$) i.e., $Z^{l+\frac{3}{2}}$.
Hope this helps.

  • 6
    $\begingroup$ "If it doesn't contain any imaginary part then it is corresponds to the orbital with zero magnetic quantum number" This is true if you're talking about the "true" orbitals / spherical harmonics $Y^m_\ell$, but not necessarily true if you're talking about the real-valued orbitals such as $\mathrm{d}_{xy}$, which are linear combinations of those and don't have an imaginary part. It's, of course, not clear from the question which OP is expected to find, since the $\mathrm{d}_{z^2}$ orbital is a member of both sets. $\endgroup$ Mar 17, 2022 at 17:02
  • $\begingroup$ It's worth pointing out that in general, in chemistry, the real-valued orbitals are used, while physics tends to stick to the "true" harmonics and not worry so much about complex / imaginary wave functions. $\endgroup$ Mar 19, 2022 at 0:30

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