# Why is there a z² label for d orbitals, but no x² and y² labels with corresponding shapes?

I know that the $$\mathrm{d}_{z^2}$$ orbitals are named as a result from the conversion of the angular component of the hydrogen-like wave function in terms of Cartesian coordinates, but how come we lack analogous orbitals along other axes, such as $$\mathrm{d}_{x^2}$$ or $$\mathrm{d}_{y^2}$$? My naive assumption would be due to the fact that $$z$$ axis is considered the principle axis in regards to chemical bonds, but I don't feel that this is the best answer, nor can I seem to come across anything in textbooks or articles online.

• See: Is there a sixth type of d-orbitals with index 2z²-x²-y²? Actually, I'm not sure that it's a duplicate question, but my answer to it should still address your problem. – Zhe Jun 24 at 11:52
• Reopening because although the answer is relevant, this is not a duplicate question. Take it up with me on meta if you don't agree. – orthocresol Jun 27 at 13:14
• @orthocresol I have no objections. – Zhe Jun 27 at 13:27

TL;DR The $$\mathrm{d}_{z^2}$$ orbital is a result of solving the Schrödinger equation for the hydrogen atom in the most mathematically convenient way.

To properly understand this, it is necessary to go back to the fundamentals. The complex d-orbitals are obtained by solution of the Schrödinger equation. In general, these d-orbitals are made up of a radial part $$R(r)$$ and an angular part $$Y(\theta,\phi)$$. The radial part is not relevant to the question, because its value does not depend on the direction in which you look, so it cannot explain why there is a $$\mathrm{d}_{z^2}$$, but not a $$\mathrm{d}_{x^2}$$. So, it is instructive to look at the angular component (the spherical harmonics), which are tabulated below. The normalisation factors are ignored here:

\begin{align} \mathrm{d}_{+2} &\propto R(r) \cdot \sin^2\theta \cdot \mathrm{e}^{\mathrm 2i \phi} \\ \mathrm{d}_{+1} &\propto R(r) \cdot (-\sin\theta \cos\theta) \cdot \mathrm{e}^{\mathrm i \phi} \\ \mathrm{d}_0 &\propto R(r) \cdot (3\cos^2\theta - 1) \\ \mathrm{d}_{-1} &\propto R(r) \cdot \sin\theta \cos\theta \cdot \mathrm{e}^{-\mathrm i \phi} \\ \mathrm{d}_{-2} &\propto R(r) \cdot \sin^2\theta \cdot \mathrm{e}^{\mathrm -2i \phi} \end{align}

These are combined in an appropriate manner to generate the real d-orbitals. It turns out that the $$\mathrm{d}_{z^2}$$ orbital is simply the same as the original $$\mathrm{d}_0$$ orbital, which is already real (the radial component $$R(r)$$ is real). Since $$\cos\theta = z/r$$ and $$r^2 = x^2 + y^2 + z^2$$ (spherical coordinates), the angular component can be rewritten as

\begin{align} 3\cos^2\theta - 1 &= 3\left(\frac{z}{r}\right)^2 - 1 \\ &= \frac{3z^2 - r^2}{r^2} \\ &= \frac{2z^2 - x^2 - y^2}{r^2} \end{align}

and conventionally we denote this as $$\mathrm{d}_{z^2}$$. There is no analogous $$\mathrm{d}_{x^2}$$ orbital because the conventional way of deriving real d-orbitals does not yield any orbital with an angular form of $$(2x^2 - y^2 - z^2)/r^2$$.

But that still does not properly answer the question. To do so, we should look more closely at how the complex d-orbitals are obtained. The Schrödinger equation, $$\hat{H}\psi = E\psi$$, only prescribes that we need to find eigenstates of $$\hat{H}$$, but does not say how. It turns out that the easiest way to accomplish this is to look for solutions which are simultaneous eigenstates of $$\hat{H}$$ (energy), $$\hat{L}^2$$ (total orbital angular momentum), and $$\hat{L}_z$$ (projection of orbital angular momentum onto the z-axis). The eigenvalues of the solutions under these operators are directly tied to the quantum numbers $$n$$, $$l$$, and $$m_l$$, respectively.

Fundamentally, the $$\mathrm{d}_{0}$$ orbital (i.e. the $$\mathrm{d}_{z^2}$$ orbital) arises because we have chosen it to be an eigenstate of the operator $$\hat{L}_z$$. If we were to solve the Schrödinger equation in a different way, by searching for simultaneous eigenstates of $$\hat{H}$$, $$\hat{L}^2$$, and $$\hat{L}_x$$ (instead of $$\hat{L}_z$$), we would indeed get a $$\mathrm{d}_{x^2}$$ orbital. This is an equally valid solution of the Schrödinger orbital (as it is an eigenstate of $$\hat{H}$$), but it is simply not the conventional solution.

So, why choose $$\hat{L}_z$$ and not $$\hat{L}_x$$ when solving the Schrödinger equation? The answer is simply mathematical convenience: the Schrödinger equation is most easily solved in spherical coordinates, and in spherical coordinates, $$\hat{L}_z$$ has a particularly nice form of being equal to $$-\mathrm{i}\hbar(\partial/\partial \phi)$$. The eigenstates of this are none other than $$\mathrm{e}^{\mathrm{i}m\phi}$$, which is exactly what we observe in the complex d-orbitals at the beginning of this post ($$m$$ is, of course, quantised, so there are only five valid solutions with $$l = 2$$, i.e. d-orbitals).

But why does $$\hat{L}_z$$ have a nice form and not $$\hat{L}_x$$? Ultimately, this mathematical convenience is a result of the definition of spherical coordinates. Because $$\theta$$ is chosen to be the angle made by the vector with the z-axis, the z-axis has a special status in spherical coordinates:

\begin{align} z &= r\cos\theta \\ x &= r\sin\theta\cos\phi \\ y &= r\sin\theta\sin\phi \end{align}

So, at the end of the day, it is a completely arbitrary convention. And it should be, too, because no axis should have an inherently special status in a spherically symmetric system like the hydrogen atom. As mentioned previously, you could obtain a $$\mathrm{d}_{x^2}$$ orbital by solving the Schrödinger equation in a different way. You could even choose to redefine the axes in such a way which gave you a $$\mathrm{d}_{x^2}$$ orbital (it's quite easy; simply rename $$(x,y,z)$$ to $$(y,z,x)$$).

But there is no way you can have both at the same time. Because $$\hat{L}_x$$ and $$\hat{L}_z$$ do not commute with one another, it is not possible to "solve" the Schrödinger equation in a way which gives you both $$\mathrm{d}_{x^2}$$ and $$\mathrm{d}_{z^2}$$ orbitals at the same time.

It’s just the naming scheme for the hydrogen atom $$l=2$$ wavefunctions in real form. When you solve the Schrodinger equation you do it in spherical coordinates and the solution you get out involve complex numbers.

Thanks to the properties of the Schrodinger equation you can transform these 5 complex valued orbitals in spherical coordinates into 5 new real values orbitals, also in spherical coordinates. They are named because of their shapes. First, the $$d_{z^2}$$ is actually short for $$d_{2z^2-x^2-y^2}$$ and and represents a wavefunction which is proportional to $$\frac{3z^2-r^2}{r^2}$$. Similarly, $$d_{xy}$$ is proportional to $$\frac{xy}{r^2}$$.

We do this because our new real d orbitals are both real and orthogonal, which are not only properties we like but back when quantum mechanics was new these properties made the math a lot easier.

• Some citations needed for the last paragraph: (1) there's nothing inherently difficult about using complex d-orbitals. The real d-orbitals are useful for discussing chemical bonding because they have convenient directional properties, i.e. it's easier to describe & visualise their overlap with other orbitals. (2) What does "converting orbitals back into Cartesian form" even mean? And how do you end up with 6 orbitals? – orthocresol Jun 23 at 18:33
• No you don't end up with six orbitals. You do have six functions $x^2, y^2, z^2, xy, xz, yz$. But the first three actually are shorthand for (functions proportional to) $2x^2-y^2-z^2, 2y^2-z^2-x^2, 2z^2-x^2-y^2$ and only two of those are linearly independent, the sum of all three being zero. So you can make only five orbitals no matter how you slice and dice space into coordinate surfaces. – Oscar Lanzi Jun 23 at 19:37
• Maybe you want to elaborate and address the comments that have been brought up. Generally speaking if you have n functions that represent an orthonormal basis spanning a space, then any unitary transformation into another orthonormal set spanning that space will generate n functions. – Buck Thorn Jun 24 at 7:24
• @OscarLanzi I get it now. I didn't realize that only two of those will actually count. I have made the edit. – Sameer Thakur Jun 24 at 11:32
• It is also worth noting that doing this conversion to purely real orbitals is entirely for convenience in visualizing them, but this does not come without a cost. In fact, the transformation to all real orbitals makes each of the orbitals that had an imaginary component no longer an eigenfunction of $l_z$. – jheindel Jun 24 at 17:18