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

*Solution:*

TL;DRThe $\mathrm{d}_{z^2}$ 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 equation (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.