$\mathrm d_{z^2}$ orbital appears to be special:

  • Although degenerate with other d orbitals, it has no nodal planes, instead it has two nodal "cones".
  • Instead of having four lobes, it has two lobes and one ring.
  • Its electron density is prominently distributed in all $x, y$ and $z$ directions unlike others.

I know the wave function is what determines the shape, but what makes $\mathrm d_{z^2}$ orbital different? Is there any fundamental reason?

  • 2
    $\begingroup$ Well, $d_{x^2-y^2}$ is sort of special too... $\endgroup$
    – Buck Thorn
    May 19, 2019 at 17:06
  • 5
    $\begingroup$ It is no more 'special' than any of the other solutions to the Schroedinger equation. $\endgroup$
    – porphyrin
    May 19, 2019 at 18:02
  • 2
    $\begingroup$ Note the degeneracy is true in absence of magnetic fields. $\endgroup$
    – Buck Thorn
    May 20, 2019 at 18:33
  • $\begingroup$ @NightWriter and electric fields too, right? $\endgroup$
    – user226375
    May 21, 2019 at 4:00
  • 1
    $\begingroup$ My understanding is that E-field interactions occur only with the right symmetry (to first order), see eg en.wikipedia.org/wiki/Stark_effect $\endgroup$
    – Buck Thorn
    May 21, 2019 at 7:12

3 Answers 3


The Wikipedia is helpful in explaining why radial variations should arise in the density of non-s orbitals:

The non radial-symmetry properties of non-s orbitals are necessary to localize a particle with angular momentum and a wave nature in an orbital where it must tend to stay away from the central attraction force, since any particle localized at the point of central attraction could have no angular momentum.

enter image description here

What's unique about the $\mathrm d_{z^2}$ orbital (see the table above, from the Wikipedia) compared to the other $l=2$ angular momentum wavefunctions is that the $z$-component is zero $(m=0).$ This further constrains the geometry of the wavefunction.

The functions describing the angular dependence of the hydrogenic wavefunctions are Spherical Harmonics $Y_{lm}(\theta,\phi)$. In the case of d-orbitals, they satisfy

$$ \hat{L}^2Y_{lm}(\theta,\phi)=\hbar^2l(l+1)Y_{lm}(\theta,\phi)$$

with $l=2$, where $ \hat{L}$ is the angular momentum operator. Since the $z$-component of the angular momentum is also quantized, the following eigenequation also holds:

$$\hat{L}_zY_{lm}(\theta,\phi)=\hbar mY_{lm}(\theta,\phi)$$

with $m=0$ in the case of the $\mathrm d_{z^2}$ orbital, and this last equation leads to the following condition:

$$\frac{\partial\psi}{\partial y^2} = \frac{\partial\psi}{\partial x^2}$$

which implies that the solutions must be cylindrically symmetric about $z.$ However, the condition $l\neq 0$ implies that the solution is not spherically symmetric. The result is the unexpected shape of the $\mathrm d_{z^2}$ orbital.


There is actually a very simple reason the $\mathrm d_{z^2}$ orbital looks so different compared to the other d orbitals. You can imagine it as being two d orbitals in one: the $x^2-z^2$ and the $y^2-z^2$, so to speak. These two "orbitals" would look just like all the other d orbitals and would lie on the x and z axis, and the y and z axis, respectively. The same way that the $x^2-y^2$ orbital lies on the x and y axis. But since there can only be 5 d orbitals, these two orbitals are combined into one. This is why the lobes on the shared z axis are preserved, but the lobes from the x and y axes combine into a ring.


It's a matter of counting, really, if you are aware of what you are counting.

We may regard the angular dependence of the four-lobe functions at any radial distance $r$ as follows:


where the nodes between the lobes are described by either factor being zero and $x,y,z$ vary according to $x^2+y^2+z^2=r^2$ for our chosen $r$. Different radial distances would have different valyes for the coefficients $a_1$, etc.

It looks like there are six independent parameters for the coefficients at any radial distance. But:

  • Changing them all by the same constant factor gives back the same orbital; this cannot change the nodal planes.

  • The nodal planes must he perpendicular to each other to match the solutions to the Schrödinger equation.

So with these limirltations there are only four independent combinations that satisfy tgehe "normal" four-lobed appearance of $d$ orbitals. To have the actual five $d$ orbitals one must be structured differently no matter what basis set you try.


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