# Why is the dz2 orbital so different from the rest?

What makes dz2 orbital so special?

Although degenerate with other d orbitals, It has no nodal planes, instead it has 2 nodal "cones".

Instead of having 4 lobes, it has 2 lobes and 1 ring.

Also, 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 this particular orbital different? Is there any fundamental reason?

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

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. What's unique about the $$d_{z^2}$$ 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 Legendre polynomials $$Y_{lm}(\theta,\phi)$$, solutions of Legendre's differential equation. 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 $$d_{z^2}$$ 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 $$d_{z^2}$$ orbital.