Why is it that for the $\mathrm{d}$ subshell we have $\ \ \mathrm{d}_{xy}, \ \ \mathrm{d}_{yz}, \ \ \mathrm{d}_{xz}, \ \ \mathrm{d}_{x^{2}-y^{2}},$ and $\mathrm{d}_{z^{2}}$ orbitals only? Why aren't there $\mathrm{d}_{y^{2}-z^{2}}$ or $\mathrm{d}_{y^{2}}$ or other orbitals similar to that?

Perhaps it is because we have an empirical evidence that there are only 5 orbitals in $\mathrm{d}$-subshell, in that case, why are $\mathrm{d}_{x^{2}-y^{2}}$ and $\mathrm{d}_{z^{2}}$ orbitals named in this particular fashion only? (as the axes are arbitrary and there could have been a different choice of axes also)

  • 1
    $\begingroup$ Well, the choice is indeed somewhat arbitrary. $\endgroup$ – Ivan Neretin Apr 21 at 13:10
  • 1
    $\begingroup$ Just armchair speculating, it may be related to the fact that within cartesian<>polar coordinate transform, axis x,y,z are not equivalent. The angle $\phi$ relates to x and y projections, the angle $\theta$ relates to z projection.. $\endgroup$ – Poutnik Apr 21 at 13:27
  • 4
    $\begingroup$ Yes, completely arbitrary physically, but biased to simpler math. Because of the funny toroid around the dz^2, the resultant electron density if all are filled is still perfectly spherical. There is no actual bias in the density along one axis. In order to discuss orbitals, it's easiest if we all use the same orthonormal basis set. We could have chosen a different one. $\endgroup$ – Andrew Apr 21 at 13:45

These orbitals represent the angular part of the wavefunction. The solution obtained directly from solving the Schrödinger equation produces equations containing complex numbers so cannot be drawn on normal $xyz$ axes and are hard to visualise.

The angular part of the wavefunction is given by functions called Spherical Harmonics these usually are given the symbol $Y$ and require two quantum numbers $\ell$ and $m_z$ making $Y_\ell^{m_z}$.

What is done to make the orbitals real functions, as opposed to complex ones, is to take linear combinations of the spherical harmonics, thus choosing the $z$-axis then the $\mathrm{d}_{z^2} = Y_2^0 = N(3\cos^2(\theta)-1)$) where $N$ is the normalisation constant. In cartesian coordinates ($xyz$) rather than polar ones (which are in $r,\;\theta,\;\phi$), $\displaystyle Y^0_2= N\frac{3z^2-r^2}{r^2}$ so you can see where the $z^2$ comes from and $r$ is the radial distance.

The other orbitals are also linear combinations e.g. $\mathrm{d}_{xz}=(Y_2^{1}+Y_2^{-1})/\sqrt{2}$ and in cartesian coordinates $\displaystyle Y_2^1 \sim \frac{z(x+iy)}{r^2},\; Y_2^{-1}\sim \frac{z(x-iy)}{r^2}$ so you can see where the notation $zx$ comes from and by making a linear combination the sum is no longer a complex number ($i=\sqrt{-1}$).

For completeness $\mathrm{d}_{yz}=(Y_2^{1}-Y_2^{-1})/\sqrt{2},\; \mathrm{d}_{x^2-y^2}=(Y_2^{2}+Y_2^{-2})/\sqrt{2},\; \mathrm{d}_{xy}=(Y_2^{2}-Y_2^{-2})/\sqrt{2}$ and in cartesians $\displaystyle Y_2^{\pm2}\sim \frac{(x\pm iy)^2}{r^2}$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.