Basics of wave-mechanical model of atom

Question

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)

Answer 1

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 $\displaystyle Y_2^{\pm2}\sim \frac{(x\pm iy)^2}{r^2}$.