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This question first popped up in my head when I learnt that $d_{z^2}$ orbital is used in $sp^3d$ hybridization and $d_{x^2-y^2}$ and $d_{z^2}$ orbitals are used in $sp^3d^2$ hybridization. If all the $d$ orbitals are said to be degenerated orbitals (i.e. they are said to have equal energy) then why these two orbitals are used in $sp^3d$ and $sp^3d^2$ hybridization?

Some said that $d_{z^2}$ is used because it is closer to the nucleus because of the ring of electron density surrounding the nucleus, but if that is the reason then why do we call the $d$ orbitals as degenerated in the first place when clearly two orbitals are closer to nucleus.

Here's what I want to know:

A) Why $d$ orbitals have same energy when its clearly apparent from their shape that two orbitals have lower energy than other three?

B) Why $d_{z^2}$ orbital is used in $sp^3d$ hybridization and $d_{x^2-y^2}$ and $d_{z^2}$ orbitals are used in $sp^3d^2$ hybridization?

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    $\begingroup$ (1) in the atom with spherical symmetry, all five d-orbitals are degenerate, there is no dispute about this, and the argument of "clearly apparent from their shape" doesn't hold up, as our eyes aren't a good measurement tool for energy of an orbital. (2) in octahedral geometry the degeneracy is partially broken. The $\{x^2-y^2, z^2\}$ set is actually higher in energy than $\{xy, yz, xz\}$, so maybe that's something else that should have been added to your previous question on why hybridisation isn't looked upon very favourably anymore. $\endgroup$ – orthocresol♦ Jul 17 '17 at 8:46
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    $\begingroup$ and the difference in energy isn't because it's "closer to the nucleus" or anything like that - it's better explained by MOT or at the very least, CFT. Anyway, for your other qn, the reason why those orbitals are chosen is because of symmetry properties - for example if you try to make an octahedral molecule using $\{xz,yz\}$ then you will end up with asymmetric hybrid orbitals (roughly speaking, they point more along the z-axis than along the x-/y-axes). Again, MOT explains it much better; the use of the $\{x^2-y^2,z^2\}$ set falls out naturally from the fact that they transform as $e_g$. $\endgroup$ – orthocresol♦ Jul 17 '17 at 8:48
  • $\begingroup$ @Orthocresol"Again, MOT explains it much better; the use of the ${x^2−y^2,z^2}$ set falls out naturally from the fact that they transform as $e_g$" I don't understand this. Can you provide me some resources about this phenomenon? $\endgroup$ – Mockingbird Jul 17 '17 at 10:14
  • $\begingroup$ @Mockingbird It means that as a ligand approaches the central atom(in case of octahedral complex) along the axes, energy of these two orbitals increase more than the remaining three orbitals; these two orbitals are now called $e_g$ orbital. orthocresol means to say that the use of these two orbitals can simply be explained by this splitting of d orbitals as the ligand approaches the central atom. $\endgroup$ – Arishta Jul 17 '17 at 10:48
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    $\begingroup$ @Mockingbird This phenomenon is at the heart of crystal field theory. You might want to read up on that. $\endgroup$ – Arishta Jul 17 '17 at 11:01
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A. The orbital shapes originate from solutions to the Schrodinger equation for hydrogen like atoms. The derivation of the orbitals come from the section for the solution to Schrodinger equation for Hydrogen-like atoms on Wikipedia: https://en.m.wikipedia.org/wiki/Hydrogen-like_atom

The solutions have the same energy because you solve the equation for specific energies. They really are the same, which can proven by doing an integral across the volume.

The different hybrid orbitals depend on what quantum numbers are possible. This question has been answered here: Why an asymmetric geometry with sp3d and sp3d3 hybridization?

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