# Why energy of all d orbitals is same?

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?

• (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. – orthocresol Jul 17 '17 at 8:46
• 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 $e_g$. – orthocresol Jul 17 '17 at 8:48
• @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? – Mockingbird Jul 17 '17 at 10:14
• @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. – Arishta Jul 17 '17 at 10:48
• @Mockingbird This phenomenon is at the heart of crystal field theory. You might want to read up on that. – Arishta Jul 17 '17 at 11:01