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?
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?