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?