My textbook says that $\ce{[CrF6]^{3-}}$ is an example of $sp^3d^2$ hybridization. But I don't really see how. Chromium has an electronic configuration of:
$$3d: \boxed{\uparrow}\boxed{\uparrow}\boxed{\uparrow}\boxed{\uparrow}\boxed{\uparrow} \ 4s: \boxed{\uparrow}$$
Give three more electrons and it might look something like this:
$$3d: \boxed{\uparrow \downarrow}\boxed{\uparrow \downarrow}\boxed{\uparrow}\boxed{\uparrow}\boxed{\uparrow} \ 4s: \boxed{\uparrow \downarrow}$$
This is the ground state of the electrons. I see no way for electrons to be promoted for the formation of six orbitals. And the electrons have to be promoted all the way up to the $4d$ orbital, which would require a significant amount of energy. What is the exact mechanism of bonding in this ion? Amazingly, I couldn't find anything on the Web about this.
EDIT: My textbook also says that $\ce{[Co(NH3)6]^{3+}}$ is an example of $d^2sp^3$ quite similar to $\ce{[CrF6]^{3-}}$. But when I try to understand the exact mechanism, I encounter a similar problem as a I did above.