Why are the bond lengths in #"CrF"_6^(4-)# not all the same?

Well, wouldn't you expect a Jahn-Teller distortion?

Each #"F"^(-)# ligand contributes a #-1# charge, so by conservation of charge, chromium has a #bb(+2)# oxidation state. That means #"Cr"^(2+)# is a so-called "#bbd^4# metal".

The #"F"^(-)# ligands are also #bbpi# donors, making them weak-field ligands.

As seen above, they slightly raise the #t_(2g)# orbital energies so that the ligand field splitting energy #Delta_o# decreases, promoting a high spin complex.

As a result, the following #d# electron configuration in the complex is predicted to be:

#uarrE" "color(white)({(" "" "color(black)(ul(uarr color(white)(darr))" "ul(color(white)(uarr darr))" "e_g^"*")),(),(color(black)(Delta_o)),(),(" "color(black)(ul(uarr color(white)(darr))" "ul(uarr color(white)(darr))" "ul(uarr color(white)(darr))" "t_(2g))):})#

In this arrangement, the #e_g^"*"# electron configuration is doubly-degenerate. It can occupy one of two orbitals spin-up. Since they are antibonding orbitals, this is going to be a strong Jahn-Teller Distortion.

To lift that degeneracy, the ligands on the #z# axis distort antisymmetrically with the equatorial ligands. One way of doing that is:

This would increase #z#-axis interactions and decrease #x//y#-axes bonding interactions. So, the orbitals with #z# components destabilize and the orbitals with #x# and #y# components stabilize:

The ligands could also stretch and compress in the other direction. That would reverse the stabilization/destabilization to alternatively give:

Either option leaves only one way to arrange the electrons, and either one is a Jahn-Teller Distortion.