It has been argued that AAlCuF6(A = K, Cs) and CuFAsF6are the only known crystals that exhibit compressed CuF64–units due to the Jahn–Teller effect. However, no grounds for this singular behavior have yet been reported. By means of first-principles calculations on such compounds and the isomorphous compounds involving Zn2+ions instead of Cu2+, we prove that neither the ground state nor the equilibrium geometry of CuF64–complexes in KAlCuF6and CuFAsF6is the result of a Jahn–Teller effect. In contrast, it is shown that the internal electric field, ER(r), created by the rest of the lattice ions upon the localized electrons in the complex, plays an important role in understanding this matter as well as the d–d transitions of these two compounds. The energy of an optical transition is shown to involve two contributions: the intrinsic contribution derived for the isolated CuF64–unit at equilibrium and the extrinsic contribution coming from the ER(r) field. Aside from reproduction of the experimental d–d transitions observed for KAlCuF6, it is found that in CuFAsF6the b1g(x2– y2) → a1g(3z2– r2) transition is not the lowest one due to the stronger effects from the internal field. Interestingly, the intrinsic contribution corresponding to that transition can simply be written as β(Req– Rax) where Reqand Raxare the equatorial and axial Cu2+–F–distances and β = 2.7 eV/Å is the same for all systems involving tetragonal CuF64–units and an average metal–ligand distance close to 2.03 Å. This shows the existence of a common point shared by the Jahn–Teller system KZnF3:Cu2+and other non-Jahn–Teller systems such as KAlCuF6, CuFAsF6, K2ZnF4:Cu2+, and Ba2ZnF6:Cu2+. Although most Jahn–Teller systems display an elongated geometry, there are however many Cu2+compounds with a compressed geometry but hidden by an additional orthorhombic instability. The lack of that instability in KAlCuF6and CuFAsF6is also discussed.