It has been argued that AAlCuF 6 (A = K, Cs) and CuFAsF 6 are the only known crystals that exhibit compressed CuF 6 4− 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 Zn 2+ ions instead of Cu 2+ , we prove that neither the ground state nor the equilibrium geometry of CuF 6 4− complexes in KAlCuF 6 and CuFAsF 6 is the result of a Jahn−Teller effect. In contrast, it is shown that the internal electric field, E R (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 CuF 64− unit at equilibrium and the extrinsic contribution coming from the E R (r) field. Aside from reproduction of the experimental d−d transitions observed for KAlCuF 6 , it is found that in CuFAsF 6 the b 1g (x 2 − y 2 ) → a 1g (3z 2 − r 2 ) 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(R eq − R ax ) where R eq and R ax are the equatorial and axial Cu 2+ −F − distances and β = 2.7 eV/Å is the same for all systems involving tetragonal CuF 6 4− 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 KZnF 3 :Cu 2+ and other non-Jahn−Teller systems such as KAlCuF 6 , CuFAsF 6 , K 2 ZnF 4 :Cu 2+ , and Ba 2 ZnF 6 :Cu 2+ . Although most Jahn−Teller systems display an elongated geometry, there are however many Cu 2+ compounds with a compressed geometry but hidden by an additional orthorhombic instability. The lack of that instability in KAlCuF 6 and CuFAsF 6 is also discussed.