Question

why SF6 and NF3 cannot be hydrolysed?

Answer 1

The average chemist does not make enough use of the ideas of Fukui (Nobel prize 1981: Check) who stated that it is the frontier orbitals that determine reactivity of two chemical species A and B. In particular the highest occupied molecular orbital (H*OMO star added to avoid censorship by you) on A must match the lowest unoccupied molecular orbital (LUMO) of B in terms of symmetry and energy for the reaction between A and B to have a low activation energy.Thus the reaction of SF₆ with H₂O is almost certainly highly allowed thermodynamically but the H*OMO and LUMO interactions are prevented. SF₄ instantly reacts with H₂O (to give HF and SO₂). The H*OMO on H₂O is the 2p (2sp³?) lone pair on O; the LUMO on SF₆ is the empty 3d AO. Consider the space-filling model top right. Can you see that the lone pair on O in an H₂O/SF₆ collision is prevented from overlapping with the 3d orbitals on S by the lone pairs of the surrounding F atoms: They literally repel the water molecules and hence there is no reaction of SF₆ with H₂O (or HCl, or KOH at 500 °C!).

With the larger Te atom there is presumably sufficient space between the F atoms to allow the H*OMO-LUMO overlap resulting in immediate HF formation and protective wall of F atoms is breached.The same explanation is for NF₃ to.