The learning objective of the post is to understand the question
SF4 when subjected to hydrolysis get hydrolysis while SF6 under the same conditions does not hydrolyze?
The electronic configuration of the Sulphur is
16S = [Ne] 3s2 3p4
now there are two unpaired electrons and Sulphur and on the basis of the principal quantum number of valence shell the allowed values of Azimuthal quantum number are 0, 1,2 thus d-orbital is allowed thus the box of presentation of Sulphur showing d orbital can be written as
In the case of Sulphur tetrafluoride, the hybridization is dsp3 and the shape of the molecule is seesaw or distorted trigonal bi-pyramidal
In the case of Sulphur hexafluoride, the hybridization is d2sp3 and the shape of the molecule is square bipyramidal and all one angles of 90°
when Sulphur tetrafluoride is subjected to hydrolysis the axial bonds and equatorial bonds possess dipole because fluorine is more electronegative than Sulphur and the overall dipole is out towards as shown in the figure
due to the dipole moment, the bond length of sulfur-fluorine bonds increases, and lesser energy is required to break the bonds so hydrolysis of SF4 occurs readily.
When Sulphur hexafluoride is subjected to hydrolysis, due to symmetry in the structure no dipole moment is observed on the molecule as they are canceled off. As there is no effective dipole over the Sulphur-fluorine bond when it is subjected to hydrolysis higher energy is required to break the bond. The Bond energy provided by the water molecule is not sufficient to break the Sulphur fluorine bond and thus hydrolysis of Sulphur hexafluoride is not possible.
So due to differences in the shapes or geometries of the molecules are responsible for the hydrolysis of the molecule.