Assertion (A) :${\text{SF}}_6$ molecule has a regular octahedral geometry.

Reason (B):  Hybridization of S in ${\text{SF}}_{\text{6}}$ is ${\text{sp}}^{\text{3}}{\text{d}}^{\text{2}}$ .

In ${\text{SF}}_6$  , the central sulphur atom has the ground state outer electronic configuration ${\text{3s}}^{\text{2}}{\text{3p}}^{\text{4}}$ .  In the second excited state the available six orbitals i.e., one s, three p and two d are singly occupied by electrons.  These orbitals hybridized to form six new ${\text{sp}}^{\text{3}}{\text{d}}^{\text{2}}$ hybrid orbitals, which are projected towards the six corners of a regular octahedron in ${\text{SF}}_{\text{6}}$  . These six ${\text{sp}}^{\text{3}}{\text{d}}^{\text{2}}$ hybrid orbitals overlap with singly occupied orbitals of fluorine atoms to form six S-F sigma bonds.  Thus ${\text{SF}}_{\text{6}}$ molecule has a regular octahedral geometry.  The structure of ${\text{SF}}_6$  molecule is as shown below.

                        

In ${\text{SF}}_6$  molecule, Sulphur (S) undergoes ${\text{sp}}^{\text{3}}{\text{d}}^{\text{2}}$ hybridization and it has octahedral geometry.