Correct Answer - A
`SiF_4` and `SF_4` are not isostructural because `SiF_4` is tetrahedral due to `sp^3` hybridisation of Si.
`._14Si=1s^2 , 2s^2,2p^6, 3s^2 3p^2` (in ground state)
`._14Si=1s^2,2s^2 2p^6,3s^1 3p^3` (in excited state)
Hence, four equivalent `sp^3` hybrid orbitals are obtained and they are overlapped by four p-orbitals of four fluorine atoms on their axis. Thus, it shows following structure.
While `SF_4` is not tetrahedral but it is arranged in trigonal bipyramidal geometry (has see saw shape) because in it S is `sp^3 d` hybrid).
`._16S=1s^2,2s^2 2p^6 , 3s^2 3p_x^2 3p_y^1 3p_z^1` (in ground state)
`=1s^2, 2s^2 , 2p^6 ,ubrace(3s^2 3p_x^1 3p_y^1 3p_z^1 3d_(xy)^1)_(sp^3d"hybridisation")`
Hence, five `sp^3d` hybrid orbitals are obtained. One orbital is already paired and rest four are overlapped with four p-orbitals of four fluorine atoms on their axis in trigonal bipyamidal form.
This structure is distorted from trigonal bipyramidal to tetrahedral due to involvement of repulsion between lone pair and bond pair.
