JEE Main · 2006 · 30 Apr 2006 · Offline · Chemical Bonding and Molecular Structure

Equal bond lengths require every bonded position to be equivalent. $\mathrm{Si}{\mathrm{F}}_4$ is tetrahedral, $\mathrm{Xe}{\mathrm{F}}_4$ square planar and $\mathrm{B}{\mathrm{F}}_4^{-}$ tetrahedral, so in each of these all the $\mathrm{M}-\mathrm{F}$ bonds are identical. $\mathrm{S}{\mathrm{F}}_4$, however, has one lone pair and adopts a see-saw shape derived from a trigonal bipyramid, in which the two axial bonds are longer than the two equatorial bonds. The molecule whose bonds are not all equal is therefore $\mathrm{S}{\mathrm{F}}_4$.