JEE Main · 2004 · 24 Apr 2004 · Offline · Chemical Bonding and Molecular Structure

A regular tetrahedron calls for four bonding pairs and no lone pairs on the central atom. $\mathrm{BF}{}_4{}^{-}$ meets this exactly: boron carries four $\mathrm{B}-\mathrm{F}$ bonds, no lone pair, and $s{p}^3$ hybridisation, giving the symmetric tetrahedral shape. $\mathrm{XeF}{}_4$ is square planar, $[\mathrm{Ni}(\mathrm{CN}){}_4]{}^{2-}$ is square planar from $ds{p}^2$ hybridisation, and $\mathrm{SF}{}_4$ is see-saw because of its one lone pair, so $\mathrm{BF}{}_4{}^{-}$ is the regular tetrahedron.