JEE Main · 2025 · 02 Apr 2025 · Shift 2 · Chemical Bonding and Molecular Structure

The molecule required must use one d orbital in bonding ($s{p}^{3}d$), carry a lone pair on the central atom, and therefore have geometrically inequivalent bonds. $\mathrm{PF}{}_5$ is $s{p}^{3}d$ but has no lone pair on phosphorus. $\mathrm{XeF}{}_4$ is $s{p}^3{d}^2$ and square planar. $\mathrm{XeF}{}_2$ is $s{p}^{3}d$ with three lone pairs but is linear with two identical bonds. $\mathrm{SF}{}_4$ is $s{p}^{3}d$ with one lone pair on sulphur, giving a see-saw shape in which the two axial S–F bonds are longer than the two equatorial S–F bonds. The species that is $s{p}^{3}d$ hybridised, has unequal bond lengths and bears a lone pair on the central atom is sulphur tetrafluoride.