QJEEAdv04Ch36f0: JEE Advanced 2004 Chemistry - Class 11 JEEAdv04Ch36f0 Solution: Draw the structure of XeF 4 and OSF 4 according to VSEPR theory …

Structure of Xenon tetrafluoride, $\mathrm{XeF}_4$ :

In $\mathrm{XeF}_4$ , the central atom $\mathrm{Xe}$ is attached to $4 \mathrm{~F}$ atoms. The central atom $\mathrm{Xe}$ undergoes $s p^3 d^2$ hybridization as shown below :

The geometry of $\mathrm{XeF}_4$ is square planar and the structure is octahedral. The bond angle is $90^{\circ}$ . The two lone pairs of electrons are shown on $\mathrm{Xe}$ atom.

Structure of Thionyl Tetrafluoride, $\mathrm{OSF}_4$

In $\mathrm{OSF}_4$ , the central atom is sulphur which is attached to a $\mathrm{O}$ atom by double bond and attached to $4 \mathrm{~F}$ atoms through a single bond. The central $S$ atom undergoes $s p^3 d$ hybridisation as shown below :

One of the $d$ orbitals will be involved in pi bond formation with $\mathrm{O}$ atom.

Thus, the molecular geometry of $\mathrm{OSF}_4$ should be trigonal bipyramidal. The geometry is distorted trigonal bipyramidal due to asymmetric charge distribution around the central atom. In the molecule, zero lone pairs are present along with 5 sigma bonds and a pi bond. The oxygen atom is at equatorial position as it is less electronegative than that of $\mathrm{F}$ .