Identify the correct set 

       Molecule                        Hybridisation                    Shape               Number of lone pairs

       a. $Xe{O}_4$                            $s{p}^3{d}^2$                         pyramidal                             1

       b. $Xe{O}_3$                               $s{p}^3$                          pyramidal                              1

       c. $Xe{F}_4$                          $s{p}^3{d}^2$                           planar                                   3

       d. $Xe{F}_2$                         $s{p}^{3}d$                              linear                                     2

     

In Xenon there are eight outermost electrons present and hence can able to form  Eight bonds with other atoms.

In ${\text{XeO}}_4$, each oxygen forms a double bond with xenon. So all eight electrons of xenon participate in bond formation. Hybridization of ${\mathrm{XeO}}_4$ is sp³ .All are bond pairs and there are no lone pairs, ${\mathrm{XeO}}_4$,  shape  is tetrahedral. 

In ${\mathrm{XeO}}_3$, each oxygen forms a double bond with xenon. Only six electrons of xenon participate in bond formation and left one lone pair. Hybridization of ${\mathrm{XeO}}_3$ is sp3, shape is pyramidal.

In ${\mathrm{XeF}}_4$, each fluorine atom forms a single bond with xenon. Only four electrons of xenon participate in bond formation and left two lone pairs. Hybridization of ${\mathrm{XeF}}_4$ is sp³d², shape is square planar.

In ${\mathrm{XeF}}_2$, each fluorine atom forms a single bond with xenon. Only two electrons of xenon participate in bond formation and left three lone pairs. Hybridization of ${\mathrm{XeF}}_4$ is sp³d, shape is linear.