(i) Structure of $\mathrm{ICl}^{-}_{4}$ : $\mathrm{I} \text{ in } \mathrm{ICl}^{-}_{4}$ has four bond pairs and two lonepairs. Therefore, according to VSEPR theory, it should be square planaras shown.Here, $\mathrm{ICl}^{-}_{4}$ has (7 + 4 × 7 + 1) = 36 valence electrons. A noblegas specieshaving 36 valence electrons isXeF$_{4}$(8 + 4 × 7 = 36). Therefore, like $\mathrm{ICl}^{-}_{4}$ , XeF$_{4}$ is also square planar.(ii) Structure of $\mathrm{IBr}^{-}_{2}$ :$\mathrm{I}$ in $\mathrm{IBr}^{-}_{2}$ has two bond pairs and three lone pairs.So,according to VSEPR theory, it should be linear.Here,$\mathrm{IBr}^{-}_{2}$ has 22 (7 + 2 × 7 + 1) valence electrons.A noble gas species having 22 valence electrons is XeF$_{2}$ (8 + 2 × 7 = 22).Thus, like $\mathrm{IBr}^{-}_{2}$ , XeF$_{2}$ is also linear :(iii)Structure of $\mathrm{BrO}^{-}_{3}$: The central atom Br has seven electrons. Four ofthese electrons form two double bonds or coordinate bonds with twooxygen atoms while the fifth electron forms a singlebond with O$^{-}$ion.The remaining two electrons form one lone pair. Hence, in all there arethree bond pairs and one lone pair around Br atom in $\mathrm{BrO}^{-}_{3}$. Therefore,according to VSEPR theory, $\mathrm{BrO}^{-}_{3}$should be pyramidal.Here,$\mathrm{BrO}^{-}_{3}$has26(7 + 3 × 6 + 1 = 26) valence electrons. A noble gas specieshaving 26 valence electrons is $\mathrm{XeO}_{3}$(8 + 3 × 6 = 26). Thus, like $\mathrm{BrO}^{-}_{3}$ , $\mathrm{XeO}_{3}$is also pyramidal.