(i) ${\left[{\mathrm{XeF}}_5\right]}^{-}$
St. No. $=(5+2)=7$ so hybridisation is$={\mathrm{sp}}^3{\mathrm{d}}^3$ and structure is pentagonal planar.

(ii) ${\mathrm{XeO}}_3{\mathrm{F}}_2$ St. No. $=5$
so hybridisation is $={\mathrm{sp}}^3\mathrm{d}$ and structure is trigonal bipyramidal.