The shape of $\mathrm{XeF}{}_5{}^{-}$ is

Xe has 8 valence electrons plus 1 from the negative charge, 9 in all; five are shared with F, and the remaining 4 form 2 lone pairs. So there are 5 bond pairs and 2 lone pairs, 7 pairs in total ($s{p}^3{d}^3$). The two lone pairs occupy the axial positions of a pentagonal bipyramid, leaving five F in one plane: pentagonal planar.