The shape of ${\mathrm{XeF}}_4$ is 

Valence electrons of ${\mathrm{XeF}}_4=8+4\times 7=36$        Lewis structure

There are 6 pairs of electrons around. These are directed to form octahedron. The four F atoms occupy apical positions, therefore, the shape of ${\mathrm{XeF}}_4$ is square planar.