The shape of $$XeF_4$$ is:
The correct option is A square planar
The shape of $$XeF_4$$ is square planar. The central $$Xe$$ atom has 4 bond pairs of electrons and two lone pairs of electrons. It undergoes $$sp^3d^2$$ hybridization which results in octahedral electron geometry and square planar molecular geometry. The two lone pairs are at opposite corners of an octahedron. This minimizes repulsion between them.