# What is the molecular shape for both NF_3 and XeCl_4?

Simple VSEPR theory predicts a trigonal pyramid for $N {F}_{3}$, and square planar for $X e C {l}_{4}$.
The electron pairs, bonding and non-bonding, are arranged around nitrogen in the shape of a tetrahedron. Because we describe molecular geometry on the basis of actual atoms, the geometry of the $N {F}_{3}$ molecule is trigonal pyramidal.
On the other hand, around the xenon atom there are 4 bonding pairs, i.e. 4 $X e - C l$ bonds, and 2 lone pairs on Xe. These are arranged in the shape of a octahedron, with the non-bonding electron pairs trans with respect to each other. Because we describe molecular geometry on the basis of actual atoms, the geometry of the $X e C {l}_{4}$ molecule is square planar.