Question

non-polar molecule, polar molecule, xef₂

Explanation:

Step1: Determine the molecular geometry of $\ce{XeF2}$

First, we calculate the number of valence electrons around the central $\ce{Xe}$ atom. Xenon has 8 valence electrons, and each fluorine has 7, so for $\ce{XeF2}$, the total valence electrons are $8 + 2\times7 = 22$. The Lewis structure of $\ce{XeF2}$ shows that $\ce{Xe}$ is the central atom with 2 bonding pairs (to $\ce{F}$ atoms) and 3 lone pairs. Using the VSEPR theory, the electron - pair geometry is trigonal bipyramidal, and the molecular geometry (considering only the bonding pairs) is linear. The three lone pairs are in the equatorial positions, and the two $\ce{F}$ atoms are in the axial positions.

Step2: Analyze the polarity of the bonds and the molecule

The $\ce{Xe - F}$ bond is polar because fluorine is more electronegative than xenon. However, in a linear molecular geometry, the two $\ce{Xe - F}$ bond dipoles are equal in magnitude and opposite in direction (since the angle between them is $180^{\circ}$). When we vectorially add these two bond dipoles, they cancel each other out. Also, the lone pairs are symmetrically arranged in the equatorial plane, which does not introduce a net dipole moment. So, the overall molecule has no net dipole moment.

Answer:

$\ce{XeF2}$ is a non - polar molecule, so it should be placed in the "Non - Polar Molecule" category.