Molecular Geometry of Xenon Dioxydifluoride Molecule

To determine the molecular geometry of the xenon dioxydifluoride ($\text{XeO}_2\text{F}_2$) molecule, we use Valence Shell Electron Pair Repulsion (VSEPR) theory:

1. Valence Electrons of Central Atom: The central atom is Xenon ($\text{Xe}$), which belongs to Group 18 and has 8 valence electrons.

2. Bonding with Surrounding Atoms:

   • Each of the two Fluorine ($\text{F}$) atoms forms a single $\sigma$-bond with Xenon using 1 electron each (total $2$ electrons).
   • Each of the two Oxygen ($\text{O}$) atoms forms a double bond ($1\ \sigma$-bond and $1\ \pi$-bond) with Xenon using 2 electrons each (total $4$ electrons).

3. Number of Lone Pairs:

   • Total valence electrons of $\text{Xe} = 8$
   • Electrons used in bonding = $2 \text{ (from F)} + 4 \text{ (from O)} = 6$
   • Remaining non-bonding electrons on $\text{Xe} = 8 - 6 = 2$ electrons, which form $1$ lone pair.

4. Steric Number and Hybridization: $\text{Steric Number (SN)} = (\text{Number of } \sigma\text{-bonds}) + (\text{Number of lone pairs})$ $\text{SN} = 4 + 1 = 5$

   A steric number of 5 corresponds to $sp^3d$ hybridization, giving a trigonal bipyramidal electron pair geometry.

5. Molecular Shape / Geometry:

   • According to VSEPR theory, larger electron domains (more repulsive), such as double bonds ($\text{Xe=O}$) and lone pairs, occupy the equatorial positions to minimize repulsions ($90^\circ$ repulsions).
   • Therefore, the $1$ lone pair and the $2$ oxygen atoms occupy the equatorial positions.
   • The $2$ fluorine atoms occupy the axial positions.

With 4 bonding positions and 1 lone pair in a trigonal bipyramidal framework, the molecular geometry (shape) of $\text{XeO}_2\text{F}_2$ is see-saw.

Correct Answer: D