Question

The shape and geometry of $Xe{F}_2$ in vapor phase are respectively:

• A. Bent and tetrahedral
• B. Linear and tetrahedral
• C. Angular and trigonal bipyramidal
• D. Linear and trigonal bipyramidal

Answer 1

The correct option is D Linear and trigonal bipyramidal

Let us look at the ground state valence electronic configuration of Xe and the first excited state:

So there are 3 lone pairs of electrons and two bond pairs (each unpaired electron shown above will pair with exactly one electron from either of the Fluorine atoms). These orbitals may be assumed to be hybridized correspond to $s{p}^{3}d$ geometry.
The lone pairs occupy three ends of a symmetrical plane while the bond pairs will be axial or perpendicular to the plane containing the 3 electron pairs.
Shape is linear (due to perfect geometry) while the geometry is trigonal bipyramidal. What will be the F – Xe – F bond angle? Yes. It will will be 180 degrees.
The geometry will be like: