Question

Draw the VSEPR structure for XeO2F2. Give the name of the molecular shape and indicate the bond angles.

Answer 1

The central atom in XeO₂F₂ is Xe.It belongs to VIII A group in the periodic table and has 8 valence e⁻ .Around this Xe atoms there are two Oxygen( divalent atom) atoms & two Fluorine(mono valent atom) atoms.

So Xe forms two single bonds with F atom & two double bonds with each O atom.

So there are 6 bond pairs & 1 lone pair on Xe atoms.It is an AX4E system based on trigonal bipyramid but with an equatorial site occupied by a lp. It has a see-saw structure.

F atoms always go axial in trigonal bipyramidal structures and the Xe=O will occupy more space so it will favor the equatorial sites & it will also occupy space than a typical bond pair.We have lp-bp>bp-bp for repulsion. So we would predict F-Xe-F < 180° and O-Xe-O to be < 120° .

The configuration of XeO2F2 is that of a trigonal bipyramid with axial F and equatorial O atoms and 1 vacant equatorial position. The O-Xe-O, O-Xe-F, and F-Xe-F bond angles are 105.7°, 91.6°, and 174.7°, respectively.