In the VSEPR structure of Xenon difluoride (), what is the total number of lone pairs on the central Xenon atom that occupy the equatorial positions of the trigonal bipyramidal geometry?

Question

In the VSEPR structure of Xenon difluoride ($XeF_2$), what is the total number of lone pairs on the central Xenon atom that occupy the equatorial positions of the trigonal bipyramidal geometry?

Prashant Kotian · Accepted Answer

Correct answer: 3. Xenon has 8 valence electrons. In , it forms 2 single bonds with fluorine, leaving 6 electrons, which pair up to form 3 lone pairs. To minimize lone pair-lone pair repulsions in the hybridized state, all 3 lone pairs occupy the equatorial plane, forcing the F atoms into the axial positions, yielding a linear molecular shape.

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