Question

The fluoride of xenon with zero dipole moment is:

• A. $Xe{F}_6$
• B. $XeO{F}_4$
• C. $Xe{F}_4$
• D. $Xe{O}_2{F}_2$

Answer 1

The correct option is C $Xe{F}_4$

$Xe{F}_4$ is symmetrical with a square planar shape so the resultant diople moment is zero.

$Xe{F}_6$ had distorted octahedron shape and it contains one lone pair of electron. So it shows dipole moment.

In $XeO{F}_4$, the xenon atom will bond with the four fluorine atoms via four single bonds, which will ensure that each of the four fluorine atoms gets a complete octet. Afterwards the xenon atom will make double bond with oxygen atom to fulfill its octet. The remaining two electrons will remain as a lone pair. It has a square pyramidal geometry. Hence, it has non-zero dipole moment.

$Xe{O}_2{F}_2$ is polar. It has 5 areas of electron density around the central Xe atom, one of which is a lone pair. The molecule is not symmetrical, and so the dipole moments can not cancel, making the molecule polar.

The structures are given below:

Hence, the correct answer is option (c).