Question

The compound of xenon with zero dipole moment is:
A. \[XeO{F_4}\]
B. \[Xe{O_2}\]
C. \[Xe{O_3}\]
D. \[Xe{F_4}\]

Answer 1

Hint: The zero dipole moment of a compound has symmetrical geometry and also has similar atoms with less distance between the charge separation.

Complete Step by Step Solution:
1. The dipole moment is the product of the magnitude of charge and the distance between the centre of the positive and negative charge.
2. It is expressed as:
\[\mu = Q \times r\]
Where, \[\mu \] =dipole moment
\[Q\] =charge
\[r\] =separation distance
3. The dipole moment is a vector quantity that is depicted on a Lewis structure as a crossed arrow.
4. A cross appears on the positive end while an arrowhead appears on the negative end.
5. The valence electrons of the xenon are eight and the valence electrons of the Fluoride are seven so the total valence electrons of the compound are thirty-six.
6. There are a total of two lone pairs on the xenon and it is attached to the four Fluoride atoms through the bond pairs.
7. The geometry of the xenon tetrafluoride is square planar with a symmetrical structure with an angle of \[{90^ \circ }\].
8. The zero dipole moment of the compound is defined as the two equal bond dipoles that point in the opposite directions and cancel the effect of each other.
9. The individual bond dipoles cancel out and leave the molecule as a nonpolar so there is no net dipole moment.
10. The Lewis dot structure of the xenon tetrafluoride is represented below:

Image: Xenon tetrafluoride

Option (D) is correct.

Additional information:
1. When the same forces of energy are applied at an angle of \[{90^ \circ }\], there is no movement.
2. The direction of the dipole moment is from a less electronegative atom to a more electronegative atom.

Note: The dipole moment arrow represents the direction of the shift of electron density in the molecule with the direction of the crossed arrow opposite the conventional direction of the dipole moment vector.