One student claims \(\mathrm{XeF}_{4}\) will be polar. Another claims it will be nonpolar. Which is right, and why?

First, we need to determine the electronic geometry and molecular geometry of the molecule. To do this, we will count the total number of valence electrons in the molecule and distribute them among the central atom and the surrounding atoms. Xenon(Xe) has 8 valence electrons and each Fluorine(F) atom has 7 valence electrons. Since there are 4 Fluorine atoms, the total number of valence electrons in the XeF4 molecule is: \( 8 +(4\times7) = 36 \) A Lewis structure of the XeF4 molecule can be drawn with the central atom Xenon surrounded by 4 Fluorine atoms forming single bonds and having two lone electron pairs on the Xe atom. Considering the central atom(Xe); - It forms 4 bonds with Fluorine atoms (4 bonding electron pairs). - There are 2 lone electron pairs in its outer shell. Now, to find the electronic geometry, we need to consider both the bonding electron pairs and the lone electron pairs. Therefore, Xe will have: \(4 + 2 = 6\) electron groups around it. The electronic geometry of a molecule with 6 electron groups is described as octahedral. However, in this case, since there are 2 lone pairs, we must account for them while determining the molecular geometry. So, the molecular geometry of the XeF4 molecule will be square planar.

Now that we know the molecular geometry of the XeF4 molecule is square planar, we can analyze if there are any net dipole moments in the molecule. In a square planar configuration, the individual bond dipoles of the Xe-F bonds cancel each other out, resulting in a net dipole moment of zero. Since the net dipole moment for the XeF4 molecule is zero, we can conclude that the molecule is nonpolar. So, the student who claims that the XeF4 molecule is nonpolar is correct.