# Why is the dipole moment of XeF$_6$ zero? [duplicate]

In Concise Inorganic Chemistry by JD Lee (4th edition; adapted by Sudarshan Guha), page 73 under the section 3.6 (VSEPR Theory chapter Chemical Bonding):

...according to this theory, the position of the lone pair in $$\ce{XeF6}$$ molecule should be in axial position and the shape should be pentagonal pyramidal. But this is not observed and the actual shape of molecule is octahedral. This implies that the lone pair is placed at equatorial position, and this explains the structure of $$\ce{XeF6}$$ shown in figures 3.7 a,b,c.

According to Figure 3.7, $$\mu_\text{expected}$$ for $$\ce{XeF6}$$ is not equal to zero. But $$\mu_\text{real}(\ce{XeF6})\sim0$$ . To explain the above facts, it is considered that the actual molecule is in dynamic equilibrium of all the three structures shown in Figure 3.7. Here it is considered that the lone pair is present in the stereochemically active $$s$$ orbital and similar structure is observed as $$\ce{IF6-}$$. But anions like $$[\ce{SbX6}]^{3-}$$, $$[\ce{TeX6}]^{2-}$$ ($$\ce{X}=\ce{Cl, Br, I}$$), $$[\ce{BrF6}]^-$$,$$[\ce{ICl6}]^-$$ have been assigned perfectly octahedral structure on the basis of X-ray crystallography and the lone pair is present in stereochemically inactive $$s$$ orbital.

It is explained the three structures given in the above diagram are in dynamic equilibrium, so dipole moment is zero. But I feel it must be towards the right since the lone pair of electrons stays mainly towards right since the vertical component cancel (structure c and d)

Could you please explain how according to the author, the dipole moment of XeF$$_6$$ be zero.

Kindly note: I have read this question related to the same passage but a different concept. My doubt is not about stereochemically active or inactive s orbitals but about the non-existence of dipole moment in the given compound.

• Carbon dioxide is not always symmetrical if you consider its vibrational modes, but the time average dipole moment is still zero. – Zhe Aug 22 '19 at 13:12
• @Zhe, I understand your case, but here only the vertical components of the dipole moment cancel and I think the net dipole moment must be towards the right since the lone pair stays mainly to the right side of the molecule as given in the figure. This is contrary to what is being said in the book. – Vishnu Aug 22 '19 at 13:15
• By symmetry, that lone pair does not stay on the right side. I would imagine that they did not draw out all 12 possible transitions. – Zhe Aug 22 '19 at 13:33
• @Zhe, If my calculations are right, there must be [ 12 (lone pair of electrons on the edge of the octahedron) + 8 (lone pair on the face of the octahedron) = 20 transitions instead of 12 ]. Am I right? – Vishnu Aug 22 '19 at 14:06
• chemistry.stackexchange.com/questions/50187/… – Mithoron Aug 22 '19 at 20:40