# Preferred Lewis structure for sulphate anion and XeO4 [closed]

In a section reviewing Lewis structure, the textbook Atkins' Physical Chemistry, 10th Edition, by Atkins, Paula, and Keeler, gives the following illustration of hypervalence and octet expansion:

I have two questions regarding this illustration:

1. With regards to the structure of the $$\ce{SO4^{2-}}$$ ion, notice that the lowest energy structure (1b) has the double covalent bonds placed at angles that are closer than had the double bonds been placed at opposing sides of the sulfur atom. I'm confused as to why this is supposedly the lowest energy state, since one would expect the lowest energy state to be when the two double bonds are at farthest angles from each other (which would be when they are on opposing sides of the Sulphur atom in the diagram), since this would be the case when there is the most distance between the repulsive negative charges of the electrons in the two double covalent bonds? Based on my research, it seems that what I am referring to here would resemble a linear molecular geometry (although, I am obviously not saying that the molecule would actually have a linear molecular geometry), where the we can imagine the red atom in the following image as being the sulfur atom, and the two white atoms as being the oxygen atoms that have the double covalent bond:

1. My next question is with regards to the $$\ce{XeO4}$$ molecule. It seems to me that the arrangement of all of the free electron pairs on the oxygen atoms would be in a more energetically favourable state (the molecule would be in a lower energy state) if they were farther away (at greater angles) from the double covalent bonds? It seems to me that their current arrangement would make sense in a scenario where there were an additional 2 free electrons on the oxygens, as is the case for two of the oxygen atoms of structure 1b, but since these oxygen atoms only possess 6 electrons instead of 8 (as is the case for two of the oxygen atoms of $$\ce{SO4^{2-}}$$), then wouldn't it be more energetically favourable to have the electron pairs at greater angles from the double covalent bonds? Based on some research, it seems that the geometry that I am referring to is a trigonal planar molecule geometry, as shown here (although, I am just referring to the Lewis structure of the molecule -- not the 3-dimensional geometry). In the following image from the same Wikipedia article, imagine the red atom as being the oxygen, and one of the white atoms as being the double covalent bond, and the other two white atoms as being the lone pairs of electrons, so that you have a $$60^\circ$$ angle between them:

I would greatly appreciate it if people could please take the time to clarify this.

• – Mithoron Dec 13 '19 at 20:32
• @Mithoron I don't see how those answer my questions? – The Pointer Dec 13 '19 at 20:42
• And what are they really? You made your post as unclear as possible. As far as I can tell your reasoning doesn't make much sense. Even book you cite is wrong - actually single bonds reflect reality much better. – Mithoron Dec 13 '19 at 21:28
• @Mithoron If my post is unclear, then it certainly isn't because I intentionally made it to be so. I am a novice when it comes to chemistry, and I'm trying to learn, which is why I am studying this textbook. The textbook itself is one of the most highly regarded, if not the most highly regarded, physical chemistry textbook of the present. If my questions do not make sense, or if there is something incorrect in the textbook, then feel free to explain. – The Pointer Dec 13 '19 at 22:01
• @Mithoron What I mean is that maybe it is my lack of understanding that has led to an absurd and/or unclear question. What about my question is unclear? I will attempt to clarify. – The Pointer Dec 13 '19 at 22:11

You are reasoning as if the ion $$\ce{SO4^2-}$$ is planar. It is not the case. It has a tetrahedral geometry. And what you see in Atkins' book is a projection of the tetrahedron on a horizontal plane. So the angles between the bonds are not 90°, but larger. It is the same for $$\ce{XeO4}$$. This reality will solve all your questions simultaneously.