0
$\begingroup$

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:

enter image description here

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:

enter image description here

  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:

    Trigonal-3D-balls

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

$\endgroup$
  • 1
    $\begingroup$ @Mithoron I don't see how those answer my questions? $\endgroup$ – The Pointer Dec 13 '19 at 20:42
  • $\begingroup$ 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. $\endgroup$ – Mithoron Dec 13 '19 at 21:28
  • 1
    $\begingroup$ @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. $\endgroup$ – The Pointer Dec 13 '19 at 22:01
  • 1
    $\begingroup$ @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. $\endgroup$ – The Pointer Dec 13 '19 at 22:11
0
$\begingroup$

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.

| improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ I don't think what you've described answers my questions. I am aware that the physical geometry of these molecules is 3-dimensional and not 2-dimensional (planar). But this is a case in which the textbook describes the molecular geometry of the molecules and their energy states using Lewis structures, which are 2-dimensional diagrams. And given the comparison the authors made between structures 1a and 1b, they are obviously using the diagrams to also convey the (lowest) energy states of the molecules; which leads to my question of why the Lewis structures are not as I wonder they should be. $\endgroup$ – The Pointer Dec 13 '19 at 20:50
  • $\begingroup$ His point is that in a tetrahedron the oxygens are equidistant from each other, so there is no distinction between the double bonds being on the Os that appear in 2D to be adjacent vs opposite. Likewise, in the XeO4 Lewis structure the position of the dots is not intended to represent the positions of the electrons in a rigorous geometrical way. Independent of your misinterpretation of the Lewis structures, expanded octets on sulfur complexes is an outdated and debunked idea. $\endgroup$ – Andrew Dec 13 '19 at 23:36
  • $\begingroup$ @Andrew Hmm, I see. Thank you for taking the time to clarify. So, out of curiosity, is the entire idea of expanded octets outdated or debunked, or just the idea of expanded octets on sulfur complexes? $\endgroup$ – The Pointer Dec 14 '19 at 0:08
  • 1
    $\begingroup$ For larger atoms, d and even f orbitals can be involved in bonding, which means the octet rule is no longer obeyed. Note, however, that the conventions for drawing Lewis structures have not changed, so the standard practice, for example, is to use formal charge rather than octet expansion on nitrogen oxides, but with sulfur and phosphorus oxides, we still draw 6 bonds. As a general statement, Lewis structures are not intended to rigorously represent the electronic or geometric structure of molecules. They are useful simple representations. $\endgroup$ – Andrew Dec 14 '19 at 13:01

Not the answer you're looking for? Browse other questions tagged or ask your own question.