27

1. Perturbations As already mentioned, the Jahn–Teller effect has its roots in group theory. The essence of the argument is that the energy of the compound is stabilised upon distortion to a lower-symmetry point group. This distortion may be considered to be a normal mode of vibration, with the corresponding vibrational coordinate $q$ labelling the "extent ...


26

General case There is indeed a mathematical theorem that deals with the number of nodes an eigenfunction corresponding to a certain eigenvalue can possess. It was laid down by Courant$^{[1, 2]}$ and it states the following: Given the self-adjoint second order (partial) differential equation \begin{equation} \left(\hat{L} + \lambda \rho(\mathbf{x}) \...


26

It's not easy to see from a diagram, because it distorts bonds and angles. I recommend building it with a balls-and-sticks model set. You can also use a molecular viewer to model it; there are a couple of open-source (or at least free) ones out there. I have calculated the molecule on the DF-BP86/def2-SVP level of theory. The point group of the molecule is ...


23

As stated in both the links Geoff and Philipp have kindly commented (1, 2) they are to do with symmetry labels we chemists like to assign to orbitals. Knowing an orbitals symmetry class can lead to a lot of simplifications down the road when you use quantum mechanical calculations and even dictate the reactivity of which orbitals are "allowed" to interact ...


23

since for every orientation of the molecule, we can reverse the orientation such that the light appears to be falling on the molecule from a direction other than the one for our original molecule. This is false. Let's take 2-butanol. For this stereoisomer, light is turning clockwise when viewed from the right side (I'm not sure of this, but we can assume). ...


19

First, you have to know the geometry of your compound. The complex $\ce{[PtCl4]^2-}$, for example, is square planar. The next step is to determine the point group of the compound from this geometry by identifying the symmetry elements that it posesses (a procedure on how to do this can be found here). The point group of the complex $\ce{[PtCl4]^2-}$, for ...


18

Introduction Instead of the usual cycloadditions (which has some extra complications in terms of the reacting orbitals), let's consider the concerted reaction $\ce{H2 + D2 -> 2HD}$. We'll come back to the cycloadditions at the end. occurring via a square planar transition state. This can be analysed with the W–H rules and you'll find that it's a ...


17

I've done some work in both symmetry detection and in distance matrix methods. I think it's a great idea in concept, but the devil will be in the details for large, more complex molecules. The first problem is that distance geometry methods are over-determined. For each atom, there are range constraints to the other $N-1$ atoms (a lower bounds and an upper ...


15

I'll start by linking to Symmetry @ Otterbein, as it's a very useful tool for looking at various structures and seeing planes an axes of symmetry. If you're having issues visualising rotations and such, then this website should certainly help. I suppose the first step would be to make sure that you're always, always, always visualising the structure in 3D. ...


15

This is not in general true Consider molecules a point group not containing inversion symmetry, e.g. $C_2$ hydrogen peroxide The $C_2$ group has only two elements, $E$ and $C_2$, and the $C_2$ rotation operation maps between two identical arrangements of atoms. Both the symmetry number and order of the group are 2. Matrix representations All rotations ...


14

Introduction Thank you for prompting me to look for square planar complexes that are not $\mathrm{d^8}$; I learnt some valuable stuff while researching the answer to this question! Do take this answer with a grain of salt, because I cannot access the paper I’m referencing here on holidays (I could when I’m back at work) but the supplementaries confirm what ...


13

I use quantum chemistry about as much as a lumberjack uses math (that is, just enough not to fell a tree on one's own head), so my answer will be short and hopefully useful as a tl;dr to someone else's more comprehensive take. In quantum chemistry, if two operators commute, their values are simultaneously observable. In other words, there is a basis of ...


12

Let's start with the basics: A representation is a set of matrices that fulfill the multiplication table for the point group. So I might be able to use a set of 3x3 matrices, since of course I can define any of the symmetry operations over Cartesian (x,y,z) space. e.g.: $$E = \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 &...


12

Have a look at the reaction mechanism of the Diels-Alder Reaction, e.g. at Wikipedia. We have talked about this reaction before on this site, and concluded that a sufficient explanation is only possible with the help of molecular orbitals: How accurate is this polar mechanism for the Diels-Alder reaction? Borrowing from Wikipedia,[1] here is the most ...


12

1. Weak-field and strong-field limits I will adopt the description used in Figgis and Hitchman's Ligand Field Theory and Its Applications (p 5), because I cannot really phrase it better: It is useful to recognise the relative importance of the terms in the Hamiltonian for different systems: First- and second-transition series: $\mathbf{H}_\mathrm{LF}...


11

For a state-to-state transition coupled by an operator to be allowed, the direct product of the irreducible representations (irreps) of all three components must contain the completely symmetric irrep of the point group you're working in. That is, we are determining whether the following integral is forced to be zero by symmetry: $$ \int \Psi_{i}(\tau) \...


11

My first thoughts are along the lines of your second-last paragraph. Here is a quick sketch of how to formalise it. From simple Hückel theory, you can obtain the coefficients of the AOs in the MOs: $$|\psi_i\rangle = \sum_a c_{n,i} |n\rangle$$ where $c_{n,i}$ denotes the coefficient of AO $|n\rangle$ in the MO $|\psi_i\rangle$. In Hückel theory the main ...


9

Even though the central atom has 4 different ligands this does not necessarily mean that the compound is optically active. The condition for that rather the following (from Wikipedia): A molecule is achiral (not chiral) when an improper rotation, that is a combination of a rotation and a reflection in a plane, perpendicular to the axis of rotation, ...


9

Ivan already stated quite briefly in the comments, that the planes you drew would transform an upward facing chlorine to a downward facing chlorine and vice versa, and that the same holds for the methyl groups. This is of course true and the first and most obvious point, why there is no symmetry plane. A more close investigation will, however, point out ...


9

If there is a molecule of $O$ (as opposed to $O_h$) or $I$ (as opposed to $I_h$) symmetry, it would be such a counter example (no mode that is both IR and raman active, and not centrosymmetric). http://symmetry.jacobs-university.de/cgi-bin/group.cgi?group=903&option=4 http://symmetry.jacobs-university.de/cgi-bin/group.cgi?group=905&option=4 ...


9

Have a look at the point groups O (not $\ce{O_h}$) I, $\ce{D_{5h} and C_{5h}}$. A pic of an O point group molecule is shown below. The $\ce{c_2}$ axes are not shown. Taken from molecule-viewer.com web site. The molecule is 'vanadium hexaoxo phosphonato' or more properly $\ce{C8H24O30P8V^+Cl^-}$, octakis(μ 3-methylphosphonato)-hexaoxo-penta- vanadium(v)-...


9

To find the ground state term symbol, you should be using symmetry and group theory arguments, you shouldn't have to resort to searching Tanabe-Sugano diagrams to get the answer. We'll start with octahedral complexes (the general idea can be extended quite easily to tetrahedral or square planar complexes). As requested in the question, I will only cover ...


9

In this complex there are two different 31P environments which are not related by symmetry: The two green phosphorus nuclei can be interconverted by a $C_2$ rotation (the rotation axis bisects the OC–Mo–CO angle), and so can the purple ones, but green and purple cannot be interconverted. As extra proof, consider that the green P is cis to both carbonyl ...


8

Diboron tetrachloride has the same symmetry as allene, which I will use to demonstrate the other two $\ce{C_2^{'}}$ axis. The back $\ce{CH_2}$ group is turned 90° in comparison to the front $\ce{CH_2}$ group. This conformation is most stable as it minimizes steric interactions between both groups. In this case, a picture tells more than I thousand words I ...


8

There seem to be two problems: the first is that your reducible representation for the $\ce{B-B}$ bond is wrong but your reducible representation for the $\ce{B-Cl}$ bonds seems to be correct and the second is that your book is apparently completely wrong. If I didn't make a mistake and my second assumption is correct then maybe you should think about ...


8

You can detect $O_{h}$, $I_{h}$ and $T_{d}$ symmetry by checking that a molecule has all of the subgroup symmetries of these point groups. According to this untitled document, which I presume1 is by W.C. Trogler, the elements are as follows: $T_{d}$: $E$, $4C_{3}$, $3C_{2}$, $3S_{4}$, $6\sigma{_d}$ $O_{h}$: $E$, $3C_4$, $4C_3$, $6C_2$, $4S_6$, $3S_4$, $i$,...


8

I refer you to the second link in my answer for How does one recognized Td/Oh symmetry in molecules?, which (according to the abstract) describes some high symmetry species of unusual point groups $T$, $O$ and $I$. Additionally, the molecule [6.6]chiralane has $T$ symmetry. Buckminsterfullerene is a well known $I_{h}$ symmetric molecule. Don't know of any $...


8

If you already know the symmetry of your site then it is quite easy. In a lot of books (e.g. this one) and on this web site you can find the character tables of the point groups supplemented with two additional columns which show the transformation properties of the basis vectors (e.g. $\ce{p}$ orbitals), their rotations and their quadratic combinations (e.g....


8

g/u is not a property of a molecule; I assume you meant molecular orbital. All functions can be written as a linear combination of even + odd functions: there is a short explanation on Wikipedia. Basically, you have a function $f(x)$; now define $$\begin{align} f_\mathrm e(x) &= \frac{1}{2}[f(x) + f(-x)] \\ f_\mathrm o(x) &= \frac{1}{2}[f(x) - f(-x)...


8

Why? Because there is no other choice. Starting from the AB6 octahedral configuration, all six vertices of the octahedron are symmetric, so it doesn't matter whichever one you “choose” to replace by the lone pair. All will yield the same final configuration.


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