# Tag Info

28

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 ...

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

Omitting j when alphabetically enumerating things has a long tradition. First of all, the alphabet did not always exist in the form we know it today. Quoting Wikipedia: After [...] the 1st century BC, Latin adopted the Greek letters ⟨Y⟩ and ⟨Z⟩ [...] Thus it was during the classical Latin period that the Latin alphabet contained 23 letters: [no J, V, W] [.....

25

The figures below show you how to navigate your way round point group tables. The irreducible representations (irreps) are shown as the row of characters. A reducible representation is a collection of the irreps that can be reduced to a number of irreps. I will post about this shortly in the meantime some reminders about definitions, important but tedious. ...

22

For the azimuthal quantum number (l) of an atom, there is no "j" because some languages do not distinguish between the letters "i" and "j". L is the total orbital quantum number in spectroscopic notation and uses capital letters. The nomenclature just follows suit with the suborbital notation and skips J since there is no corresponding j.

21

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 ...

20

It is not that we need the identity operator. It is just that things (character tables, irreducible representations, etc.) work the way they do. As to why they do so, I refer you to any decent textbook on group theory. I guess one could rebuild the entire theory from scratch (that is, reformulate all definitions, starting from the very definition of a group,...

20

What you are asking arises as a fundamental consequence of the definition of vector spaces and the operations defined for them and, specifically, symmetry groups. The identity operator does not do "nothing" (see below) and it does belong in the formal definition of a group, and it is included to satisfy said definition. From Byron and Fuller's Mathematics ...

17

By Maschke's Theorem, every direct product of representations is decomposable into a direct sum of representations, that is, the function you are integrating can be rewritten as a sum of functions with symmetry properties equal to specific irreducible representations. As you note, integrating an odd function (which has a certain symmetry, namely ...

16

As Tyberius noted, the projection formula does not work for infinite order groups (this is because the Hermitian form on characters is defined to be G-invariant by averaging over all elements in a group. That is, it works because it is possible to hit all the elements of the group in some order). How can one work in the infinite dimensional groups then? One ...

15

Group theory really just formalises the process that you're going through when you construct molecular orbital diagrams by inspection (though as the molecules get bigger it gets significantly harder to intuitively guess at what will mix and in what combinations). In order to construct an MO diagram for water, we'll take a stepwise approach: first, ...

14

Point groups are a very valuable tool for analysing molecules without knowing much about them. They rely on analysing a molecule’s symmetry and deducing both physical parameters and the shape of molecular orbitals from the information gathered. The downside is that one must know a molecule’s geometry before one can apply group theory to it, but for most ...

14

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.: $$E = \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 &... 14 The "symmetry" in symmetry-adapted perturbation theory refers to the anti-symmetry of the wave function with respect to electron exchange:$$\Psi(\mathbf{r}_1,\mathbf{r}_2) = -\Psi(\mathbf{r}_2,\mathbf{r}_1) $$The most usual way to enforce antisymmetry in electronic structure theory is to enforce orthogonality between spin-orbitals. So SAPT is ... 13 Because this is a diatomic molecule, there are no group orbitals. Put another way, the group orbitals are the molecular orbitals. Knowing the nitrogen atomic orbitals (AOs) and their irreducible representation (irrep) labels is enough. Since we'll work in the D_{\mathrm{2h}} point group, we need its character table:$$\begin{array}{c|cccccccc|cc} \hline ...

12

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 You got everything right. It is just that two of these representations would use complex numbers (\varepsilon=e^{2\pi i\over3} and \varepsilon^2), and we don't like that, so we stick them together into one representation of dimension 2, which makes the imaginary part vanish. Afterwards we treat this representation as "kinda irreducible". In fact it can ... 9 Well, according to this Math StackExchange post, you need the identity element because it is impossible to define an inverse element a if you don't know about the identity element e. The definition of a group includes the identity element (below). There exists an element e in [Group] G such that, for every element a in G, the equation e • a = a • e = a ... 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 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 Symmetry operator E is the lowest symmetry a molecule may possess; and indeed all objects possess this operator. Yet, among molecules revealing E as the only the operation are the ones exhibiting stereogenic centred chirality, like \ce{CFClBrI}. Hence E is of value, not redundant, and needed. 8 Part 1 Are there any molecules lacking plane of symmetry/center of inversion but that are achiral due to presence of an axis of improper rotation? Yes, although such examples are very rare. In general, the point groups S_{2n} (n \geq 2) are where you should look. These molecules have a S_{2n} rotation axis, but no plane of symmetry, and no inversion ... 7 At first I thought the character table was counting both positive and negative rotations, but then the order for the C_2' and C_2 rotations should be doubled if that were the case (12C_2' and 6C_2). You were very close. The missing symmetry operations are indeed ‘positive and negative’ but only those that lead to non-identical transformations. In a ... 7 General Chemistry perspective: Looking at the molecular orbitals of water, we can see that the oxygen is \mathrm{sp}^3 hybridized. No. One could say that, but \mathrm{sp}^2 is equally possible. In fact, it may seem more likely, we’ll get to that in a second. Group Theory and Quantum Chemistry perspective: Water belongs to C_{2\mathrm{v}} point group. ... 7 Distinguishing between i and j is, as others have mentioned, a rather recent phenomenon, not unlike the distinction between u and v. Especially in German when typesetting in blackletter, the glyphs for capital I and J would be identical and there are still a few (older) road signs around Germany that use a capital J where the letter is in fact an I going by ... 7 Instead of trying to use symmetry tables it is possible to calculate the vibrational normal modes directly using the equations of motion, rather as would be done for a double pendulum for example. It is normally assumed that the vibrational motion of the atoms is harmonic, i.e. Hook's law applies, and we additionally assume that the only force constants are ... 7 EDIT: Made this more rigorous. (I think. Feel free to critique). First, let's decompose \psi into components that correspond to how its representation reduces. That is, if$$\Gamma_{\psi} = \Gamma_{1}\oplus\Gamma_{2}+\ldots$$we write$$\psi = \phi_{1} + \phi_{2} + \ldots  where the symmetry of $\phi_{i}$ is captured by the $\Gamma_{i}$. Suppose $\... 7 Now, there are a few things that must be stated very clearly. Firstly, it pays to be very careful when you are talking about point groups. It's careless (and confusing to others) to write$\mathrm{T_{2u}}$when you mean$\mathrm{T_{1u}}$, or to write$\mathrm{A_{g}}$when you mean$\mathrm{A_{1g}}\$, so please pay attention to this next time, regardless of ...

7

The symmetry of a state is the MO (molecular orbital) product of its occupied molecular orbitals. Let's take the example of water molecule. We can see the symmetry label of each of the MO in the diagram. Each MO is assigned an irreducible symmetry label (associated with the point group of molecule) which can be deduced from the symmetry of MOs in space (or ...

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