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I have just started reading group theory and their applications in the MO theory. I tried to get the equations of the pi molecular orbitals of allyl system using group theory, but I am stuck at the last part.

This is what I did so far—

  1. Determined the point group of the allyl system to be $\ce{C_{2v}}$.
  2. Used the 3 p orbitals as basis.
  3. Found out the reducible representation. Reduction gives $\ce{A2 + 2B1}$

I am stuck at this point. I named the three p orbitals of C atoms as $\ce{\psi1, \psi2, \psi3}$. However, when I try to use the 'projection operator method' using the $\ce{\psi1}$ as generation vector, it gives me $\ce{\psi1 +\;\psi3}$ and $\ce{\psi1 - \psi3}$ (unnormalized). I don't know how to proceed. Pi orbitals of allyl anion [image copied from Master Organic Chemistry webpage ]

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  • $\begingroup$ The projection operator method only relates symmetry-equivalent orbitals, so if you ‘start’ from ψ1, you can’t account for ψ2. Symmetry alone cannot tell you how much of ψ2 goes with ψ1, because any linear combination of ψ2 and (ψ1 + ψ3) will still transform as B1. $\endgroup$ – orthocresol Jul 22 at 10:26
  • $\begingroup$ @orthocresol Then what do I need to do to get the component of $\ce{\psi2}$? $\endgroup$ – Shoubhik Raj Maiti Jul 22 at 10:54
  • $\begingroup$ The coefficients of the AOs in the bonding and antibonding MOs will depend on the energies of the constituent AOs. Hückel theory is the simplest (also the most simplistic) way of getting the coefficients. $\endgroup$ – orthocresol Jul 22 at 11:33
  • $\begingroup$ @orthocresol, Ok, I will try reading about Huckel theory. But can I get the same result by applying group theory? $\endgroup$ – Shoubhik Raj Maiti Jul 23 at 9:33
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    $\begingroup$ Symmetry, or group theory, can only tell you whether a coefficient is positive, zero, or negative. It cannot give you an actual number. $\endgroup$ – orthocresol Jul 23 at 9:48
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The calculation below is not exactly the same as yours but I had it to hand and it is very similar. The bit you specifically want is towards the end. In your case you add will vectors along the $\pi$ orbitals.

What projection operators operators do is to extract the symmetry adapted functions from the basis functions used to form them and so produce a linear combination L of the basis functions. Many basis functions can be chosen, but the easiest to use are usually vectors pointing along bonds or those representing a p$\pi$ orbital. To form a linear combination, a 'victim' vector is chosen and operated on with each symmetry operation in the point group, $C_2\,, \sigma(xz)$, etc. to find what vector this turns into. This vector's name (the basis function) is then multiplied with the character from the point group corresponding to that symmetry operation and symmetry species. Finally, the names are added together by moving from one symmetry operation to the next along the top row of the point group. We expect the result to be the sum or difference of the displacement vectors for a molecular vibration, such as $2v_1 + v_2 - v_3 \cdots$ and so forth. The same formula or algorithm is used if the combinations of atomic orbitals that make up molecular orbitals are sought. The equation to make a linear combination is

$$ L_M=\frac{d}{h}\sum_{j=1\cdots h}c_jS_j(v)$$

where $M$ is the symmetry species label, $A_g, B_{3g}$, and so forth, $h$ is the order of the group, $d$ the dimension of the irreducible representation, and the sum is over all the classes. Because the resulting vector will be normalized d/h can be ignored. One further point is important. In the $C_{3v}$ point group, for example, there is a heading $2C_3$ as there are two members in this class, and this must be split in the summation into $C_3^+$ and $C_3^-$. This is because moving a vector $120$ degrees to the right, say, will turn it into a different vector than turning 120 degrees to the left, similarly, any mirror planes must be separated out. The character for symmetry species $M$ is $c_j$ and $S_j(t)$ is the effect that operator $S$ in column $j$ of the point group has on our victim vector $v$. The result of this operation is to produce a vector whose name is recorded. For example, if $S$ is the identity then the result is $E_1(v) = v$, other operations may leave $v$ unchanged or change it into another vector. Finally, note that if the reduced representation of the vibrations or molecular orbitals contains two or more symmetry species of the same type, e.g. $2A_1$, then two or more different victim vectors will have to be chosen to obtain all the linear combinations.

In H$_2$O, SO$_2$%, and other triatomics with $C_{2v}$ symmetry, it is clear that two of the vibrational modes stretch the bonds and one changes the bond angle. To work out the normal mode vectors, some representative vectors are placed along the bonds as in the figure. You can choose where to put these but physically realistic choices will usually make the calculation simpler.

c2v vectors

In $C_{2v}$, the normal mode vibrations have been found to comprise the symmetry species $2A_1$ and $B_1$. The $L_M$ formula is shown as a table with $v_1$ as the victim vector. The table also shows the characters for each symmetry species.

$$\begin {array}{l|llll|l} C_{2v} & E & C_2 & \sigma(x,z) & \sigma'(y,z) & L_m \;(d=4,h=4)\\ \hline A_1& 1&1&1&1\\ L_{A1} & v_1 & v_2& v_1& v_2& 2(v_1+v_2)\\ \hline A_2& 1&1&-1&-1\\ L_{A2} & v_1 & v_2& -v_1& -v_2& 0\\ \hline B_1& 1&-1&1&-1\\ L_{B1} & v_1 & -v_2& v_1& -v_2& 2(v_1-v_2)\\ \hline B_2& 1&-1&-1&1\\ L_{B2} & v_1 & -v_2& -v_1& v_2& 0\\ \hline \end{array}$$

Notice that what the vector changes into is (obviously) the same for each symmetry species as this is determined by the symmetry operations. Only the value of the character changes in front of each term The final $L_M$ is not multiplied by $d/h$ because the resulting vectors are instead normalized giving, $L_{A1} = (v_1 + v_2)/ \sqrt{2}$ and $L_{B1} = (v_1 - v_2)/\sqrt{2}$. Note also that $v_3$ and $v_4$ do not enter into this table, they are at $90$ degrees to $v_1$ and $v_2$ and no operation in this point group can inter-convert them.

There are two $A_1$ species produced from the reducible representation but only one has been found so far. This is because the vectors $v_1$ and $v_2$ cannot produce a bend, which is the other normal mode. Using vector $v_3$ as the victim this species appears. As this mode has $A_1$ symmetry all the characters $c_j$ in the $L_M$ formula are $1$, making the calculation easy and the combination is shown below.

$$\begin {array}{l|llll|l} C_{2v} & E & C_2 & \sigma(x,z) & \sigma'(y,z) & L_m \\ \hline L_{A1} & v_3 & v_4 & v_3 & v_4 & 2(v_3+v_4) \\ \hline \end{array}$$

which describes the bending mode shown in the figure in terms of its unit vectors. When normalized this mode is $(v_3 + v_4)/\sqrt{2}$. (As the centre of mass cannot move during a vibration, there being no force to make it do so, therefore the O atom has to move a small distance away from the H atoms. Vectors could be added onto the O atom to show this and $L_M$ terms re-calculated.

In the case of degenerate symmetry species as occurs in the $C_{3v}$ and most other point groups the results of the summation $L_M$ is to produce two or more vector sums that are not orthogonal. Either these have to be made orthogonal, by choosing another victim for the degenerate term and then using the Gram-Schmidt method to make the two vectors orthogonal. R. Carter 'Molecular Symmetry and Group Theory' and A. Vincent 'Molecular Symmetry and Group Theory' both give method by which to do this.

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  • $\begingroup$ But how do I apply this to the MO problem? By using the $\psi_1$ and $\psi_3$ as generating vectors, I have got $\psi_1 + \psi_3$ and $\psi_1 -\psi_3$. However, the orbitals of allyl system as we can see from the diagram, should be $\psi_1+\psi_2 + \psi_3$, $\psi_1-\psi_2+\psi_3$ and $\psi_1 -\psi_3$. So, I am getting at least one orbital wrong, even if I use another set of generating vectors. $\endgroup$ – Shoubhik Raj Maiti Jul 23 at 9:41

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