[Vital Y]

I tried obtaining the pi SALCs of naphthalene. I got the reducible representation as 2A_1u + 3B_1u + 3B_3g + 2B_2g. I thought that if I considered the carbons (1,4,5,8) and (2,3,6,7) to belong to two D_4h groups and the carbons (9,10) belong to a C₂ group, I could find the SALCs without the projection operator (The link is an image of naphthalene with the carbons in it numbered). I tried using D_4h point group's D₂ sub group but the irreps do not have the same symmetry properties in the two point groups. Is there any way to obtain the SALCs of naphthalene without using the projection operator?

https://upload.wikimedia.org/wikipedia/commons/6/61/Naphthalene-numbering-2D-skeletal.png

[Corribus]

Not that I'm aware of. What would be the point, anyway? The projection operator is easy to use, so why complicate your life by trying to use multiple symmetry groups?

[Vital Y]

I find using the sub group approach easier. When calculating benzene's SALCs you can use the C₆ point group's character table and directly obtain them.

[mjc123]

You are aware that naphthalene doesn't have D_4h symmetry?

[Vital Y]

Yes I am aware that it doesn't have D_4h as its point group, but instead has D_2h. My question is even if it possess D_2h can we obtain its SALCs using the idea of dividing the molecule into regions and then finding those region's point groups.

[Corribus]

Even if you could reduce the complexity of the problem in the way you are attempting to, the grouping {2,3,6,7} wouldn't belong to D_4h anyway (they basically form a rectangle, not a square), so it's no surprise that your orbitals are not transforming properly.

[Vital Y]

So does this mean that (1,4,5,8) and (2,3,6,7) belong to the D₂ group? How will we relate A_1u, B_1g, B_2g, B_3g with D₂'s irreps? B_2g and B_3g can be related to B₂ and B₃ of the D₂ point group, but I couldn't find a relation for A_1u and B_1u.

[Irlanur]

What you could try is to obtain the MO's of cyclodecapentaene in the symmetry of naphtalene and then use a perturbation approach to get to the MO's of naphtalene. I doubt that this is easier than doing it directly for naphtalene.