[pm133]

What this suggests to me is that your doublet is delocalised across the conjugated pi system.
Can you confirm that the carbons with positive electron density are all in the conjugated system next to where you removed the electron? The negative densities might be across the same system but on alternating carbons. You should check this too.

Remember that when you remove electrons from a molecule, the orbitals and remaining electrons don't stay static. They "relax" by feeling the change caused by removal of the other electron and will re-adjust accordingly. That unpaired electron would need to be somewhere pretty stable to stay in that single place. If you give it a chance, it will move and I think that's what has happened here but I can't see your entire molecule.

Playing around in Gaussview with the isoval value will remove the smaller densities and reveal where the most important delocation is happening.

[darkdevil]

What this suggests to me is that your doublet is delocalised across the conjugated pi system.
Can you confirm that the carbons with positive electron density are all in the conjugated system next to where you removed the electron? The negative densities might be across the same system but on alternating carbons. You should check this too.

Remember that when you remove electrons from a molecule, the orbitals and remaining electrons don't stay static. They "relax" by feeling the change caused by removal of the other electron and will re-adjust accordingly. That unpaired electron would need to be somewhere pretty stable to stay in that single place. If you give it a chance, it will move and I think that's what has happened here but I can't see your entire molecule.

Playing around in Gaussview with the isoval value will remove the smaller densities and reveal where the most important delocation is happening.

Hi, the molecule is a conjugated system, and I also expect to have the radical or charge to be delocalized. That is how the material would become electrically conductive in reality. Going back to the original post, I am trying to find out if it is possible that the radical would have a chance to stay on certain bonds to allow the radical dissociation I suggested.. Even if there is only 1% chance that the radical cation would stay at the C-H bond in molecule A. It might also be possible for the radical cation to decay and leave the system. The molecule would then lose its radical cation (polaron) property and would be not conductive (This dissociation process could be a slow process eg. within days or weeks in reality which we observed by the conductivity change). This is what we observed in lab with a decaying conductivity for molecule A but not for molecule B ,and I am trying to propose something based on DFT calculations. Perhaps we can now have some insight on explaining why molecule A is less stable than B, but at the same time led to more questions on "How exactly it can do that?". lol

Thank you for your comments again! The DFT is a really powerful tool

[pm133]

OK so it might be worth comparing those same spin density numbers for both systems to see if there'a difference in where the unpaired electron appears.

If you want to look at the time perspective then static DFT snapshots won't help.
You might need TDDFT which is something I think Gaussian does but I have never used it so can't help you with.

I would consider running a series of single point calculations as you gradually dissociate the bond of interest from the equilibrium structure (find and optimise that transition state structure) and see how the spin density values change but really I'm not sure whether that will work accurately.