What structural features in an organic molecule would promote fast non-radiative decay to a ground state, for instance in an aromatic π-conjugated system ?
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Anything that has potential energy surfaces that cross or get close in energy. Offhand things like s-indacene and the indenofluorene family of compound, there are many others as well (azulene mentioned below). In general, if the $S_0$ and $S_1$ state for a $\pi$-conjugated organic come close in energy there is the possibility for all of that energy to be used on vibrations instead of luminescence. As far as predicive rules, that is a difficult thing to generalize. Perhaps you could argue that a low $S_0 \ce{->} S_1$ transition energy opens the possibility for non-radiative decay since these potential energy surfaces are not seperated by much energy, biradicaloid $\pi$ systems fall in this category.This is the same reason there is typically not emission from higher energy state, $S_n \ce{->} S_1$ transitions are usually quite close in energy and have crossings to get to $S_1$ efficiently through molecular vibrations. For further examples and explanations see, for example: http://www.sciencedirect.com/science/article/pii/S0009261414008847?via%3Dihub Or Josef Michl's excellent book "Excited States and Photochemistry of Organic Molecules" for more details on the topic.
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$\begingroup$ Not an expert on this by any means but azulene was the example that we were taught (ultrafast S1->S0 internal conversion). Then again, I don't know what "structural features" would lead to an intersection between the PES's. $\endgroup$ Commented Mar 20, 2017 at 14:08
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The cause fast non-radiative decay (excluding reaction, such as cis-trans isomerisation, as a non-radiative event ) is spin-orbit coupling which links singlet and triplet states and opens up a non-radiative pathway between them, in addition to fluorescence and internal conversion. The SO coupling 'breaks' the forbidden singlet-triplet selection rule. In internal conversion the operator is anharmonicity since no spin change is involved.
The second important factor, but only in the condensed phase, is the gap between the initial and final electronic states, the smaller this is the the larger non-radiative rate constant. As this is an application of a Fermi Golden Rule it is quite general and applies to all molecules not just aromatics. This answer explains it in detail Non-radiative transitions or what features of rhodamine result in it being so highly fluorescent?
Additionally heavy atoms, as a substituent or a solvent, increase non-radiative rate constants (again by spin-orbit coupling) as do paramagnetic species such as molecular oxygen dissolved in solution.
Should also mention that dipole selection rules for fluorescence /phosphorescence via excited & ground state symmetry means an additional factor is present to determine if the excited state is long or short lived.
