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All phosphorescent molecules go through the following transitions:
$$\text{excited singlet state}$$ $$\Bigg\downarrow$$
$$\text{[intersystem crossing]}$$
$$\Bigg\downarrow$$
$$\text{ excited triplet state}$$
Why does that spin change always occur, given it’s a “forbidden” transition and, thus, unfavourable and much less likely to take place?

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  • $\begingroup$ Well, if there's no easy path, unfavourable needs to be taken. $\endgroup$ – Mithoron Apr 18 '18 at 22:47
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    $\begingroup$ The triplet phosphorescence is also spin forbidden as is $S_1\to T$. As it actually occurs there is an allowed spin changing mechanism that lets this happen. This is spin-orbit coupling. See this answer for an explanation of radiationless transitions in general of which intersystem crossing is an example. chemistry.stackexchange.com/questions/28883/… $\endgroup$ – porphyrin Apr 19 '18 at 6:23
  • $\begingroup$ It doesn’t always occur,nor favorable. $\endgroup$ – Greg Apr 20 '18 at 8:19
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There are two things going on here. You have conflated with "forbidden" with unfavorable. In many systems, the first excited triplet is lower in energy than the first excited singlet. Hence, the order of energies excited singlet > triplet > ground state singlet is correct and sensible. If your question is "why is this the order?" that should be a separate question.

The forbidden part of the transition has to do with the fact that, because of conservation of (electron spin) angular momentum, you can't excite directly from a singlet ground state to a triplet excited state with a photon. The forbiddenness is a statement about not being able to get there, not unfavorability. The triplet is still lower in energy and the molecule has to do some other shenanigans (often involving spin orbit coupling) to do the intersystem crossing into the lower energy triplet state. If your question is "what is the mechanism of intersystem crossing?", that should be a separate question.

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  • $\begingroup$ Pardon my limited knowledge of quantum physics, for I really thought the first excited triplet is actually lower in energy. On spin orbit coupling, it appears it’s just the thing I needed, so I’m going to read about it, thanks a lot! $\endgroup$ – Rusty Apr 19 '18 at 17:27

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