That’s because thiophene complexes preferentially coordinate η⁵, but at least one has been isolated. It does convert to η⁵. η¹-thiophene TM complexes aren’t very persistent, but to say they don’t exist is ridiculous.

I agree, but for the sake of pedagogy I think they should have indicated only one lone pair sticking out.

If anything, thiophene should be less “aromatic” than furan, right? Worse overlap of a 3p (S) vs 2p (O) orbital with the 2p (C) orbitals? So it’s not totally absurd to draw that orbital separately from the diene. And, as mentioned, the other two electrons are symmetry-forbidden from overlapping with the pi system, so they’ve got to be in a “lone pair”.

I think the acceptor properties have more to do with the low-lying d orbitals on sulfur than the presence/absence of lone pairs.

Oxygen lone pairs in furan have to be distorted to engage in the 6 pi business. In addition, it is too electronegative to contribute to resonance stabilization properly.

As you properly noted, sulfur’s unpaired electrons are not far from d-level. They are significantly more diffuse and more polarizable. I think the low basicity of thiophene can be explained by invoking a resonance structure in which the two lone pairs take turn to be in conjugation, effectively killing any attempts to find them in “lone” state.

As a consequence, of all five-membered aromatic rings, thiophene is the closest to benzene. Compare the physical properties of ThH & PhH and benzoTh and naphthalene.

Reading my comment again, I wasn’t nearly as clear as I meant to be. Basically, as I see it, the two lone pairs don’t really have the geometry shown above: whether we’re talking about furan or thiophene, one is pretty much your typical p orbital, perpendicular to the ring and capable of resonating, while the other is more of an sp2 hybrid – coplanar with the ring, pointed out, and incapable of overlapping with the pi system. Same orbital geometries as in a carbene or water (which does not have a tetrahedral arrangement of two lone pairs and two hydrogens). So I guess I don’t understand where distortion would come into play, or alternating resonance forms. You wouldn’t need to distort the perpendicular orbital to have it resonate; it’s already ideally oriented. And I can’t really see any way for them to take turns in conjugation.

As for the electronegativity of O vs S, I’m not really sure what to think of it. From an MO standpoint, the greater electronegativity of O should mean that its AOs start lower in energy than carbon’s, which leads to worse mixing/resonance. Which seems like it could very well be true. But are the carbon (2p) orbitals really farther in energy from oxygen (also 2p) than sulfur (3p)? And then there’s the geometrical overlap – the sulfur orbitals have an extra node in them, right around where I’d expect the maximum amplitude of the carbon orbitals to be. My first thought on this was really just based on an analogy to the bonding in acetone vs. DMSO. From electronegativities alone, you’d think the C=O would be more polar than S=O. But, from what I’ve heard, there’s much more even charge distribution between C and O than S and O, because of the bad overlap between the 3p and 2p orbitals in the latter. One professor I had said it shouldn’t even be considered a pi bond. More like a zwitterion, with a negative charge on the O and positive on the S. Don’t know how much I buy that (the prof wasn’t much of an orbital guy himself), or how well it extends to this discussion. I’m just guessing at all this stuff, so don’t mind me.