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Pentalene is one of the most famous antiaromatic molecules. But I obtained its energy of $\pi$ system as $8\alpha+10.46\beta$ by Hückel method (with a bit complicated but straightforward calculation). This value is lower than that of localized form $8(\alpha+\beta)$. So Hückel method looks to be not able to explain antiaromaticity. Are there any misunderstandings? It seems that many of antiaromatic molecules are predicted lower than their all-ethylene counterparts.

(I don't think radicality is a key for antiaromaticity, and I'd like to know a general, quantum mechanical explanation about antiaromatic molecules if it exists.)

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    $\begingroup$ Aromatic or antiaromatic, it is still conjugated, so no wonder it is better off than the all-ethylene analog. The general explanation is precisely where you don't think it is. $\endgroup$
    – Ivan Neretin
    Commented Mar 21, 2019 at 15:31
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    $\begingroup$ With Huckel MOT, aromaticity / antiaromaticity should be determined with reference to a linear conjugated analogue, so e.g. cyclobutadiene is compared with butadiene, not two ethylenes. I am not sure how this carries over to pentalene. Arguably, Huckel's rules were designed for monocyclic systems, not polycyclic ones. $\endgroup$
    – orthocresol
    Commented Mar 21, 2019 at 15:40
  • $\begingroup$ I solved the secular equation of pentalene, whose matrix has two resonance integral in addition to that of cyclooctatetraene. The solutions have no degeneracy, so I think penetalene is not expected to be radical. $\endgroup$
    – Sakai
    Commented Mar 22, 2019 at 4:31
  • $\begingroup$ @sakai right. Pentalene is not a radical. But it will have a low-lying acceptor orbital and thus is highly reactive an an electrophile. Which, of course, is what actually happens. Fill the low-lying acceptor orbital and the resulting dimension is quite stable. $\endgroup$
    – Oscar Lanzi
    Commented Aug 10, 2020 at 22:08

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