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For a compound to be aromatic, each atom in the ring must be a part of conjugation. Boron here has no lone pair, so how can 1⁠H-borepine be aromatic as I was told by someone?

Structure of 1H-borepine

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1H-borepine is isoelectronic to the cycloheptatrienyl cation. In both cases, you have a planar ring of seven atoms, whose hybridisation is close enough to $\mathrm{sp^2}$ to allow for a third p-orbital to be perpendicular to the ring plane. And in both, you are then filling these seven π orbitals with 6 π electrons, resulting in a Hückel-aromatic system ($4n+2 = 6$ for $n=1$).

There is no requirement for each atom along the chain of conjugation to have a lone pair; for example, in benzene no carbon atom has a valence lone pair. The only requirement for Hückel aromaticity is through conjugation, i.e. each atom must have an accessable p-type orbital that can overlap with the neighbouring atoms’ p-orbitals in a favourable π manner. However, neutral aromatic systems with empty p-orbitals are rare if boron is not used as most non-metals do not have favourable electronic configurations.

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    $\begingroup$ We can make contributing structures in cyclopropenone where a $p$ orbital on carbon is formally empty, the carbon is positively charged (cyclopropenyl cation), and the negative charge is on a substituent (oxygen). Such an aromatic zwitterion structure is not really accurate, but cyclopropenone shows properties consistent with a zwitterionic contribution (basicity, dipole moment). $\endgroup$ – Oscar Lanzi Oct 25 '17 at 16:49
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    $\begingroup$ @OscarLanzi Interestingly, a similar (but opposite) effect is apparant in fulvene in which, when calculated, the exocyclic double bond seems to be almost more accurately represented by an exocyclic cation and a 6 pi electron ring system. $\endgroup$ – Jan Oct 26 '17 at 2:37

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