Is this heterocyclic boron molecule aromatic?

What kind of aromaticity does the following molecule have?

I read somewhere that this is an aromatic compound, but I believe that in order to be aromatic, it should have continuous resonance. I also think that there should be a break at B that causes no $\pi$ electrons to be shared among the B—B atoms...so I think this is non-aromatic. Is that correct?

That molecule is indeed aromatic.

It satisfies Huckel's rule ($4n+2$) with $n=0$. It's also planar.

Lastly, the boron atom has three electron domains, meaning it can exist with $sp^2$ hybridization. The left-over p-orbital, whilst not actively forming $\pi$ bonds, can still complete a stable $\pi$ molecular orbital.

• Okay..I get your point...but could you tell me what all points should I consider while judging if a molecule is aromatic or not ? – Piyush Kumar Dec 31 '16 at 12:49
• I mean specifically such kind of molecules wherein there is no actual pi electrons dispersion – Piyush Kumar Dec 31 '16 at 14:20
• There is some delocalisation but it doesn't necessarily mean it's really aromatic. As for answer, there are orbitals not domains, and satisfying Huckel doesn't automagically mean the compound is truly aromatic. – Mithoron Dec 31 '16 at 20:07
• Hückel's rules only work for the compounds that it is derived from: monocyclic hydrocarbons. – Martin - マーチン Jan 28 '17 at 10:42