Past exam of the chemistry B course of the University of Cambridge / 2022 / Paper 2 / B9 / c / Solution

Join three trigonal BO${}_3$ units corner to corner. The resulting alternating six-membered B–O ring has one terminal oxygen on each boron:The superscript records the total charge; the three explicit O${}^{-}$ labels each carry one share of it. Atom counting gives three B, three bridging O and three terminal O. Boron is three-coordinate and approximately trigonal planar. The original structure diagram shows the ring connectivity and terminal atoms.

A planar arrangement permits the perpendicular empty boron p orbitals to accept pi donation from suitable oxygen lone pairs. The terminal oxygens can donate effectively toward one boron, giving their B–O linkages more multiple-bond character and explaining why they are shorter than bridging B–O linkages. A bridging oxygen shares its interactions between two boron centres. An approximately planar ring is therefore expected from the trigonal units and favourable orbital alignment. This reasoning does not require describing the entire anion as benzene-like with an assumed six-electron aromatic ring; the oxygen lone pairs and terminal donations must actually be counted before making an aromaticity claim.