# Why highly unstable cyclic boron compounds should be stabilised by two electron reduction?

I am trying to figure out why highly unstable cyclic boron compounds could be stabilised by two electron reduction...

My thoughts are because each Boron has an empty p-orbital perpendicular to plane of cyclic compound, pi-orbitals can form like diagrams below.

The two electrons would then go into the pi-bonding orbital, thus stabilising the entire compound.

Is this the correct explanation? If so, how can vibrational spectroscopy be useful in examining the bonding in the dianion?

• I don't have time for a full answer, but that's not the correct picture of the MO's. Two of the orbitals are degenerate. – jerepierre Sep 16 '15 at 7:12
• you are right. The triboracyclopropenyl dianion have been reported previously. – jjyaking Feb 18 '18 at 7:58

Your energy diagram is that of a linear moiety, like an allyl cation. For a cyclic structure, it would be somewhat different, but the general idea remains the same: there is one $\pi$-bonding orbital, and when two electrons go to it, everyone wins.