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

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

enter image description here

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?

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    $\begingroup$ 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. $\endgroup$ – jerepierre Sep 16 '15 at 7:12
  • $\begingroup$ you are right. The triboracyclopropenyl dianion have been reported previously. $\endgroup$ – jjyaking Feb 18 '18 at 7:58
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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.

As for vibrational spectroscopy, well, it can help you tell the cyclic structure from linear, if that's what you want.

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