Q.Diborane, , has only 12 valence electrons but must account for 8 B–H connections. Describe its structure, distinguishing the four terminal hydrogens from the two bridging hydrogens, and explain what a three-centre two-electron (3c–2e) bond is.
You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.
Start your 14-day free trial to unlock the full solution →Diborane, , has a total of valence electrons, i.e. only 6 electron pairs. If all 8 of its B–H connections were ordinary two-centre-two-electron (2c–2e) bonds, 16 electrons (8 pairs) would be required — two more pairs than are actually available. Diborane resolves this shortfall with a structure in which the six hydrogens play two distinct roles. Four hydrogens are terminal: two bonded to each boron by conventional 2c–2e bonds lying in the plane containing both boron atoms, exactly as in a simple fragment. The remaining two hydrogens are bridging, positioned above and below the B···B axis; each bridging hydrogen is simultaneously bonded to both boron atoms at once through a single, delocalised three-centre two-electron (3c–2e) bond — one pair of electrons spread across three nuclei (the bridging H and both B atoms) rather than confined between just two. These two 3c–2e bridge bonds, often nicknamed 'banana bonds' for their curved shape, are what hold the two halves of the molecule together and account for diborane's remaining electron count exactly ( terminal 2c-2e bonds pairs, plus bridging 3c-2e bonds pairs, totalling pairs electrons). Each boron atom is best described as roughly hybridised, using two hybrid orbit …
Unlock everything free for 14 days
- Full step-by-step solutions
- Concept-first explanations
- Methods, shortcuts & mistakes
- PYQ mapping + timed mock tests
Full access for 14 days. No credit card required.