Diborane, B2H6

Diborane, $\text{B}_2\text{H}_6$, is the simplest and most important boron hydride, and is the standard textbook illustration of electron-deficient bonding. A straightforward electron count shows the difficulty: each boron atom contributes 3 valence electrons and each hydrogen contributes 1, giving $2(3) + 6(1) = 12$ valence electrons in total — only 6 electron pairs. Yet the molecule contains 8 B–H "connections" (if every hydrogen were bonded normally to just one boron by a conventional 2-electron bond), which would require 8 electron pairs, i.e. 16 electrons. There simply are not enough electrons for 8 ordinary two-centre two-electron (2c–2e) bonds — hence diborane is described as electron-deficient.

Diborane resolves this shortfall through its distinctive structure. Of the six hydrogen atoms, four are terminal: two are bonded to each boron by perfectly ordinary 2-centre-2-electron $\sigma$ bonds, lying in the plane of the two borons. The remaining two hydrogens are bridging: they sit above and below the B···B axis, each simultaneously bonded to both boron atoms through an unusual three-centre two-electron (3c–2e) bond — a single pair of electrons delocalised over three nuclei (one hydrogen and both borons) rather than confined between just two. These two bridging B–H–B bonds are often called "banana bonds" for their curved, bent shape, and together they hold the two $\text{BH}_2$ halves of the molecule together. Each boron atom in this structure is best described as roughly $sp^3$ hybridised, with two hybrid orbitals used for the terminal B–H bonds, and the remaining two hybrid orbitals from each boron combining with one hydrogen $1s$ orbital to build each 3-centre bridge bond. The bridging B–H–B angle is noticeably more acute than a regular tetrahedral angle, reflecting the geometric constraints of squeezing two boron atoms and a hydrogen into a single delocalised bonding region.

Diborane is prepared in the laboratory (or on an industrial scale via related routes) by reacting boron trifluoride with a hydride source; a standard preparation is:

$2\text{BF}_3 + 6\text{LiH} \longrightarrow \text{B}_2\text{H}_6 + 6\text{LiF}$ …