Q8(b)(i). Structure and bonding of diborane

Structure

• The two B atoms and the four terminal H lie in one plane. The two bridging H lie above and below this plane, in the plane at right angles to it (D₂ₕ).
• Terminal B–H 119 pm; bridge B–H 133 pm, longer and weaker. B…B 177 pm. Terminal H–B–H 122°, bridge H–B–H 97°. Each boron is surrounded by a distorted tetrahedron of H atoms.

Bonding

1. Boron (2s²2p¹) is promoted to 2s¹2p² and forms four sp³ hybrids: three hold one electron each, the fourth is empty.
2. On each boron two hybrids overlap with H 1s orbitals: four terminal 2c–2e bonds, 8 electrons.
3. Each boron is left with one half-filled and one empty hybrid; the two bridging H bring one electron each. Four electrons remain for the two bridges.
4. In each bridge the hybrid of one B, the H 1s orbital and the hybrid of the other B combine into three MOs: bonding, nonbonding and antibonding. Only the bonding MO is filled, with 2 electrons, which spread over three atoms: a three-centre two-electron bond (banana bond). Each B–H of a bridge has a bond order of ½.
5. Count: 4×2 (terminal) + 2×2 (bridges) =12. Fourteen atomic orbitals give enough bonding MOs for 12 electrons, so the molecule is stable although it cannot have a classical Lewis structure.

Evidence

• Electron diffraction gives the two different B–H lengths; IR and Raman spectra show two kinds of B–H stretch.
• ¹H and ¹¹B NMR show terminal and bridging hydrogens in the ratio 4: 2.
• Lewis bases cleave the bridges, which shows they are the weak point: B₂H₆ + 2NMe₃ → 2Me₃N.BH₃ (symmetric), and B₂H₆ + 2NH₃ → [BH₂(NH₃)₂]+[BH₄]⁻ (unsymmetric). The molecule is diamagnetic.

Application. 3c–2e B–H–B bridges and B–B–B bonds hold the higher boranes together; diborane itself is the hydroboration reagent (as BH₃.THF) and with NaH gives NaBH₄ (2NaH + B₂H₆ → 2NaBH₄).