Chemistry · Chemical Bonding · NEET
A banana bond is a 3-centre-2-electron bond (3c-2e). In a normal covalent bond, 2 electrons hold 2 atoms together. In a banana bond, only 2 electrons hold 3 atoms together (B-H-B). The bond curves like a banana because the electron cloud bends outward, not straight between the atoms. Diborane has TWO such banana bonds, one on each side, in the B-H-B bridges.
Count the electrons. B2H6 has 2 boron (3 valence electrons each = 6) plus 6 hydrogen (1 each = 6), so 12 valence electrons total. To draw 8 normal 2-electron bonds you would need 16 electrons, but you only have 12. There are not enough electrons for all normal bonds. So nature makes 2 banana bonds that share electrons over 3 atoms. That electron shortage is why B2H6 is electron deficient. This matters for NEET because BeCl2, BCl3 and B2H6 are classic electron-deficient examples.
There are exactly TWO 3c-2e (banana) bonds. Each banana bond is a B-H-B bridge holding one bridging hydrogen between the two boron atoms. There are also FOUR normal 2-centre-2-electron bonds (the terminal B-H bonds). So total = 4 normal bonds + 2 banana bonds.
Boron is sp3 hybridised in diborane, NOT sp2. This is a very common NEET trap. Even though BF3 and BCl3 have sp2 boron, in diborane each boron uses 4 orbitals (two for terminal H, two for the bridge bonds), so it is sp3. Remember: free BH3 would be sp2, but the diborane dimer B2H6 is sp3.
There are 4 terminal hydrogens and 2 bridging hydrogens. The 4 terminal H atoms lie in the same plane as the 2 boron atoms. The 2 bridging H atoms lie above and below that plane. So the molecule is NOT flat overall, only the 4 terminal H and 2 B are in one plane.
In the bridge, only 2 electrons are shared among 3 atoms (B-H-B), so the bonding electron density is spread thin. In the terminal bond, 2 electrons hold just 2 atoms, so it is stronger and shorter. That is why bridge B-H bonds (~133 pm) are longer than terminal B-H bonds (~119 pm).
Which of the following statements is not correct about diborane?
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Diborane is not flat. The 2 boron atoms and the 4 terminal hydrogen atoms lie in one plane. The 2 bridging hydrogen atoms sit above and below this plane, making the bridge region look like a bent B-H-B structure.
Diborane has 4 normal 2-centre-2-electron B-H bonds (terminal) and 2 three-centre-2-electron B-H-B banana bonds (bridge). Total 6 bonds but only 12 valence electrons are used.
BH3 does not have enough electrons for a stable octet on boron (only 6 electrons around B). To become more stable, two BH3 units join to form B2H6, sharing hydrogens through banana bonds. That is why diborane exists as a dimer, not as BH3.
Yes. Diborane is the textbook example of hydrogen-bridge bonding (banana bonds). Do not confuse this with hydrogen bonding in water. Here the hydrogen atom is chemically bonded in the bridge using a 3c-2e bond, not a weak intermolecular attraction.
