Q.Draw the structure of Diborane. Write a note on the nature of bonds present in it.

Diborane's structure has two BH2 groups linked by two bridging hydrogens; the terminal B–H bonds are normal 2-electron covalent bonds, but the two bridge B–H–B bonds are unusual 3-centre-2-electron bonds, since diborane simply doesn't have enough valence electrons for six normal 2-electron bonds.

Structure of diborane (B2H6):

Diborane consists of two boron atoms, each also bonded to two 'terminal' hydrogen atoms lying in a plane, and two 'bridging' hydrogen atoms that lie above and below this plane, connecting the two borons together. A schematic (planar terminal H's, with the two bridge H's out of plane) looks like:

      H   H
       \ /
H — B       B — H
       \ /
      H   H   (the two H's between the two B's are the bridging H's, out of the plane of the four terminal H's)

Each boron is bonded to: 2 terminal H atoms (normal bonds) + 2 bridging H atoms (shared with the other boron) = effectively 4 hydrogens 'around' each boron, giving each boron atom a distorted tetrahedral (sp³) local geometry.

Bonding — why it's 'electron deficient':

Each B atom contributes 3 valence electrons and each H contributes 1, giving a total of 2(3) + 6(1) = 12 valence electrons for the whole molecule — enough for only 6 normal 2-electron covalent bonds, but diborane has 8 B–H connections to account for (4 terminal + 4 boron-to-bridging-H connections, i.e., each of the 2 bridge H's connects to both borons). Boron therefore cannot form 8 ordinary 2-centre-2-electron (2c–2e) bonds with only 12 electrons — it is short of electrons for a classical Lewis structure, hence 'electron deficient'.

This is resolved as follows:

• The 4 terminal B–H bonds are perfectly normal 2-centre-2-electron (2c–2e) covalent bonds — one shared electron pair, localized between one B and one terminal H. …