STRUCTURE OF DIBORANE :

(1)  B₂H₆ contains 4-Terminal H are bonded by Sigma bond and  remaining 2-H are bridging hydrogen’s and of these are broken then dimer become monomer.

(2) Boron undergoes sp3 hybridisation 3 of its sp³ hybridised orbitals contain one( e¯) each and fourth sp³ hybrid orbital is vacant.

(3) 3-(Three) of these sp³ hybrid orbitals get overlapped by s orbitals of 3-hydrogen atoms.

(4) One of the sp3 hybrid orbitals which have been overlapped by s orbital of hydrogen gets overlapped by vacant sp³ hybrid orbital. Of 2^nd Boron atom. And it’s vice versa.

(5) By this two types of overlapping take place 4 (sp3– s) overlap bonds and 2(sp² – s – sp³) overlap bonds.

(6) H is held in this bond by forces of attraction from B and This bond is called 3 centred two electron bonds (3C-2e bond) . Also called Banana bonds. Due to repulsion between the two hydrogen nuclei, the delocalised orbitals of bridges are bent away from each other on the middle giving the shape of banana.

(7) The two bridging hydrogens are in a plane and perpendicular to the rest four hydrogen..

ILLUSTRATED EXAMPLE (1): In Diborane

(A) 4 bridged hydrogens and two terminal hydrogen are present

(B) 2 bridged hydrogens and four terminal hydrogen are present

(C) 3 bridged and three terminal hydrogen are present

(D)None of the above

ILLUSTRATED EXAMPLE (2): Which one of the following statements is not true regarding diborane?

(A) It has two bridging hydrogens and four perpendicular to the rest.

(B) When methylated, the product is Me₄B₂H₂.

(C) The bridging hydrogens are in a plane and perpendicular to the rest.

(D ) All the B–H bond distances are equal

ILLUSTRATED EXAMPLE (3): The structure of diborane (B₂H₆) contains

(A) Four (2C–2e–) bonds and two (2C–3e–) bonds

(B) Two (2C–2e–) bonds and two (3C–2e–) bonds

(C) Four (2C–2e–) bonds and four (3C– 2e–) bonds

(D )None of these

ILLUSTRATED EXAMPLE (4): The molecular shapes  of diborane is shown:

Consider the following statements for diborane:

1. Boron is approximately sp3 hybridised

2. B–H–Bangle is 180°

3. There are two terminal B–H bonds for each boron atom

4. There are only12 bonding electrons available

Of these statements:

(A ) 1, 3 and 4 are correct                  (B) 1, 2 and 3 are correct

(C) 2, 3 and 4 are correct                    (D) 1, 2 and 4 are correct

STRUCTURE OF DIAMOND :

(1) Each carbon is linked to another atom and there is very closed packing in structure of Diamond.

(2) Density and hardness is very much greater for diamond because of closed packing in diamond due to sp3 hybrid and are tetrahedrally arranged around it. And C-C distance is 154pm

(3) Diamond has sharp cutting edges that's why it is employed in cutting of glass.

(4) Diamond crystals are bad conductor of electricity because of absence of mobile electron.

(5) 1 carat of diamond = 200 mg.

(6) Diamond powder if consumed is fatal and causes death in minutes.