Question

In the structure of diborane

• A. all hydrogen atoms lie in one plane and boron atoms lie in a plane perpendicular to this plane
• B. $2$ boron atoms and $4$ terminal hydrogen atoms lie in the same plane and $2$ bridging hydrogen atoms lie in the perpendicular plane
• C. $4$ bridging hydrogen atoms and boron atoms lie in one plane and two terminal hydrogen atoms lie in a plane perpendicular to this plane
• D. all the atoms are in the same plane

Answer 1

Answer: (B)

Boron is trivalent, we would expect a simple hydride $BH _{3} .$ However $BH _{3}$ is not stable. The boron possess incomplete octet and $BH _{3}$ dimerises to form $B _{2} H _{6}$ molecule with covalent and three centre $2$ -electron bond. The simplest boron hydride is diborane $B _{2} H _{6}$.

As seen from the structure drawn, 6 electrons are required for the formation of conventional covalent bond structure by B-atom, whereas in diborane, there are $12$ valence electrons, three from each boron atoms and six from the six hydrogen atoms. The geometry of $B _{2} H _{6}$ can be represented as.



The four terminal hydrogen atoms and two boron atoms lie one plane. Above and below the plane, there are two bridging hydrogen atoms. Each boron atom forms four bonds even though it has only three electrons. The terminal $B-H$ bonds are regular bonds but the bridge $B - H$ bonds are different.

Each bridge hydrogen is bonded to the two boron atoms only by sharing of two electrons. Such covalent bond is called three centre electron pair bond or a multi centre bond or banana bond.