Question

Boron fluoride exists as $B{F}_3$ but boron hydride doesn’t exist as $B{H}_3$. Give a reason. In which form does it exist? Explain its structure.

Answer 1

$B{F}_3\text{and}B{H}_3\text{existence}$

• $B{F}_3$ is an electron deficient species, it tries to complete the octet of Boron atom by forming pπ-pπ back bonding with fluorine.
• The boron atom in $B{F}_3$ is $s{p}^2$ hybridized with a vacant $2p$ orbital, while the $F$ atom contains three lone pairs.
• These lone pairs on $F$ can overlap with the empty $2p$ orbital on $B$, thereby forming a coordinate bond. This donation of lone pair of electrons by fluorine is called back bonding. It reduces the deficiency of electrons on boron thereby increasing the stability of $B{F}_3$ molecule.

Dimer of $B{F}_3$

• In $B{F}_3$, $B$ is electron deficient and there is no lone pair on the $H$ atom to form $p\pi -p\pi$ back bonding.
• Therefore, to achieve stability it dimerises to form diborane $B_2{H}_6$.

$\text{Structure of diborane}:$

• The four terminal hydrogen atoms and the two boron atoms lie in one plane.
• Above and below this plane, there are two bridging hydrogen atoms.
• The four terminal $B$-$H$ bonds are regular two centre-two electron bonds while the two bridge $(B$-$H$-$B)$ bonds are different and can be described in terms of three centre–two electron bonds. The $3$-Centre-$2$-electron bridge bonds are also referred to as banana bonds.