• # question_answer 92)   Boron fluoride exists as $B{{F}_{3}}$ but boron hydride does not exist as$B{{H}_{3}}$. Give reason. In which form does it exist? Explain its structure.

$B{{F}_{3}}$exists as a monomer due to $p\pi -p\pi$back bonding. Fluorine transfers two electrons to vacant 2p-orbital of boron. The delocalisation reduces the deficiency of electrons on boron thereby increasing the stability of $B{{F}_{3}}$ molecule. Due to absence of lone pair of electrons on H, the back bonding does not occur in$B{{H}_{3}}$. In other words, electron deficiency of boron remains and $B{{H}_{3}}$ does not exist. To reduce electron deficiency $B{{H}_{3}}$ dimerises to form ${{B}_{2}}{{H}_{6}}$. Hence, boron hydride exists in dimeric form and known as diborane. Structure of diborane is already given in exercise 8.19(NCERT).