11th Class Chemistry The p-Block Elements-I

  • 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.

    Answer:

      \[B{{F}_{3}}\]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).


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