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A t Kumar

Jan 25, 2015

Explain the structure of diborane?

Explain the structure of diborane?

Swathi Ambati

Diborane is an electron deficient molecule. The two boron atoms and the four terminal hydrogen atoms of the molecule are all in the same plane. These four terminal B -H bonds are regular 2-centered- 2 electron bonds. The bridging hydrogen atoms lie above and below this plane. The two bridges B-H-B bonds are unusual three centered two electron bonds. The boron atoms in diborane undergo sp³ hybridisation. The overlapping of a vacant sp³ hybrid orbital of one boron atom and sp³ hybrid orbital of another boron atom containing one electron with the pure s -orbital of bridging hydrogen containing one electron results in the "banana bond" .Similarly other banana bond is formed on other side . Thus the two banana bonds formed lie above and below the plane of the boron atoms. The remaining two sp³ hybrid orbitals of each boron atom overlap with terminal hydrogen atoms to form normal sigma bonds, results in the formation of four terminal hydrogen bonds.

Syeda

Answer. Diborane is an electron deficient molecule. The two boron atoms and the four terminal hydrogen atoms of the molecule are all in the same plane. These four terminal B -H bonds are regular 2-centered- 2 electron bonds. The bridging hydrogen atoms lie above and below this plane. The two bridges B-H-B bonds are unusual three centered two electron bonds. The boron atoms in diborane undergo sp³ hybridisation. The overlapping of a vacant sp³ hybrid orbital of one boron atom and sp³ hybrid orbital of another boron atom containing one electron with the pure s -orbital of bridging hydrogen containing one electron results in the "banana bond" .Similarly other banana bond is formed on other side . Thus the two banana bonds formed lie above and below the plane of the boron atoms. The remaining two sp³ hybrid orbitals of each boron atom overlap with terminal hydrogen atoms to form normal sigma bonds, results in the formation of four terminal hydrogen bonds.
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