Specify the coordination geometry around and hybridisation of N and B atoms in a 1:1 complex of BF3 and NH3
a)N :tetrahedral,sp3 ,B :tethedral,sp3
b)N :pyramidal,sp3 , B :pyramidal,sp3
c)N :pyramidal,sp3, B :planar,sp2
d)N :pyramidal,sp3, B :tetrahedral,sp3
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1 Answers
156Boron has 3 valence electrons, which pair with single electrons from each of the hydrogen atoms to form a covalent bond. With only 3 bonds the geometry of BH3 will be trigonal planar with sp2 hybridisation.
Nitrogen has 5 valence electrons. Three electrons are added from the hydrogen to form the three covalent bonds, filling the outer shell of the nitrogen to complete its octet. Six electrons are involved in bonding, whilst two electrons are non bonding and therefore a "lone pair." For NH3 this will form into a tetrahedral geometry to keep each of the electron orbitals as far away from each other as possible. The tetrahedral geometry is sp3 hybridisation.
For a 1:1 complex of both BH3 and NH3 I believe the BH3 can use its empty valence orbitals to form larger molecular aggregates. The boron could use its empty valence orbital to accept a lone pair from the NH3. The hybridisation of the nitrogen in the NH3 would be unaffected by this bonding, however, the BH3 would switch to sp3 hybridisation making it form a tetrahedral geometry. This would also be consistent with the structure of diborane B2H6 using bridge bonds.
