Question

Specify hybridisation of N and B atoms in a 1: 1 complex of BF3 and NH3

Answer 1

Boron floride  has 3 free valence electrons. There are 2s2, 2p1 orbitals.  These (three) orbitals combine to form three Sp2 hybrid orbitals.  In each of  the hybrid orbitals, 1 electron is present.  The geometry will be planar.

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Ammonia  NH3, - in Nitrogen, we have 5 valency electrons.  The 2s2 2p3 is configuration.  these four orbitals combine to form four sp3 hybrid orbitals.  In three sp3 orbitals, we have one electron each.  They form the covalent bonds.  The remaining two electrons share the fourth sp3 hybrid orbital.  The geometry will be tetra hedral.
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However, in a complex of BF3 and NH3, it is possible that Boron will change its hybridization from sp2 to sp3 by using the empty p orbital also.  Then it will also change to tetrahedral with 4 bonds.  NH3 will have sp3 hybridization.  This is due to the attraction of lone pair of electrons in Nitrogen towards Boron atom.