Specific hybridization of N and B atoms in a 1:1 complex of BF3 and NH3 Ans is N tetrahedral sp3 and B tetrahedral sp3 Explain guyzz?

Question

Answer

NH3 and BF3 have sp3 (pyramidal) & sp2 (coplanar) hybridization respectively , having one lone pair of electron on N atom which is responsible for pyramidal shape of NH3 in spite of sp3 hybridization . However, as soon as it is coordinate to BF3, both attain tetrahedral geometry and acquire sp3 hybridization.

So, N is tetrahedral and sp3
B is tetrahedral and sp3

¶¶Hope it's help...¶¶

Answer

Due to formation of coordinate covalent bond... hybridization changes in both boron and nitrogen to TETRAHEDRAL..

Answer

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

Introduction:
In the 1:1 complex of BF3 and NH3, the boron trifluoride (BF3) molecule acts as an electron acceptor, while the ammonia (NH3) molecule acts as an electron donor. This results in the formation of a coordinate covalent bond between the two molecules, with the nitrogen (N) atom of NH3 donating a lone pair of electrons to the vacant orbital of boron (B) in BF3.

Hybridization of Nitrogen (N):
- Nitrogen in NH3 molecule is originally sp3 hybridized, with three sigma bonds and one lone pair of electrons occupying four hybrid orbitals.
- In the complex with BF3, the lone pair of electrons on N forms a coordinate bond with the vacant orbital of B.
- The formation of the coordinate bond involves the hybridization of N to accommodate the bond formation.
- In this case, the N atom undergoes sp3 hybridization, where one of the sp3 hybrid orbitals containing the lone pair overlaps with the vacant orbital of B to form the coordinate bond.
- The other three sp3 hybrid orbitals of N continue to form sigma bonds with the three hydrogen atoms in NH3.

Hybridization of Boron (B):
- Boron in BF3 molecule is originally sp2 hybridized, with three sigma bonds and an empty p orbital.
- In the complex with NH3, the vacant p orbital of B accepts the lone pair of electrons from N to form the coordinate bond.
- The formation of the coordinate bond involves the hybridization of B to accommodate the bond formation.
- In this case, the B atom undergoes sp3 hybridization, where the vacant p orbital mixes with the three sp2 hybrid orbitals to form four sp3 hybrid orbitals.
- These four sp3 hybrid orbitals of B then form sigma bonds with the three fluorine atoms in BF3 and the coordinate bond with N.

Conclusion:
In the 1:1 complex of BF3 and NH3, the nitrogen atom undergoes tetrahedral sp3 hybridization, with one sp3 hybrid orbital forming the coordinate bond with the vacant orbital of boron. The boron atom also undergoes tetrahedral sp3 hybridization, with the vacant p orbital mixing with the three sp2 hybrid orbitals to form four sp3 hybrid orbitals. The hybridization of N and B atoms allows for the formation of the coordinate bond and the stabilization of the complex.

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