Which of the following statements is correct for 1 : 1 complex of (CH3)3N and BF3?
1) Geometry around Nis pyramidal with sp3 hybridisation and that around B is planar with sp2 hybridisation
2) The B-F bond length in the complex is longer than that in BF3
3) The geometry around N and B both are tetrahedral with sp3 hybridisation
4) 2 and 3 both
1) Geometry around Nis pyramidal with sp3 hybridisation and that around B is planar with sp2 hybridisation
2) The B-F bond length in the complex is longer than that in BF3
3) The geometry around N and B both are tetrahedral with sp3 hybridisation
4) 2 and 3 both
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Answer:
I think 1 is the correct answer because the other options are obviously wrong
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Answer is 4.
Explanation:
Complex of the compound is formed by forming a Coordinate bond between the two molecules.
In the nitrogen molecule a lone pair is present and acts as a lewis base and the boron molecule which has vacant orbital acts as a lewis acid.
Due to this they join together to form a Complex Compound.
Bond Order In BF3 is 1.33.
Bond Order In the complex is 1.
We know Bond Order inversely proportional to Bond length,
Hence, option 2 is correct.
Since a coordinate bond acts as a covalent bond after formation:
The Hybridisation Changes to sp³; thus forming a tetrahedron structure.
Hence Option 3 is correct.
So the answer is '4'.