Question

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​

Answer 1

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'.

Answer 2

4) 2 and 3 both are the correct answers.

• In  1 : 1 complex of $(CH_3 ) _3N$ and $BF_{3}$ , N and B both in $(CH_3 ) _3N$ and $BF_{3}$ are $sp^{3}$ hybridized. So, the geometry around N and B both are tetrahedral.
• The B-F bond length in the complex is longer than that in $BF_{3}$ because, In $BF_3$ , back bonding is observed in between fluorine (F) and boron(B) due to presence of p-orbital in boron. 'Cause of the back bonding it acts as double bond characteristics. As B-F forms in other complex, it effects in the back bonding  and thus double bond characteristic disappears and act as single bond. Hence, bond becomes slightly longer than earlier bond length ( $1.3A^{0}$).

Hence, 4) 2 and 3 both are correct for 1 : 1 complex of $(CH_3 ) _3N$ and $BF_3$.

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