Specify the coordination geometry around and hybridisation of N and B atoms in a 1 : 1 complex of BF3 and NH3a)N : tetrahedral, sp3; B : tetrahedral, sp3 (2002S)b)N : pyramidal, sp3; B : pyramidal, sp3c)N : pyramidal, sp3; B : planar, sp2d)N : pyramidal, sp3; B : tetrahedral, sp3Correct answer is option 'A'. Can you explain this answer?

Question

Specify the coordination geometry around and hybridisation of N and B atoms in a 1 : 1 complex of BF3 and NH3a)N : tetrahedral, sp3; B : tetrahedral, sp3   (2002S)b)N : pyramidal, sp3; B : pyramidal, sp3c)N : pyramidal, sp3; B : planar, sp2d)N : pyramidal, sp3; B : tetrahedral, sp3Correct answer is option 'A'. Can you explain this answer?

Answer

H₃N → BF₃ where both N, B are attaining tetrahedral geomerty.

Answer

In BF3 the octet of B is not complete so F shares it's lone pairs with B known as back bond. Further as F is highly electro-ve so the needs of B aren't satisfied hence the lone pair on N in NH3 is shared with BF3 in the complex shown as BF3.NH3. Therefore a coordinate bond is formed b/w B & N, hence there are four bonds around B as well as around N making a tetrahedral geometry and sp3 hybridization of B and N.

Answer

Coordination Geometry around N and B atoms in the complex:
- N atom: tetrahedral geometry
- B atom: tetrahedral geometry

Hybridization of N and B atoms in the complex:
- N atom: sp3 hybridization
- B atom: sp3 hybridization

Explanation:
To determine the coordination geometry and hybridization of the N and B atoms in the BF3 and NH3 complex, we need to consider the Lewis structure and bonding of the molecules.

1. Lewis structure of BF3:
- Beryllium (B) is the central atom.
- There are three fluorine (F) atoms bonded to the central B atom.
- Each F atom contributes one valence electron, and B contributes three valence electrons.
- The total number of valence electrons in BF3 is 24 (3 x 7 + 3).
- In the Lewis structure, B forms three single bonds with F atoms, and each F atom has a lone pair of electrons.
- The Lewis structure of BF3 is represented as follows:

F F
\ /
B
/ \
F F

2. Lewis structure of NH3:
- Nitrogen (N) is the central atom.
- There are three hydrogen (H) atoms bonded to the central N atom.
- Each H atom contributes one valence electron, and N contributes five valence electrons.
- The total number of valence electrons in NH3 is 8 (3 x 1 + 5).
- In the Lewis structure, N forms three single bonds with H atoms, and N has one lone pair of electrons.
- The Lewis structure of NH3 is represented as follows:

H
|
H-N-H
|
H

3. Formation of the complex:
- When BF3 and NH3 react in a 1:1 ratio, the complex formed is BF3.NH3.
- In this complex, the N atom of NH3 acts as a Lewis base, donating its lone pair of electrons to the empty orbital of the B atom in BF3.
- The N atom forms a coordinate bond with the B atom, resulting in the formation of a dative bond.
- The complex is stabilized by the electrostatic attraction between the positively charged B atom and the negatively charged N atom.

4. Coordination geometry and hybridization:
- In the complex BF3.NH3, the N atom has four regions of electron density (three bonding pairs and one lone pair).
- Therefore, the coordination geometry around the N atom is tetrahedral.
- The tetrahedral geometry around the N atom indicates sp3 hybridization of the N atom.
- Similarly, the B atom in BF3 also has four regions of electron density (three bonding pairs and one empty orbital).
- Therefore, the coordination geometry around the B atom is tetrahedral.
- The tetrahedral geometry around the B atom indicates sp3 hybridization of the B atom.

In conclusion, the coordination geometry around the N and B atoms in the BF3.NH3 complex is tetrahedral, and the hybridization of both the N and B atoms is sp3.

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