Question 20 of 32intermediate🔧 ApplyShort Answer2 marks

Is there any change in the hybridisation of B and N atoms as a result of the following reaction? BF₃ + NH₃ → F₃B.NH₃

Correct Answer

Yes, there is a change in the hybridisation of the boron (B) atom, but no change in the hybridisation of the nitrogen (N) atom.

In BF₃, the boron atom is sp2sp^2 hybridised with a trigonal planar geometry. When it reacts with NH₃, it forms an additional bond with the nitrogen atom by accepting its lone pair. As a result, in F₃B.NH₃, the boron atom forms four bonds and becomes sp3sp^3 hybridised.

In NH₃, the nitrogen atom is already sp3sp^3 hybridised, having three N–H sigma bonds and one lone pair of electrons in the fourth sp3sp^3 hybrid orbital. In the product F₃B.NH₃, the nitrogen atom donates its lone pair to boron to form a coordinate bond, but it still uses its four sp3sp^3 hybrid orbitals. Therefore, the hybridisation of nitrogen remains sp3sp^3.

Exercise: EXERCISES | Q: 4.26 | (Page 34)
For More Understanding

Explanation

The question tests the understanding of hybridisation changes during adduct formation. In BF₃, boron is sp2sp^2 hybridised with an empty p-orbital. Upon accepting a lone pair from NH₃, boron forms four sigma bonds, requiring sp3sp^3 hybridisation. Nitrogen in NH₃ is sp3sp^3 hybridised with a lone pair. When it donates this lone pair to form a coordinate bond with boron, it still utilizes its four sp3sp^3 hybrid orbitals (three for N-H bonds, one for the N-B bond), so its hybridisation does not change.

Solution Steps

  1. Step 1: Identify the hybridisation of B in BF₃. Boron forms three bonds and has a trigonal planar geometry, so it is sp2sp^2 hybridised.

  2. Step 2: Identify the hybridisation of N in NH₃. Nitrogen forms three bonds and has one lone pair, utilizing four electron domains, so it is sp3sp^3 hybridised.

  3. Step 3: Analyze the product F₃B.NH₃. Boron now forms four bonds (3 with F, 1 with N), changing its hybridisation to sp3sp^3.

  4. Step 4: Analyze N in the product. Nitrogen still has four electron domains (3 bonds with H, 1 bond with B), so its hybridisation remains sp3sp^3.

PreviousNext