Question:
Is there any change in the hybridisation of B and N atoms as a result of the following reaction?
$\mathrm{BF}_{3}+\mathrm{NH}_{3} \rightarrow \mathrm{F}_{3} \mathrm{~B} \cdot \mathrm{NH}_{3}$
Solution:
Boron atom in BF3 is sp2 hybridized. The orbital picture of boron in the excited state can be shown as:
Nitrogen atom in NH3 is sp3 hybridized. The orbital picture of nitrogen can be represented as:
After the reaction has occurred, an adduct F3B⋅NH3 is formed as hybridization of ‘B’ changes to sp3. However, the hybridization of ‘N’ remains intact.