Is there any change in the hybridisation of $B$ and $N$ atoms as a result of the following reaction?
$BF_{3} + NH_{3} \rightarrow F_{3}B \cdot NH_{3}$

(N/A) In $BF_{3}$,the Boron atom is $sp^{2}$ hybridized. It has an empty $2p_{z}$ orbital.
In $NH_{3}$,the Nitrogen atom is $sp^{3}$ hybridized,possessing one lone pair of electrons.
During the reaction,the lone pair from the Nitrogen atom is donated to the empty $2p_{z}$ orbital of the Boron atom to form a coordinate covalent bond.
As a result,the Boron atom changes its hybridization from $sp^{2}$ to $sp^{3}$ to accommodate the four bonding pairs.
The hybridization of the Nitrogen atom remains $sp^{3}$ as it still maintains four electron domains (three bonding pairs and one lone pair).