Is there any change in hybridisation of the `B` and `N` atom as a result of the following reaction?
`BF_(3)+NH_(3)rarr F_(3)B.NH_(3)`

Explanation

Boron `(B)` is `sp^(2)` hybridised in `BF_(3)` and `N` in `sp^(3)` hybridised in `NH_(3)`. After the reaction, hybridisation of `B` changes to `sp^(3)` but that of `N` remains unchanged. `BF_(3)` is a Lewis acid and accepts a lp of `e^(-)'` from `NH_(3)`, which acts as a Lewis base.
`F_(3)B larr:NH_(3)`