Question

VSEPR theory predicts that an atom with one lone pair and three bonding pairs (such as the N-atom in aniline) will have a tetrahedral electron geometry and a trigonal pyramidal molecular geometry due to steric repulsions between H-atoms and the N-atom lone pair. However, in question 5 you observed that the N-atom in aniline is not perfectly sp3 hybridized (i.e. the hybridization is different from that predicted for a tetrahedral electron geometry). Briefly describe all of the factors that result in the calculated hybridization of the N-atom lone pair

Answer 1

Answer: The lone pair of electron on nitrogen is accommodated in a 2p orbital hence it interacts with the pi system in aniline.

Explanation:

Aniline is less basic than amines. This is because, the nitrogen atom in aniline is not purely sp3 hybridized. Its actual hybrization state is closer to sp2 because the lone pair on nitrogen is accommodated in a 2p orbital. The nitrogen atim in aniline is planar and its

lonely pair interacts with the pi electron system of aniline. This makes the lone pair unavailable for protonation hence aniline is less basic than amines.