Question

In gaseous triethyl amine the "-C-N-C-" bond angle is _________ degree.

Hint:

Use VSEPR theory to predict molecular geometry. Start with the ideal angle based on the number of electron domains (e.g., 109.5$^\circ$ for 4 domains). Then, adjust the angle based on repulsions: lone pairs compress bond angles, while bulky groups can increase them due to steric hindrance.

Answer 1

Correct Answer: 108

Triethylamine has the formula $N(CH_2CH_3)_3$.
The central nitrogen atom is bonded to three ethyl groups and has one lone pair of electrons.
According to VSEPR (Valence Shell Electron Pair Repulsion) theory, the geometry around the nitrogen atom is determined by the total number of electron domains (bonding pairs + lone pairs).
Nitrogen has 3 bonding pairs (to the three ethyl groups) and 1 lone pair. This gives a total of 4 electron domains.
The electron geometry for 4 domains is tetrahedral, with an ideal bond angle of 109.5$^\circ$.
However, the presence of a lone pair introduces repulsion. The lone pair-bond pair repulsion is stronger than the bond pair-bond pair repulsion.
This stronger repulsion from the lone pair compresses the bond angles between the bonding pairs. Therefore, the C-N-C bond angle will be slightly less than the ideal tetrahedral angle of 109.5$^\circ$.
In a similar molecule, ammonia ($NH_3$), the H-N-H bond angle is about 107$^\circ$.
In triethylamine, the ethyl groups are bulkier than hydrogen atoms. This bulkiness will cause some steric repulsion between the ethyl groups, which tends to increase the bond angle back towards the tetrahedral angle.
The actual experimentally determined bond angle for gaseous triethylamine is approximately 108$^\circ$. This value is a balance between the compression due to the lone pair and the expansion due to the steric hindrance of the bulky ethyl groups.
Given the options in such exams, a value slightly less than 109.5 is expected, and 108 is the accepted value.