Amines are derivatives of ammonia and are classified as `1^(@), 2^(@)`, and `3^(@)`, Primary and secondary (but not teritiary amines) form intermolecular hydrogen bonds and thus they boil at highter temperatures than expected. Like ammonia, all amines are basic, although they differ in their basic nature.
As amines are considered as derivatives of ammonia, quaternary ammonium salts are considered as derivatives of ammonium salts. Only the quaternary ammonium salts can shown optical activity.
When nitrogen is bounded to three different groups

Options

A. The molecule is optically inactive
B. The molecule is tetrahedral
C. The molecule is not superimposable on its mirror image
D. The amine boils at nearly similar b.p as the `1^(@)`, and `2^(@)` amine of comparable molecular weight.

Answer

To solve the question regarding the properties of amines, particularly focusing on the behavior of nitrogen when bonded to three different groups, we can break down the solution into several steps:

### Step-by-Step Solution

1. **Understanding the Structure of Amines**:
- Amines are derivatives of ammonia (NH₃) and can be classified as primary (1°), secondary (2°), and tertiary (3°) based on the number of carbon groups attached to the nitrogen atom. Tertiary amines do not form hydrogen bonds like primary and secondary amines.

**Hint**: Remember the classification of amines based on the number of carbon groups attached to nitrogen.

2. **Identifying the Nitrogen Configuration**:
- The nitrogen atom in amines has a lone pair of electrons and can form three bonds. For example, if nitrogen is bonded to a methyl group (−CH₃), an ethyl group (−C₂H₅), and a propyl group (−C₃H₇), it has one lone pair and three bond pairs.

**Hint**: Recall that nitrogen has five valence electrons, leading to one lone pair when it forms three bonds.

3. **Understanding Nitrogen Inversion**:
- When nitrogen is bonded to three different groups, it can undergo a process called nitrogen inversion, where the lone pair flips to allow for rapid interconversion between two forms (D and L forms). This process results in an equimolar mixture of both forms, leading to optical inactivity.

**Hint**: Consider how the lone pair affects the spatial arrangement of the groups around nitrogen.

4. **Determining Optical Activity**:
- Due to nitrogen inversion, the molecule does not exhibit optical activity because the D and L forms cancel each other out when plane-polarized light is passed through.

**Hint**: Optical activity depends on the non-superimposability of mirror images; in this case, the rapid interconversion leads to inactivity.

5. **Analyzing Molecular Geometry**:
- The geometry around the nitrogen atom is tetrahedral due to sp³ hybridization, but the presence of the lone pair makes the shape of the molecule pyramidal.

**Hint**: Differentiate between geometry (which is tetrahedral) and shape (which is pyramidal due to the lone pair).

6. **Evaluating Boiling Points**:
- The boiling points of primary and secondary amines differ due to the presence of hydrogen bonding, which is stronger in primary amines compared to secondary amines. Therefore, they do not boil at similar temperatures despite having comparable molecular weights.

**Hint**: Consider how hydrogen bonding influences boiling points in amines.

### Conclusion
Based on the analysis:
- The molecule is optically inactive due to nitrogen inversion.
- The geometry is tetrahedral, but the shape is pyramidal.
- The molecule is non-superimposable on its mirror image, indicating the presence of enantiomers.
- The statement regarding boiling points of primary and secondary amines being similar is incorrect.

### Final Answer
The correct statements are:
- The molecule is optically inactive (due to nitrogen inversion).
- The geometry is tetrahedral (but the shape is pyramidal).