Given th polymers,
`A =` Nylon-6,6, `B=` Buna-`S`, `C =` Polythene
Arrange these in decreasing order of their intermolecular forces:

Options

A. `AgtBgtC`
B. `BgtCgtA`
C. `CltAltB`
D. `BltCltA`

Answer

To arrange the given polymers (Nylon-6,6, Buna-S, and Polythene) in decreasing order of their intermolecular forces, we will analyze the types of intermolecular forces present in each polymer.

### Step-by-Step Solution:

1. **Identify the Polymers and Their Types**:
- **A = Nylon-6,6**: This is a type of polyamide and is known for its strong intermolecular forces due to hydrogen bonding and dipole-dipole interactions.
- **B = Buna-S**: This is a synthetic rubber (an elastomer) that primarily exhibits weak intermolecular forces, mainly Van der Waals forces.
- **C = Polythene**: This is a thermoplastic polymer that has moderate intermolecular forces, primarily due to Van der Waals forces, but stronger than those in elastomers.

2. **Determine the Strength of Intermolecular Forces**:
- **Nylon-6,6**: Strongest due to hydrogen bonding and dipole-dipole interactions.
- **Polythene**: Intermediate strength due to Van der Waals forces, stronger than elastomers but weaker than fibers.
- **Buna-S**: Weakest due to only Van der Waals forces.

3. **Arrange in Decreasing Order**:
- Based on the strength of intermolecular forces, we can arrange the polymers as follows:
- **Nylon-6,6 (A)** > **Polythene (C)** > **Buna-S (B)**

### Final Arrangement:
Thus, the final order in decreasing strength of intermolecular forces is:
**A > C > B**