1. Tardigrade
  2. Question
  3. Chemistry
  4. Arrange the following in increasing order of their intermolecular forces: Nylon-66 (I), Buna-S (II), Polythene (III)

Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. Arrange the following in increasing order of their intermolecular forces : Nylon-$66 (I)$, Buna-$S (II)$, Polythene $(III)$

AMUAMU 2010Polymers

Solution:

Nylon-$66$ is a fibre and has hydrogen bonding as intermolecular force. Buna-$S$ is an elastomer and has weak van der Waals' forces. Polythene is thermoplastic and has intermediate intermolecular forces (ie, stronger than elastomer but weaker than fibre).
Thus, the order of intermolecular forces is
$\underset{(II)}{\text{Buna-S}} < \underset{(III)}{\text{polythene}} < \underset{(I)}{\text{Nylon}-66}$