Question

The interaction energy of London force is inversely proportional to sixth power of the distance between two interacting particles but their magnitude depends upon:
A. charge of interacting particles
B. mass of interacting particles
C. polarizability of interacting particles
D. strength of permanent dipoles in the particles

Answer 1

Hint: This force of attraction was first proposed by the German scientist Fritz London. These forces are always attractive. Take into considerations the factors on which forces between the interactive particles depend. It comes into notice when the distance between the molecules is less.

Complete answer:
Let us discuss London forces, when forces are there then energy will also be present there that would be released or absorbed while the interaction will occur.
- The London forces are the temporary attractive forces and the weakest intermolecular forces that result when the electrons in two adjacent atoms occupy positions to behave as temporary dipoles.
- Nonpolar molecules are electrically neutral and have no dipole moment because their electronic charge cloud is symmetrically distributed. But a dipole may get developed momentarily in such molecules. Dispersion forces are very weak until the molecules are almost touching each other and are increased by the increment of number of contact points.
- The dipoles are formed by the unequal sharing of electrons which causes rapid polarisation and counter-polarisation of the electron cloud. Such dipoles are short-lived. Then, the dipoles formed interact with the electron clouds of their neighboring molecules forming more dipoles. This interaction between the particles is called London Dispersion forces. These forces are weaker than other intermolecular forces and extend up to short distances.
- The strength of these forces within a given molecule depends on how easily the electrons in the molecules can be polarized. Polarizability is the term used to tell about the tendency of molecules to form charge separation. Large molecules in which the electrons are far from the nucleus are easy to polarise and thus possess greater dispersion.
- The interaction energy has the relation; $\text{interaction energy}\propto \frac{1}{{{\text{r}}^{6}}}\propto \text{polarisability}$. Interaction energy is inversely proportional to the sixth power of the distance between two interacting particles. These forces are for short distances $\left( \sim \text{500 pm} \right)$ and their magnitude depends on the polarisability of the particle.

The correct option is option ‘c’.

Note:
If dispersion forces would not be there, the noble gases would have not liquefied at any temperature since no other intermolecular force exists between the noble gas atoms. The temperature at which noble gases liquefy is generally very low. The liquefying temperature of Ne is ${{2.6}^{\text{o}}}\text{C}$.