Ionic polarization & derive the expression for ionic polarizability:

• Ionic polarization is due to the displacement of cations and anions from its original position in opposite directions, in the absence of electric field.
• The displacement is independent of temperature and it occurs in ionic solids.
• Let us assume that there is one cation and one anion present int each unit cell of the ionic crystal (NaCl). When the electric field is applied, let x1 and x2 be the distances to which positive and negative ions move from their equilibrium positions.
• The resultant dipole moment per unit cell de to ionic displacement is given by: Induced dipole moment = magnitude of charge x displacement, hence$\mu_i = e(x_1+x_2) \cdots (1), \ where $ x1 is the shift of positive ion and x2 is the shift of negative ion, from their equilibrium positions.
• When the field is applied, the restoring force produced is proportional to the displacements.
• For positive ion, Restoring force F is directly proportional to x1 or F =$\beta_1x_1$
• For negative ion, Restoring force F is directly proportional to x2 or F =$\beta_2x_2$
• Here,$\beta_1 \ and \ \beta_2$ are restoring force constants, which depend on the masses of the ions and the angular frequency of the molecule in which ions are present.
• If m is mass of positive ion and M is mass of negative ion and$\omega_0 \ is$ the angular frequency then,$\beta_1=m \omega_0^2 \ and \ \beta_2=M \omega_0^2$ where$\omega_0: angular \ frequency$
• Hence, restoring force for positive ion can be written as: F =$m \ \omega_0^2 \ x_1$, we know that F = e E
  On comparison hence, e E =$m \ \omega_0^2 \ x_1 \ and \ x_1= \dfrac {e E}{m \omega_0^2}$
• Similarly for the negative ion we can write:$ x_2= \dfrac {e E}{m \omega_0^2}$
  Hence,$x_1 + x_2 = \dfrac {e E}{\omega_0^2} (\dfrac 1 m + \dfrac 1 M) \cdots (2)$
• Substitute (2) in (1) we get,
  $\mu_i= \dfrac {e^2 E}{\omega_0^2} (\dfrac 1 m + \dfrac 1 M) $
  $Where \ \alpha_i: Ionic \ polarizability, given \ by: $
  $Hence, \alpha_i= \dfrac {e^2 }{\omega_0^2} (\dfrac 1 m + \dfrac 1 M) $
• So, ionic polarizability$(\alpha_i) \ is $ directly proportional to its reduced mass which is given as$(\dfrac 1 m + \dfrac 1 M) $ and inversely proportional to the square of natural frequency of the ionic molecule. For most materials ionic polarizability is less than the electronic polarizability and typically$\alpha_i=0.1\alpha_e$.