# Calculating the ionic polarizability from the Clausius-Mossoti relation

The concentration of atoms in a crystal is $N_d = \pu{2.24e28 m-3}$, and the relative dielectric constant is $\varepsilon_{r} = 5.62$. The crystal has a refractive index of $n = 1.5$. Using the Clausius-Mossoti relation,

$$\frac{\varepsilon_{r} - 1}{\varepsilon_{r} + 2} = \frac{N\alpha}{3\varepsilon_{0}},$$

I got that

$$\alpha_\mathrm{i} = \pu{8.26e-40 F m2}.$$

$$\alpha_\mathrm{i} = \pu{3.7e-40 F m2}.$$

Was the refractive index useless info here, or have I missed something?

Yep, you missed quite an important part: Clausius-Mossoti equation in the form you've written is limited and holds true for the dielectric that consists of individual particles. However, the problem says you are dealing with a crystal, which, by the way, is most likely sodium chloride $$\ce{NaCl}$$, judging from its dielectric constant and refractive index.

In the first Clausius-Mossoti equation you've written $$\alpha$$ consists of two main components:

• electronic polarizability $$\alpha_\mathrm{e}$$ caused by the displacement of the electron shell of the atom relative to the nucleus under the action of the field;
• ionic polarizability $$\alpha_\mathrm{i}$$ caused by the displacement of ions with respect to other ions.

For the ionic crystal Clausius-Mossoti equation can be written as

$$\frac{\varepsilon_{r} - 1}{\varepsilon_{r} + 2} = \frac{N}{3\varepsilon_0}(\alpha_\mathrm{e^+} + \alpha_\mathrm{e^-} + \alpha_\mathrm{i})\, ,\label{eq:1}\tag{1}$$

where $$\alpha_\mathrm{e^+}$$ and $$\alpha_\mathrm{e^-}$$ are electron polarizabilities of positive and negative ions in crystal lattice. Their sum can be determined using a Lorentz–Lorenz equation, which is derived from the fact that the effect of an external electromagnetic field with relatively high frequencies corresponding to the visible and ultraviolet range of the spectrum results in the displacement of only the electron shells relative to atomic nuclei, while the more massive particles (atoms and ions) do not have time to shift from the occupied positions. Correspondingly, only electronic polarization contributes to the polarization of the medium (e.g. $$\alpha_\mathrm{i}\approx0$$), and the refractive index is related to the electron polarizability of the particles by the Lorentz-Lorentz formula:

$$\frac{n^2 - 1}{n^2 + 2} = \frac{N}{3\varepsilon_0}(\alpha_\mathrm{e^+} + \alpha_\mathrm{e^-})\label{eq:2}\tag{2}$$

Subtracting \eqref{eq:2} from \eqref{eq:1}, one eliminates the electronic constituent from the equation:

$$\frac{\varepsilon_{r} - 1}{\varepsilon_{r} + 2} - \frac{n^2 - 1}{n^2 + 2} = \frac{N}{3\varepsilon_0}\alpha_\mathrm{i}\tag{3}$$

and ionic polarizability is actually determined as follows:

\begin{align} \alpha_\mathrm{i} &= \frac{3\varepsilon_0}{N}\left(\frac{\varepsilon_{r} - 1}{\varepsilon_{r} + 2} - \frac{n^2 - 1}{n^2 + 2}\right)\\ &= \frac{3\cdot\pu{8.85e-12 F m-1}}{\pu{2.24e28 m-3}}\left(\frac{5.62 - 1}{5.62 + 2} - \frac{1.5^2 - 1}{1.5^2 + 2}\right)\\ &= \pu{3.7e-40 F m2}\tag{4} \end{align}