# How do chemical bonds increase molecular refractivity?

I see that the molecular refractivity of a compound consists of the refractivity indexes of its composite atoms and chemical bonds, and that the refractive index $n$ of a substance increases with its polarizability $\alpha$, as described with the Lorentz-Lorenz equation:

$${n^2-1\over n^2+2} = {4\pi N \alpha \over 3},$$

where $N$ is the number of molecules per unit volume.

I think that the formation of a covalent bond restricts the electron cloud of its composite atoms, delaying its release of an electromagnetic wave after being stimulated by a wave of the same frequency, and decreasing the wave length of the resulting created by the interference between the two electromagnetic waves affected and unaffected by the electron cloud. In the same way, I would explain the comparatively high refractivity index of a double bond between the same atoms as a stronger restriction of the shared electron cloud, delaying the release of an electromagnetic wave later than the covalent bond.

So I want to know, is there anything fundamentally mistaken with my interpretation?

## 1 Answer

The polarisability is the ability of any atom to have its electron 'cloud' displaced by an electric field. Polarisability is defined as the ratio of the induced dipole, formed by electron displacement, divided by the electric filed strength. It has units proportional to volume usually; values are in units $4\pi\epsilon_0\times 10^{-30} \mathrm{m^3}$. Larger atoms/molecules naturally have larger values, in general, than smaller ones, e.g. $\ce{CH4}\; 2.6,\; \ce{CCl4}\; 10.5$. In a general sense also the number of electrons and how tightly they are held affects the magnitude of the polarisability.

In the case of light it is its oscillating electric field that affects the atoms in the material, a solvent medium for example. In doing so the speed of light in the medium is reduced and the ratio of this to that in vacuum is the refractive index. The energy of the photon remains the same inside as it is outside the material (assuming no absorption or scattering ) so the frequency remains the same, but inside the material the wavelength changes because $c/\lambda =\nu$ is constant and c changes. In forming the refractive index, the tails of electronic absorption bands are important and are usually far into the ultraviolet.

In terms of working out the refractive index of some arbitrary molecule usually a semi-empirical method is used with the very many measured values available; values for different bonds can be extracted and parameterised. From these values then the refractive index of unknown compounds can be calculated. I found the pages www.pirika.com which has some explanation of how this is done.

In bringing atoms together to form molecules the atom/bond polarisability is used to form the refractive index and should be additive provided that in forming the bond a strong absorption band is not brought into the visible part of the spectrum. This is just what happens when a double bond or aromatic is formed so in these cases some extra empirical adjustment is needed.

• I see that I need to study electron absorption bands and the properties of light in depth. So the strength of the restriction of the electrons in a covalent bond would affect the polarizabiliy of the molecule(which is connected to its molar refraction), rather than just directly delaying the release of the electromagnetic wave of the same frequency as the incident light. Much clearer now. Thank you for your response. Jul 14 '17 at 7:14