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?

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.