A question about electronic polarizability I am having trouble with this question.
The question is:
Carbon tetrachloride contains 74 electrons in its molecule. Its relative permittivity is 2.26 when its density is 1.6810^3 kg/m^3. If the field acting on the liquid is 510^6 V/m, what is its electronic polarizability and average electron displacement. Answers: 8.5710^-33 Cm and 0.7210^-15m
I am struggling with the first part of the question. I assume the Clausius-Mosotti relation has to be used as CCl4 is non-polar. However, my answer is incorrect. This might have to do with N. I do not know to get N.
I also do not understand the second part.
Some hints and clue are needed.
Thanks
(I also apologize for the horrible notations as I do not have LaTex notation)
 A: It's true that the molecule is non-polar as a whole due to symmetry of its molecular structure, but it doesn't mean that the underlying bonds are not polarized (hence the question about its polarizability, which is a measure of how easily the polarization happens in response to an external electric field). $\vec{E} = \alpha \vec{p}$, where $\vec{p}$ is the dipole moment of the polarized bond and $\alpha$ is the polarizability. You're on the right path with the Clausius-Mossotti relation.
For the second part, consider the (classical) definition of the dipole moment, $\vec{p} = N e \vec{d}$, where $|\vec{d}|$ is the distance between the positive and negative charges in the dipole, and $e$ is the elementary charge (charge of a single electron). You can find the magnitude of the dipole moment since you're given the strength of the electric field, and you already found the polarizability in the first part. Then, consider how many electrons are involved in each $C-Cl$ bond due to covalent bonding.
