# Is there a detailed explanation as to why polarity of a molecule affects the absorption intensity of IR radiation? [closed]

Polar molecules have an electric field due to there being a net dipole moment. The electric field will be stronger with greater polarity. The stronger electric field in the polar molecule will interfere with IR radiation stronger than a weaker electric field. But how does the interference between the molecule's electric field and the IR radiation have to do with intensity of IR radiation absorbtion of the molecule? Is there a causal relationship between the two or is there simply a correlation?

## closed as unclear what you're asking by Mithoron, A.K., aventurin, Todd Minehardt, TyberiusSep 23 '18 at 19:06

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• Huh, I'd be interested why it was "favourited" twice. This question doesn't seem to make any sense... Why? False premise. – Mithoron Sep 22 '18 at 16:17

This is a phenomenological description, in a quantum description the probability of absorbing energy and moving from one vibrational level (stationary state) to another is proportional to the square of the strength of the radiation's electric field $$\epsilon$$ and the square of the transition moment $$M$$ of the vibration. Thus probability of absorption $$\sim \epsilon^2M^2$$. The electric field squared is proportional to the light intensity; no light no absorption, obviously.
The transition moment is proportional to the dipole $$\mu$$ as $$M = \int \psi_i^*\mu \psi_f dx= q\int x\psi_i^* \psi_f dx$$ where $$\psi$$ are the vibrational wavefunctions for the initial and final states in the transition, say v = $$0,1$$ and $$q$$ the charge on the electron. This shows that if the dipole is zero then there is no transition, just as in the case of homonuclear diatomic molecules. Additionally this shows that even if $$\mu$$ is large if the wavefunctions do not overlap (their product is zero) then the integral can be zero. This leads to the selection rules for transitions, which means that not all potential transitions occur.