# Why do molecules have to have a change in dipole moment in order for them to be IR active?

My chemistry textbook keeps saying that in order for a molecule to be able to absorb infrared radiation, it has to have a change in dipole moment when the bond vibrates.

I don't understand why that is. Some clarification would be appreciated.

• The actual quantum mechanical justification for this is the "Fermi Golden rule", and it can be used to justify selection rules across spectroscopy. I can post a derivation/explanation here if you like, but it is very involved (by chemistry standards) so you might not find it useful unless you've got some background in quantum. – J. LS Feb 15 '15 at 14:15
• Thank you for the information, please do post an explanation! – amiliya Feb 16 '15 at 6:06
• Seconded, do please post! – Dion Silverman Apr 4 at 22:33

As a molecule vibrates, if there is a fluctuation in its dipole moment, then this induces an electric field that interacts with the electric field associated with the infra red radiation. If there is a match in frequency of the radiation and the natural vibration of the molecule, absorption occurs.

There is an intuitive reason why a vibration is IR-active only if it involves a change in the dipole moment.

Recall that the typical wavelength of IR radiation ($\sim 10\mu\mathrm{m}$) is much larger than the typical size of a molecule ($\sim 1\mathrm{nm}$). Hence, to a very good approximation, the (time-varying) electric field of the IR radiation is spatially uniform within a molecule.

Now observe that under a spatially uniform electric field, all positive charges, regardless of their positions, are pushed to a common direction, and all negative charges are pushed to exactly the opposite direction. In this case, can the change in the total dipole moment, defined as $\Delta \vec{\mu} = \sum_{i} q_{i} \Delta \vec{r}_{i}$, be equal to zero?

Clearly, the answer is "No." The displacements $\{\Delta \vec{r}_{i}\}$ are parallel between like charges and antiparallel between opposite charges. Therefore, $\{q_{i}\Delta\vec{r}_{i}\}$ are all parallel to one another, and $\vec{\Delta\mu}$ cannot be equal to zero.

So far, I have argued that a motion induced by a spatially uniform electric field always involves a change in the dipole moment. A corollary is that any motion that does not involve a change in the dipole moment cannot be induced by a spatially uniform electric field (of IR radiation).

As a simple example, imagine applying a spatially uniform electric field to a $\ce{CO_{2}}$ molecule such that the direction of the field is parallel to the length of the molecule. Can this field ever induce a symmetric stretch, in which the two oxygen atoms move in opposite directions? Certainly no. Both oxygen atoms are negatively charged by exactly the same amount. Under a spatially uniform electric field, they are pushed to the same direction by the same magnitude, and there is no room for a symmetric stretch.

• Good argument, and it is true even for shorter wavelengths. – Greg Dec 30 '15 at 5:24
• @Greg Thanks. Indeed, that the electric field is spatially uniform within a molecule (which is equivalent to the dipole approximation) is more or less true for visible light and probably even for UV. – higgsss Dec 30 '15 at 5:33