11
$\begingroup$

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.

$\endgroup$
  • 2
    $\begingroup$ 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. $\endgroup$ – J. LS Feb 15 '15 at 14:15
  • $\begingroup$ Thank you for the information, please do post an explanation! $\endgroup$ – amiliya Feb 16 '15 at 6:06
4
$\begingroup$

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.

For more details, please see this page on the UC Davis ChemWiki website.

$\endgroup$
6
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ Good argument, and it is true even for shorter wavelengths. $\endgroup$ – Greg Dec 30 '15 at 5:24
  • 1
    $\begingroup$ @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. $\endgroup$ – higgsss Dec 30 '15 at 5:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.