In 3:50 of this video two FTIR spectras are shown. He explains that the % transmittance (double bond in his example) depends on the dipole moment, $\mu = \delta \cdot d$.

I understand that the absorbance center frequency depends on $\delta, d$, but why the percentage of transmittance depends on $\delta, d$? A private case of this would be a symmetric bond ($\delta=0$): why the transmittance is 100%?


Light has both oscillating electric and magnetic fields. In the common UV-Visible-IR spectroscopy experiments it is the interaction of the molecule with the electric field that is being observed. A molecule is very small compared to the size of a beam of light, so it is fair to view the molecule as being immersed in or surrounded by the light. In other words, it is fair to view the molecule as being immersed in a uniform electric field.

The strength of the interaction ($I_s$) of the light's electric field ($E$) with the molecule is dependent upon both the strength of the light's electric field and the vector description of the molecule's electric field. The molecule's electric field is described by the spatial arrangement of the molecule's electrons. The spatial arrangement of the molecule's electrons is described by the dipole moment ($\mu$), a vector sum. Hence, $$I_s = E \cdot \mu $$

The absorption intensity of a spectral transition (the area under the absorption curve - usually it correlates with the peak height) is given by the same equation except that rather than use the molecular dipole moment $\mu$, one uses what is called the transition dipole moment. Rather than consider the ground state distribution of electrons around the molecule, one must use the rearranged distribution of electrons around the molecule that occurs when the molecule absorbs light.

If the transition dipole moment is always similar to the ground state dipole moment, then what the video says is correct. This may be true for larger molecules when we are looking at an absorption that excites a $\ce{C-H}$ or $\ce{C-C}$ bond, or something similar. In such cases, the small perturbation in electron density around the specific bond probably doesn't radically change the overall electron distribution around the entire molecule. In smaller molecules like $\ce{CO}$ it seems possible to me that light absorption could make a significant change to the overall electron distribution and that consequently the ground state dipole moment might differ substantially from the transition dipole moment.

  • $\begingroup$ So basically what you're saying is that higher (transition) dipole moment causes, according to some unknown principle, more interaction with incident light. Why would that be the case? $\endgroup$ – Sparkler Mar 18 '15 at 1:14
  • $\begingroup$ No, I'm saying something else. The transition dipole moment may be larger or smaller. The two electric fields interact in a well-understood manner. Simplifying for a moment, the interaction of 2 point charges (attract, or repel) is really the interaction of their respective electric fields and the equations to describe this are well known. The interaction of a non-symmetric electric field generated by all of the electrons in a molecule with the electric field associated with a light wave is also understood, $\endgroup$ – ron Mar 18 '15 at 2:24
  • $\begingroup$ but much more complex, in a mathematical sense, but these are not unknown principles. These two fields may interact more or less strongly than the electric field generated by the light interacted with the ground state molecular electric field. If the interaction during the transition process is larger then the absorption is more intense than if the interaction is smaller. $\endgroup$ – ron Mar 18 '15 at 2:24
  • $\begingroup$ but why higher dipole means more interaction (=more absorbance)? or: why zero dipole means no absorbance? $\endgroup$ – Sparkler Mar 18 '15 at 2:41
  • $\begingroup$ As the equation above says, the bigger the molecule's electric field (transition dipole moment), the bigger the interaction with the lights electric field and therefore the bigger the absorption (the product of the two) $\endgroup$ – ron Mar 18 '15 at 3:24

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