Conflict Between Organic and Physical Chemists--explaining the intensities of IR absorptions I am currently a Chemistry undergrad and I love to fact check everything my O Chem professor says against my research advisor, a physical/computational chemist. Well, my O Chem professor, when discussing IR absorptions, said something which was a simplification for teaching purposes but which seemed rather unusual to me. He said that high intensity peaks are due to a large change in dipole (clealy true) which often corresponds to more electronegative bonds and then gave the example of carbonyl absorptions which are intense, or nitrile absorptions, or $\ce{CO}$ single bond absorptions, etc.

What he then said, and says he has no good explanation for, is the following:

The absorption of an $sp^3$ hybridized $\ce{CH}$ stretch is high-medium intensity.

The absorption of an $sp^2$ hybridized $\ce{CH}$ stretch is medium intensity.

The absorption of an $sp$ hybridized $\ce{CH}$ stretch is medium-low intensity.

Of course the hybridizations given refer to the hybridization state of the carbon to which the hydrogen is attached.

The reason he has no good explanation for this phenomenon is that in most other situations, an $sp$ hybridized carbon will behave as if it is more electronegative than an $sp^2$ or $sp^3$ hybridized carbon.

An example of this would be that $pK_a$ of an $sp &lt; sp^2 &lt; sp^3$. That is a characteristic of how electronegative they are in an acid-base sense.

Here, however, we see the opposite trend. That is, when the more electronegative stretch should have a higher intensity peak, it is weaker, and the less electronegative stretch has a high intensity peak.

For the record, the arbitrary assignment of more and less electronegative based on behavior isn't super satisfactory to me, and is probably why I'm digging deeper on this.

So, my advisor essentially said all that matters is this equation: $$f=\left\langle\psi_0\right| \mu\left| \psi_v\right\rangle^2$$
Basically, he said I can ignore the wavefunction details of the equation because the wavefunctions of the three stretches are not going to differ by much because they all have identical reduced masses and some other stuff I can't remember.

So, all that matters is the dipole moment change as the bond is displaced from its equilibrium position.

His physical explanation of the intensities described above is that there is a smaller dipole moment change for $sp^2$ and $sp$ hybrid carbons because the electron density is trapped in the double and triple bonds more so than in the single bond. That seems reasonable to me.

The Organic Chemist fires back, however. He says that he would expect intensity to increase with the double and triple bonds because the electron in pi bonds are in a higher energy state than the sigma bond electrons and are thus more easily displaced which ought to result in a larger dipole moment change. That also makes a lot of sense I think.

So, is my advisor correct or is there a more complicated explanation?

Sorry for being so long-winded, I just wanted to get past the simple explanations so that we can have a deeper, better answer.

Edit: A very good answer would also explain why $sp$ hybridized C-H stretches are sometimes very high intensity. Even as high as $sp^3$ stretches at times.
In considering the C-H stretch, let's consider that the C atom is attached to something else via an "immobile" bond(s).

For $sp^3$ hybridized the C would have 3 other bonds besides the immobile one.

For $sp^2$ hybridized the C would have 2 other bonds besides the immobile one.

For $sp^1$ hybridized the C would have 1 other bonds besides the immobile one.

For a non linear molecule there should be 3N-6 modes of vibration, but a linear on has 3N-5. 

   Hybridization  # Atoms  Linear   Vibration Modes
       sp3            4       No         6
       sp2            3       No         3
       sp1            2       Yes        1


The more vibration nodes, the more microstates for the molecule in both its current state and its excited state. Thus having more microstates increases the probability that there are more transitions from the current state to some excited state. 

For an analog on the microstate overlap, think about atomic spectra vs UV-VIS. Atomic spectra are line spectra, UV-VIS is band spectra.