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If there is a change in dipole moments such as absorption spectroscopy, there are infrared absorptions (and/or transmittance) corresponding to them. Is the reason change in dipole moments are in infrared frequency simply because of the size of the molecules or scale of wavelength? Or are the dipole moments producing particular frequencies simply because of their sizes just like in antenna? If not, why are they mostly in IR?

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(Note: image from https://en.wikipedia.org/wiki/Electromagnetic_spectrum)

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  • $\begingroup$ possible duplicate: chemistry.stackexchange.com/questions/73565/… $\endgroup$ – Buck Thorn Oct 14 '19 at 9:27
  • $\begingroup$ The energy of IR radiation matches the energy of the transitions between vibrational states. So it is a match in energy, not in length. $\endgroup$ – Karsten Theis Oct 14 '19 at 15:34
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    $\begingroup$ Could you edit your question to make sure it is clear what you are asking. If you look at the answers and the comments, folks are guessing what you want to ask. $\endgroup$ – Karsten Theis Oct 14 '19 at 16:03
  • $\begingroup$ Karsten. You answered it directly, I thought the energy of IR radiation is a match in length. You rightly answered it is a match in energy. $\endgroup$ – Jtl Oct 14 '19 at 22:43
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The transition energy, or wavelength if you prefer, are governed by the nature of the molecules involved. Whole body free rotational motion (in the gas phase) gives rise to microwave spectra ( fractions of wavenumbers cm$^{-1}$ or THz frequencies). The frequency is determined by the molecules moment of inertia. At higher energy comes vibrational motion, stretching and bending of chemical bonds, a few hundred to a few thousand cm$^{-1}$. These bond frequencies are approximately given by $\sqrt{k/\mu}$ where $k$ is force constant which relates force to bond extension as in Hook's law, and $\mu$ is the reduced mass. There are 3N-6 types of vibration for N atoms, (3N-5 for linear molecules). At higher frequencies are electronic transitions where and electron gets promoted from a bonding to anti-bonding orbital. Typically these transitions are in the visible and uv regions.

The exact number of transitions is difficult to estimate because each vibrational level has rotational levels superimposed on them and there are also vibrations that are combinations of levels. There are, however, fewer electronic transitions because there are generally few molecular orbitals than types of vibrational modes, and each excited state has its own set of vibrational and rotational levels.

One of a chemist's essential tools, arguably more important that any other spectroscopy, is NMR spectroscopy which is caused by transitions between nuclear spin states when in a magnetic field and occurs at very low frequency, 100-800 MHz range.

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    $\begingroup$ I think your own answer here: [chemistry.stackexchange.com/questions/73565/… is much closer to what the OP is attempting to figure out... $\endgroup$ – Buck Thorn Oct 14 '19 at 8:27
  • $\begingroup$ Thanks, I was not sure what the OP knew about this so here made general answer. $\endgroup$ – porphyrin Oct 14 '19 at 9:24
  • $\begingroup$ Your comment was very helpful, porphyrin. I was asking initialy if the IR is due to the match in length. It is not, but a match in energy. But it is true the length of the atom matches the length of x-ray? But how come atom still emit x-ray? $\endgroup$ – Jtl Oct 14 '19 at 22:46
  • $\begingroup$ As you can see in the comments, it is the energy between levels that matters. An atom can, under appropriate conditions, emit/absorb photons in the infra red, visible, uv and xray regions. Similarly an atom and a big molecule, such as C60, can have almost identical absorption and emission wavelengths. $\endgroup$ – porphyrin Oct 15 '19 at 10:08

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