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It is said rotational spectra of homonuclear diatomic molecules cannot be observed but why? In the book (Banwell, C. N. 1994. Fundamentals of Molecular Spectroscopy.) It says, for homonuclear diatomic molecules:

There will be no dipole component change if the molecule rotates so it will not interact with radiation.

What relation does the change in the dipole component have with the interaction with radiation?

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  • $\begingroup$ Which book was this in? $\endgroup$
    – Melanie Shebel
    Commented Apr 19 at 23:38
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    $\begingroup$ It is not SAID it is observed that transitions that do not have a change in charge distribution do not allow direct absorption or emission of photons ie. are IR inactive. They are active in the Raman mode or thru collisional interactions. $\endgroup$
    – jimchmst
    Commented Apr 20 at 0:51
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    $\begingroup$ @MelanieShebel I read it in Banwel spectroscopy book $\endgroup$
    – Physics
    Commented Apr 25 at 16:49

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In a classical way, think of the radiation as an oscillating dipole. If a molecule also has a dipole and it rotates or vibrates then this is also an oscillating dipole. It is a property of dipoles that they interact, and when they have the same energy (or frequency, which is the same thing) energy can move from one to the other.

In absorption, the energy from the radiation is transferred into the molecule when the dipole frequencies are the same. This can clearly only happen when the molecule has a dipole. A quantum description is needed for a proper understanding but this simple classical idea could be a start.

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  • $\begingroup$ if I am correct, the molecule will actually go to another rotational state but we will not be able to "observe " it because it will not emit light as an indication of transition. So readiation depends not only on energy difference of states but also dipole moment. $\endgroup$
    – Physics
    Commented Apr 16 at 17:08
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    $\begingroup$ No, the molecule will not be excited at all if the frequency is not right. In rotational spectroscopy there is no emission of per se. The emission is in the microwave region. $\endgroup$
    – ACR
    Commented Apr 16 at 18:47
  • $\begingroup$ Yes to another state, but as I mention the photon energy and transition energy have to be the same (Bohr condition) for a transition to be possible. In absorption we look for change in radiation intensity after passing through sample, this is how we measure the spectrum. Emission (of microwave photons if rotation) can occur and is measured against a 'dark' background, i.e. count photons if necessary by filtering out exciting light. (You will also come across other selection rules, based on conservation of angular momentum, besides having a dipole.) $\endgroup$
    – porphyrin
    Commented Apr 17 at 7:50

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