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I'm aware that the Q branch ($\Delta j = 0$) is usually not observed in rotational-vibrational spectra of diatomic molecules in the gas state, however, I've heard different things about when exactly this transition is forbidden.

I've sometimes been told that the transition is not allowed in general, for any diatomic molecule, but I've sometimes heard that this is only true if the electronic state has a $\Lambda \neq0$, so which is the correct selection rule here?

I'd love to have this clarified for me. If it is possible I would also like to know when is a Q branch observed in vibronic spectra of heteronuclear and homonuclear diatomics.

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  • $\begingroup$ You are correct in that the electronic angular momentum $\Lambda \ne 0$ for a Q branch to be observed in a diatomic molecule, i.e. the molecule is not in a $\Sigma$ state. The rule is then $\Delta J =0; \Delta v = \pm1$. (Homo-nuclear diatomic molecules do not have an ir spectrum as $d\mu/dx = 0$, as the dipole is zero) $\endgroup$ – porphyrin Sep 22 '17 at 14:32
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    $\begingroup$ ahhh, I know homo nuclear diatomics don't have an IR spectrum, there was a typo in the question, thanks. Are you sure about the $ \Lambda$ selection rule? is a Q branch always permitted if $\Lambda \neq 0$? Is there some easy derivation of this or somewhere where I could find a derivation? I can't see where this comes from from a simple transition moment integral. $\endgroup$ – Ignacio Sep 22 '17 at 15:47
  • $\begingroup$ A photon carries one unit of angular momentum which is conserved. When the photon is destroyed on absorption, this ang mom must be transferred to the molecule in one way or another, in your case $\Lambda_i +1=\Lambda_f $ where i and f are initial and final states. This is the basis of the transition moment integral and so selection rules. $\endgroup$ – porphyrin Sep 24 '17 at 7:21
  • $\begingroup$ I understand that argument, however, $L$ is not really a good quantum number in diatomic molecules and $\Lambda$ is not the angular momentum, but the projection of the angular momentum in the direction of the internuclear axis, so I don't think that the fact that the total anguar momentum is conserved helps a lot. $\endgroup$ – Ignacio Sep 25 '17 at 15:12
  • $\begingroup$ In Hydrogen, for instance, although $\Delta l$ is constrained to be $\pm 1$, $\Delta m_l $, the z projection, can be zero, and this doesn't break angular momentum conservation. $\endgroup$ – Ignacio Sep 25 '17 at 15:26
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I would like tell you what I know about the answer to this question.

So, it's known that the selection rules for a fine vibrational spectrum (rotational-vibrational spectrum) for diatomic molecules make $\Delta J = 0$ or $\pm 1$ transitions allowed between the rotational levels of two vibrational states (usually $V''=0$ to $V'=1$, where the symbols have their usual meaning). But one must be careful in realizing that the $\Delta J = 0$ transition is forbidden for all parallel vibrations in the oscillator and only allowed for perpendicular vibrations.

Now, let's see what are parallel and perpendicular vibrations!

Any vibrational mode in infrared active only when the change in the dipole moment vector is sinusoidal w.r.t. time and thus can interact with the electric field of the electromagnetic radiation. If this change in the dipole moment vector is parallel to the major axis of the molecule, then the vibrational is labelled as a parallel mode of vibration. If the change in the dipole moment is perpendicular w.r.t. the major molecular axis, it is a perpendicular mode of vibration.

Let's look at the case of diatomic molecule, as the question is focused on that.

The major molecular axis for it is the bond axis (containing both the atoms). Since a hetero-nuclear diatomic molecule has only one vibrational mode, viz., stretching, it can only occur parallel to the bond axis. Therefore, $\Delta J = 0$ transition is forbidden in the case of parallel vibrations. Thus, on observing the fine infrared spectrum of hetero-nuclear diatomic molecules, we only observe $P$ and $R$ branches. $Q$ branches are not observed.

References:

  1. Fundamentals of Molecular Spectroscopy - Banwell and McCash

  2. Introduction to Molecular Spectroscopy - Gordon M. Barrow

  3. Atomic and Molecular Spectroscopy - Rita Kakkar

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  • $\begingroup$ This doesn't really answer the question, which should say why the Q branch isn't observed, and the answer should include information about electronic states. $\endgroup$ – pentavalentcarbon Dec 16 '17 at 2:45

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