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Since I assumed that if NO has a Q-branch, then so should CO. However, this explanation of rovibrational spectroscopy says otherwise: https://pages.vassar.edu/magnes/2018/12/15/fundamental-rovibrational-spectrum-of-co/

Is there a mistake or is there something I'm missing here?

Benvz
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    "The Q-branch can be observed in polyatomic molecules and diatomic molecules with electronic angular momentum in the ground electronic state, e.g. nitric oxide" - https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Spectroscopy/Rotational_Spectroscopy/Rovibrational_Spectroscopy CO has no unpaired electron. – Ian Bush Jan 24 '24 at 09:12
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    NO has an unpaired electron, this generates electronic angular momentum about the NO bond axis. As a result the selection rules are $\Delta J=0,\pm 1,\Delta M=0,\pm 1$ and $\Delta K=0$ where $K$ is the non-zero angular momentum quantum number about the NO axis. The caveat is that the $\Delta J=0$ applies except for $J=0$. (spin orbit coupling also complicated the spectrum leading to a doubling of lines, and lambda type doubling (v.v small is als0 present)) – porphyrin Jan 24 '24 at 19:07

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