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I am studying spectroscopy and know that $\ce{CH4}$ and $\ce{CO2}$ are known as greenhouse gases because they are IR active. This is because they have asymmetric stretch or bend vibrational modes which result in electric field changes and involves emission or absorption of IR light.

But, $\ce{O2}$ is a symmetric molecule and should be IR inactive. I am wondering why there are still strong $\ce{O2}$ absorption bands in the IR range if it is expected to be transparent to IR light?

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    $\begingroup$ Oxygen has strong electronic transitions in the uv but very, very weak IR transitions as these are dipole forbidden as you mention. Perhaps you could edit and add some data to go with your question $\endgroup$ – porphyrin Apr 5 at 13:54
  • $\begingroup$ There are two low lying and 'forbidden' electronic transitions at approx 7800 ($^3\Sigma \to ^1\Delta$) and 13000 wavenumbers $^3\Sigma \to ^1\Sigma$. Additionally there are weak collision induced transitions $\ce{O2:O2}$ and $\ce{O2:N2}$ in a similar spectral region. $\endgroup$ – porphyrin Apr 5 at 16:50
  • $\begingroup$ Great question! This is related to 1) why is oxygen is blue in liquid, solid, & even high pressure gas phase, and 2) how the de-excitation of $^1\Delta$ oxygen results in red chemiluminescence at about 635 nm. This latter is a nice demo, especially in a Halloween lecture class in a large darkened lecture hall: mixing ordinary $NaClO$ bleach and 30% $H_2O_2$ gives the red emission. A VERY ADVANCED paper on the blue color is: S.C. Tsai and G.W. Robinson, "Why is Condensed Oxygen Blue?", Journal of Chemical Physics, 51 (1969) 3559-3568. $\endgroup$ – Ed V Jun 3 at 13:40

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