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S Jan 14, 2021 at 3:52 history bounty ended Karsten
S Jan 14, 2021 at 3:52 history notice removed Karsten
Jan 10, 2021 at 4:24 comment added Karsten Related: chem.libretexts.org/Bookshelves/…
Jan 10, 2021 at 2:26 history edited ron CC BY-SA 4.0
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Jan 10, 2021 at 2:25 answer added ron timeline score: 4
Jan 9, 2021 at 23:06 comment added ron These earlier answers may be helpful: How can I deduce the linearity of XeF2 from the IR spectrum? and Is carbon dioxide IR inactive?.
Jan 9, 2021 at 23:03 comment added ron @Karl You don't need a dipole moment, you just need a change in the dipole moment.
Jan 8, 2021 at 12:00 history edited Karsten
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S Jan 8, 2021 at 11:21 history bounty started Karsten
S Jan 8, 2021 at 11:21 history notice added Karsten Draw attention
Dec 10, 2020 at 5:58 comment added Jtl the shape oscillates in left and right, how can they be same shape? And why would same shape not cause change in polarizability, hope someone can Answer.
Dec 9, 2020 at 16:38 comment added porphyrin In the asymmetric stretch the molecule retains the same shape because as one extends the other contracts and vice versa.
Dec 9, 2020 at 8:27 comment added Karl The effect on both sides levels out, mostly. Remember that some selection rules are stronger than others: Spin must be retained, and if you just have no dipole moment, you get no IR interaction. Oh, remember you have to look at this from the view of an electromagnetic wavelenght that is a thousand times as large.
Dec 9, 2020 at 8:24 comment added Poutnik Is possible the polarizability change requirement applies not to particular bonds, but to the whole molecule, with the opposite bond effects nullifying each other ?
Dec 9, 2020 at 6:12 history asked Jtl CC BY-SA 4.0