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I am told that carbon dioxide is IR inactive.

This somehow strikes me as untrue or at best oversimplified, because yes, it is overall symmetrical, but can't $\ce{CO2}$ be induced to show asymmetric stretching patterns?

So this leads me to the question: in a simplified presentation of IR, why might someone present symmetrical, non-polar molecules as IR "inactive"? Could we be relying on a simpler definition of IR inactive?

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    $\begingroup$ The best way to consider IR activity of vibrations is to study symmetry and molecular point groups. It takes perhaps a few lectures, but with practice, anyone can identify vibrations and their IR or Raman activity with nothing but a pencil and paper and the symmetry character tables. $\endgroup$ – Geoff Hutchison Nov 4 '14 at 1:37
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    $\begingroup$ Incidentally, $\ce{CO2}$ is not only IR active, but a known and major greenhouse gas because of the IR absorption in the atmosphere. $\endgroup$ – Geoff Hutchison Nov 5 '14 at 0:25
  • $\begingroup$ @GeoffHutchison Interestingly the comment was made in class that because CO2 was IR inactive it didn't bounce IR radiation back into space. $\endgroup$ – Dissenter Nov 5 '14 at 0:26
  • $\begingroup$ Ack! You deserve a refund. As several people have mentioned, it's easy to confirm if you breath into an IR spectrometer while taking a spectrum. $\endgroup$ – Geoff Hutchison Nov 5 '14 at 0:28
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I am told that carbon dioxide is IR inactive.

You're right, that's not true. Since carbon dioxide is linear it has $3n-5 = 4$ vibrations and they are pictured below.

enter image description here

  • The symmetric stretch does not result in a change (of the initially zero dipole moment), so it is ir-inactive.
  • The asymmetric stretch does result in a change in dipole moment so it is ir-active.
  • The bend also results in a change in dipole moment so it too is ir-active.

We expected 4 vibrations and I've only listed 3. This is because the "bend" (let's start by placing the molecule along the x-axis) can occur in the y direction and the z direction. But these two motions are the same, just deforming in different directions, the bend is said to be degenerate, accounting for the "fourth" vibration.

To sum up, carbon dioxide has 2 ir-active vibrations.

Edit - response to example added (question d) by OP

Question d is incorrect. Either the author 1) inadvertently switched the column headings (IR active, IR inactive) or 2) meant to use some molecule other than carbon dioxide.

The first 3 rules you learn for interpreting IR and Raman spectra are

  • The number of molecular vibrational modes equals 3n-6 (3n-5 for linear molecules), where n is the number of atoms.
  • An ir active band will be observed if a vibration results in a change of the dipole moment. The initial dipole moment in the molecule's equilibrium geometry can be zero; all you need is a change. The terms "polar" and "non-polar" can be confusing, they often mean different things to different people. Leave "polar" out of the criteria for ir activity and stick with dipole moment, it is a much better understood term.
  • the rule of mutual exclusion, it states that, for centrosymmetric molecules (molecules with a center of symmetry, like carbon dioxide), vibrations that are IR active are Raman inactive, and vice versa. So for carbon dioxide there is 1 Raman band and two IR bands.

Here's a link to a recent SE Chem question: How can I deduce the linearity of XeF2 from the IR spectrum? where these rules were used to determine the structure of a molecule.

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  • $\begingroup$ but my book says that to be IR active, a bond must be polar. I'm guessing this is a true statement but the converse isn't? $\endgroup$ – Dissenter Nov 4 '14 at 1:29
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    $\begingroup$ @Dissenter to be IR active, a vibration must change the dipole moment. A polar bond can give IR active vibrations, but as the CO2 example shows, polar bonds can also give symmetric, IR inactive vibrations too. $\endgroup$ – Geoff Hutchison Nov 4 '14 at 1:35
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    $\begingroup$ And nonpolar bonds can have vibrations that change the dipole moment (for example $\ce{C=C}$ and $\ce{C#C}$). If you need proof that $\ce{CO2}$ is IR active, breath into an IR spectrometer while it is running. $\endgroup$ – Ben Norris Nov 4 '14 at 11:41
  • $\begingroup$ @ron I just spoke with the teacher; we are expected to "know" that non-polar molecules are IR inactive .... is this an acceptable simplification for a class whose main focus isn't IR or spectroscopy at all? Or is this just plain wrong? I do appreciate your answer and I see what you are saying; we can start with 0 dipole moment but we can definitely have modes of stretching and bending in which the two C=O dipoles do not mutually cancel. $\endgroup$ – Dissenter Nov 4 '14 at 21:17
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    $\begingroup$ It's just plain wrong. Look here, they analyze carbon dioxide and get the same result as posted here. They also analyze acetylene, a non-polar molecule, and again predict and observe ir-active vibrations. $\endgroup$ – ron Nov 4 '14 at 21:22
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Not true at all. The $\ce{C=O}$ bond is one of the most strongly IR active bonds there is (and the IR activity of $\ce{CO2}$ is the reason it's a greenhouse gas). I suspect the person who told you this was thinking that because $\ce{CO2}$ doesn't have a static dipole, it can't be IR active. However, IR activity is the result of dynamic dipoles (meaning the dipole changes with some type of deformation motion; in the case of $\ce{CO2}$, this occurs with bending motion and asymmetric stretching, as another answerer described), not static dipoles.

In some symmetric molecules, like $\ce{N2}$ or $\ce{O2}$, the only vibrational modes that can exist are stretching of the only bond, which because it's symmetric, doesn't lead to a dipole change. Thus, those species are not IR active.

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