Hydrogen peroxide may exist in four different forms depending on the dihedral angle, $d = \angle\,\ce{H-O-O-H}$. The four conformers are: Conformer I with $d = 0^\circ$, Conformer II with $d = 60^\circ$, Conformer III with $d = 180^\circ$, and Conformer IV with $d = 300^\circ$. Which conformer is rotationally inactive?

The answer given to this question is that conformer III is rotationally inactive.

However, what does it mean to be rotationally inactive? $\ce{H2O2}$ doesn't seem to have any steric effects that prevent it from rotating about the $\ce{O-O}$ bond. What would cause the conformer with $d = 180^\circ$ to become rotationally inactive?

  • $\begingroup$ "Rotationally inactive" may mean several totally different, unrelated things. Sure, nothing prevents the molecule from rotating about the O−O bond. Now what about rotating the plane of pol$\rm\tiny{ar}$...... $\endgroup$ Jun 21, 2021 at 7:00
  • $\begingroup$ @IvanNeretin, do you mean to say, the reason for it being called rotationally inactive is because the dipole moment is zero? I understand that the definition to be applied here is that the compound has no rotational spectrum due to symmetry. $\endgroup$ Jun 21, 2021 at 7:12
  • $\begingroup$ @SafdarFaisal It seems probable to me. The question context would help, but the original may provide no context either. $\endgroup$
    – Poutnik
    Jun 21, 2021 at 7:18
  • 2
    $\begingroup$ Rotationally inactive surely means that the discussed conformer has no rotationally active modes. Nothing is about barriers or hindered rotation. The language used is perfectly sensible in spectroscopy. That certainly comes from a section on spectroscopy. $\endgroup$
    – Alchimista
    Jun 21, 2021 at 8:09
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    $\begingroup$ BTW H2O2 doesn't have any steric effects that prevent it from rotating about the O−O bond, but has stereoelectronic. That makes rotation barrier quite high, but I guess isn't really relevant here. $\endgroup$
    – Mithoron
    Jun 21, 2021 at 12:08

1 Answer 1


"Rotationally inactive" is probably about the rotational spectrum of the molecule

I'm assuming the (unsaid) context of the question is that it is about rotational spectroscopy as this is where the term "rotationally inactive" makes sense.

Rotational spectroscopy–like its better known cousin infra-red spectroscopy–measures absorptions of electromagnetic wavelengths corresponding to transitions in molecules. IR spectra mostly involve vibrations in molecules that have energies corresponding to wavelengths of IR light. Rotational spectroscopy (usually gas phase, as molecular rotations in solids are not usually easy) measures rotational modes in molecules in a similar way (strictly speaking for both techniques absorptions happen when the wavelength/energy of the radiation matches a transition between two rotational or vibrational modes).

But for an absorption to happen the radiation has to have a mechanism to interact with the molecule. For both vibrational and rotational spectroscopy, this can only happen if the molecule has a dipole moment (or, strictly, the relevant vibration or rotation has a changing dipole). While different vibrational models in the same molecule can be "active" (eg the asymmetric stretch in carbon dioxide) or "inactive" (the symmetric stretch in carbon dioxide), in pure rotations there is no way for a molecule without a dipole to have an "active" rotational spectrum as rotation doesn't change the dipole. So only molecules sufficiently asymmetric to have a dipole moment can interact with the radiation. Conformer III of hydrogen peroxide is the only conformation with no net dipole so is the only one which is "rotationally inactive".

The terminology is confusing outside this context as it could be taken to mean something to do with the interconversion of the different conformers by rotation around the central O-O bond (which is likely fairly easy).

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    $\begingroup$ (Disclaimer: not an expert in the field.) While the terms 'IR-active', 'Raman-active', 'NMR-active', etc. are common, I've never seen 'rotationally active', not when I did this in undergrad, nor in some of the common textbooks. Hollas, for example, writes: "All regular tetrahedral molecules, which belong to the $T_\mathrm{d}$ point group, may show such a rotational spectrum. However, those spherical rotors that are regular octahedral molecules and that belong to the $O_\mathrm{h}$ point group do not show any such spectrum." $\endgroup$
    – orthocresol
    Jun 21, 2021 at 15:12
  • $\begingroup$ @orthocresol I suspect few have seen much of rotational spectroscopy as it isn't that common. And I'm not entirely sure that textbook is strictly correct (unless it is talking about 2nd order effects which are dependent on vibrations as well as rotations) as Td molecules don't have permanent dipoles so have no pure rotational spectrum. $\endgroup$
    – matt_black
    Jun 21, 2021 at 17:47
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    $\begingroup$ Oh, I quoted it a bit out of context, sorry: it's describing instantaneous deviations from Td geometry due to centrifugal distortion. IIRC this is not exactly the same effect as rotational/vibrational coupling, but similar idea in that it's not a first-order effect. So Td molecules do have very weak (but nonzero) rotational spectra; though I must confess I've never actually seen one myself. $\endgroup$
    – orthocresol
    Jun 21, 2021 at 18:20

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