Why does hydrogen peroxide exhibit a dihedral angle of $111.5^\circ$ in the gaseous state? And a dihedral angle of $90.2^\circ$ in the crystalline phase?

Dihedral angle of gaseous and crystalline HOOH

I know that in general, there is likely to be more hydrogen bonding in the crystalline phase than in the vapor phase.

But what's up with 1) the bond angles themselves in the first place and 2) the bond angle expansion in the gas phase? How can one rationalize them?

I know that there exist lone pairs on the oxygens and lone-pair/lone-pair repulsion should probably be minimized, but what it is about these conformers that minimizes energy? I'm guessing that we'd also want to minimize eclipsed and gauche interactions ... Newman projection time?

It seems that there is only one Newman projection possible for $\ce{HOOH}$; i.e. we'll have the inevitable lone-pair/lone-pair interaction, and the lone-pair/hydrogen interaction. I guess all we can do is minimize these interactions; i.e. the energy-minimized conformer is probably not the one with them all eclipsed. However, why does it seem that the energy-minimized conformer for vapor $\ce{HOOH}$ is one that puts the lone-pairs so close to each other in terms of dihedral angles?

My explanation


2 Answers 2


You've already answered your own question - hydrogen bonding.

First, don't think of it as to why is the gas state different than solid-state. Rather, think about why is the solid-state different than the gas state. In the gas state, there is very little intermolecular interaction (very, very few species do not follow the ideal gas law to many significant digits). The gas state geometry is close to what is expected for $\mathrm{sp^3}$ tetrahedral geometry and may be considered normal. The link below is a paper discussing the gas phase microwave spectroscopy of $\ce{H2O2}$ in far greater detail than I can hope to here. The description of the torsional rotation energies was not trivial.

Internal‐Rotation in Hydrogen Peroxide: The Far‐Infrared Spectrum and the Determination of the Hindering Potential by Robert H. Hunt, Robert A. Leacock, C. Wilbur Peters, Karl T. Hecht, The Journal of Chemical Physics 1965, 42, 1931-1946, https://doi.org/10.1063/1.1696228, mirror

When condensed to the solid-state, molecules get close enough to allow intermolecular forces (in this case hydrogen bonding to a much greater extent than van der Waals). Additionally, the energies of the crystal lattice must be considered. In this case, th

  • 2
    $\begingroup$ Why might the angles be what they are in the first place? $\endgroup$
    – Dissenter
    Commented Sep 1, 2014 at 7:25
  • 6
    $\begingroup$ Is this answer incomplete? $\endgroup$
    – Aritra Das
    Commented Feb 20, 2016 at 6:57

In an attempt to improve the given answer and also address other issues in the discussions of other duplicates of this question, I have optimised (PBE-D3/def2-SVP using NWChem 6.6) the $\ce{H2O2}$ structure in three different point groups: $\ce{C2}$, $\ce{C2_h}$ and $\ce{C2_v}$.

From frequency calculations, only the $\ce{C2}$ geometry is a minimum, the other two geometries having an imaginary vibrational frequency each.


Figure 1: $\ce{C2}$ geometry (angle in degrees, distances in angstrom)


Figure 2: $\ce{C2_h}$ geometry (angle in degrees, distances in angstrom)


Figure 3: $\ce{C2_v}$ geometry (distances in angstrom)

The results are summarised below.

            | Relative electronic | Dipole moment |  Dihedral
Point group |  energy (kcal/mol)  |    (Debye)    | H-O-O-H (°)
 C2         |       0.000         |     1.697     |  119.274
 C2h        |       0.022         |     0.000     |  180.000
 C2v        |       0.305         |     3.205     |    0.000

So the gauge structure is indeed the most stable found, as expected, with a dihedral angle of 119.3° (7.8° above the one cited by the original question).

NBO second-order stabilisation energies are as follows (there are two in each case, only one is shown):

            |          |            |    E(2)     | E(j)-E(i) | F(i,j)
Point group |  NBO(i)  |   NBO(j)   | (kcal/mol)  |  (a.u.)   | (a.u.)
            |  LP(O1)  |  BD*(O2-H) |    0.98     |   0.60    | 0.022
            | BD(O1-H) | BD*(O1-O2) |    0.63     |   0.68    | 0.018
            | BD(O1-H) |  BD*(O2-H) |   <0.50     |   ----    | -----
            | BD(O1-H) |  BD*(O2-H) |    1.20     |   0.99    | 0.031
            |  LP(O1)  |  BD*(O2-H) |   <0.50     |   ----    | -----
            | BD(O1-H) | BD*(O1-O2) |   <0.50     |   ----    | -----
            | BD(O1-H) | BD*(O1-O2) |    1.17     |   0.67    | 0.025
            |  LP(O1)  |  BD*(O2-H) |    0.85     |   0.95    | 0.025
            | BD(O1-H) |  BD*(O2-H) |   <0.50     |   ----    | -----

In this table, LP(O1) stands for a lone pair ($\ce{n_{O}}$), BD*(O2-H) is a anti-bonding orbital ($\sigma^*_{\ce{O-H}}$), etc.. From delocalisation energy contributions alone, we would expect the $\ce{C2_v}$ geometry to be the most stable.

Natural steric analysis shows the following for the repulsion between lone pairs in the oxygens:

            |          |          |  dE(i,j)
Point group | NLMO(i)  | NLMO(j)  | (kcal/mol)
 C2         |  LP(O1)  |  LP(O2)  |    6.84
 C2h        |  LP(O1)  |  LP(O2)  |   16.68
 C2v        |  LP(O1)  |  LP(O2)  |   16.57

This seems to be in better agreement with the energy differences found.


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