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I had my exam this week and one question asked whether $\ce{H2O2}$ was planar or not. I though that since the oxygen atoms are both sp³ hybridized and have two lone pairs each, the molecule should have a bent shape. I compared it $\ce{H2O}$ which also has the same shape and I knew was planar. But it turns out $\ce{H2O2}$ is not planar. Does anyone know the reason behind this?


marked as duplicate by Klaus-Dieter Warzecha, Todd Minehardt, Jon Custer, M.A.R. ಠ_ಠ, hBy2Py Mar 25 '16 at 16:03

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    $\begingroup$ There is repulsion of the lone pairs on the adjacent oxygens that cause a barrier to rotation around the single bond. $\endgroup$ – StevieD Mar 25 '16 at 7:59
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    $\begingroup$ $\ce{H2O}$ must be planar because three points always make a plane, that isn't a good molecule to use for comparison. $\endgroup$ – SendersReagent Mar 25 '16 at 8:01
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    $\begingroup$ related, if not possible duplicate: chemistry.stackexchange.com/q/15754/4945 $\endgroup$ – Martin - マーチン Mar 25 '16 at 8:20
  • $\begingroup$ Think: Why should the the hydrogen atoms stand at a dihedral angle of zero or 180° relative to each other? $\endgroup$ – Karl Mar 25 '16 at 11:16

Because the O atoms are sp3 hybrid and as a result they adopt a tetrahedral geometry, now the lone lairs also repel each other and also the bond pairs so according to VSEPR theory to minimize the repulsion and attain a confirmation of minimum possible energy, the molecular structure is non planar, although it deceptively looks planar in a flash! The actual structure is like that of an open book

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    $\begingroup$ Are you sure it's repulsion and not donation of the lone pairs into $\sigma ^*_{\ce{CH}}$? I know it's not a great acceptor, but it can be done. Repulsion, I would expect, would cause a dihedral angle of 180°, so that would be planar. Think of ethane: Hydrogen there oppose each other. If you went H-C-C-H, repulsion would cause a that to be planar for hydrogens further apart. $\endgroup$ – SendersReagent Mar 27 '16 at 0:44

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