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Problem

How can we qualitatively compare the bond lengths between $\ce{H2O2}$ and $\ce{O2^2-}?$

Answer

Bond length in $\ce{O2^2-}$ is slightly larger than in $\ce{H2O2}.$

Attempt

I tried using molecular orbital theory, but since both have peroxide $\ce{O2^2-},$ the differentiating factor must be some sort of repulsion, but I'm not getting it.

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  • $\begingroup$ This compilation actually gives the same bond length for both species, and also does so for ozone and ionic superoxides where both oxygen-oxygen bond orders are 1.5. $\endgroup$ – Oscar Lanzi Feb 25 at 10:45
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The way to tackle this is to look at the lone pair repulsions between the 2 molecules. It is known that $\ce{N-N};$ $\ce{O-O};$ $\ce{F-F}$ single bonds are quite unstable due to lone pair (lp) repulsions.

An evidence of this is that $\ce{N}$ exists in molecular state as $\ce{N2}$ using multiple bonds, so that it's lone pairs do not repel each other, however $\ce{P}$ doesn't do that. It is capable of forming $\ce{P4}$ because in the third period the size of the atoms are large enough to properly distance the lone pairs.

One more evidence of this act is that among diatomic halogen molecules the bond length is in the order $\ce{Cl2} < \ce{Br2} < \ce{F2} < \ce{I2}$ the bond length of $\ce{F2}$ should have been lower (2p–2p overlapping is much stronger hence the bond should have been the shortest). However, due to lp–lp repulsions it is now much longer bond.

With this we can now see that in $\ce{H2O2}$ there is only two lone pairs on each oxygen atom while in $\ce{O2^2-}$ it has three lone pairs on each atom, hence it has increased lp–lp interaction.

As a result of this $\ce{O2^2-}$ tends to elongate (relatively more than $\ce{H2O2})$ its bond to decrease this intense lp–lp interaction, hence the longer bond.

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    $\begingroup$ I don't think "lone pair repulsion" stands up as an argument for pi bonding in N2 and O2 but not in S6 and P4. Much more reasonable is that the pi bonds in second row are strong enough to overcome general electron-electron repulsion because of better side-by-side p orbital overlap than in 3rd row. For the specific question, the bond length difference in $\ce{O2^2-}$ vs $\ce{H2O2}$ is easily explained by general increased charge repulsion without invoking the specific number of lone pairs $\endgroup$ – Andrew Feb 24 at 13:54
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    $\begingroup$ @Andrew I suppose, in a sense it is equivalent here: in H2O2 we replace one lone pair with a bond pair (albeit one that is heavily polarised towards oxygen); and the bond pairs repel each other less than the lone pairs do. I certainly prefer what you say though, as it gets at the heart of the issue, and is more general, as you mentioned. $\endgroup$ – orthocresol Feb 24 at 14:53
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    $\begingroup$ @Andrew I do realize that it would be a less mis-directing in nature, however I think I should leave it and let people find out through the comments why it is misdirecting. But I must ask : should I edit it to reflect this correction? $\endgroup$ – napstablook Feb 24 at 15:33
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    $\begingroup$ Edits are totally up to you. If you feel like the answer could be improved and you want to take the time to do that, you're welcome to, but you're not obligated to. If a commenter feels strongly enough that your answer is insufficient or incorrect, they always have the option of posting a new answer and letting the community decide which is more helpful. $\endgroup$ – Andrew Feb 24 at 16:01

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