Relative O-O bond lengths in O2X2 molecules (X = H, F, Cl)

I am facing problem in the following question:



> 
> What is the order of bond length of $\ce{O-O}$ in the following
>  compounds:
> 
> 
> $\ce{H2O2, O2F2, O2Cl2}$
> 
> 
> 


In my view since fluorine is more electronegative than oxygen it will attract the bonded electrons towards itself resulting in low bond length in case of $\ce{O2F2}$ than in case of $\ce{O2Cl2}$. However if the same concept is applied to $\ce{H2O2}$ it should have the greatest $\ce{O-O}$ bond length however my book states it otherwise.




First, let me confirm your answer. The table given below shows a summary of optimised structures of $\ce{XOOX}$ compounds; experimental data when available is in parentheses. As a reference the bond length in $\ce{O2}$ is $121\ \mathrm{pm}$.[![enter image description here](https://i.stack.imgur.com/FtRnN.png)](https://i.stack.imgur.com/FtRnN.png)


Here you can clearly see that $\ce{O2F2}$ has the shortest $\ce{O-O}$ bond length of the three compounds in consideration in the original question. In fact, the fact that it is so close to $\ce{O=O}$ bond length in the dioxygen is suggestive of high double bond character between the two oxygen atoms in dioxygen difluoride. Here's another table summarising the bond order calculations:


[![enter image description here](https://i.stack.imgur.com/xZ8eM.png)](https://i.stack.imgur.com/xZ8eM.png)


Let's study the MO diagram proposed in the paper I am using as a reference for this answer.


[![enter image description here](https://i.stack.imgur.com/O6q7z.png)](https://i.stack.imgur.com/O6q7z.png)


Here, the orbitals of this $C\_2$ system have been generated by considering
the interaction of the frontier orbitals of $\ce{X ... X}$ with those of
the $\ce{O2}$ molecule. In the case of $\ce{H...H}$, they are simply the
a + b combinations of the $1s$ functions, and for the halogens they are a combination of their valence p functions.


Thus, 1a is derived from the $2p \sigma\_g$ function of $\ce{O2}$ and is 
$\ce{X–O–O–X}$ bonding. Similarly, $1b$ and $2a$ come from the $2p\pi\_u$ functions of $\ce{O2}$. 


The higher electronegativity of fluorine and bromine leads to better interaction with the quite low lying $\ce{2p\pi\_u}$ orbitals. In this scheme, we also note that the energies of 1b and 2a in these $\ce{XOOX}$ molecules decrease with the electronegativity of X. 


The orbitals $2b$ and $3a$ derive from the bonding overlap
of $\ce{X...X}$ functions with the $\ce{O2}$ $2p\pi\_g$ orbitals. 
Now, addressing the 2b and 3a functions in $\ce{FOOF}$, we note that they are similar to those in $\ce{HOOH}$. However, the high electronegativity of fluorine results in transfer of electron density from oxygen and considerable stabilisation of this $\ce{O–O}$ anti-bonding function.


The 3b and 4a functions in FOOF and BrOOBr are halogen based, non-bonding ‘lone pairs’. The higher electronegativity of fluorine, as outlined above, leads to greater bonding interaction between the $\ce{F}$ $2p$ and $\ce{O2}$ and $2p \pi\_u$ functions resulting in the low energy of $2a$ and $1b$. 


It similarly leads to the relatively low energy of the corresponding anti-bonding functions which are $4b$ and $5a$ in $\ce{FOOF}$. These are $\ce{F–O}$ anti-bonding but $\ce{O–O}$ bonding. These are the highest occupied orbitals in $\ce{FOOF}$ with the result that the $\ce{F–O}$ bond order is considerably reduced at the cost of the $\ce{O–O}$ bond.


In $\ce{FOOF}$ there is effectively a 4-orbital, 8-electron interaction
between the p orbitals of $\ce{F...F}$ and the $2p\pi\_u$ functions
of $\ce{O2}$ resulting in weak $\ce{F–O}$ bonds and little net transfer of
charge. 


This repulsive interaction is responsible for the long $\ce{F–O}$ bond. There is some bonding interaction with the $2p\sigma\_g$ and $2p\pi\_g$ functions of $\ce{O2}$. The latter results in a $\ce{O -> F}$ charge transfer and a degree of strengthening of the $\ce{O–O}$ bonds. 


In the resonance scheme proposed below, I would, based on this, call attention to resonance form 3.


Anyway, I have made an attempt to summarise the salient features of the discussing in the paper cited below in the reference section. The effects that I have called attention to here, are significant for fluorine but considerably weaker for chlorine, bromine etcetera.


[![enter image description here](https://i.stack.imgur.com/jN5cv.png)](https://i.stack.imgur.com/jN5cv.png)




---


Reference:
1] Bonding in mixed halogen and hydrogen peroxides, Adam J. Bridgeman and Joanne Rothery, J. Chem. Soc., Dalton Trans., 1999, 4077-4082 ([link](http://pubs.rsc.org/en/Content/ArticleLanding/1999/DT/a904968a#!divAbstract))


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First things first: the shorter a bond, the higher the multiple bond character. I.e. a double bond is shorter than a single bond, and a bond with a bond order of $1.5$ is in the middle of the two. Thus, the shorter the bonds in $\ce{X2O2}$, the higher the corresponding double bond character.


The key to understanding why the bond in $\ce{O2F2}$ has a very high double bond character is the *gauche* effect. The effect has its name because of the observation that $\ce{O2F2}$ prefers a *gauche* configuration rather than the expected (due to sterics) *anti* configuration. The reason is electronic in nature: a lone pair of oxygen A can interact with the $\sigma^\*(\ce{O\_B-F})$ orbital. This lowers the energy of the populated orbital at the expense of the unpopulated one. Since we have an interaction along $\ce{O\_A-O\_B}$, this increases the double bond character of the $\ce{O-O}$ bond. At the same time, since we are using the $\sigma^\*$ for stabilisation, the $\ce{O-F}$ bond decreases in bond order — it becomes less than a single bond. In Lewis resonance depictions, this can be explained by the following mesomeric structures:


$$\ce{F-O-O-F <-> F-O^+=O\bond{...}F-}$$


This effect is strongest in $\ce{O2F2}$ since fluorine is more electronegative than oxygen and therefore the $\sigma^\*$ orbital has a much higher oxygen contribution.


In hydrogen peroxide, a similar mechanism is at play that also contributes to reducing the bond length and increasing the bond order. In $\ce{Cl2O2}$, the effect is weakest.


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