This reversible reaction occurs in room condition:

$$\ce{2NO2 <=> N2O4}\quad(\Delta H=-57.23\mbox{ kJ/mol)}$$

This is the Lewis structure of $\ce{NO2}$ (courtesy Ben Mills via Wikipedia):

This is the Lewis structure of $\ce{N2O4}$ (courtesy ChemSpider):

To me, it seems that no bonds need to be broken for the forward reaction ($\ce{2NO2 -> N2O4}$) to occur, which is strange indeed.


No bonds are broken, but there are several isomers of N2O4. The unpaired electron can be on either the N or the O, for each NO2 molecule. The long bond between the monomers involves these two unpaired electrons.

See sections 7-1 to 7-3 of Bonding in Electron-Rich Molecules: Qualitative Valence-Bond Approach via Increased-valence Structures for more information.

According to First-Principles Study of the Role of Interconversion Between NO2, N2O4, cis-ONO-NO2, and trans-ONO-NO2 in Chemical Processes J. Am. Chem. Soc., 2012, 134 (31), pp 12970–12978:

there is no barrier to the bonding of two monomers to form the planar isomer with the N-N bond.

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    $\begingroup$ So the energy diagram (enthalpy) would have no uprise; only downfall? $\endgroup$ – DHMO Sep 6 '16 at 23:27
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    $\begingroup$ @user34388 There is no barrier for the formation of the isomer with the N-N bond, see the reference I just added to the answer $\endgroup$ – DavePhD Sep 7 '16 at 10:54

As stated by @DavePhD no bonds are broken. The $\ce{NO2}$ is a brown paramagnetic gas with the single electron mainly located on the nitrogen, and $\ce{N2O4}$ a colourless and diamagnetic gas. This molecule has a very long N-N bond, 0.178 nm, the O-N-O angle is 134 degrees the same as in $\ce{NO2}$. (The single bond in $\ce{H2NNH2}$ is 0.147 nm). The bond appears to be mainly of a $\sigma$ nature but delocalised over the molecule. (ref Ahlrichs & Keil J.A.C.S. 1974,96,7615)
Although there are 3 isomeric forms the most stable by far in the solid and liquid has the geometry you show in the question. The structure $\ce{O-N-O-NO2}$ is observed at 4 k in infra-red spectra. This structure may be important in some reactions even in the gas phase but this is speculative. The $\ce{N2O4}$ is planar at room temperature but twisted in an inert matrix at low temperature; this is the third isomeric form.


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