N-N bond length in Dinitrogen Tetroxide

Question

The $\ce{N-N}$ bond length in $\ce{N2O4}$ is $\pu{1.64 Å}$ which is much longer than in the case of hydrazine $\ce{N2H4}$ $\pu{(1.47 Å)}$. This means that the bond is quite weak. In another unusual case, I see that $\ce{N-N}$ bond length in $\ce{N2O3}$ (the blue asymmetric form) is $\pu{1.86 Å}$ !!

That is very weak for a nitrogen single bond. Is there a satisfactory explanation for this? And amazingly molecules with such long bonds tend to have a nonplanar structure due to bond rotation but $\ce{N2O4}$ is planar.

I found a similar question on $\ce{N2O4}$ but the answer to the planarity question wasn't so satisfactory. Also, I wanted the extend upon the weak bond question.

Answer 1

 

Images are from Wikipedia.

The first obvious thing that you can see is that $\ce{N2O4}$ is planar but $\ce{N2H4}$ is not. And the primary reason starts here.

$\ce{N2O4}$ can be viewed as two nitrogen dioxide molecules joined together. If you draw a Lewis structure of $\ce{NO2}$, you will recognize two things. Firstly, it has one unpaired electron. So imagine when two $\ce{NO2}$ molecules come near together, they will share their two unpaired electrons (like how radicals react that you might be familiar with in organic chemistry) to form a single sigma bond. This is the very bond you are asking about.

And secondly, it is sp² hybridized at the nitrogen atom. Thus, when you join the two $\ce{NO2}$ molecules together, taking the relative orientation of the sp² orbitals and how they should overlap to form a sigma bond in mind, you can see why $\ce{N2O4}$ is planar. Because the $\ce{N=O}$ bonds (I write the "$\ce{=}$" symbol for double bond for simplicity, it should be about 1.5 order due to resonance) heavily repulse each other when they are in the planar position, it weakens the $\ce{N-N}$ bond. Indeed this bond is so weak that it is homolytically dissociated at near below room temperature or above, creating the basis for the popular equilibria between $\ce{NO2}$ and $\ce{N2O4}$.

The situation is even worse for $\ce{N2O3}$. $\ce{N2O3}$ can be viewed as formed from a \ce{NO} (itself has an unpaired electron and a sp² nitrogen) and a $\ce{NO2}$ molecule, with the exact mechanism as above. And guess what? $\ce{N2O3}$ has a lone electron pair on the nitrogen (in place of the missing oxygen compared to $\ce{N2O4}$). Lone pairs repulse much, much more heavily than the bonding pairs do, and this further weakens the $\ce{N2O3}$ $\ce{N-N}$ bond. This is illustrated by the easy disproportionation of $\ce{N2O3}$ into $\ce{NO}$ and $\ce{NO2}$.

Then why $\ce{N2H4}$'s bond is stronger? Because the nitrogen atoms are sp³ hybridized in this case (note that oxygen is divalent but hydrogen is only monovalent), thus they form a tetrahedral electronic structure. Thus they can comfortably adopt a skew conformation to minimize the repulsion. Either $\ce{N2O3}$ or $\ce{N2O4}$ could achieve this because the sp² hybridization confines them to planar geometry.

Using Pauling's formula with the $\pu{1.14 A}$ triple bond in $\ce{N2}$ as a reference:

$$D(n) = D(m) - 0.6\log(n/m)$$

With $D(n, m)$ is the length of the bond with order $n$, $m$ respectively.

The $\ce{N-N}$ bond order in $\ce{N2O4}$, $\ce{N2O3}$, and $\ce{N2H4}$ can be estimated to be $0.44$, $0.19$, and $0.85$ respectively.

Answer 2

I think I can suggest an intuitive argument as to why the molecule of N2O4 is planar despite the electrostatic repulsion between the oxygen atoms (which does not favour planarity). The reason is really the sp2-hybridisation. Both pi-bonds are perpendicular to the corresponding planes OON. Consequently, the orbital magnetic momentum of each pi-bond is perpendicular to the corresponding plane. (This is really due to symmetry. Just for a moment imagine that all the atoms are equal so that the obvious difference of the N—N bond could not prevent you from seeing the symmetry: three rays in one plane making three angles of 120 degrees.) You can represent the orbital magnetic momentums as two magnets perpendicular to the corresponding planes. Now rotate the planes along the N—N bond (and the corresponding magnets, of course). The electrostatic repulsion between the oxygen atoms may decrease, but the magnetic force between the two orbitals is far more important because this force is greater than the electrostatic repulsion. (It can't be easily proved without hard calculation, but think of it this way: the magnetic attraction between the spins of two electrons is enough to build a molecule despite the electrostatic repulsion between the two nuclei and the electrostatic repulsion between the two electron clouds.)

Now, which is the stable state of the N2O4 molecule, that is the state with minimal energy? It is the state where the two magnets are antiparallel:

N — S

S — N

In this state they attract each other and the attraction is maximal. As the magnets are parallel to each other and perpendicular to their planes (each magnet to its own plane), the planes must be parallel, too (actually, they coincide).

If this proof seems too complex, it is because it cannot be explained in fewer words. A simple animation (or a static image) would make the idea obvious, but I cannot provide such an image here. I still hope that my answer can be useful.