Why only molecules with three or more atoms undergo unimolecular reaction?

Question

Only molecules with three or more atoms undergo unimolecular reactions. A diatomic molecule cannot dissociate in this way because it has a single mode of vibrational freedom. If this mode is excited by an amount equal to dissociation energy, the molecule dissociates in $10^{-12}$ seconds.

What is the connection between the dissociation of diatomic molecules and vibrational degrees of freedom ?

I think this is a false statement because molecular hydrogen dissociates into hydrogen atoms, even its dissociation energy is given alongside other diatomic molecules. Am I missing something ?

Answer 1

I'm guessing this quote is given in the context of classical transition rate theory. Although the statement may seem very strong at first, it is generally true but has some important caveats.

Energy Landscapes

Lets consider a diatomic molecule. It's geometry can entirely be described by one degree of freedom - the bond length. As we move along that degree of freedom, stretching or compressing the bond we can develop a surface of the energy of the bond at different distances. A common model used to approximate this surface is the Morse Potential.

These surfaces are often call "energy landscapes". Minima in the energy landscape are stable structures - our starting materials and different products. If you push them a bit they move back down the surface to where they started.

And maxima are our transition states between minima, push a little and they'll move down one side of the barrier or the other and into a minima.

Also, a system can easily explore the parts of the energy landscape with less energy than the system currently has - the "classically allowed region" - like a ball rolling up and down the side sides of a valley.

Over time a system tends to move along the energy landscape to positions of lower energy if allowed time to equilibrate and release energy to the outside - like a ball rolling to rest at the bottom of the valley, losing energy to friction.

So normally, below the dissociation energy, the system will roll around in the bottom of the energy landscape, vibrating the bond.

In classical transition state theory, a reaction happens when the atoms move over the energy landscape along a particular degree of freedom or group of degrees of freedom.

In order for a diatomic molecule to dissociate, it needs to have enough energy moving along the bond vibration degree of freedom to explore parts of the energy landscape at large $r$. Which means it has to have a lot of energy in the vibrational mode as mentioned in your quote, an amount about equal to the dissociation energy. If it mode has that much kinetic energy though then, as mentioned above the molecule will fall apart incredibly quickly.

Changing Energy Landscapes

However, rather that adding thermal energy we can change the shape of the energy landscape. Many different energy landscapes exist at different energies. the Difference between different landscapes is the arrangement of the electrons in the molecule.

By shining high energy light at a molecule, you can excite the electrons in it into a different configuration of orbitals, leading to a different shape . This is what happens in photodissociation. You excite the molecule into a different electronic configuration, and onto different potential energy surface. On the new surface the molecule is no longer sitting in a minimum, but near a maximum and the system releases energy by moving down the surface - this time towards extending the bond length and dissociating, rather than towards a stable minimum.

Answer 2

The statement is essentially correct. Unimolecular reactions refer to the Lindemann scheme and its improvements called RRKM theory. The Lindemann model is $\ce{A + M <=> A^* +M ;\; A^*\rightarrow P}$ where $\ce{A^*}$ is an energised molecule and P product. M is an inert gas or another A type molecule that gives up some of its energy in a collision to form $\ce{A^*}$ and this energised molecule can only react by transferring its internal energy into another normal mode. A classic example is the reaction $\ce{CH3NC \rightarrow CH3CN}$.

In a diatomic molecule there is only one mode so when energised to $\ce{A^*}$ it dissociates directly and rapidly ($\lt 10^{-13}$ s) without being able to loose its energy to reform A, thus there cannot be an equilibrium set up as is proposed by the scheme above.