3
$\begingroup$

I am trying to bridge a conceptual understanding of how light affects both breaking and forming bonds. When bonds are broken, energy is absorbed (endothermic process). Conversely, when bonds are made, energy is released (exothermic process).

We can calculate the necessary wavelength that is required to break bonds using the equation $\lambda=\hbar c/E.$ For example, light with a wavelength of about $\pu{241 nm}$ is needed to break the bonds of a $\ce{O2}$ molecule (assuming you know the bond dissociation energy).

Is it possible to use a specific wavelength to get the atoms to bond again? If so, is there an equation to calculate this for the atoms involved? I’ve been reading that lasers are used to cool atoms which leads me to think that since atoms are absorbing and reemitting the energy from the lasers to lower its kinetic energy, bonds could be formed when the it's low enough.

All I have been finding is how to break bonds using light. Is there light of a specific frequency that could be used to form a bond between given atoms in proximity?

$\endgroup$
5
  • 3
    $\begingroup$ You might find this article interesting. $\endgroup$
    – Paul
    Commented Jul 8, 2022 at 13:36
  • 3
  • 2
    $\begingroup$ Light could help to rearrange bonds by providing activation energy of reaction. OTOH, Bonding of free, unbound atom needs to release, not to absorb energy. E.g. 2 H atoms do not combine to H2, if they cannot pass energy away, like by collision. $\endgroup$
    – Poutnik
    Commented Jul 9, 2022 at 4:38
  • $\begingroup$ @NavaMoore, I think you are asking about the processes commonly known photodissociation (also photolysis) and photoassociation. $\endgroup$
    – PAEP
    Commented Jul 10, 2022 at 23:29
  • $\begingroup$ Also, regarding the photodissociation of $O_2$, it is not enough that the energy of the photon can break the bond, the photon must also be able to couple the ground state of the molecule to an excited state. I mean, there are other conditions that must be fulfilled. $\endgroup$
    – PAEP
    Commented Jul 10, 2022 at 23:32

3 Answers 3

5
$\begingroup$

Surely, if light can break bonds, it can make bonds as well.

We find some simple examples in high school textbooks as well, such as the chlorination of olefin compounds under the presence of UV light. This is an example of a photo-initiation reaction, where the first step is dependent on light. You can find more about this in this article1. Here is a schematic from the same article:

photo-halogenation of olefins

Do note that halogenation often occurs as a substitution reaction $\alpha$ to the double-bond. Under special circumstances (e.g., absence of $\alpha$-$\ce{H}$), halogenation can be made to occur as an addition reaction at the double-bond as well.

We can also consider some more interesting examples. Following reaction is from this study2

reversible dimerization reaction under light of different wavelength

Another schematic from the same article:

enter image description here

This concept is called photo-dimerization3:

The photodimerization is an example of a direct photoreaction where every step for polymer build-up is initiated by an absorbed photon, thus every single reaction step is dependent on the quantum yield of the photoreaction (generally very much smaller than one).

References

  1. Schönberg, A. (1968). Photohalogenation. In: Preparative Organic Photochemistry. Springer, Berlin, Heidelberg. DOI: 10.1007/978-3-642-87918-0_37
  2. "Light- and heat-triggered reversible luminescent materials based on polysiloxanes with anthracene groups" by Han et.al. RSC Adv., 2017,7, 56489-56495, DOI: 10.1039/c7ra12201b
  3. The Encyclopedia of Materials : Science and Technology. by K.H.J Buschow, 2001, Pages 6946-6951
$\endgroup$
1
  • 3
    $\begingroup$ From a synthetic chemist's point of view, I agree that light can make bonds ... eventually. However, the absorbance of light typically weakens or breaks bonds. This can be the first step in making a bond, as you show nicely for the radical reactions involving chlorine. Because you would not (could not) isolate the intermediate and it is happening in the same reactions mixture, it looks like the light "made the bond". Great examples in the answer! $\endgroup$
    – Karsten
    Commented Jul 10, 2022 at 8:42
5
$\begingroup$

Is it possible to use a specific wavelength to get the atoms to bond again?

Not directly. The crucial step in photochemistry is the absorption of a photon. Usually, this results in a molecule in an excited electronic state. Depending on the molecule, this will make a bond weaker, or break a bond.

An example of a simple (the simplest) molecule is the hydrogen molecular ion, $\ce{H2+}$. It has a single electron, making things so simple you can actually calculate the electronic states from scratch. If you excite the electron, the molecule will fall apart. You can read more about it on this page.

I am trying to bridge a conceptual understanding of how light affects both breaking and forming bonds. When bonds are broken, energy is absorbed (endothermic process). Conversely, when bonds are made, energy is released (exothermic process).

As Poutnik states in the comments to the question, making a bond is exothermic, so adding energy through photons is not productive for bond formation. However, for more complex photoreactions, absorbance of a photon is often the first step.

The most famous examples are cyclo-additions, or electrocyclic reactions. Some work without light (often indicated by "$\Delta$" for just heating up) while others require photoactivation (often indicated by "ℎ$ \nu$" for light with appropriate wavelength).

enter image description here

In the photochemical reaction sketched above, the absorbed photon results in a electronic excitation that does not break the molecule, but changes in electronic symmetry, allowing the reaction to happen. The product in the example above also is in an excited electronic state.

Is there light of a specific frequency that could be used to form a bond between given atoms in proximity?

Yes, but not in a single-step reaction where the bond "is made by the light", like welding two metal pieces together. Instead, using the welding analogy, you partially melt one work piece, and then attach it to the second work piece in a separate step that would not happen without the first step. Like all analogies, this does not capture the whole story, of course.

Is it possible to use a specific wavelength to get the atoms to bond again?

If the bond-breaking reaction is photochemical and there is an intermediate where the product is electronically excited before you get the ground state product, you can try to excite it to that state again to get the bond-forming reaction. This is non-equilibrium chemistry, so you can choose the wavelength (either exciting the reactant or the product) to manipulate the direction of the reaction away from the equilibrium concentrations, like the example in the answer by ananta.

$\endgroup$
0
3
$\begingroup$

Such a reaction would likely be indirect, with the bond formed by an intermediate that the electromagnetic radiation produces.

One of the simplest examples of such a reaction involves the noble gas helium. When helium gas is exposed to an electrical discharge, the electromagnetic radiation in this discharge ionizes some of the atoms, which then combine with still-neutral helium atoms to form the $\ce{He_2^+}$ cation (see Wikipedia). This is a true covalently bonded species, with bond order 1/2, because two of the valence electrons are in the bonding orbital and only one is in an antibonding orbital.

$\endgroup$
1
  • $\begingroup$ I do not say it is not equivalent. In fact, after seeing the update, I am going to delete the prior comment as obsolete. I have updated no answer. Saying one answer is good does not mean the other one is not. $\endgroup$
    – Poutnik
    Commented Jul 10, 2022 at 8:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.