# How exactly does UV light break bonds?

For example, in the photochemical decomposition of hydrogen peroxide, the weak $$\ce{O-O}$$ bond is broken by UV light. Ultraviolet light can also dissociate relatively strong bonds such as the double oxygen bond ($$\ce{O=O}$$) in molecular oxygen ($$\ce{O2}$$) and the double $$\ce{C=O}$$ bond in carbon dioxide ($$\ce{CO2}$$). My question is, how does this process occur? How exactly does UV light break bonds?

• UV breaks bonds in a fashion similar to heat. Jun 27 at 6:52
• @NisargBhavsar please elaborate. I know what dissociation energies are and that stability is inversely proportional to a species's energy, but that doesn't answer my question. Jun 27 at 6:55
• UV is just another way of providing energy to the system. It is used because sometimes it is inconvenient to heat stuff or it is very slow to do so. Amount of energy provided per package can also be controlled in case of UV and thus prevents side reactions. Jun 27 at 7:02
• Briefly an electronically excited state is produced (distinct from the ground state) and if the uv energy is sufficient it is produced above its dissociation energy , which is less than in the ground state. Alternatively in the excited state there is another excited state that the energy from the first one crosses over into and this dissociates. Jun 27 at 7:29
• All starts with quantum interaction of electrons and photons. An UV photon passes enough energy to a bond electron to break a bond, frequently together with ionisation of fragments. Jun 27 at 9:28

Briefly an electronically excited state is produced (distinct from the ground state) and if the uv energy is sufficient it is produced above its dissociation energy, which is less than in the ground state. Alternatively in the excited state there is another excited state that the energy from the first one crosses over into and this dissociates. The figure tries to illustrate this.

• Franck - Condon
– Karl
Jun 27 at 12:24

In my opinion, a good way to understand it is the molecular orbital (MO) theory. In this theory we assume that atomic orbitals (AOs) of two atoms overlap and create MOs. The number of AOs interacting is the same as the number of MOs formed. This is the representation of MOs formation:

In this case, two 1s-orbitals of hydrogen atoms overlap to form two $$\sigma$$ MOs. One MO has lower energy than the starting AOs, therefore it's called bonding, the other MO has higher energy and is called antibonding. We represent antibonding orbitals with the asterisk (*).

Now, why is this molecule formed? In this diagram we can see that the two electrons will have a state that is lower in energy if AOs overlap and MOs form. That means, if the $$\ce{H2}$$ molecule is formed, the energy of the system decreases. That's why $$\ce{H2}$$ will form.

One key thing is that the energy difference $$E(1 \sigma ^*) - E(1s) > E(1s) - E(1 \sigma)$$. That means, the antibonding orbital is slightly more antibonding than the bonding orbital is bonding. That is why if there's equal number of electrons both in $$1 \sigma ^*$$ and $$1 \sigma$$, the molecule will not form, as the overall energy of the electrons will be higher than that of the starting 1s-orbitals.

Now we can return to your question. What happens when the UV light is radiated (or heat is applied)? Of course, we add the energy to the system. And when we add that energy, if $$E(\text{UV light}) = E(1 \sigma ^*) - E(1 \sigma)$$, then one of the electrons in $$1 \sigma$$-orbital can go from its current state to $$1 \sigma ^*$$-orbital.

After this happens, we will have one electron in $$1 \sigma$$ and one electron in $$1 \sigma ^*$$. What we've said before? We've said that the overall energy in this case increases and the formation of the molecule is now not favorable. In other words, $$\ce{H-H}$$ bond will break.

This is hydrogen molecule. Same thing applies to oxygen but it will have slightly more complicated scheme.

• You need to deal with states not just orbitals and and how these states change in energy vs bond extension. Jun 27 at 8:43
• @porphyrin, sorry, what exactly do you mean? Jun 27 at 8:49
• What you say is that a single photon can bring the bond into a state where it has more energy than if there was no bond. Correct, but that does not explain why the molecule dissociates.
– Karl
Jun 27 at 12:11
• @Karl, I am confused now. Can you recommend any book or article that explains this? Jun 28 at 6:23