B What happens when ethane absorbs UV light?

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When ethane absorbs UV light, the excitation of electrons from bonding to anti-bonding orbitals can lead to the disintegration of the molecule and the formation of radicals. The discussion highlights the complexity of photodissociation, emphasizing that the molecular structure and phase (gas, liquid, solid) significantly influence the process. The nature of the excited states is also crucial in understanding the outcomes of UV exposure. While the formation of radicals is well-documented, generalizations about the behavior of ethane versus ethene should be approached with caution due to their differing energy states. Overall, the interaction of UV light with ethane is a multifaceted topic that warrants careful study.
Lotto
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I have a molecule of ethane and when I use UV light, the molecule absorb it, so ethane's electron excitates from HOMO to LUMO. But what happens to the molecule? Does the sigma bond C-C still exist?
Because when I have two carbon atoms, they create a single bond, so their two electrons are in one boding orbital, but I have also one anti-bonding orbital here. And when the electron excitates from ##\sigma## to ##\sigma^*##, how can the bond between the carbon atoms exist? It has to disappear. But what happens next? Does the UV light affect also bonds C-H?
 
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Have you conducted a search of the literature? For example on the photodissociation of ethane.
 
Hyperfine said:
Have you conducted a search of the literature? For example on the photodissociation of ethane.
I found out that ethene can disintegrate into different particles, so my ideal should be right. And it is caused by the excitation of an electron of the molecule, so it can happen that electron in bonding orbital of C-H excitates into an anti-bonding orbital as well as an electron in C-C bonding orbital excitates.

But in general, when we use UV light, the molecule disintegrates and radicals can arise for instance.
 
Lotto said:
But in general, when we use UV light, the molecule disintegrates and radicals can arise for instance.
Indeed, radicals can arise! That fact has been well documented for many decades.

However, you must realize that the topic is very broad and certainly not lacking in complexity. The structure of the initial molecule is very important as is the phase (gas, liquid, solid) in which the process takes place. And then there is the question of the nature of the excited states that are populated. All due care should be taken in considering generalizations.

As but one illustrative example, Ahmed Zewail and his group at Cal Tech pursued detailed investigations of the photodissociation of I2 in rare gas clusters utilizing femtosecond spectroscopic techniques. Who would have guessed that the dynamics would be dependent on the polarizabilty of the rare gas?

You have mentioned both ethane and ethene. Very different molecules with very different energy states. Which one of the two is particular interest to you?
 
Hyperfine said:
Indeed, radicals can arise! That fact has been well documented for many decades.

However, you must realize that the topic is very broad and certainly not lacking in complexity. The structure of the initial molecule is very important as is the phase (gas, liquid, solid) in which the process takes place. And then there is the question of the nature of the excited states that are populated. All due care should be taken in considering generalizations.

As but one illustrative example, Ahmed Zewail and his group at Cal Tech pursued detailed investigations of the photodissociation of I2 in rare gas clusters utilizing femtosecond spectroscopic techniques. Who would have guessed that the dynamics would be dependent on the polarizabilty of the rare gas?

You have mentioned both ethane and ethene. Very different molecules with very different energy states. Which one of the two is particular interest to you?
That ethene was a mistake, I meant ethane.
 
For the quantum state ##|l,m\rangle= |2,0\rangle## the z-component of angular momentum is zero and ##|L^2|=6 \hbar^2##. According to uncertainty it is impossible to determine the values of ##L_x, L_y, L_z## simultaneously. However, we know that ##L_x## and ## L_y##, like ##L_z##, get the values ##(-2,-1,0,1,2) \hbar##. In other words, for the state ##|2,0\rangle## we have ##\vec{L}=(L_x, L_y,0)## with ##L_x## and ## L_y## one of the values ##(-2,-1,0,1,2) \hbar##. But none of these...