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UV-Vis spectroscopy: non-bonding electrons and chromophores

  1. Jun 19, 2016 #1
    Hello everyone,
    I'm just learning about UV-Vis spectroscopy, I have some questions and I would be really thankful for help.

    It's stated that apsorption happens when electrons go from ground to excited state. And that transition to excited state is easier for non-bonding electrons which I imagine as ones which do not form a bond.

    But then it's said that molecule needs to have chromophores to absorb, so more unsaturated bonds, and here I imagine those electrons being inside the bond. But is that maybe wrong? My question is - do chromophores have non-bonding electrons then which allow a molecule to absorb?

    I'm a little bit confused. For sure I lack some theoretic knowledge about molecular bonds, but I hope you understood what I wanted to ask.

    Thanks :)
     
  2. jcsd
  3. Jun 20, 2016 #2
    By the definition chromophore is group of atoms responsible for selective absorption of light with wavelenght 180-800 nm. There are few types of transition (depending on molecule) that can occur after light absorption. In UV/VIS spectroscopy we are mainly talking about electronic transitions. So there are few types of transitions, for instance:
    • pi - pi anti-bonding;
    • non-bonding-sigma anti-bonding etc.
    Hence in ethylene, C=C is responsible for pi-pi anti-bonding transition (c.a. 162 nm) but in methyl bromide you can see non-bonding- sigma anti-bonding transition (c.a. 204 nm). To answer your question, chromophores may have non-bonding electrons but it is not necessary for them (see ethylene ).
     
  4. Jun 20, 2016 #3
    Thank you forestgumper! So the point is that chromophores simply absorb UV-Vis light with wavelength 180-800 nm, I don't need to specify why is that so. And I think I read that the transition to excited state is just EASIER with non-bonding electrons, but it's possible for others to get excited as well. Cool.

    Thanks again :)
     
  5. Jun 21, 2016 #4
    Take benzene for example. The lambda max is 254 nm, resulting from the transition of the non-local pi electrons to the excited pi* state. Remember, the pi electrons in benzene are in resonance throughout the ring structure. Chromophores attached to the benzene ring will shift lambda max, depending on whether the chromphore pumps electron density into the ring, resonates with the ring, or withdraws electron density from the ring.
     
  6. Jul 8, 2016 #5
    I think you are confused about transitions in molecules.

    When you say excited state, specifically, which excited state do you mean? In molecules, there may be π-π*, σ-π*, n-π*, and σ-σ* excited states. The star "*" means anti-bonding orbitals. n means non-bonding orbital.

    n-π* transitions typically have small absorption coefficient compared to π-π* transitions. Using your words, this means that n-π* transitions are "not easy". So I am not sure where you got the information that transition to excited state is "easier" for non-bonding electrons (you might have mixed it up with intersystem crossing). In technical term, n-π* transitions are "forbidden" transition. This means that the transition integral, which represent the probability that such transition occur, is zero (no transition is possible) in 0th order. To give you an intuitively comprehensible (but not necessarily the right) explanation, the non-bonding orbital and π* orbitals are orthogonal, thus these two orbitals do not overlap. When an electromagnetic wave passes through a molecule with a suitable energy, an electron in the non-bonding orbital would oscillate but since there is no overlap between this orbital and the π* orbital, the electron cannot move to the π* orbital, ending up with electromagnetic wave just passing through without being absorbed.

    n-π* transitions do occur in practice, though. This is because the vibration of the molecule alters the wavefunction of an electronic state, leading to a slight overlap of wavefunction of ground state and n-π* excited state, allowing the transition to happen.


    Meanwhile, a chromophore contain π-orbitals. Overlap between π-orbital and π*-orbital is usually large and thus transition probability is high. Benzene is actually a bad example to talk about when it comes to chromophores even though it has π-π* transitions. Benzene have 2 degenerate HOMO and 2 degenerate LUMO (4 different transition is possible), but only one of those transition is allowed due to symmetry. Even this allowed transition is not as strong as you would expect because of the mixing between the multiple excited states (configuration interaction) that "borrows" the "forbidden-ness" of the other transitions. In napthalene, anthracene, etc. this symmetry breaks down showing more stronger transitions.
     
  7. Jul 8, 2016 #6
    I think you might be confused, I didn't say anything about non-bonding electron transitions. I only mentioned pi tp pi* transitions in benzene.
     
  8. Jul 8, 2016 #7
    I know. When I was talking about benzene, I was also talking about the π-π* transition as well. But benzene is a bad example because despite having three (sorry, I said four in the above post, but it's actually three) π-π* transitions between orbitals, only one of that transition is allowed due to symmetry and the rest are symmetry forbidden. The 250 nm transition (S0→S1 transition) in benzene is actually a symmetry forbidden (but the symmetry is broken by phonon-electron interaction, hence why the transition can be observed in practice). As a result, S0→S1 only has absorption coefficient of 220 M-1 cm-1. As a comparison, for naphthalene, the absorption coefficient for the S0→S1 transition observed at 286 nm is 4000 M-1 cm-1.


    Thread starter is confused with two contradicting information she has in her mind. That is, n-π* transition is a "easier transition" (which is not correct as I mentioned in my previous post), and the requirement of chromophores (meaning π-π* transition) for efficient absorption of light. The latter is right. However, your post about benzene is misleading as an answer to her post because benzene is a special case where there are symmetry forbidden π-π* transitions. If she was to look this up on the internet and see that benzene consists of several forbidden π-π* transitions, she will be even more confused.
     
    Last edited: Jul 8, 2016
  9. Jul 8, 2016 #8
    Thank you both for your replies! :)

    I don't have much knowledge about orbitals (although I should maybe have) and I wasn't planning to go into depth with them as much (this course is about Drugs analytics so we examine lots of other techniques as well). However, what I did get is that chromophores have pi-bonds thus it's probable that electrons will get excited.

    My summary on UV-Vis:
    When we analyze something, we have a source of UV-Vis light (xenon lamp for example)...and monochromator will separate different kinds of wavelengths (why is that necessary, we already have just the ones we need?)...and then light passes through the sample...only when the wavelength/energy is equal to the energy difference between two orbitals (orbitals?space where electron spends 95% of it's time?), those bonds (between orbitals?) will absorb that certain wavelength...that's why in graph we will have absorption maxima on for example 264 nm (because that wavelength was just the right one...and maximas are the only thing that matter, right? what's with the rest of the graph which always looks kinda silly?)...and what actually happens when energy is absorbed is that electrons go to excited state (is it all the way while light is passing through the sample or only for a moment?)...and what does excited state even mean, is it manifested in physical terms, like, what does that mean for an electron?
    Anyway, even if the light that we are using is too intensive or less intensive than how it should be, absorption will not occur, it needs to be just perfect for a certain bond (why does more intensive light does not cause electrons to get excited?what does it depend on?)

    Haha I know questions are probably stupid and ''basic'', but I'd be really happy to see your answers on these too :)

    Thanks!
     
  10. Jul 8, 2016 #9
    1) why is that necessary, we already have just the ones we need?
    Well the problem is, we don't have just the ones we need. The wavelength of a xenon lamp ranges from 240 nm all the way down to 1800 nm. A molecule does not always respond the same to different wavelength of light. A molecule may have several excited states that corresponds to different energy. Sometimes these excited states behaves differently or react with other molecules differently. For analytical reason, we need to know how each wavelength affects the molecule.

    2) only when the wavelength/energy is equal to the energy difference between two orbitals
    This is actually not true, however the energy of light (wavelength) required for the transition is typically close to that energy. The energy gap is also affected by exchange interaction and coulomb interaction.

    3) those bonds (between orbitals?)
    It doesn't have to be a bond, but most of the molecules does indeed.

    4) maximas are the only thing that matter, right? what's with the rest of the graph which always looks kinda silly?
    Practically, maybe, yes. However, technically and analytically, no. The rest of the graph gives important information about the molecule. For example, where the other electronically excited states are, and how separated the vibronic levels are at excited state. Sometimes, when you take an absorption spectra of a molecule they will have a maxima, and then show several similarly shaped bands (with smaller intensity) with equal space between those bands. These bands are called "vibronic bands" and corresponds to the phonon quantum number of the molecule in the excited state. Most of the time, this band can be represented by harmonic oscillators.

    5) what actually happens when energy is absorbed is that electrons go to excited state (is it all the way while light is passing through the sample or only for a moment?)
    No, once a molecule absorbs light to become an excited state, the light (a photon) is annihilated.

    6) and what does excited state even mean?
    We have several "excited state". The one you are mainly talking about is called "electronically excited state" that is, when an electron is oscillated by incident light to an orbital with higher energy. One of the common mistakes people make about "excited state" is that they think it is the unoccupied orbital that is the excited state. This is not true. A ground state is when all of the electrons are occupied in the orbitals of low energy. So these electrons can be considered a "valance" electrons. An excited state is when one or more electron(s) are moved to the orbitals of high energy and the rest remains on the bottom orbitals.
    From the electrons point of view, the trajectory of their motion as well as their position will change. Think about π-orbitals of ethylene (http://photonicswiki.org/images/thumb/1/18/Ethylene_mo.png/300px-Ethylene_mo.png). The electron obviously occupy a different orbital.

    7) why does more intensive light does not cause electrons to get excited?what does it depend on?
    Perhaps you are confused about intensity of the light and energy of the light. As you may know, light can be considered a particle called "photons". When the intensity of the light is high, it means that there are a lot of photons. Meanwhile, when the energy of the light is high, it means that each photon have high energy. These two are different quantity that describes incident light. No matter how intense the light is, if each photon does not carry the energy required for transition of an electron in a molecule, then transition will not occur. It will just go past.
    (It should be noted that absorption still may occur with sufficient intensity of light in some very special cases such as "two-photon absorption". This is a nonlinear effect but is completely beyond the scope of our discussion so I will not explain this here.)
     
  11. Jul 9, 2016 #10
    Wow! Thank you very much for your answer, things are indeed much clearer to me now.

    Cheers! :)
     
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