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Offline onespeedgo

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IR Radiation
« on: September 30, 2014, 10:43:44 PM »
Hey guys, I'm currently in an instrumental analysis course, and we've been given a study guide or questions of sorts of things that would be good to know for the exams.
I'm more familiar with microbiology than chemistry, so I find myself doubting my answers, and well, I only have so much time, so I appreciate any help that is available.
I'd just like to check with some guys sharper than myself if I've answered the question properly, or as best as possible, hope that's okay.

The first one is
Q: Why certain molecules absorb IR and others do not
A: Molecules that absorb IR have a dipole moment. The dipole is caused by uneven sharing of the electrons, and when the molecules move/vibrate/bend, it is with such little energy, or at the same frequency of IR radiation. Because they’re at the same frequency, this is what allows the interaction.

I haven't looked up these two yet, but just in advance
Q: Differences in instrumentation and applications between UV-Vis and IR absorption spectroscopy
Looking at these two images
UV

and IR


The most noticeable difference I see is that the radiation is filter after travelling through the sample in IR, while it's filtered before in UV-Vis


Q: Why is Fourier Transform used in modern IR instruments?
P 205 Skoog
Fourier transform is used in modern IR instruments because the radiant power that reaches the detector is much greater, and results in improved signal-to-noise ratio; extremely high resolving power and wavelength reproducibility; all elements of the source reach the detector at the same time thus allowing all the data to be gathered quickly
« Last Edit: September 30, 2014, 11:06:38 PM by onespeedgo »

Offline mjc123

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Re: IR Radiation
« Reply #1 on: October 01, 2014, 04:51:46 AM »
1. Strictly, the requirement is a dipole moment that changes during the course of the vibration. It is possible to get this in a molecule that has no overall dipole moment in the equilibrium configuration. For example, CO2 has no dipole moment because the dipoles of the two C=O bonds are equal and opposite. That situation is maintained throughout the symmetric stretching vibration, so this mode is not active in IR. But the asymmetric stretching vibration, and the bending vibration, both cause oscillating transient dipole moments, and are IR active.
Molecules that cannot absorb IR include monatomic gases like Ar, and homonuclear diatomics like N2 and O2. (That's why our atmosphere doesn't produce a runaway greenhouse effect like Venus's.)
2. Another obvious difference between your diagrams is that the IR is a dual beam instrument, where you are directly comparing the sample with a reference, while in the single beam UV you would have to run the reference first, then the sample. But this isn't a fundamental difference; you can get single and dual beam UV and IR spectrometers.

Offline onespeedgo

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Re: IR Radiation
« Reply #2 on: October 01, 2014, 08:34:37 AM »
Thanks for bringing up the points of symmetry/asymmetry/bending.
For that to happen, the bending/asymmetry, is there some way to identify molecules that do this? or do all molecules do this ? I'm asking because reading the text made it seem like "no dipole = no IR spec", but that doesn't seem to be the case as you pointed out with CO2.


Offline Corribus

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Re: IR Radiation
« Reply #3 on: October 01, 2014, 12:07:47 PM »
Rather than just throwing a selection rule out there, I find it is helpful to show what the selection rule means and where it comes from. This can help students understand the nature of spectroscopic transitions. Take what you want from this.

First, a word about light:

Light can be described as an oscillating electronmagnetic field. Ignore the magnetic component for now. For the purpose of basic spectroscopy, light is an oscillating electric field and the electric field vector can be described by a sinusoidal function. The magnitude of the electric field gets large in one direction, reaches a maximum, gets smaller, passes through zero, then gets large in the opposite direction, reaches a maximum, gets smaller, passes through zero, and so on. The direction of propogation of a photon is perpindicular to the direction of the electric field oscillation. So, if the photon is moving in the z-direction, the electric field is getting larger/smaller in either the y-direction, the x-direction, or some intermediate between the y- and x-direction. Polarized light is light in which the electric field oscillations of all the individual photons in a "beam" are oriented in the same direction. So - y-polarized light is light in which every photon is travelling in the z-direction and the electric fields are oscillating in the y (and -y) direction.

Here are a few crude but very simple figures illustrating this concept:

http://www.cfht.hawaii.edu/~manset/PolarIntro_eng.html

Now a word about how a charge behaves in an electric field:

Simply put, a stationary point charge experiences a force when in the presence of an electric field. In the presence of an oscillating field, the point charge will experience an oscillating force - it will want to move back/forth in a direction parallel to the direction of the electric field vector.

With that brief background, can we understand why vibrational transitions require a change in molecular dipole moment?

In the quantum mechanical model, the energy states of molecules and atoms are quantized, which is a fancy way of saying that molecular/atomic systems can only take on certain energy values. We might define these as "states" of the system. Each various allowed state is thus defined by an energy value and is characterized by some spatial arrangement of the system's particles (electrons and nuclei) in space. Since these particles are charged, each state is characterized by a specific charge distribution. The dipole moment is a reflection of any (average) asymetry in the charges in space for a particular state - i.e., if at a certain energy state there are more negative charges one side of the molecule than the other, we'd say the state has a non-zero dipole moment.

The basic mechanism with which light can interact with matter is by moving charged particles around, a process which takes energy. This is essentially what it means for a molecule to "absorb" a photon. The fundamental question we're asking is: under what circumstances can an incident photon transfer a molecule/atom from one of these states to the other?

In a first approximation, there are two basic criteria that must be met for light to induce a molecular change of state:

1. The energy of the photon must be the same (or nearly so) as the difference in energy between the two states. In a classical system, this is the same as saying that in order to throw a ball to the top of a hill, you must provide at least enough energy to get it up there - otherwise it will fall short. The difference between the classical and quantum world is that while in a classical world you can throw the ball halfway up the hill by providing half the energy needed to get to the top, in the quantum world the ball either has to go to the top of the hill, or it goes nowhere. There are no half-measures allowed in the quantum world.

2. The photon has to be able to drive the spatial rearrangement of the particles (electrons/nuclei) between the initial and final state. This is probably best demonstrated by an rough analogy. Supposing a molecule is a sailboat and the initial state of the sailboat is New York and the final state of the sailboat is London. The light in this analogy is wind. Even if the wind is blowing with sufficient energy to get your boat from New York to London, you won't get there unless the wind is blowing east. The only change in state an east-bound wind can produce is a change in state from New York to, say, Boston. Likewise, using wind alone it is not possible to induce a change in state from New York to a cloud directly above New York, because there is no directional change that wind can induce in order to drive this change of state (ignoring the possibility that the wind can blow straight up).

Let's think about how that works in a vibrating molecule. For the moment, let's pretend the molecular and laboratory frames are both fixed. And let's assume that light is propogating in the z-direction and it is x-polarized, so that means the electric field is only oscillating in the +x and -x direction. Now let it be incident on a single molecule fixed in space but allowed to vibrate. It's hard to depict this graphically, but let the direction of propagation be indicated by the arrow and we'll have a diatomic molecule X-Y oriented along the same axis (call it z-axis - left to right on your screen) as follows:

:rarrow: X-Y

The electric field, note, is "up" and "down" in the plane of your computer screen.

The molecule X-Y has a permanent dipole moment. And when it vibrates, the dipole moment increases and decreases in an oscillatory fashion (when the bond length is long, the dipole moment increases and when the bond length is short, the dipole moment decreases; dipole moment is the difference in charge magnitude times separation distance). When the molecule absorbs energy, the vibration becomes...er... more vigrorous and the average separation between the nuclei increases. This has the effect that the average dipole moment increases between the lower state and the upper state. Dipole moment is a vector (it's oriented in a direction) and in our X-Y molecule, the dipole moment is oriented also along the z-axis. So, to induce a change in a vibrational state - that is, for the molecule to absorb the energy of a photon and have that energy go into increasing the energy of the molecular vibration - the electric field of the photon has to be in the same direction as the dipole moment change. If it is isn't, the "force" of the electric field can't move the "charges" in the molecule associated with the change in vibrational state. The wind isn't blowing in the right direction. In the orientation shown above, the electric field direction is perpendicular to the dipole moment vectors, so no transition is allowed. Light travelling in the z-direction cannot induce a change on the X-Y molecule when the bond axis is also the z-direction.

Compare that scenario to this one:

 :spinup: X-Y

Now, the direction of propogation is in the y-direction (say) and the electric field is oriented in the z-direction. The electric field can interact with the molecular vibration and the transition is allowed - provided the energy of the photon is equal to the energy difference between the lower and upper vibrational states.

Some additional but very important notes:

1. so far we've only looked at situations where the light vectors are either parallel or perpendicular to the dipole moment vector. Obviously, this doesn't have to be the case. The probability of a transition occurring is actually related to the square cosine between the two vectors.

2. In a real laboratory frame, the molecular axes are not fixed in space. Molecules are oriented in random directions. What this means in practice is that unless molecules are frozen in an aligned way, the directionality of the dipole moment vector doesn't really matter. As long as there is some dipole moment change between the states, there will be some molecules at any point in time which can interact with the light's electric field.

This second bit is the origin of the common wisdom that a vibrational transition has to involve a change in dipole moment to be allowed. Molecules with more than two nuclei have multiple possible vibrations. Only those changes in vibration state that involve a change in dipole moment are able to be induced by the oscillating electric field of the incident light. Higher vibrations of a symmetric molecule like nitrogen (N2) certainly exist, but they cannot be produced (to a first order approximation) with light, because there's no way for an electric field to "push" the nuclei in dinitrogen further apart - there's no charge asymmetry to interact properly with the electric force. As the previous poster noted, though, a permanent dipole moment is not required - carbon dioxide is symmetric and has no permament dipole moment, but some vibrations involve an asymmetric motion of the nuclei, such that higher vibration states DO have a non-zero dipole moment. There is a dipole moment change, so the "transition dipole" can couple with the light electric field in this case.

As a final note, this discussion has been completely qualitative. A quantitative assessment of "allowedness" results from solution of the transition dipole moment integral, which uses the initial and final state wavefunctions and the dipole moment operator. Solving this expression will tell you what properties of the initial and final states lead to allowed transitions (and, in principle, can quantify the probability of a transition). In the case of vibrational transitions, for example, not only is a dipole moment change required, but only transitions between adjacent states are allowed, at least in a first, harmonic oscillator approximation. In essence, though, the meaning of this integral is essentially what has been presented above: what is the ability of the light electric field to induce a change between the initial state and the final state, and this ability relates to the efficiency of coupling between the change in dipole moment between the two states and the electric field of the incident photon. Note also that these considerations also apply to other kinds of transitions, such as rotational and electronic. But the vibrational selection rules are especially easy to understand because "vibration" is a very pseudoclassical phenomenon. 

http://en.wikipedia.org/wiki/Transition_dipole_moment

Regarding your follow up question:

With the exception of symmetric diatomic molecules, virtually every molecule has IR-active modes, because there are always asymmetric vibrations (stretches and bends/wags/whatever). Therefore the question isn't what molecules are IR-active, but which vibrational modes are, where they appear, and how strong they are. Predicting all the modes of molecules requires some fairly sophisticated symmetry and/or quantum mechanical treatments that are beyond the scope of this thread. Also note that in big molecules, usually it is vibrations of functional groups that are the focus of practical study rather than vibrations that span the entire molecule, because vibrations in very large molecules aren't necessarily "globally coordinated". So we usually speak of C=O stretches and so forth rather than, "the stretching mode of trypsin".
« Last Edit: October 01, 2014, 12:31:57 PM by Corribus »
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Offline mjc123

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Re: IR Radiation
« Reply #4 on: October 01, 2014, 12:12:14 PM »
Apart from monatomics (which have no vibrations) and homonuclear diatomics (which only have a symmetric stretch), all molecules, even those with no permanent dipole moment, will have some vibrational modes which produce an oscillating dipole moment and are IR active. "no dipole = no IR spec" is definitely wrong. "No oscillating dipole = no IR absorption" is the rule.

Offline onespeedgo

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Re: IR Radiation
« Reply #5 on: October 01, 2014, 08:05:59 PM »
Wow, that'll be a lot to digest, thanks Corribus. You too mjc.

When you say "polarized light", is that the same as coherence? (and then incoherent and non-polarized? From what I read about coherence, it sounded like they may be the same thing, in that the photons are entirely synchronized.
https://au.answers.yahoo.com/question/index?qid=20080730114522AAywxla
 
Regarding your first point Corribus, that's how it was explained today in class. Although I did I have a question that I wasn't able to ask my prof. The way he said it, and this is probably just my misunderstanding, he made it sound like if the photon provided too much energy, that it wouldn't jump to the next level. My thinking is that if each level of the excited state say required 100 J, then a photon with 110 J would get you to the first level but not the second. Is that correct?

Your second point is understood. Thank you.

When you say
Quote
The molecule X-Y has a permanent dipole moment. And when it vibrates, the dipole moment increases and decreases in an oscillatory fashion (when the bond length is long, the dipole moment increases and when the bond length is short, the dipole moment decreases; dipole moment is the difference in charge magnitude times separation distance).

Also, when you speak about the importance of directionality, is it that the photon and the oscillation of the molecule have to be in line? or could a photon with enough energy that isn't traveling in the same direction but passes by be enough to cause the electron to reach a higher energy?

Is it that the oscillating movement requires the same energy as the IR photon in order for it to be detected? or that the oscillation is the same distance as the wavelength the photon is moving? I hope that makes sense, feeling pretty newbie right now.

I really wish this course was two terms because of all it covers. It's interesting, but so fast-paced that it's hard to keep up and gain a true appreciation. :(

We were talking about the spectral fingerprint of different elements today, and it seems like based on each element's proton-electron configuration, and how they are in relation to each other, eg how tightly/loosely they're held, that this affects the energy required to make an electron reach a higher energy level, is why each element has a unique "fingerprint". Hope that made sense, and if so, is my thinking correct?

Apologies for all the questions, I am meeting with my prof this Friday, but it's always nice to have multiple interpretations of the same information.

Offline Corribus

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Re: IR Radiation
« Reply #6 on: October 01, 2014, 09:57:56 PM »
@onespeedgo

Quote
When you say "polarized light", is that the same as coherence? (and then incoherent and non-polarized? From what I read about coherence, it sounded like they may be the same thing, in that the photons are entirely synchronized.
Not typically. Coherence usually relates to the relative phases of several waves; polarization refers to the relative orientation of the (in the case of light wave) vector electric field magnitude.  So, I presume you know that the energy of the light wave is related to the wavelength - the distance between adjacent peaks. Well, suppose you have two waves travelling side by side. They can have the same energy (same wavelength) but be out of phase, meaning the peak maxima reach a certain point in space at different times. Sin (x) and Cos (x) have the same wavelengths and amplitudes (and vector orientation, if we consider 3 space) but are out of phase - in the cos(x) function, the first peak is at x = 0 and in the sin (x) function, the first peak is at x = pi/2. This kind of temporal coherence is usually encountered in things like lasers, where constructive and destructive interference of pulsed light are important.

Sometimes polarization is called "polarization coherence", although this isn't very common. This may be causing your confusion.

Quote
Regarding your first point Corribus, that's how it was explained today in class. Although I did I have a question that I wasn't able to ask my prof. The way he said it, and this is probably just my misunderstanding, he made it sound like if the photon provided too much energy, that it wouldn't jump to the next level. My thinking is that if each level of the excited state say required 100 J, then a photon with 110 J would get you to the first level but not the second. Is that correct?
Officially, no, it wouldn't get you anything. The photon wouldn't be absorbed at all, because the resonance condition is not met. The resonance (or Bohr) condition is that the energy of the incident photon has to match exactly the energy gap between the two states, or the light simply doesn't interact at all with the system. Because of the uncertainty principle and some other physical effects, however, there is often some "wiggle-room". This is why all absorption bands have a natural linewidth - they are not infinitely narrow lines.

Quote
Also, when you speak about the importance of directionality, is it that the photon and the oscillation of the molecule have to be in line? or could a photon with enough energy that isn't traveling in the same direction but passes by be enough to cause the electron to reach a higher energy?
The electric field and the oscillating dipole vectors have to share some commonality for there to be any non-zero probability of absorption. If the angle between them is 90 degrees, the probability of absorption is exactly zero. If the angle between them is 0.0000001 degrees, there is a small but nonzero probability of absorption. Do note that it is not the direction of propagation that matters - it's the direction in which the electric field is oriented, which is perpendicular to the propagation direction.

Quote
Is it that the oscillating movement requires the same energy as the IR photon in order for it to be detected? or that the oscillation is the same distance as the wavelength the photon is moving? I hope that makes sense, feeling pretty newbie right now.
I am not sure I quite understand the question. In some way, these two alternatives are the same. The energy of the IR photon has to match the energy difference between the ground and excited vibrational state. The spatial magnitude of the oscillation of the atoms during a vibration is related to the vibration frequency/energy (because it's related to the force constant, if you are familiar with Hooke's law); and the wavelength of the photon is related to its energy.

Quote
We were talking about the spectral fingerprint of different elements today, and it seems like based on each element's proton-electron configuration, and how they are in relation to each other, eg how tightly/loosely they're held, that this affects the energy required to make an electron reach a higher energy level, is why each element has a unique "fingerprint".
Yes, this is correct. It's the basis, for example, of how astronomers determine the composition of stars. In fact, atomic/stellar spectral lines were among the earliest unexplained observations in classical physics that led to the formulation of quantum mechanics and was historically important in determining the structure of the atom... and the ability of quantum mechanics to explain spectral lines was an early crowning achievement of the theory. Your statement above is a very broad simplification of what's going, but in general it's correct.

You might be interested in this article:

http://en.wikipedia.org/wiki/Balmer_series

To bring the discussion back to analytical chemistry, atomic spectral lines are also the basis of atomic absorption spectroscopy (AAS) and inductively-coupled plasma atomic emission spectroscopy (ICP-AES), which are routinely used today for trace metal analysis, which has applications in forensics, environmental and food safety, and a lot of basic scientific research.

http://en.wikipedia.org/wiki/Atomic_absorption_spectroscopy
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Offline onespeedgo

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Re: IR Radiation
« Reply #7 on: October 01, 2014, 10:27:41 PM »
Coherence
Ok, thank you for clearing that up.

Quote
Resonance condition

Quote
Electric field propagation

Quote
Hooke's law

Thank you for clearing these up for me. Wish I knew a fraction of what you do.

Quote
To bring the discussion back to analytical chemistry, atomic spectral lines are also the basis of atomic absorption spectroscopy (AAS) and inductively-coupled plasma atomic emission spectroscopy (ICP-AES), which are routinely used today for trace metal analysis, which has applications in forensics, environmental and food safety, and a lot of basic scientific research.

http://en.wikipedia.org/wiki/Atomic_absorption_spectroscopy

I'll be using an ICP-OES in a week.

What field did you work in ? I saw this on wiki by the same author as you go by, I'm assuming it's you.
https://en.wikipedia.org/wiki/User:Corribus

Offline Corribus

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Re: IR Radiation
« Reply #8 on: October 01, 2014, 10:44:26 PM »
LOL

Well, I went through a 5 minute period of my life where I intended to edit some Wikipedia articles and then I promptly decided, "The hell with that," and haven't done it since. :)

Since you asked, I'm a physical chemist trained in spectroscopy. But I've done a little of everything, from atmospheric chemistry to electronic materials to nanoscience. Currently I work with nanocomposites. I've also developed a keen interest in, and published a little bit on, public risk perceptions and how they relate to development of emerging technologies.  As you can see, I don't like to stick in one area for long!
What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman

Offline onespeedgo

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Re: IR Radiation
« Reply #9 on: October 01, 2014, 11:14:43 PM »
I'm finding it pretty hard to decide which direction to go with all the things going on in the scientific world.
I'm in my 4th of a 5 year program (it's actually a 4 year degree, but too many labs to take 5 courses/term) in biotechnology. It's a joint chemistry-microbiology honours program.
I find that I'm getting burnt out from the university educational system. I've been told university's are for research and teaching is a thing on the side, but this might just be because it's a research-based university.

I don't like how many courses I've done where the most difficult part is simply the sheer volume of material.
It's like the courses have continued to put things in, while never trimming the course.

What was your university experience like?

Offline Corribus

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Re: IR Radiation
« Reply #10 on: October 02, 2014, 09:06:17 AM »
I don't want to start ranting about university hiring practices. But in my personal experience there are many professors who view teaching as an obligation - and hence they either aren't good enough at it or don't care enough about it to do it well. While teaching is allegedly a major part of the professor's job description, in practice bringing in research dollars is treated as far more important by Universities. You can't be a teacher first and researcher second. If you're lucky you'll get a teaching assistant who likes it, but they're as much a mixed bag as professors are. Unfortunately.
What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman

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