For more information, please see full course syllabus of Physical Chemistry

For more information, please see full course syllabus of Physical Chemistry

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## Download Lecture Slides

## Table of Contents

## Transcription

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### Vibration-Rotation

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

- Intro
- Vibration-Rotation
- What is Molecular Spectroscopy?
- Microwave, Infrared Radiation, Visible & Ultraviolet
- Equation for the Frequency of the Absorbed Radiation
- Wavenumbers
- Diatomic Molecules: Energy of the Harmonic Oscillator
- Selection Rules for Vibrational Transitions
- Energy of the Rigid Rotator
- Angular Momentum of the Rotator
- Rotational Term F(J)
- Selection Rules for Rotational Transition
- Vibration Level & Rotational States
- Selection Rules for Vibration-Rotation
- Frequency of Absorption
- Diagram: Energy Transition
- Vibration-Rotation Spectrum: HCl
- Vibration-Rotation Spectrum: Carbon Monoxide

- Intro 0:00
- Vibration-Rotation 0:37
- What is Molecular Spectroscopy?
- Microwave, Infrared Radiation, Visible & Ultraviolet
- Equation for the Frequency of the Absorbed Radiation
- Wavenumbers
- Diatomic Molecules: Energy of the Harmonic Oscillator
- Selection Rules for Vibrational Transitions
- Energy of the Rigid Rotator
- Angular Momentum of the Rotator
- Rotational Term F(J)
- Selection Rules for Rotational Transition
- Vibration Level & Rotational States
- Selection Rules for Vibration-Rotation
- Frequency of Absorption
- Diagram: Energy Transition
- Vibration-Rotation Spectrum: HCl
- Vibration-Rotation Spectrum: Carbon Monoxide

### Physical Chemistry Online Course

### Transcription: Vibration-Rotation

*Hello and welcome back to www.educator.com, welcome back to Physical Chemistry.*0000

*Today, we are going to start on the next major broad topic of Physical Chemistry which is molecular spectroscopy.*0004

*We have done the thermodynamics, we have done the quantum mechanics, and now, we are going to bring*0013

*the quantum mechanics to bare and talk about probably the most important topic for practicality*0016

*is concerned for the chemist because pretty much everything you do as a chemist is going to be some spectroscopic technique.*0024

*We will be discussing the theory behind molecular spectroscopy, let us get started.*0031

*Molecular spectroscopy studies the interaction between a matter and electromagnetic radiation.*0039

*Molecular spectroscopy studies the interaction of molecules with electromagnetic radiation.*0053

*We get molecules with certain amount of radiation in different regions of the electromagnetic spectrum and we see what happens.*0079

*That information gives us most of the information that we have about what is happening inside the molecules and how molecules behave.*0085

*We will be concerned with three regions of the E and M spectrum.*0095

*We are going to be concerned with the microwave region.*0115

*Microwave region is related to rotations.*0120

*When a molecule is hit with microwave radiation, there are changes in the rotational quantum state of the molecule.*0125

*And we are also going to be concerned with the infrared, the infrared vibrations.*0133

*When a molecule is hit with some infrared radiation, changes in the vibration levels of the molecules take place.*0138

*We have already done a fair amount of IR spectroscopy from your course in organic chemistry.*0147

*And then of course, the last is visible and ultraviolet.*0152

*Visible and ultraviolet range electronic transitions, that is when it is in this range,*0157

*the energy of the visible and ultraviolet range, where electrons actually move to higher states themselves, electronic states.*0165

*Rotation vibration electronic, rotational spectroscopy, vibration spectroscopy, and electronic spectroscopy.*0175

*Let me see how I want to do this.*0184

*The absorption of microwave radiation, as we just said, we have rotational transitions.*0202

*Now in the IR, not only do we have vibrational transitions but accompany those vibrational transitions there are also rotational transitions.*0212

*I’m going to say vibrational + rotational transitions.*0224

*As you would expect in the visible UV range, we have not only electronic transitions*0232

*but there is much energy there that that energy causes an electronic transition,*0240

*it causes vibrational transitions and rotational transitions.*0244

*+ vibrational + rotational transition, all transitions take place when you are hitting it with energy and the visible UV range.*0249

*Most of the information that we get from spectroscopy, we actually get from electronic spectroscopy.*0263

*Electronic spectroscopy allows us to, it is difficult to analyze but everything that we need is there.*0268

*It gives us information on electronic states, on vibrational states, on rotational states.*0275

*The rotational spectra, the vibrational spectra tend to be easier, but they do not give is as much information.*0280

*Electronic spectra give us all the information that we want.*0288

*Let us see, the frequency of the absorb radiation comes from this δ E = H ν,*0297

*Planck's constant × the frequency is a change in energy from one energy state to another.*0318

*If we solve for the frequency, what you would end up with is the frequency of absorption is going to equal*0324

*the change in the energy between the 2 states divided by Planck's constant,*0330

*or we can say the energy of the upper state - the energy of the lower state divided by Planck's constant.*0335

*That gives us the actual frequency that we observe in the spectra.*0345

*That is what we are going to see on the spectrum.*0348

*When we see a line of the spectrum or a peak, it is this number right here.*0350

*That is what that is, it come from this relation.*0356

*What we are going to do, we would use the energies for the rigid rotator, the harmonic oscillator,*0359

*and electronic energies, to find this difference.*0365

*Because we want to find the equation for what is the absorption number.*0369

*In general, spectroscopy frequencies, I will put in parentheses.*0376

*Frequencies are listed in something called wave numbers, we have seen them before.*0387

*Frequencies are given in wave number.*0391

*That is what we would be working in,*0394

*are given in wave numbers which is just the inverse of the wavelength to inverse cm.*0396

*In general, it is going to be inverse cm.*0404

*The wave number, anything that is in a wave numbers is going to have a ~ over it.*0408

*That is equal to 1/ λ, or my preference it is equal to the actual frequency*0412

*that we got from the other equation just divided by the speed of light.*0420

*If you get a frequency, if you divide that frequency by the speed of light, you are going to end up getting your wave number.*0424

*We said transitions between vibrational states are accompanied by transitions and rotational states as well.*0431

*The following discussion is going to apply to diatomic molecules.*0480

*We will discuss polyatomic molecules but for right now,*0484

*we are just going to talk about diatomic molecules, homo nuclear and hetero nuclear.*0487

*The following discussion and for several more lessons, the following discussion applies to diatomic molecules.*0495

*Let us begin with the energy of the harmonic oscillator.*0514

*The energy of the harmonic oscillator HO is energy, that space on the quantum number R,*0518

*the vibrational quantum number is equal to H ν × R + ½, or R takes on the values 0, 1, 2, and so on.*0532

*When R is equal to 0, we have ½ H ν, that is the energy of the ground state.*0545

*When the quantum number is 0, put it in the equation for the energy which gives the ground state energy.*0551

*We see that there is always some vibrational energy, it is never a 0.*0557

*Where ν, this frequency, is equal to 2 π × the force constant of the molecule divided by the reduced mass ^½.*0564

*And this ν is called the fundamental vibration frequency.*0583

*You will sometimes see it as ν sub 0, something like that.*0604

*K is the force constant of the bond and how springy the bond is.*0611

*Is it really tight or is it really loose?*0621

*And μ is the reduced mass, we have seen the reduced mass before.*0625

*I just want to make sure that we understand what all of the parameters are.*0632

*There are selection rules for vibrational transition.*0638

*The selection rules for vibrational transitions and the selections rules are that δ R = + or -1.*0642

*In other words, if it is going to make a transition from one vibrational state to another,*0660

*it is going to go from 1 to 2, 2 to 3, 3 to 4, or 4 to 3, 3 to 2, 2 to 1.*0664

*In the case of emission, absorption up emission down, as far as the harmonic oscillator is concerned,*0669

*it is only going to go 1 or 2 steps.*0677

*I’m sorry, 1 step up or 1 step down.*0681

*Its not going to go from 1 to 5, from 1 to 4.*0683

*If it is going to make a transition from 1 to 5, it is going to pass to 2, 3, 4, and then to 5.*0685

*Those are the transitions that are allowed.*0691

*The other transition rule is the dipole moment of the molecule must vary during a vibration.*0694

*I would not worry too much about this idea of the selection rule.*0715

*Now the dipole moment is related to the selection rule, this is often called gross selection rule.*0722

*The name itself does not really matter.*0728

*The dipole moment of the molecule must vary during a vibrational transition.*0729

*This does not mean that the molecule has to have a permanent dipole.*0737

*It can or cannot have a permanent dipole, in the case of molecule like hydrogen chloride, it has a permanent dipole.*0741

*In the case of a molecule like N2, they are both the same.*0748

*It is homo nuclear so there is no permanent dipole.*0752

*Now, these are actually not the best examples to use but it has to change during the vibrations.*0755

*It does not have to have a permanent one.*0764

*But if the molecule, if somewhere during the vibration the dipole shows up then it is capable of a vibrational transition.*0766

*The one that we would be concert with most of all, is the changes in the quantum numbers itself.*0779

*In this case, δ R = + or -1, but it is good to know this because some of your classes depending on what they are going to cover,*0783

*what they are not going to cover, they may actually go into the mathematics behind this.*0791

*I do not know, we ourselves are not.*0794

*We are going to be concerned more with just the spectroscopic aspects.*0795

*As we said earlier, spectroscopy is conducted primarily in wave numbers.*0801

*Spectroscopy is conducted in inverse cm, in wave numbers.*0817

*We want to take this equation for energy and convert them to wave numbers.*0825

*For vibration, the symbol that we use is G.*0832

*G is a function of R, it is equal to the energy of R that we had from the previous page divided by HZ.*0837

*The energy that you have, the energy that is given in Joules divided by Planck's constant divided by the speed of light.*0847

*Any energy in Joules divided by those by HC is going to give you the number in inverse cm.*0856

*This is called the vibrational term and this is what you will see in the literature.*0861

*This G of R called the vibrational term and it is just a symbol for the vibrational energy expressed in terms of wave numbers,*0868

*expressed in inverse cm as opposed to joules or anything else.*0882

*G as a function of R is equal to, the E of R is H ν R + ½.*0884

*We will go ahead and divide by HC.*0895

*The H will cancel, what we end up with is this ν ~.*0898

*It is the frequency expressed in wave numbers, R + ½.*0903

*Everything else is the same, R is going to take on the values of 0, 1, 2, 3, and so on.*0908

*The reason is here we are left with N/ C.*0916

*That is just equal to N/ C.*0921

*G of R is equal to ν ~, the fundamental vibration frequency expressed in wave numbers, × R + ½,*0930

*where ν ~ is equal to this ν/ C, which is equal to 1/ 2 π C.*0944

*That is it, just a little bit of mathematical manipulation.*0955

*This is the important thing right here.*0958

*This is what we have, this gives us the energy of a particular vibrational state depending on the quantum number R.*0961

*Let me go ahead and put R = 0, 1, 2, and so n.*0968

*It gives the energy of that particular vibrational state expressed in inverse cm.*0974

*That is all this.*0979

*That is the energy of the harmonic oscillator.*0983

*Let us go ahead and talk about the energy of the rigid rotator.*0986

*The energy of the rigid rotator, in case you are wondering why we are talking about the harmonic oscillator and the rigid rotator,*0994

*these things are for spectroscopic movement, spectroscopic transitions.*1004

*The harmonic oscillator is this way, it is the molecule that is vibrating.*1011

*The mathematics behind this is what we are doing.*1014

*The rigid rotator is something that rotates like this.*1018

*When you have a diatomic molecule that is rotating, we modeled it with the mathematics of the rigid rotator.*1021

*That is all, that is what we are doing here.*1026

*The energy of the rigid rotator is, in terms of what it is that we studied earlier, EJ = H ̅² / 2I × J × J + 1,*1028

*where J is the rotational quantum number that takes on the values 0, 1, 2, and so on.*1043

*Here I is the rotational inertia of the molecule.*1050

*It is equal to the reduced mass × the equilibrium bond length, the radius length between the two nuclei.*1054

*This E stands for its equilibrium², that is it.*1066

*This is the moment of inertia.*1069

*This I is the moment of inertia or the rotational inertia of the molecule.*1071

*Our E is the equilibrium bond length.*1080

*One of the parameters that you will see when you look at a table of constants for spectroscopic data is you are going to see the R sub E.*1092

*You are going to see the equilibrium bond length for that molecule.*1099

*The rotational say this gives the energy of the rotational state, the degeneracy of each level.*1104

*In other words, the number of levels that actually has this energy.*1111

*The degeneracy is equal to 2 J + 1.*1115

*When we discussed the rigid rotator earlier, I do not think I explained why this degeneracy exists.*1121

*I may have, but I do not believe that I did.*1144

*I think I just throw it out there as a number.*1146

*I do not think I explained where this degeneracy in the rotational states comes from,*1149

*explained that nature of this degeneracy.*1155

*Here is what is going on, I will tell you and then I will write it all out.*1170

*Let us say J is 1, that is going to have certain energy.*1175

*It is going to rotate with certain energy.*1182

*However, for each quantum number, in this case 1, there are 2 × 1 + 1.*1186

*There are going to be three actual orientations in space, fundamental orientations where the rotation is going to have that energy.*1194

*That is what degeneracy means.*1202

*It is going to be a particular quantum state that has that same energy, that is what the degeneracy.*1204

*We know that by definition.*1209

*The orientation in space of the molecule does not affect its rotational energy.*1211

*In other words, if I have a certain molecule that is oriented this way and it is rotating like this,*1215

*or if it is this way rotating like this, or if it is this way rotating like this, they have the same energy.*1221

*That is what this degeneracy means.*1228

*In the case of J = 1, it can be this way, it can be this way, it can be this way.*1230

*If J = 2, that is 2 × 2 + 1.*1238

*It means there are going to be 5 fundamental orientations in space that give you that same energy.*1241

*It is going to be 1, 2, 3, 4, 5.*1246

*For 3, you are going to end up with 7 levels of degeneracy, 7 fundamental orientations in space that all have the same energy.*1255

*That is the nature of this degeneracy of the rotational states.*1262

*Let us write it all out and give you a little bit of a quantitative aspect of it.*1265

*In molecules orientation in space has no affect on this rotational energy.*1274

*J is the quantum number that represents the angular momentum of the rigid rotator.*1298

*In other words, we know that anything that rotates has an angular momentum.*1325

*That angular momentum is going to be perpendicular to the direction of rotation.*1329

*If the molecule is rotating like this, its angular momentum is that way.*1332

*The magnitude of that angular momentum that is what J is.*1337

*J² actually.*1340

*That is what it represents.*1343

*It represents the magnitude of the angular momentum.*1345

*If it is rotating this way, the angular momentum vector is pointing that way.*1347

*If it is rotating this way, it is pointing that way.*1351

*This is the Z axis, it can be rotating like this, like this like this.*1356

*Angular momentum pointing that way, angular momentum vector pointing this way,*1363

*angular momentum vector pointing this way.*1366

*For each value of J, there is a J sub Z.*1369

*It is the component of the angular momentum vector along the Z axis, J sub Z is, we already seen this before.*1381

*For angular momentum, for rotational angular, we have seen this for spin angular momentum already.*1391

*JZ is the component of the angular momentum vector along the Z axis.*1395

*JZ takes on the values 0, + or -1, + or -2, all way to + or - J.*1422

*If J is 1 then what we have is 1, 0, -1.*1444

*If J is 2, we have 2, 1, 0, -1, 2.*1449

*If J is 3, we have 3, 2, 1, 0, -1, -2, -3.*1454

*It is the projection of the angular momentum vector along the Z axis.*1461

*All of those orientations in space carry the same energy given by the,*1466

*Here is where 2J + 1 come from.*1473

*There are 2J + 1 value in J sub Z.*1475

*For each J, or each J, there are 2J + 1.*1478

*I will just call that fundamental orientation with the same energy.*1490

*When we discuss the hydrogen atom, we call the J we called it L, the angular momentum quantum number.*1519

*The rotational quantum number.*1537

*And we called this J sub Z, we call it M sub L.*1544

*This was the magnetic quantum number, that is all that is going on here.*1551

*Remember for each value of L, you have 0, + or -1, + or – 2, all the way to ± L.*1558

*That is the magnetic quantum number which takes on the values 0, + or -1, + or -2, all the way to + or – L.*1567

*We will do the rotational term.*1589

*We went ahead and expressed the vibrational energy in terms of wave numbers.*1596

*We are going to express the rotational energy in terms of wave numbers.*1600

*The rotational term is symbolized as F of J.*1604

*F of J is equal to E of J divided by HZ equal to H ̅² / 2I HZ × J × J + 1.*1612

*H ̅ is equal to H/ 2 π, that implies that H ̅² is equal to H²/ 4 π².*1630

*When we put all of this back in to here, we get that the rotational term,*1644

*in other words the rotational energy expressed in terms of wave numbers is going to be equal to H² / 8 I π² HZ × J × J + 1.*1648

*We get some cancellation with the H and one of these.*1668

*What we are left with is F of J is equal to H/ 8I π² ψ × J × J + 1.*1671

*And again, J takes on the value 0, 1, 2, 3, 4.*1690

*I will circle the whole thing not just the bottom part.*1696

*This whole thing this is called the rotational constant and is symbolized as a B with a ~ on it.*1699

*This is another one of the spectroscopic parameters that you find in a table of spectroscopic data.*1722

*Just like you see in the equilibrium bond length, you will also see the rotational constant.*1730

*And as we go on with the lessons, you will see that there are more and more constants that are actually tabulated.*1734

*Since that is symbolized that way, we will go ahead and write it as F of J is equal to B ~ × J × J + 1,*1742

*where J takes on the values of 0, 1, 2, and so on.*1758

*And B is of course what we just said F of J is now in inverse cm.*1763

*The selection rules for rotational transitions δ J = + or -1,*1771

*that means it can only go from one rotational state to the next, either up or down.*1790

*It is not going to jump 5 levels.*1793

*In this case, the molecule must have a permanent dipole.*1795

*In the case of a rotational transition, it has to have a permanent dipole.*1807

*In the case of the vibrational transition, there has to be a change in the dipole moment during the vibration, during the transition.*1810

*If the transition is to happen, it must have a permanent dipole.*1821

*That is the difference between the two.*1825

*We have the rigid rotator energy, we have the harmonic oscillator energy, therefore like we said,*1828

*the transitions in the infrared, the vibrational transitions are accompanied by rotational transitions.*1835

*The combined energy of the transition is going to be the energy of the rotation + the energy of the vibration.*1841

*The harmonic oscillator rigid rotator approximation for the energy of the molecule is*1849

*therefore, the sum of the vibrational rotational energy.*1873

*Therefore, the energy R J is equal to the vibrational term + the rotational term.*1892

*The vibrational energy + the rotational energy.*1901

*Let me make my J a little bit more clear so it is not connected like that.*1903

*E sub RJ is equal to, this one is ν prime R + ½ +, now we have B~ J × J + 1.*1908

*Here R takes on the values 0, 1, 2, and so on.*1924

*J takes on the values 0, 1, 2, and so on.*1928

*Notice, there are two quantum numbers here in the total expression for the energy.*1932

*Once again, this ν ~ is equal to 1/ 2 π Z × the force constant divided by the reduced mass ^½.*1937

*This is the fundamental frequency of the vibration.*1949

*B~ is equal to Planck's constant divided by 8I π² Z.*1966

*All the parameters are taken care of, this is the equation of the harmonic oscillator rigid rotator approximation.*1976

*The mathematical equation that approximates what we see when we look at vibration rotation spectra is this.*1983

*The energy level diagram looks like this.*1997

*Let me draw this one by hand, actually.*2001

*What we have is my harmonic oscillator R = 0, R = 1, R = 2.*2004

*Remember the spacing between energy levels is the same for the harmonic oscillator.*2021

*R = 3, this is R = 0.*2027

*R = 1, R = 2, R = 3.*2030

*Within each vibrational level, there is a series of rotational levels.*2036

*We have J = 0, J = 1, J = 2, J = 3, and so on.*2042

*Here we have this one, this one, this one, this one, this one.*2053

*For R = 2, for each vibrational level there is a series of rotational levels.*2060

*The spacing of the rotation levels is not the same, the energies.*2072

*Between each vibration level, or for each vibrational level*2080

*which is the vibrational quantum number R there is a progression of rotational states.*2090

*When a photon of infrared is absorbed,*2119

*not only does a vibrational transition take place from R to R + 1*2140

*but several rotational transitions take place.*2155

*Several rotational transitions take place from J to either J + 1 or to J-1.*2165

*Right now, we are talking just about absorption.*2182

*When we are talking about absorption, we are going to go from R to R + 1.*2184

*If we are talking about emission, we would be talking about going from some level R to R -1.*2189

*For the sake of absorption, we are going up one vibrational level but the J level can actually go down or up.*2194

*You can go from J1 to J2 or J1 to J0, that is what this means.*2202

*But several rotation transitions take place when a photon of IR not only does the transition take place*2211

*from R to R + 1 but several rotational transitions take place from J to J + 1.*2219

*The selection rules, this is called vibration rotation.*2230

*When we to look a spectrum, it is called a vibration rotation spectrum.*2242

*We can get pure rotational spectra, we can get information of pure vibrational spectra*2246

*but when we run out of vibrational spectrum what we will get is a vibration rotation spectrum.*2253

*They are the combination of the R jump and the J jumps, that is what is happening.*2256

*The selection rules for vibration rotation are*2267

*δ R = + or -1, δ J = + or -1.*2285

*If R is + 1 that means it is going from a lower or to a higher state of absorption.*2293

*If R is -1, it means it is coming from a higher to a lower state, that is emission.*2298

*In the case of δ J + or -1, in the case of absorption, because you are going from one vibrational state to another vibrational state,*2304

*the rotational state might go up 1 or down 1.*2316

*But it is still absorption because it is actually going up an entire vibrational level.*2322

*You are still looking at a higher level.*2331

*Let us look at what the mathematics behind absorption.*2337

*What we want to do now is to derive an equation for the frequencies, for the spectra that we see.*2341

*Let us look at absorption, in the case of absorption δ R = + or -1.*2352

*Δ R is + 1, it is absorption, sorry about that.*2365

*Δ J is + or -1.*2370

*For R = + 1 and J = + 1, we would have two cases, 1 + 1 and 1 -1, + 1, the frequency of absorption.*2374

*The frequency of absorption is the difference between two energy levels.*2389

*The frequency of absorption is ν, what we observe.*2399

*It is the energy of the R + 1 J + 1 state - the lower state which is the energy of the R and J.*2408

*Let us go ahead and do the mathematics here.*2434

*This is going to equal, this energy term – this.*2437

*What we are going to have is prime × R + 1 + ½ + B~ × J + 1 × J + 2.*2441

*This is the upper state - the lower state which is ν~ × R + ½ + B ~ × J × J + 1.*2455

*I will not actually go through all the algebra here.*2471

*I think I actually will, I will do it for this one, that is not a problem.*2481

*This is equal to ν × R + 3/2 + B ~ × J² + 3J + 2 - ν × R + ½ - B ~ × J² + J,*2484

*I will distribute the - or both, what you end up with is ν ~ R + 3/2 ν ~ + B~ J² + 3 B ~ J + 2 B ~ – ν R - ½ ν - B ~ J.*2511

*This is just algebra, that is all it is.*2538

*It is always the worst part of mathematics.*2540

*Algebra has always been the worst and it will always be the worst.*2542

*Do not let it get to you.*2546

*A combined term, like for example I can cancel this one and this one.*2548

*I can cancel BJ² and that one I can combine the 3/2 ν - ½ ν.*2553

*When I'm left with is ν ~ + 2B ~ J + 2B ~.*2559

*Let me simplify this a little bit.*2570

*ν observed is going to equal ν + 2B.*2572

*I’m going to factor out the 2B ~, that is going to be J + 1, where J is going to take on the values 0, 1, 2, 3.*2580

*This is very important, J here is the value of the lower rotational state.*2590

*J is the value of the lower rotational state.*2598

*In other words, the smaller lower quantum number.*2608

*The quantum number of lower state.*2611

*It is the value of the quantum number in the lower state.*2613

*This is one of the equations.*2637

*This equation for the different values that J takes on.*2640

*What I'm going to see is the spectrum but I expect to see in the spectrum is this.*2644

*I expect to see a frequency at this number.*2649

*Its fundamental vibrational frequency + 2 × this rotational constant × whatever the J value happens to be in the lower state.*2655

*I expect to see a line there.*2663

*For the other case, for R = + 1.*2668

*This time J = -1, the observed frequency that I expect is going to be the energy of the R + 1 state J-1,*2674

*that is the upper state is J -1 – ERJ.*2688

*I go through the same algebra and what I end up with is, ν observe is going to equal ν -2 BJ.*2697

*Here, J takes on the values 1, 2, 3.*2710

*And again, J here is the value of the quantum number in the lower rotational state.*2716

*That is why there is a difference between these two.*2725

*Let us actually see what this looks like.*2730

*This is the other equation that I'm interested in right here.*2731

*This gives me one set of lines for different values of J.*2739

*This gives me another set of lines for different values of J.*2742

*Let us go ahead and go to a picture of the energy transitions to see what is happening first,*2757

*then we will take a look at the spectrum.*2761

*These right here, the blue lines, this blue level lower vibrational state,*2765

*Let me actually erase this.*2771

*Most books tend to use the symbol V or sort of a variation on ν as the vibrational quantum number.*2774

*I do not like that because it looks a lot like the frequency and it tends to get really confusing*2782

*which is why I use R for the vibrational quantum number.*2788

*This is R = 0, the 0 vibrational quantum state.*2792

*Here, this is the R = 1 vibrational quantum state.*2795

*Notice, within each vibration level, there are several rotational levels.*2801

*J = 0, J = 1, J = 2, J = 3, 4, 5.*2806

*And of course, in the upper level it has its own rotational quantum states 0, 1, 2, 3, 4, 5.*2810

*What we see in the spectrum is a series of lines.*2817

*For right now, let us not worry about with this Q branch is.*2821

*I will tell you in a second.*2825

*What is happening is R, the vibrational state is going up by 1.*2827

*We are jumping up from this vibrational state to this vibrational state.*2832

*Let us go to the R branch first, the R branch of the spectrum.*2837

*Here, the δ R = + 1 and the δ J = + 1.*2845

*That is fine, I will stick with blue.*2856

*I'm going from the J = 0, this one right here, this black line is J equal 0 to the J = 1.*2859

*The vibrational quantum state is going up by 1.*2866

*The rotational quantum state is going up by 1.*2871

*The molecules that are in the state of J = 1, they transition to the J = 2.*2877

*The 2 go up to 3, the 3 go up to 4, the 4 go up to 5.*2885

*That is the R branch.*2894

*The L branch represents δ R = + 1, we are going up a vibrational state.*2896

*But for the δ J = -1, in other words on the spectrum, I see a line for this transition,*2905

*a line for this transition, a line for this transition, and a line for this transition.*2912

*Now for the P branch, sorry about that.*2918

*It represents a vibrational state of going from a lower vibration to a higher vibration R = + 1,*2926

*but the rotational state drops by 1.*2931

*Here we are going from 1 to 0, J = 2 to J = 1.*2933

*J = 3 to J = 2, J = 4 to J = 3, J = 5 to J = 4.*2942

*There is a line for each of these transitions.*2950

*What we see is 1, 2, 3, 4, 5 and so on lines.*2954

*To the left we see 1, 2, 3, 4, 5, and so on lines.*2958

*This Q branch, notice where the lines go.*2963

*We are jumping up from a vibrational state to a vibrational state R = 0 to R = 1.*2967

*That is fine, that is not a problem.*2971

*The selection rule can handle that but we said that δ J has to be + or -1.*2973

*Δ J cannot be 0, that is a forbidden transition.*2977

*Notice this line right here, that is going from J = 0 to J = 0.*2983

*J = 1 to J = 1, because δ J = 0 is a forbidden transition, these transitions we do not see them in the spectrum.*2988

*What you see in the spectrum is this, no line at ν, the fundamental vibration frequency.*2999

*Remember what we had just a second ago, we had a couple of equations.*3009

*We said that for the R branch, we would end up seeing a bunch of lines at ν + 2 B × J + 1.*3014

*And for the P branch, we would see a bunch at ν – 2BJ.*3028

*There are going to be lines to the left, lines to the right of this number.*3037

*But we do not actually see a line in there because the transition from the δ J = 0 transition is forbidden.*3042

*There is no actual Q branch.*3051

*There are molecules with Q branch, they will show up*3053

*Your teacher may or may not decide not to talk about it.*3056

*We, ourselves, we will not talk about it.*3060

*It was not altogether that important, at least for what we are doing.*3061

*But know that there is a P branch, there is a Q branch, there is an R branch.*3064

*But in general, for the vibration rotation spectra of diatomic molecule,*3068

*because δ G = 0 is a forbidden transition, δ J has to equal + or -1, we do not see a Q branch.*3072

*It is just that nothing there.*3079

*Let us go ahead and take a look at the vibration rotation spectra.*3082

*This is a vibration rotation spectrum for HCL.*3089

*Do not worry about this, it should not be in here.*3093

*This is a vibration rotation of HCL.*3097

*You will sometimes see them this way, in terms of the peaks to be pointing down.*3101

*Other times you are going to see them when the peaks are pointing up.*3107

*The R branch here, right here, see these lines?*3110

*This right here, this little gap, this is the ν sub 0.*3113

*This is the fundamental frequency.*3122

*It to this point where the J = 0 to J = 0 transitions should happen.*3123

*But because they are forbidden, it is not going to happen, but that is where we see it.*3128

*If we know what ν 0 is, reading of the spectrum we just find that middle point and we go down and we mark it.*3134

*That is our ν sub 0, that is our fundamental vibration frequency.*3143

*The R branch we said is where you do,*3148

*Let us write it down.*3156

*It is equal to that, + 2B × J + 1.*3159

*Notice, it is where the absorption frequency is going to be increasing from the fundamental frequency.*3168

*The fundamental frequency + a certain amount.*3175

*The fundamental frequency + a certain amount.*3178

*The fundamental frequency + a certain amount + a certain amount, this is a second transition.*3180

*+ a certain amount, this is the third transition.*3184

*+ a certain amount, this is the fourth transition.*3186

*+ a certain amount, it is the fifth transition.*3187

*As the frequencies go up, that is the R branch.*3190

*This right here is the R branch.*3193

*Over here, the frequencies go down.*3196

*The fundamental frequency of the P branch is the fundamental frequency - some value as J increases.*3199

*- - -, these are going down.*3206

*This is the P branch.*3209

*Like I said, sometimes you see it the other way.*3212

*When you look at the spectra, you are not going to look at left and right.*3214

*You are looking at R and P.*3220

*R is for the frequencies that increase, P is for the frequencies that decrease from the fundamental frequency.*3222

*That is what is happening here.*3228

*Let us look at these lines, the fundamental frequency is here.*3230

*This one represents the transition from 0 to J = 1.*3233

*This one represents J = 1 to J = 2.*3237

*This one represents J = 2 to J = 3.*3240

*J = 3 to J = 4, and so on.*3244

*Here represents J = 1 to J = 0, J = 2 to J = 1, J = 3 to J = 2.*3246

*This is the δ J = + 1, here is the δ J = -1.*3260

*Let us look at the spectra that actually look the other way.*3267

*This is the spectrum for carbon monoxide.*3271

*Let me write it down here.*3274

*This is the spectrum from carbon monoxide.*3276

*In this particular case, the peaks are pointing up.*3278

*This is interesting.*3280

*The left is actually increasing, to the right it is actually decreasing.*3288

*Here is the fundamental frequency increase.*3291

*This is the R branch decreasing, that is the P branch.*3294

*That is all that is going on here.*3299

*Because the equation R ν + 2B × J + 1 and ν - 2 BJ, the difference in the spacing, this difference right here,*3303

*the difference between the lines is 2B.*3322

*The difference between them is 2B.*3328

*We see absorptions.*3334

*That is what is going on there.*3340

*Let us go ahead and see what else we have to say about this.*3342

*Notice there is a nonexistent Q branch here.*3349

*Let us talk about the intensities.*3359

*Notice that all of these lines they have different intensities.*3361

*Some are very very intense, some not so intense.*3364

*This is the last page.*3370

*Let me go ahead and write it over here and continue it down here.*3374

*The intensities of the transitions are related to the populations of each J level.*3386

*In other words, if I had a bunch of molecules, let us just say I have 100 molecules in level 2.*3405

*A hundred molecules in level 2, the more molecules that you have, the more transitions are going to take place.*3413

*The more molecules you have, each one of those molecules is going to make a transition.*3428

*Each one of those transitions is represented by, it is going to contribute to the size of the peak.*3432

*If you only have 10 molecules making the transition, let us say from level 3 to level 4,*3437

*it is going to have a certain height of a peak.*3443

*If you have 100 molecules making the transition from level 2 to level 3, that peak is going to be higher.*3445

*In this particular case, notice that the level 2 to level 3 to level 4, those tend to be the peaks of highest intensity.*3452

*That tells us that the rotational levels 2, 3, 4, those are the ones that are most highly populated at normal temperatures.*3461

*As far as the rotational state level molecule, most molecules are not in there,*3472

*0, 1, 2 rotational states at room temperature, most of them tend to be in the 3, 4, 5 rotational states.*3476

*That is where most of the molecules are.*3483

*Therefore, the transitions that are going to take place, they are going to have the ones of higher intensity.*3485

*Let me say that again.*3491

*The intensities of the transitions are related to the populations of each J level.*3492

*The greater the population of the level, the greater the number of transitions,*3504

*therefore, the more intense the line.*3532

*We see from these spectra that the J = 3, 4, 5, are the most populated.*3542

*Once again, notice that the Q branch does not exist because δ J = 0 is a forbidden transition.*3574

*Thank you so much for joining us here at www.educator.com.*3582

*We will see you next time for a continuation of molecular spectroscopy, bye.*3584

2 answers

Last reply by: Kaye Lim

Sat Apr 8, 2017 3:06 PM

Post by Kaye Lim on April 5, 2017

Greeting sir,

You said around 5:20: 'Most of the information that we get from spectroscopy, we actually get from electronic spectroscopy. Electronic spectroscopy allows us to, it is difficult to analyze but everything that we need is there. It gives us information on electronic states, on vibrational states, on rotational states. The rotational spectra, the vibrational spectra tend to be easier, but they do not give is as much information.'

-a UV/Vis spectrum that I obtain from running UV/Vis spectroscopy on a sample gives me maximum absorption wavelength of that sample. And then I can create calibration curve and quantify the concentration of the sample. That is my experiences with UV/Vis spectroscopy which I think of it mostly as a quantitative method.

-From IR spectroscopy, it is more of a qualitative method since I know what functional group presented in my sample. I know IR could also be used as quantitative method as well.

-So to me, I think IR gives me more information then UV/Vis spectroscopy. Why did you say UV/Vis spectroscopy or electronic transitions give more information than IR spectroscopy? Do you mean when you do very high resolution UV/Vis spectroscopy inwhich you can zoom in the UV/Vis absorption peak to get sub-peaks of vibrational transitions and sub-subpeaks of rotational transitions?

-If that is the case, then what is the difference between vibrational and transitional transitions in high energy UV/Vis region compared to those in IR region? are they vibrational and transitional transitions between different electronic states instead of in the same electronic state as in the case of IR?