We now turn to our final quantum mechanical topic, molecular spectroscopy.

Transcription

1 07 Lecture 34 We now turn to our fnal quantum mechancal topc, molecular spectroscopy. Molecular spectroscopy s ntmately lnked to quantum mechancs. Before the md 190's when quantum mechancs was developed, people had begun to develop the scence of molecular spectroscopy. However, ts early use was purely qualtatve, based solely on the observaton that molecules and atoms, when exposed to lght, absorbed and emtted t n patterns that were unque to each speces. The nterpretaton of these spectra n terms of the physcal parameters that descrbe the geometres and bondng of the molecules that produce them had to wat untl the development of quantum mechancs. The close nterrelaton between spectroscopy and quantum mechancs s shown by the fact that Robert Mullken, who was the foremost practtoner and proponent of Molecular Orbtal theory n ts early years, was spurred on by an attempt to understand the detals of molecular spectra. It s further shown by the fact that Gerhard Herzberg, the foremost practtoner of molecular spectroscopy from the mddle of the '30s to the 70's, used to keep a bound copy of Mullken's papers by hs desk. In fact, I keep a bound copy of Mullken's major papers, edted by Norman Ramsey, Herzberg's collaborator, by my desk. Molecular spectroscopy, whch s the study of the nteracton of lght wth atoms and molecules, s one of the rchest probes nto molecular structure. Radaton from varous regons of the electromagnetc spectrum yelds dfferent nformaton about molecules. For example, mcrowave radaton s used to nvestgate the rotaton of molecules and yelds moments of nerta and therefore bond lengths. Infrared radaton s used to study the vbratons of molecules, whch yelds nformaton about the stffness or rgdty of chemcal bonds. Vsble and ultravolet radaton s used to nvestgate electronc states of molecules, 07

2 08 and yelds nformaton about ground and excted state vbratons, electronc energy levels, and bond strengths. We wll do a farly detaled treatment of rotatonal and vbratonal spectroscopy and a somewhat brefer treatment of electronc spectroscopy. In our treatment we wll begn wth the basc nterpretaton of molecular spectra, and fnsh wth the quantum mechancal bass of these nterpretatons. The features of the electromagnetc spectrum that are of nterest to us are summarzed below. The absorpton of mcrowave radaton s due to transton between rotatonal energy levels; the absorpton of nfrared radaton s due to transtons between vbratonal levels, Regon Frequency/Hz Wavelength/m Wave number/cm -1 Energy/ J *molecule -1 Molecular Process Mcrowave x x x Rotaton of 6.6x10-3 polyatomc molecules Far Infrared x x x x10-1 Rotaton of small molecules Infrared x x x x10-0 Vbraton of flexble bonds Vsble and Ultravolet x x , x10 5.x x10-18 Electronc transtons accompaned by transtons between rotatonal levels; and the absorpton of vsble and ultravolet radaton s due to transtons between electronc energy levels, accompaned by smultaneous transtons between vbratonal and rotatonal levels. The frequency of radaton absorbed s gven by E E E h u l where E u and E l are the energes of the upper and lower states respectvely. 08

3 09 For example, f we have an absorpton wth wavenumber, = 1.00 cm -1, we calculate E as follows. Remember that 1/ ( cm). Therefore E s related to by hc E h hc and so E = hc = (6.66 x Js)(3.00 x cm/s)(1.00cm -1 ) = 1.99 x 10-3 J. TO WHAT TYPE OF MOLECULAR PROCESS WILL THIS RADIATION CORRESPOND? We wll begn wth molecular rotaton. A rgd rotor s the smplest model of molecular rotaton. We dscussed the quantum-mechancal propertes of a rgd rotor before, but wll revew the pertnent results here. The energy of rotaton of a rgd rotor s all knetc energy. For a moton wth crcular symmetry, t s convenent to express ths energy n terms of the angular varables L, angular momentum, and I, moment of nerta. The moment of nerta for a datomc molecule s gven by I R 0 where s the reduced mass of the molecule and R 0 s the bond length of the molecule. The angular momentum L s gven by L I, where s the angular frequency. When the Schrödnger equaton for ths problem s solved, t s found that the angular momentum s quantzed accordng to the equaton J = J(J +1) where J s the spectroscopc notaton for the total angular momentum of a molecule, and that the angular momentum n the z drecton s quantzed accordng to the equaton 09

4 10 Jz M, where M = J, J-1,..., -J. The energy of the rgd rotor, n the absence of an external feld, depends on J only, s also quantzed and s gven by E J = J(J +1), J = 0,1,,... I Snce there are J+1 values of M for each energy E J, the energy states are J+1 fold degenerate. When mcrowave or far nfrared radaton shnes on a rotatng molecule, the J value can ether ncrease or decrease, and the energy of the lght absorbed s gven by E Eu El. An mportant characterstc of all molecular spectra s that the choce of upper and lower states s not completely free but s n fact extremely lmted. The rules that govern the choces of upper and lower states are called selecton rules. These selecton rules are determned by tme dependent perturbaton theory. For the pure rotatonal transtons of a datomc rgd rotor the selecton rule s J = 1. In other words, when a datomc rgd rotor absorbs lght, ts rotatonal quantum number can ncrease by 1 or decrease by 1. These are the only possble transtons that occur. (The reason that I specfy datomc rgd rotors s that the selecton rules depend n part on symmetry, and molecules wth other symmetres have dfferent selecton rules.) In addton, for pure rotatonal transtons the molecule must have a permanent dpole moment. Thus molecules lke H and I wll have no pure rotatonal spectra, whle HI or CO, whch have permanent dpole moments, wll exhbt rotatonal spectra. The consequence of ths selecton rule s that the rotatonal absorpton spectrum of a rgd rotor s a seres of evenly spaced lnes n the mcrowave or far nfrared regon. To see ths we begn wth the energy egenvalues, 10

5 11 E J = J(J +1), J = 0,1,,... I Accordng to our selecton rule, for absorpton J = +1. The energy change for ths transton s E = E - E = (J +1)(J +)- J(J +1) I I J 1 J = (J +1), J = 0,1,,... I The wavenumber change for ths absorpton s gven by = E hc = h (J +1), J = 0,1, 4 ci It s typcal n spectroscopy to wrte the energy n the form E = hcbj(j +1) J where B, the rotatonal constant of the molecule, has unts of wavenumbers and s gven by B = h 8 ci In ths form the wavenumber of the rotatonal absorptons s gven by = B(J +1), J = 0,1,,... Thus for the lowest energy rotatonal transton, J = 0 to J = 1, the absorpton occurs at a wavenumber of B. The second transton, J = 1 to J =, occurs at a wavenumber of 4 B. The thrd transton occurs at a wavenumber of 6 B and so on. Thus the rotatonal spectrum conssts of a seres of equally spaced lnes separated by B. [Illustrate] Lets check on our asserton that the rotatonal spectrum s n the mcrowave regon, by calculatng the value of B for the rotaton of HI. 11

6 B = 1 h 8 ci The moment of nterta I s gven by R 0, where s the reduced mass n kg and R 0 s the bond length n meters. For HI, the reduced mass s gven by = m m m +m = 1AMUx17AMU 1AMU +17AMU x1.66x kg 10 AMU = 1.65x 10 HI H I H I 7 7 kg and the bond length s pm, so the moment of nterta s I = R 0 = 1.65 x 10-7 kg (160.4 x 10-1 m) = 4.5 x kg m, x10 Js and B = (8 )(3.00x cm / s)(4.5x10 kg m ) = 6.58cm Our frst rotatonal transton wll occur at a wavenumber of B = cm -1, whch s n the far nfrared regon, as predcted. A more typcal applcaton s to determne the value of B from the spacng of a rotatonal spectrum, and use ths to determne the moment of nerta and therefore the bond length. For example, the mcrowave spectrum of 39 K 17 I conssts of a seres of lnes whose spacng s almost constant at 3634 MHz. Calculate the bond length of 39 K 17 I. The rotatonal spacng for KI s gven n Hz, a unt of frequency. Our equaton for the rotatonal wavenumber s = B(J +1) The relaton between wavenumbers and frequency s = c where c s the speed of lght n cm/s. Thus our rotatonal absorpton frequency becomes cb J 1, 1 1

7 13 and our rotatonal spacng becomes cb. Thus B = = 3634x 10 s c x3.00x = 6.05x10 cm 1 From ths we can use our formula for B to calculate the moment of nterta for KI, I = h 34 8 cb = 6.66x10 45 = 4.618x10 kgm. 10 (8 )(3.00x 10 )(.06057) The reduced mass of 39 K 17 I s KI = 39AMUx17AMU 39AMU +17AMU x1.67x kg 10 AMU = 4.983x 10 Thus R 0 = (I/) 1/ = 3.04 x M. 7 6 kg. 13

8 14 Lecture The smplest model of a vbratng molecule s the harmonc oscllator. The harmonc oscllator assumes that the potental energy functon for the vbraton s gven by V 1 kx, where k, the vbratonal force constant, s proportonal to the stffness of the bond. The energy egenvalues for the harmonc oscllator are quantzed and are gven by E = (n+ 1 n )h, n = 0,1,,... where = 1 F HG Transtons between vbratonal levels are subject to the selecton rules n = 1 and that the dpole moment must change durng a vbraton. Once agan for datomc molecules ths k I KJ 1 /. means that homonuclear datomcs wll have no vbratonal absorpton spectrum. The mplcaton of ths selecton rule together wth the formula for the energy egenvalues of the harmonc oscllator s that the pure vbratonal absorpton spectrum of a datomc molecule s a sngle absorpton whose frequency s gven by = 1 absorpton n = +1, so F HG k I KJ 1 / E = E n+1 - E n = (n /) h - (n + 1/) h = h. To see ths note that for Thus the spectrum conssts of a sngle lne wth frequency. Determnng the nfrared frequency allows us to determne the force constant k of the bond. For example, the nfrared spectrum of 39 K 35 Cl has a sngle ntense lne at cm -1. What s the force constant? Our frst step s to convert the wavenumber to frequency, usng c. Ths yelds 14

9 15 = 378 cm -1 x 3.00 x cm/s = x s -1. The reduced mass of 39 K 35 Cl s KCl = 35AMUx39AMU 35AMU +39AMU x1.67x kg 10 AMU = 3.08x Rewrtng our frequency equaton to solve for force constant yelds kg k = () = ( x ) x 3.08 x 10-6 = 156 N/m. All of the quantum mechancal systems we have studed so far have the common property that ther energes are quantzed. Suppose we have a system where the avalable energes are the set {E J }. An example of ths would be a datomc harmonc oscllator, whose energes belong to the set {1/h, 3/h, 5/h,...} = {E 0, E 1, E,...}. A queston of practcal mportance s the fracton of the molecules n a gaseous sample that are at a gven energy. In other words f we have N molecules, how many are n E 0, how many n E 1, how many n E and so on. The answer to ths queston comes from statstcal mechancs and says that f N s the number of molecules wth energy E, then N = ce E / kt where k s Boltzmann's constant, T s the absolute temperature and c s a proportonalty constant. We can evaluate ths constant c by summng both sdes over all possble energes and recognzng that N N Ths yelds or N N c e E / kt 15

10 c = 16 N e -E / kt whch yelds fnally E / kt N = Ne E / kt e Ths equaton s called a Boltzmann dstrbuton after the German physcst Ludwg Boltzmann, and governs how the energy levels are populated at any temperature. Snce most of the tme we won't know the absolute number of molecules n our sample, a more practcal quantty to calculate s the fracton of molecules n a gven state. Ths s gven by f = N N = e e E / kt E / kt Let s use ths to fgure out the fracton of molecules n the ground state of a harmonc oscllator. For the harmonc oscllator, E s gven by E = ( + 1/)h so our equaton for the f becomes f = ( h 1/ ) kt e e ( h 1/ ) kt 0 The nfnte seres n the denomnator can be evaluated to be h kt 1 e h 0 kt 1 e 16

11 17 Thus for the harmonc oscllator our equaton for f becomes f = (1-e )e h / kt h / kt Lets apply ths equaton to calculate the fracton of molecules n the ground vbratonal state for HBr at 300 K and 000 K. The fundamental vbratonal frequency of HBr s 7.7 x Hz. The easest way to begn ths s to calculate the quantty h/kt. At 98 K ths s h (7.7x10 s )(6.66x10 Js) = kt (98K)( x10 JK ) = At 000 K the same type of calculaton shows that h/kt = 1.8. If we substtute = 0, the quantum number for our ground vbratonal state nto our equaton for f, we fnd that f = 1-e at 98K 1. 8 whle f 0 = 1- e =.83 at 000K. Thus we see that at room temperature practcally all of the HBr molecules are n the ground vbratonal state, whle even at 000 K the bulk of the molecules are stll n the ground vbratonal state. Let s look at the effect of vbratonal frequency on the ground state populaton by calculatng f 0 for I, whch has a ground state vbratonal frequency of 6.4 x 10 1 Hz, over a factor of 10 smaller than HBr. Here h/kt = 1.034, so f = 1- e =.6444 at 98K. 0 We see that even at room temperature a substantal fracton of odne molecules wll be n excted vbratonal states. We conclude that ether ncreasng the temperature or decreasng the vbratonal frequency wll result n an ncrease n the populaton of vbratonal excted states. 17

12 18 Molecules smultaneously vbrate and rotate. If we treat the rotatons usng the rgd rotor approxmaton, and the vbratons usng the harmonc oscllator approxmaton, then the energy of ths vbratng, rotatng molecule s gven by E vb,rot = (n+1 / )h 0 +hcbj(j +1) where n = 0, 1,,... and J = 0, 1,,... An mportant feature of these energes s that the rotatonal energes are generally much smaller than the vbratonal energes, so each vbratonal state s accompaned by a large number of rotatonal states. For example, for HCl at room temperature, we wll fnd fnte populatons of molecules wth v= 0 and J rangng from 0 to approxmately 0. The lowest energy state wll be the pure vbraton wth E nj = 0 /. The next energy wll be 0 / + B, the next wll be 0 / + 6B, the next 0 / + 1B and so on. For HCl, the vbratonal frequency s 885 cm -1, whle B s 10.6 cm -1. Thus the frst few energes are cm -1, cm -1, 1505 cm -1 and so on. Because the energes of the rotatons are small compared to the vbratons, vbratonal rotatonal spectra have the appearance of wdely separated clusters of rotatonal lnes. When a datomc molecule absorbs nfrared radaton, the vbratonal transton s accompaned by a rotatonal transton. The selecton rules for absorpton of nfrared radaton are n = 1 J = 1. In other words, when an nfrared photon s absorbed, the vbratonal quantum number goes up by one, and the rotatonal quantum number ether goes up by one or down by one. wavenumber obs assocated wth the absorpton s The 18

13 obs 19 = 0 + B[J (J +1)- J(J +1)] where J s the ntal rotatonal quantum number, and J', the fnal rotatonal quantum number, can be ether J + 1 or J - 1. If J' = J + 1, then If J' = J - 1, then obs ( J =+1)= 0 +B(J +1), J = 0,1,,... obs ( J = -1)= 0 -BJ, J =1,,3,... WHY CAN'T J BE ZERO IN THIS CASE? Typcally, B s on the order of 1 cm -1 or less, and 0 ranges from a few hundred to a few thousand wavenumbers, so the spectrum produced by these two equatons typcally contans lnes between 3700cm -1 and 350 cm -1 ntegral multples of 1 cm -1. Notce that accordng to these equatons there s no absorpton at 0. Before the quantum mechancal explanaton of molecular spectroscopy, ths was consdered an nexplcable anomaly. Its explanaton was one of the early successes of quantum mechancs. The nfrared spectrum that you measured for HCl s typcal of these spectra. [Draw] The gap centered around 885 cm -1 corresponds to the mssng lne at 0. On each sde of the gap s a seres of lnes whose spacng s about 10 cm -1. The seres toward the hgh frequency sde s called the R branch and s due to rotatonal transtons wth J = +1. The seres toward the low frequency sde s called the P branch and s due to rotatonal transtons wth J = -1. Let s do an example wth rotatonal-vbratonal spectra. The bond length n 1 C 14 N s 117 pm and ts force constant s 1630 N m -1. Let s predct the vbraton-rotaton spectrum of 1 C 14 N. Our frst step s to calculate 0 and B. Both quanttes requre the reduced mass, whch s 19

14 0 = (1.0AMU)(14.0AMU) ( )AMU (1.66x kg 10 )= 1.07x10 AMU kg The fundamental frequency 0 s gven by c ( k 1 1/ 1 ) = (3.00x cm / s) ( 1630Nm 1/ ) =.07x10 cm x10 kg The rotatonal constant B s gven by B = h 8 ci = h 8 cr (6.66x10 Js) = 8 (3.00x cm / s)(1.07x kg)(117x m ) = 1.91cm The vbraton-rotaton spectrum wll consst of lnes at 0 BJ, where J = 1,,3,... There wll 1 be no lne at 0 and the separaton between the lnes n the P and R branches wll be even at B = 3.8 cm -1. If we compare these results wth expermental data, we see that there are several features we are not able to explan wth ths combnaton of the rgd rotor and the harmonc oscllator alone. The ntenstes, or heghts, of the lnes n the P and R branches show a defnte pattern and the spacng of the lnes s not equal. Close examnaton shows that the lnes n the R branch are more closely spaced wth ncreasng frequency, and that the lnes of the P branch become farther apart wth decreasng frequency. We now turn to an explanaton of these detals. We turn frst to the ntensty pattern of the P and R branches. We know that Beer's law, A = lc, holds for all optcal spectra, and that therefore the ntenstes of the rotatonal lnes n a vbraton rotaton spectra depend on the populaton n any gven rotatonal state. 0

15 1 Snce ths populaton s proportonal to the fracton of molecules n that rotatonal state, we can use the Boltzmann dstrbuton of rotatonal energes to explan the general nature of the observed ntenstes. We need to modfy our Boltzmann equaton a bt snce the rotatonal energy levels are J + 1 degenerate. If an energy level N has a degeneracy g, then the number of molecules wth energy becomes N = cg e E / kt where c s a constant. For rotatonal energes ths becomes N = N (J +1)e J T hcbj ( J 1) kt where N T s the total populaton. Ths allows us to calculate the rato of the populaton of the state wth quantum number J, to the populaton of the lowest energy level, and therefore to calculate the ratos of the ntenstes of the peaks correspondng to these two states. If we plot ths equaton, we fnd that t yelds the characterstc shape of the P and R branches. Thus we see that the shape of the P and R branches reflects the populaton of the rotatonal energy levels. So far n our treatment of the rotatonal and vbratonal spectra of datomc molecules, we have used two approxmatons, that the bond length of the molecule was fxed durng rotaton, and that the vbratons were governed by a harmonc potental. These approxmatons yeld a crude approxmaton of the expermental results, but closer examnaton of these results shows that they are not entrely correct. Let s consder rotatons agan. The expermental observaton whch shows that our rgd rotor model needs to be refned s that the lnes n a rotatonal spectrum are not equally spaced, as the rgd rotor model predcts. You have all seen ths wth the vbratonal-rotatonal spectrum of HCl, but t s 1

16 The Rotatonal bsorpton Spectrum of 35 Cl also true for the pure rotatonal spectrum of HCl. If we examne the frst few transtons n the rotatonal absorpton spectrum we can see ths clearly. The Rotatonal Spectrum of H 35 Cl Transton 1 obs / cm obs / cm 1 The spacng between the rotatonal lnes vares n these few ponts between 0. cm -1 and 1.07 cm -1. The dscrepancy can be resolved by realzng that a chemcal bond s not truly rgd. As the molecule rotates faster and faster,.e., as J ncreases, the centrfugal force causes the bond to stretch. The extent of the stretchng of the bond can be determned by balancng the centrfugal force, R, whch causes the stretchng, wth the Hooke's law force, k(r - R 0 ) whch ressts the stretchng. Thus we have k(r - R 0 ) = R. where R 0 s the bond length when J = 0 and there s no rotaton. Solvng for R, we fnd that the stretched length of the rotatng molecule s R = kr 0 R

17 3 In order to determne the effect of the centrfugal stretchng on our rotatonal spectra, we need to fnd how ths changes our rotatonal energy. As usual we begn wth the classcal case and use the correspondence prncple to fnd the quantum analog. As a result of the stretchng the total classcal energy of the rotator s made up of a knetc energy term and a potental energy term: E = 1 I + 1 k(r - R 0 ) Substtutng our value for R nto ths equaton yelds E = 1 I + R 4 k Snce L = I = R, ths n turn equals E = L I + L 4 I kr We obtan the quantum mechancal energy for ths stretchng rotor by substtutng the quantum value for L, J(J+1): 4 E J = J(J +1)+ J (J +1 ) I I kr All we have left to do now s express the stretched radus R n terms of the ntal bond length and we are done. Ths part s a bt of a mess. The basc approach s to take our classcal results for the stretched bond length R, rewrte t n terms of 1/R, and then substtute the quantum terms for the L terms. When ths s worked through the result s 4 E = J(J +1)- J (J +1 ) I I kr J Ths s usually wrtten n wavenumbers as 3

18 4 E = E J = BJ(J +1)- DJ (J +1 ) hc J By comparng these two equatons we fnd that D = 3 = 4B 4 ci kr 0 D s called the centrfugal dstorton constant, and results n a lowerng of the rotatonal energy compared wth the rgd rotor for all states wth J > 0. Typcal values of D are much smaller than B. For H 35 Cl, for example, B = cm -1 and D =.0004 cm -1, and therefore the centrfugal dstorton term represents a small correcton to the rgd-rotator approxmaton. Note also because the centrfugal dstorton term s a functon of J (J + 1), that the sze of the correcton ncreases wth J. The frequences of absorpton due to a transton from J J + 1 are gven by = E - E = B(J +1)-4D(J +1 ) 3, J = 0,1,,... J 1 J If we use ths equaton and the values of B and D for H 35 Cl, we fnd that the new calculated values are closer to the expermental values, but are stll off by a bt. Ths s due to the fact that the bond length also changes because the molecule s vbratng, whch affects the rotatonal energy. Ths phenomenon s called vbraton - rotaton nteracton. We wll cover ths n a moment, but frst I would lke to consder correctons to our basc pcture of molecular vbraton. 0 There s expermental evdence that our pcture of vbratng molecules s only approxmately correct. The most obvous problem s our observaton, whch we made before, that the harmonc potental does not allow for the possblty of dssocaton of the molecule. The devaton of a true potental from a harmonc potental s called the anharmoncty. Snce 3 4

19 5 the potental s correct only toward the bottom of the potental well, and vbratonal frequences are dependent on the shape of the potental, vbratonal frequences whch are based on a harmonc potental wll have to be approxmate. Note that the harmonc potental s close to the true molecular potental for much of the lower energy part of the curve. Ths suggests that the easest way to fnd the true vbratonal energy s to begn wth the harmonc frequency and fnd a correcton term. The easest way to represent the true potental n a way that lends tself to fndng such a correcton term s to expand the potental functon U(R) n a Taylor seres about R e, the equlbrum bond length, whch s also the mnmum of U(R). Ths expanson yelds U(R)=U(R )+( du dr ) (R - R )+ 1! ( d U dr ) (R - R + 1 3! ( d 3 U 3 ) ) (R - R ) dr e R=Re e R=Re e R=Re e The second term of the expanson s zero, because the frst dervatve of U wth respect to R wll be zero at the mnmum of the potental. The dervatve n the second term ( d U dr ) R=R e s smply our force constant k. Thus the dfference n energy from the potental mnmum can be wrtten as U(R)-U(R )= k x + 6 x + 4 x e where x = (R - Re), γ 3 = ( d 3 U dr ) 3 R=R e, and γ 4 = ( d 4 U dr ) 4 R=R e. Notce that f we gnore the last two terms we have a harmonc potental. Therefore the harmonc oscllator approxmaton conssts of keepng only the quadratc term n the Taylor expanson. As we showed earler, the harmonc oscllator approxmaton predcts that there wll 5

20 6 be only one lne n the vbratonal spectrum of a datomc molecule. Expermentally t s found that there s ndeed one domnant lne, called the fundamental, but n addton there are lnes of weaker ntensty at almost ntegral multples of the fundamental. These lnes are called overtones. For example for H 35 Cl, the fundamental occurs at cm -1, whle successvely weaker bands appear at , , and cm -1. These bands are called the frst, second, thrd and fourth overtones respectvely. They represent transtons from v = 0, v = 0 3, v = 0 4, and v = 0 5. Note that all of these transtons are n volaton of the harmonc oscllator selecton rule of n = ± 1. The wavenumbers of these bands also do not match the predcton of the harmonc oscllator, whch predcts that all of the overtones wll occur exactly at ntegral multples of the fundamental frequency. If the terms of hgher order than the harmonc terms, called the anharmonc terms, are ncluded n the Hamltonan operator for the vbratonal moton of a datomc molecule, then perturbaton theory yelds E = (n+1/ )+ x (n+1/ ) +... n e e e where x e s called the anharmoncty constant. The anharmonc correcton s much smaller than the harmonc term because x e << 1. For example, for H 35 Cl, e = 989 cm -1, whle e x e = 5.05 cm -1. For another example, CO has e = 170. cm -1, whle t has e x e = cm -1. The effect of the anharmonc terms s that the spacng of vbratonal energy levels s not even, and that the spacng decreases wth ncreasng n. [Draw] Notce that the harmonc oscllator approxmaton s best for small values of n, whch are the states whch have the hghest populaton at room temperature. 6

21 7 The selecton rule for an anharmonc oscllator s that n can have any ntegral value, although the ntenstes of the n = ±, ±3,... transtons are much less than for the n = ± 1 transtons. Earler we showed that for datomc molecules, most of the populaton at room temperature s n the v = 0 state. Ths mples that most vbratonal transtons wll orgnate n the v = 0 state. The wavenumbers of the observed v = 0 n transtons wll be gven by = E - E = n - x n(n+1) n=1,,... obs n 0 e e e, Applcaton of ths equaton to the spectrum of H 35 Cl results n an agreement wth experment to the ffth sgnfcant fgure. Let s work an example of calculatng the wavenumbers for the absorptons of an anharmonc oscllator. Gven that e = cm -1 and e x e = 3.83 cm -1 for 3 Na 19 F, calculate the frequences of the fundamental and the frst and second overtones. We use our equaton for the wavenumber for a transton of an anharmonc oscllator for all three of these transtons. The fundamental s calculated by lettng n = 1 and the overtones are calculated by lettng n = and 3. For the fundamental, For the frst overtone For the second overtone 1 obs = e - ex e = 58.44cm. obs = e -6 ex e = cm 1 1 obs = 3 e -1 ex e = 156.3cm. Note that the overtones are not ntegral multples of the fundamental frequency. 7

22 8 Now we turn to the fnal correcton to our vbratng rotatng molecule. Recall that the rgd rotor harmonc oscllator model leads to E n, j = 0 (n+1 / )+ BJ(J +1). Because the ampltude of vbraton ncreases wth the vbratonal state, we expect that R 0 should ncrease slghtly wth n, and that therefore B should decrease wth ncreasng n. Snce B depends on the vbratonal quantum number we now wrte E = (n+1 / )+ B J(J +1). n, j 0 The dependence of B on the vbratonal quantum number s called vbraton-rotaton nteracton. The vbraton-rotaton nteracton affects the P and R branches of a rotatonal vbratonal spectrum dfferently. The frequences of the P and R branches are gven by ( J =+1)= +(B - B )J +(B - B )J, J = 0,1,,... R and p( J = -1)= -(B + B )J +(B - B )J , J = 1,,3,... In both cases J s the ntal rotatonal quantum number. Because B 1 < B 0, the spacng between the lnes n the P branch ncrease wth ncreasng J. You should all ether have observed ths or wll shortly observe ths n your spectra of H 35 Cl. The lnes n the R and P branches are labeled by the ntal value of the rotatonal quantum number gvng rse to the lnes. Thus, the lnes for the R branch are labeled as R(0), R(1), R(),... and those of the P branch are labeled P(1), P(), P(3),... etc. Gven the followng data, n Lne Wavenumber/ cm -1 8

23 9 R(0) R(1) P(1) P() lets calculate B 0, B 1, R 0 and R 1. The reduced mass of the molecule s 1.58 x 10-7 kg. If we apply our equatons for the R and P brances we fnd that for the R branch cm = + B and 418.3cm = +6B - B. For the P branch we have cm = - B and cm = + B -6B If we subtract the frst lne of the P branch from the second lne of the R branch, we fnd cm = 6B1 or B 1 = 0.4 cm -1. If we subtract the second lne of the P branch from the frst lne of the R branch, we fnd cm = 6B0 or B 0 = 0.81 cm -1. Usng our usual equaton for B we fnd that R 0 = 9.3 pm and R 1 = 93.6 pm. The dependence of B n on n s usually expressed as 9

24 30 B n = Be - e(n+1 / ). The vbratonal spectra of polyatomc molecules are more complcated. Ths s prmarly because of one factor: the larger number of atoms yelds a larger number of vbratons, any of whch may or may not absorb nfrared radaton. The bggest factor s the number of vbratons. There are two smple rules whch tell us how many dfferent vbratons a gven molecule has. For a lnear molecule, there wll be 3N-5 vbratons, where N s the number of atoms n the molecule. For a nonlnear molecule there wll be 3N-6 vbratons. Where do these rules come from? Consder a sngle partcle. It can move n space three dfferent ndependent ways, n the x-drecton, the y-drecton and the z- drecton. We call these ndependent ways the partcle can move degrees of freedom, and say that a sngle partcle has three degrees of freedom. Now suppose that we have two dfferent partcles movng through space. HOW MANY DEGREES OF FREEDOM WILL THESE TWO PARTICLES HAVE? [6] Now suppose that we have N ndependent partcles movng through space. How many degrees of freedom wll these N partcles have? [3N] Now suppose that we take these N partcles and bnd them together n a molecule. The nterestng thng s that the number of degrees of freedom s conserved, but now t s dvded among three dfferent types of through-space moton: translaton, rotaton and vbraton. The way the 3N degrees of freedom are dvded between the types of moton depend slghtly on whether the molecule s lnear or not. Let s consder a nonlnear molecule frst. The molecule can move through space only three ndependent ways n the x, y and z drectons, so of the 3N total degrees of freedom, three are taken up by translaton. Ths leaves 3N-3 for rotaton and vbraton. A three dmensonal molecule can rotate n three dfferent ways 30

25 31 (llustrate). Thus there are three rotatonal degrees of freedom. Ths leaves 3N-6 degrees of freedom, whch means that a nonlnear molecule wll have 3N-6 ndependent vbratons. A lnear molecule dffers because t has only rotatonal degrees of freedom (llustrate), and thus wll have 3N -5 vbratonal degrees of freedom. As examples, acetylene, whch s a lnear polyatomc molecule has, 3N-5 or 7 vbratons, whle ammona, whch s nonlnear, has 3N-6 or 6 vbratons. These polyatomc vbratons are crudely dvded nto two types, stretchng vbratons and bendng vbratons. There s a smple rule whch determnes the number of stretchng and bendng vbratons - the number of stretchng vbratons s equal to the number of bonds and the rest are bendng vbratons. For example n the case of acetylene, there are three bonds, the two CH bonds and the C C bond, and therefore there wll be three stretchng vbratons and four bendng vbratons. What do these vbratons look lke? The man requrement s that the vbratonal motons conserve lnear momentum (.e. don t get the molecule translatng faster) and conserve angular momentum (.e. don t get the molecule rotatng faster). Vbratons whch satsfy these requrements are called normal modes of vbraton or normal vbratons. Let me llustrate for water and acetylene. [Symmetrc and asymmetrc stretches, bendng, degeneracy for C H ] The way that these vbratons are dentfed s to group them by symmetry, from most symmetrc to least symmetrc, and wthn each symmetry group, number the vbratons startng from the hghest vbratonal frequency and movng to the lowest. Thus for water, we have two symmetrc vbratons, the stretch and the bend. Snce these have the hghest symmetry they are numbered 31

26 3 frst. The stretch has a hgher vbratonal frequency, so t becomes ν 1, and the symmetrc bend s labeled ν. The asymmetrc stretch s labeled last and s called ν 3. Wll we see all of these vbratons? The answer s no. A crtcal requrement for a vbraton to absorb lght s the presence of a dpole dervatve, a change n the dpole moment of the molecule due to the vbraton. For datomc molecules, t s trval to determne whether a the dpole moment of a molecule wll change on vbraton - f the molecule s homonuclear, there wll be no dpole moment change and no nfrared absorpton, whle f the molecule s heteronuclear, there wll be a dpole moment change and an nfrared absorpton wll be observed. Thus HCl, CO or KF wll all have dpole dervatves and nfrared spectra, whle H, N, and O wll not absorb n the nfrared. Ths latter fact s very convenent, snce t means that we can take nfrared spectra n ar wthout nterference from the enormous numbers of ntrogen and oxygen molecules present. For polyatomc molecules all polar molecules wll have dpole dervatves for all ther vbratons. However, unlke datomc molecules, nonpolar polyatomc molecules can stll have nfrared absorptons from some of ther vbratons. The general prncple here s that for most nonpolar molecules, the lack of polarty s from the dpole moments n polar bonds cancellng due to symmetry. [Illustrate wth CO ] Ths lack of polarty s retaned durng a vbraton f the vbraton doesn t change the symmetry of the molecule. Ths s easest to see f you draw the molecule before and after the vbraton. [Illustrate wth CO ] Thus the CO symmetrc stretch retans the orgnal symmetry, and wll not result n an nfrared spectrum, whle the asymmetrc stretch and the bend both break the symmetry and wll result n nfrared absorptons. 3

27 33 In addton to absorbng electromagnetc radaton as a result of rotatonal and vbratonal transtons, molecules can absorb electromagnetc radaton as a result of electronc transtons. The dfference n energes between electronc levels are usually such that the radaton absorbed falls n the vsble or ultravolet regons. Just as rotatonal transtons accompany vbratonal transtons, both rotatonal and vbratonal transtons accompany electronc transtons. If we draw a seres of electronc potentals, we see that each electronc state has a seres of vbratonal states assocated wth t. The energes of these vbratonal states are small compared to the energes of the electronc states, so the vbratonal levels wll appear as fne structure n an ultravolet-vsble (UV-vs) spectrum. Each of these vbratonal levels, wll, n turn, have a set of rotatonal levels assocated wth t. 33

28 34 Lecture 38 Accordng to the Born-Oppenhemer approxmaton, the electronc energy s ndependent of the vbratonal-rotatonal energy. If we use the anharmonc oscllator to descrbe the vbratons of the molecule, and the rgd rotor to descrbe the rotatons, the total nternal energy of a datomc molecule,.e., the total energy excludng translaton, s n wavenumbers E = + (n+ 1 )- x (n+ 1 total el e e e ) + BJ(J +1) where el s the energy at the mnmum of the electronc potental energy curve. The selecton rules for vbronc transtons ( vbratonal transtons n electronc spectra) allows n to take on any ntegral value, unlke the case of vbratonal-rotatonal transtons where n = ±1. Because rotatonal energes are much smaller than vbratonal energes, we shall gnore the rotatonal term n our equaton for the nternal energy and nvestgate only the vbratonal substructure of electronc spectra. In electronc absorpton spectroscopy, the vbronc transtons usually orgnate from the n = 0 state because that s the only state whch s apprecably populated at room temperature. The frequences of an electronc transton orgnatng n the n = 0 state are gven by obs = elel +( 1 e - 1 ex e )-( 1 e - 1 ' ' ' '' '' '' ' ' ' ' ' ' ex e )+ en - exen (n +1) 4 4 The term el, el s the dfference n energes of the mnma of the two electronc potental energy curves [llustrate]. The sngle prmes ndcate the upper electronc state and the double prmes ndcate the lower electronc state. Both the fundamental vbratonal frequency, e, and the 34

29 35 anharmoncty, e x e, depend on the shape of the electronc potental energy curve at ts mnmum and so should dffer for each electronc state. 35