Harmonic Oscillator Molecular Vibrations (16.9) Diatomic Molecules I

Section VIII: Vibrational Spectroscopy Vibrational Spectroscopy

Simple Harmonic Oscillator Molecular Vibrations (16.9) Diatomic molecules i. IR Spectra (16.9, 16.10) ii. Raman Spectra (16.13) iii. Anharmonicity (16.11) iv. Vibration-Rotation (Rovibrational) Spectroscopy (16.12) Polyatomic molecules (16.14) i. Vibrational Selection Rules ii. Vibration-Rotation (Rovibrational) Spectroscopy (16.15) iii. Raman Spectra (16.16) iv. Anharmonicity Applications Harmonic Oscillator We can treat the vibrations in a diatomic molecule as an single oscillator of mass µ that obeys Hooke’s Law

m1 m2 re r re x r µ x µ

Suppose the bond between the two atoms behaves as a spring obeying Hooke’s law: F = -kx where k is a force constant, with units of N m-1, and x is the vibrational coordinate, a measurement of displacement away from equilibrium, x = r - re. If µ is the reduced mass, we rewrite Hooke’s law as: d 2x µ '&kx dt 2 and rearrange into a second-order differential equation:

d 2x k d 2x k % x ' % $2x = 0 , $ ' dt 2 µ dt 2 µ Classical Harmonic Oscillator One may use the series method to solve this equation for differential equations, or just try some simple trial functions. In fact, x(t) = Asin(Tt) is a solution, yielding: dx k 1/2 p(t) ' µ ' µ TAcos(Tt), where T ' dt µ Thus, the position of the particle varies in time harmonically (i.e., sin Tt) with frequency < = T/2B. The particle is also stationary (i.e., p = 0) when the displacement x has its maximum value, A, which is the amplitude of the motion. The potential energy is written as dV '&Fdx '%kxdx 1 1 V ' kx 2 ' kA 2sin2 Tt 2 2 The kinetic energy is written as p 2 (µ TAcosTt)2 1 E ' ' ' µ T2A 2cos2Tt K 2µ 2µ 2 and since T = (k/µ)1/2, it can be rewritten: 1 E ' kA 2cos2Tt K 2 and the total energy is therefore constant! 1 1 1 E ' kA 2sin2Tt % kA 2cos2Tt ' kA 2 2 2 2 Q.M. Harmonic Oscillator Using the Schroedinger equation with the known potential energy for the harmonic oscillator, we write 2 2 £ d 1 2 & % kx R ' ELR 2µ dx 2 2 Solving this Schroedinger equation is a bit messy, as it involves the use of Hermite polynomials. Through some effort, it can be shown that the wavefunctions for this equation are found to be of the form: Tµ R ' N H ( "x)exp[(&"x 2)/2] " ' L L £

NL is a normalization factor 1 " 1/2 1/2 NL ' 2LL! B and HL is a Hermite polynomial L 2 L 2 d exp(y ) HL(y) ' (&1) exp(y ) dy 2 Hermite Polynomials: H0(y) = 1 H1(y) = 2y 2 H2(y) = 4y -2 3 H3(y) = 8y -12y 4 H4(y) = 16y -48y2 +12 5 3 H5(y) = 32y -160y +120y 6 4 2 H6(y) = 64y -480y +720y -120 Q.M. Harmonic Oscillator, 2 The wavefunction for the vibrational ground state is: 2 R0 ' N0H0( "x)exp[(&"x )/2] Substituting in: Tµ " ' H ( "x) ' 1 £ 0 1 " 1/2 1/2 " 1/4 N0 ' ' 200! B B we finally have: " 1/4 "x 2 R ' exp & 0 B 2 which is a gaussian function. A table of harmonic oscillator wavefunctions are shown below, along with a graphical representation: Energies of the Harmonic Oscillator Now, the eigenvalue comes from the Schroedinger equation, with the energy of the state with the vibrational quantum number, L = 0, 1, 2, ..., 4 equal to: £ k k E ' (2L % 1) ' £ (L % ½) L 2 µ µ or if we let k T ' £ e µ we have:

EL ' Te (L % ½) or in more traditional frequency units: 1 k E ' h< (L % ½), < ' L 0 0 2B µ

# Even if v = 0, energy is not equal to zero (i.e., this is the zero point energy -there is always some vibrational contribution to energy).

# Wavefunctions also extend outside of the potential (-ve EK is implied by this!) - this leads to the concept of tunneling Introduction: Vibrational Spectroscopy In this section, we will focus upon the region of the spectrum ranging from 100 - 5000 cm-1 (1.2 - 60 kJ mol-1). The energy of light in this region excites vibrations of molecules which absorb it (i.e., molecule is excited to a state with increased vibrational energy). Since rotational energies of molecules are smaller than vibrational energies, rotational transitions are often simultaneously excited. In infra-red spectroscopy, light of all different frequencies passes through a sample and the intensity of the transmitted light is measured at each frequency. Samples can be gaseous, suspended in a Nujol mull or pressed into a KBr pellet. NaCl or KBr windows are typically used on the sample cells, since there is no absorbance in the IR region. In Raman spectroscopy, scattered light is observed as opposed to transmitted light. The sample if often held in a capillary tube, and the anti-Stokes and Stokes scattering are detected at right angles to the sample with various types of photon counters. The intensity of the scattered light is about 1/1000 of the incident beam.

sample cell ir detector ir sousource reference cell

incident light sample tube visible or ir laser scattered light visible photon counter Vibrational Spectra of Diatomics We have already seen that under the simple harmonic oscillator approximation, vibrational energy levels, Ev, of a diatomic molecule are given by 1 k E ' h<0 (L % ½), <0 ' L 2B µ where the reduced mass is µ = m1m2/(m1 + m2) and k is the force constant, which typically has units of aJ Å-2. Force constants increase with increasing bond order. O2, NO, CO and N2 have bond orders of 2, 2.5, 3 and 3, respectively. Molecule k (aJ Å-2) Molecule k (aJ Å-2) Molecule k (aJ Å-2) HCl 5.16 F2 4.45 CO 18.55 HF 9.64 O2 11.41 N2 22.41 Cl2 3.20 NO 15.48 note that 1 aJ = 10-18 J (a signifies atto / 10-18) The strength of the spring representing the bond results from a subtle balance of nuclear repulsions, electron repulsions and nuclear-electron attractions - none of which are affected by isotopic substitution (so neither is k). A common shape for the potential energy is a parabola, V = ½kx2 Infra-Red Vibrational Spectroscopy Similar to rotational spectroscopy, term values are used to describe the spectrum. The vibration term G(v) has dimensions of wavenumber in most cases: E L ' G(L) ' T(L % ½) hc N.B.: another unfortunate & inconsistent convention is that T is the vibration wavenumber in cm-1 (not frequency) A transition between lower and upper states described by the vibrational wavefunctions RLO and RLN is given by ( R ' R) µR)) dx L m L L x represents displacement from equilibrium (i.e., r - re) For a molecule without a permanent electric dipole moment

(i.e., homonuclear diatomic), µ = 0, so RL = 0, and IR active vibrational transitions are forbidden. For a heteronuclear diatomic, µ … 0 and varies with x. The variation goes as a Taylor series expansion (e = equilibrium configuration):

dµ 1 d 2µ µ ' µ % x % x 2 % ... e dx 2! 2 e dx e

(dµ/dr)e µ

re r IR Spectroscopy, 2 This means the transition moment integral can be written as ) ( )) dµ ) ( )) R ' µ R R dx % R x R dx % ... L e m L L m L L dx e

Since RvO and RvN are eigenfunctions of the same Hamiltonian, they are orthogonal, and when LO … LN ( R) R)) dx ' 0 m L L which means that

dµ ) ( )) R ' R x R dx % ... L m L L dx e The first term in this series will be non-zero if

)L ' ±1 which is the vibrational selection rule. Once again, the transition is always )L = L(upper) - L(lower) = +1 Vibrational spectra are typically observed as absorption spectra. At normal temperatures, the intensities of the transitions decrease as LO increases, since the population NL of the vth vibrational level is related by N0 by N E L ' exp & L N0 kT IR Spectroscopy, 3 Each vibrational transition gives rise to a band. All bands with LO … 0 are referred to as hot bands, since the populations of the lower levels of such transitions, and therefore transition intensities, increase with temperature. The term line is reserved for transitions between rotational levels associated with the two vibrational levels. The lines are referred to as fine structure of a band, which is commonly observed in vibrational-rotational spectra of molecules in the gas phase. 2 Transition intensities are proportional to *Rv* , and 2 therefore to (dµ/dx)e

(dµ/dr)e µ

re r # As r 6 0, µ 6 0, since the nuclei are in the same position. # As r 6 4, µ 6 0, since the nuclei dissociate into atoms. # The maximum values of µ occurs at r < re in the figure above, meaning that there is a negative slope at this point, and a negative slope at re. If the maximum of µ occurs at r > re, then dµ/dr is positive, and so is the slope at re. # If the maximum µ is at re, then dµ/dr = 0. In this case, the )L = 1 transitions are still allowed, but will have zero intensity (selection rules tell us which transitions may occur, but nothing about their intensities!!!) Vibrational Raman Spectroscopy Both homonuclear and heteronuclear molecules are Raman active, since the polarizability varies during vibrational motion, giving the vibrational Raman effect. Consider the polarizability ellipsoids in the CO molecule:

L1, symmetric stretching mode O CO OCO OCO

L2, bending mode O O C OCO C O O

L3, asymmetric stretching mode OCO OCO OC O Vibrational Raman Spectroscopy, 2 In a similar manner to rotational Raman spectroscopy, the variation of the dipole moment which is induced by irradiation of the sample with intense monochromatic radiation at wavenumber L is

µ ' "0,vAsin2Bc¯Lt & ½"1,vAcos2Bc(¯L % T)t

% ½"1,vAcos2Bc(¯L & T)t where "0,v is the average polarizability during the vibration, "1,v is the amplitude of the change to polarizability due to vibration, A is the amplitude of the oscillating EM field and T is the vibration wavenumber. The second and third terms correspond to Stokes (L - T) and anti-Stokes (L + T) Raman scattering. Unlike for rotational Raman spectra, the vibration which alters the polarizability only goes through one cycle per vibration; thus " changes at the same frequency as vibration (note: missing factor of 2 compared to rotation) The polarizability can be expanded like the dipole: d" 1 d 2" " ' " % x % x 2 % ... e dx 2! 2 e dx e meaning that the vibrational Raman transition moment is d" ) ( )) R ' A R x R dx % ... v m v v dx e The first term is non-zero if )v = +1 , which is the vibrational Raman selection rule. Vibrational Raman Spectroscopy, 3 Intensities of the Raman transitions are proportional to 2 2 *Rv* , and therefore to (d"/dx)e . Since " is a tensor, we can demonstrate the variation of the mean or isotropic polarizability, ", where: 1 "¯ ' (" % " % " ) 3 xx yy zz

(d"/dr)e

re r Unlike µ, " varies little with r, and d"/dr is usually positive. Raman vibrational intensities are therefore less sensitive to the environment of the molecule (i.e., solvent) than corresponding IR vibrational intensities. The origin of the vibrational Raman bands

V1 is analogous to the rotational transitions. Monochromatic EMR promotes a V0 molecule from v = 0 to a virtual state V0, which may return to v = 0 (Rayleigh) or v = 1 (Stokes Raman). Or, a molecule may be promoted from v = 1 to virtual state V1, v and return to which may return to v = 0 1 (Rayleigh) or v = 0 (anti-Stokes Raman). ***At normal T, the v = 1 state may have 0 such a low population that the anti-Stokes Stokes anti-Stokes transitions may not be observed. Anharmonicity We have only considered the first terms in the expansions of the equations for µ and ". If µ and " were linear in x, then they would be said to vary harmonically with x. However, the higher order terms make contributions (albeit smaller ones) to how µ and " vary with x. This effect is known as anharmonicity, or specifically, in the case of electrical and mechanical properties of the molecule, electrical & mechanical anharmoncity. One effect of electrical anharmonicity is to modify the selection rule )v = ±1 of IR and Raman spectroscopy to )v = ±1, ±2, ±3, .... However, since electrical harmonicity is relatively small, the transition intensities of the )v = ±2, ±3, ... transitions are very small: vibrational overtones.

The mechanical behaviour of the diatomic molecule is also not completely harmonic. If vibrational motion obeys Hooke’s law exactly, all of the vibrational energy levels are spaced by hL, and there is a single peak in the spectrum. But it does not, there is mechanical anharmonicity!

The parabola only approximates the mechanical behaviour near the bottom of the potential well (i.e., where r and re are not much different, and x is small). For instance, we know that when r is large, the molecule dissociates, k = 0, and V is no longer influenced. Overtones To get a strong transition we need: # A large transition dipole moment: i.e., electric dipole moment changes in the molecule upon excitation # )v = ±1 # Population in the lower state must be higher: using Boltzmann distribution one can show that the population in v = 0 state is much higher than for v = 1, 2, 3, ... states for almost all small molecules: hot bands are comparably weak! fundamental first overtone transition second overtone

)E = hTe

T Or, if there is rotational fine structure:

So at this point, it seems as if we only have a single peak in all of our vibrational spectra!! Origins of Overtones If anharmonic wavefunctions are used, then the vibrational selection rules expand to include )L = ±2, ±3, ... transitions. Part of the reason for this is that since the anharmonic wavefunction is represented as an expansion of harmonic oscillator functions: R ' c R vib,anh j i iHO i The dipole moment function which determines the vibrational selection rules is now not terminated after the linear term, and we have integrals of the type:

) 2 )) ) 3 )) +L |(r & re) |L ,, +L |(r & re) |L ,,etc. in the Taylor series expansion. Generally speaking, these overtone transitions are weak, but can be of appreciable intensity. The oscillator is said to be “electrically anharmonic” if terms higher than linear are used in the representation of µe. The “mechanical anharmonic” from the expansion of µ(r) in terms of x = r - re (slide “Infra-Red Vibrational Spectroscopy”).

The LN=1 - LO=0 transition, or shorthand: L = 1-0 or L = 170 transition, is called the fundamental, and any transition with LO … 0 is called a hot band. The first overtone is the 2-0, the second overtone is the 3-0, etc. Mechanical Anharmonicity The potential energy curve flattens out when V = De, where De is the dissociation energy relative to the equilibrium potential energy: at r > re, the p.e. curve becomes shallow.

At small values of r < re, there is mutual repulsion between the nuclei, and the p.e. curve becomes very steep.

The deviations away from the simple harmonic oscillator described above result from mechanical anharmonicity. The effects of mechanical anharmonicity are larger than electrical anharmoncity. Recall, electrical anharmonicity modifies the selection rule of )L = ±1 of IR and Raman spectroscopy to )L = ±1, ±2, ±3, .... and produce consecutively weaker vibrational overtones )L = ±2, ±3, ... But perhaps more importantly, mechanical anharmonicity modifies the vibrational term values and wavefunctions. The simple term values G(L) = T(L + ½), becomes modified in a power series in (L + ½): 2 3 G(L) ' Te(L % ½) & Texe(L % ½) % Te ye(L % ½) % ... where Te is the wavenumber for an infinitessimal displacement from equilibrium, and Texe and Teye are anharmonic constants (Texe is always +ve) Mechanical Anharmonicity, 2

The anharmonic constants along the series, Texe, Teye, Teze, 1 35 rapidly decrease in magnitude. For example: in H Cl, Te -1 -1 -1 = 2990.946 cm , Texe = 52.8186 cm , Teye = 0.2244 cm -1 and Teze = -0.0122 cm .

The positive Texe value has the effect of closing up the energy levels with increasing values of L. The harmonic oscillator has the evenly spaced energy levels, but the anharmonic oscillator levels converge at the dissociation limit, De, above which there is a continuum of levels. simple harmonic oscillator anharmonic oscillator

Unlike the S.H.O., the value of Te cannot be measured directly. The wavenumbers )GL+1/2 for (L+1) - L transitions:

)GL%1/2 ' G(L % 1) & G(L) 13 ' T & T x (2L % 2) % T y (3L 2 % 6L % ) % ... e e e e e 4

So to accurately determine Te and Texe, two transition wavenumbers must be measured (e.g., G(1) - G(0) = T0 and G(2) - G(1) = T1, must be obtained). Dissociation Energy

The dissociation energy, De, can be approximated by 2 Te De • 4Texe Experimentally, one can measure the dissociation energy relative to the zero-point level, D0. Examination of the anharmonic energy levels makes it clear that

D0 ' j )GL% 1/2 L If anharmonic constants are neglected, then

)GL+1/2 is a linear function of L and D0 is the energy under the

plot of )GL+1/2 vs. L. Typically only the first few )G values can be observed, extrapolation

to )GL+1/2 = 0 is made.

The area under the Birge-Sponer plot gives an approximate value of D0 - however, most plots deviate from linearity at high values of L, so D0 is often overestimated.

Experimental values of )GL+1/2 for higher values of L are normally not obtained from IR or Raman spectroscopy because of the low intensities of the )L = ±2, ±3 transitions and low populations of excited vibrational levels (usually fine structure of electronic emission spectroscopy) Birge-Sponer Plot

As mentioned, if all of the vibrational intervals )GL+1/2 are available, then the dissociation energy, D0, can be calculated from the sum of the intervals.

Birge-Sponer plot H2 ground state

The Birge-Sponer plot is )GL+1/2 as a function of L, and the area under the curve is the dissociation energy. If the vibrational expression has two terms (e.g., Morse):

2 G(L) ' Te(L % ½) & Texe(L % ½) then

)GL%1/2 ' Te & 2Texe & 2TexeL 2 and the equilibrium dissociation energy is De = Te /4Texe It is rare that all of the vibrational levels are known for a given state, so it is possible to extrapolate to approximate the last few levels to an unobserved dissociation limit with quantum number LD (which can be non-integer). The Birge-Sponer plot for the ground state of H2 above shows curvature at high values of L; one can make a linear extrapolation to approximate LD. Dissociation Energy, 2

De is unaffected by isotopic substitution, since the potential energy curve (and therefore k) is not affected by the number of neutrons in each nucleus (no charge change)

D0 is affected by isotopic substitution, since T is dependent -1/2 upon the change in µ (proportional to µ ), and D0 < De. The term value, G(0), for the Z.P.E. is 1 1 1 G(0) ' T & T x % T y % ... 2 e 4 e e 8 e e 2 1 For example, Te is less for H2 than for H2, so 2 1 D0( H2)>D0( H2) If a chemical reaction involves bond dissociation in the rate determining step, the rate of reaction is decreased by substitution of a heavier isotope at either end of the bond. Finally, the wavefunctions are different for anharmonic oscillator, with higher magnitude on the shallow side of the curve. In 1929, Morse suggested the potential function: 2 V(x) ' De [1 & exp(&ax)]

D0 De Vibration-Rotation Spectroscopy Each vibrational energy level has a stack of smaller spaced rotational levels associated with it. Rotational spectroscopy: transitions between rotational energy levels associated with a single vibrational level Vibrational spectroscopy: transitions between vibrational energy levels Vibration-rotation spectroscopy: transitions between rotational energy levels associated with two different vibrational energy levels (rovibrational spectroscopy) Vibration-rotation spectroscopy is normally only conducted in the gas phase at low pressures (where lines are not broad, i.e., high resolution spectra), whereas broadened bands are observed in spectra of solids or liquids and rotational fine structure is not typically observed. Experiments are absorption mode (IR) or scattering experiments (Raman) The selection rules are typically: )J = ±1, )L = ±1 with an R-branch, )J = +1 and a P- branch, )J = -1. Transitions are labelled R(J) and P(J), where J represents the lower state, JO. )J = 0 is forbidden, and pure vibrational transitions are not observed! Vibration-Rotation Spectroscopy, 2 For vibrational-rotational spectroscopy, the term values are

S ' G(L) % FL(J) 2 2 2 ' Te(L% ½) & Texe(L% ½) % ... % BLJ(J% 1) & DLJ (J% 1) The rotational selection rule is )J = ±1 in all cases, except in molecules that have electronic angular momentum in the ground state like NO, where the rotational selection rule is )J = 0, ±1. The (JN = 0)-(JO = 0) transition is the first line of the Q-branch ()J = 0) marks the centre of the spectrum. Rotational transitions accompanying a vibrational transition in (a) IR and (b) Raman spectra of a diatomic molecule are shown below. Vib-rotational Spectrum of HCl The v = 1-0 IR spectrum of 1H35Cl and 1H37Cl showing the P-branch and R-branch rotational structure is shown below:

The 35Cl and 37Cl isotopes occur naturally in 3:1 abundance and the band due to 1H37Cl is displaced to lower wavenumbers relative to 1H35Cl because of the higher µ. Each band is fairly symmetrical about the band centre and there is roughly equal spacing between adjacent P-branch lines, with twice the spacing between R(0) and P(1) (this is called the zero gap).

The approximate symmetry comes from B1 • B0 (the vibration-rotation interaction constant, ", is very small). If we assume B1 = B0 = B, and neglect centrifugal distortion:

¯<[R(J)] ' T0 % B(J% 1)(J% 2) & BJ(J% 1)

' T0 % 2BJ % 2B

¯<[P(J)] ' T0 % B(J& 1)J & BJ(J% 1)

' T0 & 2BJ and the zero gap (difference btw. R(0) and P(1) ) is 4B, with 2B spacing between all of the other lines. Combination Differences Actually, close examination reveals some asymmetry, with a slight convergence and divergence of the R- and P- branches, respectively, which results from the small difference between B0 and B1. The method of combination differences can be used to determine these quantities. The idea is that differences in the wavenumbers between transitions with the same upper states are dependent upon the lower states only, and the wavenumbers between transitions with the same lower states are dependent upon the upper states only. For example, since R(0) and P(2) have a common upper state with JN = 1, then <[R(0)] - <[P(2)] must be a function of BO only. Also, since R(1) and P(3) have a common upper state with JN = 2, then <[R(1)] - <[P(3)] must be a function of BO only. In general, <[R(J-1)] - <[P(J+1)] , which is written as )2OF is a function of BO only. )) )2 F(J) ' ¯<[R(J& 1)] & ¯<[P(J+1)] ) )) ' T0 % B J(J% 1) & B (J& 1)J ) )) & [T0 % B J(J% 1) & B (J% 1)(J% 2)] ' 4B ))(J % ½)

So a plot of )2OF(J) as a function of J + ½ is a straight line of slope 4BO. Combination Differences, 2 Similarly, all pairs of transitions R(J) and P(J) have a common lower state, and <[R(J)] - <[P(J)] is a function of BN only: ) )2F(J) ' ¯<[R(J)] & ¯<[P(J)] ) )) ' T0 % B (J% 1)(J% 2) & B J(J% 1) ) )) & [T0 % B (J& 1)J & B J(J% 1)] ' 4B )(J % ½)

So a plot of )2NF(J) as a function of J + ½ is a straight line of slope 4BN. The band centre is not quite midway between R(0) and P(1) but the wavenumber T0 can be obtained from

T0 ' <¯[R(0)] & 2B ' <¯[P(1)] % 2B Any centrifugal distortion effects show up as a curvature of 2 2 )2F(J) vs. (J + ½) graphs. Including the term -DJ (J+1) :

)) )) )) )) 3 )2 F(J) ' (4B & 6D )(J % ½) & 8D (J % ½) ) ) ) ) 3 )2F(J) ' (4B & 6D )(J % ½) & 8D (J % ½)

Thus, if BL can be obtained for two vibrational levels, then Be and " can be obtained from BL ' Be & "(L%1/2) Intensities of lines are calculated as usual: )) )) )) NJ )) hcB J (J % 1) ' (2J )) % 1)exp & N0 kT Raman Vibration-Rotation Spectra The Raman selection rule for diatomic molecules holds: )J ' 0, ±2 with Q, S and O branches for )J = 0, +2 & -2, respectively. The L=1 - L=0 Stokes Raman transition of CO is shown below with fine rotational structure. Assuming B1 = B0 = B:

¯<[S(J)] ' T0 % B(J% 2)(J% 3) & BJ(J% 1)

' T0 % 4BJ % 6B

¯<[O(J)] ' T0 % B(J& 2)(J& 1) & BJ(J% 1)

' T0 & 4BJ % 2B

¯<[Q(J)] ' T0

And, combination differences can be used to obtain BO and BN, so transitions with a common upper state are dependent on BO only and with a common lower state upon BN only: )) )) )4 F(J) ' ¯<[S(J& 2)] & ¯<[O(J% 2)] ' 8B (J % ½) ) ) )4F(J) ' ¯<[S(J)] & ¯<[O(J)] ' 8B (J % ½) As discussed earlier, nuclear statistical spin weights can also influence the appearance of these spectra. Key Concepts 1. Vibrational spectrscopy is observed with IR absorption or Raman scattering techniques. 2. Oscillation of the molecule electric dipole moment, µ, leads to an observed IR signal - so homonuclear diatomic molecules with µ = 0 are IR inactive. The vibrational selection rule is )v = +1. 3. Oscillation of the mean polarizability, ", leads to obseravation of a vibrational Raman spectrum. The Raman vibrational selection rule is )v = +1 as well. Both homo- and heteronuclear molecules are active. 4. Vibrational transitions give rise to bands, some of which may display rotational fine structure. Bands where the lower level is vO … 0 are hot bands. 5. Mechanical anharmonicity is important since it (1) alters the shape of the potential energy curve and (2) results in converging in the vibrational energy levels at high values of v. A Birge-Sponer plot, constructed by adding all of the vibrational energies, can be used to calculate the dissociation energy. 6. In high-resolution IR and Raman spectra on gaseous samples, it is possible to observe rotational fine structure, which may be analyzed to obtain accurate rotational constants and bond lengths for different vibrational states. 7. If rotational transitions btw. different vibrational states are observed, the method of combination differences is used to calculate rotational constants.