Vibrations in Polyatomic Molecules3N - 6 normal modes of vibration in non-linear molecules3N - 5 normal modes of vibration in linear molecules

1 2

3

r1 r2

2 1 2

3

Classically, the internuclear forces may be visualized as Hooke’slaw springs. Strong forces between O and H are represented bystrong springs (resist stretching of bonds) and between non-bonded H nuclei by a weaker spring (resistance to increase ordecrease of HOH angle)

However, it is clear that even if one nucleus undergoes some sortof displacement, the whole molecule undergoes a very complexset of motions (known as Lissajous motion). This motion is amix of bond stretching and angle-bending, and be can brokendown in terms of the normal modes which are then superimposedon one another.

z

y

C2 Fv(xz)L1 1 1

z

C2 Fv(xz)L2 1 1

z

C2 Fv(xz)L3 -1 -1

y y y

'(RL(1)) = A1 '(RL(2)) = A1 '(RL(3)) = B2

Term Symbols & TransitionsEach of the vibrations of a polyatomic molecule can beapproximated as harmonic, and the vibrational term valuesassociated with each vibration i are given by:

G(Li) ' Ti(Li % ½)

where Ti is the vibrational wavenumber (in cm-1) and Li thevibrational quantum number. For vibrations with somedegree of degeneracy di, we can write:

G(Li) ' Ti Li %di

2

As for diatomics, the vibrational selection rule is )Li = ±1,and overtone transitions are allowed, )Li = ±2, ±3, ...,which are very weak and only present when anharmonicityis also taken in account.Combination tones are also possible, involving transitionsto vibrationally excited states where more than one normalvibration is excited.

0

1

2

3Li

1

2

3Lj

Li Lj1 1

(A), (B): Fundamental and overtonetransitions for vibrations Li and Lj, and(C): combination tone transition.

(A)

(B)(C)

Low Symmetry & Localized VibrationsAs for diatomics, for vibrational transitions to be observed in anIR spectrum there must be a changing dipole moment during thevibration, and for observable Raman transitions there must be achange of the induced dipole moment.

In molecules of low symmetry (e.g., C1, CS, etc.) the selectionrules are unrestrictive, and usually all vibrational modes lead tosome sort of observable transition.

H

HH

H

ClF

For example, in C6H4FCl, N = 12,3N - 6 = 30 vibrational modes. Since themolecule is CS symmetry, all 30 modesinvolve a change in dipole moment &amplitude of induced dipole moment, so areIR and Raman active. Some transitions maybe too weak to be observed (how dipolemoments change, magnitudes, etc.) but arestill allowed.

In general, a mode of vibration involves all of the atoms in themolecule, though there are certain circumstances where thevibration is localized in one part of the molecule.

These are typically called group vibrations, characterized by agroup wavenumber (or small range of cm-1), and reflect the veryslight dependence of the vibration on the molecule in the vicinityof the chemical group.- terminal -X-Y group, X heavier than Y (e.g., -OH)- C-C, C=C, C/C groups (all different bending T)- skeletal vibrations (in chains, branched chains, or rings)- terminal double-bonded oxygens (e.g., C=O, S=O, etc.)

Group Vibrational WavenumbersBelow are some typical bond-stretching and angle-bendinggroup vibration wavenumbers, T (in cm-1)

Names of Group Vibrations

C

H

H

C

H

H

C

H

H

C

H

H

C C

H

H H

H

NH H

H

rocking twisting scissoring

wagging torsional inversion/umbrella

ring breathing

Some descriptive names for group vibrations other thanstretch, bend and deformation include rock, twist, scissors,wag, torsion, ring breathing and inversion.

O H O

Et

Et

solvent interactions

Some group vibration wavenumbers can change as theresult of solvent interactions. Phenol in hexane has T(OH,str) = 3622 cm-1, in diethylether T1(OH, str) = 3344 cm-1

IR & Raman: Complimentarys-trans-crotonaldehyde has been analyzed byboth IR and Raman spectroscopy, and has 27vibrational modes. The IR spectrum is of thesample dissolved in CCl4, and the Ramanspectrum is of the neat liquid.

Most vibrations show up in both spectra, whereas v15 (the C-CH3stretch), is strong in the IR but weak in the Raman, whereas v3(the CH3 antisymmetric stretch) is very strong is the Raman butweak in the IR.

Normal Vibrations & Symmetry SpeciesThere are simple rules for determining the number of vibrationsbelonging to each symmetry species of a point group to which amolecule belongs. Nuclei form a set if they can be transformedinto one another by symmetry operations of the point group(equivalent nuclei).

For the C2v point group:(1) Does not lie on a symmetryelement, then it must be 1 of 4equivalent nuclei, so the moleculeis symmetric w.r.t. all symmetryelements.(2) Lies on a Fv(xz) element, it is 1of set of 2(3) Lies on a Fv(yz) element, it is 1of set of 2(4) Lies on all of the symmetryelements, it is 1 of a set of 1

For non-degenerate vibrations of a particular symmetry species,all of the nuclei in a set move in the same way. Each nucleus has3 degrees of freedom, and can contribute a maximum of threedegrees of freedom to each non-degenerate species (maximumobtained if nuclei do not lie on any symmetry element).

If there are m sets of this kind they contribute 3m degrees offreedom to each symmetry type. Nuclei which lie on 1, 2 or 3(all) symmetry elements contribute 2×m, 1×m, 0×m degrees offreedom to the symmetry type in question.

Non-degenerate Vibrations, C2vConsider the C2v point group again. For set of nuclei:(1) Each set of this type (designated m) contributes 3 degrees offreedom to each symmetry species.

(2) If the motions of this set are a1symmetry, in order to be symmetric to alloperations they must move only in the xzplane and have 2 degrees of freedom. Ifthey are a2 symmetry, they are anti-symmetric w.r.t. reflection in both planes,and the atoms move along oppositedirections but lines perpendicular to the xzplane. Thus they only contribute 1 degreeof freedom to the a2 species. Similarly, 2and 1 degrees of freedom each arecontributed to b1 and b2, respectively. The number of sets of nuclei of this typeare designated as mxz.

(3) Motion of this set of nuclei are analagous to (2), anddesignated myz.

(4) There is only one nucleus in this set, and its motion issymmetric to all operations (designated as mo). Thus, it canonly move along the C2 axis and have 1 degree of freedom for thea1 symmetry species. It cannot move to be antisymmetric to bothplanes, so contributes 0 degrees of freedom to a2. It has 1 degreeof freedom for both the b1 and b2 symmetry species - to beantisymmetric to reflection in the plane it must move along a lineperpendicular to that plane.

Non-degenerate Vibrations, C2v, 2A summary of all of the normal vibrations of each symmetryspecies in the C2v point group is given in the table below (fromHollas, 3rd ed. p. 145).

For example, consider the water molecule again.

z It has 1 nucleus (oxygen) on all symmetryelements, so m0 = 1.It has 1 set of 2 nuclei (hydrogen) on the yzplane, so myz = 1.All other m’s are equal to zero.

According to the table above, the number of normal vibrationsare distributed across the symmetry species as follows:

m(a1) = 3m + 2mxz + 2myz + m0 - 1 = 2m(a2) = 3m + mxz + myz + - 1 = 0m(b1) = 3m + 2mxz + myz + m0 - 2 = 0m(b2) = 3m + mxz + 2myz + m0 - 2 = 1

So of the 3N - 6 = 3 possible vibrational modes, the totalrepresentation (which can also be obtained from normal modesanalysis) is 'vib = 2a1 + b2

Non-degenerate VibrationsFortunately the formulae for the vibrational modes havebeen derived for all of the point groups - the numbers ofvibrations for point groups with non-degenerate vibrationsare given below:

Exercise: Naphthalene belongs to the D2h point group. Assigning the short in-plane axis as the z axis, and the longin-plane axis as the y-axis, prove that the 48 (3N - 6)normal vibrations are distributed as follows: 9ag, 4au, 3b1g,8b1u, 4b2g, 8b2u, 8b3g, 4b3u.

Degenerate Vibrations, C3v

When a molecule belongs to a degeneratepoint group, e.g., C3v, the non-degeneratevibrations can be treated as previouslydescribed. In the C3v point group there are 3types of sets of nuclei (diagram shows viewdown C3 axis)

(1) These nuclei lie on no symmetry element, must be 6 nucleibelonging to each set. The 6 nuclei have 18 degrees of freedomof which 3 belong to each of the a1 and a2 species. In a doublydegenerate degree of freedom, displacement of one of thenuclei of a set corresponds to two different displacements ofeach of the other nuclei of the set. Thus, this set of nuclei have12 degrees of freedom comprising six doubly-degeneratedegrees of freedom (i.e., belong to the e symmetry species). Ifthere are m such sets, they contribute 6m degrees of freedom.

(2) These nuclei lie on the Fv planes, 3 nuclei belong to each set.3 nuclei have 9 degrees of freedom, two for the a1 and one forthe a2 species. These nuclei have 6 degenerate degrees offreedom, since the displacement of one corresponds to twodifferent displacements of the other two nuclei. 6 degrees offreedom comprise 3 doubly degenerate degrees of freedom, andare designated mv, contributing 3mv doubly degenerate degreesof freedom.

(continued next page)

(3) One nucleus lying on all symmetry elements. 3 degrees offreedom, one is a1 (along C3) and two are perpendicular to theC3 axis (doubly degenerate). If there are m0 such sets theycontribute m0 doubly degenerate degrees of freedom.

Degenerate Vibrations, C3v, 2

Degenerate vibrations are treated similarly in other pointgroups.

Exercise: Benzene belongs to the D6h point group. Show thatm2 = 2, all other m’s are zero, and prove that the normalvibrations are distributed as follows: 2a1g, 0au, 1a2g, 1a2u, 0b1g,2b1u, 2b2g, 1e1g, 3e1u, 4e2g, 2e2u (where each doubly degeneratevibration counts as two vibrations)

Vibrational Selection RulesThe vibrational selection rules are extremely useful for analyzingvibrations in polyatomics, and here we formally outline the rulesand application.

For instance, H-C/C-H has 7 normal modes of vibration,including v4 and v3 (trans and cis bending vibrations) which aredoubly degenerate. In general it is difficult to find all of themodes of vibrations and define them by inspection as polyatomicmolecules get larger and more complicated.

The vibration transition intensity is proportional to *Rv*2:

R L ' mR)

L

( µ R))

L dJL

where RN and RO represent the upper and lower states of thetransition (transitions are usually denoted LN-LO) and µ is theelectric dipole transition moment moment operator.

If R L = 0, transition is forbidden andIf R L … 0, transition is allowed.

If both vibrational levels are non-degenerate, the requirement fora transition is that the symmetry species of the quantity to beintegrated (the integrand) is totally symmetric:

Vibrational Selection Rules, 2

'(R)

L) × '(µ) × '(R))

L ) ' A

where ' is the “representation of” and A denotes the totallysymmetric species of any non-degenerate point group.

If either or both of the vibrational states are degenerate then theequation above is not quite satisfactory, since for example: E × E= A1 + A2 + E, or A × A = E+ + E- + ). So for degenerate states:

'(R)

L) × '(µ) × '(R))

L ) Ag

where we regard e as the Boolean logic symbol meaning“contains” (e.g., (E × E) e A1), whereas g means “contains or isequal to”

The transition moment is a vector with components

RL,x ' mR)

L

( µ x R))

L dJv RL,y ' mR)

L

( µ y R))

L dJL RL,z ' mR)

L

( µ z R))

L dJL

along the x, y and z axes since

|R L| ' (RL,x)2% (RL,y)

2% (RL,z)

22

So the transition is allowed if any of the R components above isnon-zero.

This means that the integrands to evaluate become:

Vibrational Selection Rules, 3

'(R)

L) × '(Tx) × '(R))

L ) ' A

'(R)

L) × '(Ty) × '(R))

L ) ' A

'(R)

L) × '(Tz) × '(R))

L ) ' A

and/or

and/or

and/or implies one or more components of Rv may be non-zero. If the lower state of the transition is LO = 0, as it often is, then'(RLO) = A (totally symmetric wavefunction), and the aboveequations become:

Translation and electric dipole moments share the samesymmetry species, so we can write:

'(µ x) ' '(Tx) '(µ y) ' '(Ty) '(µ z) ' '(Tz)

'(R)

L) × '(Tx) ' A

'(R)

L) × '(Ty) ' A

'(R)

L) × '(Tz) ' Awhich implies that:

'(R)

L) ' '(Tx) and/or '(Ty) and/or '(Tz)and:

'(R)

L) e '(Tx) and/or '(Ty) and/or '(Tz)so allowed transitions with LO as the lower state are:

'(Tx) & A; '(Ty) & A; '(Tz) & A

For H2O, the transitions are allowed since L1, L2and L3 are a1, a1 and b2 vibrations, respectively. We hadfigured this our before by determining that all threevibrations had a changing dipole moment. However, nowwe can derive rules for overtone and combinationtransitions as well.

Vibrational Selection Rules, 4110, 21

0, 310

If H2O is vibrating with two quanta of L3 then '(RLN) is A1(i.e., B2 × B2 = A1) and in general, if it is vibrating with nquanta of symmetry species S, then

'(R)

L) ' S n

This means that the symmetry species of the vibrationaland fundamental overtones for v3 alternate (A1 for L evenand B2 for L odd), so the transitions are polarizedalong the y, z, y, ... axes.

310, 32

0, 330

H2O vibrating with onequantum each of L1 or L2 andL3 has '(R LN) = B2, so boththe transitionsare allowed.

Selection rules for a C2vmolecule with a2 and b1vibrations (like CH2F2) couldbe applied in an analogousmanner.

Transitions in a C3v MoleculeNH3 belongs to the C3v point group:L1 and L2 are a1 vibrations, '(Tz) = A1L3 and L4 are degenerate E vibrations, '(Tx,Ty) = ESo the are allowed and polarized along the z axis(the C3 axis), and are allowed and polarized in thexy plane.

110 21

0and31

0 410and

If the upper state is a combination or overtone level then'(R LN) may not be a single symmetry species. Forexample, the combination transition in NH3 we have:31

0 410

'(R)

L) ' E × E ' A1 % A2 % E

since '(Tx,Ty) = E and '(Tz) = A1, it is clear that '(R LN) e'(Tx,Ty) and '(Tz), and the transition is allowed. Transitions involving A1 and E of the combination state areallowed, but that involving the A2 component is forbidden.For the overtone transition we have:32

0'(R)

L) ' (E)2' A1 % E

where transitions to both components are allowed.

Note that E × E corresponds to excitation by one quantumeach of two separate e vibrations (combination level) and(E)2 corresponds to excitation by two quanta of the samevibration level (overtone level).(E)2 corresponds to the symmetric part of E × E (i.e., this isthe part symmetric w.r.t. particle exchange).

C2v E C2 σv(xz) σNv(xz)A1 1 1 1 1 Tz αxx, αyy, αzz

A2 1 1 -1 -1 Rz αxy

B1 1 -1 1 -1 Tx, Ry αxz

B2 1 -1 -1 1 Ty, Rx αyz

Raman Selection RulesRaman selection rules work in an analagous fashion to thevibrational selection rules. For an allowed transition:

'(R)

v) × '("ij) × '(R))

v ) ' A

which applies between non-degenerate states. Here, "ij isany one of the components of the polarizability tensor(whose symmetry species are given at the side of thecharacter table). For transitions between vibrational stateswhere one of the vibrational states is degenerate:

'(R)

v) × '("ij) × '(R))

v ) e A

If the lower level is the zero-point level, then '(RvO) = A,and the requirement for a Raman transition is

'(R)

v) ' '("ij)

for degenerate or non-degenerate vibrations.

Inspection of the C2v character table above shows that all ofthe fundamental vibrations of H2O are Raman active,(recall, all but A2 are IR active).

D4h E 2C4φ ... 4σv i 2S4

φ ...

A1g/Σg+ 1 1 ... 1 1 1 ... x2+y2, z2

A2g/Σg- 1 1 ... -1 1 1 ... Rz

E1g/Πg 2 2cosφ ... 0 2 -2cosφ ... (Rx, Ry) (xz, yz)E2g/∆g 2 2cos2φ ... 0 2 2cos2φ ... (x2-y2, xy)

E3g/Φg 2 2cos3φ ... 0 2 -2cos3φ ...

! ! ! ! ! ! ! !

A1u/Σu+ 1 1 ... 1 -1 -1 ... Tz

A2u/Σu- 1 1 ... -1 -1 -1 ...

E1u/Πu 2 2cosφ ... 0 -2 2cosφ ... (Tx, Ty)

E2u/∆u 2 2cos2φ ... 0 -2 -2cos2φ ...E3u/Φu 2 2cos3φ ... 0 -2 2cos3φ ...

! ! ! ! ! ! ! !

Vibrations of a D4h Molecule

The v3 vibration in acetylene is Fu+, and '(Tz) = Eu

+. So,the transition is allowed and polarized along z. Similarly, since v5 is a Bu transition, the transition isallowed with the transition moment in the xy plane.

310

510

Vibrations of a D4h Molecule, 2The symmetry species of the 4151 combination state is

'(R)

v) ' Ag × Au ' E%

u % E&

u % )u

so one component of the transition involving Eu+ is

allowed.41

0 510

The symmetry species of the 52 overtone state is

'(R)

v) ' (Au)2' E

%

g % )g

so the transition is forbidden.520

Final points:1. Remember that these selection rules do not say anythingabout the intensity of the transition, just whether thetransition is allowed or forbidden.2. These rules apply to the free molecule, and there may besome breakdown is the liquid or solid state (e.g., because ofintermolecular interactions such as hydrogen bonding).

Mutual exclusion rule:Molecules with a centre of inversion which havefundamentals active in the Raman spectrum (e.g., g) will beinactive in the IR spectrum (e.g., u), while those active inIR will be inactive in Raman.

Key Concepts1. Vibrational motion is a mix of bond stretching and

angle-bending, broken down in terms of the normalmodes which are superimposed on one another.

2. The vibrational selection rule is )vi = ±1, andovertone transitions are allowed, )vi = ±2, ±3, ...,which are very weak and only present whenanharmonicity is also taken in account. Combinationtones are also possible, involving transitions tovibrationally excited states where more than onenormal vibration is excited.

3. A mode of vibration involves all of the atoms in themolecule, though there are certain circumstanceswhere the vibration is localized in one part of themolecule: these are typically called group vibrations.

4. Nuclei form a set if they can be transformed into oneanother by symmetry operations of the point group(equivalent nuclei). These can be used to performvibrational analysis on the molecule.

5. Vibrational selection rules can be used to determinewhich transitions are allowed or forbidden, forfundamental, overtone and combination transitions.