Determination of the normal modes

Normal mode coordinates are linear combinations of the atomic displacements {xi yi zi}, which are the components of a set of vectors {Q} in a 3N-dimensional vector space called

displacement vector space, with basis vectors {e01 e₀₂ . . . eN
1, 3}¼ {eij}, where i¼ 0, 1, . . . , N
1 labels the atoms and j ¼ 1, 2, 3 is the orientation of each one of the subset of three orthogonal unit vectors parallel to OX, OY, OZ and centered at the equilibrium positions of the atoms (see, for example, Figure9.1). A symmetry operation induced by the symmetry operator R2 G, which interchanges like particles, transforms the basis heijj to heij0j,

Rheijj ¼ heij0j ¼ heijj
dispðRÞ (1) The MR of R,
disp(R), is a 3N3N matrix which consists of N 33 blocks labeled
^lm which are non-zero only when R transforms atom l into atom m, and then they are identical with the MR for an orthonormal basis {e1 e2 e3} in 3-D space. Since a 33 matrix
^lm occurs on the diagonal of
disp(R) only when l¼ m, it is a straightforward matter to determine the character system for
dispand hence the direct sum of IRs making up
disp

and which give the symmetry of the atomic displacements in displacement vector space, in which we are describing the motion of the atomic nuclei. This basis {eij} is not a convenient one to use when solving the equations of motion since both the potential energy  and the kinetic energy T contain terms that involve binary products of different coordinates {xi yi zi} or their time derivatives. However, an orthogonal transformation

heð
Þj ¼ heijjðA
^Þ
¹ (2)

to a new basis set {e(
)} can always be found in which both T and  are brought to diagonal form so that Q(
) referred to the new basis {e(
)} in which
denotes one of the IRs and  denotes the component of the IR
when it has a dimension greater than unity. The particle masses do not appear in T and  because they have been absorbed into the Qkby the definition of the normal coordinates. A displacement vector Q is therefore

C C

9.4 Determination of the normal modes 163

Q¼ heijjqiji ¼ heijjðA
^Þ
¹A
^jqiji ¼ heð
ÞjQki, (4) where {qij}implies the whole set of mass-weighted displacements. Similarly {eij} in eq. (4) implies the set of Cartesian unit vectors on each of the i¼ 0, 1, . . . , N
1 atoms. Note that A
^is the orthogonal matrix which transforms the coordinates qijinto the normal coordi-nates {Qk}. The normal coordinates {Qk} form bases for the IRs, and therefore they will now be called {Q(
)}. We do not need to evaluateA
^explicitly since the {Q(
)} may be found by projecting an arbitrary one of the {qij} into the appropriate
subspace,

Qð
Þ ¼ N ð
ÞP

R

ðRÞ^Rq^ ij: (5)

Here qij¼ M^½_i xij, Miis the mass of atom i, and xijis the jth component of the displacement of atom i. The procedure must be repeated for each of the IRs (labeled here by
); N(
) is a normalization factor. The projection needs to be carried out for a maximum of three times for each IR, but in practice this is often performed only once, if we are able to write down by inspection the other components Q(
) of degenerate representations. It is, in fact, common practice, instead of using eq. (5), to find the transformed basis

(4) jeð
Þi ¼ A
^jeiji (6)

by projecting instead one of the {ei} and then using the fact that Q(
) is given by the same linear combination of the {qij} as e(
) is of the {eij} (cf. eqs. (6) and (4)). The absolute values of the displacements are arbitrary (though they are assumed to be small in com-parison with the internuclear separations) but their relative values are determined by symmetry.

There is a complication if the direct sum of IRs contains a particular representation
more than once, for then we must take linear combinations of the e(
) for this IR by making a second orthogonal transformation. This second transformation is not fully determined by symmetry, even after invoking orthogonality conditions. This is a common situation when bases for different representations of the same symmetry are combined: the linear combinations are given by symmetry, but not the numerical coefficients, the deter-mination of which requires a separate quantum or classical mechanical calculation. (We met a similar situation when combining linear combinations of ligand orbitals with central-atom atomic orbitals (AOs) that formed bases for the same IRs in the molecular orbital (MO) theory described in Chapter6.) Because of the assumed quadratic form for the potential energy  (by cutting off a Taylor expansion for  at the second term, valid for small-amplitude oscillations) the time dependence of the normal coordinates is simple harmonic.

Example 9.4-1 Determine the normal coordinates for the even parity modes of the ML₆ molecule or complex ion with Ohsymmetry.

A diagram of the molecule showing the numbering system used for the atoms is given in Figure9.3. Set up basis vectors {ei1 e_i2 e_i3}, i¼ 0, 1, . . . , 6, on each of the seven atoms as shown in the figure. Table9.2shows the classes cpof the point group of the molecule, the number of atoms NR that are invariant under the symmetry operator R2 cp, the

Table9.2.DerivationofthesymmetryofthenormalmodesofvibrationfortheML6moleculeorcomplexionwithOhsymmetry. Therowsofthistablecontain(1)theclassesofOh¼OCi;(2)oneelementRfromeachclass;(3)thenumberofatomsoftypeMinvariant underR;(4)thenumberofatomsNRoftypeLthatareinvariantunderR;(5)thematrixrepresentativeofRforthebasis{ei1ei2ei3};(6)the character
rofthisMR
r;(7)thecharacter
¼NR
rofthe1818MRofL6;(8)thecharacterofthe2121MRforthewholebasis.¼2p/3, n¼3
½ [111],a¼2
½ [110]. E3C28C36C46C20I3h8S66S46d EC2zR(n)C^{þ 4z}C2aIzIR(n)S
^4za 1111111111 6202004002 1 1 1

2 4

3 51 1 1

2 4

3 51 1 1

2 4

3 51 1 1

2 4

3 51 1 1

2 4

3 51 1 1

2 4

3 51 1 1

2 4

3 51 1 1

2 4

3 51 1 1

2 4

3 51 1 1

2 4

3 5 3
101
1
310
11 18
202004002 21
303
1
350
13

submatrices
r, the character set {
r}, the characters for the 18 18 reducible matrix representation for L6, and finally the characters for the 21 21 reducible matrix representa-tion
. This representarepresenta-tion may be reduced in the usual way to yield the direct sum

¼ A1g Eg T1g T2g 3T1u T2u, (7) with a total degeneracy of 21. The character table for Ohshows that the three rotations form a basis for T1gand subtracting off T1gfrom the direct sum in eq. (7) leaves

_v
_t¼ A_1g E_g T_2g 3T_1u T_2u (8) as the representations to which the 3N
6 ¼ 15 normal modes and the three translations belong. We may not separate off
tsince there are in all three degenerate modes of T1u

symmetry, two vibrational modes and one translational mode. We now apply the projection operator in eq. (5) for the three even-parity representations A1g Eg T2gto obtain

A1g ¼¹= ffiffip6

½x1
x4þ y2
y5þ z3
z6, (9) E_gðuÞ ¼¹= ffiffi^p₁₂½2z₃
2z₆
x₁þ x₄
y₂þ y₅, (10) EgðvÞ ¼ ½½x1
x4
y2þ y5, (11) T2gðÞ ¼ ½½z2
z5þ y₃
y₆, (12) T_2gðÞ ¼ ½½z1
z4þ x3
x6, (13) T2gðÞ ¼ ½½y₁
y₄þ x2
x5: (14)

e₀₁ e₂₁

e₃₁

e₃₂

e₂₂ 0

1

2 3

4

5

6 e₀₂ e₀₃

e₁₁ e₁₂ e₁₃

e₂₃ e₃₃

Figure 9.3. Basis vectors used to describe the atomic displacements in the ML₆molecule, showing the numbering system used for the seven atoms. (Labeling of the unit vectors at atoms 4, 5, and 6 is not shown explicitly to avoid overcrowding the figure.)

The normal mode displacements are sketched in Figure 9.4. The notation u, v for the degenerate pair of Egsymmetry and , ,  for the T2gtriplet is standard. Actually, these projections had already been done in Section6.4, but this example has been worked in full here to illustrate the projection operator method of finding normal modes.

Similarly, for the odd-parity T2umodes,

(6) or (6.4.42) T2uðÞ ¼ ½½x3þ x6
x2
x5 , (15) (6) or (6.4.43) T_2uðÞ ¼ ½½y₁þ y₄
y₃
y₆, (16) (6) or (6.4.44) T_2uðÞ ¼ ½½z1þ z4
z2
z5: (17) Equation (7) tells us that there are, in all, three independent motions of T1usymmetry, the three independent components of each set being designated T1u(x), T1u(y), and T1u(z).

Omitting normalization factors, the ligand contributions to the z components of the normal coordinates of two of these modes are

(5) QðT1u, z, 1Þ ¼ q13þ q23þ q43þ q53, (18)

(5) QðT1u, z, 2Þ ¼ q33þ q63: (19)

The third one is just the z component for the central atom

QðT1u, z, 3Þ ¼ q03: (20)

X 1

6

2 Y 4 3 Z

5

A_1g E_g(v) E_g(u)

T_2g(η)

T_2g(ζ) T_2g(ξ)

Figure 9.4. Normal modes of vibration of the even-parity modes of ML₆in O_hsymmetry. Arrows show the phases and relative magnitudes of the displacements, but the actual displacements have been enlarged for the sake of clarity.

9.4 Determination of the normal modes 167

Superimposing the normal coordinates in eqs. (18), (19), and (20) with equal weight and phase gives

QðT1u, z, IIIÞ ¼ ½q03þ q13þ q23þ q33þ q43þ q53þ q63: (21) Two more linear combinations of eqs. (18), (19), and (20) give the coordinates

QðT1u, z, IÞ ¼ ½
q03þ aðq13þ q23þ q43þ q53Þ
bðq33þ q63Þ, (22) QðT1u, z, IIÞ ¼ ½þq03þ a⁰ðq13þ q23þ q43þ q53Þ
b⁰ðq33þ q63Þ: (23) There are three orthogonality conditions between the Q(T1u, z) normal coordinates but four unknown constants in eqs. (22) and (23), so this is as far as one can go without a model for the adiabatic potential.

Problems

9.1 The observed infra-red spectrum of ozone contains three fundamental bands at fre-quencies 705 cm
¹, 1043 cm
¹, and 1110 cm
¹. Use this information to decide which of I, II, and III in Figure9.5are possible structures for ozone. Predict what you would expect to find in the Raman spectrum of ozone.

9.2 The chromate ion CrO²
₄ has the shape of a tetrahedron. Deduce the symmetries of the normal modes and explain which of these are infra-red active and which are Raman active.

9.3 The following bands were found in the region of the spectrum of OsO4N (N denotes pyridine) associated with the stretching of Os––O bonds:

infra-red n=cm
¹¼ 926, 915, 908, 885,

Raman n=cm
¹¼ 928ðpÞ, 916ðpÞ, 907ðpÞ, 886ðdpÞ,

where p indicates that the scattered Raman radiation is polarized and therefore can only be due to a totally symmetric vibration, and similarly dp indicates that the Raman band at 886 cm
¹is depolarized and therefore not associated with a totally symmetric vibration. Four possible structures of OsO4N are shown in Figure9.5, in each of which the four arrows indicate unit vectors along the direction of the Os––O stretching mode.

State the point group symmetry of each of the four structures and determine the number of allowed infra-red and Raman bands associated with Os––O stretching in each structure, the number of coincidences, and whether the Raman bands are polarized.

Hence decide on the structure of OsO4N. [Hint: It is not necessary to determine the symmetries of all the normal modes.]

9.4 When XeF4was first prepared it was thought to be highly symmetrical, but it was not known whether it was a tetrahedral or a square-planar molecule. The infra-red absorp-tion spectrum of XeF4consists of three fundamental bands and the vibrational Raman spectrum also has three bands. Determine the symmetry of the normal modes of a

square planar AB4molecule, and hence show that the above evidence is consistent with a square-planar configuration for XeF4.

9.5 The infra-red spectrum of Mo(CO)3[P(OCH3)3]3(VIII) shows three absorption bands at 1993, 1919, and 1890 cm
¹in the region in which CO stretching frequencies usually appear. But Cr(CO)3(CNCH)3 (IX) has two absorption bands in the C––O stretch region at 1942 and 1860 cm
¹. Octahedral ML3(CO)3complexes can exist in either the mer or fac isomeric forms (Figure 9.2). Assign the structures of the above two molecules. How many bands would you expect to see in the vibrational Raman spectra of these two molecules, and for which of these bands would the scattered Raman radiation be polarized?

9.6 Two important geometries for seven-coordinate complex ions are the mono-capped trigonal prism (X) and the pentagonal bipyramid (XI) (Figure9.5). Infra-red spectra have been measured for the seven-coordinate complex ion MoðCNÞ⁴
₇ as solid K4Mo(CN)7.2H2O and in aqueous solution. In the C––N stretching region the infra-red spectrum shows six bands at 2119, 2115, 2090, 2080, 2074, and 2059 cm
¹for the solid, and two bands at 2080 and 2040 cm
¹ for solutions. How many Raman and infra-red bands would you expect for (X) and (XI)? What conclusions can be drawn from the experimental data given? How many Raman bands are to be expected for the solid and the solution?

9.7 The NO
₃ ion is planar (like CO²
₃ ), but when NO
₃ is dissolved in certain crystals (called type 1 and type 2) it is observed that all four modes become both Raman active

N

Figure 9.5. Three possible structures of ozone, I, II, and III. Four possible structures of OsO4N (N = pyridine), IV, V, VI, and VII. Two possible structures for Mo(CN)₇⁴
, X and XI.

Problems 169

and infra-red active. In crystals of type 1 there is no splitting of degenerate modes, but in type 2 crystals the degenerate modes of NO
₃ are split. Suggest an explanation for these observations. [Hint: Character tables are given in AppendixA3. Table9.3is an extract from a correlation table for D3h.]

9.8 Two likely structures for Fe(CO)5are the square pyramid and the trigonal bipyramid.

Determine for both these structures the number of infra-red-active and Raman-active C––O stretching vibrations and then make use of the data given in Table9.4to decide on the structure of Fe(CO)5.

Table 9.4.

CO stretching frequencies/cm
¹

Infra-red absorption 2028, 1994

Vibrational Raman scattering 2114, 2031, 1984 Table 9.3.

D

_3h C3v Cs

A₁⁰ A₁ A⁰

A₂⁰⁰ A₁ A⁰⁰

E⁰ E 2A⁰

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