Ideas of Quantum Chemistry P98 potx

Now only those ψlare allowed in optical transitions from the ground state Ag that are labelled by B3u, because only the direct product B3u× B3u may contain the fully symmetric irreducibl

Now only those ψlare allowed in optical transitions (from the ground state Ag) that are labelled by B3u, because only the direct product B3u× B3u may contain the fully symmetric irreducible representation Ag Thus, the transitions Ag⇒ B3u

as well as B3u⇒ Agare allowed, if the light is polarized along x, i.e perpendicular

to the ring of the molecule

Now let us take light polarized along y, i.e within the molecular plane,

per-polarization y

pendicularly to the N–N line This time we are interested in the irreducible repre-sentations that arise from Ag× B2u, because y transforms according to B2u Very similarly [by analyzing a(α)] we find that

Ag× B2u= B2u This means that the allowed states are now of the B2utype

Similarly, for polarization along z (z belongs to B1u), i.e along the

nitrogen-polarization z

nitrogen direction, we have

Ag× B1u= B1u Thus for polarization parallel to the NN axis of the molecule, absorption may occur from the ground state to any state of the B1utype (and vice versa).

Nothing more can be said when relying solely on group theory We will not get any information about the energies of the transitions, or about the corresponding intensi-ties To get this additional (and important) information we have work hard to solve

the Schrödinger equation, rather than count on some easy profits obtained by the primitive multiplication of integers (as in group theory) To obtain the intensities,

we have to calculate the transition moment integrals μkl However, group theory,

by excluding from the spectrum many transitions (forbidden ones), provides a lot

transition

moment of important information on the molecule Table C.7 collects the calculated light

frequencies41(¯ν in wavenumbers, or cm−1, ν= c¯ν, where ν is the usual frequency),

the oscillator strengths fkl(in a.u.)

fkl=4πc

as well as the polarization of light for excitations from the electronic ground state for pyrazine and the pyrazine monocation It is seen that the left-hand side of Ta-ble C.7 is consistent with the selection rules derived above Indeed, a large fklonly corresponds to those transitions from the ground state of the pyrazine that have been predicted as allowed (B1u, B2u and B3u) The predicted polarization also agrees with the observed polarization

oscillator

strength

Excitations from an excited state

Calculations for absorption from the ground-state were particularly simple Now let us see whether anything will be more complicated for the transitions from

an excited state of the B2g type of symmetry We are going to calculate a(α) (for every α) for the following representations:

41 J Koput, unpublished results.

Table C.7. Wave numbers (¯ν), oscillator strengths (f kl ) and light polarization (in parentheses)

for polarization along x: B2g× B3u

for polarization along y: B2g× B2u

for polarization along z: B2g× B1u

The characters of the representation B2g× B3uare the following (Table C.6, the

first finger goes along B2g, the second – along B3u, etc.)

and are identical with the characters of B1u Hence, even without any calculation

of a(α), we have B2g× B3u= B1u Thus the transitions (for the polarization along

x) are allowed only for the states labelled by B1u because otherwise there is no

chance of obtaining a fully symmetric integrand Similarly, by multiplying B2gand

B2uwe obtain the following characters of B2g× B2u:

and these are identical to the characters of Au, therefore B2g× B2u= Au If the

polarization of light is along y, the only excitations (or deexcitations) possible are

for states belonging to Au Finally, for polarization along z, we find the characters

of B2g× B1u:

that turn out to be those of B3u This means that B2g× B1u= B3u and that the

transitions are possible only for states belonging to B3u

Example 14 Pyrazine monocation. As to the selection rules, nothing was said so

far about the pyrazine monocation We will be interested in excitations from the

electronic ground state (as in Table C.7) The pyrazine monocation corresponds to

symmetry group C2v(Table C.8)

The ground state belongs to the fully symmetric irreducible representation A1

Since (as before) we begin by excitations from the ground state, let us see which

irreducible representations arise from A1× B1 (for the x polarization of light,

see Table C.8, x transforms according to B1), A1× B2 (for the y polarization)

Table C.8. C2vgroup characters

C2v E C2 σ v (xz) σ v (yz)

and A1× A1(for the z polarization) We calculate the characters of A1× B1 by multiplying 1 by

and checking in Table C.8 that these correspond to B1(it has to be like this, because the characters of A1are all equal to 1), i.e A1× B1= B1 Similarly, even without immediate checking, we see that A1× B2= B2and A1× A1= A1 In this way the following allowed transitions from the ground state (A1) have been predicted: for polarization along x: A1→ B1

for polarization along y: A1→ B2

for polarization along z: A1→ A1 This agrees with fkl = 0 values of Table C.7

Now we are able to compare the spectrum for pyrazine and for its monocation, Table C.7 Attaching a proton to the pyrazine (creating its monocation) does not look like something that would ruin the UV-VIS spectrum We might expect that the frequencies of the bands, even their intensities should be somehow similar in both molecules As we can see from the Table, the frequencies are similar indeed For both molecules there are forbidden (fkl= 0) and allowed (fkl = 0) transitions Note that what is allowed for the pyrazine is also allowed for its cation, the light po-larization coincides, even the values of fklare similar (we have taken into account that the transition to B1u in pyrazine with frequency 49800 cm−1corresponds to the transition to A1in the monocation with frequency 49720 cm−1) In the mono-cation there are some additional transitions allowed: to B1and to B2 This is quite understandable, because the number of symmetry operations for the monocation

is smaller, and the higher molecular symmetry the more numerous are forbidden transitions If a molecule had no symmetry operations at all (except of course the identity symmetry), then all transitions would be allowed

Thus, practically with zero effort, we find the selection rules in UV-VIS for any molecule we want

Selection rules in IR and Raman spectra

The selection rules derived above pertain to electronic transitions, when the posi-tions of the nuclei are fixed in space Now a vibrational excitation of the molecule will be considered, while the electronic state is assumed to be unchanged The vi-brations of a molecule are related to its vibrational levels (each corresponding to

an irreducible representation) and the corresponding vibrational wave functions,

Fig C.6.Small amplitude harmonic vibrations of a molecule (N atoms) are described by 3N − 6

in-dependent harmonic oscillators (normal modes) Each normal mode is characterized by an irreducible

representation A diagram shows the vibrational energy levels of three normal modes corresponding to

the irreducible representations 1 2 3 The modes have different frequencies, hence the interlevel

separations are different for all of them (but equal for a given mode due to the harmonic potential).

On the right-hand side all these levels are shown together.

and the IR spectrum results from transitions between such levels Fig C.6 shows

the energy levels of three normal modes

In the harmonic approximation the problem of small amplitude vibrations

(Chapters 6 and 7) reduces to the 3N− 6 normal modes (N is the number of

atoms in the molecule) Each of the normal modes may be treated as an

indepen-dent harmonic oscillator A normal mode moves all the atoms with a certain

fre-quency about their equilibrium positions in a concerted motion (the same phase)

The relative deviations (i.e the ratios of the amplitudes) of the vibrating atoms

from equilibrium are characteristic for the mode, while the deviation itself is

ob-tained from them by multiplication by the corresponding normal mode coordinate

Q∈ (−∞ ∞) The value Q = 0 corresponds to the equilibrium positions of all

the atoms, Q and−Q correspond to two opposite deviations of any atom from its

equilibrium position

Each normal mode belongs to an irreducible representation of the symmetry

group of the molecule What does it really mean? In any mode the displacements

of the equivalent atoms from equilibrium have the same absolute value, although

they may differ in sign

We assume that small atomic deviations satisfy the symmetry requirements

of the symmetry group of the molecule (valid for all atoms in equilibrium po-sitions) and transform according to the irreducible representation, to which the normal mode belongs Squaring the deviations destroys information about their signs, i.e the absolute values of the deviations of the equivalent atoms are the same This means that the squares of deviations transform according to the fully symmetric representation of the group

To establish the vibrational selection rules, let us first define the vibrational states of 3N− 6 harmonic oscillators (normal modes) The ground state of the

system is no doubt the state in which every normal mode i is in its ground state,

ψi 0 The ground-state wave function of the i-th normal mode reads as (p 166)

ψi 0= N0exp

−aiQ2i

where ai> 0 is a constant, and Qiis the normal mode coordinate Whatever this

normal mode is, the wave function contains the square of Qi, i.e the sign of the

deviations of the equivalent atoms is irrelevant.

The squares of the deviations, and therefore function ψi 0itself, transform independently of i

Let us denote this fully symmetric irreducible representation by A1 The

wave-function of the first excited state of a normal mode has the form (p 166)

ψi 1= N1Qiexp

−aiQ2i

(C.30) and we see that ψi 1transforms exactly as the coordinate Qidoes, i.e according to the irreducible representation to which the normal mode belongs (because Q2i in the exponent and therefore the exponent itself both belong to the fully symmetric representation) In the harmonic approximation the total vibrational wavefunction

of the system of 3N− 6 normal (i.e independent) oscillators can be written as:

ψosc0 = ψ1 0ψ2 0ψ3 0   ψ3N−6 0 (C.31)

the zeros in the indices mean that all the modes are in their ground states This means that ψosc0 transforms according to the representation being the direct prod-uct A1× A1× A1× · · · × A1= A1(a banality, all the characters of A1are equal 1) Now let us focus on the excited states of the 3N− 6 vibrational modes The excited states may be quite complex, but the most important (and the simplest) are those with all the normal modes in their ground states, except a single mode that is in its first excited state A transition from the many-oscillator ground state to such an

excited state is called a fundamental transition The intensities of the fundamental

fundamental

transition transitions are by at least one order of magnitude larger than the others This is

why we will focus on the selection rules for such transitions Let us take one such

singly excited state (with the first mode excited):

ψosc1 = ψ1 1ψ2 0ψ3 0   ψ3N−6 0 (C.32)

The function ψ1 1 corresponding to the first excited state transforms according

to the irreducible representation , to which the normal mode 1 belongs Thus,

ψosc1 transforms according to × A1× A1× A1× · · · × A1= , i.e it belongs

to the same irreducible representation as ψ1 1 Of course, if the only excited mode

were the i-th one, then the many-oscillator wavefunction would belong to the same

irreducible representation as the wavefunction of the particular oscillator does We

will need this result later

IR selection rules. Let us consider a molecule with a fixed position in a Cartesian

coordinate system To excite the molecule, IR light (because the separation of the

vibrational levels corresponds to the infrared region) is used, which is polarized

along the x axis Theory of electromagnetism says the transition integral42decides

the intensity of the absorption



where ˆμx stands for the dipole moment component x The selection rules

estab-lish which integrals of this kind will be zero for symmetry reasons To this end we

need information about the irreducible representations to which ψosc0 ˆμx ψosc1

belong.43 Since ψosc0 transforms according to A1, for the integral to survive, the

function ψosc1 has to belong to the same irreducible representation as ˆμx (and

therefore x itself) We showed above that ψosc1 belongs to the same irreducible

representation to which the normal mode 1 belongs In other words, the rule is:

SELECTION RULE IN IR

the transition from the ground state is allowed for those normal modes that

transform as x, where x is the direction of light polarization, and similarly

for light polarization along y and z

Raman selection rules. The physics of Raman spectra44is different: rather than

direct absorption this is light scattering (in the UV-VIS region) on molecules It

turns out, that beside the light the source is emitting, we also detect quanta of

energy lower or higher by hν, where ν is the vibrational frequency of the molecule

42 The integration goes over the coordinates of the nuclei.

43 We are going to analyze the direct product of these three representations If it contains the fully

symmetric representation, the integral is not zero.

44 Chandrasekhar Venkata Raman (1888–1970), Indian physicist, professor at the University of

Cal-cutta and at the Indian Scientific Institute in Bangalore In 1928 Raman discovered light scattering that

is accompanied by a change of frequency (by frequency of the molecular vibrations) Raman received

the Nobel prize in 1930 “for his work on the scattering of light and for the discovery of the effect named

after him”.

For the Raman scattering to be non-zero, at least one of the following integrals should be non-zero



where ˆαqq  with q q= x y z is a component of the polarizability tensor which transforms as one of the following (cf eq (12.40), p 636): qq= x2, y2, z2, xy,

xz, yz or their linear combinations (this information is available in the tables of characters) Identical reasoning leads to the conclusion that

the normal mode excited in a fundamental transition has to belong to the same irreducible representation as the product qq

It remains to be seen to which irreducible representations the normal modes be-long The procedure consists of two stages.

Stage 1 First, the global Cartesian coordinate system is chosen, Fig C.7 In

this system we draw the equilibrium configuration of the molecule, with numbered atoms A local Cartesian coordinate system is located on each atom with axes par-allel to the axes of the global coordinate system For each atom, we draw the arrows

of its displacements along x, y and z, oriented towards the positive values (alto-gether 3N displacements), assuming that the displacements of equivalent atoms have to be the same When symmetry operations are applied, these displacements transform into themselves45and therefore form a basis set of a (reducible) repre-sentation  of the symmetry group of the molecule (in its equilibrium position) This representation will be decomposed into the irreducible representations

Stage 2 The reducible representation describes the genuine (internal)

vibra-tions as well as the six apparent vibravibra-tions (three translavibra-tions and three rotavibra-tions) The apparent vibrations correspond to those irreducible representations that are associated to x, y, z (translations) and Rx, Ry, Rz (rotations) We know from the corresponding table of characters what the later ones are Summing up: the re-ducible representation mentioned above has to be decomposed into irrere-ducible representations The decomposition yields = a(1)1+ a(2)2+ a(3)3   From this decomposition we have to subtract (in order to eliminate the appar-ent vibrations) all the irreducible represappar-entations the x y, z, Rx Ry and Rz be-long

After these two stages we are left with the number of the irreducible represen-tations which pertain to the genuine vibrations.46Only after this can we establish the vibrational selection rules according to the procedure used before All this will

be shown by a simple example of the carbonate anion CO2−3 that in its equilibrium configuration corresponds to the D3hsymmetry group, Fig C.7

45 For example, a displacement of an atom along x under a symmetry operation turns out to be a displacement of another equivalent atom.

46 Rather internal motions Note that some of these genuine vibrations may correspond to rotations of the functional groups in the molecule.

Fig C.7.The carbonate anion CO 2−

3 , the coordinate system used and the versors

de-scribing the displacements of the atoms.

Example 15 IR and Raman spectra of the carbonate anion. To decompose a

re-ducible representation into irrere-ducible ones, we do not need the rere-ducible

rep-resentation be given in full details It is sufficient to know its characters (p 920)

These characters are easy to deduce by considering what happens to the

displace-ment vectors along xi, yi, zi(for atom i) under all the symmetry operations What

will greatly simplify our task is that only the diagonal elements of the matrices of

the reducible representation contribute to the characters How it looks in practice

is shown in Table C.9

Thus, the characters of the reducible representation have been found To

de-compose the representation, we have to know the table of characters for the D3h

symmetry group, Table C.10

Let us write (in the same order as in Table C.10) the characters of the reducible

representation just found:

Now, let us find (p 920) how many times[a(α)] the irreducible

representa-tion α is present in the reducible representarepresenta-tion (the sum over classes: number

of operations in class× the calculated character × the character of irreducible

representation):

a

A

12[1 × 12 × 1 + 2 × 0 × 1 + 3 × (−2) × 1 + 1 × 4 × 1 + 2 × (−2) × 1

+ 3 × 2 × 1] = 1

Similarly, we find (only needing to know how to multiply such numbers as 1 2 3)

that

a(A)= 1 aE

= 3 aA1

= 0 aA2

= 2 a(E) = 1

This means that the reducible representation in question decomposes into

= A

1+ A

2+ 3E+ 2A

Table C.9.

Class The character of the corresponding matrix

E χ(E) = 12

Justification: each versor transforms into itself Hence each diagonal element is equal to 1,

and the number of them is equal to 3 times the number of atoms = 12 2C3 χ(C3) = 0

cos 120 ◦(from x

4 ) + cos 120◦(from y4) = 0 3C2 χ(C2) = −2

Justification: it is sufficient to consider only one of the operations of the class – others will

have the same character Let us take the rotation about the C2axis going through O1and C Then the only unit vectors that transform into themselves (eventually changing sign – then the contribution to the character is −1) are those related to O 1 and C We have χ(C2) =

−1(from z 4 ) + (−1)(from z 1 ) − 1(from x 1 ) − 1(from x 4 ) + 1(from y 1 ) + 1(from y 4 ) = −2

σh χ(σh) = 4

Justification: the contribution from each atom will be the same, i.e χ will be equal to

4 times the contribution from a single atom, the latter equals: −1(from z) + 1(from x) + 1(from y) = 1

2S3 χ(S3) = −2

2 (from x4) −

1

2 (from y4) = −2 3σ v χ(σ v ) = 2

Justification: Let us take only a single operation from the class, the one, which represents the

reflection in the plane going through O1and C4 Then the contributions to χ are the same for both atoms, and one gives: −1(from x) + 1(from z) + 1(from y) = 1.

Table C.10. Characters of the irreducible representations of symmetry group D3h

A 

A 

A 

A 

From the table of characters, we see that the apparent vibrations (see the irre-ducible representations corresponding to x y z Rx Ry Rz) belong to A2, E,

A2, E After subtracting them from , we obtain the irreducible representations that correspond to the genuine vibrations:

A1 A2 2E i.e one vibration of symmetry A1(and a certain frequency ν1), two vibrations (each doubly degenerate) of symmetry E (they differ by frequency ν3 = ν4) and one vibration of A2symmetry (corresponding to frequency ν2)

SELECTION RULES FOR IR:

Therefore, we expect the following selection rules for the fundamental

tran-sitions in the IR spectrum for the CO2−3 anion:

1 x and y belong to representation E, and therefore frequencies ν3and ν4

are active in IR;

2 z belongs to representation A2, and therefore frequency ν2 is active in

IR

SELECTION RULES FOR RAMAN SPECTRA

For the Raman spectra we expect the following selection rules The

vibra-tions with the frequency will be active:

1 ν1, because x2+ y2and z2belong to A1;

2 ν3and ν4, because x2− y2and xy belong to E,

while the vibration of the frequency ν2 will be inactive in Raman spectroscopy,

because none of the polarizability components (symbolized by x2 y2, etc.) belongs

to A2

The results are collected in Table C.11 (sign “+” = active vibration, sign “–”

= inactive vibration, the polarization of the light is shown in parentheses)

As seen from Table C.11, in case of the carbonate anion the vibration ν1is

inac-tive in IR, but acinac-tive in Raman spectroscopy, while the opposite is true for ν2 The

vibrations with frequencies ν3and ν4are active both in IR and in Raman spectra

EXCLUSION RULE

If the molecule under study has a centre of symmetry, the exclusion rule

is valid, i.e the vibrations that are active in IR are inactive in the Raman

spectrum, and vice versa.

This follows from the fact that, in this case, x y z belong to different irreducible

representations than x2 y2 z2 xy xz yz Indeed, the x y z are

antisymmet-ric with respect to the inversion operation, whereas x2 y2 z2 xy xz yz or

their combinations are symmetric with respect to inversion This guarantees that

they belong to different irreducible representations, therefore for a molecule with

Table C.11. Transitions in CO 2−

3 active ( +) and inactive (−)

in IR and in Raman spectra Representation ν IR (polarization) Raman

A 

A