Symmetry - Louisiana Tech Universityupali/chem481/slides/GHW3-chem481-chapter-6.pdf · Molecular symmetry An introduction to symmetry analysis 6.1 Symmetry operations, elements and

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Chapter 6-1Chemistry 481, Spring 2017, LA Tech

Instructor: Dr. Upali Siriwardane

e-mail: [email protected]: CTH 311 Phone 257-4941

Office Hours:

M,W 8:00-9:00 & 11:00-12:00 am;

Tu,Th, F 9:30 - 11:30 a.m.

April 4 , 2017: Test 1 (Chapters 1, 2, 3, 4)

April 27, 2017: Test 2 (Chapters (6 & 7)

May 16, 2016: Test 3 (Chapters. 19 & 20)

May 17, Make Up: Comprehensive covering all Chapters

Chemistry 481(01) Spring 2017


Chapter 6. Molecular symmetryAn introduction to symmetry analysis

6.1 Symmetry operations, elements and point groups 179

6.2 Character tables 183

Applications of symmetry

6.3 Polar molecules 186

6.4 Chiral molecules 187

6.5 Molecular vibrations 188

The symmetries of molecular orbitals

6.6 Symmetry-adapted linear combinations 191

6.7 The construction of molecular orbitals 192

6.8 The vibrational analogy 194

Representations

6.9 The reduction of a representation 194

6.10 Projection operators 196


SymmetryM.C. Escher


Symmetry Butterflies


Fish and Boats Symmetry


Symmetry elements and operations

• A symmetry operation is the process of

doing something to a shape or an object so

that the result is indistinguishable from the

initial state

• Identity (E)

• Proper rotation axis of order n (Cn)

• Plane of symmetry (s)

• Improper axis (rotation + reflection) of order

n (Sn), an inversion center is S2

2


2) What is a symmetry operation?


E - the identity element

The symmetry operation corresponds to

doing nothing to the molecule. The E

element is possessed by all molecules,

regardless of their shape.

C1 is the most common element leading to E,

but other combination of symmetry

operation are also possible for E.


Cn - a proper rotation axis of order n

• The symmetry operation Cn corresponds to

rotation about an axis by (360/n)o.

• H2O possesses a C2 since rotation by 360/2o = 180o

about an axis bisecting the two bonds sends the

molecule into an

• indistinguishable form:


s - a plane of reflectionThe symmetry operation corresponds to reflection in

a plane. H2O possesses two reflection planes.

Labels: sh, sd and sv.


i - an inversion center

The symmetry operation corresponds to inversion

through the center. The coordinates (x,y,z) of every

atom are changed into (-x,-y,-z):


Sn - an improper axis of order nThe symmetry operation is rotation by (360/n)o and

then a reflection in a plane perpendicular to the

rotation axis.

operation is

the same as

an inversion

is S2 = i

a reflection

so S1 = s.

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2) What are four basic symmetry elements and

operations?


3) Draw and identify the symmetry elements in:

a) NH3:

b) H2O:

c) CO2:

d) CH4:

e) BF3:


Point Group

AssignmentThere is a systematic

way of naming most

point groups

C, S or D for

the principal

symmetry axis

A number for the

order of the

principal axis

(subscript) n.

A subscript h, d, or v

for symmetry planesChapter 6-16Chemistry 481, Spring 2017, LA Tech

4) Draw, identify symmetry elements and assign the

point group of following molecules:

a) NH2Cl:

b) SF4:

c) PCl5:

d) SF6:

e) Chloroform

f) 1,3,5-trichlorobenzene


Special Point Groups

Linear molecules have a C∞ axis - there are an

infinite number of rotations that will leave a linear

molecule unchanged

If there is also a plane of symmetry perpendicular to

the C∞ axis, the point group is D∞h

If there is no plane of symmetry, the point group is

C∞v

Tetrahedral molecules have a point group Td

Octahedral molecules have a point group Oh

Icosahedral molecules have a point group Ih


Point groups

It is convenient to classify molecules

with the same set of symmetry elements

by a label. This label summarizes the

symmetry properties. Molecules with the

same label belong to the same point

group.

For example, all square molecules

belong to the D4h point group

irrespective of their chemical formula.

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5) Determine the point group to which each of

the following belongs:

a) CCl4

b) Benzene

c) Pyridine

d) Fe(CO)5

e) Staggered and eclipsed ferrocene, (η5-C5H5)2Fe

f) Octahedral W(CO)6

g) fac- and mer-Ru(H2O)3Cl3Chapter 6-20Chemistry 481, Spring 2017, LA Tech

Character tablesSummarize a considerable amount of information

and contain almost all the data that is needed to

begin chemical applications of molecule.

C2v E C2 sv sv'

A1 1 1 1 1 z x2, y2, z2

A2 1 1 -1 -1 Rz xy

B1 1 -1 1 -1 x, Ry xz

B2 1 -1 -1 -1 y, Rx yz


Character Table Td


Information on Character TableThe order of the group is the total number of

symmetry elements and is given the symbol h. For

C2v h = 4.

First Column has labels for the irreducible

representations. A1 A2 B1 B2

The rows of numbers are the characters (1,-1)of the

irreducible representations.

px, py and pz orbitals are given by the symbols x, y

and z respectively

dz2, dx2-y2, dxy, dxz and dyz orbitals are given by the

symbols z2, x2-y2, xy, xz and yz respectively.


H2O molecule belongs to C2v point

group


Symmetry Classes

The symmetry classes for each point group and are

labeled in the character table

LabelSymmetry Class

A Singly-degenerate class, symmetric with respect to the principal axis

B Singly-degenerate class, asymmetric with respect to the principal axis

E Doubly-degenerate class

T Triply-degenerate class

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Molecular Polarity and Chirality

Polarity:

Only molecules belonging to the point groups Cn,

Cnv and Cs are polar. The dipole moment lies along

the symmetry axis for molecules belonging to the

point groups Cn and Cnv.

• Any of D groups, T, O and I groups will not be

polar


ChiralityOnly molecules

lacking a Sn axis

can be chiral.

This includes

mirror planes

and a center of

inversion as

S2=s , S1=i and Dn

groups.

Not Chiral: Dnh,

Dnd,Td and Oh.


Meso-Tartaric Acid


Optical Activity


Symmetry allowed combinations

• Find symmetry species spanned by a set of

orbitals

• Next find combinations of the atomic orbitals on

central atom which have these symmetries.

• Combining these are known as symmetry adapted

linear combinations (or SALCs).

• The characters show their behavior of the

combination under each of the symmetry

operations. several methods for finding the

combinations.


Example: Valence MOs of Water

• H2O has C2v symmetry.

• The symmetry operators of the C2v group all

commute with each other (each is in its own

class).

• Molecualr orbitals should have symmetry

operators E, C2, sv1, and sv2.

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Building a MO diagram for H2O

x

z

y


a1 orbital of H2O


b1 orbital of H2O


b1 orbital of H2O, cont.


b2 orbital of H2O


b2 orbital of H2O, cont.

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[Fe(CN)6]4-


Reducing the Representation

Use reduction formula


MO forML6 diagram Molecules


Group Theory and Vibrational

Spectroscopy

• When a molecule vibrates, the symmetry of the

molecule is either preserved (symmetric

vibrations) or broken (asymmetric vibrations).

• The manner in which the vibrations preserve or

break symmetry can be matched to one of the

symmetry classes of the point group of the

molecule.

• Linear molecules: 3N - 5 vibrations

• Non-linear molecules: 3N - 6 vibrations (N is the

number of atoms)

Chapter 6-41Chemistry 481, Spring 2017, LA Tech Chapter 6-42Chemistry 481, Spring 2017, LA Tech

Reducible Representations(3N) for

H2O: Normal Coordinate Method

• If we carry out the symmetry operations of C2v on this

set, we will obtain a transformation matrix for each

operation.

• E.g. C2 effects the following transformations:

• x1 -> -x2, y1 -> -y2, z1 -> z2 , x2 -> -x1, y2 -> -y1, z2 ->

z1, x3 -> -x3 , y3 -> -y3, z3 -> z3.

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Summary of Operations by a set of four

9 x 9 transformation matrices.


Use Reduction Formula


Example H2O, C2v


Use Reduction Formula:

R

pp )R()R(g

1a

to show that here we have:

G3N = 3A1 + A2 + 2B1 + 3B2

This was obtained using 3N cartesian coordinate vectors.

Using 3N (translation + rotation + vibration) vectors would

have given the same answer.

But we are only interested in the 3N-6 vibrations.

The irreducible representations for the rotation and

translation vectors are listed in the character tables,

e.g. for C2v,


GT = A1 + B1 + B2

GR = A2 + B1 + B2

i.e. GT+R = A1 + A2 + 2B1 + 2B2

But Gvib = G3N - GT+R

Therefore Gvib = 2A1 + B2

i.e. of the 3 (= 3N-6) vibrations for a molecule

like H2O, two have A1 and one has B2 symmetry


INTERNAL COORDINATE METHOD

We used one example of this earlier - when we used

the "bond vectors" to obtain a representation

corresponding to bond stretches.

We will give more examples of these, and also the other

main type of vibration - bending modes.

For stretches we use as internal coordinates changes

in bond length, for bends we use changes in bond angle.

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Deduce G3N for our triatomic molecule, H2O

in three lines:

E C2 sxz syz

unshifted atoms 3 1 1 3

/unshifted atom s 3 -1 1 3

\ G3N 9 -1 1 3

For more complicated molecules this shortened

method is essential!!

Having obtained G3N, we now must reduce it to find

which irreducible representations are present.


Example H2O, C2v


Use Reduction Formula:

R

pp )R()R(g

1a

to show that here we have:

G3N = 3A1 + A2 + 2B1 + 3B2

This was obtained using 3N cartesian coordinate vectors.

Using 3N (translation + rotation + vibration) vectors would

have given the same answer.

But we are only interested in the 3N-6 vibrations.

The irreducible representations for the rotation and

translation vectors are listed in the character tables,

e.g. for C2v,


GT = A1 + B1 + B2

GR = A2 + B1 + B2

i.e. GT+R = A1 + A2 + 2B1 + 2B2

But Gvib = G3N - GT+R

Therefore Gvib = 2A1 + B2

i.e. of the 3 (= 3N-6) vibrations for a molecule

like H2O, two have A1 and one has B2 symmetry


Further examples of the determination of Gvib, via G3N:

NH3 (C3v) N

HH

H

C3v E 2C3 3sv

12 0 2

\ G3N 12 0 2

Reduction formula G3N = 3A1 + A2 + 4E

GT+R (from character table) = A1 + A2 + 2E,

\ Gvib = 2A1 + 2E

(each E "mode" is in fact two vibrations (doubly degenerate)


CH4 (Td)

H

C

HH

H

Td E 8C3 3C2 6S4 6sd

15 0 -1 -1 3

\ G3N 15 0 -1 -1 3

Reduction formula G3N = A1 + E + T1 + 3T2

GT+R (from character table) = T1 + T2,\ Gvib = A1 + E + 2T2

(each E "mode" is in fact two vibrations (doubly degenerate),

and each T2 three vibrations (triply degenerate)

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XeF4 (D4h)Xe

F

F F

F

D4h E 2C4 C2 2C2' 2C2" i 2S4 sh 2sv 2sd

15 1 -1 -3 -1 -1 -1 5 3 1

\G3N 15 1 -1 -3 -1 -1 -1 5 3 1

Reduction formula G3N = A1g + A2g + B1g + B2g + Eg + 2A2u + B2u + 3Eu

GT+R (from character table) = A2g + Eg + A2u + Eu,

\ Gvib = A1g + B1g + B2g + A2u + B2u + 2Eu

For any molecule, we can always deduce the overall symmetry

of all the vibrational modes, from the G3N representation.

To be more specific we need now to use the

INTERNAL COORDINATE method.


INTERNAL COORDINATE METHOD

We used one example of this earlier - when we used

the "bond vectors" to obtain a representation

corresponding to bond stretches.

We will give more examples of these, and also the other

main type of vibration - bending modes.

For stretches we use as internal coordinates changes

in bond length, for bends we use changes in bond angle.


Let us return to the C2v molecule:

H

O

H

r1 r2

Use as bases for stretches:

Dr1, Dr2.

Use as basis for bend:

D

C2v E C2 sxz syz

Gstretch 2 0 0 2

Gbend 1 1 1 1

N.B. Transformation matrices for Gstretch :

E, syz: 1 0

0 1

C2, sxz : 0 1

1 0

i.e. only count UNSHIFTED VECTORS (each of these +1 to ).


Gbend is clearly irreducible, i.e. A1.

Gstretch reduces to A1 + B2

We can therefore see that the three vibrational

modes of H2O divide into two stretches (A1 + B2)

and one bend (A1).

We will see later how this information helps

in the vibrational assignment.


Other examples:

NH3 N

HH

Hr1

r2

r31 opposite to r1

2 opposite to r2

3 opposite to r3

Bases for stretches: Dr1, Dr2, Dr3.

Bases for bends: D1, D2, D3.

C3v E 2C3 3sGstretch 3 0 1

Gbend 3 0 1

Reduction formula Gstretch = A1 + E

Gbend = A1 + E

(Remember Gvib (above) = 2A1 + 2E)


CH4H

C

HH

H

r1

r2r3

r46 angles 1,.....6, where 1

lies between r1 and r2 etc.

Bases for stretches: Dr1, Dr2, Dr3, Dr4.

Bases for bends: D1, D2, D3, D4, D5, D6.

Td E 8C3 3C2 6S4 6sd

Gstretch 4 1 0 0 2

Gbend 6 0 2 0 2

Reduction formula Gstretch = A1 + T2

Gbend = A1 + E + T2

But G3N (above) = A1 + E + 2T2

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IR Absorptions

Infra-red absorption spectra arise when a molecular

vibration causes a change in the dipole moment of

the molecule. If the molecule has no permanent

dipole moment, the vibrational motion must create

one; if there is a permanent dipole moment, the

vibrational motion must change it.

Raman Absorptions

Deals with polarizability


Raman Spectroscopy• Named after discoverer, Indian physicist C.V.Raman (1927).

It is a light scattering process.

• Irradiate sample with visible light - nearly all is transmitted; of the rest, most scattered at unchanged energy (frequency) (Rayleigh scattering), but a little is scattered at changed frequency (Raman scattering). The light has induced vibrational transitions in molecules (ground excited state) - hence some energy taken from light,

• scattered at lower energy, i.e. at lower wavenumber. Raman scattering is weak - therefore need very powerful light source - always use lasers (monochromatic, plane polarised, very intense).

• Each Raman band has wavenumber:where n = Raman scattered wavenumber

n0 = wavenumber of incident radiation

nvib = a vibrational wavenumber of the molecule

(in general several of these)


Molecular Vibrations

• At room temperature almost all molecules are in

their lowest vibrational energy levels with

quantum number n = 0. For each normal mode, the

most probable vibrational transition is from this

level to the next highest level (n = 0 -> 1). The

strong IR or Raman bands resulting from these

transitiions are called fundamental bands. Other

transitions to higher excited states (n = 0 -> 2, for

instance) result in overtone bands. Overtone

bands are much weaker than fundamental bands.


If the symmetry label of a normal mode corresponds

to x, y, or z, then the fundamental transition for this

normal mode will be IR active.

If the symmetry label of a normal mode corresponds

to products of x, y, or z (such as x2 or yz) then the

fundamental transition for this normal mode will be

Raman active.


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Symmetry - Louisiana Tech Universityupali/chem481/slides/GHW3-chem481-chapter-6.pdf · Molecular symmetry An introduction to symmetry analysis 6.1 Symmetry operations, elements and