chemistry 481(01) spring 2014

Chapter 6-1Chemistry 481, Spring 2015, LA Tech

Instructor: Dr. Upali Siriwardane

e-mail: [email protected]

Office: 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 7 , 2015: Test 1 (Chapters 1, 2, 3)

April 30, 2015: Test 2 (Chapters 5, 6 & 7)

May 19, 2015: Test 3 (Chapters. 19 & 20)

May 19, Make Up: Comprehensive covering all Chapters

Chemistry 481(01) Spring 2015


Chapter 6. Molecular symmetry An introduction to symmetry analysis 6.1 Symmetry operations, elements and point groups 1796.2 Character tables 183Applications of symmetry 6.3 Polar molecules 1866.4 Chiral molecules 1876.5 Molecular vibrations 188The symmetries of molecular orbitals 6.6 Symmetry-adapted linear combinations 1916.7 The construction of molecular orbitals 1926.8 The vibrational analogy 194Representations 6.9 The reduction of a representation 1946.10 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) 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 centerThe 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.


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 GroupAssignment

There 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

planes


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 GroupsLinear molecules have a C∞ axis - there are an infinite number of rotations that will leave a linear molecule unchangedIf 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.


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)3Cl3


Character tables Summarize 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 respectivelydz2, 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 ClassesThe symmetry classes for each point group and are labeled in the character tableLabelSymmetry Class

A Singly-degenerate class, symmetric with respect to the principal axisB Singly-degenerate class, asymmetric with respect to the principal axisE Doubly-degenerate classT Triply-degenerate class


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 planesand 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.


Building a MO diagram for H2OO

H HO

H H

a1

a1

a1

a1

a1

b1b1

b2

b2

b2

b2

a1

x

z

y


a1 orbital of H2O

E(1a1) (+1)(1a1)C2(1a1) (+1)(1a1)sv1(1a1) (+1)(1a1)sv2(1a1) (+1)(1a1)

C2

sv1

sv2


b1 orbital of H2O

sv2

sv2(1b1) (+1)(1b1)E(1b1) (+1)(1b1)


b1 orbital of H2O, cont.

C2

sv1

C2(1b1) (-1)(1b1)sv1(1b1) (-1)(1b1)


b2 orbital of H2O

sv1

sv1(1b2) (+1)(1b2)E(1b2) (+1)(1b2)


b2 orbital of H2O, cont.

C2

C2(1b2) (-1)(1b2)sv2(1b2) (-1)(1b2)

sv2


[Fe(CN)6]4-


Reducing the RepresentationUse 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)


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.


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(g1a

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.


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(g1a

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)


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 00 1

C2, sxz :

0 11 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 3s

Gstretch 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

r4 6 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


IR AbsorptionsInfra-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.

chemistry 481(01) spring 2014

Documents

molecular symmetry

boats symmetry chapter

symmetry operations

symmetry analysis

applications of symmetry

principal symmetry axis

chapters chemistry

basic symmetry elements