molecularsymmetry raman ir

8/18/2019 MolecularSymmetry Raman Ir

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Symmetry helps us understand molecular structure, some chemical

properties, and characteristics of physical properties (spectroscopy) – used with group theory to predict vibrational spectra for theidentification of molecular shape, and as a tool for understandingelectronic structure and bonding.

Molecular Symmetry

Symmetrical : implies the species possesses anumber of indistinguishable configurations.


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Element Operation Symbol

Identity Identity E

Symmetry plane Reflection in the plane σ

Inversion center Inversion of a point x,y,z to -x,-y,-z i

Proper axis Rotation by (360/n)° C n

Improper axis 1. Rotation by (360/n)°2. Reflection in plane perpendicular to rotation axis S n

Group Theory : mathematical treatment of symmetry.

symmetry operation – an operation performed on an objectwhich leaves it in a configuration that is indistinguishablefrom, and superimposable on, the original configuration.

symmetry elements – the points, lines, or planes to which asymmetry operation is carried out.

Proper a es of rotation !" n#

$otation with respect to a line !a is of rotation#.%"n is a rotation of !&'()n#*.

•C2 = 180°rotation, C 3 = 120°rotation, C 4 = 90°rotation, C 5 = 72°rotation, C 6 = 60°rotation…

%+ach rotation brings you to an indistinguishable state from the original.

owever, rotation by -(* aboutthe same a is does not give bacthe identical molecule.

Therefore / 0 does 10T possessa " 2 symmetry a is.

3e4 2 is s5uare planar.

6t has four different " / a es.

7 " 2 a is out of the page iscalled the principle axisbecause it has the largest n .

8y convention, the principlea is is in the z 9direction


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If reflection of all parts of a molecule through a plane

produced an indistinguishable configuration, the symmetryelement is called a mirror plane or plane of symmetry .σ v(vertical): plane colinear with principal axis

$eflection through a planes of symmetry !mirror plane#

σ d(dihedral) Vertical, parallel to principal axis , σ parallel to C nand bisecting two C 2' axes



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σ h(horizontal): plane perpendicular to principal axis


Inversion, Center of Inversion (i)

7 center of symmetry: 7 point at the center of the molecule. ! ,y, # !9 ,9y,9 #.

6t is not necessary to have an atom in the center ! e.g. ben ene#.

Tetrahedrons, triangles, and pentagons don;t have a center of inversion symmetry.

$u!"0# '" / ' " 2 2"l / 4 /


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$otation9reflection, 6mproper rotation!S n#

This is a compound operation combining a rotation 360 ° /n (C n) with areflection through a plane perpendicular to the C n axis σ h.(C n followed by σ h)

σ Cn=S n

An improper rotation (or rotation–reflection), S n , involves rotation about 360 ° /nfollowed by reflection through a plane that is perpendicular to the rotation axis.


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6dentity !+#

Simplest symmetry operation. 7ll molecules have this element. 6f themolecule does have no other elements, it is asymmetric.

The identity operation amounts to doing nothing to a molecule andso leaves any molecule completely unchanged.

" 4"l8r S04"l

Successive 0perations


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Symmetry Point Groups

%Symmetry of a molecule located on symmetry a es, cut byplanes of symmetry, or centered at an inversion center is

nown as point symmetry .

%"ollections of symmetry operations constitute mathematicalgroups .

%+ach symmetry point group has a particular designation.

Cn , C nh , C nv Dn , Dnh , D nd S 2n C∞ v , D∞ hIh , I Td , Th ,T Oh ,O C1,C i , C s


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Tetrahedral Geometry

P ! Cl

0ctahedral Geometry 6cosahedral Geometry

"#(C$) %& "! '2 H '2 &2−−−−


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Identify the symmetry elements that are present in benzene.

Scheme for assigning point groups of molecules and molecular ions


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" n Point Groups

PBrClF C1 H2O2 C2

C3 M(NH2CH 2CO 2)4 C4

E C nn =

As(C 6H5)3

" nh Point Groups

The direction of the C n axis is take as vertical, so a symmetryplane perpendicular to it is a horizontal plane, σ h.

1 / 40"l88r"l4

The point group is called C s (C1h)

1 / 4 /

C2h

8!0 # &

C3h


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" nv Point Groups

H2O

C2v

If a mirror plane contains the rotational axis, the group is called a C nv group.

SF 4

C2v

NF 3C3v

CHCl 3C3v

SF 5Cl

C4v

= n and = nh Point Groups

Adding a C 2 axis perpendicular to a C n axis generates one of the dihedral groups.

= &

There must be n C2 axesperpendicular toCn

Adding a σ h to a D n group generates a D nh group.

" / 2

= /h

>Pt"l 2?/9

= 2h

= /n le *et+een rin s not - or -


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84 &

D3h

= nd Point Groups

Adding a vertical mirror plane to a D n group in such a fashionto bisect adjacent C 2 axes generates a D nd group.

FerroceneFe(C 5H5)2

D5d

D2d


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1,3,5,7-tetrafluorocyclooctatetraene

S n groups

For odd n , (S n)n = σ h

For even n , (S n)n = E

The S n for odd n is the same as the C nh .

Absence of mirror planes distinguish S ngroups from D nd groups.


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The letter ‘A’.

GeH 3F

AsBr 5 (trigonal bipyramid)

trans rotamer of Si 2H6

Crown-shaped S 8

tetrachloroplatinate(II)

HOCl

O

H

H

B

O

O

H

B(OH) 3

H

B

H

H

BH3

Ni(en) 3

SF 6

SeH 3F


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"haracteristic symmetry elements ofsome important classes of point groups.


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"haracter Tables

%"haracter tables contain, in a highly symbolic form, informationabout how something of interest !an bond, an orbital, etc.# is affectedby the operations of a given point group .

%+ach point group has a uni5ue character table, which is organi edinto a matri .

%"olumn headings are the symmetry operations , which are groupedinto classes .

% ori ontal rows are called irreducible representations of the pointgroup.

%The main body consists of characters !numbers#, and a section onthe right side of the table provides information about vectors andatomic orbitals .

C 2v E C 2 σ v(xz) σ v’(yz)

A1 1 1 1 1 z x2, y2, z2A2 1 1 -1 –1 R z xy

B1 1 -1 1 –1 x, R y xz

B2 1 –1 –1 1 y, R x yz

C 3v E 2C 3 3σ vA1 1 1 1 z x2 + y2, z2

A2 1 1 -1 R zE 2 -1 0 (x,y), (R x , R y) (x2-y2 , xy)(xz, yz)

%Symmetry elements possessed by the point group are in the top row

%


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Selection $ules: 6nfrared and $aman Spectroscopy6nfrared energy is absorbed for certain changes in vibrationalenergy levels of a molecule.

9for a vibration to be infrared active , there must be a change inthe molecular dipole moment vector associated with the vibration.

4or a vibration mode to be $aman active , there must be a changein the net polari ability tensor

9polari ability is the ease in which the electron cloud associatedwith the molecule is distorted

4or centrosymmetric molecules, the rule of mutual e clusionstates that vibrations that are 6$ active are $aman inactive, andvice versa

The transition from the vibrational ground state to the firste cited state is the fundamental transition .

The two bending modes re5uire the same amountof energy and are therefore degenerate.


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A1 1 1 1 1 z x2, y2, z2

A2 1 1 -1 –1 R z xy

B1 1 -1 1 –1 x, R y xz

B2 1 –1 –1 1 y, R x yz

6f the symmetry label !e.g. 7 B, 8 B, +# of a normal mode ofvibration is associated with x , y , or z in the character table, thenthe mode is 6$ active .

6f the symmetry label !e.g. 7 B, 8 B, +# of a normal mode ofvibration is associated with a product term ! x 2 , xy # in thecharacter table, then the mode is $aman active .


A1 1 1 1 1 z x2, y2, z2

A2 1 1 -1 –1 R z xyB1 1 -1 1 –1 x, R y xz

B2 1 –1 –1 1 y, R x yz


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D 3h E 2C 3 3C 2 Ξh 2S 3 3Ξ v1 1 1 1 1 1

x2 +

y2,

z2

1 1 -1 1 1 -1 R z2 -1 0 2 -1 0 ( x , y ) ( x 2 – y 2, 2 xy )1 1 1 -1 -1 -11 1 -1 -1 -1 1 z2 -1 0 -2 1 0 ( R x , R y ) ( xy , yz )

1A ′

2A ′

E ′

1A ′′

2A ′′

E ′′

σ h 3σ v


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The vibrational modes of " 2 ! T d#, only two of which are 6$ active.

CCl http/00fy.chalmers.se0$1 3S45S0*ro6in07olecular7otions0CCl spectra.html


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vibrational modes of >Pt"l 2 ?/ − ! D2h #


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Point groups of octahedral metal carbonyl comple es


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+nantiometers

7 pair of enantiomers consists of two molecular species which aremirror images of each other and are nonsuperposable.


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Group 8heory

Supplemental 7aterial

E

Cp to this point, we haveconsidered symmetry operationsonly insofar as they affect atomsoccupying points in molecules.

"onsider a water molecule ! / 0#.

"haracter Table =evelopment

y

" /

σ vA ! yz #

σv

! xz #

"oordinates are assigned according to the convention that the highestfold a is of rotation is aligned with the z 9a is, and the x a is isperpendicular to the plane of the molecule.


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The set of four labels !DB, 9B, 9B, DB# generated in the analysisconstitutes one irreducible representation within the " /v pointgroup. 6t is irreducible in the sense that it cannot be decomposed

into a simpler or more fundamental form.%1ot only does it describe the effects on the y translation but alsoon other F y 9vector functionsA such as a p y orbital.

%Therefore, y is understood to serve as a basis function for thisirreducible representation within the " /v point group.

E y

0

0

C 2

0

0

σ v (xz)

0

0

σ v (yz)

0

0

A / E DB DB 9B9B

+ffect of a symmetry rotation about the z 9a is.

E y

0

0

C 2

0

0

σ v (xz)

0

0

σ v (yz)

0

0

BB E DB 9B 9BDB

Translation of the molecule in the D x direction

E y

0

0

C 2

0

0

σ v (xz)

0

0

σ v (yz)

0

0

A B EDB DB DBDB

Translation of the molecule in the D z direction


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C 2v E C 2 σ v(xz) σ v(yz)

A1 1 1 1 1 z x2, y2, z2

A2 1 1 -1 –1 R z xy

B1 1 -1 1 –1 x, R y xz

B2 1 –1 –1 1 y, R x yz

The resulting character table for " /v is shown above.

The column heading are classes of symmetry operations for thegroup, and each row depicts one irreducible representation .

The DB and 9B numbers, which correspond to symmetric andantisymmetric behavior, are called characters .

"olumns on the right are some of the basis functions which havethe symmetry properties of a given irreducible representation.

x , y , z stand for rotations about the specified a es.

Symbols in the column on the far left are @ulli en


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∑= x

x xi

xr nh

N χ χ 1

$educible $epresentationsThe reduction can be achieved using the reduction formula. 6t is amathematical way of reducing that will always wor when the answercannot be spotted by eye. 6t is particularly useful when there arelarge numbers of bonds involved.

The vibrational modes of the molecule are reduced to produce areducible representation into the irreducible representations. Thismethod uses the following formula reduction formula:

! is the number of times a symmetry species occurs in the reduciblerepresentation,h is the Forder of the groupA: simply the total number of symmetry operations inthe group.

The summation is over all of the symmetry operations. 4or each symmetryoperation, three numbers are multiplied together. These are: " r is the character for a particular class of operation in the irreduciblerepresentation

" i is the character of the irreducible representation.n is the number of symmetry operations in the class

The characters of the reducible representation can be determinedby considering the combined effect of each symmetry operation onthe atomic vectors .

7tomic contributions, by symmetry operations,to the reducible representation for the & ! degrees of freedom for a molecule.

Operation Contributionper atom*

E 3

C 2 -1

C 3 0

C 4 1

C 6 2σ 1i -3

S 3 -2

S 4 -1

S 6 0

HC n E B D /cos!&'() n #I S n E 9B D /cos!&'() n #

y y

σ ! #

yy

+

y y

i


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a b

0

=erivation of reducible representation for degrees of freedom in / 0

+

a b

0

Cnshifted atoms

&

a b

0

a b

0

a b

0

" /

b a

0

B

σ ! #

b a

0

a b

0

y

B

σ !y #&

0btain the reducible representation !for / 0#by multiplying the number of unshifted atomstimes the contribution per atom.

E C 2 σ v(xz) σ v(yz)Unshifted Atoms 3 1 1 3

Contribution per atom 3 -1 1 1Γ tot 9 -1 1 3


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∑= x

x xi

xr nh

N χ χ 1

$educible $epresentationsThe reduction can be achieved using the reduction formula . 6t is amathematical way of reducing that will always wor when the answer

cannot be spotted by inspection. 6t is particularly useful when thereare large numbers of atoms and bonds involved.

The vibrational modes of the molecule are reduced to produce areducible representation into the irreducible representations. Thismethod uses the following formula reduction formula :

! is the number of times a symmetry species occurs in the reduciblerepresentation,h is the Forder of the groupA: simply the total number of symmetry operations inthe group.

The summation is over all of the symmetry operations. 4or each symmetryoperation, three numbers are multiplied together. These are: " r is the character for a particular class of operation in the reduciblerepresentation

" i is the character of the irreducible representation .n is the number of symmetry operations in the class

∑= x

x xi

xr nh

N χ χ 1

Tabulate our nown information.

E C 2 σ v(xz) σ v(yz)Γ r 9 -1 1 3

7 B: !B)h#>! χ r Ε #! χ i Ε #!n Ε # D ! χ r C 2 #! χ i C 2 #!n C 2 # D ! χ rσ v! ##! χ i

σ v! ##!n σ v! ##

D ! χ rσ v!y #

#! χ i σ v!y #

#!nσ v!y #

#

B#

/#C 2v E C 2 σ v(xz) σ v(yz)

A1 1 1 1 1 z x2, y2, z2

A2 1 1 – 1 –1 R z xy

B1 1 -1 1 –1 x, R y xz

B2 1 –1 –1 1 y, R x yz

$educible $epresentation !for / 0#

"haracter Table

7 B: !B)2#>! - #! χ i Ε #!n Ε # D ! 9B#! χ i C 2 #!n C 2 # D ! B#! χ i σ v! ##!n σ v! ## D

! ! χ i σ v!y # #!n σ v!y # #

h E 2

7 B: !B)2#>!-#! B#!n Ε # D !9B#! B#!nC 2 # D !B#! B#!n σ v! # # D !! B#!n σ v!y # #

7 B: !B)2#>!-#!B#! B# D !9B#!B#! B# D !B#!B#! B# D !!B#! B# E 9

irrep


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"alculate irreducible representation 7 /

7 / : !B)h#>! χ r Ε #! χ i Ε #!n Ε # D ! χ r C 2 #! χ i C 2 #!n C 2 # D ! χ rσ v! # #! χ i

σ v! # #!n σ v! # #

D ! χ rσ v!y #

#! χ i σ v!y #

#!nσ v!y #

#7 / : !B)2#>!-#!B#!B# D !9B#!B#!B# D !B#!9B#!B# D !!9B#!B# E '

"alculate irreducible representation 8 B

8 B: !B)2#>!-#!B#!B# D !9B#!9B#!B# D !B#!B#!B# D !!9B#!B# E 2

"alculate irreducible representation 8 /

8 / : !B)2#>!-#!B#!B# D !9B#!9B#!B# D !B#!9B#!B# D !!B#!B# E 9

E C 2 σ v(xz) σ v(yz)

Γ r 9 -1 1 3

The reducible representationJ

Jis resolved into three 7 B, one 7 / , two 8 B, and three 8 / species.

C 2v E C 2 σ v(xz) σ v(yz)

A1 1 1 1 1 z x2, y2, z2

A2 1 1 – 1 –1 R z xy

B1 1 -1 1 –1 x, R y xz

B2 1 –1 –1 1 y, R x yz

Γ tot E & A B D A / D / BB D &B/9>Γ trans E A B D BB D B/ ?

9>Γ rot E A / D BB D B/ ?

Γ vib E / A B D 8 /

1otice this is the same result weobtained by analy ing the symmetriesof the vibrational modes.

9each mode is 6$ active


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=erive Γ tot for 8"l & given the character table for = &h=erive the number of vibrational modes and assign modes for 8"l &.

E 2C 3 3C 2 σ h 2S 3 3σ vCnshifted atoms 4 1 2 4 1 2

"ontribution per atom 3 0 -1 1 -2 1Γ tot 12 0 -2 4 -2 2

D 3h E 2C 3 3C 2 Ξh 2S 3 3Ξ v1 1 1 1 1 1 x 2 + y 2, z 2

1 1 -1 1 1 -1 R z2 -1 0 2 -1 0 ( x , y ) ( x 2 – y 2, 2 xy )1 1 1 -1 -1 -11 1 -1 -1 -1 1 z2 -1 0 -2 1 0 ( R x , R y ) ( xy , yz )

1A ′

2A ′

E ′

1A ′′

2A ′′

E ′′

σ h 3σ v

$esults of using the reduction formula.

Χ rx*Χ i

x*n x 1/h Sum Total

12 0 -6 4 -4 6 0.083333 12 1

12 0 6 4 -4 -6 0.083333 12 1

24 0 0 8 4 0 0.083333 36 3

12 0 -6 -4 4 -6 0.083333 0 0

12 0 6 -4 4 6 0.083333 24 2

24 0 0 -8 -4 0 0.083333 12 1

1A ′

2A ′

E ′

1A ′′

2A ′′

E ′′

Therefore, we have determinedΓ tot E D D& D/ D1A ′ 2A ′ E ′ 2A ′′ E ′′

but, subtract off the translational representations.

9>Γ trans E D ?E ′ 2A ′′

and subtract off the rotational representations.

9>Γ rot E D ?2A ′ E ′′

Γ vib E D/ D ?1A ′ E ′ 2A ′′

+ach +Arepresentationdescribes twovibrationalmodes of e5ualenergy.


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Symmetricalstretching.

0ut9of9planebending mode.

Cnsymmetricalstretching.

6n9planebending mode.

$aman active. 6$ active. $aman and 6$ active.

e can use isotopic substitution to interpret spectra, since thecharacteristic fre5uency of the mode will depend on the masses ofthe atoms #oving in that #ode .

5evie+/ #hat 6o I 6o +hen I nee6 to;<

7ssign symmetry labels to vibrational modesK%6f the vibrational mode is nown and illustrated, s etch the resultingvibrational mode before and after each symmetry operation of thepoint group. Csing the character table, assign the symmetry label andidentify if the mode is 6$ and)or $aman active.=etermine the symmetries of all vibrational modes and if the modesare 6$ and)or $aman activeK%=etermine how many atoms are left unchanged by each symmetryoperation. 4ind the reducible representation and reduce into theirreps. Subtract translational and rotational modesJ 6dentify whichmodes are 6$ and)or $aman active.J=etermine the symmetries of only the stretching modes and if themodes are 6$ and)or $aman activeK%=etermine how many bonds are left unchanged by each symmetryoperation. 4ind the reducible representation and reduce into theirreps. 6dentify which are 6$ and)or $aman active.J=evelop a character tableK%=etermine the effect of each symmetry operation on the , y, translation and the rotation $ , $ y, and $ . The resulting set ofcharacters correspond to an irrep in the character table.


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B. 1 &/. " 2

&. >Pt"l 2?/9

2. S4 '

L. S4 L"l

=etermine number of "0 stretching modes in

B. @n!"0# '/. @n!"0# L"l

$. trans 9@n!"0# 2"l /

%. cis 9@n!"0# 2"l /&. 'ac 9@n!"0# &"l &

. #er 9@n!"0# &"l &

=etermine the number of and assign the vibrational modes of the following:ow many pea s in the !B# 6$ spectra and !/# $aman spectra

molecularsymmetry raman ir

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