Inorganic Chapter4

7/22/2019 Inorganic Chapter4

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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 the

identification of molecular shape, and as a tool for understanding

electronic structure and bonding.

Chapter 4 - Molecular Symmetry

Symmetrical: implies the species possesses anumber of indistinguishable configurations.


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Sn1. Rotation by (360/n)

2. Reflection in plane perpendicular to rotation axisImproper axis

CnRotation by (360/n)Proper axis

iInversion of a point x,y,z to -x,-y,-zInversion center

Reflection in the planeSymmetry plane

EIdentityIdentity

SymbolOperationElement

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 axes of rotation (Cn)

Rotation with respect to a line (axis of rotation).Cn is a rotation of (360/n).

C2 = 180 rotation, C3 = 120 rotation, C4 = 90 rotation, C5 = 72 rotation, C6 = 60 rotation

Each rotation brings you to an indistinguishable state from the original.

However, rotation by 90 aboutthe same axis does not give backthe identical molecule.

Therefore H2O does NOT possessa C4 symmetry axis.

XeF4 is square planar.

It has four different C2 axes.

A C4 axis out of the page iscalled the principle axisbecause it has the largest n.

By convention, the principleaxis is in thez-direction


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If reflection of all parts of a molecule through a plane

produced an indistinguishable configuration, the symmetry

element is called a mirror plane or plane of symmetry.

h(horizontal): plane perpendicular to principal axis

d(dihedral), v(vertical): plane colinear with principal axis

v: Vertical, parallel to principal axis

d: parallel to Cn and bisecting two C2' axes

Planes and Reflection ()


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Inversion, Center of Inversion (i)

A center of symmetry: A point at the center of the molecule. (x,y,z) --> (-x,-y,-z).

It is not necessary to have an atom in the center (e.g. benzene).

Tetrahedral, triangles, pentagons don't have a center of inversion symmetry.

Ru(CO)6C2H6

Rotation-reflection, Improper axis (Sn)

This is a compound operation combining a rotation (Cn) with a reflection

through a plane perpendicular to the Cn axis h.(Cn followed by h)

Cn=Sn


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Identity

Simplest symmetry operation. All molecules have this element. If 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.


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Successive Operations

Symmetry Point GroupsSymmetry Point Groups

Symmetry of a molecule located on symmetry axes, cut byplanes of symmetry, or centered at an inversion center isknown as point symmetry.

Collections of symmetry operations constitute mathematicalgroups.

Each symmetry point group has a particular designation.

C1,Ci, CsOh,OTd , Th ,TIh, I

Cv , DhS2nDn, Dnh, DndCn, Cnh, Cnv


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Identify the symmetry elements that are present in benzene.


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Cn Point Groups

PBrClF C1 H2O2 C2

C3 M(NH2CH2CO2)4 C4

ECnn =

As(C6H5)3

Cnh Point Groups

The direction of the Cn axis is take as vertical, so a symmetry planeperpendicular to it is a horizontal plane, h.

NH2FHOClBBrClF

The point group is called Cs (C1h)

N2F2

C2h

B(OH)3

C3h


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Cnv Point Groups

H2O

C2v

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

SF4

C2v

NF3

C3v

CHCl3

C3v

SF5Cl

C4v

Dn and Dnh Point Groups

Adding a C2 axis perpendicular to a Cn axis generates one of the dihedral groups.

D3

There must be n

C2 axes

perpendicular to

Cn

Adding a h to a Dn group generates a Dnh group.

C2H4

D2h

[PtCl4]2-

D4h

D2

Angle between rings not 0 or 90


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BF3

D3h

Dnd Point Groups

Adding a vertical mirror plane to a Dn group in such a fashion to bisect

adjacent C2 axes generates a Dnd group.

Ferrocene

Fe(C5H5)2

D5d


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1,3,5,7-tetrafluorocyclooctatetraene

Sn groups

S2n

For odd n, (Sn)n = h

For even n, (Sn)n = E

The Sn for odd n is the same as the Cnh.

Absence of mirror planes distinguish Sngroups from Dnd groups.

Linear Groups

Cv

H-Cl O=O

Dh

Has a h

High Symmetry Molecules

CCl4

Td

SF6

Oh

C60

Ih


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The letter A.

GeH3F

AsBr5 (trigonal bipyramid)

trans rotamer of Si2

H6

Crown-shaped S8

tetrachloroplatinate(II)

HOCl

O

H

H

B

O

O

H

B(OH)3

H

B

H

H

BH3

Ni(en)3

SF6

SeH3F


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Characteristic symmetry elements of

some important classes of point groups.


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Character Tables

Character tables contain, in a highly symbolic form, informationabout how something of interest (an bond, an orbital, etc.) is affected

by the operations of a given point group.

Each point group has a unique character table, which is organizedinto a matrix.

Column headings are the symmetry operations, which are groupedinto classes.

Horizontal 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.

yzy, Rx1111B2

xzx, Ry11-11B1

xyRz1-111A2x

2

,y2

,z2

z1111A1

v(yz)v(xz)C2EC2v

(x2-y2, xy)(xz, yz)(x,y), (Rx, Ry)0-12E

Rz-111A2

x2 +y2,z2z111A1

3v2C3EC3v

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

Left hand column gives a list of symmetry labels

Gives information about degeneracies (A and B indicate non-degenerate, E refers to doubly degenerate, T means triplydegenerate)

Main part of table contains characters (numbers) to label thesymmetry properties (of MOs or modes of molecular vibrations)


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To obtain from this total set the representations for vibration only,it is necessary to subtract the representations for the other two

forms of motion: rotation and translation.

Cartesian displacement vectors for a water moleculex

y

z

H H

O

xx

x

y

y

y

z

z

z

Translational Modes Rotational Modes

A mode in whichall atoms aremoving in thesame direction,

equivalent tomoving themolecule.

A mode in whichatoms move torotate (changethe orientation

of) the molecule.There are 3rotational modesfor nonlinearmolecules, and 2rotational modesfor linearmolecules.


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Selection Rules: Infrared and Raman Spectroscopy

Infrared energy is absorbed for certain changes in vibrational

energy levels of a molecule.-for a vibration to be infrared active, there must be a change inthe molecular dipole moment vector associated with the vibration.

For a vibration mode to be Raman active, there must be achange in the net polarizability tensor

-polarizability is the ease in which the electron cloud associatedwith the molecule is distorted

For centrosymmetric molecules, the rule of mutual exclusionstates that vibrations that are IR active are Raman inactive, andvice versa

The transition from the vibrational ground state to the firstexcited state is the fundamental transition.


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yzy, Rx1111B2

xzx, Ry11-11B1

xyRz1-111A2

x2,y2,z2z1111A1

v(yz)v(xz)C2EC2v

If the symmetry label (e.g. A1, B1, E) of a normal mode ofvibration is associated withx, y, orzin the character table, thenthe mode is IR active.

If the symmetry label (e.g. A1, B1, E) of a normal mode ofvibration is associated with a product term (x2,xy) in thecharacter table, then the mode is Raman active.

yzy, Rx1111B2

xzx, Ry11-11B1

xyRz1-111A2

x2,y2,z2z1111A1

v(yz)v(xz)C2EC2v


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D 3h E 2C3 3C2 h 2S3 3v

1 1 1 1 1 1x

2

+y

2

,z

2

1 1 -1 1 1 -1 Rz

2 -1 0 2 -1 0 (x ,y ) (x2y

2, 2xy )

1 1 1 -1 -1 -1

1 1 -1 -1 -1 1 z

2 -1 0 -2 1 0 (Rx ,Ry ) (xy ,yz)

1A

2A

E

1A

2A

E

h 3v


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CCl4 http://fy.chalmers.se/OLDUSERS/brodin/MolecularMotions/CCl4spectra.html


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Enantiometers


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Group Theory

Supplemental Material

E

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

Consider a water molecule (H2O).

Character Table Development

x

y

zC2

v(yz)

v(xz)

Coordinates are assigned according to the convention that the highestfold axis of rotation is aligned with thez-axis, and thexaxis isperpendicular to the plane of the molecule.

Let the translation of the molecule in the +ydirection be representedby unit vectors on the atoms, and consider how they change whenundergoing the C2v symmetry operations.

x

y

z

H H

O

H H

O

C2

H H

O

H H

O

v (xz)

H H

O

H H

O

v (yz)

H H

O

H H

O

B2 = +1 -1 +1-1


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The set of four labels (+1, -1, -1, +1) generated in the analysisconstitutes one irreducible representation within the C2v pointgroup. It is irreducible in the sense that it cannot be decomposed

into a simpler or more fundamental form.Not only does it describe the effects on the ytranslation but alsoon other y-vector functions such as a py orbital.

Therefore, yis understood to serve as a basis function for thisirreducible representation within the C2v point group.

E

x

y

z

H H

O

H H

O

C2

H H

O

H H

O

v (xz)

H H

O

H H

O

v (yz)

H H

O

H H

O

A2 = +1 +1 -1-1

Effect of a symmetry rotation about thez-axis.

E

x

y

z

H H

O

H H

O

C2

H H

O

H H

O

v (xz)

H H

O

H H

O

v (yz)

H H

O

H H

O

B1 = +1 -1 -1+1

Translation of the molecule in the +xdirection

E

x

y

z

H H

O

H H

O

C2

H H

O

H H

O

v (xz)H H

O

H H

O

v (yz)H H

O

H H

O

A1 =

+1 +1 +1+1

Translation of the molecule in the +zdirection


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yzy, Rx1111B2

xzx, Ry11-11B1

xyRz1-111A2

x2,y2,z2z1111A1

v(yz)v(xz)C2EC2v

The resulting character table for C2v is shown above.

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

The +1 and -1 numbers, which correspond to symmetric andantisymmetric behavior, are called characters.

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

Rx, Ry, Rzstand for rotations about the specified axes.

Symbols in the column on the far left are Mulliken Labels.

Reducible Representations

When applying the methods of group theory to problemsrelated to molecular structure or dynamics, the procedurethat is followed usually involves deriving a reduciblerepresentation for the phenomenon of interest, such as amolecular vibration, the then decomposing it into itsirreducible components.

A reducible representation will always be a sum ofirreducible representations.

In some cases (simple molecules with few bonds) we can

perform the decomposition by inspection, for the moregeneral case (complicated molecule with many bonds),we can use the reduction formula.


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=x

xx

i

x

r nh

N 1

Reducible Representations

The reduction can be achieved using the reduction formula. It is amathematical way of reducing that will always work when the answer

cannot be spotted by eye. It 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:

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

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

Atomic contributions, by symmetry operations,to the reducible representation for the 3Ndegrees of freedom for a molecule.

-2S3

-1S4

0S6

-3i

1

2C6

1C4

0C3

-1C2

3E

Contribution

per atom*

Operation

*Cn = 1 + 2cos(360/n); Sn = -1 + 2cos(360/n)

x

y

z

x

y

z

(xz)

x

y

z

x

y

z

E

x

y

z

x

y

z

i


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Ha Hb

O

Derivation of reducible representation for degrees of freedom in H2O

E

Ha Hb

O

Unshifted atoms

3

Ha Hb

O

Ha Hb

O

Ha Hb

O

C2

Hb Ha

O

1

(xz)

Hb Ha

O

Ha Hb

O

x

y

z

1

(yz)

3

Obtain the reducible representation (for H2O)by multiplying the number of unshifted atomstimes the contribution per atom.

31-19tot11-13Contribution per atom

3113Unshifted Atoms

v(yz)v(xz)C2E


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=x

xx

i

x

r nh

N 1

Reducible RepresentationsThe reduction can be achieved using the reduction formula. It is amathematical way of reducing that will always work when the answer

cannot be spotted by inspection. It 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:

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

The summation is over allof the symmetry operations. For each symmetryoperation, three numbers are multiplied together. These are:r is the character for a particular class of operation in the reduciblerepresentationi is the character of the irreducible representation.n is the number of symmetry operations in the class

=x

xx

i

x

r nh

N 1

Tabulate our known information.

31-19r

v(yz)v(xz)C2E

A1: (1/h)[(r)(i

)(n) + (rC2)(i

C2)(nC2) + (rv(xz))(i

v(xz))(nv(xz))

+ (rv(yz)

)(iv(yz)

)(nv(yz)

)

1)

2)

yzy, Rx1111B2

xzx, Ry11-11B1

xyRz1 111A2

x2,y2,z2z1111A1

v(yz)v(xz)C2EC2v

Reducible Representation (for H2O)

Character Table

A1: (1/4)[(9)(i)(n) + (-1)(i

C2)(nC2) + (1)(iv(xz))(nv(xz)) +

(3)(iv(yz))(nv(yz))

h = 4

A1: (1/4)[(9)(1)(n) + (-1)(1)(nC2) + (1)(1)(nv(xz)) + (3)(1)(nv(yz))

A1: (1/4)[(9)(1)(1) + (-1)(1)(1) + (1)(1)(1) + (3)(1)(1) = 3

irrep


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Calculate irreducible representation A2

A2: (1/h)[(r)(i

)(n) + (rC2)(i

C2)(nC2) + (rv(xz))(i

v(xz))(nv(xz))

+ (rv(yz)

)(iv(yz)

)(n

v(yz)

)A2: (1/4)[(9)(1)(1) + (-1)(1)(1) + (1)(-1)(1) + (3)(-1)(1) = 1

Calculate irreducible representation B1

B1: (1/4)[(9)(1)(1) + (-1)(-1)(1) + (1)(1)(1) + (3)(-1)(1) = 2

Calculate irreducible representation B2

B2: (1/4)[(9)(1)(1) + (-1)(-1)(1) + (1)(-1)(1) + (3)(1)(1) = 3

31-19r

v(yz)v(xz)C2E

The reducible representation

is resolved into three A1, one A2, two B1, and three B2 species.

yzy, Rx1111B2

xzx, Ry11-11B1

xyRz1 111A2

x2,y2,z2z1111A1

v(yz)v(xz)C2EC2v

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

-[trans = A1 + B1 + B2]

-[rot = A2 + B1 + B2]

vib = 2A1 + B2

Notice this is the same result weobtained by analyzing the symmetriesof the vibrational modes.

-each mode is IR active


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Consider only the OH stretches in H2O

-consider the number of unchanged O-H bonds under the symmetryoperations of the point group

2002Unchanged OH bonds

v(yz)v(xz)C2EC2v

yzy, Rx1111B2

xzx, Ry11-11B1

xyRz1 111A2

x2,y2,z2z1111A1

v(yz)v(xz)C2EC2v

CharacterTable

ReducibleRepresentation

By inspection, the reducible representation is composed of the A1and B2 representation.

Sum of rows

B2

A1 1111

1-1-11

2002

v(yz)

v(xz)C2E

General Method: Determine OH stretching in H2O using

reducible representations and reduction formula

2002Unchanged bonds (OH)

4Order of group

1111Coefficient

v(yz)v(xz)C2EC2v

yzy, Rx

1111B2

xzx, Ry11-11B1

xyRz1 111A2

x2,y2,z2z1111A1

v(yz)v(xz)C2EC2v

Using reducible representations and the reductionformula, one obtains A1 + B2 modes.


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Derive tot for BCl3 given the character table for D3h

Derive the number of vibrational modes and assign modes for BCl3.

4

1

4

h

-2

-2

1

2S3

2-2012tot

1-103Contribution per atom

2214Unshifted atoms

3v3C22C3E

D 3h E 2C3 3C2 h 2S3 3v

1 1 1 1 1 1 x2

+y2,z

2

1 1 -1 1 1 -1 Rz

2 -1 0 2 -1 0 (x ,y ) (x2y

2, 2xy )

1 1 1 -1 -1 -1

1 1 -1 -1 -1 1 z

2 -1 0 -2 1 0 (Rx ,Ry ) (xy ,yz)

1A

2A

E

1A

2A

E

h 3v

Results of using the reduction formula.

rx

*ix

*nx

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 determined

tot = + +3 +2 +1A 2A E 2A E

but, subtract off the translational representations.

-[trans = + ]E 2A

and subtract off the rotational representations.

-[rot = + ]2A E

vib = +2 + ]1A E 2A

Each Erepresentationdescribes twovibrationalmodes of equalenergy.


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Symmetricalstretching.

Out-of-planebending mode.

Unsymmetricalstretching.

In-planebending mode.

Raman active. IR active. Raman and IR active.

We can use isotopic substitution to interpret spectra, since thecharacteristic frequency of the mode will depend on the masses ofthe atoms moving in that mode.

Review: What do I do when I need to?

Assign symmetry labels to vibrational modes?If the vibrational mode is known and illustrated, sketch the resultingvibrational mode before and after each symmetry operation of thepoint group. Using the character table, assign the symmetry label andidentify if the mode if IR and/or Raman active.Determine the symmetries of all vibrational modes and if the modesare IR and/or Raman active?Determine how many atoms are left unchanged by each symmetryoperation. Find the reducible representation and reduce into theirreps. Subtract translational and rotational modes Identify whichmodes are IR and/or Raman active.Determine the symmetries of only the stretching modes and if themodes are IR and/or Raman active?Determine how many bonds are left unchanged by each symmetryoperation. Find the reducible representation and reduce into theirreps. Identify which are IR and/or Raman active.Develop a character table?Determine the effect of each symmetry operation on the x, y, ztranslation and the rotation Rx, Ry, and Rz. The resulting set ofcharacters correspond to an irrep in the character table.


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1. Determine the number of and assign the vibrational modes of NH3

2. Determine the number of and assign the vibrational modes of CH4

3. Determine the number of and assign the vibrational modes of [PtCl4]2-

4. Determine the number of and assign the vibrational modes of SF6

5. Determine the number of and assign the vibrational modes of SF5Cl

6. Determine number of CO stretching modes in Mn(CO)6

7. Determine number of CO stretching modes in Mn(CO)5Cl

8. Determine number of CO stretching modes in trans-Mn(CO)4Cl2

9. Determine number of CO stretching modes in cis-Mn(CO)4Cl2

10.Determine number of CO stretching modes infac-Mn(CO)3Cl3

11.Determine number of CO stretching modes inmer-Mn(CO)3Cl3

Inorganic Chapter4

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