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Molecular Symmetry helps us understand molecular structure, some chemical properties, and characteristics of physical properties () – used with theory to predict vibrational spectra for the identification of molecular , and as a tool for understanding electronic structure and bonding.

Symmetrical : implies the species possesses a number of indistinguishable configurations.

1 : mathematical treatment of symmetry.

symmetry operation – an operation performed on an object which leaves it in a that is indistinguishable from, and superimposable on, the original configuration.

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

Element Operation Symbol Identity Identity E Symmetry in the plane σ Inversion center Inversion of a x,y,z to -x,-y,-z i

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

Proper axes of rotation (C n) Rotation with respect to a (axis of rotation).

•Cn is a rotation of (360/n)°.

•C2 = 180° rotation, C 3 = 120° rotation, C 4 = 90° rotation, C 5 = 72° rotation, C 6 = 60° rotation… •Each rotation brings you to an indistinguishable state from the original.

However, rotation by 90° about the same axis does not give back the identical . XeF 4 is planar. Therefore H 2O does NOT possess It has four different C 2 axes. a C 4 symmetry axis. A C 4 axis out of the page is called the principle axis because it has the largest n . By convention, the principle axis is in the z-direction

2 3 Reflection through a planes of symmetry (mirror plane) If reflection of all parts of a molecule through a plane produced an indistinguishable configuration, the symmetry is called a mirror plane or plane of symmetry . σ v(vertical): plane colinear with principal axis

Reflection through a planes of symmetry (mirror plane)

σ σ d(dihedral) Vertical, to principal axis , parallel to C n and bisecting two C2' axes

4 Reflection through a planes of symmetry (mirror plane) σ h(horizontal): plane perpendicular to principal axis

Inversion, Center of Inversion (i)

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

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

Tetrahedrons, , and don't have a center of inversion symmetry.

C2H6 Ru(CO) 6 C4H4Cl 2F2

5 Rotation-reflection, (S n) ° This is a compound operation combining a rotation 360 /n (C n) with a σ σ reflection through a plane perpendicular to the C n axis h.(C n followed by h) σ Cn=S n

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

6 Identity (E)

Simplest symmetry operation. All have this element. If the molecule does have no other elements, it is asymmetric.

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

CHFClBr SOFCl

Successive Operations

7 Symmetry Point Groups

•Symmetry of a molecule located on symmetry axes, cut by planes of symmetry, or centered at an inversion center is known as point symmetry .

•Collections of symmetry operations constitute mathematical groups .

•Each symmetry has a particular designation.

Cn, C nh , C nv Dn, D nh , D nd S2n C∞v ,D∞h

Ih, I Td , Th ,T Oh,O C1,C i, C s

8 HCl F2

C D ∞∞∞v ∞∞∞h

Linear molecular species can be classified according to whether they possess a centre of symmetry (inversion centre) or not.

9 Tetrahedral Geometry

P4 B4Cl 4

Octahedral Geometry Icosahedral Geometry

2−−− [W(CO) 6] [B 12 H12 ]

10 Identify the symmetry elements that are present in benzene.

Scheme for assigning point groups of molecules and molecular ions

11 Cn Point Groups n = Cn E

PBrClF C1 H2O2 C2

As(C 6H5)3

C3 M(NH 2CH 2CO 2)4 C4

Cnh Point Groups

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

HOCl NH 2F BBrClF

The point group is called C s (C 1h )

B(OH) N2F2 3 C C2h 3h

12 Cnv Point Groups

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

H2O SF 4

C2v C2v

NF SF Cl 3 CHCl 3 5 C3v C3v C4v

Dn and Dnh Point Groups

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

There must be n

C2 axes perpendicular to

Cn between rings not 0 ° or 90° D2 D3 σ Adding a h to a D n group generates a D nh group.

2- C2H4 [PtCl 4]

D2h D4h

13 BF 3

D3h

Dnd Point Groups

Adding a vertical mirror plane to a Dn group in such a fashion to bisect adjacent C 2 axes generates a Dnd group.

D2d

Ferrocene

Fe(C 5H5)2

D5d

14 Sn groups

n σ For odd n, (S n) = h

n For even n, (S n) = E

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

1,3,5,7-tetrafluorocyclooctatetraene Absence of mirror planes distinguish S n groups from D nd groups.

Linear Groups

H-Cl O=O

C∞v D∞h

σ Has a h

High Symmetry Molecules

CCl 4 SF 6 C60

Td O h Ih

15 The letter ‘A’. tetrachloroplatinate(II)

GeH 3F HOCl SF 6

H O SeH 3F B(OH) 3 B H

O O

AsBr 5 (trigonal bipyramid) H

H BH 3

B trans rotamer of Si H 2 6 H H

Ni(en) 3 Crown-shaped S 8

16 Characteristic symmetry elements of some important classes of point groups.

17 Character Tables

•Character tables contain, in a highly symbolic form, information about 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 , which is organized into a . •Column headings are the symmetry operations , which are grouped into classes . •Horizontal rows are called irreducible representations of the point group. •The main body consists of characters (), and a section on the right side of the table provides information about vectors and atomic orbitals . σ σ C2v EC2 v(xz) v’ (yz) 2 2 2 A1 1 1 1 1 z x , y , z

A2 1 1 -1 –1 Rz xy

B1 1 -1 1 –1 x, R y xz

B2 1 –1 –1 1 y, R x yz

σ C3v E 2C 3 3 v 2 2 2 A1 1 1 1 z x + y , z

A2 1 1 -1 Rz 2 2 E 2 -1 0 (x,y), (R x, R y) (x -y , xy)(xz, yz)

•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 triply degenerate) •Main part of table contains characters (numbers) to label the symmetry properties (of MO’s or modes of molecular vibrations)

18 To obtain from this total the representations for vibration only, it is necessary to subtract the representations for the other two forms of : rotation and . z z O y z x z

y H y H y

x x x Cartesian vectors for a molecule Translational Modes Rotational Modes A mode in which A mode in which all atoms are atoms move to moving in the rotate (change same direction, the orientation equivalent to of) the molecule. moving the There are 3 molecule. rotational modes for nonlinear molecules, and 2 rotational modes for linear molecules.

19 Selection Rules: Infrared and 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 in the molecular moment vector associated with the vibration.

For a vibration mode to be Raman active , there must be a change in the net polarizability tensor -polarizability is the ease in which the electron cloud associated with the molecule is distorted

For centrosymmetric molecules, the states that vibrations that are IR active are Raman inactive, and vice versa

The transition from the vibrational ground state to the first excited state is the fundamental transition .

The two bending modes require the same amount of energy and are therefore degenerate.

20 σ σ C2v EC2 v(xz) v’ (yz) 2 2 2 A1 1 1 1 1 z x , y , z

A2 1 1 -1 –1 Rz xy

B1 1 -1 1 –1 x, R y xz

B2 1 –1 –1 1 y, R x yz

If the symmetry label (e.g. A 1, B 1, E) of a mode of vibration is associated with x, y, or z in the character table, then the mode is IR active .

If the symmetry label (e.g. A 1, B 1, E) of a normal mode of vibration is associated with a product term ( x2, xy ) in the character table, then the mode is Raman active .

σ σ C2v EC2 v(xz) v’ (yz) 2 2 2 A1 1 1 1 1 z x , y , z

A2 1 1 -1 –1 Rz xy

B1 1 -1 1 –1 x, R y xz

B2 1 –1 –1 1 y, R x yz

21 σΞ 3σΞ D 3h E 2C 3 3C 2 hh 2S 3 3 v v ′ 2 2 2 A1 1 1 1 1 1 1 x + y , z ′ A 2 1 1 -1 1 1 -1 R z ′ E 2 -1 0 2 -1 0 (x , y ) (x 2 – y 2, 2 xy ) ′′ A 1 1 1 1 -1 -1 -1 ′′ A 2 1 1 -1 -1 -1 1 z ′′ E 2 -1 0 -2 1 0 (R x , R y )(xy , yz )

22 The vibrational modes of CH 4 (Td), only two of which are IR active.

CCl 4 http://fy.chalmers.se/OLDUSERS/brodin/MolecularMotions/CCl4spectra.html

23 2− vibrational modes of [PtCl 4] (D4h )

24 Point groups of octahedral metal carbonyl complexes

25 Enantiometers

A pair of enantiomers consists of two molecular species which are mirror images of each other and are nonsuperposable.

26 Group Theory Supplemental Material

Character Table Development σ Up to this point, we have C2 v z considered symmetry operations (xz ) only insofar as they affect atoms occupying points in molecules. σ v’ (yz ) y Consider a water molecule (H 2O). x Coordinates are assigned according to the convention that the highest fold axis of rotation is aligned with the z-axis, and the x axis is perpendicular to the plane of the molecule. Let the translation of the molecule in the + y direction be represented by unit vectors on the atoms, and consider how they change when

undergoing the C 2v symmetry operations. z O O O O H H H H H H H H σ σ E C2 v (xz) v (yz) y O O O O x H H H H H H H H B2 = +1 -1-1 +1

27 The set of four labels (+1, -1, -1, +1) generated in the analysis

constitutes one irreducible representation within the C 2v point group. 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 y translation but also

on other ‘ y-vector functions’ such as a p y orbital. •Therefore, y is understood to serve as a basis for this

irreducible representation within the C 2v point group. Effect of a symmetry rotation about the z-axis.

z O O O O H H H H H H H H σ σ E C2 v (xz) v (yz) y x O O O O

H H H H H H H H A2 = +1 +1-1 -1

Translation of the molecule in the + x direction

z O O O O H H H H H H H H σ σ E C2 v (xz) v (yz) y O O O O x H H H H H H H H B = 1 +1 -1+1 -1

Translation of the molecule in the + z direction

z O O O O H H H H H H H H σ σ E C2 v (xz) v (yz) y

x O O O O

A1 = H H H H H H H H +1 +1+1 +1

28 σ σ C2v EC2 v(xz) v(yz) 2 2 2 A1 1 1 1 1 z x , y , z

A2 1 1 -1 –1 Rz xy

B1 1 -1 1 –1 x, R y xz

B2 1 –1 –1 1 y, R x yz

The resulting character table for C 2v is shown above. The column heading are classes of symmetry operations for the group, and each row depicts one irreducible representation . The +1 and -1 numbers, which correspond to symmetric and antisymmetric behavior, are called characters . Columns on the right are some of the basis functions which have the symmetry properties of a given irreducible representation.

Rx, Ry, Rz stand for 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 problems related to molecular structure or dynamics, the procedure that is followed usually involves deriving a reducible representation for the phenomenon of interest, such as a , the then decomposing it into its irreducible components. A reducible representation will always be a sum of irreducible representations. In some cases (simple molecules with few bonds) we can perform the decomposition by inspection, for the more general case (complicated molecule with many bonds), we can use the reduction formula .

29 Reducible Representations

The reduction can be achieved using the reduction formula. It is a mathematical way of reducing that will always work when the answer cannot be spotted by eye. It is particularly useful when there are large numbers of bonds involved.

The vibrational modes of the molecule are reduced to produce a reducible representation into the irreducible representations. This method uses the following formula reduction formula: = 1 χ x χ x x N ∑ r i n h x N is the number of a symmetry species occurs in the reducible representation, h is the ‘order of the group’: simply the total number of symmetry operations in the group. The summation is over all of the symmetry operations. For each symmetry operation, three numbers are multiplied together. These are:

Χr is the character for a particular class of operation in the irreducible representation

Χi 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 determined by considering the combined effect of each symmetry operation on the atomic vectors . z z Atomic contributions, by symmetry operations, E to the reducible representation for the 3 N for a molecule. y y Operation Contribution x per atom* x E 3 z z C -1 2 σ(xz) C 0 3 y y C4 1 x x C6 2 σ 1 z i -3 i x S3 -2 y y S4 -1

S6 0 x z

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

30 z Derivation of reducible representation for degrees of freedom in H 2O

y Unshifted atoms x O O E 3

Ha Hb Ha Hb

O O C 2 1

Ha Hb Hb Ha O O σ(xz) 1

Ha Hb Hb Ha

O O σ(yz) 3

Ha Hb Ha Hb

Obtain the reducible representation (for H 2O) by multiplying the number of unshifted atoms times the contribution per atom.

σ σ EC2 v(xz) v(yz) Unshifted Atoms 3 1 1 3 Contribution per atom 3 -1 1 1 Γ tot 9 -1 1 3

31 Reducible Representations The reduction can be achieved using the reduction formula . It is a mathematical way of reducing that will always work when the answer cannot be spotted by inspection. It is particularly useful when there are large numbers of atoms and bonds involved.

The vibrational modes of the molecule are reduced to produce a reducible representation into the irreducible representations. This method uses the following formula reduction formula :

= 1 χ x χ x x N ∑ r i n h x N is the number of times a symmetry species occurs in the reducible representation, h is the ‘order of the group’: simply the total number of symmetry operations in the group. The summation is over all of the symmetry operations. For each symmetry operation, three numbers are multiplied together. These are:

Χr is the character for a particular class of operation in the reducible representation

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

Tabulate our known information.

Reducible Representation (for H 2O) σ σ 1) EC2 v(xz) v(yz) Γ r 9 -1 1 3 Character Table 2) σ σ C2v EC2 v(xz) v(yz) h = 4 A 1 1 1 1 z x2, y2, z2 1 1 A 1 1 – 1 –1 R xy = χ x χ x x 2 z N ∑ r i n B1 1 -1 1 –1 x, R y xz h x B 1 –1 –1 1 y, R yz irrep 2 x σ σ σ χ Ε χ Ε Ε χ C2 χ C2 C2 χ v(xz) χ v(xz) v(xz) A1: (1/h)[( r )( i )( n ) + ( r )( i )( n ) + ( r )( i )( n ) σ σ σ χ v(yz) χ v(yz) v(yz) + ( r )( i )( n ) σ σ χ Ε Ε χ C2 C2 χ v(xz) v(xz) A1: (1/4)[( 9)( i )( n ) + ( -1)( i )( n ) + ( 1)( i )( n ) + σ σ χ v(yz) v(yz) (3)( i )( n )

Ε C2 σv(xz) σv(yz) A1: (1/4)[(9)( 1)( n ) + (-1)( 1)( n ) + (1)(1)( n ) + (3)(1)( n )

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

32 Calculate irreducible representation A 2 σ σ σ χ Ε χ Ε Ε χ C2 χ C2 C2 χ v(xz) χ v(xz) v(xz) A2: (1/h)[( r )( i )( n ) + ( r )( i )( n ) + ( r )( i )( n ) σ σ σ χ v(yz) χ v(yz) v(yz) + ( r )( i )( n )

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

Calculate irreducible representation B 1

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

Calculate irreducible representation B 2

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

The reducible representation… σ σ EC2 v(xz) v(yz) Γ r 9 -1 1 3

…is resolved into three A 1, one A 2, two B 1, and three B 2 species.

σ σ C2v EC2 v(xz) v(yz) 2 2 2 A1 1 1 1 1 z x , y , z

A2 1 1 – 1 –1 Rz xy

B1 1 -1 1 –1 x, R y xz

B2 1 –1 –1 1 y, R x yz

Γ tot = 3 A1 + A2 + 2 B1 + 3 B2 Γ -[ trans = A1 + B1 + B2] Γ -[ rot = A2 + B1 + B2] Γ vib = 2 A1 + B 2

Notice this is the same result we obtained by analyzing the of the vibrational modes. -each mode is IR active

33 Consider only the OH stretches in H 2O -consider the number of unchanged O-H bonds under the symmetry operations of the point group σ σ C2v EC2 v(xz) v(yz) Reducible Unchanged OH bonds 2 0 0 2 Representation σ σ C2v EC2 v(xz) v(yz) 2 2 2 A1 1 1 1 1 z x , y , z Character A2 1 1 – 1 –1 R xy Table z B1 1 -1 1 –1 x, R y xz

B2 1 –1 –1 1 y, R x yz

By inspection , the reducible representation is composed of the A1 and B representation. 2 σ σ EC2 v(xz) v(yz)

A1 1 1 1 1

B2 1 -1 -1 1 Sum of rows 2 0 0 2

General Method: Determine OH stretching in H 2O using reducible representations and reduction formula σ σ C2v EC2 v(xz) v(yz) Coefficient 1 1 1 1 Order of group 4 Unchanged bonds (OH) 2 0 0 2

σ σ C2v EC2 v(xz) v(yz) 2 2 2 A1 1 1 1 1 z x , y , z

A2 1 1 – 1 –1 Rz xy

B1 1 -1 1 –1 x, R y xz

B2 1 –1 –1 1 y, R x yz

Using reducible representations and the reduction

formula, one obtains A 1 + B 2 modes.

34 Γ Derive tot for BCl 3 given the character table for D 3h

Derive the number of vibrational modes and assign modes for BCl 3. σ σ E 2C 3 3C 2 h 2S 3 3 v Unshifted atoms 4 1 2 4 1 2 Contribution per atom 3 0 -1 1 -2 1 Γ tot 12 0 -2 4 -2 2

σΞ 3σΞ D 3h E 2C 3 3C 2 hh 2S 3 3 v v ′ 2 2 2 A1 1 1 1 1 1 1 x + y , z ′ A 2 1 1 -1 1 1 -1 R z ′ E 2 -1 0 2 -1 0 (x , y ) (x 2 – y 2, 2 xy ) ′′ A 1 1 1 1 -1 -1 -1 ′′ A 2 1 1 -1 -1 -1 1 z ′′ E 2 -1 0 -2 1 0 (R x , R y )(xy , yz )

Results of using the reduction formula.

x x x Χr *Χi *n 1/h Sum Total ′ A1 12 0 -6 4 -4 6 0.083333 12 1 ′ A 2 12 0 6 4 -4 -6 0.083333 12 1 ′ E 24 0 0 8 4 0 0.083333 36 3 ′′ A1 12 0 -6 -4 4 -6 0.083333 0 0 ′′ A 2 12 0 6 -4 4 6 0.083333 24 2 E′′ 24 0 0 -8 -4 0 0.083333 12 1 Therefore, we have determined Γ ′ ′ ′ ′′ ′′ tot = A 1 + A 2 +3 E +2 A 2 + E

but, subtract off the translational representations. Γ ′ ′′ Each E’ -[ trans = E + A 2 ] representation and subtract off the rotational representations. describes two ′ ′′ vibrational -[Γ = A 2 + E ] rot modes of equal Γ ′ ′ ′′ vib = A 1 +2 E + A 2 ] energy.

35 Symmetrical Out-of-plane Unsymmetrical In-plane stretching. bending mode. stretching. bending mode.

Raman active. IR active. Raman and IR active.

We can use isotopic substitution to interpret spectra, since the characteristic frequency of the mode will depend on the masses of the 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 resulting vibrational mode before and after each symmetry operation of the point group. Using the character table, assign the symmetry label and identify if the mode is IR and/or Raman active. Determine the symmetries of all vibrational modes and if the modes are IR and/or Raman active? •Determine how many atoms are left unchanged by each symmetry operation. Find the reducible representation and reduce into the irreps. Subtract translational and rotational modes… Identify which modes are IR and/or Raman active.… Determine the symmetries of only the stretching modes and if the modes are IR and/or Raman active? •Determine how many bonds are left unchanged by each symmetry operation. Find the reducible representation and reduce into the irreps. Identify which are IR and/or Raman active.… Develop a character table? •Determine the effect of each symmetry operation on the x, y, z translation and the rotation R x, R y, and R z. The resulting set of characters correspond to an irrep in the character table.

36 Determine the number of and assign the vibrational modes of the following: How many peaks in the (1) IR spectra and (2) Raman spectra

1. NH 3

2. CH 4 2- 3. [PtCl 4]

4. SF 6

5. SF 5Cl

Determine number of CO stretching modes in

1. Mn(CO) 6

2. Mn(CO) 5Cl

3. trans -Mn(CO) 4Cl 2

4. cis -Mn(CO) 4Cl 2

5. fac -Mn(CO) 3Cl 3

6. mer -Mn(CO) 3Cl 3

37