Symmetry operation motion of molecule which leaves it indistinguishable from before.

Symmetry element plane, line (axis) or point (centre) about which symmetry operation is performed.

Point group quantitative measure of amount of symmetry possessed by a molecule.

Molecular Symmetry Revision

Element (symbol) Operation (symbol)

Types of Symmetry Operation/Element

Axis of rotation (Cn) Rotation by m (360/n)° (Cnm)

Plane of symmetry () Reflection in plane ()

Centre of symmetry (i) Inversion at centre (i)

Rotation/reflection Rotation by m (360/n)followed axis (Sn) by m reflection in

perpendicular plane (Snm)

(None) Identity (do nothing) (E)

Note – main axis drawn vertical by convention. Plane containing main axis= v (vertical plane). Plane perpendicular to main axis = h (horizontal plane). Sometimes two different types of vertical plane differentiated as v and d (dihedral plane) – examples and explanation later. If no rotation axis symbol is simply .

Step 1 Look for Cn axis – if found go to Step 2.Otherwise look for (i) plane of symmetry Cs

(ii) centre of inversion Ci

(iii) only identity C1

Assignment of Molecules to Point Groups

Step 2 Look for nC2 to Cn – if found go to Step 3.Otherwise look for. (i) h Cnh

(ii) n v Cnv

(iii) no Cn

Step 3 Look for (i) h Dnh

(ii) n v Dnd

(iii) no Dn

'Special' Groups(i) Linear molecules – with no centre of symmetry Cv

- with centre of symmetry Dh

(ii) Tetrahedral molecules: Td

(iii) Octahedral molecules: Oh

Effects of carrying out successive symmetry operations

Definition: MULTIPLICATION of symmetry operations= sequential operation

Simple examples of multiplication of symmetry operations:

C41 C4

1 = C42 = C2

1

C21 C2

1 = C22 = = i i

What if we want to multiply together (i.e. perform successively)different symmetry operations?

Take as an example the ammonia molecule, NH3:

C3

N

H1H2

H3

Looking down the C3 axis:

v

v'

v"

H3

N H1

H2

Now carry out two successive operations - first C31, then v

C31

v [NOT shifted by the rotation]

v'

N

H1H2

H3 N

H3H1

H2

N

H3H2

H1

[A] [B]

[C]

Thus we could have gonefrom [A] to [C] in one step,simply by reflecting in v’

vC31 = v’

Note - start at right and workbackwards

The final result of carrying outsuccessively (multiplying) two or

more symmetry operations is always equivalent to a single

symmetry operation characteristic of the molecule

What if we multiply the same two operations, but in thereverse order?

v

C31

v"

N

H1H2

H3 N

H1H3

H2

N

H23

H1

H3

[A] [B]

[C]

Thus: C31v = v” and C3

1v vC31

Compare ‘ordinary’ multiplication - which is COMMUTATIVE.Multiplication of symmetry operations can be NON-COMMUTATIVE, i.e. the order of multiplication is important. Try more examples using NH3.

INVERSE OPERATIONSIf any object has a symmetry operation A, then it musthave another one, B, such that:

AB = BA = E

i.e. carrying out second exactly undoes the effects of the first.

B is said to be the INVERSE of A, and is given the symbol A-1.

Therefore: A A-1 = A-1 A = E

For reflections and inversions, simply repeating the operationreturns everything to its original place, i.e.

2 = E and i2 = E

Thus a reflection or inversion is its own inverse, and

= -1 and i = i-1

Not so simple for rotations. How do we get back tothe original position?

For C31 (rotation by 120o) we can get back by a further

rotation by 240o (C32) -

C32 C3

1 = E = C31 C3

2

In general: Cnn-1 Cn

1 = E = Cn1 Cn

n-1

and: Cnn-m Cn

m = E = Cnm Cn

n-m

We can therefore say that Cnn-1 = Cn

-1 etc.

For improper rotations (rotation/reflections) the position is similar, but Sn

n = E only when n is even,while Sn

2n = E when n is odd.

Therefore Snn-1 Sn

1 = E = Sn1 Sn

n-1 when n is even

Sn2n-1 Sn

1 = E = Sn1 Sn

2n-1 when n is odd

Final (trivial) example: EE = E, therefore E = E-1

Hence - all symmetry operations possess inverses

Why is this important?

We are going to use the symmetry properties of moleculesto solve problems in spectroscopy, bonding theory etc.

This will use a branch of maths called GROUP THEORY

We need to show that collections of symmetry operations doconstitute mathematical groups

BASIC PROPERTIES OF A GROUPAny collection of objects (A, B, C, D,….) is a group if it obeys the following four rules:

(1) One element combines with all of the others and leaves them unchanged. This is the IDENTITY (E) for which AE = EA = A for all members of the group.

(2) Every element must possess an INVERSE (X), such that AX = XA = E. We have just seen that this is true for symmetry operations.

(3) Any combination of two or more elements in the collection must be equivalent to a single element which is also a member of the collection, AB = C. Shown earlier to be true for symmetry operations.

(4) Although multiplication need not be COMMUTATIVE, it must beASSOCIATIVE, i.e. A(BC) = (AB)C. This can also be proved to betrue for symmetry operations.

We have seen that multiplication of symmetry operations is oftennon-commutative.

Therefore, for any molecule,its collection of symmetryoperations constitutes a

GROUP

Effects of symmetry operations on the water molecule:

O

21

O

21

O

2 1

O

12

O

21

z

x

y

EC2 xz

yz

Multiplication Table for H2O

Using the diagram on the previous overhead, we see that:

C2v E C2 xz yz

E E C2 xz yz

C2 C2 E yz xz

xz xz yz E C2

yz yz xz C2 E

Operate first

Operatesecond

In this case, all multiplications arecommutative

Multiplication Table for NH3

z

y

x

1

2

3

Note direction of rotation

Looking down z-axis

1

2 3

v

v' v"

Remember the order of operations (start at right),and that the symmetry elements are left unshifted:

C3v E C31 C3

2 v v' v”

E E C31 C3

2 v v' v"C3

1 C31 C3

2 E v" v v'C3

2 C32 E C3

1 v' v" v

v v v' v" E C31 C3

2

v'v" etc.

The completion of this table is left to you - see TutorialProblem sheet.

Note that the multiplications are oftennon-commutative - as seen earlier.

Classes of symmetry operations

Use the C3v multiplication table.

Carry out multiplication C32vC3

1 - remember that C32

is the INVERSE of C31, so that this could be written as

C3-1vC3

1.

Use multiplication table to show that C3

2vC31 = C3

2v’ = v”

Now carry out similar multiplications X-1vX (whereX = any symmetry operation of C3v).

We find that the answer is always v, v’ or v”.

This is also true if the ‘central’ operation is v’ or v”

The answer is never E, C31 or C3

2.

To summarise these results: in the point group C3v, the operations v, v’ and v” are in the same CLASS.

Similarly X-1EX = E for all X in group, and X-1 C31 X or

X-1 C32 X = C3

1 or C32 for all X in group.

Hence, in the point group C3v, the 6 symmetry operationscan be grouped into 3 classes:

E C31, C3

2 v, v’ , v”

The importance of this will be revealed later!

Representations of Groups

Drawing diagrams to show effects of symmetry operationson molecules is too cumbersome. Need numericalrepresentation of these effects.

One way to obtain such representations is by attaching oneor more VECTORS to the molecule. The effects ofsymmetry operations on these can be expressednumerically.

Simple example : SO2 molecule (triatomic, C2v)

S

OO

z

yx

z

yxy

z

x

Set up cartesian axes as shown, and anarrow pointing along the y-axis for eachatom. These arrows correspond to a translation of the whole molecule in they-direction.

These arrows are therefore a TRANSLATION VECTOR,

given the symbol Ty.

Remember - symmetry operations for this molecule are E, C2, xz and yz.

Carrying out E or yz leaves Ty unshifted, but C2 or xz reverses its direction,

e.g for C2

S

OO

S

O O

C2 rotation

The effects of E and yz can be symbolised by +1, the effects of C2 and xz by -1.

E(Ty) = (+1)(Ty)C2(Ty) = (-1)(Ty)xz(Ty) = (-1)(Ty)yz(Ty) = (+1)(Ty)

Thus we have a numerical representation of the effects

of the symmetry operations on SO2, with Ty as the

BASIS VECTOR.

We can draw Tx and Tz vectors in exactly same way - giving different representations. It is also possible to draw ROTATION VECTORS, to generate representations, e.g.

S

OO

z

yx

z

yx

z

yx

Looking down the z-axis we see:

O S OThis vector corresponds to rotationabout the z-axis, i.e. Rz. Using Rz as a basis vector we can see that E and C2 leave it unchanged (+1), while xz and yz reverse its direction (-1).

Similar vectors for rotation about the x and y axes (Rx and Ry) can be drawn, and we can summarise the representations produced by the 6 translation and rotation vectors :

C2v E C2 xz yz

+1 +1 +1 +1 Tz

+1 +1 -1 -1 Rz

+1 -1 +1 -1 Tx, Ry

+1 -1 -1 +1 Ty, Rx

How can we prove that each of these is a true (faithful)representation of the group? The requirement for this is thatthe multiplication properties of the components of therepresentation should mirror exactly the operations that theyare said to represent.

The representation must reproduce the multiplication table

From C2v multiplication table we know that

xz yz = C2

Consider the corresponding relationships in our representations of C2v:

Tz representation : (+1)(+1) = (+1)Rz representation : (-1)(-1) = (+1)Tx/Ry representation : (+1)(-1) = (-1)Ty/Rx representation : (-1)(+1) = (-1)

In each case, the numbers representing the two reflectionsmultiply to give the number representing C2 - just like theoperations themselves.

All are therefore FAITHFUL REPRESENTATIONS of the C2v point group.

This was an easy example - all operations simply convertedthe basis vectors to themselves (+1) or to themselves reversed(-1). What happens when that is no longer the case?

Consider a molecule with threefold symmetry, e.g. NH3 (point group C3v)

N

HH

H

z y

x

Use translation vectors (only need to consider those on N) along x and y directions.

Look down the z (C3) axis :

N H

H

H

y

x

Ty

Tx

If we now rotate the Tx and Ty vectors by 120o about C3, what is the result?

N H

H

H

Ty

Tx

Ty'Tx'

120o

What is the relationship between the "new" vectors, Tx' and Ty'. and the "old" vectors, Tx and Ty? This can be expressed by a simple trigonometric relationship (i.e. determine the components of new vectors along directions of old).

Components of "new" vectors in directions of 'old' vectors :

Tx' = (cos 120o)Tx - (sin 120o)Ty

Ty' = (sin 120o)Tx + (cos 120o)Ty

Inserting the numerical values:

Tx' = -(1/2)Tx - (3/2)Ty

Ty' = +(3/2)Tx - (1/2)Ty

Because symmetry operations of C3v mix Tx and Ty together, they cannot be separated.

These equations can be written more conveniently in terms of MATRICES :

)2/1()2/3(

)2/3()2/1(TT'T'T yxyx

We now need a digression to explain whatmatrices are, and how we can multiply themtogether.

MATRICES

For more details - see "Group Theory for Chemists" Chapter 3.

Matrices are arrays of numbers (rectangular or square). Their multiplication properties are as follows:

XY = Z

When the matrices are written out in full wehave the arrangement shown on the next slide.

x x

x x

x x

y y y

y y y

z z z

z z z

z z z

11 12

21 22

31 32

11 12 13

21 22 32

11 12 13

21 22 23

31 32 33

where:

z x y x y

z x y x y

z x y x y

11 11 11 12 21

12 11 12 12 22

13 11 13 12 23

z x y x y

z x y x y

z x y x y

21 21 11 22 21

22 21 12 22 22

23 21 13 22 23

z x y x y

z x y x y

z x y x y

31 31 11 32 21

32 31 12 32 22

33 31 13 32 23

ROW

COLUMN

Notes: 1. No. of columns in X = No. of rows in Y 2. If X is n x p, and Y is p x m, then Z is n x m 3. Summarise process as multiplying ith row of X by kth column of Y to give zik.

T T T Tx y x y

( / ) ( / )

( / ) ( / )

1 2 3 2

3 2 1 2

For each symmetry operation of C3v, we can write outan equivalent matrix equation.

Each time there is a 2 x 2 TRANSFORMATION MATRIXwhich transforms Tx and Ty into a new pair of vectors.

They are as follows:

E 1 0

0 1

C3

1 1 2 3 2

3 2 1 2

( / ) ( / )

( / ) ( / )C3

2 1 2 3 2

3 2 1 2

( / ) ( / )

( / ) ( / )

v 's 1 0

0 1

( / ) ( / )

( / ) ( / )

1 2 3 2

3 2 1 2

( / ) ( / )

( / ) ( / )

1 2 3 2

3 2 1 2

It is also possible to draw out rotation vectorsRx and Ry, and it can be shown that they transform in exactly the same way as do Tx and Ty (i.e. same set of transformationvectors). N.B. Not easy to visualise Rx and Ry!

For Tz and Rz vectors of NH3:

N

HH

H

z

y

x

Tz

and N

HH

H

z y

x

Rz

Looking down C3 axis for Rz :

N H

H

HTz gives +1 for every symmetry operation.

Rz gives +1 for E, C31, C3

2; -1 for 3 v's,

i.e. the situation is like that for the C2v examples (earlier).

We can summarise the transformation matrices for all of the T and R vectors for NH3

Note that 1 vector gives 1 x 1 matrix, i.e. a number, 2 vectors give 2 x 2 matrices.

Representations generated by translation and rotation vectors for NH3

E C31 C3

2 v v v

Tz +1 +1 +1 +1 +1 +1

Rz +1 +1 +1 -1 -1 -1

(Tx,Ty)or(Rx,Ry)

1 0

0 1

( / ) ( / )

( / ) ( / )

1 2 3 2

3 2 1 2

( / ) ( / )

( / ) ( / )

1 2 3 2

3 2 1 2

1 0

0 1

( / ) ( / )

( / ) ( / )

1 2 3 2

3 2 1 2

( / ) ( / )

( / ) ( / )

1 2 3 2

3 2 1 2

In each case the set of 1 x 1 (=numbers) or 2 x 2transformation matrices is a REPRESENTATIONof C3v - easy to show for Tz and Rz; for others we can show that the matrices multiply in the same wayas the operations themselves.

Thus - can use translation or rotation vectors for any molecules to generate representations.

BASIS VECTORS REPRESENTATIONS