1) Symmetry Elements & Point Groups
Symmetry is a very fascinating phenomenon in nature.
It is found in geometrical figures such as cube, a sphere, an equilateral triangle, a rectangle, a square, regular pentagon, a regular hexagon, etc.
However, we shall be concerned with the symmetry found in molecules and solids. Mathematical study of symmetry is called Group theory.
Symmetry Element and Symmetry Operations:
ELEMENT OF SYMMETRY:
Geometrical entity on the basis of that we can define the symmetry of an object is known as element of symmetry like, Axis, Plane, Point etc.
OPERATION OF SYMMETRY ELEMENT:
Operation of symmetry element is the process which we apply on the element to define the symmetry.
There are five types of symmetry elements:
E, Cn, σ, Sn and i.
E is called the identify;
Cn is then-fold proper axis of rotation;
σ is the plane of symmetry;
Sn is then-fold improper axis of rotation
i the centre of symmetry ( or the inversion centre).
These symmetry elements and the corresponding operations are listed in the table:
All the symmetry operations in a molecule can be combined to form a molecular group.
This group is called point group since all the symmetry elements of the molecule intersect at a common points which remains fixed under all the symmetry operations.
SOME MOLECULAR POINT GROUPS:
CLASSIFICATION OF AXIS OF SYMMETRY:
Axis are divided into two types :
(i) Principal axis – Axis of highest order (n=order of axis)
(ii) Subsidiary (Proper) axis - Axis other than principal axis
CLASSIFICATION OF PLANE OF SYMMETRY:
(i) Vertical Plane (σv): The plane parallel to principal axis
(ii) Horizontal plane (σh): Plane perpendicular to principal axis
(iii) Dihedral Plane ((σd): Plane parallel to principal axis and bisecting angle between C2 axis in the molecule.
Example of point group
Having 3 structures –Cis, Trans, Open book
Angle formed between Cis and Trans all known as Dihedral angle.
Maximum time H2O2 exist as open book
Structure having C2 point group.
Cis H2O2 having C2v point group.
Trans H2O2 having C2h point group.
Matrix representation of rotation axis along x, y and z axis:
MATRIX FOR Sn(z) AXIS:
2) Groups Representations:
Representation is a set of matrices for a group each corresponding to a single operation in the group, that can be combine among themselves in a manner parallel to the way in which the group elements.
Representation of higher dimension which can reduce to a lower dimension representation is known as reducible representation.
The character of identity is known as dimension.
To derive a reducible representation first of all we have to choose the basis set according to our need.
The characters of the reducible representation can be obtained by considering the combined effect of each symmetry operation on the atomic vectors
Where n is order of rotation axis.
A group is a collection of elements of symmetry which are interrelated according to the following rules:
(1) The products of any two elements in the group and square of each element must be an element of the same group.
e.g. In C2v the elements are E, C2, σxz , σyz
C2 x C2 = E
C2 x σxz = σyz
σxz x σyz = C2
(2) In some groups the elements are commutable to each other.
σxz x σyz = C2
σyz x σxz = C2
There groups are known as Abelian group.
But in the majority of groups the elements are not cammutable.
These groups are non Abelian group. e.g. C3v point group.
3) In a group there should be an element which should be commutable with all other elements and should leave unchanged i.e. E.
The identity E is that element.
4) For the set of elements in a group associative law of combination should be followed by the elements.
(a x b) x c = a x ( b x c )
(σxz σyz) C2 = (σyz C2) σxz
C2 C2 = σxz σyz
E = E
(5) Each elements of a group must have its reciprocal and this should also be element of the same group.
e.g. Plane will be reciprocal of the same plane.
ORDER OF THE GROUP:
The number of total operations in a group is called order of group.
C2v → E , C22σv = 4
C3v → E , C31C32 , 3 σv = 6
Cn = 2n order ; Td = 24
Dn = 4n order ; Oh = 48
CLASSES OF THE GROUP:
"A complete set of elements in a group which are conjugated to each other similarity transformation, are said to be in the same class."
The elements which are conjugate to each other:
"The elements which are in the same class always shows the same matrix"
e.g. C2V point group.
C2v = E , C2, σxz , σyz
In C2v all 4 elements ( E, C2, σxz , σyz ) are in different class.
GENERAL RULE FOR CLASS:
(i) lnversion (i) will be always in the separate class, if it is present.
(ii) σh will be always in the separate class.
(iii) In the case of proper axis Cnm and Cnn-m will be in the same class.
C31 and C32
C41 and C43
C81 and C87
C83 and C85
(iv) There will be separate different class for σd and σv
(v) In the case of improper axis Snm and Snn-m will be in the same class.
The elements which are reciprocal of each other then both lying in the same class.
A representation of lower dimension which cannot be further reduce is known as irreducible representation.
The irreducible representation for a group can be derive using great orthogonality theorem (G.O.T.)
Postulates of G.O.T. and Derivation of lrreducible representation of C2v point group.
(i) Number of irreducible representation in a group are always equal to the number of classes of the group.
Here 4 irreducible representation because C2v has 4 class.
(ii) The sum of square of dimension of all the irreducible representation in the group will be equal to the order of the group.
The character related to the identity known as dimension according to Rule (ii)
(iii) The sum of squares of the characters of an irreducible representation will be equal to the order of group.
(iv) Orthogonality Rule: The characters of any 2 irreducible representation in the same group will be always orthogonal to each other i.e. on multiply any 2 Irreducible representation the result should be 0.
1R1 x 1R2 = 0
1 x 1 + 2(1 x L2) + 3(1 x m2) = 0
1R1 x 1R3 = 0
1R2 x 1R3 = 0
1R1 x 1R3 = 0
1 x 2 + 2(1 x l3) + 3(1 x m3) = 0
1R1 + 1R3 = 2 + 2l3 + 3m3 = 0 .... (1)
1R2 x 1R3 = 0
= 1 x 2 + 2(1 x l3) + 3(-1 x m3) = 0
1R2 x 1R3 = 2 + 2l3 - 3m3 = 0 ....(2)
From equation (1) and (2), we get:
4l3 = -4
l3 = -4/4 = -1
Put l3 value in equation (1) then:
2 x -1 + 3m3 = -2
-2 + 3m3 = -2
3m3 = -2 + 2
m3 = 0
3) CHARACTER TABLE & ITS DERIVATION
A character table is a way which can be utilised for application of group theory for the molecular symmetry.
Each character table has 4 columns and different symbols in them which can be explain by taking example of C2v point group.
COLUMN – 1:
First column of character table shows mulliken symbol for irreducible representation of the group.
A,B,E & T : The mulliken symbol can be decided by the following rule:
If dimension 1 then put A or B .
If dimension 2 then put E.
If dimension 3 then put T
Alphabet A, B, E and T are notation for dimension of irreducible representation.
Subscript 1 and 2 :
If representation w.r.t. subsidiary axis symmetric -1
If representation w.r.t. subsidiary axis Antisymmetric-2.
Example 1: What is the meaning of E'2g Mulliken symbol?
This is 2 dimensional irreducible representation which is Antisymmetric w.r.t. P-axis
Symmetric w.r.t. inversion
Symmetric w.r.t. σh in the molecule.
Second column of charater table shows irreducible representation of the groups which can be derive using G.O.T.
Third column explain transformation properties of 3-cartesion coordinate x,y, z and 3 rotational axis i.e. Rx , Ry, Rz
Fourth column explain transformation properties of quadratic functions of Cartesian coordinates like transformation properties of x2 , y2 , z2
x2 – y2 , xy , yz , xz , x2 – y2 – z2
RELATION BETWEEN REDUCIBLE & IRREDUCIBLE REPRESENTATION
Reduction of a reducible representation:
Conversion of a higher dimension to a lower dimension representation.
Example 2: How many times A1 comes given by the formula?
nA1 = 1/4 (9 x 1 + -1 x 1 + 3 x 1 + 1 x 1)
Check: Number comes always as complete no.
nA2 = 1/4 ( 9 x 1 + -1 x 1 + 3 x -1 + 1 x -1) = 1
nB1 = 1/4 ( 9 x 1 + -1 x - 1 + 3 x 1 + 1 x -1) = 3
nB2 = 1/4 ( 9 x 1 + -1 x - 1 + -1 x 3 + 1 x 1) = 2
then 3A1 + 1A2 + 3B1 + 2B2
It is count the dimension = 9 (Character of identity in water)
4) FORMATION OF HYBRID ORBITALS
EXPLANATION OF HYBRIDIZATION (BONDING)
ABn TYPE MOLECULE:
The hybrid orbitals of an atom are obtained by mixing atomic orbitals of the same atom.
The bonding in a molecule ABn may be described in terms of hybrid orbitals of the central atom A.
The constituent atomic orbitals in a hybrid orbitals can be predicted from the reducible representation of the molecule of which bond vectors form its basis.
Example of Planar AB3 :
The planar AB3 molecule belongs to the point group D3h.
The bond vector coincides with the three hybrid orbitals of the atom A.
These orbitals point towards the atom B with each bond angle equal to 1200.
The character of reducible representation based on vectors may be worked out by following the guideline states as follows:
The character of reducible representation of a symmetry operation is equal to the number of unshifted bond vectors.
The set of characters for the symmetry operations of the point group D3h is given below:
Example 1:The character table of Tetrahedron (Td) is given below. Answer the following questions?
(i) What is the order of group?
(ii) What is the number of classes in the group?
(iii) What is the number of irreducible representation in the group?
(iv) Prove that irreducible representation A1 and E are orthogonal to each other
(v) Find out the value of direct product
• A1 x E
• E x E
(vi) Consider a AB4 type Tetrahedral molecule with s, p and d-orbitals on the central atom and find out the combinations of orbitals for σ bonding.
(vii) Explain the best fit combination of orbitals for CH4 and MnO2-2
(i) order of group= 24
(ii) number of classes in the group = 5
(iii) irreducible representation always equal to the number of classes present in the group = 5
(iv) Proof of irreducible representation of A1 x E are orthogonal to each other
A1 x E = [1 x 2 + 8(1 x -1) +3(1 x 2) + 6(1 x 0) + 6(1 x 0) ]
= 2 – 8 + 6 = 0
So the irreducible representation of A1 and E are orthogonal to each other.
(v) The direct product.
For E x E:
nA1 = 1/24 [( 4 x 1) + 8 (1 x 1) + 3(4 x 1) + 6(0 x 1) + 6(0 x 1)] = 1
nA2 = 1/24 [( 4 x 1) + 8 (1 x 1) + 3(4 x 1) + 6(0 x -1) + 6(0 x -1)] = 1
nE = 1/24 [( 4 x 2) + 8 (1 x -1) + 3(4 x 2) + 6(0 x 0) + 6(0 x 0)] = 1
nT1 = 1/24 [(4 x 3) + (1 x 0) + 3(4 x -1)] = 0
nT2 = 1/24 [(4 x 3) + 8(1 x 0) + 3(4 x -1)] = 0
Total R.R for E x E = A1 + A2 + E.
For A1 x E:
nA1 = 1/24 [( 2 x l ) + 8 ( -l x l) + 3 ( 2 x 1 )] = 0
nA2= 1/24 [( 2 x l ) + 8 ( -l x l) + 3 ( 2 x 1 )] = 0
nE = 1/24 [( 2 x 2 ) + 8 ( -l x -l) + 3 ( 2 x 2 )] = 1
nT1 = 1/24 [( 2 x 3 ) + 8 ( -l x 0) + 3 ( 2 x -1 )] = 0
nT2 = 1/24 [( 2 x 3 ) + 8 ( -l x 0) + 3 ( 2 x -1 )] =0
(vi) Step (i) Derive reducible representation using bond vector as basis.
Step (ii) Reduce the R. R. obtained
Step (iii) Find out the transformation properties of orbitals according to mulliken symbols
Step (iv) Make the combinations
nA1 = 1/24 [(4 x l) + 8(l x l) + 6 (2 x 1)] = 1
nA2 = 1/24 [(4 x 1)+ 8(l x l) + 6( 2 x -1 )] = 0
nE = 1/24 [( 4 x 2)+ 8(1 - 1) +6 (2 x 0)] = 0
nT1 = 1/24 [( 4 x 3) + 8 ( l x 0 ) + 6(2 x -1)] = 0
nT2 = 1/24 [ 4 x 3 + 8 ( 1 x 0) + 6 ( 2 x 1)] = 1
Total R.R. = A1 + T2
(vii) s orbital having always A1 symmetry
• s orbital will transform according to A1 symmetry
s orbital = ( x2 + y2 + z2)
• Px , Py and Pz orbital will transform according to x, y, z axis that means T2 symmetry.
Px, Py, Pz orbital= x, y, z
A1 = s orbital
T2 = Px , Py , Pz and dxy , dxz , dyz
Here 2 combinations are possible i.e.
First Combination: s , Px , Py , Pz= sp3
Second Combination: = s , dxy , dxz , dyz = sd3
5) I.R. and Raman Spectroscopy Explain by Group Theory
Example 2: Consider a AB2 type bent molecule (H2O) i.e. v-shape molecule. Find out the I.R. and Raman active vibration using the symmetry rules. The character table of C2v point group given.
(1) Derivation of Reducible representation using 3N coordinate as basis set.
(2) Reduction of Reducible representation. This will give 3N degree of freedom.
(3) Removal of Translational and Rotational degree of freedom to get vibration degree of freedom
(4) Translational properties of vibrations according to I.R. and Raman spectroscopy
(5) Drawing of vibrations
(1 & 2):
nA1 = 1/4 ( 9 x 1) + ( -1 x 1) + ( 3 x 1) + ( 1 x 1) = 3
nA2 = 1/4 ( 9 x 1) + ( -1 x 1) + ( 3 x -1) + ( 1 x -1) = 1
nB1 = 1/4 ( 9 x 1) + ( -1 x -1) + ( 3 x 1) + ( 1 x -1) = 3
nB2 = 1/4 ( 9 x 1) + ( -1 x -1) + ( 3 x -1) + ( 1 x -1) = 2
Total = 3A1 + A2 + 3B1 + 2B2
(3) Rotational degree of freedom transform according to Rx , Ry & Rz
(4) Translational degree of freedom transform according to x, y & z.
Rx – B2 x = B1
Rx – B1 y = B2
Rz – A2 z = A1
(5) Vibrational= 2A1 + B1 : Three degree of freedom
I.R. active vibrations transforms according to 3 Cartesian coordinate x, y, z.
Raman active vibration transforms according to the quadratic function of x, y, z.
A1 = z , x2 , y2 , z2 I.R. as well as Raman active
B1 = x , xz I.R. and Raman active both
Means all vibrations are asssymmetric