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Symmetry operation motion of molecule which leaves it indistinguishable from before. Symmetry element plane, line (axis) or point (centre) about which.

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Presentation on theme: "Symmetry operation motion of molecule which leaves it indistinguishable from before. Symmetry element plane, line (axis) or point (centre) about which."— Presentation transcript:

1 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

2 Element (symbol)Operation (symbol) Types of Symmetry Operation/Element Axis of rotation (C n )Rotation by m  (360/n)° (C n m ) Plane of symmetry (  )Reflection in plane (  ) Centre of symmetry (i)Inversion at centre (i) Rotation/reflection Rotation by m  (360/n )  followed axis (S n ) by m  reflection in perpendicular plane (S n m ) (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 .

3 Step 1Look for C n axis – if found go to Step 2. Otherwise look for (i)plane of symmetry  C s (ii)centre of inversion  C i (iii)only identity  C 1 Assignment of Molecules to Point Groups Step 2Look for nC 2  to C n – if found go to Step 3. Otherwise look for.(i)  h  C nh (ii)n   v  C nv (iii) no   C n Step 3Look for (i)  h  D nh (ii)n   v  D nd (iii) no   D n 'Special' Groups (i) Linear molecules – with no centre of symmetry C  v - with centre of symmetry D  h (ii) Tetrahedral molecules: T d (iii) Octahedral molecules: O h

4 Effects of carrying out successive symmetry operations Definition:MULTIPLICATION of symmetry operations = sequential operation Simple examples of multiplication of symmetry operations: C 4 1  C 4 1 = C 4 2 = C 2 1 C 2 1  C 2 1 = C 2 2 =  = i  i  What if we want to multiply together (i.e. perform successively) different symmetry operations? Take as an example the ammonia molecule, NH 3 :

5 Looking down the C 3 axis: Now carry out two successive operations - first C 3 1, then  v Thus we could have gone from [A] to [C] in one step, simply by reflecting in  v ’  v C 3 1 =  v ’ Note - start at right and work backwards

6 The final result of carrying out successively (multiplying) two or more symmetry operations is always equivalent to a single symmetry operation characteristic of the molecule

7 What if we multiply the same two operations, but in the reverse order? Thus: C 3 1  v =  v ” and C 3 1  v  v C 3 1 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 NH 3.

8 INVERSE OPERATIONS If any object has a symmetry operation A, then it must have 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 operation returns everything to its original place, i.e.  2 = E and i 2 = E Thus a reflection or inversion is its own inverse, and  =  -1 andi = i -1

9 Not so simple for rotations. How do we get back to the original position? For C 3 1 (rotation by 120 o ) we can get back by a further rotation by 240 o (C 3 2 ) - C 3 2 C 3 1 = E = C 3 1 C 3 2 In general:C n n-1 C n 1 = E = C n 1 C n n-1 and: C n n-m C n m = E = C n m C n n-m We can therefore say that C n n-1 = C n -1 etc.

10 For improper rotations (rotation/reflections) the position is similar, but S n n = E only when n is even, while S n 2n = E when n is odd. ThereforeS n n-1 S n 1 = E = S n 1 S n n-1 when n is even S n 2n-1 S n 1 = E = S n 1 S n 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?

11 We are going to use the symmetry properties of molecules to 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 do constitute mathematical groups

12 BASIC PROPERTIES OF A GROUP Any 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 be ASSOCIATIVE, i.e. A(BC) = (AB)C. This can also be proved to be true for symmetry operations. We have seen that multiplication of symmetry operations is often non-commutative.

13 Therefore, for any molecule, its collection of symmetry operations constitutes a GROUP

14 Effects of symmetry operations on the water molecule:

15 Multiplication Table for H 2 O Using the diagram on the previous overhead, we see that: C 2v EC 2  xz  yz EEC 2  xz  yz C 2 C 2 E  yz  xz  xz  xz  yz EC 2  yz  yz  xz C 2 E Operate first Operate second In this case, all multiplications are commutative

16 Multiplication Table for NH 3 Note direction of rotation Looking down z-axis Remember the order of operations (start at right), and that the symmetry elements are left unshifted:

17 C 3v EC 3 1 C 3 2  v  v '  v ” EEC 3 1 C 3 2  v  v '  v " C 3 1 C 3 1 C 3 2 E  v "  v  v ' C 3 2 C 3 2 EC 3 1  v '  v "  v  v  v  v '  v "EC 3 1 C 3 2  v '  v "etc. The completion of this table is left to you - see Tutorial Problem sheet. Note that the multiplications are often non-commutative - as seen earlier.

18 Classes of symmetry operations Use the C 3v multiplication table. Carry out multiplication C 3 2  v C 3 1 - remember that C 3 2 is the INVERSE of C 3 1, so that this could be written as C 3 -1  v C 3 1. Use multiplication table to show that C 3 2  v C 3 1 = C 3 2  v ’ =  v ” Now carry out similar multiplications X -1  v X (where X = any symmetry operation of C 3v ).

19 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, C 3 1 or C 3 2. To summarise these results: in the point group C 3v, the operations  v,  v ’ and  v ” are in the same CLASS. Similarly X -1 EX = E for all X in group, and X -1 C 3 1 X or X -1 C 3 2 X = C 3 1 or C 3 2 for all X in group. Hence, in the point group C 3v, the 6 symmetry operations can be grouped into 3 classes: EC31, C32C31, C32  v,  v ’,  v ” The importance of this will be revealed later!

20 Representations of Groups Drawing diagrams to show effects of symmetry operations on molecules is too cumbersome. Need numerical representation of these effects. One way to obtain such representations is by attaching one or more VECTORS to the molecule. The effects of symmetry operations on these can be expressed numerically.

21 Simple example : SO 2 molecule (triatomic, C 2v ) Set up cartesian axes as shown, and an arrow pointing along the y-axis for each atom. These arrows correspond to a translation of the whole molecule in the y-direction. These arrows are therefore a TRANSLATION VECTOR, given the symbol T y. Remember - symmetry operations for this molecule are E, C 2,  xz and  yz. Carrying out E or  yz leaves T y unshifted, but C 2 or  xz reverses its direction, e.g for C 2

22 The effects of E and  yz can be symbolised by +1, the effects of C 2 and  xz by -1. E(T y ) = (+1)(T y ) C 2 (T y ) = (-1)(T y )  xz (T y ) = (-1)(T y )  yz (T y ) = (+1)(T y ) Thus we have a numerical representation of the effects of the symmetry operations on SO 2, with T y as the BASIS VECTOR.

23 We can draw T x and T z vectors in exactly same way - giving different representations. It is also possible to draw ROTATION VECTORS, to generate representations, e.g. Looking down the z-axis we see: This vector corresponds to rotation about the z-axis, i.e. R z. Using R z as a basis vector we can see that E and C 2 leave it unchanged (+1), while  xz and  yz reverse its direction (-1).

24 Similar vectors for rotation about the x and y axes (R x and R y ) can be drawn, and we can summarise the representations produced by the 6 translation and rotation vectors : C 2v E C 2  xz  yz +1 +1 +1 +1 T z +1 +1 -1 -1 R z +1 -1 +1 -1 T x, R y +1 -1 -1 +1 T y, R x How can we prove that each of these is a true (faithful) representation of the group? The requirement for this is that the multiplication properties of the components of the representation should mirror exactly the operations that they are said to represent.

25 The representation must reproduce the multiplication table From C 2v multiplication table we know that  xz  yz = C 2 Consider the corresponding relationships in our representations of C 2v : T z representation : (+1)(+1) = (+1) R z representation : (-1)(-1) = (+1) T x /R y representation : (+1)(-1) = (-1) T y /R x representation : (-1)(+1) = (-1) In each case, the numbers representing the two reflections multiply to give the number representing C 2 - just like the operations themselves.

26 All are therefore FAITHFUL REPRESENTATIONS of the C 2v point group. This was an easy example - all operations simply converted the 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. NH 3 (point group C 3v )

27 Use translation vectors (only need to consider those on N) along x and y directions. Look down the z (C 3 ) axis : If we now rotate the T x and T y vectors by 120 o about C 3, what is the result?

28 What is the relationship between the "new" vectors, T x ' and T y '. and the "old" vectors, T x and T y ? 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 : T x ' = (cos 120 o )T x - (sin 120 o )T y T y ' = (sin 120 o )T x + (cos 120 o )T y Inserting the numerical values: T x ' = -(1/2)T x - (  3/2)T y T y ' = +(  3/2)T x - (1/2)T y Because symmetry operations of C 3v mix T x and T y together, they cannot be separated.

29 These equations can be written more conveniently in terms of MATRICES : We now need a digression to explain what matrices are, and how we can multiply them together.

30 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 we have the arrangement shown on the next slide.

31 where: 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 z ik.

32 For each symmetry operation of C 3v, we can write out an equivalent matrix equation. Each time there is a 2 x 2 TRANSFORMATION MATRIX which transforms T x and T y into a new pair of vectors. They are as follows:

33 It is also possible to draw out rotation vectors R x and R y, and it can be shown that they transform in exactly the same way as do T x and T y (i.e. same set of transformation vectors). N.B. Not easy to visualise R x and R y !

34 For T z and R z vectors of NH 3 : and Looking down C 3 axis for R z : T z gives +1 for every symmetry operation. R z gives +1 for E, C 3 1, C 3 2 ; -1 for 3  v 's, i.e. the situation is like that for the C 2v examples (earlier).

35 We can summarise the transformation matrices for all of the T and R vectors for NH 3 Note that 1 vector gives 1 x 1 matrix, i.e. a number, 2 vectors give 2 x 2 matrices.

36 Representations generated by translation and rotation vectors for NH 3 E C 3 1 C 3 2  v  v  v T z +1 +1 +1 +1 +1 +1 R z +1 +1 +1 -1 -1 -1 (T x,T y ) or (R x,R y )

37 In each case the set of 1 x 1 (=numbers) or 2 x 2 transformation matrices is a REPRESENTATION of C 3v - easy to show for T z and R z ; for others we can show that the matrices multiply in the same way as the operations themselves. Thus - can use translation or rotation vectors for any molecules to generate representations. BASIS VECTORS  REPRESENTATIONS


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