Group Theory and Miller Indices (MSCCH-504) PDF

MSCCH-504 M. Sc. I Semester GROUP THEORY, INSTRUMENTATION CHEMISTRY & COMPUTER FOR CHEMIST SCHOOL OF SCIENCES DEPARTMENT OF CHEMISTRY UTTARAKHAND OPEN UNIVERSITY  MSCCH-504 GROUP THEORY, INSTRUME...

MSCCH-504 M. Sc. I Semester GROUP THEORY, INSTRUMENTATION CHEMISTRY & COMPUTER FOR CHEMIST SCHOOL OF SCIENCES DEPARTMENT OF CHEMISTRY UTTARAKHAND OPEN UNIVERSITY  MSCCH-504 GROUP THEORY, INSTRUMENTATIONCHEMISTRY & COMPUTER FOR CHEMIST SCHOOL OF SCIENCES DEPARTMENT OF CHEMISTRY UTTARAKHAND OPEN UNIVERSITY Phone No. 05946-261122, 261123 Toll free No. 18001804025 Fax No. 05946-264232, E. mail [email protected] htpp://uou.ac.in  Expert Committee Prof. B.S.Saraswat Prof. A.K. Pant Department of Chemistry Department of Chemistry Indira Gandhi National Open University G.B.Pant Agriculture, University Maidan Garhi, New Delhi Pantnagar Prof. A. B. Melkani Prof. Diwan S Rawat Department of Chemistry Department of Chemistry DSB Campus, Delhi University Kumaun University, Nainital Delhi Dr. Hemant Kandpal Dr. Charu C. Pant Assistant Professor Academic Consultant School of Health Science Department of Chemistry Uttarakhand Open University, Haldwani Uttarakhand Open University, Board of Studies Prof. A.B. Melkani Prof. G.C. Shah Department of Chemistry Department of Chemistry DSB Campus, Kumaun University SSJ Campus, Kumaun University Nainital Nainital Prof. R.D.Kaushik Prof. P.D.Pant Department of Chemistry Director I/C, School of Sciences Gurukul Kangri Vishwavidyalaya Uttarakhand Open University Haridwar Haldwani Dr. Shalini Singh Dr. Charu C. Pant Assistant Professor Academic Consultant Department of Chemistry Department of Chemistry School of Sciences School of Science Uttarakhand Open University, Haldwani Uttarakhand Open University, Programme Coordinator Dr. Shalini Singh Assistant Professor Department of Chemistry Uttarakhand Open University Haldwani Unit Written By Unit No. 1. Dr. Priyanka Tiwari 01 Department of Chemistry M.B. Govt. P.G. College, Haldwani 2. Dr. Mahesh Chandra Vishwakarma 02 Department of Chemistry, L.S.M. Govt. PG. College, Pithoragarh 3. Dr. Renu Loshali 03 Department of Chemistry, Govt. Degree College, Bazpur 4. Dr. Jyoti Singh 04 Department of Applied Sciences Amarpali Group of Institute Haldwani, Nainital 5. Dr. Ajay Pali 05 Department of Applied Sciences Amarpali Group of Institute Haldwani, Nainital 6. Dr. Devendra Singh Dhami 06 Department of Chemistry SSJ Campus Kumaun University, Almaora 7. Mr. Gaurav Joshi 07 & 11 Assistant Professor Department o f Computer Science MB Govt. PG College, Haldwani 8. Mr. Shekhar Kumar 08, 09 & 10 Assistant Professor Department o f Computer Science MB Govt. PG College, Haldwani Course Editors 1. Prof. (Dr.) Narendra Singh Bhandari Member, Uttarakhand Public Service Commission Gurukul Kangri, Haridwar 2. Dr. Neeta Joshi Associate Professor Department of Chemistry, PNG Govt. PG. College, Ramnagar 3. Mr. Harsh Vardhan Pant Department of Computer Science Amrapali Group of Institute, Haldwani Title : Group Theory, Instrumentation Chemistry& Computer for Chemist ISBN No. : Copyrigh : Uttarakhand Open University Edition : 2021 Published by : Uttarakhand Open University, Haldwani, Nainital- 263139  CONTENTS BLOCK 1: GROUP THEORY IN CHEMISTRY AND X-RAY DIFFRACTION Unit 1: Symmetry and Group Theory 01-28 Unit 2: Matrix 29-58 Unit 3: X-ray Diffraction Methods 59-82 BLOCK 2: CHROMATOGRAPHIC AND RADIO ANALYTICAL METHODS Unit 4: Chromatographic methods 83-112 Unit 5: Gas Chromatography 113-143 Unit 6: Radio Analytical Methods 144-159 BLOCK 3: INTRODUCTION TO COMPUTERS Unit 1 Data Analysis 160-188 Unit 2 History of development of computers 189-207 Unit 3 Introduction to Computers and Computing 208-255 BLOCK 4 : PROGRAMMING IN CHEMISTRY Unit 4 Programming in Chemistry 256-270 Unit 5 Use of Computer Programmes 271- 304  UNIT 1- SYMMETRY AND GROUP THEORY CONTENTS: 1.0 Objectives 1.1 Introduction 1.2 Element of symmetry 1.1.1 Identity (E) 1.1.2 Axis of symmetry (Cn) 1.1.3 Plane of symmetry 1.1.4 Point of symmetry 1.1.5 Identity Operation 1.2 A complete Set of symmetry operations as mathematical group 1.2.1 Abelian group 1.2.2 Non-Abelian Group 1.3 Generator 1.4 Summary 1.5 Terminal questions 1.0 OBJECTIVES In this chapter we will discuss about following: Symmetry and symmetry operations Basic definition of group theory A complete set of symmetry operations as mathematical group To discuss about Generator Flow chart for point group determination Characteristics of symmetry operations Determination of point groups UTTARAKHAND OPEN UNIVERSITY Page 1 1.1 INTRODUCTION Symmetry is a phenomenon of geometrical property of the world in which we live. In nature many type of flowers and plants, snowflakes, insects, certain fruits vegetables and various microscopic organism exhibit characteristic symmetry (Fig. 1.1). Symmetry concepts are extremely useful in Chemistry. On the basis of symmetry we can predict infrared spectra, type of orbital used in bonding, predict the optical activity and interpret electronic spectra and study of molecular properties. 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. Figure-1.1 The term symmetry is derived from the Greek word “symmetria” which means “measured together” UTTARAKHAND OPEN UNIVERSITY Page 2  A group consists of a set of symmetry elements (and associated symmetry operations) that completely describe the symmetry of an object. Point groups have symmetry about a single point at the center of mass of the system. 1.2 ELEMENT OF SYMMETRY Symmetry elements are geometric entities as point, line, plane of molecule about which a symmetry operations as rotations, reflections, inversions and improper rotations can be performed. In a point group, all symmetry elements must pass through the center of mass (the point). A symmetry operation is the action that produces an object identical (i.e. position indistinguishable from the original position) to the initial object (even though atoms and bonds may have been moved. In generally we can say that the element of symmetry is geometrical tools of symmetry or entity of symmetry tools i.e. point, line, plane in the molecule. The actual reflection, rotation or inversion is called the symmetry operation. A molecule possesses a symmetrical element, which is unchanged in appearance after applying the symmetry operation, correspond to the element. The element of symmetry divided into five operations. 1. Identity symmetry (E) 2. Axis of symmetry (Cn where n=360/angle of rotation in θ) 3. Plane of symmetry (σ) 4. Improper axis of symmetry ( Sn and then ┴ Cn ) 5. Point of symmetry (i) 1.1.1 Identity (E): The operation, which brings back the molecule to the original orientation, is called identity operation. It is represented as E from the German word Einhert meaning unity. The identity operation in effect means doing nothing on the molecule and hence does not seen to be of much consequence but it is important lies in considering the molecule as a group. Each molecule has this type of symmetry. UTTARAKHAND OPEN UNIVERSITY Page 3 1.1.2 Axis of symmetry (Cn): The symmetry element in which the molecule can represent the identical image after 0 rotation of molecule by 360 /n with respect to any imaginary axis called as axis of symmetry or rotational axis. n= - Where n is always an integer. This axis defined is an n-fold rotation axis, Cn. - In water there is a C2 axis so we can perform a 2-fold (180°) rotation to get the identical arrangement of atoms Fig 1.2 - In BF3 there is a C3 axis so we can perform 3-fold (120°) rotations to get identical arrangement of atoms Fig 1.2 1800 O C2 O H H H H F F - C3 C3 + C3 B B F F F F Figure 1.2 - Rotations are considered positive in the counter-clockwise direction. - Each possible rotation operation is assigned using a superscript integer m of the form Cnm. - The rotation Cnn is equivalent to the identity operation (nothing is moved) Classification of axis of symmetry Many molecules have more than one Cn axis It can be divided into two different types: UTTARAKHAND OPEN UNIVERSITY Page 4  a) Principal axis of symmetry b) Secondary/ subsidiary axis of symmetry (a) Principal axis of symmetry The Principal axis in an object is the highest order rotation axis i.e. having largest value of n. It is usually easy to identify the principle axis and this is typically assigned to the z- axis if we are using Cartesian coordinates.If there are more than one axis of same order ,then the axis passing through maximum number of atoms is called Principle axis. Example: In BF3 molecule three fold rotation axis (C3) and two fold rotational axis (C2) are present. The three fold axis is coincident with the perpendicular to the plane and the rotation angle is =1200. After two C3 operations molecule comes into identity operation. Three-fold axis have higher order than two fold axis, therefore C3 is the principal axis of BF3 molecule Figure 1.3 and Figure 1.4. In Benzene there is six fold axes of symmetry C6 as principal axes and it has six axes of two fold symmetry three passing through centre of Benzene and two opposite carbon atoms and three passing through centre of symmetry and centre of two opposite edges Figure 1.5. In NH3 molecule only three-fold axis of symmetry Figure 1.6 and in water only two fold symmetry is present Figure 1.7. C31=120o , C32= 240o , C33=E=360o Figure: 1.3 UTTARAKHAND OPEN UNIVERSITY Page 5  Figure: 1.4 Figure: 1.5 Figure: 1.6 Figure: 1.7 UTTARAKHAND OPEN UNIVERSITY Page 6 If the order of rotation is similar, the principal axis is that in which the higher number of atoms present across in particular axis. Example: In ethene molecule there are three C2 axis are present shown in Figure C’2 is the principal axis of ethene molecule the two carbon atom present across this axis, while another two axis have no atom is present Figure 1.8. In naphthalene there are three C2 axis are present where C2’ is the principal axis shown in Figure: 1.8 Figure: 1.8 Figure: 1.9 1800 C2 O O H H H H 900 C4 O H O H H H However, rotation by 90° about the same axis does not give back the identical molecule. Therefore H2O does not possess a C4 symmetry axis. UTTARAKHAND OPEN UNIVERSITY Page 7  C2 F C4 F Xe F F Figure: 1.10 XeF4 is square planar. It has four different C2 axes. A C4 axis out of the page is called principal axis. (b) Subsidiary or secondary axes of Symmetry: Lower fold of rotation axis (lower order of axis of symmetry) is the secondary or subsidiary axis of the molecule. Example: In BF3 molecule C3 axis is principal axis and C2 axis is secondary axis. In ethane molecule C2’ is the principal axis and C2 is subsidiary axis. In NH3 molecule three- fold of axis present that is principal axis, here subsidiary axis is absent. 1.1.3 Plane of symmetry: Plane of symmetry have generate only one operation, on repeating the reflection operation molecule comes into initial structure that is σ2 = E. The imaginary plane bisects the molecule into two halves which are mirror image to each other. σn = E (if n is even) σn = σ (if n is odd) The molecules possessing only plane of symmetry but no rotation axis other than C1 (=E), there can be no definition vertical and horizontal planes and the symbol for this plane is simply σ. The point group for this type of molecule is Cs. Rotational axis along with plane, the planes can be classified into three types. a) Vertical Plane (σv: σ ǁ principal axis) b) Horizontal plane (σh : σ ┴ principal axis) c) Dihedral plane (σd bisect plane with respect to two C2 axis) UTTARAKHAND OPEN UNIVERSITY Page 8 1.1.4 (a) Vertical plane: The plane of operation undergoes parallel with respect to principal axis is vertical plane of symmetry and denoted by σv. The subscript “v” in σv, indicates a vertical plane of symmetry. This indicates that the mirror plane passing through the principle axis and one of the subsidiary axis ( if present). 1.1.3 (b) Horizontal plane: The plane reflection is perpendicular with respect to principal axis is called horizontal plane and denoted by σh. Example: In bent molecule of H2O have σv(yz) and σv(xz) C plane, in σv(xz) all the atoms of H and O are bisect in the same plane and in σv(yz), two hydrogen atoms are reflected to each other. Example: In pyramidal NH3 molecule, three plane of reflections are present. Which are parallel with respect to C3 principal axis, i.e. there are three vertical planes are present. Example: In trigonal planar BF3 molecule have three reflection of plane which are parallel with respect to three fold (C3) axis. One reflection of plane which is perpendicular with respect to principal axis. Example: In square planar complex [PtCl4]-2 have a horizontal plane, two vertical plane and two dihedral plane (which bisect the two C2 axis) shown in figure: 1.11, and 1.12 Figure: 1.11 Figure 1.12 UTTARAKHAND OPEN UNIVERSITY Page 9 1.1.3 (c) Dihedral plane: The plane passing through the principal axis but passing between two subsidiary axis is called dihedral plane and denoted by σd. Example: In allene compound there are two C2 axis which are different type one of them have three atoms are passed though the axis and principle axis. There are two dihedral angle which are bisecting to two C2 axis. H H C C C H H Figure: 1.13 Figure: 1.14 Figure 1.15 Rotational axes and mirror planes in Benzene UTTARAKHAND OPEN UNIVERSITY Page 10 1.1.3 Improper axis of symmetry (Sn): An improper rotation operation is one that comprises of a proper rotation operation around an axis followed by reflection through a plane perpendicular to it. It is denoted by Sn. n order of rotational axis. It can be also defined as, imaginary axis passing through molecule rotation in which perpendicular reflection to give equivalent orientation. Sn= Cn ┴ σn Example: In CH4 has tetrahedral geometry. If we place a tetrahedral in a cube then the four atom touches the four corners of cubical box. CH4 having S4 it doesn’t mean that C4 symmetry is present. Figure: 1.16 UTTARAKHAND OPEN UNIVERSITY Page 11  IMPROPER AXIS IN OCTAHEADRAL STRUCTURE Sn = 4 even Improper axis in trigonal bipyramidal structure sn=3 odd UTTARAKHAND OPEN UNIVERSITY Page 12  n Sn σh order id odd Sn2n E 4 4 4 4 1 1 1 1 1 C3 2 2 C3 2 3 C3 2 2 3 3 3 3 5 5 5 5 σh σ h + 1C = 1S 3 3 5 5 4 1 1 1 2 2 C3 4 C3 2 σh 2 2 S3 3 3 3 4 4 5 2 2 S3 C3 4 5 1 4 4 1 1 1 4 -h 2 C3 2 2 σh 2 2 S3 4 S3 3 3 5 3 3 4 5 5 5 2 1 C3 S3 C3 3 σh S3 4 5 4 1 5 1 1 1 2 σh 2 6 1 C3 C3 2 σh 2 5 S3 6 S3 3 3 5 3 3 4 4 5 6 S3 E 3 C3 Number of operation in Sn axis: S2= C2 ┴ σ UTTARAKHAND OPEN UNIVERSITY Page 13  S3= C3 ┴ σ S4 = C4┴ σ σ (odd) = σ σ(even S21= C2× σ1= S21= i S22= C22× σ2= E×E=E S3= C31 1.1.4 Point of symmetry: It is a point within the molecule through which all the atoms can inverted within a point. The element corresponds to this operation is a center of symmetry. This is designated by i. In this operation, every part of the object is reflected through the inversion center, which must be at the center of mass of the object UTTARAKHAND OPEN UNIVERSITY Page 14  Figure: 1.17 Example: In ethene molecule each atoms are gives the mirror image at the center of the molecule. That center is called an inversion center. Example: In benzene each carbon and hydrogen gives mirror image in a point. We should remember that C2 ≠ Ci Figure: 1.18 Figure 1.18 UTTARAKHAND OPEN UNIVERSITY Page 15 We should remember C2 × σ h = S2 = C i For example improper axis in Trans Dichloroethylene Figure :2.7 Cl H' Cl' H C2 H' Cl σv C C C C C C Cl' Cl H H H' Cl' H Cl' C C H' Cl σv Cl H' C C Cl' H Figure: 1.19 1.1.5 Identity Operation: This operation gives no change in the molecule. It is included mathematical completeness. An identity operation is characteristic of every molecule, even if it has no other symmetry. Difference between symmetry operation and symmetry element: S.N. Symmetry operation Symmetry element 1. No change Identity 2. Rotation by 360/θ about A n-fold axis of an axis of symmetry (n) rotation 3. The reflection in any A plane of UTTARAKHAND OPEN UNIVERSITY Page 16  plane of symmetry symmetry parallel w.r.t. to principle containing the axis of symmetry (σv). principle axis of symmetry (σv). 4. The reflection of plane of A plane of symmetry perpendicular symmetry to the principle axis of perpendicular to symmetry (σh). the principle axis of symmetry. 5. Reflection of a plane of A plane of symmetry containing symmetry parallel w.r.t. principle containing the axis of symmetry and principle axes of bisecting between two C2 symmetry and axes of symmetry (σd). bisecting angle between two C2 of symmetry (σd). 6. Rotation by 360/θ about The n fold axis followed by alternating axis reflection in a plane (Sn). perpendicular to that axis (Sn) 7. Inversion in a centre of The Centre of symmetry (i) symmetry 1.2 A COMPLETE SET OF SYMMETRY OPERATIONS AS MATHEMATICAL GROUP UTTARAKHAND OPEN UNIVERSITY Page 17  Group was defined by Arthor and Cayley in 1854. “Groups are a collection or set of element and obey the certain rules. It should denote number, matrix, vector, symmetry operation, element, addition-matrix and multiplication etc. are called group.” There are four postulate of the group: I) The product (multiplication of any two elements) in the group and the square of each element must be an element in the group. Example: In H2O, element of symmetry C2 (z), E, σvxz, σvyz are present. C2.C2= E Since the product of C2.C2= E which is the element present in group. C2. σvxy= C2 In this, the product of C2. σvxy= C2, which is also present in the group. The product multiplication of elements means combination of elements of the group. The square of an element means multiplication an element with itself. The product of two group of elements A and B of the group can be AB or BA these type of groups are known as Abelian group. Case1- A.B=B.A C2.E=E.C2 Case 2 Quite often the product A.B may give the element C, the group where B.A may give another element D of the same group. II) There is one element in each mathematical group which must commute (A.B=B.A) with all other elements of the group and leave them unchanged such element is known as identity element and this operation is called identity operation. A.E=E.A=A Example: In H2O UTTARAKHAND OPEN UNIVERSITY Page 18 C2.E=E.C2=C2 Example: In NH3 C3.E=E.C3=C3 III) The element of mathematical group obeys the associative law of multiplication. A.B (C.D.E)=A (B.C) (D.E)=(A.B)(C.D). E IV) Each element of mathematical group must have a reciprocal which is also an element of the group. When an element is multiplied by its reciprocal we get identity. -1 -1 A.A =A.A=E Type of group- 1) Abelian group 2) Non-abelian 1.2.1 Abelian group: “The groups in which all elements commute to each other are called Abelian group.” (Matrix of plane of symmetry) (Matrix of plane of symmetry) 1.2.2 Non-Abelian Group: “If all the elements do not commute to each other known as Non-abelian group. Shown in Figure: 2.1 UTTARAKHAND OPEN UNIVERSITY Page 19 ? 6v+ 1. BF3- Element of symmetry - 2C3 ,3C2 6v, 6v1, 6v'',6h for C3', 6v = 6v , C3' = ? F1 6v B F3 F2 6v' 6v'' (a) For C3'. 6v = ? F1 F1 6v F1 C 3' 6v B B B F2 F3 F2 F3 F2 F3 6v'' C3. 6v = 6v'' (b) For 6v. C3'. = ? F1 F2 6v F3 6v C 3' B B B F2 F3 F1 F1 F2 F3 6v' Since, C3'. 6v = 6v'' Non- abilian group 6v. C3' = 6v' Figure: 2.1 UTTARAKHAND OPEN UNIVERSITY Page 20 1.3 GENERATOR For arriving at the point group of a molecule we need to have knowledge of the generator of the symmetry operation. Generator can be defined as: “The minimum number of symmetry operation combined to each other be give the product of the group is known as point group.” Classification of point group: 1) Non-axial point group 2) Axial point group 1.3 (a) Non-axial point group: No axis of symmetry present in this type of point group. A) If only identity (E) present as generator. Only C1( n=360/360) or e operator is possible so that the point group is C1. B) Only plane of symmetry is present in a group, the point group is Cs. C) If only point of symmetry present the point group is Ci 1.3 (b) Axial point group: Basic generator as well as additional generators are present in the molecule those are called axial group. Figure: 2.2 UTTARAKHAND OPEN UNIVERSITY Page 21  Figure: 2.3 UTTARAKHAND OPEN UNIVERSITY Page 22  Figure: 2.4 Flow chart for point group determination UTTARAKHAND OPEN UNIVERSITY Page 23 D3h Figure: 2.5 BF3 molecules the point group is D3h Example: Cnh Point Groups: N2F2 B(OH)3 C2h C3h Figure: 2.6 Cnh Point Groups Example: Cnv Point Groups AX2 bent type molecule. If a mirror plane contains the rotational axis, the group is called a Cnv group. SF4 C2v Figure: 2.7 C2v point group Example: Dnh type point group, in CO2 UTTARAKHAND OPEN UNIVERSITY Page 24  O O There are an infinite number of possible Cn axes and σv mirror planes in addition to the σh. Figure 2.8 D2h Figure 2.9 Examples of Ci, C2, and C3 Figure 2.10 Examples of Cnv UTTARAKHAND OPEN UNIVERSITY Page 25 1.4 SUMMARY After reading this unit learner becomes familiar with group theory. The unit is summarized as -The experimental chemist in his daily work and thought is concerned with observing and, to as great an extent as possible, understanding and interpreting his observations on the nature of chemical compounds. In this chapter we have discuss about absolute and non redundant list of the elements of a finite group and we know what all of the possible products are, then the group is absolutely and uniquely defined at least in an abstract sense. 1.5 TERMINAL QUESTIONS 1. What is the meant by a symmetry operation? 2. What is the relation between principal axis and horizontal plane? 3. Determine the symmetry elements in BF3 molecule. 4. Determine the symmetry elements in BF3 molecule. 5. What is abelian group. Explain with an example. UTTARAKHAND OPEN UNIVERSITY Page 26 6. Give the example of molecules where their molecular plane are identified with (a) vertical plane and (b) horizontal plane. 7. Define the dihedral plane of symmetry? 8. Write down the point group of the following molecules. How many and which symmetry operations are possible in each case? H O O O H H B H O (b) H (a) H H H Cl C C C C H H Cl H (c) (d) H H H H H H (e) (f) 9. What is symmetry? 10. What are symmetry elements? 11. What are symmetry operations? 12. How is axis of symmetry represented? 13. What is represented by point group ‘Cn’? 14. What is plane of reflection? 15. What do you understand by S4 ? UTTARAKHAND OPEN UNIVERSITY Page 27 REFERENCE: 1. K Veera Reddy, Symmetry and Spectroscopy of molecule, New Age International publishers, 2014. 2. Gurdeep Raj, Ajay Bhagi and Vinod Jain, Group Theory and Symmetry in chemistry, Krishna publication, 2017. 3. B.R. Puri, L.R Sharma and K.C. Kalia, Principles of Inorganic Chemistry, Vishal publication, 2016 4. Group Theory and its Chemical Applications - P.K. Bhattacharya (Himalaya Publishing House) 2003 UTTARAKHAND OPEN UNIVERSITY Page 28  UNIT-2 MATRIX CONTENTS: 2.1 Objectives 2.2 Introduction 2.3 Matrix 2.4 Order of group: denoted by ‘h’ 2.5 Character table of symmetry operation 2.6 Application of group theory in I.R. and Raman Spectroscopy 2.7 Terminal questions 2.1 OBJECTIVES This chapter deals with the Matrix, order of group: denoted by ‘h’, Character table of symmetry operation and also discuss about the applications of group theory in I.R. and Raman Spectroscopy. 2.2 INTRODUCTION Matrix can use for the study of symmetry. “A matrix is rectangular array of number or symbol for number.” The symbol written down in rows and columns the square bracket is used. 2.3 MATRIX The method of wave mechanics developed by Schrodinger can be applied to the solution of spectroscopic problems. Matrix can use for the study of symmetry. “A matrix is rectangular array of number or symbol for number.” The symbol written down in rows and columns the square bracket is used. Matrix representation of symmetry elements: Element of symmetry- E,Cn, Sn, i and σ UTTARAKHAND OPEN UNIVERSITY Page 29 2.3.1 Matrix for identity (E): E or Contribution of E=1+1+1= 3 2.3.2 Matrix for Rotation (Cn(z)) : Anticlockwise Rotation Clockwise Rotation θ 0o 30o 45o 60o 90o 120o 180o 360o Sin 0 ½ 1/√2 √3/2 1 √3/2 0 0 Cos 1 √3/2 1/√2 ½ 0 -1/2 -1 1 In C4(z) θ =90o = = = Contribution of C4 =0+0+1=1 For C2, θ=1800 = = Contribution= -1-1+1= -1 Contribution of Cn can calculated with the help of above matrix chart. 2.3.3 Matrix for Reflection (σ): This is the reflection plane σh, σv and σd (a) For σ xz operation: = x2 = +x1 y2 = -y1 UTTARAKHAND OPEN UNIVERSITY Page 30 z2 = +z1 = Contribution of σ =+1-1+1=+1 For each type of plane contribution is +1. 2.3.4 Matrix for inverse (Ci): = x2 = -x1 y2 = -y1 z2 = -z1 = Contribution of i= (-1)+(-1) +(-1) = -3 2.3.5 Matrix for improper rotation (Sn(z)) Sn= Cn┴ σ Contribution for S2= (-1) +(-1)+ (-1)= -3 Table 1 Contribution of atoms in different symmetry Symmetry element Contribution E 3 C2 -1 C3 0 C4 +1 I -3 S3 -2 S4 -1 Σ 1 UTTARAKHAND OPEN UNIVERSITY Page 31 2.4 ORDER OF GROUP: DENOTED BY ‘h’ The total number of symmetry operations of point group is known as order of group. Point Cs Cn Cnv Cnh Dn Dnh Dnd Sn Td Oh Ih group Order of 2 n 2n 2n 2n 4n 4n N 24 48 120 group 2.4.1 Class of operation in the group: Class of Group has sorted elements of point group. Arrangement or sorting of elements require action of similarity transformation If A, B AND X are elements of a group and X-1 A X = B ‘B’ is similarity transform of A by X. If this relationship is satisfied we say A & B as conjugate elements. If elements conjugate then 1. Every element is conjugate to itself i.e. X-1 A X = A 2. If A is conjugate with B , then B must conjugate with A then X-1 A X = B and then there must be some other element Y in the group that Y-1 B Y=A 3. If A conjugates with B and C and then B & C conjugate with each other then, X-1 A X = B Y-1 A Y = C X-1 B X = C Y-1 C Y = B The set of symmetry operations of a group, which are conjugate to each other form conjugate (reciprocal) class or symmetry class. 1) Inversion (i) will be always in the separate class, if it is present. 2) σh will be always in the separate class. 3) In the case of proper axis Cnm and Cnn-m will be in the same class. Example: C31 and C32 C41 and C43 C83and C85 UTTARAKHAND OPEN UNIVERSITY Page 32 E C31 E = C31 C32 C31 C31 = C31 C31 C31 C32 = C31 C31 and C32 are inverse of each other. σv C31 σv = C32 σv’ C31 σv’ = C32 σv’’ C31 σv’’ = C32 So C31and C32 are mutually conjugate, hence they can be in one class. Order is 2. 4) There will be separate different class for σd and σv 5) In the case of improper axis Snm and Snn-m will be in the same class. Example: (1) In H2O (C2v) Element of symmetry=E, C2(z), σv(xz), σv(yz) Number of class= 4 (2) In NH3 (C3v) Element of symmetry=E, 2C3, 3σv. Here all three planes σv’, σv’’, σv’’’ are in one class. C31= C32= C33=E C31and C32 in the same class. 2.4.2 Representation of group: Representation is defined as a set of matrix which corresponding to a single operation in the group that can be combine among themselves in a manner to the way in which the group elements. 1) Reducible representation 2) Irreducible representation 2.4.2.1 Reducible Representation: Representation of higher dimension which can reduce to a lower dimension representation is known as reducible representation. The character or identity (E) is known as dimension. To derive a reducible representation first of all we have the basis set according x, y and z direction to our need Example: Reducible representation for H2O i.e. C2v Basis set= 3 Cartesian coordinate (x, y, z) UTTARAKHAND OPEN UNIVERSITY Page 33  Symmetry operation No. Unshifted atom 3 1 3 1 C2v E C2(z) σv(xz) σv(yz) No. of unshifted atom 3 1 1 3 Contribution of atom 3 -1 1 1 Reducible representation 3×3 1×-1 1×1 3×1 9 -1 1 3 Dimension E=9 i.e. 9 dimension reducible representation Example: Reducible representation for BF3 Basis set= 3 Cartesian coordinate (x, y, z) D3h E 2C3 3C2 3σv σh 2S3 UTTARAKHAND OPEN UNIVERSITY Page 34  No. of un-shifted atom 4 1 2 2 4 1 Contribution of atom 3 0 -1 1 1 2 Reducible representation 4×3 1×0 2×-1 2×1 4×1 1×2 12 0 -2 2 4 2 Dimension E=12 12 dimension reducible representation. Example: Reducible representation for NH3 Basis set= 3 Cartesian coordinate (x, y, z) C3v E 2C3 3σv No. of un-shifted atom 4 1 2 Contribution of atom 3 0 1 Reducible representation 4×3 1×0 2×1 12 0 2 2.4.3 Irreducible representation: A representation of lower dimension which cannot be further reduced is known as irreducible representation. The irreducible representation for a group can derive by using great orthogonality theorem (GOT). Postulates of Great Orthogonality Theorem (GOT) and derivation of Irreducible representation: Ist Postulate: Number of irreducible representation is always equal to the number of class. No. of IR = No. of class Example: In H2O molecule there are E, C2(z), σv(xz), σv(yz) elements are present. The number of order = 4 The number of class =4 C2v E C2(z) σv(xz) σv(yz) I.R. 1 L1 M1 N1 O1 I.R. 2 L2 M2 N2 O2 I.R. 3 L3 M3 N3 O3 I.R. 4 L4 M4 N4 O4 UTTARAKHAND OPEN UNIVERSITY Page 35 Here four IR representations occur because C2v has four classes. IInd postulate: The sum of square of dimension of all the IR in the group will equal to the order of group. L12+ L22+L32+L42=4 12+12+12+12=4 IIIrd postulate: The sum of square of the characters of IR will be equal to order of the group. L12+M22+N32+O42 = 4 C2v E C2(z) σv(xz) σv(yz) I.R. 1 1 1 1 1 I.R. 2 1 M2 N2 O2 I.R. 3 1 M3 N3 O3 I.R. 4 1 M4 N4 O4 IVth postulate: The characters of any two IR in the same group will be always orthogonal to each other (on multiplying any two IR the result should be zero). IR1× IR2 = 0 IR1× IR3 = 0 IR1× IR4 = 0 IR3× IR2 = 0 IR4× IR2 = 0 IR1× IR2 = 0 1×1×L2+1×1×M2+1×1×N2+1×1×O2 = 0 1×1×1+1×1×1+1×1×(-1)+1×1×(-1) = 0 C2v E C2(z) σv(xz) σv(yz) I.R. 1 1 1 1 1 I.R. 2 1 1 -1 -1 I.R. 3 1 M3 N3 O3 I.R. 4 1 M4 N4 O4 IR1× IR3 = 0 1×1×L3+1×1×M3+1×1×N3+1×1×O3 = 0 1×1×1+1×1×(-1)+1×1×1+1×1×(-1) = 0 UTTARAKHAND OPEN UNIVERSITY Page 36  C2v E C2(z) σv(xz) σv(yz) I.R. 1 1 1 1 1 I.R. 2 1 1 -1 -1 I.R. 3 1 -1 1 -1 I.R. 4 1 M4 N4 O4 IR1× IR4 = 0 1×1×L4+1×1×M4+1×1×N4+1×1×O4 = 0 1×1×1+1×1× (-1)+1×1×(-1)+1×1×1 = 0 C2v E C2(z) σv(xz) σv(yz) I.R. 1 1 1 1 1 I.R. 2 1 1 -1 -1 I.R. 3 1 -1 1 -1 I.R. 4 1 -1 -1 1 Ist Postulate: Number of irreducible representation is always equal to the number of class. No. of IR = No. of class Example: In NH3 molecule there are E, 2C3, 3σv elements are present. The number of order = 6 The number of class = 3 C3v E 2C3 3σv I.R. 1 L1 M1 N1 I.R. 2 L2 M2 N2 I.R. 3 L3 M3 N3 IInd postulate: The sum of square of dimension of all the IR in the group will equal to the order of group. L12+ L22+L32 =6 12+12+22 = 6 UTTARAKHAND OPEN UNIVERSITY Page 37  C3v E 2C3 3σv I.R. 1 1 1 2 I.R. 2 L2 M2 N2 I.R. 3 L3 M3 N3 IIIrd postulate: The sum of square of the characters of IR will be equal to order of the group. I.R.1× I.R.2 = 0 1×1×L2+1×1×M2+1×1×N2+1×1×O2 = 0 1×1×1+2×1×1+3×1×(-1) = 0 C3v E 2C3 3σv I.R. 1 1 1 1 I.R. 2 1 1 -1 I.R. 3 L3 M3 N3 IR1× IR3 = 0 1×1×L3+1×1×M3+1×1×N3+1×1×O3 = 0 1×1×2+2×1×(-1)+3×1×0 = 0 C3v E 2C3 3σv I.R. 1 1 1 1 I.R. 2 1 1 -1 I.R. 3 2 1 0 2.5 CHARACTER TABLE OF SYMMETRY OPERATION IR and their transformation properties which can be utilized for the chemical operation of group theory. Each character table has four columns. Character tables contain, in a highly symbolic form, information about how something of interest (a bond, an orbital, etc.) is affected by the operations of a given point group. Each point group has a unique character table, which is organized into a matrix. Column headings are the symmetry operations, which are grouped into classes. Horizontal rows are called irreducible representations of the point group. UTTARAKHAND OPEN UNIVERSITY Page 38  The main body consists of characters (numbers), and a section on the right side of the table provides information about vectors and atomic orbitals For H2O molecule the character table is given bellow. Point Class of the symmetry h=4 group elements Ist IInd Column IIIrd IVth Column Column Column Column Ist: first column of character table shows Mullikon symbols for I.R. of the group. (i) Dimension 1. 1D (A or B) 2. 2D (E) 3. 3D (T) (ii) Principal axis (+) symmetry – A (-) symmetry – B (iii) If subsidiary axis is present (+) gerade-1 (-) gerade- 2 (iv) If subsidiary axis absent then molecular plane (+) subscript - 1 (-) subscript – 2 (v) If σh present (+) single prime – ’ (-) double prime- ’’ (vi) If Ci present (+) gerade- g (-) ungerade- u Example: For H2O molecule UTTARAKHAND OPEN UNIVERSITY Page 39 C2v E C2(z) σv(xz) σv(yz) A1 1 1 1 1 A2 1 1 -1 -1 B1 1 -1 1 -1 B2 1 -1 -1 1 Column II has dimensions for each irreducible representation Example: For H2O molecule C2v E C2(z) σv(xz) σv(yz) I.R. 1 1 1 1 1 I.R. 2 1 1 -1 -1 I.R. 3 1 -1 1 -1 I.R. 4 1 -1 -1 1 C2v E C2(z) σv(xz) σv(yz) R.R. 9 -1 3 1 Example: For NH3 molecule C3v E 2C3 3σv A1 1 1 1 A2 1 1 -1 E 2 1 0 C3v E 2C3 3σv R.R. 12 0 2 Column III: Explain transformation properties of three Cartesian coordinate (x,y,z) and three rotational axis (Rx, Ry, Rz). Column IV: It explains transformation properties of quadratic function of coordinates like UTTARAKHAND OPEN UNIVERSITY Page 40 transformational properties of x2, y2, z2. Example: The character table for water molecule (for bent XY2 type molecule) h=4 C2v E C2(z) σv(xz) σv(yz) Basis functions or Symmetry of functions A1 1 1 1 1 Z x2, y2, z2 A2 1 1 -1 -1 Rz Xy B1 1 -1 1 -1 (x, Ry) Yz B2 1 -1 -1 1 (y, Rx) Zx The character table for ammonia molecule (for pyramidal XY3 type molecule) C3v E 2C3 3σv h=6 Basis functions or Symmetry of functions A1 1 1 1 z x2+ y2, z2 A2 1 1 -1 Rz E 2 1 0 (x2- y2, xy) (xz, yz) (x, y)(Rx, Ry) 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: N = 1/h [ ∑ nx Χr x Χix ] UTTARAKHAND OPEN UNIVERSITY Page 41 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 The characters of the reducible representation can be determined by considering the combined effect of each symmetry operation on the atomic vectors nA1 = ∑ operation of group ×R.R. ×I.R.] nA1 = ×1×9+1×1×(-1)+1×1×3+1×1×1] nA1= 12/4 nA1 = 3 The value of n is 3 3A1 nA2 = ×1×9+1×1×(-1)+1×(-1)×3+1×(-1)×1] nA2 = 9-1-3-1] nA2 =4/4 nA2 =1 The value of n is 1 A2 nB1 = ×1×9+1×(-1)×(-1)+1×1×3+1×(-1)×1] nB1= 12/4 nB1= 3 The value of n is 3 3B1 nB2 = ×1×9+1×(-1)×(-1)+1×(-1)×3+1×1×1] UTTARAKHAND OPEN UNIVERSITY Page 42 nB2 = 8/4 nB2 = 2 The value of n is 2 2B2 Relation between I.R. and R.R. in H2O is; 3A1+1A2+3B1+2B2 Example: For NH3 molecule C3v E 2C3 3σv A1 1 1 1 A2 1 1 -1 E 2 1 0 C3v E 2C3 3σv R.R. 12 0 2 nA1 = ∑ operation of group ×R.R. ×I.R.] nA1 = ×1×12+2×1×0+3×1×2] nA1 = 18/6 nA1 = 3 The value of n is 3 3A1 nA2 = ×1×12+2×1×0+3×(-1)×2] nA2 =6/6 nA2 =1 The value of n is 1 1A2 nE= ×2×12+2×(-1)×0+3×0×2] nE= 24/6 nE= 4 The value of n is 4 UTTARAKHAND OPEN UNIVERSITY Page 43 4E Relation between I.R. and R.R. = 3A1+A2+4E Column III: Explain transformation properties of three Cartesian coordinate (x,y,z) and three rotational axis (Rx, Ry, Rz). Column IV: It explain transformation properties of quadratic function of coordinates like transformational properties of x2, y2, z2. We know that Example: The character table for water molecule (for bent XY2 type molecule) h=4 C2v E C2(z) σv(xz) σv(yz) Basis functions or Symmetry of functions A1 1 1 1 1 Z x2, y2, z2 A2 1 1 -1 -1 Rz Xy B1 1 -1 1 -1 (x, Ry) Yz B2 1 -1 -1 1 (y, Rx) Zx The character table for ammonia molecule (for pyramidal XY3 type molecule) C3v E 2C3 3σv h=6 A1 1 1 1 z x2+ y2, z2 A2 1 1 -1 Rz E 2 1 0 (x2- y2, (x, y)(Rx, Ry) xy) (xz, yz) 2.6 APPLICATION OF GROUP THEORY IN I.R. AND RAMAN SPECTROSCOPY UTTARAKHAND OPEN UNIVERSITY Page 44 AB2 type molecule can exist in two geometry: S.N. AB2 geometry Degree of freedom=3N- (Rotational +Translational) 1. Linear 3N- (2+3) 2. Non-linear 3N- (3+3) To obtain from this total set the representations for vibration only, it is necessary to subtract the representations for the other two form of motion: rotation and translation. One of the most practical uses of point groups and group theory for the inorganic chemist in is predicting the number of infrared and Raman bands that may be expected from a molecule. Alternatively, given the IR or Raman spectrum, the symmetry of a molecule may be inferred. In both IR and Raman spectroscopy the molecule is viewed as containing moving vectors. How these vectors are affected by symmetry will provide a means to determine how many bands would be expected in these spectra. 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 in the molecular dipole 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 centro symmetric molecules, the rule of mutual exclusion 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. Quadratic function: (1) x2, y2, z2, x2-y2, x2- y2- z2, xy, yz, zx- Raman active In which these quadratic function occurs in IV column are Raman active. (2) Rx, Ry, Rz, x, y, z- IR active In which these quadratic function occurs in III column UTTARAKHAND OPEN UNIVERSITY Page 45 Character tables can be used in Determining the orbitals involved in bonding in a molecule. Determine the point group of the molecule. Determining the Reducible Representation, Γ, for the type of bonding you wish to describe (e.g. σ, π, π⊥, π//). The Reducible Representation indicates how the bonds are affected by the symmetry elements present in the point group. Identifying the Irreducible Representation that provides the Reducible Representation; there is a simple equation to do this. The Irreducible Representation (e.g. 2A1 + B1 + B2) is the combination of symmetry representations in the point group that sum to give the Reducible Representation. Identifying orbitals that are involved from the Irreducible Representation and the character table. Cartesian displacement vectors for a water molecule: Translational Modes: A mode in which all atoms are moving in the same direction, equivalent to moving the molecule. Rotational Modes: A mode in which atoms move to rotate (change the orientation of) the molecule. There are three rotational modes for nonlinear molecules, and two rotational modes for linear molecules. Figure: 2.12 UTTARAKHAND OPEN UNIVERSITY Page 46  Figure: 2.13 Example: In H2O IR and Raman active mode can explained with the help of group Theory.The relation between Irreducible and Reducible representation in H2O is: 3N=3A1+ A2+ 3B1+2B2 C2v E C2(z) σv(xz) σv(yz) h=4 A1 1 1 1 1 Z x2, y2, z2 A2 1 1 -1 -1 Rz xy B1 1 -1 1 -1 (x, Ry) yz B2 1 -1 -1 1 (y, Rx) zx Translational: A1= z, B1= x, B2= y Rotational: B2 = Rx , B1 = Ry, A2=Rz Degree of Freedom = 3N – (Rotational + Translational) = 3A1+ A2+ 3B1+2B2- (B2+ B1+ A2+ A1+ B1+ B2) =2A1+B1 There are three modes of vibrations. Rotations have the same symmetry as Rx, Ry and Rz. So in C2v The three vibrational modes remain. Two have A1 symmetry, and one has B1 symmetry For a molecular vibration to be seen in the Raman spectrum (Raman active), it must change the polarizability of the molecule. The polarizability has the same symmetry properties as the quadratic functions: xy, yz, xz, x2, y2 and z2 UTTARAKHAND OPEN UNIVERSITY Page 47  A1= Raman and IR active (due to presence of x2, y2 and z present) B1= Raman and IR active (due to presence of x, Ry and yz present) Example: In NH3 molecule the 3N is equal to 3A1+A2+4E Translational: A1= z, E= x, y Rotational: A2 = Rz, E= Ry, Rx Degree of Freedom = 3N – (Rotational + Translational) =3A1+A2+4E-(A1+ E+ A2+ E) =2A1+2E A1= Raman and IR active E= Raman and IR active Polarity of Molecules can be studied with group theory A molecule cannot be polar if it has 1. A centre of inversion i.e. any group with Ci 2. A ny electric dipole moment perpendicular to any mirror planes i.e. any of the groups D and their derivatives. 3. A ny electric dipole moment perpendicular to any axis of rotation.i.e. the cubic groups T,O,the Icosohesdral I, and their modifications. Group theory is useful in studying Chirality of molecules Amolecule is not chiral if 1. It posses an improper rotation axis Sn. 2. It belongs to the group Dnh or Dnd 3. It belongs Td or Oh 4. It possess centre of symmetry 5. It posses mirror plane. Examples 1 Cs ,Cnv , Dnh, Dnd , and Td posses plane of symmetry 2 Ci has centre of symmetry 3 Oh,and Ih has Centre and plane of symmetry UTTARAKHAND OPEN UNIVERSITY Page 48  4 Sn has improper axis Appendix A: Character Tables for selected point groups. Cs E H A' A" 1 1 x,y,R z x2,y2,z2,xy 1 -1 z,R x,Ry yz,xz C2 E C2 AB 1 1 z,Rz x2+y2,z2 1 -1 x,y,Rx,Ry yz,xz D2 E C 2(z) C 2(y) C 2(x) D3 E 2C 3 3C2 A B1 1 1 1 1 x2,y2,z2, xy A1 1 1 1 x2+y2, z2 1 1 -1 -1 z,Rz Xy 1 1 -1 z,Rz B2 A2 1 -1 1 -1 y, Ry Xz B3 1 -1 -1 1 x,Rx Yz E 2 -1 0 (x,y);(Rx,Ry) (xz,yz); C2v E C2 v(xz) v(yz) A1 1 1 1 1 z x2,y2,z2 1 1 -1 -1 Rz Xy A2 1 -1 1 -1 x,Ry Xz B1 1 -1 -1 1 y,Rx Yz C3v E 2C 3 3 v A1 1 1 1 z x2+y2, z2 1 1 -1 Rz A2 2 -1 0 (x,y), (Rx,Ry) (x2-y2,xy), (xz,yz) E C4v E 2C4 C2 2 v 2 d A1 1 1 1 1 1 z x2+y2, z2 A2 1 1 1 -1 -1 Rz B1 1 -1 1 1 -1 x2-y2 B2 1 -1 1 -1 1 Xy E 2 0 -2 0 0 (x,y) (R x,Ry) (xz,yz) Ci E i Ag 1 1 Rx,Ry,Rz x2,y2,z2,xy,xz,yz Au 1 -1 x,y,z UTTARAKHAND OPEN UNIVERSITY Page 49  E E C2(z) C2(y) C2(x) I (xy) (xz) (yz) Ag 1 1 1 1 1 1 1 1 x2, y2,z2 B 1g 1 1 -1 -1 1 1 -1 -1 Rz xy 1 -1 1 -1 1 -1 1 -1 Ry xz B 2g 1 -1 -1 1 1 -1 -1 1 Rx yz B 3g 1 1 1 1 -1 -1 -1 -1 1 1 -1 -1 -1 -1 1 1 z Au 1 -1 1 -1 -1 1 -1 1 y B 1u 1 -1 -1 1 -1 1 1 -1 x D3h E 2C 3 3C 2 2S 3 3 v H A1' 1 1 1 1 1 1 x2+y2, z2 A2' E' 1 1 -1 1 1 -1 Rz 2 -1 0 2 -1 0 (x,y) (x2-y2, xy) A1" 1 1 1 -1 -1 -1 A2" 1 1 -1 -1 -1 1 z 2 -1 0 -2 1 0 (Rx,Ry) (xz,yz) E" D4h E 2C4 C2 2C2' 2C2" I 2S 4 h 2 v 2 d A1g 1 1 1 1 1 1 1 1 1 1 x2+y2, z2 A2g 1 1 1 -1 -1 1 1 1 -1 -1 Rz 1 -1 1 1 -1 1 -1 1 1 -1 x2-y2 B 1g 1 -1 1 -1 1 1 -1 1 -1 1 xy 2 0 -2 0 0 2 0 -2 0 0 (Rx, Ry) (xz, yz) D5h 1 1 1 1 1 -1 -1 -1 -1 -1 A1' 1 1 1 -1 -1 -1 -1 -1 1 1 z A2' E1' 1 -1 1 1 -1 -1 1 -1 -1 1 1 -1 1 -1 1 -1 1 -1 1 -1 E2' 2 0 -2 0 0 -2 0 2 0 0 (x,y) UTTARAKHAND OPEN UNIVERSITY Page 50  D6h E 2C6 2C 3 C2 3C 2' 3C 2" i 2S 3 2S 6 h 3 d 3 v A1g 1 1 1 1 1 1 1 1 1 1 1 1 x2+y2, z2 1 1 1 1 -1 -1 1 1 1 1 -1 -1 Rz A2g 1 -1 1 -1 1 -1 1 -1 1 -1 1 -1 x2-y2 B1g 1 -1 1 -1 -1 1 1 -1 1 -1 -1 1 xy B2g 2 1 -1 -2 0 0 2 1 -1 -2 0 0 (Rx,Ry) (xz,yz) 2 -1 -1 2 0 0 2 -1 -1 2 0 0 E1g 1 1 1 1 1 1 -1 -1 -1 -1 -1 -1 E2g 1 1 1 1 -1 -1 -1 -1 -1 -1 1 1 z A1u 1 -1 1 -1 1 -1 -1 1 -1 1 -1 1 1 -1 1 -1 -1 1 -1 1 -1 1 1 -1 A2u 2 1 -1 -2 0 0 -2 -1 1 2 0 0 (x,y) B1u 2 -1 -1 2 0 0 -2 1 1 -2 0 0 D2d E 2S 4 C2 2C2' 2 d A1 1 1 1 1 1 x2+y2, z2 1 1 1 -1 -1 Rz A2 1 -1 1 1 -1 x2-y2 B1 1 -1 1 -1 1 z xy B2 2 0 -2 0 0 (x,y); (Rx,Ry) (xz,yz) A-2 D3d E 2C 3 3C 2 i 2S 6 3 d A1g 1 1 1 1 1 1 x2+y2, z2 1 1 -1 1 1 -1 Rz A2g Eg 2 -1 0 2 -1 0 (Rx,Ry) (x2-y2,xy);(xz,yz) A1u 1 1 1 -1 -1 -1 A2u Eu 1 1 -1 -1 -1 1 z 2 -1 0 -2 1 0 (x,y) S4 E S4 C2 S 43 A B 1 1 1 1 Rz x2+y2, z2 E 1 -1 1 -1 z x2-y2, xy 1 i -1 -( i) (x,y); (Rx,Ry) (xz,yz) Td E 8C3 3C2 6S 4 6 d A1 1 1 1 1 1 x2+y2+z2 1 1 1 -1 -1 A2 2 -1 2 0 0 (2z2-x2-y2, x2-y2) E T1 3 0 -1 1 -1 (Rx, Ry,Rz ) T2 3 0 -1 -1 1 (x,y,z) (xz,yz.xy) UTTARAKHAND OPEN UNIVERSITY Page 51 Appendix B: Constants & Useful Energy Conversions: Planck's constant, h = 6.626x10-34 J-s Boltzman's Constant, k = 1.381x10-23 J/K = 0.6950 cm-1/K Speed of light, c= 2.998x108 m/s 1 eV = 1.60219 x10-19 J = 96.485 kJ/mol = 22.58 kcal/mol = 8065.5 cm-1 1 cm-1 = 11.96 J/mol = 2.859 cal/mol = 0.1240 meV Appendix C: Some Direct Products Note that is some instances, g and u must be added (gxg=uxu=g; gxu=u), some subscripts must be omited and ' and " must be added (' x ' = " x" = '; ' x " = ") C2v A1 A2 B1 B2 A1 A1 A2 B1 B2 A2 A1 B2 B1 B1 A1 A2 B2 A1 C6v, D6, D6h A1 A2 B1 B2 E1 E2 A1 A1 A2 B1 B2 E1 E2 A2 A1 B2 B1 E1 E2 B1 A1 A2 E2 E1 B2 A1 E2 E1 E1 A1+[A2]+E2 B 1+B2+E1 E2 A1+[A2]+E2 C5v, , C6, D5, A A E E C6h A B A1 A1 A2 E1 E2 A2 A1 E1 E2 A A BB E1 A1+[A2]+E2 E1+E2 A E1 E2 A1+[A2]+E1 E C4v, D4, D2d , A1 A2 B1 B2 E D A1 A1 A2 B1 B2 E A2 A1 B2 B1 E B1 A1 A2 E B2 A1 E E A1+[A2]+B1+B2 UTTARAKHAND OPEN UNIVERSITY Page 52  Oh , Td A1 A2 E T1 T2 A1 A1 A2 E T1 T2 A2 A1 E T2 T1 E A1+[A2]+E T1+T2 T1+T2 T1 A1+E+[T1]+T2 A2+E+T1+T2 T2 A1+E+[T1]+T2 D2, D2h A B1 B2 B3 A A B1 B2 B3 B1 A B3 B2 B2 A B1 B3 A C4, C4h , S4 A A B E A B E B A E E [A]+A+E C3v, D3, D A1 A2 E A1 , D A1 A2 E A2 A1 E E A1+[A2]+E UTTARAKHAND OPEN UNIVERSITY Page 53 Appendix D: Standard Valence Orbital H ii values (eV). Atom ns np (n-1)d n H -13.6 1 B -15.2 -8.5 C -21.4 -11.4 N -26.0 -13.4 O -32.3 -14.8 2 F -40.0 -18.1 Si -17.3 -9.2 P -18.7 -14.0 3 S -20.0 -13.3 Cl -26.3 -14.2 Sc -8.9 -2.8 -8.5 Ti -9.0 -5.4 -10.8 V -8.8 -5.5 -11.0 Cr -8.7 -5.2 -11.2 Mn -9.8 -5.9 -11.7 Fe -9.1 -5.3 -12.6 Co -9.2 -5.3 -13.2 Ni -9.2 -5.2 -13.5 Cu -11.4 -6.1 -14.0 Zn -12.4 -6.5 Ga -14.6 -6.8 Ge -16.0 -9.0 As -16.2 -12.2 4 Se -20.5 -13.2 Br -22.7 -13.1 Mo -8.3 -5.2 -10.5 Ru -10.4 -6.9 -14.9 Rh -3.09 -4.6 -12.5 Pd -7.3 -3.8 -12.0 5 Sb -18.8 -11.7 I -18.0 -12.7 Te -20.8 -13.2 W -8.3 -5.2 -10.4 Re -9.36 -6.0 -12.7 Os -8.5 3.5 -11.0 6 Pt -9.1 -5.5 -12.6 Au -10.9 -5.6 -15.1 Page 54 Point Groups and their Detailed List of Symmetry Elements are Included in the Below Table. Point group Order of group, h Type of symmetry elements C1 1 E (=C1) C1 2 E, I (=S2 ) C1 2 E, σ Cn – groups: ( h = n ) C2 2 E, C2 C3 3 E C31, C32 C4 4 E, C41, C42 (=C2), C43 C5 5 E, C41, C42 ,C43, C44 Cnv – groups: ( h = 2n ) C2v 4 E, C2 , σ C3v 6 E,C31, C32 , 3σv C4v 8 E,C41,C42(=C2), C43, 2σv, 2σv’ Cnh – groups: ( h = 2n ) C2h 4 E, C2 ,i=( S2 ), σh C3h 6 E, C31, C32 , S31, S35, σh C4h 8 E,C41,C42(=C2), C43, S41, S43,σh, i=( S2 ) Dn – groups: ( h = 2n ) D2 4 E, C2, C2’ D3 6 E, C31, C32, 3C2 D4 8 E,2C4, C2, 4C2 Page 55 Point group Order of group, h Type of symmetry elements Dnh – groups: ( h = 4n ) D2h 8 E, C2 , 2C2’ , i=( S2 ), σh , 2σv D3h 12 E, 2C3 ,3C2 , σh ,3σv , 2S3, (S31, S35) D4h 16 E, 2C4,( C41,C42 ), C2=( C42), 2C2’,2C2”, σh ,2σv ,3σd, i , 2S4 (S41, S43 ) Dnd– groups: ( h = 4n ) D2d 8 E, C2 , 2C2’, 2 σd, 2S4 D3d 12 E, 2C3 (C31, C32), 3C2 ,i, 3σd, 2S6(S61, S63 ) D4d 16 E, 2C4, (C41, C43), C2=( C42), 4C2’,4σd, 4S8(S81, S83 , S85, S87) Sn (n=even)– groups: ( h = n ) S4 4 E, S41, S43 , C2 S6 6 E, S61, S45, C31, C32, i S8 8 E, S8 ,S81, S83 , S85, S87, C41, C43, C2=( C42) Infinite- point group (h=∞) C∞v ∞ E, ∞, C∞, ∞σv D∞v ∞ E, ∞, C∞, ∞σv, σh, i Page 56 2.7 Terminal Questions 1. Draw trans, cis and gauche structures of H2O2, and assign these structures to point groups. trans cis gauche C2h C2v C2 2. Write point group for Ethane staggered and Ethane eclipsed, Acetylene, Cis and Trans dichloroethylene. Ethane in the staggered conformation has 2 C3 axes,3 perpendicular C2 axes bisecting the C–C line, in the plane of the two C’s and the H’s on opposite sides of the two C’s. 3d, i, S6.There is no σh ,So Point group is D3d. Ethane in eclipsed conformation has two C3 axes (the C–C line), three perpendicular C2 axes bisecting the C–C line, in the plane of the two C’s and the H’s on the same side of the two C’s. Mirror planes include σh and 3d. So point group is D3h. Acetylene has a C∞ axis through all four atoms, an infinite number of perpendicular C2 axes, a σh plane, and an infinite number of σd planes through all four atoms. D∞h. cis-1,2-dichloroethylene has a C2 axis perpendicular to the C–C bond, two mirror planes (one the plane of the molecule and one perpendicular to the plane of the molecule and perpendicular to Page 57 the C–C bond). C2v. trans-1,2-dichloroethylene has a C2 axis perpendicular to the C–C bond and perpendicular to the plane of the molecule σh a mirror plane in the plane of the molecule, and an inversion center. C2h. 3. Distinguish between the concepts of (a) symmetry operation, (b) symmetry element, and (c) symmetry point group. (a) Symmetry operation: an action, such as a rotation through a certain angle, that leaves the molecule apparently unchanged. (b) Symmetry element: a point, a line, or a plane with respect to which the symmetry operation is performed. (c) Symmetry point group: All of the operations that leave at least one point of the molecule unmoved. 4. For the following molecules and ions, provide the Lewis structure, VSEPR geometry, and assign the point group (the central atom(s) are underlined) and write symmetry operations. 2– – SF5Cl (C4v), BeH2 (D∞h), OCS (C∞v), SO4 (Td), CH3F (C3v), NO2F (C2v), IF4 (D4h), 2– – [TeCl6] (Oh), NNO (C∞v) , NHF2 (Cs), XeF6 (Oh), XeO3 (C3v), TeF6 (Oh), [BrBr2] – (D∞h), BrNO (Cs), IF5 (C4v), OSO (C2v), ICl3 (C2v), SeCl4 (C2v), KrF2 (D∞h), SCN (C∞v) 6. Write point groups and symmetry operations of followings a) NH3; b) SF6; c) CCl4; d) H2C=CH2; e) H2C=CF2. Page 58  UNIT 3: X-RAY DIFFRACTION METHODS CONTENTS: 3.1 Objectives 3.2 Introduction 3.3 Bragg condition 3.3.1 Derivation of Bragg’s equation 3.4 Miller indices 3.5 Method of X-ray structural analysis of crystals 3.5.1 Laue’s method 3.5.2 Bragg’s method 3.5.3 Debye- Scherrer method 3.6 Procedure for an X-ray structure analysis 3.7 Procedure for absolute configuration of molecules 3.8 Ramchandran diagram 3.9 Electron Diffraction: Measurement technique 3.9.1 Scattering intensity vs scattering angle 3.10 Wierl equation 3.11 Elucidation of structure of simple gas phase molecules 3.12 Summary 3.13 Self-Assessment Questions (SAQ) 3.14 Glossary Page 59 3.15 Answers to SAQ’s 3.16 References 3.17 Terminal Questions 3.1 OBJECTIVES After reading this unit, we will able to: Explain the concept of X-ray diffraction. Define Bragg’s law. Describe Bragg’s method of determining the structure of crystals. Describe Laue’s method of determining the structure of crystals. Explain Debye- Scherrer method of determining the structure of crystals. Give the concept of Ramchandran diagram or plots. Write Wierl equation. Know about scattering intensity and scattering angle. Give the application of electron diffraction. 3.2 INTRODUCTION In lower classes, we have discussed about the electromagnetic radiations. In these radiation, electric field and magnetic field are present and possess speed as the speed of light (3 × 108 m/s). X-ray is one of the electromagnetic radiation that possess energy of 9.0 × 103 kJ/mol. X-Rays have wavelengths in the range of 0.01 to 10 nanometers, corresponding to frequencies in the range 30 petahertz to 30 exahertz (3×1016 Hz to 3×1019 Hz) Now moving to diffraction, it is defined as the bending (slight) of light when it passes around the edge of an object. The phenomenon of diffraction is shown by electromagnetic radiations like X-rays, gamma rays and microscopic particles like electron, neutron. In this unit we will discuss about X-ray and its various method of determining the structure of crystals by using the concept of X-ray diffraction that involves Laue’s method, Bragg’s method, Page 60 Debye- Scherrer method. In addition to these methods, we will also discuss about the procedure for an X-ray structure analysis. In order to understand the three dimensional structure of protein in a clear way, Ramchandran diagram or plots were taken into consideration. General Introduction of Electron Diffraction that includes scattering intensity, scattering angle, Wierl equation, and measurement technique molecules are also discussed in this unit. The entire unit will be clearer after going through each and every topic in detail given below. After reading the unit we will came to know that how on passing X-ray on the surface of crystals we can determines the structure of the crystal. Let us now discuss these topics one by one. 3.3 X- Ray and BRAGG’S CONDITION Before discussing the Bragg’s condition, let us have some general idea of X-rays. X-ray as we know are the electromagnetic radiations having wavelength of 0.1 to 150oA. X rays are produced whenever high-speed electrons collide with a metal target in a vacuum tube. A hot tungsten filament is source of electrons. A current is passed through the tungsten filament and heats it up. As it is heated up the increased energy enables electrons to be released from the filament (thermionic emitter). High voltage (in kilo volts) is applied between cathode and anode. A small increase in voltage causes large increase in tube current which accelerates high speed electrons from negative cathode ( tungsten filament cathode ) to positive anode ( target tungsten metal anode). The electrons are attracted towards the positively charged anode, the tungsten target with a maximum energy (Energy can be controlled or determined by the tube potential). Figure 1: X-ray Tube Crossection Ref: Univ of Queensland Physics Museum Page 61 When an high energy electron collides with electron present in the inner shell of tungsten then both are ejected out from tungsten atom leaving a hole in inner shell which is filled by electron in outer shell accompanying loss of energy emitted out as an X-ray photon. More than 99% of the electron energy is converted into heat and less than 1% of energy is converted into x-rays. When high-energy electron interacts with K-shell electron leads to the production of X-ray photons, K characteristic radiation.. When high-energy electron interacts with nucleus i.e. it passes near nucleus it is slowed and its path is deflected. Energy lost is emitted as bremsstrahlung X ray photon leads to bremsstrahlung radiations. When high-energy electron interacts with outer shell electrons causes line spectrum Figure 2 : X-ray Spectrum Ref :Wikibooks, Basic Physics of Digital Radiography X-ray diffraction method is used for determining the structure of a molecule. In X-ray diffraction method, X-rays are allowed to pass through the crystal there is interaction between X-rays and atoms of the crystal. The atomic planes of crystal are responsible for interference of incident beam of X-rays with one another when they leave the crystal. So there is scattering of X-rays (electromagnetic radiations). These emitted radiations either overlap with each other to form strong peaks in which bright spots (constructive interference) are developed on the photographic Page 62 plate or get subtracted from each other resulting in dark spot (destructive interference) on the photographic plate (interference). This results in diffraction pattern. This is X-ray diffraction or Bragg’s diffraction. From this diffraction pattern we can determine the distribution of electrons (position of particles), nature of interactions within the molecule. Study of X ray diffraction pattern helps to measure the average spacing between layers or rows of atoms. We can measure the size, shape and internal stress of small crysta