Presentation on theme: "Construction of Symmetry Coordinates for XY2 bent type molecules"— Presentation transcript:
1 Construction of Symmetry Coordinates for XY2 bent type molecules by Dr.D.UTHRA Head Department of Physics DG Vaishnav College Chennai-106
2 This presentation has been designed to serve as a self- study material for Postgraduate Physics students pursuing their programme under Indian Universities, especially University of Madras and its affiliated colleges. If this aids the teachers too who deal this subject, to make their lectures more interesting, the purpose is achieved D.Uthra
3 I acknowledge my sincere gratitude to my teacher Dr.S.Gunasekaran, for teaching me group theory withso much dedication and patience & for inspiring meand many of my friends to pursue research.- D.Uthra
4 Pre-requisites, before to you get on Knowledge of symmetry elements and symmetry operationsKnowledge of various point groupsLook-up of character tablesIdea of the three dimensional structure of the molecule chosen
5 Normal modes of vibration of a molecule 3N-6 degrees of freedom are required to describe vibration of a non-linear molecule.3N-5 degrees of freedom are required to describe vibration of a linear molecule.Hence,For an XY2 bent moleculeN=3 ; 3N-6 = 3 modes of vibrationFor an XY3 pyramidal moleculeN=4 ; 3N-6 = 6 modes of vibration
6 Internal CoordinatesChanges in the bond lengths and in the inter-bond angles.Types of internal coordinates and their notationbond stretching (Δr )angle deformation (ΔΦ)torsion (Δτ )out of plane deformation (ΔΦ')
7 How to calculate number of vibrations using internal coordinates? nr = bnΦ= 4b-3a+a1nτ = b-a1nΦ’= no.of linear subsectionsin a moleculewhere,b - no.of bonds (ignoring the type of bond)a - no.of atomsa1- no.of atoms with multiplicity oneMultiplicity means the number of bonds meeting the atom , ignoring the bond type.In H2O, multiplicity of H is 1 and that of O is 2. In NH3, multiplicity of N is 3, while H is 1.In C2H2, it is 1 for H and 2 for C (not 4 b’coz you ignore bond type).In CH4, it is 1 for H and 4 for C. In CO2, it is 2 for C and 1 for O.
8 Let us calculate the number of vibrations for a XY2 bent molecule In XY2 bent molecule,b = 2 (no.of bonds)a = 3 (no.of atoms)a1= 2 (no.of atoms with multiplicity one)Hence,nr = 2nΦ = 4*2-3*3+2 =1nτ = 2-2 = 03N-6 = nr+nΦ+nτ = 2+1= 3
9 Let us calculate the number of vibrations for a XY3 bent molecule In XY3 bent molecule,b = 3 (no.of bonds)a = 4 (no.of atoms)a1= 3 (no.of atoms with multiplicity one)Hence,nr = 3nΦ = 4*3-3*4+3 =3nτ = 3-3 = 03N-6 = nr+nΦ+nτ = 3+3=6
10 Symmetry CoordinatesSymmetry coordinates describe the normal modes of vibration of the moleculesThey are the linear combinations of the internal coordinates related to various vibrations of the molecule
11 Basic properties of Symmetry Coordinates must be normalisedmust be orthogonal in the given specieshas the formSj = ∑k Ujkrknot necessarily unique
12 How are Symmetry Coordinates formed? Generated by Projection operator method,S ij = ∑R χi(R)RLkR- symmetry operation of the point groupχi(R)-character of the species (A1,A2,B1,etc) covering the symmetry operation R (C2, C3,σ,etc)Lk-generating coordinate( may be single internal coordinate or some linear combination of internal coordinates)RLk-new coordinate which Lk becomes, after the symmetry operation is performedYou are projecting the operator, that is doing symmetry operationson various internal coordinates of the chosen molecule, find what happensto those internal coordinates after projection and sum up the result!
13 Caution : Brush up your knowledge and then proceed To proceed further…You need to have the knowledge ofStructure of the chosen moleculeSymmetry operations it undergoesPoint group symmetry it belongs toCharacter table of that point group symmetryCaution : Brush up your knowledge and then proceed
14 Point Group and Symmetry Coordinates Identify the point group symmetry of the molecule.By using its Character table, find the distribution of vibrations among its various species , ie find Γvib.Find symmetry coordinates defining each vibration! Do you realise the importance of understanding symmetryoperations and point groups and place the molecules correctly in theirpoint group symmetry. If You are incorrect in finding the point groupsymmetry of the molecule, things will not turn out correctly!!
15 XY2 bent moleculeXY2 bent molecule belongs to C2v point group symmetryΓvib = 2A1 + B2 = 3Hence, we need 3 symmetry coordinatesThis gives you an idea how thevarious normal modes of vibration(calculated with 3N-6, then checkedwith internal coordinates) aredistributed among various species.C2vEC2σvσv ’A11A2-1B1B2
16 How to arrive at the symmetry coordinates? As seen earlier, these are generated by Projectionoperator method,Step 1 : For every internal coordinate Lk, find RLk.Step 2 : Use character table and find the product χi(R)RLk for every RStep 3 : Sum them upBefore that, keep a note ofinternal coordinates of the chosen moleculewhat changes take place in them after every symmetry operationcharacter table to which the molecule belongs toS ij = ∑R χi(R)RLk
17 S ij = ∑R χi(R)RLk In case of XY2 bent molecule Γvib = 2A1 + B2 = 3 Internal coordinates (Lks) are Δd1 , Δd Δd1, ΔαSymmetry operations (Rs) are E, C2 , σv , σv‘ andtheir respective χis are 1 or -1 (from character table ),while i = A1, A2, B1,B2In A1 speciesS1 = ?S2 = ?In B2 speciesS3 = ?C2vEC2σvσv ’A11A2-1B1B2Γvib = 2A1 + B2 = 3
18 Generating the symmetry coordinate with Δd1 as Lk After operartionR Lk R LkE :C2 :σ v :σv ’ :C2vEC2σvσv ’A11A2-1B1B2Δd1Δd1d1d2Δd1Δd2Δd1Δd2C2vEC2σvσv ’A1Δd1Δd2A2B1B2Δd1Δd1
19 S i = ∑R χi(R)RLk C2v E C2 σv σv ’ A1 A2 B1 B2 SB2 = Δd1 - Δd2 Now check your filled uptable having χi(R)RLk entriesFind summation of the χi(R)RLkentries for every RC2vEC2σvσv ’A1Δd1Δd2A2-Δd2-Δd1B1- Δd2B2Sd1A1= Δd1+Δd2+Δd2+Δd1 = 2(Δd1+Δd2)Sd1A2= Δd1+ Δd2- Δd2- Δd1 = 0Sd1B1= Δd1- Δd2+ Δd2- Δd1 = 0Sd1B2= Δd1- Δd2- Δd2+ Δd1 = 2(Δd1-Δd2)From above, you getSA1 = Δd1+Δd2SB2 = Δd1 - Δd2
20 Generating the symmetry coordinate with Δd2 as Lk Similarly generate the symmetrycoordinates with Δd2 as LkS i = ∑R χi(R)RLkSd2A1=Δd2+Δd1+Δd1+Δd2 =2(Δd2+Δd1)Sd2A2= Δd2+Δd1-Δd1-Δd2 = 0Sd2B1= Δd2-Δd1+Δd1-Δd2 = 0Sd2B2=Δd2-Δd1-Δd1+Δd2 = 2(Δd2–Δd1)SA1 = Δd1+ Δd2SB2 = Δd2–Δd1=-(Δd1-Δd2)C2vEC2σvσv ’A1Δd2Δd1A2-Δd1-Δd2B1B2
21 Generating the symmetry coordinate with Δα as Lk Now generate the symmetrycoordinates with Δ α as LkS i = ∑R χi(R)RLkSαA1 = 4 ΔαSαA2 = 0SαB1 = 0SαB2 = 0SA1 = ΔαC2vEC2σvσv ’A1ΔαA2-ΔαB1B2
22 SALCS- Symmetry Adapted Linear Combinations By projection operator method, employingSALCs for XY2 bent molecule are found to beSA1 = Δd1+Δd2SB2 = Δd1 - Δd2SA1 = ΔαS ij = ∑R χi(R)RLk
23 Condition to normalise and orthogonalise SALCs For normalisationFor singly degenerate species,Uak2 = 1/q or Uak = ±(1/q)½For doubly degenerate species,Uak2 + Ubk2 = 2/qFor triply degenerate species,Uak2 + Ubk2 + Uck2 = 3/qq - total no.of symmetry equivalent internal coordinates involved in that SALCFor orthogonalisation∑k UakUbk =0
24 For XY2 bent moleculeAll three SALCs belong to singly degenerate species.Applying first condition,In SA1 = Δd1+Δd2 and SB2 = Δd1 - Δd2,q=2SA1 = Δαq=1
25 Obtain Orthonormal SALCs After NormalisationSA1 = 1/√2 (Δd1+ Δd2)SB2 = 1/√2 (Δd1 - Δd2)SA1 = ΔαNext, check for orthogonality
27 Recap For an XY2 bent molecule N=3 ; 3N-6 = 3 modes of vibration In XY2 bent molecule,b = 2a = 3a1= 2Hence,nr = 2nΦ = 4*2-3*3+2 =1nτ = 2-2 = 03N-6 = 2nr+1nΦ = 3XY2 bent moleculebelongs to C2v point group symetry.Γvib. = 2A1 + B2 = 3In case of XY2 bent moleculeLks are Δd1 , Δd2 , ΔαRs are E, C2 , σv , σv‘ and their respectiveχis are 1 or -1 (from character table ), whilei = A1, A2, B1,B2Γvib. = 2A1 + B2 = 3 impliesCheck for orthogonalityNormalisationA1 SpeciesS1 = 1/√2 (Δd1+ Δd2)S2 = ΔαB2 SpeciesS3 = 1/√2 (Δd1 - Δd2)
28 All the Best ! uthra mam Hope You enjoyed learning to form symmetry coordinates !See U in the next session of group theory!