Download presentation

Published byColt Hansberry Modified over 2 years ago

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 with so much dedication and patience & for inspiring me and many of my friends to pursue research. - D.Uthra

4
**Pre-requisites, before to you get on**

Knowledge of symmetry elements and symmetry operations Knowledge of various point groups Look-up of character tables Idea 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 molecule N=3 ; 3N-6 = 3 modes of vibration For an XY3 pyramidal molecule N=4 ; 3N-6 = 6 modes of vibration

6
Internal Coordinates Changes in the bond lengths and in the inter-bond angles. Types of internal coordinates and their notation bond stretching (Δr ) angle deformation (ΔΦ) torsion (Δτ ) out of plane deformation (ΔΦ')

7
**How to calculate number of vibrations using internal coordinates?**

nr = b nΦ= 4b-3a+a1 nτ = b-a1 nΦ’= no.of linear subsections in a molecule where, b - no.of bonds (ignoring the type of bond) a - no.of atoms a1- no.of atoms with multiplicity one Multiplicity 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 = 2 nΦ = 4*2-3*3+2 =1 nτ = 2-2 = 0 3N-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 = 3 nΦ = 4*3-3*4+3 =3 nτ = 3-3 = 0 3N-6 = nr+nΦ+nτ = 3+3=6

10
Symmetry Coordinates Symmetry coordinates describe the normal modes of vibration of the molecules They are the linear combinations of the internal coordinates related to various vibrations of the molecule

11
**Basic properties of Symmetry Coordinates**

must be normalised must be orthogonal in the given species has the form Sj = ∑k Ujkrk not necessarily unique

12
**How are Symmetry Coordinates formed?**

Generated by Projection operator method, S ij = ∑R χi(R)RLk R- symmetry operation of the point group χi(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 performed You are projecting the operator, that is doing symmetry operations on various internal coordinates of the chosen molecule, find what happens to 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 of Structure of the chosen molecule Symmetry operations it undergoes Point group symmetry it belongs to Character table of that point group symmetry Caution : 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 symmetry operations and point groups and place the molecules correctly in their point group symmetry. If You are incorrect in finding the point group symmetry of the molecule, things will not turn out correctly!!

15
XY2 bent molecule XY2 bent molecule belongs to C2v point group symmetry Γvib = 2A1 + B2 = 3 Hence, we need 3 symmetry coordinates This gives you an idea how the various normal modes of vibration (calculated with 3N-6, then checked with internal coordinates) are distributed among various species. C2v E C2 σv σv ’ A1 1 A2 -1 B1 B2

16
**How to arrive at the symmetry coordinates?**

As seen earlier, these are generated by Projection operator method, Step 1 : For every internal coordinate Lk, find RLk. Step 2 : Use character table and find the product χi(R)RLk for every R Step 3 : Sum them up Before that, keep a note of internal coordinates of the chosen molecule what changes take place in them after every symmetry operation character table to which the molecule belongs to S 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, Δα Symmetry operations (Rs) are E, C2 , σv , σv‘ and their respective χis are 1 or -1 (from character table ), while i = A1, A2, B1,B2 In A1 species S1 = ? S2 = ? In B2 species S3 = ? C2v E C2 σv σv ’ A1 1 A2 -1 B1 B2 Γvib = 2A1 + B2 = 3

18
**Generating the symmetry coordinate with Δd1 as Lk**

After operartion R Lk R Lk E : C2 : σ v : σv ’ : C2v E C2 σv σv ’ A1 1 A2 -1 B1 B2 Δd1 Δd1 d1 d2 Δd1 Δd2 Δd1 Δd2 C2v E C2 σv σv ’ A1 Δd1 Δd2 A2 B1 B2 Δd1 Δd1

19
**S i = ∑R χi(R)RLk C2v E C2 σv σv ’ A1 A2 B1 B2 SB2 = Δd1 - Δd2**

Now check your filled up table having χi(R)RLk entries Find summation of the χi(R)RLk entries for every R C2v E C2 σv σv ’ A1 Δd1 Δd2 A2 -Δd2 -Δd1 B1 - Δd2 B2 Sd1A1= Δd1+Δd2+Δd2+Δd1 = 2(Δd1+Δd2) Sd1A2= Δd1+ Δd2- Δd2- Δd1 = 0 Sd1B1= Δd1- Δd2+ Δd2- Δd1 = 0 Sd1B2= Δd1- Δd2- Δd2+ Δd1 = 2(Δd1-Δd2) From above, you get SA1 = Δd1+Δd2 SB2 = Δd1 - Δd2

20
**Generating the symmetry coordinate with Δd2 as Lk**

Similarly generate the symmetry coordinates with Δd2 as Lk S i = ∑R χi(R)RLk Sd2A1=Δd2+Δd1+Δd1+Δd2 =2(Δd2+Δd1) Sd2A2= Δd2+Δd1-Δd1-Δd2 = 0 Sd2B1= Δd2-Δd1+Δd1-Δd2 = 0 Sd2B2=Δd2-Δd1-Δd1+Δd2 = 2(Δd2–Δd1) SA1 = Δd1+ Δd2 SB2 = Δd2–Δd1 =-(Δd1-Δd2) C2v E C2 σv σv ’ A1 Δd2 Δd1 A2 -Δd1 -Δd2 B1 B2

21
**Generating the symmetry coordinate with Δα as Lk**

Now generate the symmetry coordinates with Δ α as Lk S i = ∑R χi(R)RLk SαA1 = 4 Δα SαA2 = 0 SαB1 = 0 SαB2 = 0 SA1 = Δα C2v E C2 σv σv ’ A1 Δα A2 -Δα B1 B2

22
**SALCS- Symmetry Adapted Linear Combinations**

By projection operator method, employing SALCs for XY2 bent molecule are found to be SA1 = Δd1+Δd2 SB2 = Δd1 - Δd2 SA1 = Δα S ij = ∑R χi(R)RLk

23
**Condition to normalise and orthogonalise SALCs**

For normalisation For singly degenerate species, Uak2 = 1/q or Uak = ±(1/q)½ For doubly degenerate species, Uak2 + Ubk2 = 2/q For triply degenerate species, Uak2 + Ubk2 + Uck2 = 3/q q - total no.of symmetry equivalent internal coordinates involved in that SALC For orthogonalisation ∑k UakUbk =0

24
For XY2 bent molecule All three SALCs belong to singly degenerate species. Applying first condition, In SA1 = Δd1+Δd2 and SB2 = Δd1 - Δd2, q=2 SA1 = Δα q=1

25
**Obtain Orthonormal SALCs**

After Normalisation SA1 = 1/√2 (Δd1+ Δd2) SB2 = 1/√2 (Δd1 - Δd2) SA1 = Δα Next, check for orthogonality

26
**Orthonormalised Symmetry Coordinates of XY2 bent molecule**

3N-6 = 3 distributed as Γvib. = 2A1 + B2 = 3 A1 Species S1A1 = 1/√2 (Δd1+ Δd2) S2A1 = Δα B2 Species S3B2 = 1/√2 (Δd1 - Δd2)

27
**Recap For an XY2 bent molecule N=3 ; 3N-6 = 3 modes of vibration**

In XY2 bent molecule, b = 2 a = 3 a1= 2 Hence, nr = 2 nΦ = 4*2-3*3+2 =1 nτ = 2-2 = 0 3N-6 = 2nr+1nΦ = 3 XY2 bent molecule belongs to C2v point group symetry. Γvib. = 2A1 + B2 = 3 In case of XY2 bent molecule Lks are Δd1 , Δd2 , Δα Rs are E, C2 , σv , σv‘ and their respective χis are 1 or -1 (from character table ), while i = A1, A2, B1,B2 Γvib. = 2A1 + B2 = 3 implies Check for orthogonality Normalisation A1 Species S1 = 1/√2 (Δd1+ Δd2) S2 = Δα B2 Species S3 = 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!

Similar presentations

Presentation is loading. Please wait....

OK

NORMAL COORDINATE ANALYSIS OF XY2 BENT MOLECULE – PART 3

NORMAL COORDINATE ANALYSIS OF XY2 BENT MOLECULE – PART 3

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Ppt on online shopping cart system Supply and demand elementary ppt on cells Ppt on preservation of public property auction Ppt on economy of india Sap procure to pay ppt online Ppt on brand marketing strategy Ppt on job rotation spreadsheet Electrical print reading ppt on ipad Ppt on phonetic transcription translator Ppt on information technology act