Presentation on theme: "By Dr.D.UTHRA Head Department of Physics DG Vaishnav College Chennai-106 Construction of Symmetry Coordinates for XY 2 bent type molecules."— Presentation transcript:
by Dr.D.UTHRA Head Department of Physics DG Vaishnav College Chennai-106 Construction of Symmetry Coordinates for XY 2 bent type molecules
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
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
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
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 XY 2 bent molecule N=3 ; 3N-6 = 3 modes of vibration For an XY 3 pyramidal molecule N=4 ; 3N-6 = 6 modes of vibration Normal modes of vibration of a molecule
Internal Coordinates Changes in the bond lengths and in the inter-bond angles. Types of internal coordinates and their notation bond stretching ( Δ r ) angle deformation ( Δ Φ) torsion ( Δτ ) out of plane deformation ( Δ Φ')
How to calculate number of vibrations using internal coordinates? n r = b n Φ = 4b-3a+a 1 n τ = b-a 1 n Φ’ = no.of linear subsections in a molecule where, b - no.of bonds (ignoring the type of bond) a - no.of atoms a 1 - no.of atoms with multiplicity one Multiplicity means the number of bonds meeting the atom, ignoring the bond type. In H 2 O, multiplicity of H is 1 and that of O is 2. In NH 3, multiplicity of N is 3, while H is 1. In C 2 H 2, it is 1 for H and 2 for C (not 4 b’coz you ignore bond type). In CH 4, it is 1 for H and 4 for C. In CO 2, it is 2 for C and 1 for O.
Let us calculate the number of vibrations for a XY 2 bent molecule In XY 2 bent molecule, b = 2 (no.of bonds) a = 3 (no.of atoms) a 1 = 2 (no.of atoms with multiplicity one) Hence, n r = 2 n Φ = 4*2-3*3+2 =1 n τ = 2-2 = 0 3N-6 = n r +n Φ +n τ = 2+1= 3
Let us calculate the number of vibrations for a XY 3 bent molecule In XY 3 bent molecule, b = 3 (no.of bonds) a = 4 (no.of atoms) a 1 = 3 (no.of atoms with multiplicity one) Hence, n r = 3 n Φ = 4*3-3*4+3 =3 n τ = 3-3 = 0 3N-6 = n r +n Φ +n τ = 3+3=6
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
Basic properties of Symmetry Coordinates must be normalised must be orthogonal in the given species has the form S j = ∑ k U jk r k not necessarily unique
How are Symmetry Coordinates formed? Generated by Projection operator method, S i j = ∑ R χ i (R)RL k R- symmetry operation of the point group χ i (R)-character of the species (A 1,A 2,B 1,etc) covering the symmetry operation R (C 2, C 3,σ,etc) L k -generating coordinate( may be single internal coordinate or some linear combination of internal coordinates) RL k -new coordinate which L k 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!
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
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 Γ vib. 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!!
XY 2 bent molecule XY 2 bent molecule belongs to C 2v point group symmetry Γ vib = 2A 1 + B 2 = 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. C 2v EC2C2 σvσv σv ’σv ’ A1A1 1111 A2A2 11 B1B1 1 1 B2B2 1 1
How to arrive at the symmetry coordinates? As seen earlier, these are generated by Projection operator method, Step 1 : For every internal coordinate L k, find RL k. Step 2 : Use character table and find the product χ i (R)RL k 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 i j = ∑ R χ i (R)RL k
In case of XY 2 bent molecule Internal coordinates (L k s) are Δd 1, Δd Δd 1, Δα Symmetry operations (Rs) are E, C 2, σ v, σ v ‘ and their respective χ i s are 1 or -1 (from character table ), while i = A 1, A 2, B 1,B 2 In A 1 species S 1 = ? S 2 = ? In B 2 species S 3 = ? C 2v EC2C2 σvσv σv ’σv ’ A1A1 1111 A2A2 11 B1B1 1 1 B2B2 1 1 Γ vib = 2A 1 + B 2 = 3
Generating the symmetry coordinate with Δd 1 as L k C 2v EC2C2 σvσv σv ’σv ’ A1A1 Δd1Δd1 Δd 2 Δd1Δd1 A2A2 B1B1 B2B2 After operartion R L k E : C 2 : σ v : σ v ’ : C 2v EC2C2 σvσv σv ’σv ’ A1A1 1 1 1 1 A2A2 1 1 B1B1 1 1 B2B2 1 1 Δd1 Δd2 Δd1
S i = ∑ R χ i (R)RL k Find summation of the χi(R)RLk entries for every R C 2v EC2C2 σvσv σv ’σv ’ A1A1 Δd1Δd1 Δd2Δd2 Δd 2 Δd1Δd1 A2A2 Δd1Δd1 Δd2Δd2 -Δd2-Δd2 -Δd1-Δd1 B1B1 Δd1Δd1 - Δd 2 Δd 2 -Δd1-Δd1 B2B2 Δd1Δd1 - Δd 2 -Δd2-Δd2 Δd1Δd1 Now check your filled up table having χ i (R)RL k entries S d1 A 1 = Δd 1 +Δd 2 +Δd 2 +Δd 1 = 2(Δd 1 +Δd 2 ) S d1 A 2 = Δd 1 + Δd 2 - Δd 2 - Δd 1 = 0 S d1 B 1 = Δd 1 - Δd 2 + Δd 2 - Δd 1 = 0 S d 1 B 2 = Δd 1 - Δd 2 - Δd 2 + Δd 1 = 2(Δd 1 -Δd 2 ) From above, you get S A 1 = Δd 1 +Δd 2 S B 2 = Δd 1 - Δd 2
Generating the symmetry coordinate with Δd 2 as L k S i = ∑ R χ i (R)RL k S d2 A 1 =Δd 2 +Δd 1 +Δd 1 +Δd 2 =2(Δd 2 +Δd 1 ) S d2 A 2 = Δd 2 +Δd 1 -Δd 1 -Δd 2 = 0 S d2 B 1 = Δd 2 -Δd 1 +Δd 1 -Δd 2 = 0 S d2 B 2 =Δd 2 -Δd 1 -Δd 1 +Δd 2 = 2(Δd 2 –Δd 1 ) S A 1 = Δd 1 + Δd 2 S B 2 = Δd 2 –Δd 1 =-(Δd 1 -Δd 2 ) C 2v EC2C2 σvσv σv ’σv ’ A1A1 Δd 2 Δd 1 Δd 2 A2A2 Δd 1 -Δd 1 -Δd 2 B1B1 Δd 2 -Δd 1 Δd 1 -Δd 2 B2B2 Δd 2 -Δd 1 Δd 2 Similarly generate the symmetry coordinates with Δd 2 as L k
Generating the symmetry coordinate with Δα as L k S i = ∑ R χ i (R)RL k S α A 1 = 4 Δα S α A 2 = 0 S α B 1 = 0 S α B 2 = 0 S A 1 = Δα C 2v E C 2 σv σv σv ’ σv ’ A1A1 ΔαΔα Δ α A2A2 -Δα B1B1 Δα-Δα Δα-Δα B2B2 Δα-Δα Δα Now generate the symmetry coordinates with Δ α as L k
SALCS- Symmetry Adapted Linear Combinations By projection operator method, employing SALCs for XY 2 bent molecule are found to be S A1 = Δd 1 +Δd 2 S B2 = Δd 1 - Δd 2 S A1 = Δα S i j = ∑ R χ i (R)RL k
Condition to normalise and orthogonalise SALCs For normalisation For singly degenerate species, U ak 2 = 1/q or U ak = ±(1/q) ½ For doubly degenerate species, U ak 2 + U bk 2 = 2/q For triply degenerate species, U ak 2 + U bk 2 + U ck 2 = 3/q q - total no.of symmetry equivalent internal coordinates involved in that SALC For orthogonalisation ∑ k U ak U bk =0
For XY 2 bent molecule All three SALCs belong to singly degenerate species. Applying first condition, In S A1 = Δd 1 +Δd 2 and S B2 = Δd 1 - Δd 2, q=2 S A1 = Δα q=1
Obtain Orthonormal SALCs After Normalisation S A1 = 1/√2 (Δd 1 + Δd 2 ) S B2 = 1/√2 (Δd 1 - Δd 2 ) S A1 = Δα Next, check for orthogonality
Orthonormalised Symmetry Coordinates of XY 2 bent molecule 3N-6 = 3 distributed as Γ vib. = 2A 1 + B 2 = 3 A 1 Species S 1 A1 = 1/√2 (Δd 1 + Δd 2 ) S 2 A1 = Δα B 2 Species S 3 B2 = 1/√2 (Δd 1 - Δd 2 )
Recap For an XY 2 bent molecule N=3 ; 3N-6 = 3 modes of vibration In XY 2 bent molecule, b = 2 a = 3 a 1 = 2 Hence, n r = 2 n Φ = 4*2-3*3+2 =1 n τ = 2-2 = 0 3N-6 = 2n r +1n Φ = 3 XY 2 bent molecule belongs to C 2v point group symetry. Γ vib. = 2A 1 + B 2 = 3 In case of XY 2 bent molecule L k s are Δd 1, Δd 2, Δα Rs are E, C 2, σ v, σ v ‘ and their respective χ i s are 1 or -1 (from character table ), while i = A 1, A 2, B 1,B 2 Γ vib. = 2A 1 + B 2 = 3 implies Check for orthogonality Normalisation A 1 Species S 1 = 1/√2 (Δd 1 + Δd 2 ) S 2 = Δα B 2 Species S 3 = 1/√2 (Δd 1 - Δd 2 )
All the Best ! uthra mam Hope You enjoyed learning to form symmetry coordinates ! See U in the next session of group theory!