by

Dr.D.UTHRAHead

Department of PhysicsDG Vaishnav College

Chennai-106

Construction of Symmetry Coordinates

for XY2 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 XY2 bent molecule

N=3 ; 3N-6 = 3 modes of vibration For an XY3 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 (Δr ) angle deformation (ΔΦ) torsion (Δτ ) out of plane deformation (ΔΦ')

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 moleculewhere, 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.

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

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

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

Sj = ∑k Ujkrk

not necessarily unique

How are Symmetry Coordinates formed?

Generated by Projection operator method,S i

j = ∑R χi(R)RLkR- 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 operationson 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 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!!

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 thevarious normal modes of vibration(calculated with 3N-6, then checkedwith internal coordinates) aredistributed among various species.

C2v E C2 σv σv ’

A1 1 1 1 1

A2 1 1 -1 -1

B1 1 -1 1 -1

B2 1 -1 -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 Lk, find RLk.Step 2 : Use character table and find the product χi(R)RLk for

every RStep 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

S ij = ∑R χi(R)RLk

In case of XY2 bent molecule

Internal coordinates (Lks) are Δd1 , Δd Δd1, ΔαSymmetry 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 speciesS3 = ?

C2v E C2 σv σv ’

A1 1 1 1 1

A2 1 1 -1 -1

B1 1 -1 1 -1

B2 1 -1 -1 1

Γvib = 2A1 + B2 = 3

Generating the symmetry coordinate with Δd1 as Lk

C2v E C2 σv σv ’

A1 Δd1 Δd2 Δd2 Δd1

A2

B1

B2

After operartion

R Lk R Lk

E :

C2 :

σ v :

σv ’ :

C2v E C2 σv σv ’

A1 1 1 1 1

A2 1 1 -1 -1

B1 1 -1 1 -1

B2 1 -1 -1 1

Δd1Δd1

Δd1 Δd2

Δd2Δd1

Δd1Δd1

S i = ∑R χi(R)RLk

Find summation of the χi(R)RLkentries for every R

C2v E C2 σv σv ’

A1Δd1 Δd2 Δd2 Δd1

A2Δd1 Δd2 -Δd2 -Δd1

B1Δd1 - Δd2 Δd2 -Δd1

B2Δd1 - Δd2 -Δd2 Δd1

Now check your filled uptable having χi(R)RLk entries

Sd1A1= Δd1+Δd2+Δd2+Δd1 =

2(Δd1+Δd2) Sd1

A2= Δd1+ Δd2- Δd2- Δd1 = 0 Sd1

B1= Δd1- Δd2+ Δd2- Δd1 = 0

Sd1B2= Δd1- Δd2- Δd2+ Δd1 = 2(Δd1-

Δd2)

From above, you get

SA1 = Δd1+Δd2

SB2 = Δd1 - Δd2

Generating the symmetry coordinate 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 ’

A1Δd2 Δd1 Δd1 Δd2

A2Δd2 Δd1 -Δd1 -Δd2

B1Δd2 -Δd1 Δd1 -Δd2

B2Δd2 -Δd1 -Δd1 Δd2

Similarly generate the symmetry

coordinates with Δd2 as Lk

Generating the symmetry coordinate with Δα 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 Δα Δα

Δα

A2 Δα Δα

-Δα

-Δα

B1 Δα -Δα

Δα

-Δα

B2 Δα -Δα

-Δα

Δα

Now generate the symmetry coordinates with Δ α as Lk

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

Condition to normalise and orthogonalise SALCs

For normalisationFor singly degenerate species,

Uak2 = 1/q or Uak = ±(1/q)½

For doubly degenerate species,Uak

2 + Ubk2 = 2/q

For triply degenerate species,Uak

2 + Ubk2 + Uck

2 = 3/q

q - total no.of symmetry equivalent internal coordinates involved in that SALC

For orthogonalisation∑k UakUbk =0

For XY2 bent molecule

All three SALCs belong to singly degenerate species. Applying first condition,

In SA1 = Δd1+Δd2 and SB2 = Δd1 - Δd2,

q=2SA1 = Δα

q=1

Obtain Orthonormal SALCs

After Normalisation SA1 = 1/√2 (Δd1+ Δd2) SB2 = 1/√2 (Δd1 - Δd2)

SA1 = Δα

Next, check for orthogonality

Orthonormalised Symmetry Coordinates of XY2 bent molecule

3N-6 = 3 distributed asΓvib. = 2A1 + B2 = 3

A1 Species S1

A1 = 1/√2 (Δd1+ Δd2)

S2A1 = Δα

B2 Species S3

B2 = 1/√2 (Δd1 - Δd2)

RecapFor 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 moleculebelongs 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

NormalisationA1 SpeciesS1 = 1/√2 (Δd1+ Δd2)S2 = ΔαB2 SpeciesS3 = 1/√2 (Δd1 - Δd2)

All the Best ! uthra mam

Hope You enjoyed learning to form symmetry coordinates !

See U in the next session of group theory!