Transcript Construction of Sym. coordinates for XY2 molecule

Construction of Symmetry Coordinates
for XY2 bent type molecules
by
Dr.D.UTHRA
Head
Department of Physics
DG Vaishnav College
Chennai-106
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
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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
Normal modes of vibration of a molecule

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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,
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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
Internal Coordinates

Changes in the bond lengths and in the
inter-bond angles.
Types of internal coordinates and their notation
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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 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.
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,
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
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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,
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nr = 3
nΦ = 4*3-3*4+3 =3
nτ = 3-3 = 0
3N-6 = nr+nΦ+nτ = 3+3=6
Symmetry Coordinates
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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
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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 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!
To proceed further…
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You need to have the knowledge of
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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
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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 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
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,
S ij = ∑R χi(R)RLk
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
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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
In case of XY2 bent molecule
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
Γvib = 2A1 + B2 = 3
In
S1
S2
In
S3
A1 species
= ?
= ?
B2 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
Generating the symmetry coordinate with
Δd1 as Lk
After operartion
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
R
C2v
E
C2
σv
σv ’
A1
Δd1
Δd2
Δd2
Δd1
A2
B1
B2
Lk
R Lk
E :
Δd1
C2 :
Δd1
Δd2
σv:
Δd1
Δd2
σv ’ :
Δd1
Δd1
Δd1
S
i
= ∑R χi(R)RLk
Now check your filled up
table having χi(R)RLk entries
C2v
E
C2
σv
A1
A2
B1
B2
Δd1
Δd2
Δd2
Δd1
Δd2
-Δd2
-Δd1
Δd1
- Δd2
Δd2
-Δd1
Δd1
- Δd2
-Δd2
Find summation of the χi(R)RLk
entries for every R
σv ’
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)
Δd1
From above, you get
Δd1


SA1 = Δd1+Δd2
SB2 = Δd1 - Δd2
Generating the symmetry coordinate with
Δd2 as Lk
Similarly generate the symmetry
coordinates with Δd2 as Lk
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
C2v
E
C2
σv
σv ’
A1
A2
Δd2
Δd1
Δd1
Δd2
Δd2
Δd1
-Δd1
-Δd2
B1
Δd2
-Δd1
Δd1
-Δd2
B2
Δd2
-Δd1
-Δd1
Δd2



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)
Generating the symmetry coordinate with
Δα as Lk
Now generate the symmetry
coordinates with Δ α as Lk

A1
A2
Δα
Δα
C2
σv
σv ’
Δα
Δα Δα
Δα -Δα -Δα
B1
Δα -Δα
Δα -Δα
B2
Δα -Δα -Δα
Δα
= ∑R χi(R)RLk
i

SαA1 = 4 Δα
SαA2 = 0
SαB1 = 0
SαB2 = 0

SA = Δα

C2v E
S


1
SALCS- Symmetry Adapted Linear
Combinations
By projection operator method, employing
S ij = ∑R χi(R)RLk
SALCs for XY2 bent molecule are found to be
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
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SA1 = Δd1+Δd2
SB2 = Δd1 - Δd2
SA1 = Δα
Condition to normalise and
orthogonalise SALCs
For normalisation
For singly degenerate species,
2
½
Uak = 1/q or Uak = ±(1/q)
For doubly degenerate species,
2
2
Uak + Ubk = 2/q
For triply degenerate species,
2
2
2
Uak + Ubk + Uck = 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=2
SA1 = 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 as
Γvib. = 2A1 + B2 = 3

A1 Species
S1A1 = 1/√2 (Δd1+ Δd2)

S2A1 = Δα

B2 Species
S3B2 = 1/√2 (Δd1 - Δd2)
Recap
For an XY2 bent molecule
N=3 ; 3N-6 = 3 modes of
vibration

In XY2 bent molecule,
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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)
All the Best !
uthra mam
Hope You enjoyed learning to form
symmetry coordinates !
See U in the next session of group theory!