Symmetry

Report
Molecular Symmetry
Symmetry Elements
Group Theory
Photoelectron Spectra
Molecular Orbital (MO) Diagrams for Polyatomic Molecules
The Symmetry of Molecules



The shape of a molecule influences its physical properties,
reactivity, and its spectroscopic behavior
Determining the symmetry of a molecule is fundamental to gaining
insight into these characteristics of molecules
The chemist’s view of symmetry is contained in the study of group
theory



This branch of mathematics classifies the properties of a molecule into
groups, defined by the symmetry of the molecule
Each group is made up of symmetry elements or operations, which are
essentially quantum operators disguised as matrices
Our goal is to use group theory to build more complex molecular
orbital diagrams
CHEM 3722
Symmetry
2
Symmetry Elements

You encounter symmetry every day





A ball is spherically symmetric
Your body has a mirror image (the left and right side of your body)
Hermite polynomials are either even (symmetric on both sides of the
axis) or odd (symmetric with a twist).
Symmetry operations are movements of a molecule or object such
that after the movement the object is indistinguishable from its
original form
Symmetry elements are geometric representations of a point, line,
or plane to which the operation is performed





Identity element (E)
Plane of reflection (s)
Proper rotation (Cn)
Improper rotation (Sn)
Inversion (i)
CHEM 3722
Symmetry
3
Symmetry Elements II

Identity


If an object (O) has coordinates (x,y,z),
then the operation E(x,y,z)
(x,y,z)
The object is unchanged
H
H
H
B
B
H
H
E
H
H

H
H
B
B
H
H
H
Plane of reflection

s(xz) (x,y,z) = (x,-y,z)

s(xy) (x,y,z) = (x,y,-z)

s(yz) (x,y,z) = (-x,y,z)
s(xz)
z
y
s(xy)
x
CHEM 3722
Symmetry
4
Symmetry Elements III

Proper Rotation

Cn where n represents angle of rotation out of 360 degrees


C2 = 180°, C3 = 120°, C4 = 90° …
C2(z) (x,y,z) = (-x,-y,z)
6
C6
5
1
4
2
3
C6
1
How many C2
operations are there
for benzene?
5
6
2
5
3
C3
4
6
3
1
4
2
C6
4
C2
3
5
2
6
1
CHEM 3722
Symmetry
5
Symmetry Elements IV

Inversion
Takes each point through the center in a straight line to the exact distance on
the other side of center
i (x,y,z) = (-x,-y,-z)


1
4
6
2
3
5
5
3
2
6
4

1
Improper Rotation


A two step operation that first does a proper rotation and then a reflection
through a mirror plane perpendicular to the rotational axis.
S4(z) (x,y,z) = (y,-x,-z)


Same as (s)(C4) (x,y,z)
Note, symmetry operations are just quantum operators: work from right to left
1
6
2
5
3
4
CHEM 3722
6
sh C6 = S6
5
1
4
2
3
Symmetry
6
A Few To Try
Determine which symmetry elements are applicable for each of the
following molecules

O
H
H
H
H
N
H
Cl
Cl
Cl
Ru
Cl
CHEM 3722
Cl
Cl
O
Cl
Ru
Cl
Cl
Cl
Symmetry
7
Point Groups

We can
systematically
classify molecules
by their symmetry
properties


Special groups:
a) Cv,Dh (linear groups)
b) T, Th, Td, O, Oh, I, Ih
(1)
Start
(2)
No proper or improper axes
(3)
Call these point
groups
Use the flow
diagram to the
right
Only Sn (n = even) axis: S4, S6…
(4)
Cn axis
(5)
No C2’s  to Cn
CHEM 3722
n C2’s  to Cn
sh
n sv’s
No s’s
sh
n sd’s
No s’s
Cnh
Cnv
Cn
Dnh
Dnv
Dn
Symmetry
8
Some Common Groups
AB3

D3h Point Group


AB3

C3v Point Group


AB5





C2, 2C2’, 2C2’’, C4, S4, S42, 2sv, 2sd, sh, i
Square planar
C4v Point Group


CHEM 3722
C3, C32, 3C2, S3, S35, 3sv, sh
Trigonal bi-pyramid
C4h Point Group

AB5
C3, 3sv
Trigonal pyramid
D3h Point Group

AB4
C3, C32, 3C2, S3, S35, 3sv, sh
Trigonal planar
C2, C4, C42, 2sv, 2sd
Square pyramid
Symmetry
9
Character Tables

Character tables hold
the combined
symmetry and effects
of operations

For example,
consider water (C2v)
Point Group
Symmetry operations available
C 2v
E
C2
s ( xz )
s ( yz )
A1
1
1
1
1
z
x ,y ,z
A2
1
1
1
1
Rz
xy
B1
1
1
1
1
x, R y
xz
B2
1
1
1
1
y,Rx
yz
2
2
2
Effect of this operation
on an orbital of this
Coordinates and rotations
symmetry.
of this symmetry
Symmetry of states
A, B = singly degenerate
E = doubly degenerate
T = triply degenerate
1 = symmetric to C2 rotation*
2 = antisymmetric to C2 rotation*
CHEM 3722
Symmetry
10
The Oxygen’s px orbital

For water, we can look at any orbital and see which symmetry it is
by applying the operations and following the changes made to the
orbital



If it stays the same, it gets a 1
If it stays in place but gets flipped, -1
If it moves somewhere else, 0
z
H
O
x
H
y
E
C2
s (x z )
s (y z )
CHEM 3722
1
-1
1
C 2v
E
C2
s ( xz )
s ( yz )
A1
1
1
1
1
z
x ,y ,z
A2
1
1
1
1
Rz
xy
B1
1
1
1
1
x,R y
xz
B2
1
1
1
1
y,R x
yz
2
2
The px orbital
has B1 symmetry
-1
Symmetry
11
2
The Projection Operator

Since we want to build MO diagrams, our symmetry needs are
simple


Each point group has many possible symmetries for an orbital, and
thus we need a way to find which are actually present for the
particular molecule of that point group


Only orbitals of the same symmetry can overlap to form bonds
We’ll also look at collections of similar atoms and their collective orbitals
as a group
The projection operator lets us find the symmetry of any orbital or
collection of orbitals for use in MO diagrams



It will also be of use in determining the symmetry of vibrations, later
We just did this for the px orbital for water’s oxygen atom
l
It’s functional form is
j
j
ˆ
P    ( R ) Rˆ
h

j
But it’s easier to use than this appears
CHEM 3722
Symmetry
12
Ammonia

Let’s apply this mess to ammonia



First, draw the structure and determine the number of s and p bonds
Then look at how each bond changes for the group of hydrogen atoms,
building a set of “symmetry adapted linear combinations of
atomic orbitals” (SALC) to represent the three hydrogen's by
symmetry (not by their individual atomic orbitals)
Finally, we’ll compare these symmetries to those of the s and p atomic
orbitals of the nitrogen to see which overlap, thus building our MO
diagram from the SALC
z
N
H
H
x
H
y
3 sigma bonds and no pi bonds, thus we’ll build our
SALC’s from the projection operator and these 3 MO’s
Note, ammonia is in C3v point group (AB3)
CHEM 3722
Symmetry
13
Ammonia II


The Character Table
for C3v is to the right
E
2C 3
3s v
A1
1
1
1
z
B1
1
1
1
Rz
E
2
1
0
( x , y ), ( R x , R y )
x  y ,z
2
2
2
( x  y ), xy , xz , yz
2
2
Take the s bonds through the operations & see how many stay put
s3
H
H
s3
H
H
s3
H
H

C 3v
N s1
s2
H
C3
H
N s1
s2
H
s2
H
H
sv
N s1
s2
H
H
N s1
s2
s3
E
H
N s3
s1
s2
H
N s1
H
s3
H
H
(3)
(0)
(1)
We now have a representation (G) of this group of orbitals that
has the symmetry
Gs
CHEM 3722
E
2C 2
3s V
3
0
1
Symmetry
14
Ammonia III

This “reducible representation” of the hydrogen’s s-bonds must be a
sum of the symmetries available



C 3v
E
2C 3
3s v
A1
1
1
1
z
B1
1
1
1
Rz
E
E
2
2C12
30
sV
( x , y ), ( R x , R y )
Gs
3
0
1
x  y ,z
2
2
2
( x  y ), xy , xz , yz
2
2
From the character table, we now can get the symmetry of the
orbitals in N




Only one possible sum will yield this reducible representation
By inspection, we see that Gs = A1 + E
N(2s) = A1 -- x2 + y2 is same as an s-orbital
N(2pz) = A1
N(2px) = N(2py) = E
So, we can now set up the MO diagram and let the correct
symmetries overlap
CHEM 3722
Symmetry
15
Ammonia, The MO Diagram
3A1
2Ex
2Ey
s*
2p’s
A 1 , E x, E y
2A1
Ex, Ey
A1
H
H
H
2s
1A1
z
N
H
H
CHEM 3722
x
H
y
Symmetry
16
Methane

The usual view of methane is one where four equivalent sp3 orbitals
are necessary for the tetrahedral geometry
H
sp3-s overlap for a s MO
H
sp3-s overlap for a s MO
C
H
H
Photoelectron spectrum (crude drawing) adapted from
Roy. Soc. Chem., Potts, et al.

However, the photoelectron spectrum shows two different orbital
energies with a 3:1 population ratio


Maybe the answer lies in symmetry
Let’s build the MO diagram using the SALC method we saw before
CHEM 3722
Symmetry
17
Methane: SALC Approach

Methane is a tetrahedral, so use Td point group


The character table is given below
Td
E
8C 3
3C 2
6S 4
6s d
A1
1
1
1
1
1
A2
1
1
1
1
1
E
2
1
2
0
0
T1
3
0
1
1
1
(R x , R y , R z )
T2
3
0
1
1
1
x, y , z
2
2
2
2z  x  y , x  y
2
2
2
2
2
xy , yz , xz
To find the SALC’s of the 4H’s, count those that do not change position for
each symmetry operation and create the reducible representation, GSALC.
H H
C
H

x y z
H
Td
E
8C 3
3C 2
6S 4
6s d
G SALC
4
1
0
0
2
GSALC = A1 + T2
The character table immediately gives us the symmetry of the s and p
orbitals of the carbon:


C(2s) = A1
C(2px, 2py, 2pz) = T2
CHEM 3722
Symmetry
18
The MO Diagram of Methane

Using the symmetry
of the SALC’s with
those of the carbon
orbitals, we can
build the MO
diagram by letting
those with the same
symmetry overlap.
s7* s8*
A1
2px
2py
T2
s9*
T2
s6*
2pz
1s
A 1 + T2
2s
A1
1s
A1
C
CHEM 3722
s3
s4
s5
s2
A1
s1
A1
CH4
Symmetry
T2
4H’s
19
An Example: BF3

BF3 affords our first look at a molecule where p-bonding is possible


The “intro” view is that F can only have a single bond due to the
remaining p-orbitals being filled
We’ll include all orbitals
F
F
B
F


The point group for BF3 is D3h, with the following character table:
D3h
E
2C 3
3C 2
sh
2S 3
3s v
A1
1
1
1
1
1
1
A 2
1
1
1
1
1
1
Rz
E
2
1
0
2
1
0
x, y
A1
1
1
1
1
1
1
A 2 
1
1
1
1
1
1
z
E 
2
1
0
2
1
0
R x ,R y
x  y ,z
2
xy , x  y
2
2
2
2
xz , yz
We’ll begin by defining our basis sets of orbitals that do a certain
type of bonding
CHEM 3722
Symmetry
20
F3 Residue Basis Sets

Looking at the 3 F’s
as a whole, we can
set up the s-orbitals
as a single basis set:
s s o rb ita ls
E
2C 3
3C 2
sh
2S 3
3s v
G ss
3
0
1
3
0
1
G s s  A1  E 
B
Regular
character
table
Worksheet
for
reducing
Gss
CHEM 3722
D3h

D3h
E
2C 3
3C 2
1s h
2S 3
3s v
G ss
3
0
1
3
0
1
A1
1
1
1
1
1
1
A 2
1
1
1
1
1
1
E
2
1
0
2
1
0
A1
1
1
1
1
1
1
A 2 
1
1
1
1
1
1
E 
2
1
0
2
1
0
A1
3
0
3
3
0
3
12
1
A 2
3
0
3
3
0
3
0
0
E
6
0
0
6
0
0
12
1
A1
3
0
3
3
0
3
0
0
A 2 
3
0
3
3
0
3
0
0
E 
6
0
0
6
0
0
0
0
Symmetry

12
21
The Other Basis Sets
p s o rb ita ls
B
B
D3 h
E
2C 3
3C 2
sh
2S 3
3s v
G ps
3
0
1
3
0
1
G ps  A1  E 
p n b o rb ita ls
B
p p o rb ita ls
B
B
D3h
E
2C 3
3C 2
sh
2S 3
3s v
D3h
E
2C 3
3C 2
sh
2S 3
3s v
G pp
3
0
1
3
0
1
G pnb
3
0
1
3
0
1
G pp  A 2   E 
CHEM 3722
G p n b  A 2  E 
Symmetry
22
The MO Diagram for BF3
No s-orbital interaction
from F’s is included in
this MO diagram!
B (2 p )
E
A 2 
p nb
pp
ps
E
E 
E
A 2
B(s)
A1
A 2  A1
ss
E
A1
CHEM 3722
Symmetry
23
The MO Diagram for BF3
s-orbital interaction
from F’s allowed
B (2 p )
E
A 2 
p nb
pp
ps
E
E 
E
A 2
B(s)
A1
A 2  A1
ss
E
A1
CHEM 3722
Symmetry
24
Using Hybrid Orbitals for BF3

If we use sp2 hybrids and the remaining p-orbital (pz) of the boron,
we see how hybridization yields the same exact picture.


Build our sp2 hybrids and take them through the operations
Find the irreducible representations using the worksheet method and
the reducible representation
B
D3h
E
2C 3
3C 2
sh
2S 3
3s v
G sp 2
3
0
1
3
0
1
G sp 2  A1  E 



pz is found in the character table to be A2”.
Result: identical symmetries for boron’s orbital’s in both cases
This is how it should be, since hybridization is an equivalent set of
orbitals that are simply oriented in space differently.
CHEM 3722
Symmetry
25
p-Bonding in Aromatic Compounds
The p-bonding in C3H3+1 (aromatic)
1 node
0 nodes
The p-bonding in C4H4+2 (aromatic)
2 nodes
1 node
0 nodes
CHEM 3722
Symmetry
Aromatic compounds must
have a completely filled set
of bonding p-MO’s.
This is the origin of the
Hückel (4N+2) p-electron
definition of aromaticity.
26
Cyclopentadiene

As the other examples showed, the actual geometric structure of
the aromatic yields the general shape of the p-MO region
The p-bonding in C5H5-1 (aromatic)
2 nodes
1 node
0 nodes
CHEM 3722
Symmetry
27
Benzene
3 nodes
The p-bonding in C6H6 (aromatic)
2 nodes
1 node
0 nodes
CHEM 3722
Symmetry
28

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