molecular symmetry
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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 - PowerPoint PPT PresentationTRANSCRIPT
Molecular Symmetry
Symmetry ElementsGroup TheoryPhotoelectron SpectraMolecular Orbital (MO) Diagrams for Polyatomic Molecules
Symmetry 2CHEM 3722
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
Symmetry 3CHEM 3722
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)
Symmetry 4CHEM 3722
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
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)
s(xy)x
y
z
BBH
H
HHH H
E BBH
H
HHH H
Symmetry 5CHEM 3722
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
5
4
3
2
1
5
4
3
2
1
6
4
3
2
1
6
5
3
2
1
6
5
4
C6
C3
C2
C6
C6
How many C2 operations are
there for benzene?
Symmetry 6CHEM 3722
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)
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 left12
34
6
5
61
23
5
4
12
34
6
5
45
61
3
2
sh C6 = S6
Symmetry 7CHEM 3722
A Few To Try Determine which symmetry elements are applicable for
each of the following molecules
Ru O Ru
Cl Cl
ClClCl
Cl Cl
ClClCl
NHH
H
OHH
Symmetry 8CHEM 3722
Point Groups We can
systematically classify molecules by their symmetry properties Call these point
groups Use the flow
diagram to the right
Start
Cn axis
Special groups: a) Cv,Dh (linear groups)b) T, Th, Td, O, Oh, I, Ih
(1)
(2) No proper or improper axes
Only Sn (n = even) axis: S4, S6…(3)
(4) (5)
No C2’s to Cn 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 9CHEM 3722
Some Common Groups D3h Point Group
C3, C32, 3C2, S3, S3
5, 3sv, sh Trigonal planar
C3v Point Group C3, 3sv Trigonal pyramid
D3h Point Group C3, C3
2, 3C2, S3, S35, 3sv, sh
Trigonal bi-pyramid
C4h Point Group C2, 2C2’, 2C2’’, C4, S4, S4
2, 2sv, 2sd, sh, i Square planar
C4v Point Group C2, C4, C4
2, 2sv, 2sd Square pyramid
AB3
AB3
AB5
AB4
AB5
Symmetry 10CHEM 3722
Character Tables Character tables
hold the combined symmetry and effects of operations For example,
consider water (C2v)
Point Group
s s
22 2
1
2
1
2
22
1 1 1 1 , ,1 1 1 11 1 1 1 ,1 1 1 1 ,
( ) ( )v
z
y
x
ECz x y z
R xyA
C xz yz
x R xzy
BB R
A
yz
Symmetry operations available
Symmetry of statesA, B = singly degenerateE = doubly degenerateT = triply degenerate1 = symmetric to C2 rotation*2 = antisymmetric to C2 rotation*
Coordinates and rotations of this
symmetry
Effect of this operation on an
orbital of this symmetry.
Symmetry 11CHEM 3722
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
s s
22 2
1
2
1
2
22
1 1 1 1 , ,1 1 1 11 1 1 1 ,1 1 1 1 ,
( ) ( )v
z
y
x
ECz x y z
R xyA
C xz yz
x R xzy
BB R
A
yz1
-1
1
-1
The px orbital has B1 symmetry
HO
H
z
x
yE
s(yz)
s(xz)
C2
Symmetry 12CHEM 3722
The Projection Operator Since we want to build MO diagrams, our symmetry needs
are simple Only orbitals of the same symmetry can overlap to form bonds
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 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 It’s functional form is
But it’s easier to use than this appears
ˆ ˆ( )j j
j
lP R Rh
Symmetry 13CHEM 3722
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
3 sigma bonds and no pi bonds, thus we’ll build our SALC’s from the projection operator and these 3 MO’sNote, ammonia is in C3v point group (AB3)
z
x
y
NHH
H
Symmetry 14CHEM 3722
Ammonia II The Character
Table for C3v is to the right
s
3 32 2 2
1
12 2
2 31 1 1 ,1 1 12 1 0 ( , ),( , ) ( ), , ,
v v
z
x y
C E CA z x y zB RE x y R R x y xy xz yz
NHH
H s1s2
s3 NHH
H s3s1
s2C3
NHH
H s1s2
s3 NHH
H s1s3
s2sv
NHH
H s1s2
s3 NHH
Hs3 s1E
s2(3)
(0)
(1)
Take the s bonds through the operations & see how many stay put
We now have a representation (G) of this group of orbitals that has the symmetry
s
sG
22 33 0 1
VE C
Symmetry 15CHEM 3722
Ammonia III This “reducible representation” of the hydrogen’s s-bonds
must be a sum of the symmetries available Only one possible sum will yield this reducible representation By inspection, we see that Gs = A1 + E
From the character table, we now can get the symmetry of the orbitals in N 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
s
3 32 2 2
1
12 2
2 31 1 1 ,1 1 12 1 0 ( , ),( , ) ( ), , ,
v v
z
x y
C E CA z x y zB RE x y R R x y xy xz yz
s
sG
22 33 0 1
VE C
Symmetry 16CHEM 3722
Ammonia, The MO Diagram
s*
A1
Ex, Ey
A1, Ex, Ey
2p’s
2s
2Ex2Ey
3A1
2A1
1A1
z
x
y
NHH
H
HHH
Symmetry 17CHEM 3722
Methane The usual view of methane is one where four equivalent sp3
orbitals are necessary for the tetrahedral geometry
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
CHH
H Hsp3-s overlap for a s MOsp3-s overlap for a s
MO
Photoelectron spectrum (crude drawing) adapted from Roy. Soc. Chem., Potts, et al.
Symmetry 18CHEM 3722
Methane: SALC Approach Methane is a tetrahedral, so use Td point group
The character table is given below
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.
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
s
3 2 42 2 2
1
22 2 2 2 2
1
2
8 3 6 61 1 1 1 11 1 1 1 12 1 2 0 0 2 ,3 0 1 1 1 ( , , )3 0 1 1 1 , , , ,
d d
x y z
T E C C SA x y zAE z x y x yT R R RT x y z xy yz xz
CHH
H Hs
G3 2 48 3 6 6
4 1 0 0 2d d
SALC
T E C C SGSALC = A1 + T2
Symmetry 19CHEM 3722
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.
1sA1 + T22s
2px 2py 2pz
1s
T2
A1
A1
s2
s1
s4 s5s3
s6*
s7* s8* s9*
A1
A1
T2
T2
A1
C CH4 4H’s
Symmetry 20CHEM 3722
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
The point group for BF3 is D3h, with the following character table:
We’ll begin by defining our basis sets of orbitals that do a certain type of bonding
F BF
F
s s
3 3 2 3
2 2 21
22 2
1
2
2 3 2 3
1 1 1 1 1 1 ,
1 1 1 1 1 12 1 0 2 1 0 , ,
1 1 1 1 1 1
1 1 1 1 1 12 1 0 2 1 0 , ,
h h v
z
x y
D E C C S
A x y z
A RE x y xy x y
A
A zE R R xz yz
Symmetry 21CHEM 3722
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:
Bss orbitals
3 2
1
3 32 3 2 33 0 1 3 0 1
h
s
h v
s
D C
A E
E C S
s
s
s sG
G
3 3 2 3
1
2
1
2
1
2
1
2
2 2 3 123 0 0 1
1 1 1 1 1
1 1 1 1 1 12 1 0 1 0
1 1 1 1 1 1
1 1 1 1 1 1
1
1
1
2 1 0 2 1 0
3 0 3 0 3 12
3 0 3 3 0 3 0 06 0 0 0 0 12
3 0 3 3 0 3 0 0
3 0 3 3 0 3 0 06 0 0 6 0 0 0 0
3
6
1
3
2
1
3
h h v
s
D E C C S
A
AE
A
AE
A
AE
A
AE
s
s s
G
Regular character table
Worksheet for
reducing Gss
Symmetry 22CHEM 3722
The Other Basis Sets
Bpp orbitals
3 3
2
2 32 3 2 33 0 1 3 0 1
h
p
h v
p
D E C C S
A E
p
p
s s
G
G
BB
ps orbitals
3 2
1
3 32 3 2 33 0 1 3 0 1
h
p
h v
p
D C
A E
E C S
s
s
s sG
G
B B
pnb orbitals
3 3 2
2
32 3 2 33 0 1 3 0 1
h h v
pn
n
b
p b
D E C C
A E
Ss s
G
G
Symmetry 23CHEM 3722
The MO Diagram for BF3
No s-orbital interaction from F’s is
included in this MO diagram!
nbpE
pE
p
pE
s
2A 2A
1A
sE
s
1A
1A
(2 )B pE
2A
B(s)
Symmetry 24CHEM 3722
The MO Diagram for BF3
s-orbital interaction from F’s allowed
nbpE
pE
p
pE
s
2A 2A
1A
sE
s
1A
1A
(2 )B pE
2A
B(s)
Symmetry 25CHEM 3722
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
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.
Bs s
G
G
2
2
3 3 2 3
1
2 3 2 33 0 1 3 0 1
h h v
s
sp
p
D E C C S
A E
Symmetry 26CHEM 3722
p-Bonding in Aromatic CompoundsThe p-bonding in C3H3
+1 (aromatic)
0 nodes
1 node
The p-bonding in C4H4+2 (aromatic)
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.
0 nodes
1 node
2 nodes
Symmetry 27CHEM 3722
Cyclopentadiene As the other examples showed, the actual geometric
structure of the aromatic yields the general shape of the p-MO region
0 nodes
1 node
2 nodesThe p-bonding in C5H5
-1 (aromatic)
Symmetry 28CHEM 3722
Benzene
The p-bonding in C6H6 (aromatic)
0 nodes
1 node
2 nodes
3 nodes