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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

BB

H

H

HHH H

EBB

H

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 left1

2

3

4

6

5

6

1

2

3

5

4

1

2

3

4

6

5

4

5

6

1

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

22 2

1

2

1

2

22

1 1 1 1 , ,

1 1 1 1

1 1 1 1 ,

1 1 1 1 ,

( ) ( )v

z

y

x

EC

z x y z

R xy

A

C xz yz

x R xz

y

B

B 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.


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

22 2

1

2

1

2

22

1 1 1 1 , ,

1 1 1 1

1 1 1 1 ,

1 1 1 1 ,

( ) ( )v

z

y

x

EC

z x y z

R xy

A

C xz yz

x R xz

y

B

B R

A

yz1

-1

1

-1

The px orbital has B1 symmetry

HO

H

z

x

yE

(yz)

(xz)

C2


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 R

h


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


Ammonia II The Character

Table for C3v is to the right

3 32 2 2

1

12 2

2 3

1 1 1 ,

1 1 1

2 1 0 ( , ),( , ) ( ), , ,

v v

z

x y

C E C

A z x y z

B R

E x y R R x y xy xz yz

NHH

H 1

2

3 NHH

H 3

1

2C3

NHH

H 1

2

3 NHH

H 1

3

2 v

NHH

H 1

2

3 NHH

H 3 1E

2(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

22 3

3 0 1VE C


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

3 32 2 2

1

12 2

2 3

1 1 1 ,

1 1 1

2 1 0 ( , ),( , ) ( ), , ,

v v

z

x y

C E C

A z x y z

B R

E x y R R x y xy xz yz

22 3

3 0 1VE C


Ammonia, The MO Diagram

s*

A1

Ex, Ey

A1, Ex, Ey

2p’s

2s

2Ex

2Ey

3A1

2A1

1A1

z

x

y

NHH

H

HHH


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.


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

3 2 42 2 2

1

22 2 2 2 2

1

2

8 3 6 6

1 1 1 1 1

1 1 1 1 1

2 1 2 0 0 2 ,

3 0 1 1 1 ( , , )

3 0 1 1 1 , , , ,

d d

x y z

T E C C S

A x y z

A

E z x y x y

T R R R

T x y z xy yz xz

CHH

H H

3 2 48 3 6 6

4 1 0 0 2d d

SALC

T E C C SGSALC = A1 + T2


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 + T2

2s

2px 2py 2pz

1s

T2

A1

A1

s2

s1

s4 s5s3

s6*

s7* s8* s9*

A1

A1

T2

T2

A1

C CH4 4H’s


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

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 1

2 1 0 2 1 0 , ,

1 1 1 1 1 1

1 1 1 1 1 1

2 1 0 2 1 0 , ,

h h v

z

x y

D E C C S

A x y z

A R

E x y xy x y

A

A z

E R R xz yz


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:

Bs orbitals

3 2

1

3 32 3 2 3

3 0 1 3 0 1h

s

h v

s

D C

A E

E C S

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 1

2 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 0

6 0 0 0 0 12

3 0 3 3 0 3 0 0

3 0 3 3 0 3 0 0

6 0 0 6 0 0 0 0

3

6

1

3

2

1

3

h h v

s

D E C C S

A

A

E

A

A

E

A

A

E

A

A

E

Regular character table

Worksheet for

reducing Gss


The Other Basis Sets

Bp orbitals

3 3

2

2 32 3 2 3

3 0 1 3 0 1h

p

h v

p

D E C C S

A E

BB

p orbitals

3 2

1

3 32 3 2 3

3 0 1 3 0 1h

p

h v

p

D C

A E

E C S

B B

pnb orbitals

3 3 2

2

32 3 2 3

3 0 1 3 0 1h h v

pn

n

b

p b

D E C C

A E

S


The MO Diagram for BF3

No s-orbital interaction from F’s is

included in this MO diagram!

nbp

Ep

E

p

E

2A 2A

1A

s

E

1A

1A

(2 )B p

E

2A

B(s)


The MO Diagram for BF3

s-orbital interaction from F’s allowed

nbp

Ep

E

p

E

2A 2A

1A

s

E

1A

1A

(2 )B p

E

2A

B(s)


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.

B

2

2

3 3 2 3

1

2 3 2 3

3 0 1 3 0 1h h v

s

sp

p

D E C C S

A E


p-Bonding in Aromatic CompoundsThe -bonding in C3H3

+1 (aromatic)

0 nodes

1 node

The -bonding in C4H4+2 (aromatic)

Aromatic compounds must have a completely filled set of bonding -

MO’s.This is the origin of the

Hückel (4N+2) -electron definition of aromaticity.

0 nodes

1 node

2 nodes


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 -bonding in C5H5

-1 (aromatic)


Benzene

The -bonding in C6H6 (aromatic)

0 nodes

1 node

2 nodes

3 nodes