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Molecular Symmetry

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Molecular Symmetry

Symmetry Elements

Group Theory

Photoelectron Spectra

Molecular Orbital (MO) Diagrams for Polyatomic 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

- 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

- Identity
- If an object (O) has coordinates (x,y,z), then the operation E(x,y,z) (x,y,z)
- The object is unchanged

E

- 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

s(xy)

y

x

Symmetry

- 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)

- Cn where n represents angle of rotation out of 360 degrees

C6

C6

C3

How many C2 operations are there for benzene?

C6

C2

Symmetry

- 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

6

2

5

1

sh C6 = S6

5

3

4

2

4

3

Symmetry

- Determine which symmetry elements are applicable for each of the following molecules

Symmetry

- We can systematically classify molecules by their symmetry properties
- Call these point groups
- Use the flow diagram to the right

Special groups:

a) Cv,Dh (linear groups)

b) T, Th, Td, O, Oh, I, Ih

(1)

Start

(2)

No proper or improper axes

(3)

Only Sn (n = even) axis: S4, S6…

Cn axis

(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

AB3

- D3h Point Group
- C3, C32, 3C2, S3, S35, 3sv, sh
- Trigonal planar

- C3v Point Group
- C3, 3sv
- Trigonal pyramid

- D3h Point Group
- C3, C32, 3C2, S3, S35, 3sv, sh
- Trigonal bi-pyramid

- C4h Point Group
- C2, 2C2’, 2C2’’, C4, S4, S42, 2sv, 2sd, sh, i
- Square planar

- C4v Point Group
- C2, C4, C42, 2sv, 2sd
- Square pyramid

AB3

AB5

AB4

AB5

Symmetry

Point Group

- Character tables hold the combined symmetry and effects of operations
- For example, consider water (C2v)

Symmetry operations available

Effect of this operation on an orbital of this symmetry.

Coordinates and rotations 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*

Symmetry

- 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

1

-1

1

-1

The px orbital has B1 symmetry

Symmetry

- 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

Symmetry

- 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’s

Note, ammonia is in C3v point group (AB3)

Symmetry

- The Character Table for C3v is to the right

- 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

(3)

(0)

(1)

Symmetry

- 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

Symmetry

3A1

2Ex

2Ey

s*

2p’s

A1, Ex, Ey

2A1

Ex, Ey

A1

2s

1A1

Symmetry

- The usual view of methane is one where four equivalent sp3 orbitals are necessary for the tetrahedral geometry

sp3-s overlap for a s MO

sp3-s overlap for a s MO

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

Symmetry

- 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.

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

Symmetry

- 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*

s9*

T2

A1

s6*

2px

2py

2pz

T2

1s

A1 + T2

2s

s3

s4

s5

A1

T2

s2

A1

s1

1s

A1

A1

C CH4 4H’s

Symmetry

- 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

Symmetry

- Looking at the 3 F’s as a whole, we can set up the s-orbitals as a single basis set:

Regular character table

Worksheet for reducingGss

Symmetry

Symmetry

No s-orbital interaction from F’s is included in this MO diagram!

Symmetry

s-orbital interaction from F’s allowed

Symmetry

- If we use sp2hybrids and the remaining p-orbital (pz) of the boron, we see how hybridization yields the same exact picture.
- Build our sp2hybrids 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.

Symmetry

The -bonding in C3H3+1 (aromatic)

1 node

0 nodes

The -bonding in C4H4+2 (aromatic)

2 nodes

1 node

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

Symmetry

- As the other examples showed, the actual geometric structure of the aromatic yields the general shape of the p-MO region

The -bonding in C5H5-1 (aromatic)

2 nodes

1 node

0 nodes

Symmetry

3 nodes

The -bonding in C6H6 (aromatic)

2 nodes

1 node

0 nodes

Symmetry