Chem IV - Symmetry and Group Theory- Symmetry and Group Theory

Chapter 7

Part 1 - Introduction to symmetry elements,symmetry operations and point groups

Symmetry in Nature

The Platonic Solids

ChristianArt

Islamic Art

Art Deco

Architecture

Photography

Ndebele

Zulu

Symmetry analysisSymmetry analysis

• A symmetry operation is an action that leaves a molecule apparently unchanged

• Each symmetry operation is associated with a symmetry element

Symmetry operation: rotation

Symmetry element:axis of rotation

H2O

Point, line, or plane

A symmetry operation leaves at least one point in the molecule unmoved – they are operations of point group symmetry

The identity operation, E – do nothing.

All molecules have at least E, and some have only the symmetry element E

An n-fold rotation is a symmetry operation that leaves a molecule apparently unchanged after rotation by 360o/n.

The symmetry element is an n-fold axis of rotation, Cn

BH3

Note that C33 = E

BrF5

Note that C42 = C2

Associated with the symmetry element C4 we have the symmetry operations

C4 C42 ( = C2) C4

3 C44 ( = E)

XeF4

The axis with the highest order, here the C4 axis, is called the principal axisThe principal axis defines the z axis

When assigning axis of the same order, we give preference to those that go through atoms, C2’, followed by those bisecting the bond angle, C2’’.

Identify the axes of rotation of this snowflake.

List the symmetry operations associated with each of these symmetry elements

Symmetry operation reflection through the symmetry element mirror plane, σ

The mirror planes contain the principal C2 axis. They are therefore vertical mirror planes (subscript “v”).

z

This mirror plane is perpendicular to the principal C4 axis. It is therefore a horizontal mirror (subscript “h”)

This mirror plane is perpendicular to the principal C4 axis. It is therefore a vertical mirror (subscript “v”). Vertical mirror planes are those that usually go through atoms, and again, you give those that go through atoms as higher priority, i.e. v. Those mirror planes that do not go through atoms are sometimes more accurately called dihedral mirror planes.

This mirror plane is parallel to (contains) the principal C4 axis and bisects the two C2' axes. It is a dihedral mirror (subscript “d”).

σh(horizontal): plane perpendicular to principal axis

σd(dihedral), σv(vertical): plane colinear with principal axis

σd: σ parallel to Cn and bisecting two C2' axes or two σv

σv: σ parallel to Cn and are often coincident with lower rotation axes of high priority.

Mirror planes in molecules

Examples of difference between vertical and dihedral planes.

The vertical planes lie parallel with the C2’ axes. The dihedral planes are bisecting the C2’ axes and in this case also contain the C2’’ axes.

Examples of dihedral planes in the absence of C2 axes.

Compare staggered ethane with the complex ML4ClBr. In the former molecule, there are only the three "horizontal" C2 axes and the planes bisect the angle between them. In the Newman projection the reason for calling these planes "dihedral" is clear. Finally we consider the complex in figure 1.15c where there are no "horizontal" C2 axes but there are two sets of planes containing the principal axis. Now the planes on the atoms take priority and are labelled v while the planes between the ligands bisect the angle between the vertical planes and become dihedral. Again which way round you do this is arbitrary but having the atoms as the highest priority pleases most chemists

Identify the mirror planes (σh, σd,σv) in the following examples

acetylene

Identify the mirror planes (σh, σd,σv) in the following examples

(Don’t forget the double bonds are delocalised)

Identify the mirror planes (σh, σd,σv) in the following examples

Os(cp)2

The symmetry operation inversion, i, involves projecting each atom through a point, the centre of inversion i, that is located at the centre of the molecule

Staggered form of ethane

i

Do these molecules have a centre of inversion?

The symmetry operations i and C2 should not be confused

The symmetry operation improper rotation occurs about the symmetry element Improper Axis, Sn

This is a compound operation combining a rotation (Cn) with a reflection through a plane perpendicular to the Cn axis σh:

Cn followed by σh orσhCn = Sn

Neither the 90o rotation nor the reflection is itself a symmetry operation for a tetrahedral molecule, but their combined effect is the symmetry operation S4

Read: Do Cn followed by σh

Do you see that S1 = σ and that S2 = i?

Use a CH4 molecule to verify that 2S4 = C2

Identify a S3 symmetry operation in BF3. Whatis it equal to in this molecule?

Identify all the symmetry elements in the

(i) eclipsed

(ii) staggered

conformation of ethane

E, C3, C2, σh, σv, S3

E, C3, C2, σd, i, S6

Sketch the S4 axis of NH4+. Is there a C4 axis? How many S4

axes are there in the ion?

The Point Groups of MoleculesThe Point Groups of Molecules

The set of symmetry elements of a molecule constitute a group.

Since all symmetry operations leave at least one point in the molecule unchanged, the group is called a point group

The point group is identified by it Schoenflies symbol

The process used to assign a molecule to a point group is straightforward with a few exceptions. Use this schematic to guide you.

CO2 OCS

SiIBrClFLinear – No2 of Cn, n>2 – NoCn – Noσh – Noi - No

So…

Determine the point groups of the following molecules

CH2CBrCl C2H2Cl2Br2 H2O2 H2O

1,3,5,7-tetrafluoro-cyclooctatetrane [Ni(en)3]2+

S8

naphthalene

XeF4

Convention for orientation of a molecule in a coordinate system

1. Centre of mass (or centre of symmetry) at origin.

2. z axis = highest order axis and principal axis.If several axes of highest order, the z axis is the one that passes through most atoms.

3. If a non-planar molecule contains a plane which contains more atoms than any other plane, this is treated as it if where the molecular plane.

4. Then assign x axis. If molecule is planar and z axis in plane, then the x axis is plane.

5. If molecule planar and z axis plane, then x axis will be in the plane and chosen to pass through greatest number of atoms.

6. The y axis is to other two axes. Use right-hand rule, where your thumb = x, index = y and middle = z.

Place the following in a cartesian coordinate system:NH3 H2O cis-[Pt(NH3)2Cl2] trans-C2H2Cl2

Determine the point groups of the following molecules – Homework

CCl4ChloroformDichloromethaneSF6

BF5 (sq pyramidal)1,2-dichloronaphthalene1,5-dichloronapthaleneEthane (staggered)Ethane (eclipsed)Os(cp)2 (staggered)Os(cp)2 (eclipsed)PPh3

Part 2 - Group Theory and Character Tables

Groups and Group Theory

Group: a collection of elements that obey the following rules:

• Any combination of two elements must be equivalent to another element in the group (the group shows closure)

Show that σv(xz)C2 = σv’(yz) in the C2v point group using H2O as a representation of that point group

• There must exist an element that commutes with all other elements and leaves them unchanged

For molecules this is the identity E since EX = XE = X

• The associative law must hold

Use the symmetry elements C2, σv and σv’ of the water molecule to show that the associative law holds true

X(YZ) = (XY)Z

• Each X must have a reciprocal X-1 such that

Find the reciprocals of C2 and σv’ of the water molecule

XX-1 = X-1X = E

Note the absence of the general requirement that the commutative law holds (XY = YX). If it holds, the group is an Abelian group

The set of symmetry operations constitute a group, called a point group because at least one point is unchanged.

Once we identify the point group → can look up all the symmetry elements in character tables (see below).

The number of elements in a group: order of group, h.

Elements A, B, C, ..., M belong to the same class if, and only if

X-1AX X-1BX X-1CX ▪ = A, B, C, ..., or M▪ ▪ X1-MX

Show that all the symmetry elements of NH3 belong to three classes. Hint: Identify ALL symmetry elements first.

As will be seen, need only consider properties associated with one representative from each class rather than all operations.

So, for NH3, where we have operations

E, C3, C32, σv, σv’, σv”

we can write as E, 2C3, 3σv

The symmetry of orbitals

Central importance for chemists: effect of symmetry on properties of orbitals

Example: The effect of the symmetry elements of the C2v point group (E, C2, σv, σv’) on the valence orbitals of O (2s, 2p)

σv’ (yz)

σv (xz)

2s 2px 2py 2pz

xy

z

C2

E

Operators either

• change sign of wavefunction (≡ ×(-1))• leave sign of wavefunction unchanged (≡ ×(+1))

For example:

(C2)(2px) = (-1)×(2px)

The operation C2 on the wavefunction

2px

may be represented by multiplying the

wavefunction by a matrix

Here a 1 × 1 matrix

Such a matrix is called a

transformation matrix

So

Ψ´ = R(Ψ) = [M]×Ψ

Final wavefunction

Initial wavefunction

Symmetry operator

Transformation matrix

Initial wavefunction

An aside

Multiplying two matrices

P = MN

5043

2219

84637453

82617251

87

65

43

21

1 0 0

0 1 0 0 0 1

x y z

=

In linear algebra, the trace of an n-by-n square matrix is defined to be the sum of the elements on the main diagonal (the diagonal from the upper left to the lower right)

Let a suitable set of functions exist that are the basis set of a point group.

There exists a collection of matrices that represents the operation of the symmetry operators.

If AB(Ψ) = C(Ψ)

(there must exist)

[MA], [MB] and [MC]

such that

[MA][MB](Ψ) = [MC](Ψ)

A collection of matrices that obeys the multiplication laws of the symmetry operators of a point group is a representation of that group, symbol Γ (Gk., Gamma).

x

y

z

C2

Let’s return to this:

2px

Take the 2px wavefunction as a basis set for the C2v point

group

Then Γ = (1, -1, 1, -1) is a representation of the C2v point group

E

What if our basis set is the set of valence orbitals on O (2s, 2p)?

Then we need a set of 4 × 4 matrices.

z

y

x

z

y

x

p

p

p

s

p

p

p

s

C

1000

0100

0010

0001

2

z

y

x

z

y

xv

p

p

p

s

p

p

p

s

xz

1000

0100

0010

0001

)(

Write down the matrices corresponding to the symmetry operators E and σv’(yz)

Use the two H 1s orbitals in H2O as basis set and show that the transformation matrices are

and determine the trace of each one.

10

01)('

01

10)(

01

10

10

012 yzxzCE vv

In general, for n basis functions of a point group we will need a set of n × n matrices to represent the point group

IMPORTANT Beware of how many elements your basis set consists off.

For a point group with a symmetry operator greater than two-fold a complication arises – the rotation will “mix” elements of the basis set.

x´ = x cos θ – y sin θy´ = x sin θ + y cos θ

y

x

y

xCn

cossin

sincos

Example. Take the valence orbitals of N in ammonia as the basis set for the C3v point group. Then the transformation matrix for the C3 operation is

1000

021

230

023

210

0001

z

y

x

p

p

p

s

Write down the transformation matrices for all other symmetry operators of the C3v point group

If X, A and B of a group G:

... then X-1 G (why?)

Suppose X-1AX = B

We say - A is converted to B by a similarity transformation- B is the similarity transform of A by X- A and B are conjugate

Reducing matrices

For symmetry operators A, B and C square transformation matrices [A], [B] and [C].

Suppose AB = C; then [A][B] = [C]

Let [A]´ and [B]´ be similarity transforms of [A] and [B]

So[A]´ = [X]-1[A][X][B]´ = [X]´[B][X]

Now[A]´[B]´ = [X]-1[A][X][X]-1[B][X]

= [X]-1[A][B][X] = [X]-1[C][X] = [C]´

So [A]´, [B]´ and [C]´ are true representations of G

Suppose as the result of suitable transformations we end up with matrices of the form

|||0||0|

|0||||0|

|0||0|||

][

|||0||0|

|0||||0|

|0||0|||

][

3

2

1

3

2

1

b

b

b

B

a

a

a

A

where |0| is a region of the matrix where all elements are 0.

These are said to be reduced matrices.

Submatrices |an| and |bn| have the same order.

It can be shown that if [A]´[B]´ = [C]´ then |an||bn| = |cn|

The submatrices are also true representations of G

Thus smaller representations can be found within larger ones.

If this is impossible, then the original matrix is irreducible.

The character of a matrix

Character χ = sum diagonal elements

χ is unaltered by similarity transforms

Larger matrices may be reduced to irreducible components by similarity transforms alone. So

n

iiiR Ra

1

)(

- χR is the character of a reducible matrix for the

Rth operation of the point group- χi(R) is the character of the ith irreducible representation of the Rth operation- ai is the number of times χi(R) is contained in χR.

Character Tables

This is a typical character table

Character table: tabulation of χi(R) i irreducible representations of G.

Symmetry elements of the same class related by a similarity transform matrices have the same character.

Mulliken symbols give information about symmetry elements of the irreducible representations

This is a 2-dimensional representation (2×2 matrix)

This is a 1-dimensional representation (1×1 matrix)

One dimensional representations symmetric w.r.t. rotation about the principal axis (+1 under Cn) are designated A; those that are antisymmetric (-1 under Cn) are designated B.

If the one dimensional representation is symmetric w.r.t. C2 Cn subscript “1” is added to A or B. If C2 is absent then the symmetry w.r.t. a vertical plane is used instead.

Two dimensional representations are designated E and three dimensional ones are designated T

Primes and double primes indicate representations that are respectively symmetric and antisymmeric w.r.t. σh

If i G then subscript “g” (Ger., gerade, even) is added if the representation is symmetric w.r.t. inversion; otherwise “u” (Ger., ungerade, odd) is used

Direct products

As we shall see, it is often necessary to multiply together representations of two sets of functions. This is a direct product of the two representations.

Example: B1 × B2 in C2v

B1 × B2 = (1×1 -1×-1 1×-1 -1×1)= (1 1 -1 -1)= A2

Reduction of matrices using character tables

It can be shown that the number of times, aj, which is the irreducible representation j, occurs in a reducible representation is given by:

g

jRj RRnh

a )()(1

• h = order of group (= number of elements in group = sum of squares of characters entered under E in the character table;

• g = number of classes;• nR = number of elements in a class (number in front of class symbol)• χ(R) = character of reducible representation;• χj(R) = character of the irreducible representation.

Usually not necessary to write out a full transformation matrix for each symmetry operator – merely the character.

Diagonal elements describes to what extent a function remains in its original position after a symmetry operation: 0 – moves completely away; 1 – remains unchanged; -1 – changes sign; and so on.

Example: The p orbitals of C in the carbonate anion. How to they transform in the D3h point group? (So we use the 2p orbitals as a basis set.)

Operator Effect Contribution to χ

χ

E Nothing 3

C3 z → zx → x cos 120o – y sin 120o

y → x in 120o + y cos 120o

1-1/2-1/2

0

C2’ z → -zx → xy→ -y

-11-1

-1

σh z → -zx,y unchanged

-12

1

S3 z → -zx → x cos 120o – y sin 120o

y → x in 120o + y cos 120o

-1-1/2-1/2

-2

σv z → zx → xy→ -y

11-1

1

on diagonaloff diagonal

Hence Γ2p = (3, 0, -1, 1, -2, 1) is a reducible representation of the C3v point group.

Let’s reduce this representation and express this in terms of the irreducible representations of the point group.

Γ2p = (3 0 -1 1 -2 1)

g

jRj RRnh

a )()(1

a(a1’) = 1/12[1.3.1 + 2.0.1 + 3.-1.1 + 1.1.1 + 2.-2.1 + 3.1.1]= 1/12[3 + 0 + -3 + 1 + -4 + 3]= 0

This means that a1’ does not appear at all in Γ2p

a(a2’) = 1/12[1.3.1 + 2.0.1 + 3.-1.-1 + 1.1.1 + 2.-2.1 + 3.1.-1]= 1/12[ 3 + 0 + 3 + 1 - 4 - 3 ]= 0

a(e’) = 1/12[1.3.2 + 2.0.-1 + 3.-1.0 + 1.1.2 + 2.-2.-1 + 3.1.0]= 1/12[ 6 + 0 + 0 + 2 + 4 + 0 ]= 1/12[12]= 1

a(a1”) = 1/12[1.3.1 + 2.0.1 + 3.-1.1 + 1.1.-1 + 2.-2.-1 + 3.1.-1]= 1/12[ 3 + 0 - 3 - 1 + 4 - 3 ]= 0

Γ2p = (3 0 -1 1 -2 1)

g

jRj RRnh

a )()(1

a(a2”) = 1/12[1.3.1 + 2.0.1 + 3.-1.-1 + 1.1.-1 + 2.-2.-1 + 3.1.1]= 1/12[ 3 + 0 + 3 - 1 + 4 + 3 ]= 1/12[12]= 1

a(e”) = 1/12[1.3.2 + 2.0.-1 + 3.-1.0 + 1.1.-2 + 2.-2.1 + 3.1.0]= 1/12[ 6 + 0 + 0 - 2 - 4 + 0 ]= 0

Hence Γ2p = a2” + e’. Two of the p orbitals (2px and 2py are degenerate) and the other (2pz) has a different energy.

NOTE: When describing the symmetry species of orbitals, lower

case letters are used.

We could have told this straight away by consulting the 2nd last column of the character table, where we see the functions (x,y) are degenerate and transform as e’ whilst the function (z) transforms as a2”.

Recall that ψ2px = n1(x).f(r)ψ2py = n2(y).f(r)

ψ2pz = n3(z).f(r)Hence ψ2px has the same symmetry properties as (x), and so on.

How do the d orbitals transform under the D3h point group? And the 4s orbital?

Notice this typo in Atkins: should be (xz, yz)

Show that the valence orbitals of nitrogen in ammonia transform as 2a1 + e

How do the metal d orbitals transform in trans-tetra-ammineaquachlorocobalt(III)?

Summary of important character table conceptsTaken from

http://mutuslab.cs.uwindsor.ca/macdonald/Teaching/0359-250.htm

Part 3 - Some Applications of Group Theory in Chemistry

Polarity of molecules

Polar molecule: permanent electric dipole moment

δ-

δ+

Cannot be polar if molecule posses a centre of inversion iDipole moment cannot be to a mirror planeDipole moment cannot be to a Cn axis

A polar molecule cannot belong to a group that includes i, any of the groups D and their derivatives, the cubic groups (Oh,, Td, Ih) and their modifications

Are these molecules polar?

Chirality of molecules

A molecule cannot be chiral if it possesses an improper axis of rotation, Sn. These are groups such as Dnd, Dnh, Td, Oh, etc

A mirror plane σ ≡ S1

A centre of inversion i ≡ S2

So molecules with σ and i are not chiral too.

Are these chiral molecules (assume locked in these conformations)?

Symmetry adapted linear combination of atomic orbitals

Molecular orbitals are built up from the combination of atomic orbitals of the correct symmetry. This is a symmetry adapted linear combination (SALC) of atomic orbitals.

Example. The diagram shows a molecular orbital made up of a linear combination of H 1s orbitals in ammonia. To what symmetry species does φ belong?

φ1 = ψA1s + ψB1s + ψC1s

Example. Identify the symmetry label of the orbital φ = ψ0 – ψ0’ in the C2v molecule NO2 where ψ0 is an O2px orbital on one oxygen atom and ψ0’ is an O2px orbital on the other O atom.

The construction of molecular orbitals

Now that we know something about group theory let’s return to a topic in Chem III Inorganic: the MO description of the bonding in transition metal complexes

Consider a metal from the first row of the d block in an Oh complex such as high-spin [Fe(H2O)6]2+.

Available orbitals: 3d, 4s, 4p

4s: a1g

4p: t1u

3d: eg, t2g

Γmetal = a1g + t1u + eg + t2g

Now the ligands. Assume for the moment σ bonding only (appropriate for H2O as ligands).

Exact nature of ligand orbital use unimportant – only the symmetry of the orbitals is important.

+

+

+

+

+

+

Six ligand-based orbitals to be combined – in this case as it sigma bonding only, can use spheres.

+

+

+

+

+

+

Now apply the symmetry operations of the Oh point group to these 6 orbitals (Your basis set is 6 now).

E 8C3 6C2 6C4 3C2 i 6S4 8S6 3σh 6σd

6 0 0 2 2 0 0 0 4 2

and we find that Γσ(ligand)= (6 0 0 2 2 0 0 0 4 2) which is a reducible representation of the Oh point group. Apply the methods we have learned to express as the sum of irreducible representations and findΓσ(ligand) = a1g + eg + t1u

Γσ(ligand) = a1g + eg + t1uΓmetal = a1g + t1u + eg + t2g

and notice that there are no ligand orbitals with symmetry t2g.

Therefore the t2g orbitals (xy, xz, yz) will be non-bonding.

The MO diagram is therefore the familiar one we met in Chem III – but hopefully you will now have a deeper understanding of its origins.

What if we include π bonding?

This will involve overlap of ligand orbitals which are to the M–L σ bond.

σπ

π

The p orbitals have the same symmetry properties as a vector orthogonal to the main σ bond framework.

E 8C3 6C2 6C4 3C2 i 6S4 8S6 3σh 6σd

12 0 0 0 -4 0 0 0 0 0

and reducing this into irreducible representations we find

Γπ(ligand) = t1g + t2g + t1u + t2u

Γπ(ligand) = t1g + t2g + t1u + t2u

t1g and t2u – non bonding because no metal orbitals with this symmetry

t1u – largely non-bonding because metal t1u orbitals (4p) already involved in σ bonding

t2g orbitals can overlap with non-bonding metal t2g orbitals (xy, xz, yz) to form M–L π bonds

If the ligands are π donors such as Cl– the ligand π orbitals with be lower in energy that the metal t2g orbitals.

Ligand such as Cl– that act as π donors are low down in the spectrochemical series.

The ligand field splitting is now between t2g* and eg*. This splitting has decreased relative to the σ-only case.

If the ligands are π acceptors such as CO the ligand π orbitals with be higher in energy that the metal t2g orbitals.

The ligand field splitting between t2g and eg*. has increased relative to the σ-only case. Ligand such as CO that act as π acceptors are high up in the spectrochemical series and produce low-spin complexes.

Molecular vibrations – Infrared and Raman spectroscopy

IR absorption occurs when a vibration results in change in electric dipole moment of a molecule.

Raman transition occurs when the polarisability of a molecule changes during a vibration

Exclusion rule

If a molecule has a centre of inversion none of its normal vibrational modes are both IR and Raman active

Example

PtH3N

H3N Cl

ClPt

Cl

H3N Cl

NH3

C2v D2h

Pd-Cl stretching occurs between 200 and 400 cm-1.

Trans isomer: modes cannot be both IR and Raman active

Molecular vibrations – Infrared and Raman spectroscopy

No. of atoms

degrees of freedom

Translational modes

Rotational modes

Vibrational modes

N (linear) 3N 3 2 3N-5

Example

3 (HCN)

9 3 2 4

N (non- linear)

3N 3 3 3N-6

Example

3 (H2O)

9 3 3 3

A molecular vibration is IR active only if it results in a change in the dipole moment of the molecule

A molecular vibration is Raman active only if it results in a change in the polarisability of the molecule

In group theory terms:

A vibrational motion is IR active if it corresponds to an irreducible representation with the same symmetry as an x, y, z coordinate (or function)

and it is Raman active if the symmetry is the same as x2, y2, z2, xy, etc

symmetric stretch anti-symmetric stretch

The vibration will have the same symmetry properties as these vectors. The two vectors together constitute the molecular vibration.

Γ = (1 1 1 1) A1

Γ = (1 -1 -1 1) B2

Both IR andRaman active

Both IR andRaman active

symmetric stretch anti-symmetric stretch

Γ = (1 1 1 1 1 1 1 1) Ag

Γ = (1 -1 1 -1 -1 1 1 -1 ) B2u

x

y

z

Raman active

IR active

In H2O, how many vibrational modes belong to each irreducible representation?

You need the point group and the character table

• Centre of mass (or centre of symmetry) at origin.

• z axis = highest order axis and principal axis. If several axes of highest order, the z axis is the one that passes through most atoms.

• Then assign x axis. If molecule is planar and z axis in plane, then the x axis is plane

In H2O, how many vibrational modes belong to each irreducible representation?

You need the point group and the character table

Use the translation vectors of the atoms as the basis of a reducible representation.

Since you only need the trace recognise that only the vectors that are either unchanged or have become the negatives of themselves by a symmetry operation contribute to the character.

degrees of freedom

Note that a vector will either be unchanged or the negative of itself if the atom does not move.

A reflection can eitherleave the vector unchanged (multiply by +1)

orchange direction (multiply by -1)

A rotation can either invert the direction of the vector (multiply by -1)

or leave it unchanged (multiply by +1)

Now consider only the atoms that don’t move

Nothing moves

9

x -1y -1z 1-1

x 1y -1z 11

x -1,-1,-1y 1,1,1,z 1,1,1

3

Γ

Now express the reducible representation as the sum of the irreducible representations of the point group.

You should find that

Γ = 3A1 + A2 + 2B1 + 3B2

Three are translations (t): A1, B1, B2.

Three are rotations (r): A2, B1, B2

The remaining three are vibrations (v): 2A1 + B2

Three are translations (t): A1, B1, B2.

Three are rotations (r): A2, B1, B2

Vibrational modes of water

Which of these vibrations with A1 or B2 symmetry are IR and/or Raman active?

IR activeRaman active

Suppose we were only interested in the stretching modes and not all the vibrational modes. Then a reasonable basis set would be the following two vectors:

O

H H

It should be easy to see that

Γ = (2 0 0 2) = A1 + B2

Don’t confuse with the following:

To what symmetry species* does the following vibration belong? Is it IR- or Raman-active or both?

*or “how does the following vibration transform”“what us the symmetry of”etc.

Note that this is one single vibration not two vectors representing themotion of the molecule as in the previous example.

N

O O

Review question

a) Draw a Lewis structure of XeF3+ and use VSEPR theory to

predict two plausible structures.

b) Use group theory to explain how vibrational spectroscopy could be used to determine the actual structure