4b: Symmetry and Group theory

Question 1

What is the difference between a proper rotation and an improper rotation?

Answer 1

Proper rotations can be done with a model, improper cannot.

Question 2

Name all the proper an improper rotations.

Answer 2

E (C₁), C_n, σ (S₁), i (S₂), S_n

Question 3

Name and define the labels for reflections.

Answer 3

h = plane perpendicular to the principle axis

v = in the plane of the principle axis and passes through the bonds

d (or v’) = in the plane of the principle axis and passes between the bonds

Question 4

How are anticlockwise rotations show?

Answer 4

With a minus, C_n^-

Question 5

How many unique operations exist for a S₃ axis?

Answer 5

2, S₃ and S₃⁵.

Question 6

What do the labels gerade and ungerade mean?

Answer 6

gerade = g = orbital phases don’t invert when molecule inverts

ungerade = u = orbital phases invert with the molecule

Question 7

How can you work out if you have missed some operations?

Answer 7

The number of proper and improper operations is always equal unless there is no improper operations.

Question 8

Define the principle axis.

Answer 8

The rotational axis with the highest symmetry which defines the z axis.

Question 9

Define a point group of symmetry operations and its order.

Answer 9

A set of symmetry operations that form a closed set where succesive applications of the operations in the set is equivalent to another operation in the set.

The order is the total number of symmetry operations in the group.

Question 10

How can eigenvalues define orbital changes after symmetry operations?

How are these values used

Answer 10

O^ψ = aψ

O^ is the operator

ψ is the property

a is the eigenvalue

a = 1 = unchanged, a = -1 = inversion, a = 0 = shifted

Question 11

What do the irreducible labels, A, B, E and T mean?

Answer 11

A = single orbital, totally symmetric about the principle axis

B = single orbital, antisymmetric about the principle axis

E = double orbital

T = triple orbital

Question 12

What do each of the subscript 1, 2, g and u and superscript ‘ and ‘’ labels mean for a irreducible representation?

Answer 12

₁ = property is symmetric under C₂’ (perpendicular to principle axis) (or σ_v if no C₂’)

₂ = property is antisymmetric under C₂’ (perpendicular to principle axis) (or σ_v if no C₂’)

_g = symmetric under i

_u = antisymmetric under i

’ = symmetric under σ_h

’’ = antisymmetric under σ_h

Question 13

How can matrix notation be used to describe an orbital rotation?

Answer 13

By describing how the orbtial changes postion upon rotation, [p_x] → [p_y]

This can also be described for 2 orbitals with a 2x2 matrix.

Question 14

What is the general matrix form for a clockwise rotation?

Answer 14

Question 15

How are basic and complex reducible representations reduced?

Answer 15

Basic can be simply reduced by looking at the eigenvalues and matching the combined reducible values to individual values from the point group.

The equation is 1/h ∑ x_R x_i N = n (Γ)

n (Γ) = the number of times the irreducible appears in the reducible

h = order of the group

xR = character in the reducible

xi = character in the irreducible

N = no. of symmetry operations in the class

Question 16

How are orbitals that can’t be directly represented shown on MO diagrams?

Answer 16

Using LCAO such as for H2O, the H atoms on the water cannot be individually represented so they are split into an in phase and out of phase pair. These have their own irreducible assignments so they can both be placed on the MO diagram.

Question 17

Draw how does mixing affects the orbitals of H₂O.

Answer 17

Question 18

How can you tell how orbitals allign when the MO scheme becomes very complicated?

Answer 18

Projection operator method/SALCs - symmetry adapted linaer combinations.

Take BF₃s F, 2p_z orbitals. Transform one orbtial by each of the symmetry operations and find the number of each type of orbital made (σ₁ with E = σ₁, σ₁ with C₃ = σ₂)

For BF3 the 2p_z orbitals combine all in phase.

Question 19

How are double degenerate orbitals represented by SALCs?

Answer 19

2 different combinations are produced, one from the swaping of orbital phases to give the in phase, out of phase outcome. The other is produced from the fact that all the orbitals are orthogonal so they must all add up to 0.

For BF3: a1’ = σ1 + σ2 + σ3 = 0 (a + b + c = 0)

1st e’ = 2σ1 - σ2 - σ3 = 0 (2a - b - c = 0)

Summing these gives 3a = 0

Therefore b = -c hence you can find the orthogonal combination by b - c = 0