Group Theory

Question 1

symmetry operations and symmetry elements

Answer 1

Question 2

point group

Answer 2

denotes the number and nature of the symmetry elements of a given molecule

labelled by its Schoenflies symbol

assigning a molecule to its point group depends on identifying its symmetry elements

Question 3

the composition of some common groups

Answer 3

it is not essential to memorise the symmetry elements of a point group - as these will be listed in the character table

Question 4

character tables

Answer 4

lists the symmetry elements of a particular point group

symmetry elements arranged by class - a specific grouping of symmetry operations of the same general type

Question 5

group order, h

Answer 5

total number of symmetry operations that can be carried out in the same point group

Question 6

irreducible representation (irrep)

Answer 6

each row of characters on character table corresponds to a particular irrep of the group

symmetry species - label of irrep (corresponds to first column of character table)

Question 7

symmetry species labels

Answer 7

A - function is symmetric under that rotation (its character is 1)

B - function changes sign under that rotation (its character is -1)

subscript 1 - symmetric under principle reflection (its character is 1)

subscript 1 - antisymmetric under principle reflection (its character is -1)

Question 8

characters, χ

Answer 8

corresponds to entries in main part of the table

each character shows how an object or mathematical function (such as atomic orbital), is affected by the corresponding symmetry operation

Question 9

character table - the last two columns

Answer 9

contain examples of functions that exhibit the characteristics of each symmetry species

one column contains functions defined by a single axis, such as:
(x,y,z) p-orbitals
(Rx, Ry, Rz) rotations around an axis

other column contains quadratic functions (defined by two axes), such as those that represent d orbitals (xy etc.)

Question 10

consider vector r1 under the symmetry operations of C2v

Answer 10

under identity operator (E)
-vector is unchanged
-appears as r1

under C2 rotation
-vector roatates 180° (sign changes)
-appears as -r1

under σ(zx) operation
-vector is unchanged
-appears as r1

under σ’(yz) operation
-vector is mirrored (sign changes)
-appears as -r1

Question 11

consider matrix representative that is required to transform the vector r1 under each of the symmetry operations in C2v

Answer 11

r1 → r1
unchanged after identity operator, E

r1 → -r1
180 degrees rotation around z-axis, C2

r1 → r1
reflection on zx mirror plane, σv(zx)
- r1 sits on this mirror plane

r1 → -r1
reflection on yx mirror plane, σv(yx)

Question 12

compare characters with the C2v character table

Answer 12

r1 spans B1 in C2v (function r1 has the same character as B1 irrep)

The first of these two columns contains functions defined by a single axis, such as translations in space or p-orbitals (x, y, z) or rotations around an axis (Rx, Ry, Rz).

We could have instantly determined the irreps spanned by our function r1, which is defined by a single axis, simply by referring to the character table
- this is only the case when the function being operated on sits at the intersection of all the symmetry elements (the point of the point group)

The second of these two columns contains quadratic functions, i.e. functions defined by two axes, such as d orbitals (e.g. xy, yz, xz).

Question 13

reducible representation can be broken down into sum of irreps - broken down into more simple components that are irreps within the point grop

Answer 13

consider point P located at coordinates (x,y,z) - transform point P according to symmetry operations in C2v using matrix representations

leading diagonal = trace
trace of matrix provides its character - this is NOT an irreducible representation in C2v

representations of Pxyz broken down into its componenets Px, Py and Pz

Question 14

sum of irreps

Answer 14

Γ = n1Γ1 + n2Γ2 + …

Γi - denotes the various symmetry species of the group

ni - denotes how many times each symmetry species appears in the reducible representation

for this example:
Γ = 1xA1 + 0xA2 + 1xB1 + 1xB2

Question 15

reduction formula

Answer 15

ni = number of times a irrep i appears in the reducible representation

R = symmetry class

h = the group order of the point group (read from character table)

g(R) = the number of symmetry operations in the symmetry class R (read from the character table)

χi(R) = the character of symmetry class R for the irrep i (read from the character table)

χ(R) = the character of the reducible representation Γ under the symmetry class R

Question 16

reduction formula used to determine the irreps sanned by our reducible basis in the C2v - confirms what we found by inspection of the matrix representative

Answer 16

Question 17

vibrational spectroscopy

Answer 17

concerned with the observation of the degrees of vibrational freedom

to determine the number of degrees of vibrational freedom, whether vibration can be observed in spectroscopy and to predict ordering of energies - requires analysis of symmetry associated with molecule

Question 18

consider O-H stretching vibrations in water (C2v)

Answer 18

two vectors along the O-H bond
-consider how many basis functions are left unchanged by each symmetry operation in C2v

this is not an irreducible representation - apply the reduction formula to our reducible basis Γ in the C2v point group to determine irreps spanned by Γ

Γstretch spans A1 and B1

hence, there are two O-H stretching vibrations in an H2O molecule which span A1 and B1 in C2v

NOTE: number of input functions must equal to number of answers returned - started with two input functions, determined two stretching vibrations

Question 19

projector operator method

Answer 19

to visualise what the A1 and B1 stretches look like

Ψa = combination of the basis functions that gives a function of irrep ‘a’

χa(i) = the character of symmetry operation i for the irrep ‘a’ (read from the character table

Pi(Φ) = the projection operator of one of the basis functions under the symmetry operation i

Question 20

use vectors r1 and r2 to generate the projection operator

Answer 20

observe effects of r1 only:
- identity and reflection through σv(xz) leave r1 unchanged
- C2 rotation and reflection through σv’(yz) transforms r1 into r2

multiply these results by characters of each irrep spanned by the basis (r1, r2)
-Γr1+r2 span A1 and B1 in C2v

A1 stretching vibration appears as (r1+r2)
- the two H move into O together (motion of the two Hs is in sync)

B1 stretching vibration appears as (r1-r2)
- antisymmetric vibration

Question 21

IR and Raman Spectroscopy

Answer 21

records the frequency of a molecular vibration, but not all modes of vibration of a particular molecule give rise to observable absorption bands in IR or Raman spectra

Question 22

intensity, I

Answer 22

in any form of spectroscopy, the intensity of a transition from an initial state described mathematically by Ψi to a final state Ψf is given by:

T = a transition moment operator (depends on type of spectroscopy)
dτ = integration over all space

Question 23

infrared spectroscopy

Answer 23

absorption of electromagentic radiation by molecules occurs only if the charge distributions in Ψi and Ψf differ in a manner corresponding to an electric dipole

transition between Ψi and Ψf is called an electric dipole transition
T is therefore the electric dipole moment operator, μ

Question 24

application of group theory in spectroscopy

Answer 24

group theory tells us if I is zero or non-zero
- whether a particular molecular vibration appears in IR spectrum
- we need to know the irreps spanned by μ, Ψi and Ψf

at room temperature, all molecules will be in their ground vibrational state - which spans the totally symmetric irrep of its point group (A1 ~ all characters =1)

Question 25

Ψi and Ψf

Answer 25

Ψi spans A1 Ψf spans the irreps determined by a group theory analysis of the molecular vibrations μ spans the same irreps as the Cartesian x, y and z axes (directly read from the character table) requirement for non-zero I: Ψf spans the same irreps as at least one componenent of μ (i.e. x, y or z) FOR INSTANCE: in H2O, Γstretch spans A1 + B1 (i.e. Ψf spans A1 + B1) in C2v, μ spans A1 (z), B1 (x) and B2 (y) therefore, both stretching vibrations of water visible in its spectrum - as they span the same irrep as a component of μ (both vibrations involve a change in dipole moment)

Question 26

Raman spectroscopy

Answer 26

molecular vibration is detectable in Raman spectroscopy if there is a change in the polarisability of the molecule between the initial Ψi and final Ψf states for Raman sepctroscopy, T is known as the polarisability operator, α FOR INSTANCE: in H2O, at least one component of α spans every irrep in C2v therefore, both stretching vibrations of water appear in Raman spectrum

Question 27

molecular orbital (MO) theory

Answer 27

group theory allows us to construct molecular orbital energy level diagrams to understand chemical bonding MO arises from interaction between AO in MO theory, we assume that the wavefunction (Ψ) of the electrons in a molecule can be written as a product of one-electron functions Ψ = ψ(1).ψ(2)...ψ(N) wavefunction squared gives the propability of finding that electron in a molecule when an electron is close to the nucleus of an atom, its wavefunction closely resembles an atomic orbital of that atom - when an electron is close to nucleus of H atom in a molecule, its wavefunction is like a 1s orbital of that atom

Question 28

linear combination of atomic orbitals (LCAO)

Answer 28

we can construct a reasonable approximation of a MO by combining the atomic orbitals contributed by each atom in the bond combine H 1s orbitals on each atoms to approximate the MO formed in an H2 molecule - consider each 1s orbital as a wavefunction (wave) - it will have a phase (+ or -) and interfere in a constructive (in-phase) or destructive (out-of-phase) manner

Question 29

basis set of atomic orbitals

Answer 29

ΨH2(in-phase) = ΦHa + ΦHb ΨH2(out-of-phase) = ΦHa - ΦHb an electron in MO is likely to be found where the orbital has a large amplitude - will not be found at node

Question 30

bonding orbital

Answer 30

electron that occupies the in-phase combination has an enhanced probability of being found in the internuclear region - can interact strongly with both nuclei, binding the two atoms together

Question 31

antibonding orbital

Answer 31

electron that occupies the out-of-phase combination is largely excluded from the internuclear region due to the formation of a nodal plane between the atoms - reduced electron density between atoms results in increased internuclear repulsion

Question 32

In general, a MO arises from the interactions between the orbitals on atomic centres and such interactions are:

Answer 32

1. allowed if the atomic orbitals span the same irrep of the point group (have the same symmetry that allows them to overlap in a constructive way) 2. efficient if the atomic orbitals are relatively close in energy 3. efficient if there is significant spatial overlap of the orbitals numer of MOs that can be formed must equal to the number of AO of the constituent atoms

Question 33

MOs in BH3 (polyatomic molecule) - the ligand group orbital (LGO) approach BORON

Answer 33

BH3 point group: D3h consider 2s atomic orbital of B atom in BH3 - apply each symmetry operation of D3h - irrep spanned by B 2s orbital is A1 (label as a1) B atom sits on the point at which all symmetry operations meet (the point of the point group) - irreps spanned by B 2px, 2pz and 2py orbitals can be read straight from character table - x and y span e' (doubly degenerate in D3h) z spans a2'' 2px and 2py are not divisible by individual irreps - after C3 rotation, 2px has some component of x and some component of y - 2px commutes to xy - 2py commutes to xy - they span e'

Question 34

MOs in BH3 (polyatomic molecule) - the ligand group orbital (LGO) approach 3 x HYDROGEN

Answer 34

for an interaction between the orbitals on B and for orbitals on the H atoms to be allowed, the orbitals must span the same irreps start with 3 input AOs and construct 3 output LGOs - to determine the symmetry of the H--H LGOs, count the unmoved H 1s orbitals when each symmetry operation is performed characters of reducible representation in a basis of the H 1s AOs does not correspond to an irrep in D3h - reduce basis (three H 1s orbitals) into a basis of LGOs that correspond to a combination of irreps in D3h - apply reduction formula to reducible basis ΓH 1s in the D3h point group ΓH3 spans a1' and e'

Question 35

constructing qualitative MO diagram for BH3 - overlap efficient if AOs relatively close in E

Answer 35

in molecules containing atoms of different elements, consider the relative energies of the orbitals involved - estimate obtained from electronegativity values - B less electronegative than H, approximate that 2p AOs of B higher in E than AOs of H - s/p separation in B not large, B 2s and H 1s are therefore close in E as the effective nuclear charge (B to F) increases the energy separation between 2s and 2p increases considerably.

Question 36

constructing qualitative MO diagram for BH3 - overlap efficient if there is significant spatial overlap of the orbitals

Answer 36

can estimate whether spatial overlap is likely to be significant by considering projections of the orbitals which overlap B 2s orbital is relatively large (low Zeff) - expected to have good overlap with H 1s B 2p orbitals dont directly overlap with H 1s - poorer overlap with H 1s compared to B 2s (2pz orbital has zero overlap with H 1s)

Question 37

drawing the MO diagram (BH3)

Answer 37

to form MOs, match the symmetry of B valence orbitals to H SALCs, drawing resultant MOs in the centre of the diagram ignore the B core 1s state (lablled as 1a1) - too low in energy to be involved in bonding B 2s and B 2p have relatively small spacing between them - H orbitals between the B 2s and B 2p combine B 2s orbital (a1) with the a1 H SALC to form a bonding and antibonding combination - good energy overlap and good spatial overlap - forms strongly bonding and strongly antibonding a1 MOs combine doubly degenerate px and py orbitals with doubly degenerate H SALCS - good energy overlap but poorer spatial overlap - reasonably set of bonding and antibonding double degenerate e' MOs B 2pz (a2') is non-bonding antibonding destabilising interaction = bonding stabilising interaction

Question 38

constructing qualitative MO diagram for NH3

Answer 38

1) SYMMETRY OVERLAP NH3 point group: C3v N lies on point of point group - therefore, irreps spanned by N 2s and N 2p orbitals can be directly read from character table N 2s spans a1 N 2pz spans a1 N 2px and 2py span E (i.e. px and py are degenerate) apply reducing formula to find that H 1s LGO span a1 and e in C3v 2) ENERGY OVERLAP N more electronegative than H - N 2p orbitals lower in E than H 1s 3) SPATIAL OVERLAP estimate whether spatial overlap is likely to be good by considering projections of orbitals which overlap

Question 39

drawing MO diagram - NH3

Answer 39

N 2s interacts with the a1 LGO - large energy gap between N 2s and H 1s - relatively weak interaction N 2px and 2py interact with the two H 1s LGOs of e symmetry - relatively well matched in energy - form two relatively bonding and two relatively antibonding MOs N 2pz interacts with a1 LGO - forms higher energy bonding a1 orbital - poor spatial overlap with the H 1s orbitals due to orientation of ther orbital - will be principally N 2pz in character

Question 40

visualise MOs in NH3

Answer 40

use projection operator method on our basis of H 1s AOs to generate set of Symmetry Adapted Linear Combinations (SALCs) visualise what a1 and e H 1s LGO look like - label the three H 1s orbitals as Ψ1, Ψ2, Ψ3 - observe effects on Ψ1 only e is doubly degenerate, therefore, there must be 2 LGOs for e - first consider LGO(2), the condition for orthogonality means that the second nodal plane must pass through Ψ1 - hence, Ψ1 = 0 for the second e LGO (LGO(3))

Question 41

hybridisation of atomic orbitals

Answer 41

AOs can overlap to from hybridised orbitals if they span the same irrep - the extent to which this happens will be dictated by the energy difference between the orbitals when s orbital hybridised with p orbital, two hybrid sp orbitals are formed (an in-phase and out-of-phase combination) orbitals overlap in-phase: constructive interference orbitals overlap out-of-phase: destructive interference for s and p orbitals of similar energy, this results in two sp hybrid orbitals with increased electron density on one side for orbitals of different energies, the degree of interaction decreases for orbitals with a large separation, the overlap which results wll retain the majority of their initial character, but is slightly polarised

Question 42

consider the NH3 MO diagram using hybrid AOs

Answer 42

For nitrogen, the s/p separation is quite large - hybrid orbitals are not equivalent orbital with principally s character is skewed down towards the H atoms, providing stronger overlap with H 1s AO orbital with principally pz character has decreased electron density below the N atom and a large lobe projecting above the N atom - this orbital has very poor spatial overlap - neglected in bond order calculations ~ it is the lone pair of NH3

Question 43

Molecular structure - two molecules of formula AH3 planar BH3 (D3h) and pyramidal (C3v) - why do these molecules adopt different structures

Answer 43

the structure adopted by any molecule will be that which minimised its total energy estimate the electronic contribution to the total energy by considering the energy of the occupied MOs - molecular shape is a function of the total energy of molecule - lowest energy configuration of molecule will be adopted considering steric interactions and electronic energy of molecule as a whole

Question 44

consider trigonal pyramidal and trigonal planar geometries

Answer 44

1a1 lower in energy than 1a1' - the s/p hybridisation that is symmetry allowed in C3v allows stronger overlap (more stable) between central s orbital and H 1s AOs 1e orbital higher in energy than 1e1' - in C3v, H 1s below the plane of the molecule - decreased overlap (less stable) with the H 1s AO than in D3h where the central px and py AO are in the same plane pz orbital (1a2') is entirely non-bonding in D3h but in C3v hybridisation on the central atom stabilises the orbital compared to the unhybridised orbtial (2a1) - there are no other valence orbitals on the central atom with a2' symmetry in D3h 6 electrons ~ D3h energetically favourable (BH3) 8 electrons ~ C3v energetically favourable (NH3)

Question 45

ligands in metal complexes

Answer 45

different ligands influence the magnitude of the octahedral splitting experienced by metal centres in metal-ligand complexes - apparant by changes in the colours of complexes

Question 46

spectrochemical series

Answer 46

ligands low in series (π-donors) - provide a weak field (small splitting) ligands high in series (π-acceptors) - provide strong field large splitting) ligands in middle (σ-donors)

Question 47

use MO theory to understand why ligands (H2O) appear at different locations in spectrochemical series

Answer 47

MO of H2O frontier orbitals in H2O involved in bonding with M - close in energy to oxygen 2p - electronegativity of oxygen is a reasonable proxy (typical transition metals are significantly more electropositive than H ~ orbitals much higher in MO diagram) 2a1 MO of H2O too low in energy - unlikely to have significant overlap with metal 1b1 has no significant elecron density available for bonding - nothing above the oxygen atom, poor spatial overlap 3a1 and 1b2 MO remain for bonding

Question 48

consider symmetry overlap of 3a1 and 1b1 MO of H2O for bonding to metal

Answer 48

σ - orbital UNCHANGED on 180° rotation around the internuclear axis π - orbital CHANGES on 180° rotation around the internuclear axis 3a1 donates electron density in a σ fashion 1b2 donates electron density in a π fashion H2O is both a σ and π donor

Question 49

use MO theory to understand why ligands (NH3) appear at different locations in spectrocchemical series

Answer 49

2a1 MO in NH3 too low in energy to bond to metal 1e MO in NH3 would not have significant spatial overlap 3a1remains for bonding

Question 50

consider symmetry overlap of 3a1 MO of NH3 for bonding to metal

Answer 50

3a1 donates electron density in a σ fashion there are no available donor orbitals of π-symmetry in NH3 unlike H2O (1b2) NH3 is a σ-only ligand H2O has (weak) π-donating character H2O is lower in the spectrochemical series (weaker field ligand) than NH3

Question 51

use MO theory to compare positon of PH3 ligand to NH3 in spectrochemical series

Answer 51

P is less electronegative than N - P 3p orbitals similar in energy to H 1s - s/p separation in P is much smaller than in N P 3s and H 1s much closer in energy, forming stronger bonding and antibonding a1 orbitals - pushing the a1 antibond above e antibond both NH3 and PH3 have a1 HOMO which will donate electron densiy in a σ fashion in NH3, LUMO is the a1* orbital (σ symmetry) - not the correct symmetry to accept back donation of electron density from a metal centre - NH3 is σ-donor only in PH3, LUMO is the e* orbital - correct symmetry to accept back donation - PH3 is a π-acceptor occupying antibonding MO weakens molecular bonding - the MOs higher in energy than the LUMO are not available for back-bonding - orbitals have a strongly destabilising effect on molecule

Question 52

hypervalence

Answer 52

hypervalent compounds commonly form in row 3 and later - requires s and p orbital participation - 3d orbitals is so much higher in energy than 3s and 3p, their interaction is unlikely

Question 53

Oh point group character table for S6

Answer 53

Question 54

bonding scheme for hypervalent SH6 (Oh)

Answer 54

consider interaction between the AO of S and LGOs of H6 construct MO diagram 1) S sits on point of point group - irreps spanned by its valence orbitals (3s, 3p) directly read from character table - 3s spans a1g - 3p spans t1u use basis of 6 H 1s AOs to obtain reducible representation - apply reducing formula to determine irreps spanned - Γ(H 1s) spans a1g, eg and t1u 2) consider spatial overlap - use projection operator to understand the H 1s SALCs 3) consider energy overlap - S more electronegative than H, S 3p orbital lower in energy than H 1s - relatively small s/p gap for S - S 3d orbitals too high in energy to be involved in bonding with H 1s

Question 55

drawing MO diagram for SH6

Answer 55