Group Theory Flashcards

1
Q

symmetry operations and symmetry elements

A
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2
Q

point group

A

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

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3
Q

the composition of some common groups

A

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

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4
Q

character tables

A

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

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5
Q

group order, h

A

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

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6
Q

irreducible representation (irrep)

A

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)

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7
Q

symmetry species labels

A

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)

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8
Q

characters, χ

A

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

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9
Q

character table - the last two columns

A

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.)

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10
Q

consider vector r1 under the symmetry operations of C2v

A

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

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11
Q

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

A

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)

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12
Q

compare characters with the C2v character table

A

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).

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13
Q

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

A

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

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14
Q

sum of irreps

A

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

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15
Q

reduction formula

A

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

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16
Q

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

A
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17
Q

vibrational spectroscopy

A

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

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18
Q

consider O-H stretching vibrations in water (C2v)

A

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

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19
Q

projector operator method

A

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

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20
Q

use vectors r1 and r2 to generate the projection operator

A

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

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21
Q

IR and Raman Spectroscopy

A

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

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22
Q

intensity, I

A

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

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23
Q

infrared spectroscopy

A

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

24
Q

application of group theory in spectroscopy

A

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)

25
Q

Ψi and Ψf

A

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)

26
Q

Raman spectroscopy

A

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

27
Q

molecular orbital (MO) theory

A

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

28
Q

linear combination of atomic orbitals (LCAO)

A

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

29
Q

basis set of atomic orbitals

A

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

30
Q

bonding orbital

A

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

31
Q

antibonding orbital

A

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

32
Q

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

A
  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

33
Q

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

BORON

A

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’

34
Q

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

3 x HYDROGEN

A

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’

35
Q

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

A

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

36
Q

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

A

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)

37
Q

drawing the MO diagram (BH3)

A

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

38
Q

constructing qualitative MO diagram for NH3

A

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

39
Q

drawing MO diagram - NH3

A

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
- realatively 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

40
Q

visualise MOs in NH3

A

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))

41
Q

hybridisation of atomic orbitals

A

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

42
Q

consider the NH3 MO diagram using hybrid AOs

A

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

43
Q

Molecular structure - two molecules of formula AH3

planar BH3 (D3h) and pyramidal (C3v)
- why do these molecules adopt different structures

A

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
44
Q

consider trigonal pyramidal and trigonal planar geometries

A

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)

45
Q

ligands in metal complexes

A

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

46
Q

spectrochemical series

A

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)

47
Q

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

A

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 and 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

48
Q

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

A

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

49
Q

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

A

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

50
Q

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

A

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

51
Q

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

A

P is less electronegative than N
- P 3p orbitals similar in enery 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 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

52
Q

hypervalence

A

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

53
Q

Oh point group character table for S6

54
Q

bonding scheme for hypervalent SH6 (Oh)

A

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

55
Q

drawing MO diagram for SH6