Chem 59-651 Symmetry and Introduction to Group Theory
Symmetry is all around us and is a fundamental property of nature. Chem 59-651 Symmetry and Introduction to Group Theory
The term symmetry is derived from the Greek word “symmetria” which means “measured together”. An object is symmetric if one part (e.g. one side) of it is the same* as all of the other parts. You know intuitively if something is symmetric but we require a precise method to describe how an object or molecule is symmetric.
Group theory is a very powerful mathematical tool that allows us to rationalize and simplify many problems in Chemistry. A group consists of a set of symmetry elements (and associated symmetry operations) that completely describe the symmetry of an object.
We will use some aspects of group theory to help us understand the bonding and spectroscopic features of molecules. Chem 59-651 We need to be able to specify the symmetry of molecules clearly.
F H No symmetry – CHFClBr
Br Cl F H
Some symmetry – CHFCl2 H H Cl Cl
More symmetry – CH2Cl2 Cl Cl H Cl
More symmetry ? – CHCl3 Cl
Cl
What about ?
Point groups provide us with a way to indicate the symmetry unambiguously. Chem 59-651 Symmetry and Point Groups Point groups have symmetry about a single point at the center of mass of the system. Symmetry elements are geometric entities about which a symmetry operation can be performed. In a point group, all symmetry elements must pass through the center of mass (the point). A symmetry operation is the action that produces an object identical to the initial object. The symmetry elements and related operations that we will find in molecules are: Element Operation
Rotation axis, Cn n-fold rotation
Improper rotation axis, Sn n-fold improper rotation Plane of symmetry, σ Reflection Center of symmetry, i Inversion Identity, E
The Identity operation does nothing to the object – it is necessary for mathematical completeness, as we will see later. Chem 59-651 ∞ n-fold rotation - a rotation of 360°/n about the Cn axis (n = 1 to )
O(1) 180° O(1)
H(2) H(3) H(3) H(2)
In water there is a C2 axis so we can perform a 2-fold (180°) rotation to get the identical arrangement of atoms.
H(3) H(4) H(2)
120° 120°
N(1) N(1) N(1)
H(4) H(2) H(4) H(3) H(2) H(3)
In ammonia there is a C3 axis so we can perform 3-fold (120°) rotations to get identical arrangement of atoms. Chem 59-651 Notes about rotation operations: - Rotations are considered positive in the counter-clockwise direction. - Each possible rotation operation is assigned using a superscript integer m m of the form Cn . n - The rotation Cn is equivalent to the identity operation (nothing is moved).
H(3) H(2) H(4)
1 C 2 C3 3
N(1) N(1) N(1)
H(2) H(4) H(4) H(3) H(3) H(2)
H(2)
3 C3 = E
N(1)
H(4) H(3) Chem 59-651 m Notes about rotation operations, Cn : - If n/m is an integer, then that rotation operation is equivalent to an n/m - fold rotation. 2 1 2 1 3 1 e.g. C4 = C2 , C6 = C3 , C6 = C2 , etc. (identical to simplifying fractions)
Cl(5) Cl(2) Cl(3)
1 2 1 C4 C4 = C2 Cl(2) Ni(1) Cl(3) Cl(4) Ni(1) Cl(5) Cl(5) Ni(1) Cl(4)
Cl(4) Cl(3) Cl(2)
3 C4
Cl(4)
Cl(3) Ni(1) Cl(2)
Cl(5) Chem 59-651
m Notes about rotation operations, Cn : - Linear molecules have an infinite number of rotation axes C∞ because any rotation on the molecular axis will give the same arrangement.
C(1) O(2) O(2)C(1)
O(3) C(1) O(2)
N(1)N(2) N(1) N(2) Chem 59-651 The Principal axis in an object is the highest order rotation axis. It is usually easy to identify the principle axis and this is typically assigned to the z-axis if we are using Cartesian coordinates.
Ethane, C2H6 Benzene, C6H6
The principal axis is the three-fold The principal axis is the six-fold axis axis containing the C-C bond. through the center of the ring.
The principal axis in a tetrahedron is a three-fold axis going through one vertex and the center of the object. Chem 59-651 Reflection across a plane of symmetry, σ (mirror plane)
σ O(1) v O(1)
H(2) H(3) H(3) H(2)
These mirror planes are called “vertical” mirror σ planes, v, because they contain the principal axis. σ O(1) v O(1) The reflection illustrated in the top diagram is through a mirror plane H(2) H(3) H(2) H(3) perpendicular to the plane of the water molecule. The plane shown on the bottom is in the same plane as the Handedness is changed by reflection! water molecule. Chem 59-651 Notes about reflection operations: - A reflection operation exchanges one half of the object with the reflection of the other half. σ - Reflection planes may be vertical, horizontal or dihedral (more on d later). - Two successive reflections are equivalent to the identity operation (nothing is moved). σ σ h A “horizontal” mirror plane, h, is perpendicular to the principal axis. This must be the xy-plane if the z- axis is the principal axis. σ σ σ In benzene, the h is in the plane d d of the molecule – it “reflects” each atom onto itself.
σ h
Vertical and dihedral mirror σ σ planes of geometric shapes. v v Chem 59-651 Inversion and centers of symmetry, i (inversion centers) In this operation, every part of the object is reflected through the inversion center, which must be at the center of mass of the object.
1 F Cl 22F Cl 1
2 Br i Br 1 1 221 1Br 2 Br
1 Cl F2 2 Cl F1
i [x, y, z] [-x, -y, -z]
We will not consider the matrix approach to each of the symmetry operations in this course but it is particularly helpful for understanding what the inversion operation does. The inversion operation takes a point or object at [x, y, z] to [-x, -y, -z]. Chem 59-651 m n-fold improper rotation, Sn (associated with an improper rotation axis or a rotation-reflection axis) This operation involves a rotation of 360°/n followed by a reflection perpendicular to the axis. It is a single operation and is labeled in the same manner as “proper” rotations. F1 F2 H1 F F4 1 1 S4 H4 C H2
H3 F F 2 F 3 1 2 S4 F3 F4 F1 H2 σ 90° h H C H 1 1 3 C2
F3 H4 2 F4 S4
H3
H2 C H4
H1
σ Note that: S1 = , S2 = i, and sometimes S2n = Cn (e.g. in box) this makes more sense when you examine the matrices that describe the operations. Chem 59-651 Identifying point groups
We can use a flow chart such as this one to determine the point group of any object. The steps in this process are:
1. Determine the symmetry is special (e.g. octahedral).
2. Determine if there is a principal rotation axis.
3. Determine if there are rotation axes perpendicular to the principal axis.
4. Determine if there are mirror planes.
5. Assign point group. Chem 59-651 Identifying point groups Chem 59-651 Identifying point groups Special cases:
Perfect tetrahedral (Td) e.g. P4, CH4
-2 Perfect octahedral (Oh) e.g. SF6, [B6H6]
-2 Perfect icosahedral (Ih) e.g. [B12H12] , C60 Chem 59-651 Identifying point groups Low symmetry groups:
Only* an improper axis (Sn)
e.g. 1,3,5,7-tetrafluoroCOT, S4
F1
F4
F2
F3
Only a mirror plane (Cs) e.g. CHFCl2
F H
Cl Cl Chem 59-651 Identifying point groups Low symmetry groups:
Only an inversion center (Ci) e.g. (conformation is important !)
F Cl
Br
Br
Cl F
No symmetry (C1) e.g. CHFClBr
F H
Br Cl Chem 59-651 Identifying point groups
Cn type groups: σ A Cn axis and a h (Cnh)
e.g. B(OH)3 (C3h, conformation is important !)
H O
B H
O O
H
e.g. H2O2 (C2h, conformation is important !)
H
O O
H
Note: molecule does not have to be planar e.g. B(NH2)3 (C3h, conformation is important !) Chem 59-651 Identifying point groups
Cn type groups:
Only a Cn axis (Cn) e.g. B(NH2)3 (C3, conformation is important !)
H H N
B H
N N H H
H e.g. H2O2 (C2, conformation is important !)
H
O O
H Chem 59-651 Identifying point groups
Cn type groups: σ A Cn axis and a v (Cnv) e.g. NH3 (C3v)
H
N
H H e.g. H2O2 (C2v, conformation is important !)
O O
H H Chem 59-651 Identifying point groups
Cn type groups: σ A Cn axis and a v (Cnv) e.g. NH3 (C3v, conformation is important !)
H H e.g. carbon monoxide, CO (C∞v) H There are an infinite number of possible
N C axes and σ mirror planes. H H n v H H
H H C O
- e.g. trans-[SbF4ClBr] (C4v) Cl F
F F Sb OC F SbBrCl F F F
Br F Chem 59-651 Identifying point groups
Dn type groups:
A Cn axis, n perpendicular C2 axes σ and a h (Dnh) e.g. BH3 (D3h)
H
H HB H
B
H H
e.g. NiCl4 (D4h)
Cl(2) Cl ClNi Cl
Cl(4) Ni(1) Cl(5)
Cl Ni Cl Cl(3) Chem 59-651 Identifying point groups
Dn type groups: e.g. pentagonal prism (D5h) A Cn axis, n perpendicular C2 axes σ and a h (Dnh)
η5 e.g. Mg( -Cp)2 (D5h in the eclipsed conformation)
Mg
Mg View down the C5 axis
e.g. square prism (D4h)
e.g. carbon dioxide, CO2 or N2 (D∞h)
There are an infinite number of possible σ Cn axes and v mirror planes in addition σ to the h.
O O OC Chem 59-651 Identifying point groups
Dn type groups:
A Cn axis, n perpendicular C2 axes and no mirror planes (Dn) -propellor shapes
e.g. Ni(CH2)4 (D4) H H H H
H H Ni
H H H H
H H
Ni
H H
H H
H H H H
Ni
H H H H Chem 59-651 e.g. (SCH CH ) (D conformation is important!) 2 2 3 3 e.g. propellor (D3)
e.g. Ni(en)3 (D3 conformation is important!) en = H2NCH2CH2NH2 Chem 59-651 Identifying point groups
Dn type groups:
A Cn axis, n perpendicular C2 axes σ and a d (Dnd) e.g. ethane, H3C-CH3 (D3d in the staggered conformation)
H
H H
H H
H
H
H
H
H dihedral means between sides or planes – this is where you find the C2 axes Chem 59-651 η5 e.g. Mg( -Cp)2 and other metallocenes in the staggered conformation (D5d)
Fe Mg Al M
View down the C5 axis These are pentagonal antiprisms
e.g. square e.g. triagular e.g. allene or a tennis ball (D2d) antiprism (D ) antiprism (D3d) 4d Chem 59-651 Identifying point groups
We can use a flow chart such as this one to determine the point group of any object. The steps in this process are:
1. Determine the symmetry is special (e.g. tetrahedral).
2. Determine if there is a principal rotation axis.
3. Determine if there are rotation axes perpendicular to the principal axis.
4. Determine if there are mirror planes and where they are.
5. Assign point group. Chem 59-651 Character Tables for Point Groups Each point group has a complete set of possible symmetry operations that are conveniently listed as a matrix known as a Character Table. As
an example, we will look at the character table for the C2v point group.
Point Group Label Symmetry Operations – The Order is the total number of operations
In C2v the order is 4: σ σ σ σ 1 E, 1 C2, 1 v and 1 ’v C2V E C2 v (xz) ’v (yz)
A1 1 1 1 1
A2 1 1 -1 -1 Character
B1 1 -1 1 -1
B2 1 -1 -1 1 Representation of B2
Symmetry Representation Labels Representations are subsets of the complete point group – they indicate the effect of the symmetry operations on different kinds of mathematical functions. Representations are orthogonal to one another. The Character is an integer that indicates the effect of an operation in a given representation. Chem 59-651 Character Tables for Point Groups
The effect of symmetry elements on mathematical functions is useful to us because orbitals are mathematical functions! Analysis of the symmetry of a molecule will provide us with insight into the orbitals used in bonding. Symmetry of Functions
σ σ C2V E C2 v (xz) ’v (yz) 2 2 2 A1 1 1 1 1 z x ,y ,z
A2 1 1 -1 -1 Rz xy
B1 1 -1 1 -1 x, Ry xz
B2 1 -1 -1 1 y, Rx yz
Notes about symmetry labels and characters:
“A” means symmetric with regard to rotation about the principle axis. “B” means anti-symmetric with regard to rotation about the principle axis. Subscript numbers are used to differentiate symmetry labels, if necessary. “1” indicates that the operation leaves the function unchanged: it is called “symmetric”. “-1” indicates that the operation reverses the function: it is called “anti-symmetric”. Chem 59-651 Symmetry of orbitals and functions
A pz orbital has the same symmetry as an arrow pointing along the z-axis.
z z
≡ y y E x x C2 σ v (xz) σ ’v (yz) No change ∴ symmetric ∴ 1’s in table
σ σ C2V E C2 v (xz) ’v (yz) 2 2 2 A1 1 1 1 1 z x ,y ,z
A2 1 1 -1 -1 Rz xy
B1 1 -1 1 -1 x, Ry xz
B2 1 -1 -1 1 y, Rx yz Chem 59-651 Symmetry of orbitals and functions
A px orbital has the same symmetry as an arrow pointing along the x-axis. z z
y y E No change ≡ x σ (xz) x ∴ symmetric v ∴ 1’s in table z z
y y Opposite C2 x σ’ (yz) x ∴ anti-symmetric v ∴ -1’s in table σ σ C2V E C2 v (xz) ’v (yz) 2 2 2 A1 1 1 1 1 z x ,y ,z
A2 1 1 -1 -1 Rz xy
B1 1 -1 1 -1 x, Ry xz
B2 1 -1 -1 1 y, Rx yz Chem 59-651 Symmetry of orbitals and functions
A py orbital has the same symmetry as an arrow pointing along the y-axis. z z
y y E No change ≡ x σ’ (yz) x ∴ symmetric v ∴ 1’s in table z z
y y Opposite C2 x σ (xz) x ∴ anti-symmetric v ∴ -1’s in table σ σ C2V E C2 v (xz) ’v (yz) 2 2 2 A1 1 1 1 1 z x ,y ,z
A2 1 1 -1 -1 Rz xy
B1 1 -1 1 -1 x, Ry xz
B2 1 -1 -1 1 y, Rx yz Chem 59-651 Symmetry of orbitals and functions
Rotation about the n axis, Rn, can be treated in a similar way. y y The z axis is pointing out of the screen! x x If the rotation is still in No change the same direction (e.g. E ∴ symmetric counter clock-wise), C2 ∴ 1’s in table then the result is y y considered symmetric.
If the rotation is in the opposite direction (i.e. x x clock-wise), then the σ (xz) Opposite result is considered v σ’ (yz) ∴ anti-symmetric anti-symmetric. v ∴ -1’s in table σ σ C2V E C2 v (xz) ’v (yz) 2 2 2 A1 1 1 1 1 z x ,y ,z
A2 1 1 -1 -1 Rz xy
B1 1 -1 1 -1 x, Ry xz
B2 1 -1 -1 1 y, Rx yz Chem 59-651 Symmetry of orbitals and functions d orbital functions can also be treated in a similar way y y The z axis is pointing out of the screen! x x E No change ∴ symmetric C2 ∴ 1’s in table y y
x x σ Opposite v (xz) σ’ (yz) ∴ anti-symmetric v ∴ -1’s in table σ σ C2V E C2 v (xz) ’v (yz) 2 2 2 A1 1 1 1 1 z x ,y ,z
A2 1 1 -1 -1 Rz xy
B1 1 -1 1 -1 x, Ry xz
B2 1 -1 -1 1 y, Rx yz Chem 59-651 Symmetry of orbitals and functions d orbital functions can also be treated in a similar way y y The z axis is pointing out of the screen! So these are representations of the x x view of the dz2 orbital and dx2-y2 orbital down the z-axis. No change y y ∴ symmetric ∴ 1’s in table E
C2 σ x v (xz) x σ ’v (yz)
σ σ C2V E C2 v (xz) ’v (yz) 2 2 2 A1 1 1 1 1 z x ,y ,z
A2 1 1 -1 -1 Rz xy
B1 1 -1 1 -1 x, Ry xz
B2 1 -1 -1 1 y, Rx yz Chem 59-651 Symmetry of orbitals and functions Note that the representation of orbital functions changes depending on the point group – thus it is important to be able to identify the point group correctly. σ σ C2V E C2 v (xz) ’v (yz) 2 2 2 A1 1 1 1 1 z x ,y ,z
A2 1 1 -1 -1 Rz xy
B1 1 -1 1 -1 x, Ry xz
B2 1 -1 -1 1 y, Rx yz
σ σ D3h E 2 C3 3 C2 h 2 S3 3 v 2 2 2 A’1 1 1 1 1 1 1 x + y , z
A’2 1 1 -1 1 1 -1 Rz E’ 2 -1 0 2 -1 0 (x,y) (x2 -y2, xy)
A’’1 1 1 1 -1 -1 -1
A’’2 1 1 -1 -1 -1 1 z
E’’ 2 -1 0 -2 1 0 (Rx, Ry) (xz, yz) Chem 59-651 Symmetry of orbitals and functions σ σ D3h E 2 C3 3 C2 h 2 S3 3 v 2 2 2 A’1 1 1 1 1 1 1 x + y , z
A’2 1 1 -1 1 1 -1 Rz E’ 2 -1 0 2 -1 0 (x,y) (x2 -y2, xy)
A’’1 1 1 1 -1 -1 -1
A’’2 1 1 -1 -1 -1 1 z
E’’ 2 -1 0 -2 1 0 (Rx, Ry) (xz, yz) More notes about symmetry labels and characters: -“E” indicates that the representation is doubly-degenerate – this means that the functions grouped in parentheses must be treated as a pair and can not be considered individually. -The prime (‘) and (“) double prime in the symmetry representation label indicates σ “symmetric” or “anti-symmetric” with respect to the h. cos(120°) = -0.5 y y y sin(120°) = 0.87 +0.87 C C -0.5 +1 2 (x) 3 x x x -1 -0.87 -0.5 (+1) + (-1) = 0 (+0.87) + (-0.87) + (-0.5) + (-0.5) = -1 Chem 59-651 Symmetry of orbitals and functions σ σ Oh E 8 C3 6 C2 6 C4 3 C2 i 6 S4 8 S6 3 h 6 d 2 (C4 ) 2 2 2 A1g 1 1 1 1 1 1 1 1 1 1 x + y + z
A2g 1 1 -1 -1 1 1 -1 1 1 -1 2 2 2 Eg 2 -1 0 0 2 2 0 -1 2 0 (2z -x -y, x2 -y2)
T1g 3 0 -1 1 -1 3 1 0 -1 -1 (Rx,Ry,Rz)
T2g 3 0 1 -1 -1 3 -1 0 -1 1 (xz, yz, xy)
A1u 1 1 1 1 1 -1 -1 -1 -1 -1
A2u 1 1 -1 -1 1 -1 1 -1 -1 1
Eu 2 -1 0 0 2 -2 0 1 -2 0
T1u 3 0 -1 1 -1 -3 -1 0 1 1 (x, y, z)
T2u 3 0 1 -1 -1 -3 1 0 1 -1 More notes about symmetry labels and characters: -“T” indicates that the representation is triply-degenerate – this means that the functions grouped in parentheses must be treated as a threesome and can not be considered individually. -The subscripts g (gerade) and u (ungerade) in the symmetry representation label indicates “symmetric” or “anti-symmetric” with respect to the inversion center, i. Chem 59-651 Character Tables and Bonding We can use character tables to determine the orbitals involved in bonding in a molecule. This process is done a few easy steps.
1. Determine the point group of the molecule.
2. Determine the Reducible Representation, Γ, for the type of bonding you σ π π π wish to describe (e.g. , , ⊥, //). The Reducible Representation indicates how the bonds are affected by the symmetry elements present in the point group.
3. Identify the Irreducible Representation that provides the Reducible Representation; there is a simple equation to do this. The Irreducible
Representation (e.g. 2A1 + B1 + B2) is the combination of symmetry representations in the point group that sum to give the Reducible Representation.
4. Identify which orbitals are involved from the Irreducible Representation and the character table. Chem 59-651 Character Tables and Bonding σ H H Example, the bonding in dichloromethane, CH2Cl2.
The point group is C2v so we must use the appropriate character table Γ Cl for the reducible representation of the sigma bonding, σ. To Cl determine Γσ all we have to do is see how each symmetry operation affects the 4 σ bonds in the molecule – if the bond moves, it is given a value of 0, if it stays in the same place, the bond is given a value of 1. Put the sum of the 1’s and 0’s into the box corresponding to the symmetry operation.
The E operation leaves everything where it is so all four bonds stay in the same place and the character is 4 (1+1+1+1).
The C2 operation moves all four bonds so the character is 0. σ Each v operation leaves two bonds where they were and moves two bonds so the character is 2 (1+1).
Overall, the reducible representation is thus: σ σ C2V E C2 v (xz) ’v (yz)
Γσ 4 0 2 2 Chem 59-651 Character Tables and Bonding We now have to figure out what combination of symmetry representations will add up to give us this reducible representation. In this case, it can be done by inspection, but there is a simple equation that is useful for more complicated situations. σ σ C2V E C2 v (xz) ’v (yz)
Γσ 4 0 2 2
σ σ C2V E C2 v (xz) ’v (yz) 2 2 2 A1 1 1 1 1 z x ,y ,z
A2 1 1 -1 -1 Rz xy
B1 1 -1 1 -1 x, Ry xz
B2 1 -1 -1 1 y, Rx yz
Because the character under E is 4, there must be a total of 4 symmetry representations (sometimes called basis functions) that combine to make Γσ. Since the character under C2 is 0, there must be two of A symmetry and two of B symmetry. The irreducible representation is (2A1 + B1 + B2), which corresponds to: s, pz, px, and py orbitals – the same as in VBT. You can often use your understanding of VBT to help you in finding the correct basis functions for the irreducible representation. Chem 59-651 Character Tables and Bonding σ σ C2V E C2 v (xz) ’v (yz)
Γσ 4 0 2 2
σ σ C2V E C2 v (xz) ’v (yz) 2 2 2 A1 1 1 1 1 z x ,y ,z
A2 1 1 -1 -1 Rz xy
B1 1 -1 1 -1 x, Ry xz
B2 1 -1 -1 1 y, Rx yz
The formula to figure out the number of symmetry representations of a given type is: 1 n=×× ∑[]()()# of operations in class (character of RR) character of X X order Thus, in our example: 1 1 n=+++ []()()()141101121121 ()()() ()()() ()()()n=+−++− []()()()141 10 ()()( 1 121 ) ()()() 12 1 ()()( ) A1 4 B1 4
=++−+−1 =1 + −+ −+ n []()()()141 101 ()()() 12 1 ()()( 12 1 )n ()()(B []()()()141 ) 10 ()()( 1 12 ) 1 ()()( 121 ) ()()() A2 4 2 4
Which gives: 2 A1’s, 0 A2’s, 1 B1 and 1 B2. Chem 59-651 Character Tables and Bonding O σ π Example, the and bonding in SO3. SO
The point group is D3h so we must use the appropriate character table to find the reducible representation of the sigma bonding, Γσ first, then we can O go the representation of the π bonding, Γπ. To determine Γσ all we have to do is see how each symmetry operation affects the 3 σ bonds in the molecule.
σ The E and the h operations leave everything where it is so all three bonds stay in the same place and the character is 3 (1+1+1).
The C3 and S3 operations move all three bonds so their characters are 0.
The C2 operation moves two of the bonds and leaves one where it was so the character is 1.
σ Each v operation leaves one bond where it was and moves two bonds so the character is 1.
Overall, the reducible representation for the sigma bonding is: σ σ D3h E 2 C3 3 C2 h 2 S3 3 v
Γσ 3 0 1 3 0 1 σ σ Chem 59-651 D3h E 2 C3 3 C2 h 2 S3 3 v
Γσ 3 0 1 3 0 1 σ σ D3h E 2 C3 3 C2 h 2 S3 3 v 2 2 2 A’1 1 1 1 1 1 1 x + y , z
A’2 1 1 -1 1 1 -1 Rz E’ 2 -1 0 2 -1 0 (x,y) (x2 -y2, xy)
A’’1 1 1 1 -1 -1 -1
A’’2 1 1 -1 -1 -1 1 z
E’’ 2 -1 0 -2 1 0 (Rx, Ry) (xz, yz)
=+++++1 ==12 nA' []()()()131 201 ( )()() 311 ( 131 )()() 201 ()()() 311 ( )()()n ( )()()1 1 12 A'1 12
=++−+++−1 ==0 nA' []()()()131 201 ( )()() 31 1 ( )()( 131 ) 201 ()()() 31 ( 1 )()()n ( )()(0 ) 2 12 A'2 12 1 12 n= []()()()13 2 + 2 0 ()()() −+ 1 310 + ()()()()()() 13 2 + 2 0 −+ 1 310 ()()()n== ()()()1 E' 12 E' 12 Γ We can stop here because the combination (A’1 + E’) produces the σ that we determined. None σ of the other representations can contribute to the bonding (i.e. nA”1,nA”1 and nE” are all 0). The irreducible representation (A’1 + E’) shows us that the orbitals involved in bonding are the s and 2 the px and py pair; this corresponds to the sp combination we find in VBT. Chem 59-651 Character Tables and Bonding O Γ π Now we have to determine for the bonding in SO3. SO
To determine Γπ we have to see how each symmetry operation affects the π systems in the molecule. The treatment is similar to what we did O for sigma bonding but there are a few significant differences:
1) Pi bonds change sign across the inter-nuclear axis. We must consider the effect of the symmetry operation on the signs of the lobes in a π bond.
2) There is the possibility of two different π type bonds for any given σ bond (oriented 90° from each other). We must examine each of these.
This means that we have to find reducible representations for both the π system perpendicular to the molecular plane (π⊥, vectors shown in red) π and the pi system in the molecular plane ( // , vectors shown in blue).
Note: These are just vectors that are associated with each sigma bond (not with any particular O atom) – they could also be placed in the middle of each SO bond. The vectors should be placed SOto conform with the symmetry of the point group (e.g. the blue vectors conform to the C axis). O 3 σ π Chem 59-651 Example, the and bonding in SO3. First determine the reducible representation for the pi bonding perpendicular to the molecular plane, Γπ⊥. O SO The E operation leaves everything where it is so all three vectors stay in the same place and the character is 3. O
σ The C3 and S3 operations move all three vectors so their characters h ,C2 are 0.
The C operation moves two of the vectors and reverses the sign of the 2 O other one so the character is -1. SO The σ operation reverses the sign of all three vectors so the character h O is -3.
σ Each v operation leaves one vector where it was and moves the two others so the character is 1.
Overall, the reducible representation for the perpendicular π bonding is: σ σ D3h E 2 C3 3 C2 h 2 S3 3 v
Γπ⊥ 3 0 -1 -3 0 1 σ σ Chem 59-651 D3h E 2 C3 3 C2 h 2 S3 3 v
Γπ⊥ 3 0 -1 -3 0 1 σ σ D3h E 2 C3 3 C2 h 2 S3 3 v 2 2 2 A’1 1 1 1 1 1 1 x + y , z
A’2 1 1 -1 1 1 -1 Rz E’ 2 -1 0 2 -1 0 (x,y) (x2 -y2, xy)
A’’1 1 1 1 -1 -1 -1
A’’2 1 1 -1 -1 -1 1 z
E’’ 2 -1 0 -2 1 0 (Rx, Ry) (xz, yz)
=++−+−++1 ==0 nA' []()()()131 2 01 ( )()() 3 11 ( )( 1 31 )() 2 ()( 01 )() 311 ( )()()n ( )()() 0 1 12 A'1 12
=++−−+−−+−+1 ==12 nA'' []()()()131201311131 ( )()() ( )( )( ) 201 ()( )( 311 ) ( )()(n ) (1 )()() 2 12 A''2 12 1 12 n=+−+−+−−++ []()()()132201310132 ()()() ()()() 201310 ()()() ()()()n== ()()()1 E'' 12 E'' 12
Going through all the possibly symmetry representations, we find that the combination (A”2 + E”) produces the Γπ⊥that we determined. The irreducible representation shows us that the possible π orbitals involved in perpendicular bonding are the pz and the dxz and dyz pair. This is in agreement with the π bonding we would predict using VBT. σ π Chem 59-651 Example, the and bonding in SO3. First determine the reducible representation for the π bonding in the Γ O molecular plane, π//. SO The E operation leaves everything where it is so all three vectors stay in the same place and the character is 3. O
σ The C3 and S3 operations move all three vectors so their characters v ,C2 are 0.
The C2 operation moves two of the vectors and reverses the sign of the other one so the character is -1. O
σ SO The h operation leaves all three vectors unchanged so the character is 3. O σ Each v operation reverses the sign one vector where it was and moves the two others so the character is -1.
Overall, the reducible representation for the parallel π bonding is: σ σ D3h E 2 C3 3 C2 h 2 S3 3 v Γ π// 3 0 -1 3 0 -1 σ σ Chem 59-651 D3h E 2 C3 3 C2 h 2 S3 3 v Γ π// 3 0 -1 3 0 -1 σ σ D3h E 2 C3 3 C2 h 2 S3 3 v 2 2 2 A’1 1 1 1 1 1 1 x + y , z
A’2 1 1 -1 1 1 -1 Rz E’ 2 -1 0 2 -1 0 (x,y) (x2 -y2, xy)
A’’1 1 1 1 -1 -1 -1
A’’2 1 1 -1 -1 -1 1 z
E’’ 2 -1 0 -2 1 0 (Rx, Ry) (xz, yz) 1 0 n=++−+++− []()()()131 2 01 ( )()() 3 11 ( )( 131 )() 2 ()()() 01 31 ( 1 )()()n== ( )()(0 ) A'1 12 A'1 12 1 12 n=++−−+++−− []()()()131 2 01 ( )()() 3 1 ( 1 )( 131 )( ) 2 01 ()()() 3 1 ( 1 )()()n== ( )(1 )( ) A'2 12 A'2 12 1 12 n=+−+−++−+− []()()()132201310132201 ()()() ()()() ()()() 310 ()()()n== ()()()1 E' 12 E' 12
Going through all the possibly symmetry representations, we find that the combination (A’2 + E’) Γ π produces the π// that we determined. The possible orbitals involved in parallel bonding are only the dx2-y2 and dxy pair. The A’2 representation has no orbital equivalent. Note: Such analyses do NOT mean that there is π bonding using these orbitals – it only means that it is possible based on the symmetry of the molecule. Chem 59-651 Character Tables and Bonding O σ π - Example, the and bonding in ClO4 . O Cl O
The point group is Td so we must use the appropriate character table to find the reducible representation of the sigma bonding, Γσ first, then we can go O the representation of the π bonding, Γπ.
The E operation leaves everything where it is so all four bonds stay in the same place and the character is 4.
Each C3 operation moves three bonds leaves one where it was so the character is 1.
The C2 and S4 operations move all four bonds so their characters are 0.
σ Each d operation leaves two bonds where they were and moves two bonds so the character is 2.
σ Td E 8 C3 3 C2 6 S4 6 d
Γσ 4 1 0 0 2 Chem 59-651 σ Td E 8 C3 3 C2 6 S4 6 d
Γσ 4 1 0 0 2
σ Td E 8 C3 3 C2 6 S4 6 d 2 2 2 A1 1 1 1 1 1 x + y + z
A2 1 1 1 -1 -1 E 2 -1 2 0 0 (2z2 -x2 -y2, x2 -y2)
T1 3 0 -1 1 -1 (Rx,Ry,Rz)
T2 3 0 -1 -1 1 (x, y, z) (xy, xz, yz)
σ The irreducible representation for the bonding is (A1 + T2), which corresponds to the s orbital and the (px, py, pz) set that we would use in VBT to construct a the sp3 hybrid orbitals suitable for a tetrahedral arrangement of atoms. To get the representation for the π bonding, we must do the same procedure that we did for
SO3, except that in the point group Td, one can not separate the representations into parallel and perpendicular components. This is because the three-fold symmetry of the bond axis requires the orthogonal vectors to be treated as an inseparable pair. σ π - Chem 59-651 Example, the and bonding in ClO4 . The analysis of how the 8 vectors are affected by the symmetry operations gives:
σ Td E 8 C3 3 C2 6 S4 6 d
Γπ 8 -1 0 0 0
σ Td E 8 C3 3 C2 6 S4 6 d 2 2 2 A1 1 1 1 1 1 x + y + z
A2 1 1 1 -1 -1 E 2 -1 2 0 0 (2z2 -x2 -y2, x2 -y2)
T1 3 0 -1 1 -1 (Rx,Ry,Rz)
T2 3 0 -1 -1 1 (x, y, z) (xy, xz, yz)
π The irreducible representation for the bonding is (E + T1 + T2), which corresponds to the dx2-y2 and dxy pair for E and either the (px, py, pz) set or the (dxy, dxz, dyz) set for T2, since T1 does not correspond to any of the orbitals that might be involved in bonding. Because the (px,py,pz) set has already been σ π used in the bonding, only the (dxy,dxz,dyz) set may be used for bonding.