group theory and symmetry...group theory and symmetry o n o n z z dj q give the irreducible...
TRANSCRIPT
Group Theory and Symmetry
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dj q
Give the irreducible representation(s) for the d
orbitals of the P atom in PF5 (use the appropriate
character table in your text)
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O zz
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symmetry of orbitals and
vibrations
how do things that are part of a molecule
change under allowed symmetry operations?
how can we determine the symmetry types of
more more complex systems
that is, how can one get the irreducible
representations from reducible representations?
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O zz
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group theory analysis for
bonding in H2O
orbitals available
O 1s
O 2s
O 2px, 2py, 2pz
H 1s’s
7 valence atomic orbitals
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O
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O zz
z
reducible representation
for C2
matrix
representation of C2
= 1
for E, = 7
Find for v(xz),
v(yz) and reps and
irred reps in groups
1 0 0 0 0 0 0
0 1 0 0 0 0 0
0 0 1 0 0 0 0
0 0 0 1 0 0 0
0 0 0 0 1 0 0
0 0 0 0 0 0 1
0 0 0 0 0 1 0
O1s
O2s
O2 px
O2 py
O2 pz
H11s
H21s
=
O1s
O2s
O2 px
O2 py
O2pz
H21s
H11s
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O
ON
O zz
z
N: number of times irred rep, x, appears in the reducible
representation
h is the order of the group(sum of all E characters)
r is the character of the reducible representation for the
operation, x
i is the character of the irreducible representation for the
operation, x
n is the number of operations in the class, x
C4 would be a class (not in C2v)
In C4v there are 2 C4
N =1
h rx• i
x
x
•nx
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O
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O zz
z
irreducible
representations of orbitals
molecule lies in yz plane
4 A1
2 B2
1 B1
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O
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O zz
z
character table
C2v E C2 v(xz) v(yz)
A1 1 1 1 1 z, z2, x2,y2
A2 1 1 -1 -1 Rz, xy
B1 1 -1 1 -1 x, Ry,xz
B2 1 -1 -1 1 y, Rx, yz
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OON
O
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O zz
z
irreducible
representations of O atom
orbitals
molecule lies in yz plane and O sits at the
origin…
4 A1
O1s, O2s, O2pz
2 B2
O2py
1 B1
O2px
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O
ON
O zz
z
but what about H’s?
note that in most operations, the H 1s orbitals are
interchanged, so a Linear Combination of the two
Atomic Orbitals will have to be used…
later in P. Chem. you will learn about projection operators
Will use symmetry adapted orbitals
two valid combinations are
1 = H11s + H21s (A1 symmetry)
2 = H11s - H21s (B2 symmetry)
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H
HO
H
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molecular orbitals?
form from combining orbitals (atomic or
LCAO’s (symmetry adapted linear
combinations) of like irreducible
representations
degree of mixing is determined by similarity
of energies of orbitals and the effectiveness
of overlap to form bonds…
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O
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O zz
z
irreducible representations of
orbitals and SALC’s
4 A1
O1s, O2s, O2pz, 1 = H11s + H21s
2 B2
O2py , 2 = H11s - H21s
1 B1
O2px
Note that seven AO’s give 7 SALC’s of AO’s
Mixing can occur between the differentorbitals of the same symmetry
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schematic and ab initio
results for A1 orbitals
HO
H
HO
H
HO
H
O 1s
O2s+1
O2pz+1
1a1
2a1
3a1
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O
ON
O zz
z
now have 5 molecular orbitals
(have not examined antibonding
combs)
HO
H
HO
H 1b1
1 b2O2py + 2
O2px
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O
ON
O zz
z
all orbitals and relative
energies
1a1
2a1
3a1
1b1
1b2
||
||
||
||
||
E
Just the 1s orbital
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O zz
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all orbitals and relative
energies
1a1
2a1
3a1
1b1
1b2
||
||
||
||
||
E
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OON
O
ON
O zz
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Bond order?
Note that there are 4 electrons in bonding
orbitals
No electrons in anti-bonding orbitals (the
3a1and 1b1 are nonbonding)
BO = 2
Like the 2 O-H bonds from the Lewis
structure!
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O zz
z
Compare with
linear (D h)BeH2
2a1
3a1
1b1
1b2
||
||
||
|| ||
||
u
u+
g+
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ON
O zz
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Correlation (Walsh)
Diagrams
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u
u+
g+
2a1
3a1
1b1
1b2
||
||
||
||
C2vD h
bond angle 180°90°
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today
Bonding and MOT with group theory
Today’s question
Consider BF3, and exam type question…
Write the matrix representations for the 2s orbitals
on the four atoms for the C3 and reflection
operations.
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CO2 and NO2-
CO2 D h
NO2- C2v
what are the valence orbitals?
central C/N: 2s, and 2px,y,z
O: 2s, and 2px,y,z
18 valence electrons for NO2-
We will deal with only the “p”
orbital bonding to simplify the
work.
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will work out the
bonding in the NO2- case
operations are
E
C2
v(yz)
v (xz)
the orbitals are laid out
in the array to right.
Sigma bonding can be
thought of as similar to
H2O
Npx
Npy
Npz
O1px
O1py
O1pz
O2px
O2py
O2pz
O
N
O
z
O
N
O
z
O
N
O
z
ON
OON
O
ON
O zz
z
C2 matrix:
R = -1
1 0 0 0 0 0 0 0 0
0 1 0 0 0 0 0 0 0
0 0 1 0 0 0 0 0 0
0 0 0 0 0 0 1 0 0
0 0 0 0 0 0 0 1 0
0 0 0 0 0 0 0 0 1
0 0 0 1 0 0 0 0 0
0 0 0 0 1 0 0 0 0
0 0 0 0 0 1 0 0 0
Npx
Npy
Npz
O1px
O1py
O1pz
O2px
O2 py
O2 pz
=
Npx
Npy
Npz
O2px
O2 py
O2pz
O1px
O1py
O1pz
ON
OON
O
ON
O zz
z
v (xz)
R = 1
1 0 0 0 0 0 0 0 0
0 1 0 0 0 0 0 0 0
0 0 1 0 0 0 0 0 0
0 0 0 0 0 0 1 0 0
0 0 0 0 0 0 0 1 0
0 0 0 0 0 0 0 0 1
0 0 0 1 0 0 0 0 0
0 0 0 0 1 0 0 0 0
0 0 0 0 0 1 0 0 0
Npx
Npy
Npz
O1px
O1py
O1pz
O2px
O2py
O2 pz
=
Npx
Npy
Npz
O2px
O2 py
O2 pz
O1px
O1py
O1pz
ON
OON
O
ON
O zz
z
v (yz)
R= 3
1 0 0 0 0 0 0 0 0
0 1 0 0 0 0 0 0 0
0 0 1 0 0 0 0 0 0
0 0 0 1 0 0 0 0 0
0 0 0 0 1 0 0 0 0
0 0 0 0 0 1 0 0 0
0 0 0 0 0 0 1 0 0
0 0 0 0 0 0 0 1 0
0 0 0 0 0 0 0 0 1
Npx
Npy
Npz
O1px
O1py
O1pz
O2px
O2py
O2 pz
=
Npx
Npy
Npz
O1px
O1py
O1pz
O2px
O2 py
O2 pz
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ON
O zz
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N: number of times irred rep, x, appears in the reducible
representation
h is the order of the group(sum of all E characters)
r is the character of the reducible representation for the
operation, x
i is the character of the irreducible representation for the
operation, x
n is the number of operations in the class, x
N =1
h rx• i
x
x
•nx
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reducible representation, and
irreducible reps obtained using
formula
rep. E C2 v v’
9 -1 1 3
3 A1 3 3 3 3
1 A2 1 1 -1 -1
2 B1 2 -2 2 -2
3 B2 3 -3 -3 3
total 9 -1 1 3
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We have found the symmetries of the x,
y and z orbitals for each atom(like x, y
and z vectors)
Symmetries of “p” orbitals are:
3 A1 , 1 A2 , 2 B1 , 3 B2
How about the “s” orbitals?same as the “s” orbitals in water (C2v as well!)
2 A1 and 1 B2
Symmetries of all valence orbitals are:
5 A1, 1 A2, 2 B1 and 4 B2
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O zz
z
C2v E C2 v(xz) v(yz)
A1 1 1 1 1 z, z2, x2,y2
A2 1 1 -1 -1 Rz, xy
B1 1 -1 1 -1 x, Ry,xz
B2 1 -1 -1 1 y, Rx, yz
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O
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O zz
z
combinations?
confirm these have the
right symmetry
5A1 (group 1)
Npz, N2s, O(1s1+1s2), O(2py1-2py2),O(2pz1+2pz2)
only 1 A2: must be nonbonding (group 2)
O(2px1-2px2)
2 B1 (group 3)
N2px, O(2px1+2px2)
4 B2 (group 4)
Npy, O(1s1- 1s2), O(2py1+ 2py2), O(2pz1- 2pz2)
Draw some of these out!
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NO2-,
18 v.
e-’s
1a1,
1 b2
2a1
1b1
2b2
3b2
4a1
1a2
3a1
bonding
bonding
bonding
nb( )
nb( )
nb( )
nb( )
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O zz
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CO2
CO2 D h
central C: 2s, and 2px,y,z
O: 2s, and 2px,y,z
16 valence electrons
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O
ON
O zz
z
u*
g
g+
u+
u+ *
g+*
16 valence
electrons in
CO2
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ON
O zz
z
u*
g
g+
u+
u+ *
g+*
1a1,
1 b2
2a1
1b1
2b2
3b2
4a1
1a2
3a1
bonding
bonding
bonding
nb( )
nb( )
( )
nb( )
C2v D h
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O zz
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PtCl42-: “s” orbitals only
Next, analyze the bonding picture in PtCl42-
We ignore all pi bonding that is possible with
“p” and “d” orbitals
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s orbitals in PtCl42- (D4h)
some symmetry elements shown to
right
all the ops are listed in the table on
the following slide
need to see how the 5 orbitals
transform
know E will have red rep char = 5
PtCl
Cl Cl
Cl
C4,C2
C2'
C2''
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character table for D4h
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the characters for the
reducible representation
E, character = 5
C4, character = 1
C2, character = 1
C2’, character = 3
C2”, character = 1
i, character = 1
S4, character = 1
h , character = 5
v , character = 3
d , character =1
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O
ON
O zz
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N: number of times irred rep, x, appears in the
reducible representation
h is the order of the group(sum of all E characters)
r is the character of the reducible representation
for the operation, x
i is the character of the irreducible representation
for the operation, x
n is the number of operations in the class, x
N =1
h rx• i
x
x
• nx
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O
ON
O zz
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irreducible representations
for the “s” orbitals only are
2 A1g irreducible reps
1 B1g irreducible rep
1 Eu irred rep
the Eu rep is 2 fold degenerate (identity character
is 2)
indicates that some “s” orbitals are interchanged
under the symmetry operations of the D4h point
group
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O zz
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Orbital salc’s? start with Pt
2 A1g irreducible reps
Pt s*
Pt dz2
1 B1g irreducible rep
Pt dx2-y2
1 Eu irred rep
Pt px, py
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O
ON
O zz
z
Now need to deal with Cl
s orbitals
A1g
B1g
Eu
Pt orbitalsSALC's ofH 1s orbitals
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O
ON
O zz
z
Bonding combinations
A1g
B1g
Eu
ON
OON
O
ON
O zz
z
Antibonding combinations
A1g
B1g
Eu
ON
OON
O
ON
O zz
z
Constructing an energy
diagram
Less nodes, lower E
More nodes, higher E
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O zz
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Rank Bonding
combinations
A1g
B1g
Eu
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OON
O
ON
O zz
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Rank the Antibonding
combinations
A1g
B1g
Eu
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OON
O
ON
O zz
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Give It a try!
Pt H (1s)
5d
6p
6s
1a1g
2a1g
3a1g
1a1u
1eu
2eu
1eg (dxz, dyz)
1b2g (dxy)
2b1g
2b1g
Cl 3s
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O
ON
O zz
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thoughts
Did we leave out any orbitals?
What would make this more complex?
pi bonding
sigma bonding using “p” orbitals