Answers to Questions on Point Group Representations and molecular properties.
1) For each of the following molecular properties, identify the point group of the
molecule and the representation of the property.
a) the lone pair orbital of NH3
b) the anti-symmetric stretch vibration of CO2
c) the π orbital of acetone
d) the H – C – H angle bend vibration of formaldehyde
e) the all-anti-bonding orbital of benzene
f) the d xy and d x2 – y2 atomic orbitals of Pt in [Pt Cl4 2- ]
(In a-e the property is a single ‘thing’ and applying the group operations (correctly)
will give you the result. f will need a little more thought – but not much!)
a) C3v, lone pair axial along z axis, a1 symmetry (same symmetry as z)
b) D∞h, vibration moves atoms OCO so clearly axial, character +1 under C∞,
must be ∑ of some sort. Clearly it changes sign, character -1, under i to OCO
therefore ungerade. It does not change sign under σv. character +1, therefore ∑+u
.(Actually it is fairly obvious that it has the same properties as the z vector, so ∑+u,
x
from the character table.)
c) C2v
d) C2v
E
1
C2
-1
σ(xz) σ(yz)
1
-1
therefore b1
E
1
C2
1
σ(xz) σ(yz)
1
1
therefore a1
C
z
O
y
C
e) D6h
E 2C6 2C3 C2 3C2' 3C2'' i 2S3 2S6 σh 3σd 3σv
1 -1 1
-1 -1 1
1 -1 1
-1 -1 1
O
z
x
b2g
z
f) D4h:Ah – actually I made a mistake here. The character table shows that d xy
transforms as b1g and d x2 – y2 as b2g. I meant to ask about d xz and d yz which are a
degenerate pair transforming as Eg.
2) trans-glyoxal (HCO – COH ), symmetry C2h , has two O atom n type molecular
orbitals, two carbonyl π type molecular orbitals and two carbonyl π* type molecular
orbitals. What are the representations of these molecular orbitals?
(Draw the 2 components in each case; operate with the group operations to get the
reducible representation; reduce the representation by inspection of the character
table.)
Sketch the probable appearances of the orbitals. (Making sure that your sketches
match the representations.)
O
n orbitals - perpendicular to C=O bond:
C2h
E
2
C2 i σh
0 0 2 (both components move under i and σh)
O
Inspection of character table show this reduces to ag + bu
O
O
O
O
ag 1 1 1 1
bu 1 -1 -1 1
Understand that the first diagram shows the 2 atomic orbitals whilst these two
diagrams show the 2 molecular orbitals constructed from the two atomic orbitals.
two carbonyl π
C2h
E
2
C2 i σh
0 0 -2
au + bg
au 1 1 -1 -1
bg 1 -1 1 -1
and exactly the same for the two carbonyl π* orbitals. (Could have started with whole
thing with the 4 2pz aos of the 4 atoms in the manner of problem 3.)
3) The π molecular orbitals of benzene (symmetry D6h , molecule in the xy plane) are
formed from the 6, 2pz atomic orbitals of the 6 carbon atoms. What are the
irreducible representations of the 6 molecular orbitals?
(Same approach as above but now with the 6 a.o.s as components, possibly better to
use the reduction procedure rather than inspection.)
x
z
These are the 6 2pz aos of the 6 C atoms in benzene
drawn in an arbitrary arrangement. (In fact I have
copied the diagram form 1e to save time. But any
other arrangement would do. It will give the same
reducible representation.
Apply point group operations to the 6 ‘things’.
E 2C6 2C3 C2 3C2' 3C2'' i 2S3 2S6 σh 3σd 3σv
Γ 6 0
0 0 0
-2 0 0 0
-6 0
2
To get this you need to note that C2' and σd go through the bonds in the plane whilst
C2'' and σv go through the atoms. Count 1 for orbitals that do not move, 0 for orbitals
that move, -1 for orbitals that do not move but change sign.
as calculating the number of times of each irreducible representation ΓI is involved
uses the formula:
the number of times that Γ contains ΓI is: n (ΓI) = (Σa= 1 N na χRa χIa ) / h
then all the operations involving χRa = 0 can be ignored.
(Note h = 24)
n ( a1g)
E 3C2'' σh 3σv
(use any one of C2'' or σv to get character)
Γ
6 -2 -6 2
Γ(a1g)
1
1
1 1
na χRa χIa 6 -6 -6 6 therefore (Σa= 1 N na χRa χIa ) / h = 0/24 = 0
n ( b2g) (as we know there are at least one of these!)
E 3C2'' σh 3σv
Γ
6 -2 -6 2
Γ(a1g)
1 -1 -1 1
R
I
na χ a χ a 6 6
6 6 therefore (Σa= 1 N na χRa χIa ) / h = 24/24 = 1
and so on to show Γ = b2g + e1g + a2u + e2u
4) The molecule CS2, D∞h , has the vibrational modes:
symmetric stretch v1, energy 658 cm-1 , symmetry σ+g
bend v2, energy 397 cm-1 , symmetry π u
anti-symmetric stretch v1, energy 1532 cm-1 , symmetry σ+u
What symmetry vibrational levels arise from the following combinational states (and
approximately what are the energies of the levels):
v1 = 0, v2 = 0 , v3 = 0
(that is the vibrationless ground state)
v1 = 0, v2 = 2 , v3 = 0
v1 = 2, v2 = 0 , v3 = 0
v1 = 1, v2 = 1 , v3 = 2
(these three are slightly occupied at room temperature)
v1 = 1, v2 = 1 , v3 = 1
To get the symmetry of the combination levels you multiply the symmetries of the
components together.
v1 = 0, v2 = 0 , v3 = 0 is totally symmetric as the molecule has no excited vibrational
modes σ+g in this level. Its energy is 0 cm-1 (or ~ 0.5(658+397+1532) cm-1 = 1288.5
cm-1 if considering zero point energies – ignored from here on in this question.)
v1 = 0, v2 = 2 , v3 = 0; π u x π u (2 quanta of v2) = σ+g, σ-g, δg 3 levels with energy
~ 2x397 cm-1 = 794 cm-1
v1 = 2, v2 = 0 , v3 = 0; σ+g x σ+g = σ+g
energy ~ 1316 cm-1
v1 = 1, v2 = 1 , v3 = 2; σ+g x π u x σ+u x σ+u = π u
~ 4119 cm-1
v1 = 1, v2 = 1 , v3 = 1; σ+g x π u x σ+u = π g
~ 25878 cm-1
5) Show that the representations of the vibrations of the following molecules are:
In each case attach x, y, z vectors to each of the N atoms and consider the effect of
the group operations on the 3N vectors to obtain the reducible representation. Reduce
it and take out the representations of translation and rotation.
HCOOH, Cs , 7a', 2 a''
5 atoms, in x, y plane
E
σ(xy)
15
5
(x,y unchanged, z reverses)
Inspection shows this reduces to 10 a' and 5 a''
Character table shows: translation 2 a' and a'', rotation a' and 2 a''
Therefore vibration: 7 a' and 2 a''
Cs
H2CCO (ketene – if you don’t know this molecule), C2v,
4a1, 2b1, 3b2
E
all atoms stay put: 5x3 = 15
C2
H’s move, CCO z’s stay, x,y’s reverse: 3x-1 = -3
σ(xz) H’s move, CCO x,z’s stay, y’s reverse: 3x1 = 3
σ(yz) HHCCO y,z’s stay, x’s reverse: 5x1 = 5
E
C2 σ(xz) σ(yz)
Γ
15
-3
3
5
y
H
C
C
O
H
n ( a1)
E
15
1
15
C2
-3
1
-3
σ(xz) σ(yz)
3
5
1
1
3
5 therefore (Σa= 1 N na χRa χIa ) / h =20/4 = 5
E
Γ
15
Γ(a1)
1
na χRa χIa 15
C2
-3
1
-3
σ(xz) σ(yz)
3
5
-1
-1
-3
-5 therefore (Σa= 1 N na χRa χIa ) / h =4/4 = 1
C2
-3
-1
3
σ(xz) σ(yz)
3
5
1
-1
3
-5 therefore (Σa= 1 N na χRa χIa ) / h =16/4 = 4
C2
-3
-1
3
σ(xz) σ(yz)
3
5
-1
1
-3
5 therefore (Σa= 1 N na χRa χIa ) / h =20/4 = 5
Γ
Γ(a1)
na χRa χIa
n ( a2)
n ( b1)
E
Γ
15
Γ(a1)
1
R
I
na χ a χ a 15
n ( b2)
Γ
Γ(a1)
na χRa χIa
E
15
1
15
T = a1, b1, b2; R = a2, b1, b2: therefore V = 4 a1, 2 b1, 3 b2 (Note 3N-6 = 9)
z
CH4 , Td , a1, e, t1, t2 (careful with the x,y,z coordinates of the C atom – best to read
their characters off the table)
z
C3 – consider C3 about z; 3 H’s move, the C and one H stay on z
axis; can read their character from the table under C3, look at x,y,z
its 0.
Or consider matrix
0
 x  1
  
C  y   0 1 2
3  
 z  0  3 2
C2
y
0   x
 
3 2  y 
 1 2   z 
which has character 0.
C2 (bisects 2 CH bonds) all H’s move and for C read it off the table -1 (look at x,y,z)
S4 (bisects 2 CH bonds) all H’s move and for C read it off the table -1 (look at x,y,z)
you could do these two by the trigonometry to get the same result.
σd one of these lies in the yz plane, 2 H’s move, C and 2H’s stay in plane, y, z stay, x
reverses, 1 per atom
Γ
E
15
8C2 3C2
0
-1
6S4
-1
6σd
3
n ( a1)
E
8C2 3C2 6S4
6σd
Γ
15
0
-1
-1
3
Γ(a1)
1
1
1
1
1
na χRa χIa 15
0
-3
-6
18
therefore (Σa= 1 N na χRa χIa ) / h =24/24 = 1
n ( a2)
E
8C2 3C2 6S4
6σd
Γ
15
0
-1
-1
3
Γ(a2)
1
1
1
-1
-1
na χRa χIa 15
0
-3
6
-18
therefore (Σa= 1 N na χRa χIa ) / h =0/24 = 0
n (e)
E
8C2 3C2 6S4
6σd
Γ
15
0
-1
-1
3
Γ(e)
2
-1
2
0
0
na χRa χIa 30
0
-6
0
0
therefore (Σa= 1 N na χRa χIa ) / h =24/24 = 1
h = 24
n ( t1 )
E
8C2 3C2 6S4
6σd
Γ
15
0
-1
-1
3
Γ(t1)
3
0
-1
1
-1
na χRa χIa 45
0
3
-6
-18
N
R
I
therefore (Σa= 1 na χ a χ a ) / h =24/24 = 0
n ( t2 )
E
8C2 3C2 6S4
6σd
Γ
15
0
-1
-1
3
Γ(t2)
3
0
-1
-1
1
R
I
na χ a χ a 45
0
3
6
18
therefore (Σa= 1 N na χRa χIa ) / h =72/24 = 3
Γ = a1 + e + t1 + 3 t2
From table T transforms as T2 and R transforms as T1 therefore V transforms as:
a1 + e + 2 t2 (3N – 6 = 9, 1 (a1) + 2 (e)+ 2x3 (t2)= 9)
(You could guess that the a1 is the all symmetric stretch of the C-H bonds.)
6) Determine the representations of the vibrational motions of the C5H5 – ion.
Clearly all the atoms move for C5 C52 S5 S52 operators as these pass through the
middle of the molecule. Therefore their character is zero and can be ignored.
C5H5- N=10 atoms, h = 20 operations
E
5C2' σh
5σv
Γ 30
-1
10
1
n ( a'1)
E
5C2' σh
5σv
Γ
30
-1
10
1
Γ(a'1)
1
1
1
1
na χRa χIa 30
-5
10
5
therefore (Σa= 1 N na χRa χIa ) / h =40/20 = 2
etc. to show: Γ = 2 a'1 + 2 a'2 + 4 e1' + 4 e2' + a''1 + a''2 + 2 e1'' + 2 e2''
T transforms as a''2 , e1', Rotation transforms as a'2 , e1''
therefore V transforms as: 2 a'1 + a'2 + 3 e1' + 4 e2' + a''1 + e1'' + 2 e2''