Answers to Questions on Representations

7/27/2019 Answers to Questions on Representations

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Answers to Questions on Point Group Representations and molecular properties.

1) For each of the following molecular properties, identify the point group of themolecule and the representation of the property.

a) the lone pair orbital of NH3b)

the anti-symmetric stretch vibration of CO2c) the orbital of acetone

d) the HCH angle bend vibration of formaldehydee) the all-anti-bonding orbital of benzenef) the dxy and d x2y2 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 thoughtbut not much!)

a) C3v, lone pair axial along z axis, a1 symmetry (same symmetry as z)b) Dh, vibration moves atoms OCO so clearly axial, character +1 under C,

must be of some sort. Clearly it changes sign, character -1, under i to OCO

therefore ungerade. It does not change sign under v. character +1, therefore

+

u.(Actually it is fairly obvious that it has the same properties as the z vector, so +

u,

from the character table.)

c) C2v E C2 (xz) (yz)1 -1 1 -1 therefore b1

d) C2v E C2 (xz) (yz)

1 1 1 1 therefore a1

e) D6hE 2C6 2C3 C2 3C2' 3C2'' i 2S3 2S6 h 3d 3v

1 -1 1 -1 -1 1 1 -1 1 -1 -1 1 b2g

f) D4h:Ahactually I made a mistake here. The character table shows that dxy

transforms as b1g and d x2y

2as b2g. I meant to ask about dxz and dyz which are a

degenerate pair transforming as Eg.

C O z

x

z

y

C O

z

x


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2) trans-glyoxal (HCOCOH ), symmetry C2h , has two O atom n type molecular

orbitals, two carbonyl type molecular orbitals and two carbonyl *

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 thereducible representation; reduce the representation by inspection of the character

table.)Sketch the probable appearances of the orbitals. (Making sure that your sketchesmatch the representations.)

n orbitals - perpendicular to C=O bond:

C2h E C2 i h

2 0 0 2 (both components move under i and h)

Inspection of character table show this reduces to ag + bu

O

O

ag 1 1 1 1 bu 1 -1 -1 1Understand that the first diagram shows the 2 atomic orbitals whilst these twodiagrams show the 2 molecular orbitals constructed from the two atomic orbitals.

two carbonyl

C2h E C2 i h

2 0 0 -2 au + bg

au 1 1 -1 -1 bg 1 -1 1 -1

and exactly the same for the two carbonyl *orbitals. (Could have started with whole

thing with the 4 2pz aos of the 4 atoms in the manner of problem 3.)

O

O

O

O


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3) The 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 touse the reduction procedure rather than inspection.)

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 anyother 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 h 3d 3v

6 0 0 0 0 -2 0 0 0 -6 0 2

To get this you need to note that C2' and d go through the bonds in the plane whilst

C2'' and v 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 Iis involved

uses the formula:the number of times that contains

Iis: n(

I) = (a=

1N

naR

aIa ) / h

then all the operations involving R

a = 0 can be ignored. (Note h = 24)

n ( a1g

)E 3C2'' h 3v (use any one of C2'' orv to get character)

6 -2 -6 2

(a1g) 1 1 1 1na

Ra

Ia 6 -6 -6 6 therefore (a=

1N

naR

aIa ) / h = 0/24 = 0

n ( b2g) (as we know there are at least one of these!)E 3C2'' h 3v

6 -2 -6 2

(a1g) 1 -1 -1 1

naR

aIa 6 6 6 6 therefore (a=

1N

naR

aIa ) / h = 24/24 = 1

and so on to show = b2g + e1g + a2u + e2u

z

x


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4) The molecule CS2, Dh , has the vibrational modes:

symmetric stretch v1, energy 658 cm-1

, symmetry +

g

bend v2, energy 397 cm-1

, symmetry uanti-symmetric stretch v1, energy 1532 cm

-1, symmetry

+u

What symmetry vibrational levels arise from the following combinational states (andapproximately 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 thecomponents together.

v1 = 0, v2 = 0 , v3 = 0 is totally symmetric as the molecule has no excited vibrationalmodes +

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 energiesignored from here on in this question.)

v1 = 0, v2 = 2 , v3 = 0; u x u (2 quanta of v2) = +

g, -g, g 3 levels with energy

~ 2x397 cm-1

= 794 cm-1

v1 = 2, v2 = 0 , v3= 0; +

g x +

g= +

g energy ~ 1316 cm-1

v1 = 1, v2 = 1 , v3= 2; +

g x u x +

u x +

u = u ~ 4119 cm-1

v1

= 1, v2

= 1 , v3

= 1; +

gx

ux

+

u=

g~ 25878 cm

-1


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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 Cs E (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''

H2CCO (keteneif you dont know this molecule), C2v,

4a1, 2b1, 3b2E all atoms stay put: 5x3 = 15

C2 Hs move, CCO zs stay, x,ys reverse: 3x-1 = -3

(xz) Hs move, CCO x,zs stay, ys reverse: 3x1 = 3(yz) HHCCO y,zs stay, xs reverse: 5x1 = 5

E C2 (xz) (yz)

15 -3 3 5

n ( a1)

E C2 (xz) (yz)

15 -3 3 5(a1) 1 1 1 1

naR

aIa 15 -3 3 5 therefore (a=

1N

naR

aIa ) / h =20/4 = 5

n ( a2)

E C2 (xz) (yz)

15 -3 3 5

(a1) 1 1 -1 -1na

Ra

Ia 15 -3 -3 -5 therefore (a=

1N

naR

aIa ) / h =4/4 = 1

n ( b1)E C2 (xz) (yz)

15 -3 3 5

(a1) 1 -1 1 -1

naR

aIa 15 3 3 -5 therefore (a=

1N

naR

aIa ) / h =16/4 = 4

n ( b2)

E C2 (xz) (yz)

15 -3 3 5(a1) 1 -1 -1 1

naR

aIa 15 3 -3 5 therefore (a=

1N

naR

aIa ) / h =20/4 = 5

T = a1, b1, b2; R = a2, b1, b2: therefore V = 4 a1, 2 b1, 3 b2 (Note 3N-6 = 9)

H

C

H

C O z

y


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CH4 , Td , a1, e, t1, t2 (careful with the x,y,z coordinates of the C atombest to read

their characters off the table)

C3consider C3about z; 3 Hs move, the C and one H stay on z

axis; can read their character from the table under C3, look at x,y,zits 0.Or consider matrix

z

y

x

z

y

x

C

21230

23210

001

3

which has character 0.

C2(bisects 2 CH bonds) all Hs move and for C read it off the table -1 (look at x,y,z)S

4 (bisects 2 CH bonds) all Hs 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.

done of these lies in the yz plane, 2 Hs move, C and 2Hs stay in plane, y, z stay, xreverses, 1 per atom

E 8C2 3C2 6S4 6d 15 0 -1 -1 3 h = 24

n ( a1)

E 8C2 3C2 6S4 6d 15 0 -1 -1 3(a1) 1 1 1 1 1

naR

aIa 15 0 -3 -6 18

therefore (a=1N

naR

aIa ) / h =24/24 = 1

n ( a2)

E 8C2 3C2 6S4 6d 15 0 -1 -1 3(a2) 1 1 1 -1 -1

naR

aIa 15 0 -3 6 -18

therefore (a=1N

naR

aIa ) / h =0/24 = 0

n (e)

E 8C2 3C2 6S4 6d 15 0 -1 -1 3(e) 2 -1 2 0 0

naR

aIa 30 0 -6 0 0

therefore (a=1N

naR

aIa ) / h =24/24 = 1

z

y

C2


7/7

n ( t1)

E 8C2 3C2 6S4 6d 15 0 -1 -1 3(t1) 3 0 -1 1 -1

na

R

a

I

a 45 0 3 -6 -18therefore (a=1N

naR

aIa ) / h =24/24 = 0

n ( t2)

E 8C2 3C2 6S4 6d 15 0 -1 -1 3(t2) 3 0 -1 -1 1

naR

aIa 45 0 3 6 18

therefore (a=1N

naR

aIa ) / h =72/24 = 3

= a1 + e + t1 + 3 t2

From table T transforms as T2 and R transforms as T1 therefore V transforms as:

a1 + e + 2 t2 (3N6 = 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.

C5H

5

-N=10 atoms, h = 20 operations

E 5C2' h 5v

30 -1 10 1

n ( a'1)

E 5C2' h 5v

30 -1 10 1(a'1) 1 1 1 1

naR

aIa 30 -5 10 5

therefore (a=1N

naR

aIa ) / h =40/20 = 2

etc. to show: = 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''

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