Chemistry 481 Answer Set #1Question #1 Identify the symmetry elements and operations present in each of the following molecules:

Molecule

(a) Chloroform

(b)1,3,5-trichlorobenzene

(c) chair cyclohexane

(c) boat cyclohexane

(d) B2H6 (diborane)

(e) planar, trans-H2O2

(f) Re2Cl82-

Structure Pointgroup

SymmetryElements

SymmetryOperations

C3v C3, 3σv E, 2C3, 3σv

D3h C3, 3C2, 3σv,

σh, S3

E, 2C3, 3C2,

3σv, σh, 2S3

D3d S6, C3, i,

3C2, 3σd

E, 3S6, 2C3, i,

3C2, 3σd

C2v C2, 2σv E, C2, 2σv

D2h 3C2, 3σ, i E, 3C2, 3σ, i

C2h C2, σh, i E, C2, σh, i

D4h C4, C2, 2σv,

2σd,σh, 2C2',

2C2'', i, S4

E, 2C4, C2, 2σv,

2σd, σh, 2C2',

2C2'', i, 2S4

(g) Co(NH3)5(H2O)3+

(without thehydrogens)

(h) S8

C4v C4, C2, 2σv',

2σv''

E, 2C4, C2, 2σv',

2σv''

D4d S8, C4, C2,

4C2', 4σd

E, 2S8, 2S83,

2C4, C2, 4C2',

4σd

Note: The red atomsare all in the sameplane, and the blueatoms are all in thesame plane.

Question #2Identify the single symmetry operations that gives the same result.

2a. C2z.σz(a,b,c) = C2

z(a,b,-c) = (-a,-b,-c) = i(a,b,c)

2b. C2z.i(a,b,c) = C2

z(-a,-b,-c) = (a,b,-c)=σz(a,b,c)

2c. C2z.σy(a,b,c) = C2

z(a,-b,c) = (-a,b,c) = σx(a,b,c)

2d. C2z.C4

z(a,b,c) = C2z(b,-a,c) = (-b,a,c) = (C4

z)3(a,b,c)

2e. C4z.C2

x(a,b,c) = C4z(a,-b,-c) = (-b,-a,-c) = C2

x=-y(Note: C2x=-y is a C2 rotation about the

line x=-y.)

2f. C4z.σx(a,b,c) = C4

z(-a,b,c) = (b,a,c) = σx=y(a,b,c) (Note: σx=y is a mirror plane containing

the line x=y and the z-axis.)

2g. σx.C4z(a,b,c) = σx(b,-a,c) = (-b,-a,c) = σx=-y(a,b,c) (Note: σx=-y is a mirror plane containing

the line x=-y and the z-axis.)

2h. S4z.σz(a,b,c) = S4

z(a,b,-c) = (b,-a,c) = C4z(a,b,c) (Recall that S4

z = C4z.σz )

2i. Two C2 axes at an angle of 2π/ 3. For simplicity, we will choose one of the axes as the x-axis

(denoted C2a); the other axis is denoted as C2

b.

C2a.C2

b C2b.C2

a

C2a.C2

b = C3z C2

b.C2a = (C3

z)2

They do not commute.

Question #3constructing group multiplication tables, find the groups with following symmetry operations.

Note: For the group multiplication tables, the operation listed in the top rowwas done first, followed by the operation in the first column. Remember thatorder is important.

3a 3b

C2vE C2

z σσσσx σσσσy

E E C2z σx σy

C2z C2

z E σy σx

σσσσx σx σyE C2

z

σσσσy σy σx C2z E

C2hE C2

z i σσσσz

E E C2z i σz

C2z C2

z E σzi

i i σzE C2

z

σσσσz σzi C2

z E

3c

D2hE C2

x C2y C2

z i σσσσx σσσσy σσσσz

E E C2x C2

y C2z i σx σy σz

C2x C2

x E C2z C2

y σxi σz σy

C2y C2

y C2z E C2

x σy σzi σx

C2z C2

z C2y C2

x E σz σy σxi

i i σx σy σzE C2

x C2y C2

z

σσσσx σxi σz σy C2

x E C2z C2

y

σσσσy σy σzi σx C2

y C2z E C2

x

σσσσz σz σy σxi C2

z C2y C2

x E

3d

C4vE C4

z C2z (C4

z)3 σσσσx σσσσy σσσσd(1) σσσσd

(2)

E E C4z C2

z (C4z)3 σx σy σd

(1) σd(2)

C4z C4

z C2z (C4

z)3 E σd(2) σd

(1) σx σy

C2z C2

z (C4z)3 E C4

z σy σx σd(2) σd

(1)

(C4z)3 (C4

z)3 E C4z C2

z σd(1) σd

(2) σy σx

σσσσx σx σd(1) σy σd

(2) E C2z C4

z (C4z)3

σσσσy σy σd(2) σx σd

(1) C2z E (C4

z)3 C4z

σσσσd(1) σd

(1) σx σd(2) σy (C4

z)3 C4z E C2

z

σσσσd(2) σd

(2) σy σd(1) σx C4

z (C4z)3 C2

z E

Note: All of the rotations were done clockwise. σd(1) is a mirror plane that contains the

line x=y and the z-axis, while σd(2) is a mirror plane that contains the line x=-y and the

z-axis.

3e 3f

S4E S4

z C2z (S4

z)3

E E S4z C2

z (S4z)3

S4z S4

z C2z (S4

z)3 E

C2z C2

z (S4z)3 E S4

z

(S4z)3 (S4

z)3 E S4z C2

z

D3E C3

z (C3z)2 C2

a C2b C2

c

E E C3z (C3

z)2 C2a C2

b C2c

C3z C3

z (C3z)2 E C2

c C2a C2

b

(C3z)2 (C3

z)2 E C3z C2

b C2c C2

a

C2a C2

a C2b C2

c E C3z (C3

z)2

C2b C2

b C2c C2

a (C3z)2 E C3

z

C2c C2

c C2a C2

b C3z (C3

z)2 E

Question 4 4a. See column 3 of Question 1 answers. 4b. See top left cell of Question 3 answers. 4c. CCl4: Td 4d. Benzene: D6h 4e. Pyridine: C2v 4f. Fe(CO)5 D3h 4g. Ferrocene Staggered: D5d; Eclipsed: D5h 4h. IrCl3(PMe3)3 fac: C3v mer: C2v

4i. [Ni(CN)4]2- D4h

Question #5a. cyclopropane, D3h b. SO2, C2v c. CO2, D4h

d. B2H6, D2h e. P4, Td f. Cl2C=C=CCl2, D2d

g. BF3, D3h h. PH3, C3v i. OSCl2, CS

j. O2SCl2, C2v1 k. B(OH)3, C3h l. P2I4, C2h

Question #6

The symmetry elements in an octahedron are: four C3 axes (there are eight faces--the C3axes pass through opposite faces); three C4 axes (there are six vertices--the C4 axes pass

through opposite vertices); six C2 axes (there are twelve edges--the C2 axes pass

through opposite edges); three additional C2 axes that are colinear with the three C4axes; a center of symmetry, i; three S4 axes that are colinear with the three C4 axes; four

S6 axes that are colinear with the four C3 axes; three planes of symmetry that are

perpendicular to the three C4 axes; six planes of symmetry that bisect the twelve edges.

Question #7 :What are vibrational modes in BF3?

Nitrogen trifluoride, a nonlinear molecule, will have six vibrational degrees of freedom(3 N - 6). It has C3v symmetry. The character table for this point group shows three

classes of symmetry operations: E, 2C3, and 3σv. The number of atoms that are

unshifted when these operations are carried out are counted and multiplied by thecontribution per atom to give a reducible representaion:

E 2C3 3σUnshifted atoms 4 1 2

Contributions per atom 3 0 1

Γtotal 12 0 1

Next we find the irreducible components:

N(A1) = (1/6)[12(1)(1) + 0(1)(2) + 2(1)(3)] = 3

N(A2) = (1/6)[12(1)(1) + 0(1)(2) + 2(-1)(3)] = 1

NE = (1/6)[12(1)(2) + 0(-1)(2) + 2(0)(3)] = 4

Thus Γtotal = 3A1 + A2 + 4E. Subtracting the translational modes (A1 + E) leaves 2A1+ A2 + 3E. Subtracting the rotational modes (A2 + E) gives 2A1 + 2E. The symmetric

stretching mode and the symmetric bending mode are both nondegenerate aand belongto the totally symmetric representation, A1. The asymmetric stretch and the

asymmetric bend are both doubly degenerate and belong to the irreduciblerepresentation E.

What are vibrational modes in OF2? OF 2 has the same number of atoms and the same symmetry as H2O.

E C2 σv(xz) σv(yz)

Unshifted atoms 3 1 1 3

Contribution per atom 3 -1 1 3

Γtotal 9 -1 1 3

The irreducible components (page 63 of the text) are 3A1 + A2 + 2B1 + 3B2.

Subtracting the translational and rotational modes (page 68) gives the irreduciblerepresentations of the vibrational modes, 2A1 + B2. The symmetric stretch and the

symmetric bend both transform as the totally symmetric representation, A1, and the

asymmetric stretch belongs to the irreducible representation B2.

Question #8 :What are vibrational modes in XeF4?

The molecule has D4h symmetry. A reducible representation can be obtained as

follows:

E 2C4 C2 2C2’ 2C2” i 2S4 σh 2σv 2σd

Unshifted Atoms 5 1 1 3 1 1 1 5 3 1

Contribution per atom 3 1 -1 -1 -1 -3 -1 1 1 1

Γtotal 15 1 -1 -3 -1 -3 -1 5 3 1

The irreducible representaions are found as follows:

N(A1g) = (1/16)[(15)(1)(1) + (1)(1)(2) + (-1)(1)(1) + (-3)(1)(2) + (-1)(1)(2) + (-3)(1)(1)

+ (-1)(1)(2) + (5)(1)(1) + (3)(1)(2) + (1)(1)(2)] = 1

N(A2g) = (1/16)[(15)(1)(1) + (1)(1)(2) + (-1)(1)(1) + (-3)(-1)(2) + (-1)(-1)(2) +

(-3)(1)(1) + (-1)(1)(2) + (5)(1)(1) + (3)(-1)(2) + (1)(-1)(2)] = 1

N(B1g) = (1/16)[(15)(1)(1) + (1)(-1)(2) + (-1)(1)(1) + (-3)(1)(2) + (-1)(-1)(2) +

(-3)(1)(1) + (-1)( -1)(2) + (5)(1)(1) + (3)(1)(2) + (1)(-1)(2)] = 1

N(B2g) = (1/16)[(15)(1)(1) + (1)(-1)(2) + (-1)(1)(1) + (-3)(-1)(2) + (-1)(1)(2) +

(-3)(1)(1) + (-1)(-1)(2) + (5)(1)(1) + (3)(-1)(2) + (1)(1)(2)] = 1

N(Eg) = (1/16)[(15)(2)(1) + (1)(0)(2) + (-1)(-2)(1) + (-3)(0)(2) + (-1)(0)(2) + (-3)(2)(1)

+ (-1)(0)(2) + (5)(-2)(1) + (3)(0)(2) + (1)(0)(2)] = 1

N(A1u) = (1/16)[(15)(1)(1) + (1)(1)(2) + (-1)(1)(1) + (-3)(1)(2) + (-1)(1)(2) +

(-3)(-1)(1) + (-1)(-1)(2) + (5)(-1)(1) + (3)(-1)(2) + (1)(-1)(2)] = 0

N(A2u) = (1/16)[(15)(1)(1) + (1)(1)(2) + (-1)(1)(1) + (-3)(-1)(2) + (-1)(-1)(2) +

(-3)(-1)(1) + (-1)(-1)(2) + (5)(-1)(1) + (3)(1)(2) + (1)(1)(2)] = 2

N(B1u) = (1/16)[(15)(1)(1) + (1)(-1)(2) + (-1)(1)(1) + (-3)(1)(2) + (-1)(-1)(2) +

(-3)(-1)(1) + (-1)(1)(2) + (5)(-1)(1) + (3)(-1)(2) + (1)(1)(2)] = 0

N(B2u) = (1/16)[(15)(1)(1) + (1)(-1)(2) + (-1)(1)(1) + (-3)(-1)(2) + (-1)(1)(2) +

(-3)(-1)(1) + (-1)(1)(2) + (5)(-1)(1) + (3)(1)(2) + (1)(-1)(2)] = 1

N(Eu) = (1/16)[(15)(2)(1) + (1)(0)(2) + (-1)(-2)(1) + (-3)(0)(2) + (-1)(0)(2) + (-3)(-2)(1)

+ (-1)(0)(2) + (5)(2)(1) + (3)(0)(2) + (1)(0)(2)] = 3

Subtracting rotational modes A2g and Eg, and translational modes A2u and Eu, leaves

the following vibrational modes: A1g + B1g + B2g + A2u + B2u + +2Eu. All of the

gerade modes are Raman active but not IR active. The B2u mode is neither Raman nor

IR active. The A2u and Eu modes are IR active but not Raman active. Note that none

of the fundamental vibrations are simultaneously Raman and IR active, consistent withthe rule of mutual exclusion which tells us that a molecule with a center of symmetrycannot have fundamental vibrations that are both Raman and IR active.