Atkins, Solution, 7th Ed

15

Molecular symmetry

Solutions to exercises Discussion questions E15.1(b) Symmetry operations 1. Identity, E 2. n-fold rotation 3. Reflection 4. Inversion 5. n-fold improper rotation

Symmetry elements 1. The entire object 2. n-fold axis of symmetry, Cn 3. Mirror plane, σ 4. Centre of symmetry, i 5. n-fold improper rotation axis, Sn

E15.2(b)

A molecule may be chiral, and therefore optically active, only if it does not posses an axis of improper rotation, Sn . An improper rotation is a rotation followed by a reflection and this combination of operations always converts a right-handed object into a left-handed object and vice-versa; hence an Sn axis guarantees that a molecule cannot exist in chiral forms.

E15.3(b)

See Sections 15.4(a) and (b).

E15.4(b)

The direct sum is the decomposition of the direct product. The procedure for the decomposition is the set of steps outlined in Section 15.5(a) on p. 471 and demonstrated in Illustration 15.1.

Numerical exercises E15.5(b)

CCl4 has 4 C3 axes (each C–Cl axis), 3 C2 axes (bisecting Cl–C–Cl angles), 3 S4 axes (the same as the C2 axes), and 6 dihedral mirror planes (each Cl–C–Cl plane).

E15.6(b)

Only molecules belonging to Cs , Cn , and Cnv groups may be polar, so . . . (a) CH3 Cl (C3v ) may be polar along the C–Cl bond; (b) HW2 (CO)10 (D4h ) may not be polar (c) SnCl4 (Td ) may not be polar

E15.7(b)

The factors of the integrand have the following characters under the operations of D6h px z pz Integrand

E 2 1 1 2

2C6 1 1 1 1

2C3 −1 1 1 −1

C2 −2 1 1 −2

3C2
0 −1 −1 0

3C2

0 −1 −1 0

i −2 −1 −1 −2

2S3 −1 −1 −1 −1

2S6 1 −1 −1 1

σh 2 −1 −1 2

3σd 0 1 1 0

3σv 0 1 1 0

The integrand has the same set of characters as species E1u , so it does not include A1g ; therefore the integral vanishes E15.8(b)

We need to evaluate the character sets for the product A1g E2u q, where q = x, y, or z A1g E2u (x, y) Integrand

E 1 2 2 4

2C6 1 −1 1 −1

2C3 1 −1 −1 1

C2 1 2 −2 −4

3C2
1 0 0 0

3C2

1 0 0 0

i 1 −2 −2 4

2S3 1 1 −1 −1

2S6 1 1 1 1

σh 1 −2 2 −4

3σd 1 0 0 0

3σv 1 0 0 0

INSTRUCTOR’S MANUAL

244

To see whether the totally symmetric species A1g is present, we form the sum over classes of the number of operations times the character of the integrand c(A1g ) = (4) + 2(−1) + 2(1) + (−4) + 3(0) + 3(0) + (4) +2(−1) + 2(1) + (−4) + 3(0) + 3(0) = 0 Since the species A1g is absent, the transition is forbidden for x- or y-polarized light. A similar analysis leads to the conclusion that A1g is absent from the product A1g E2u z; therefore the transition is forbidden. E15.9(b)

The classes of operations for D2 are: E, C2 (x), C2 (y), and C2 (z). How does the function xyz behave under each kind of operation? E leaves it unchanged. C2 (x) leaves x unchanged and takes y to −y and z to −z, leaving the product xyz unchanged. C2 (y) and C2 (z) have similar effects, leaving one axis unchanged and taking the other two into their negatives. These observations are summarized as follows xyz

E 1

C2 (x) 1

C2 (y) 1

C2 (z) 1

A look at the character table shows that this set of characters belong to symmetry species A1 E15.10(b) A molecule cannot be chiral if it has an axis of improper rotation. The point group Td has S4 axes and mirror planes (= S1 ) , which preclude chirality. The Th group has, in addition, a centre of inversion (= S2 ) . E15.11(b) The group multiplication table of group C4v is C4− C2 σv (x) σv (y) σd (xy) σd (−xy) E C4+ + − E E C4 C4 C2 σv (x) σv (y) σd (xy) σd (−xy) C4+ C4+ C2 E C4− σd (xy) σ (−xy) σv (y) σv (x) C4− C4+ σd (−xy) σ (xy) σv (x) σv (y) C4− E C2 C2 C2 C4− C4+ E σv (y) σv (x) σd (−xy) σd (xy) σv (x) σv (x) σd (−xy) σd (xy) σv (y) E C2 C4− C4+ σv (y) σv (y) σd (xy) σd (−xy) σv (x) C2 E C4+ C4− + − σd (xy) σd (xy) σv (x) σv (y) σd (−xy) C4 C4 E C2 σd (−xy) σd (−xy) σv (y) σv (x) σd (xy) C4− C4+ C2 E

E15.12(b) See Fig. 15.1. (a) Sharpened pencil: E, C∞ , σv ; therefore C∞v (b) Propellor: E, C3 , 3C2 ; therefore D3 (c) Square table: E, C4 , 4σv ; therefore C4v ; Rectangular table: E, C2 , 2σv ; therefore C2v (d) Person: E, σv (approximately); therefore Cs E15.13(b) We follow the flow chart in the text (Fig. 15.14). The symmetry elements found in order as we proceed down the chart and the point groups are (a) Naphthalene: E, C2 , C2
, C2

, 3σh , i; D2h (b) Anthracene: E, C2 , C2
, C2

, 3σh , i; D2h

MOLECULAR SYMMETRY

245

(a)

(b)

(c)

(d)

Figure 15.1 (c) Dichlorobenzenes: (i) 1,2-dichlorobenzene: E, C2 , σv , σv
; C2v (ii) 1,3-dichlorobenzene: E, C2 , σv , σv
; C2v (iii) 1,4-dichlorobenzene: E, C2 , C2
, C2

, 3σh , i; D2h E15.14(b) (a) H F (b)

(c) F I F

OC

F

F

F

(d)

F

F

Fe

Xe O

F

CO CO

OC

F OC

(e)

CO

OC

O

(f) Td

Fe

OC F F

F

F

The following responses refer to the text flow chart (Fig. 15.14) for assigning point groups. (a) HF: linear, no i, so C∞v (b) IF7 : nonlinear, fewer than 2Cn with n > 2, C5 , 5C2
perpendicular to C5 , σh , so D5h

CO

INSTRUCTOR’S MANUAL

246

(c) XeO2 F2 : nonlinear, fewer than 2Cn with n > 2, C2 , no C2
perpendicular to C2 , no σh , 2σv , so C2v (d) Fe2 (CO)9 : nonlinear, fewer than 2Cn with n > 2, C3 , 3C2 perpendicular to C3 , σh , so D3h (e) cubane (C8 H8 ): nonlinear, more than 2Cn with n > 2, i, no C5 , so Oh (f) tetrafluorocubane (23): nonlinear, more than 2Cn with n > 2, no i, so Td E15.15(b) (a) Only molecules belonging to Cs , Cn , and Cnv groups may be polar. In Exercise 15.13b ortho-dichlorobenzene and meta-dichlorobenzene belong to C2v and so may be polar; in Exercise 15.10b, HF and XeO2 F2 belong to Cnv groups, so they may be polar. (b) A molecule cannot be chiral if it has an axis of improper rotation—including disguised or degenerate axes such as an inversion centre (S2 ) or a mirror plane (S1 ). In Exercises 15.9b and 15.10b, all the molecules have mirror planes, so none can be chiral. E15.16(b) In order to have nonzero overlap with a combination of orbitals that spans E, an orbital on the central atom must itself have some E character, for only E can multiply E to give an overlap integral with a totally symmetric part. A glance at the character table shows that px and py orbitals available to a bonding N atom have the proper symmetry. If d orbitals are available (as in SO3 ), all d orbitals except dz2 could have nonzero overlap. E15.17(b) The product f × (µ) × i must contain A1 (Example 15.7). Then, since i = B1 , (µ) = (y) = B2 (C2v character table), we can draw up the following table of characters B2 B1 B 1 B2

E 1 1 1

C2 −1 −1 1

σv
1 −1 −1

σv −1 1 −1

= A2

Hence, the upper state is A2 , because A2 × A2 = A1 . E15.18(b) (a) Anthracene H

H

H

H

H

H

H

D 2h H

H

H

The components of µ span B3u (x), B2u (y), and B1u (z). The totally symmetric ground state is Ag . Since Ag ×  =  in this group, the accessible upper terms are B3u (x-polarized), B2u (y-polarized), and B1u (z-polarized). (b) Coronene, like benzene, belongs to the D6h group. The integrand of the transition dipole moment must be or contain the A1g symmetry species. That integrand for transitions from the ground state is A1g qf , where q is x, y, or z and f is the symmetry species of the upper state. Since the ground state is already totally symmetric, the product qf must also have A1g symmetry for the entire integrand to have A1g symmetry. Since the different symmetry species are orthogonal, the only way qf can have A1g symmetry is if q and f have the same symmetry. Such combinations include zA2u , xE1u , and yE1u . Therefore, we conclude that transitions are allowed to states with A2u or E1u symmetry.

MOLECULAR SYMMETRY

E15.19(b) A1 A2 E sin θ cos θ Product

247

E 1 1 2 1 1 1

2C3 1 1 −1 Linear combinations of sin θ and cos θ 1

3σv 1 −1 0 1 −1 −1

The product does not contain A1 , so yes the integral vanishes.

Solutions to problems P15.3

Consider Fig. 15.2. The effect of σh on a point P is to generate σh P , and the effect of C2 on σh P is to generate the point C2 σh P . The same point is generated from P by the inversion i, so C2 σh P = iP for all points P . Hence, C2 σh = i , and i must be a member of the group.

Figure 15.2 P15.6

Representation 1 D(C3 )D(C2 ) = 1 × 1 = 1 = D(C6 ) and from the character table is either A1 or A2 . Hence, either D(σv ) = D(σd ) = +1 or −1 respectively. Representation 2 D(C3 )D(C2 ) = 1 × (−1) = −1 = D(C6 ) and from the character table is either B1 or B2 . Hence, either D(σv ) = −D(σd ) = 1 or D(σv ) = −D(σd ) = −1 respectively.

P15.8

A quick rule for determining the character without first having to set up the matrix representation is to count 1 each time a basis function is left unchanged by the operation, because only these functions give a nonzero entry on the diagonal of the matrix representative. In some cases there is a sign change, (. . . −f . . .) ← (. . . f . . .); then −1 occurs on the diagonal, and so count −1. The character of the identity is always equal to the dimension of the basis since each function contributes 1 to the trace.

INSTRUCTOR’S MANUAL

248

E: all four orbitals are left unchanged; hence χ = 4 C3 : One orbital is left unchanged; hence χ = 1 C2 : No orbitals are left unchanged; hence χ = 0 S4 : No orbitals are left unchanged; hence χ = 0 σd : Two orbitals are left unchanged; hence χ = 2 The character set 4, 1, 0, 0, 2 spans A1 + T2 . Inspection of the character table of the group Td shows that s spans A1 and that the three p orbitals on the C atom span T2 . Hence, the s and p orbitals of the C atom may form molecular orbitals with the four H1s orbitals. In Td , the d orbitals of the central atom span E + T2 (character table, final column), and so only the T2 set (dxy , dyz , dzx ) may contribute to molecular orbital formation with the H orbitals. P15.9

(a) In C3v symmetry the H1s orbitals span the same irreducible representations as in NH3 , which is A1 + A1 + E. There is an additional A1 orbital because a fourth H atom lies on the C3 axis. In C3v , the d orbitals span A1 + E + E [see the final column of the C3v character table]. Therefore, all five d orbitals may contribute to the bonding. (b) In C2v symmetry the H1s orbitals span the same irreducible representations as in H2 O, but one “H2 O” fragment is rotated by 90◦ with respect to the other. Therefore, whereas in H2 O the H1s orbitals span A1 + B2 [H1 + H2 , H1 − H2 ], in the distorted CH4 molecule they span A1 + B2 + A1 + B1 [H1 + H2 , H1 − H2 , H3 + H4 , H3 − H4 ]. In C2v the d orbitals span 2A1 + B1 + B2 + A2 [C2v character table]; therefore, all except A2 (dxy ) may participate in bonding.

P15.10

The most distinctive symmetry operation is the S4 axis through the central atom and aromatic nitrogens on both ligands. That axis is also a C2 axis. The group is S4 .

P15.12

(a) Working through the flow diagram (Fig. 15.14) in the text, we note that there are no Cn axes with n > 2 (for the C3 axes present in a tetrahedron are not symmetry axes any longer), but it does have C2 axes; in fact it has 2C2 axes perpendicular to whichever C2 we call principal; it has no σh , but it has 2σd . So the point group is D2d . (b) Within this point group, the distortion belongs to the fully symmetric species A1 , for its motion is unchanged by the S4 operation, either class of C2 , or σd . (c) The resulting structure is a square bipyramid, but with one pyramid’s apex farther from the base than the other’s. Working through the flow diagram in Fig. 15.14, we note that there is only one Cn axis with n > 2, namely a C4 axis; it has no C2 axes perpendicular to the C4 , and it has no σh , but it has 4σv . So the point group is C4v . (d) Within this point group, the distortion belongs to the fully symmetric species A1 . The translation of atoms along the given axis is unchanged by any symmetry operation for the motion is contained within each of the group’s symmetry elements.

P15.14

(a) xyz changes sign under the inversion operation (one of the symmetry elements of a cube); hence it does not span A1g and its integral must be zero (b) xyz spans A1 in Td [Problem 15.13] and so its integral need not be zero (c) xyz → −xyz under z → −z (the σh operation in D6h ), and so its integral must be zero

MOLECULAR SYMMETRY

P15.16

249

We shall adapt the simpler subgroup C6v of the full D6h point group. The six π -orbitals span A1 + B1 + E1 + E2 , and are 1 a1 = √ (π1 + π2 + π3 + π4 + π5 + π6 ) 6 1 b1 = √ (π1 − π2 + π3 − π4 + π5 − π6 ) 6  1   √ (2π1 − π2 − π3 + 2π4 − π5 − π6 ) 12 e2 =  1 2 (π2 − π3 + π5 − π6 )  1  √ (2π1 + π2 − π3 − 2π4 − π5 + π6 ) 12 e1 =  1 2 (π2 + π3 − π5 − π6 )



The hamiltonian transforms as A1 ; therefore all integrals of the form

ψ
H ψ dτ vanish unless ψ


and ψ belong to the same symmetry species. It follows that the secular determinant factorizes into four determinants  1 A1 : Ha1 a1 = 6 (π1 + · · · + π6 )H (π1 + · · · + π6 ) dτ = α + 2β  1 B1 : Hb1 b1 = 6 (π1 − π2 + · · ·)H (π1 − π2 + · · ·) dτ = α − 2β He1 (a)e1 (a) = α − β, He1 (b)e1 (b) = α − β, He1 (a)e1 (b) = 0

E1 : Hence

α−β −ε 0

He2 (a)e2 (a) = α + β, He2 (b)e2 (b) = α + β, He2 (a)e2 (b) = 0

E2 : Hence P15.17

0 = 0 solves to ε = α − β (twice) α−β −ε

α+β −ε 0

0 = 0 solves to ε = α + β (twice) α+β −ε

Consider phenanthrene with carbon atoms as labeled in the figure below a c

a⬘ b⬘

b

c⬘ d⬘

d g e

f

g⬘ f⬘

e⬘

(a) The 2p orbitals involved in the π system are the basis we are interested in. To find the irreproducible representations spanned by this basis, consider how each basis is transformed under the symmetry operations of the C2v group. To find the character of an operation in this basis, sum the coefficients of the basis terms that are unchanged by the operation.

E C2 σv σv


a a −a
a
−a

a
a
−a a −a


b b −b
b
−b

b
b
−b b −b


c c −c
c
−c

c
c
−c c −c


d d −d
d
−d

d
d
−d d −d


e e −e
e
−e

e
e
−e e −e


f f −f
f
−f

f
f
−f f −f


g g −g
g
−g

g
g
−g g −g


χ 14 0 0 −14

INSTRUCTOR’S MANUAL

250

To find the irreproducible representations that these orbitals span, multiply the characters in the representation of the orbitals by the characters of the irreproducible representations, sum those products, and divide the sum by the order h of the group (as in Section 15.5(a)). The table below illustrates the procedure, beginning at left with the C2v character table. A1 A2 B1 B2

E 1 1 1 1

C2 1 1 −1 −1

σv 1 −1 1 −1

σv
1 −1 −1 1

product

E 14 14 14 14

C2 0 0 0 0

σv 0 0 0 0

σv
−14 14 14 −14

sum/h 0 7 7 0

The orbitals span 7A2 + B2 . To find symmetry-adapted linear combinations (SALCs), follow the procedure described in Section 15.5(c). Refer to the table above that displays the transformations of the original basis orbitals. To find SALCs of a given symmetry species, take a column of the table, multiply each entry by the character of the species’ irreproducible representation, sum the terms in the column, and divide by the order of the group. For example, the characters of species A1 are 1, 1, 1, 1, so the columns to be summed are identical to the columns in the table above. Each column sums to zero, so we conclude that there are no SALCs of A1 symmetry. (No surprise here: the orbitals span only A2 and B1 .) An A2 SALC is obtained by multiplying the characters 1, 1, −1, −1 by the first column: 1 4 (a

− a
− a
+ a) = 21 (a − a
).

The A2 combination from the second column is the same. There are seven distinct A2 combinations in all: 1/2(a − a
), 1/2(b − b
), . . . , 1/2(g − g
) . The B1 combination from the first column is: 1 4 (a

+ a
+ a
+ a) = 21 (a + a
).

The B1 combination from the second column is the same. There are seven distinct B1 combinations in all: 21 (a + a
), 21 (b + b
), . . . , 21 (g + g
) . There are no B2 combinations, as the columns sum to zero. (b) The structure is labeled to match the row and column numbers shown in the determinant. The H¨uckel secular determinant of phenanthrene is: a b c d e f g g
f
e
d
c
b
a


a α−E β 0 0 0 0 0 0 0 0 0 0 0 β

b β α−E β 0 0 0 β 0 0 0 0 0 0 0

c 0 β α−E β 0 0 0 0 0 0 0 0 0 0

d 0 0 β α−E β 0 0 0 0 0 0 0 0 0

e 0 0 0 β α−E β 0 0 0 0 0 0 0 0

f 0 0 0 0 β α−E β 0 0 0 0 0 0 0

g 0 β 0 0 0 β α−E β 0 0 0 0 0 0

g
0 0 0 0 0 0 β α−E β 0 0 0 β 0

f
0 0 0 0 0 0 0 β α−E β 0 0 0 0

e
0 0 0 0 0 0 0 0 β α−E β 0 0 0

d
0 0 0 0 0 0 0 0 0 β α−E β 0 0

This determinant has the same eigenvalues as as in exercise 14.16(b)b.

c
0 0 0 0 0 0 0 0 0 0 β α−E β 0

b
0 0 0 0 0 0 0 β 0 0 0 β α−E β

a
β 0 0 0 0 0 0 0 0 0 0 0 β α−E

MOLECULAR SYMMETRY

251

(c) The ground state of the molecule has A1 symmetry by virtue of the fact that its wavefunction is the product of doubly occupied orbitals, and the product of any two orbitals of the same symmetry has A1 character. If a transition is to be allowed, the transition dipole must be non-zero, which in turn can only happen if the representation of the product /f∗ µ/i includes the totally symmetric species A1 . Consider first transitions to another A1 wavefunction, in which case we need the product A1 µA1 . Now A1 A1 = A1 , and the only character that returns A1 when multiplied by A1 is A1 itself. The z component of the dipole operator belongs to species A1 , so z-polarized A1 ← A1 transitions are allowed. (Note: transitions from the A1 ground state to an A1 excited state are transitions from an orbital occupied in the ground state to an excited-state orbital of the same symmetry.) The other possibility is a transition from an orbital of one symmetry (A2 or B1 ) to the other; in that case, the excited-state wavefunction will have symmetry of A1 B1 = B2 from the two singly occupied orbitals in the excited state. The symmetry of the transition dipole, then, is A1 µB2 = µB2 , and the only species that yields A1 when multiplied by B2 is B2 itself. Now the y component of the dipole operator belongs to species B2 , so these transitions are also allowed (y-polarized). P15.21

(a) Following the flow chart in Fig. 15.14, not that the molecule is not linear (at least not in the mathematical sense); there is only one Cn axis (a C2 ), and there is a σh . The point group, then, is C2h . b a

d

f

c

h

e

j

g

i

k⬘ k

i⬘

g⬘

j⬘

h⬘

e⬘ f⬘

c⬘ d⬘

a⬘ b⬘

(b) The 2pz orbitals are transformed under the symmetry operations of the C2h group as follows.

E C2 i σh

a a a
−a
−a

a
a
a −a −a


b b b
−b
−b

b
b
b −b −b


c c c
−c
−c

c
c
c −c −c


... ... ... ... ...

j j j
−j
−j

j
j
j −j −j


k k k
−k
−k

k
k
k −k −k


χ 22 0 0 −22

To find the irreproducible representations that these orbitals span, we multiply the characters of orbitals by the characters of the irreproducible representations, sum those products, and divide the sum by the order h of the group (as in Section 15.5(a)). The table below illustrates the procedure, beginning at left with the C2h character table.

Ag Au Bg Bu

E 1 1 1 1

C2 1 1 −1 −1

i 1 −1 1 −1

σh 1 −1 −1 1

product

E 22 22 22 22

C2 0 0 0 0

i 0 0 0 0

σh −22 22 22 −22

sum/h 0 11 11 0

The orbitals span 11Au + 11Bg . To find symmetry-adapted linear combinations (SALCs), follow the procedure described in Section 15.5(c). Refer to the above that displays the transformations of the original basis orbitals. To find SALCs of a given symmetry species, take a column of the table, multiply each entry by the character of the species’ irreproducible representation, sum the terms in the column, and divide by the order of the group. For example, the characters of species Au are 1, 1, 1, 1, so the

INSTRUCTOR’S MANUAL

252

columns to be summed are identical to the columns in the table above. Each column sums to zero, so we conclude that there are no SALCs of Ag symmetry. (No surprise: the orbitals span only Au and Bg ). An Au SALC is obtained by multiplying the characters 1, 1, −1, −1 by the first column: 1 4 (a

+ a
+ a
+ a) = 21 (a + a
).

The Au combination from the second column is the same. There are 11 distinct Au combinations in all: 1/2(a + a
), 1/2(b + b
), . . . 1/2(k + k
) . The Bg combination from the first column is: 1 4 (a

− a
− a
+ a) = 21 (a − a
).

The Bg combination from the second column is the same. There are 11 distinct Bg combinations in all: 1/2(a − a
), 1/2(b − b
), . . . 1/2(k − k
) . There are no Bu combinations, as the columns sum to zero. (c) The structure is labeled to match the row and column numbers shown in the determinant. The H¨uckel secular determinant is: a b c ... i j k k
j
i
... c
b
a


a α−E β 0 ... 0 0 0 0 0 0 ... 0 0 0

b β α−E β ... 0 0 0 0 0 0 ... 0 0 0

c 0 β α−E ... 0 0 0 0 0 0 ... 0 0 0

... ... ... ... ... ... ... ... ... ... ... ... ... ... ...

i 0 0 0 ... α−E β 0 0 0 0 ... 0 0 0

j 0 0 0 ... β α−E β 0 0 0 ... 0 0 0

k 0 0 0 ... 0 β α−E β 0 0 ... 0 0 0

k
0 0 0 ... 0 0 β α−E β 0 ... 0 0 0

j
0 0 0 ... 0 0 0 β α−E β ... 0 0 0

i
0 0 0 ... 0 0 0 0 β α−E ... 0 0 0

... ... ... ... ... ... ... ... ... ... ... ... ... ... ...

c
0 0 0 ... 0 0 0 0 0 0 ... α−E β 0

b
0 0 0 ... 0 0 0 0 0 0 ... β α−E β

a
0 0 0 ... 0 0 0 0 0 0 ... 0 β α−E

The energies of the filled orbitals are α +1.98137β, α +1.92583β, α +1.83442β, α +1.70884β, α + 1.55142β, α + 1.36511β, α + 1.15336β, α + 0.92013β, α + 0.66976β, α + 0.40691β, and α + 0.13648β. The π energy is 27.30729β. (d) The ground state of the molecule has Ag symmetry by virtue of the fact that its wavefunction is the product of doubly occupied orbitals, and the product of any two orbitals of the same symmetry has Ag character. If a transition is to be allowed, the transition dipole must be nonzero, which in turn can only happen if the representation of the product /f ∗ µ/i includes the totally symmetric species Ag . Consider first transitions to another Ag wavefunction, in which case we need the product Ag µAg . Now Ag Ag = Ag , and the only character that returns Ag when multiplied by Ag is Ag itself. No component of the dipole operator belongs to species Ag , so no Ag ← Ag transitions are allowed. (Note: such transitions are transitions from an orbital occupied in the ground state to an excited-state orbital of the same symmetry.) The other possibility is a transition from an orbital of one symmetry (Au or Bg ) to the other; in that case, the excited-state wavefunction will have symmetry of Au Bg = Bu from the two singly occupied orbitals in the excited state. The symmetry of the transition dipole, then, is Ag µBu = µBu , and the only species that yields Ag when multiplied by Bu is Bu itself. The x and y components of the dipole operator belongs to species Bu , so these transitions are allowed.