In Td, thed orbitals of the central atom span E+ T2character table, final column, and so only the T2set d xy,d yz,d zx may contribute to molecular orbital formation with the H orbitals..
Trang 1Solutions to exercises
Discussion questions
E15.1(b)
Symmetry operations Symmetry elements
1 Identity,E 1 The entire object
2.n-fold rotation 2.n-fold axis of symmetry, C n
3 Reflection 3 Mirror plane,σ
4 Inversion 4 Centre of symmetry,i
5.n-fold improper rotation 5 n-fold improper rotation axis, S n
E15.2(b) A molecule may be chiral, and therefore optically active, only if it does not posses an axis of improper
rotation, S n 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
S naxis 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) CCl4has 4C3axes (each C–Cl axis), 3C2axes (bisecting Cl–C–Cl angles), 3S4axes (the same
as theC2axes), and 6 dihedral mirror planes (each Cl–C–Cl plane)
E15.6(b) Only molecules belonging toCs,C n, andC nvgroups may be polar, so .
(a) CH3Cl(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 ofD6h
E 2C6 2C3 C2 3C
2 3C
2 i 2S3 2S6 σh 3σd 3σv
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 A1gE2uq, where q = x, y, or z
E 2C6 2C3 C2 3C
2 3C
2 i 2S3 2S6 σh 3σd 3σv
Trang 2To 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 forx- or y-polarized light A similar
analysis leads to the conclusion that A1gis absent from the product A1gE2uz; therefore the transition
is forbidden
E15.9(b) The classes of operations forD2are: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
andz 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
E C2(x) C2(y) C2(z)
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
S4axes 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 groupC4vis
4 C2 σv(x) σv(y) σd(xy) σd(−xy)
4 C2 σv(x) σv(y) σd(xy) σd(−xy)
C+
4 σd(xy) σ (−xy) σv(y) σv(x)
C−
4 σd(−xy) σ (xy) σv(x) σv(y)
4 E σv(y) σv(x) σd(−xy) σd(xy)
σv(x) σv(x) σd(−xy) σd(xy) σv(y) E C2 C−
4
σv(y) σv(y) σd(xy) σd(−xy) σv(x) C2 E C+
4
σd(xy) σd(xy) σv(x) σv(y) σd(−xy) C+
σd(−xy) σd(−xy) σv(y) σv(x) σd(xy) C−
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; thereforeC2v
(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
Trang 3(a) (b)
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)
F
F
F
F
F
F
F
F F F
F I
(a)
(b)
(e)
(f)
Xe
Td
OC OC
Fe
Fe OC
O
H F
The following responses refer to the text flow chart (Fig 15.14) for assigning point groups
(a) HF: linear, noi, so C∞v
(b) IF7: nonlinear, fewer than 2C nwithn > 2, C5, 5C2 perpendicular toC5, σh, so D5h
Trang 4(c) XeO2F2: nonlinear, fewer than 2C nwithn > 2, C2, noC2 perpendicular toC2, noσh, 2σv,
so C2v
(d) Fe2(CO)9: nonlinear, fewer than 2C nwithn > 2, C3, 3C2perpendicular toC3, σh, so D3h
(e) cubane (C8H8): nonlinear, more than 2C nwithn > 2, i, no C5, so Oh
(f) tetrafluorocubane (23): nonlinear, more than 2C nwithn > 2, no i, so Td
E15.15(b) (a) Only molecules belonging to Cs, C n, and C nv groups may be polar In Exercise 15.13b
ortho-dichlorobenzene and meta-dichlorobenzene belong to C2vand so may be polar; in Exercise 15.10b, HF and XeO2F2 belong toC nvgroups, 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 inte-gral with a totally symmetric part A glance at the character table shows that p x andp y orbitals available to a bonding N atom have the proper symmetry Ifd orbitals are available (as in SO3), alld orbitals except d2
z could have nonzero overlap.
E15.17(b) The product f× (µ) × imust contain A1(Example 15.7) Then, since i = B1, (µ) = (y) =
B2(C2vcharacter table), we can draw up the following table of characters
E C2 σv σ
v
Hence, the upper state is A2 , because A2× A2= A1
E15.18(b) (a)
H H H H H
H H
Anthracene
D2h
The components ofµ 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 theD6hgroup The integrand of the transition dipole moment must be or contain the A1gsymmetry species That integrand for transitions from the ground state
is A1gqf , 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 productqf must also have A1gsymmetry for the entire integrand to have A1gsymmetry Since the different symmetry species are orthogonal, the only wayqf can have A1g symmetry is ifq and f have the same symmetry Such combinations
includezA2u,xE1u, andyE1u Therefore, we conclude that transitions are allowed to states with
A2uor E1u symmetry
Trang 5sinθ
cosθ
1 1
Linear combinations of sinθ and cos θ
1
−1
The product does not contain A1, so yes the integral vanishes
Solutions to problems
P15.3 Consider Fig 15.2 The effect ofσhon a pointP is to generate σhP , and the effect of C2onσhP is
to generate the pointC2σhP The same point is generated from P by the inversion i, so C2σhP = iP
for all pointsP Hence, C2σh= 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 B1or 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
Trang 6E: all four orbitals are left unchanged; hence χ = 4
C3: One orbital is left unchanged; henceχ = 1
C2: No orbitals are left unchanged; henceχ = 0
S4: No orbitals are left unchanged; henceχ = 0
σd: Two orbitals are left unchanged; henceχ = 2
The character set 4, 1, 0, 0, 2 spans A1+ T2 Inspection of the character table of the groupTdshows thats spans A1and that the threep 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, thed orbitals of the
central atom span E+ T2(character table, final column), and so only the T2set (d xy,d yz,d zx) may contribute to molecular orbital formation with the H orbitals
P15.9 (a) InC3vsymmetry the H1s orbitals span the same irreducible representations as in NH3, which is
A1+ A1+ E There is an additional A1orbital because a fourth H atom lies on theC3axis In
C3v, thed orbitals span A1+ E + E [see the final column of the C3vcharacter table] Therefore, all fived orbitals may contribute to the bonding.
(b) InC2v symmetry the H1s orbitals span the same irreducible representations as in H2O, but one “H2O” fragment is rotated by 90◦ with respect to the other Therefore, whereas in H
2O the H1s orbitals span A1+ B2[H1+ H2, H1− H2], in the distorted CH4molecule they span
A1 + B2+ A1+ B1 [H1+ H2, H1 − H2, H3+ H4, H3 − H4] In C2v the d orbitals span
2A1+ B1+ B2+ A2[C2vcharacter table]; therefore, all except A2(d xy ) 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 noC naxes
withn > 2 (for the C3axes present in a tetrahedron are not symmetry axes any longer), but it does haveC2axes; in fact it has 2C2axes perpendicular to whicheverC2we call principal; it has noσh, 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 theS4operation, either class ofC2, orσd
(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
C naxis withn > 2, namely a C4axis; it has noC2axes perpendicular to theC4, 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 A1gand its integral must be zero
(b) xyz spans A1inTd[Problem 15.13] and so its integral need not be zero
(c) xyz → −xyz under z → −z (the σhoperation inD6h), and so its integral must be zero
Trang 7P15.16 We shall adapt the simpler subgroup C6v of the full D6h point group The six π-orbitals span
A1+ B1+ E1+ E2, and are
a1=√1
6(π1+ π2+ π3+ π4+ π5+ π6)
b1=√1
6(π1− π2+ π3− π4+ π5− π6)
e2=
1
√
12(2π1− π2− π3+ 2π4− π5− π6)
1
2(π2− π3+ π5− π6)
e1=
1
√
12(2π1+ π2− π3− 2π4− π5+ π6)
1
2(π2+ π3− π5− π6)
The hamiltonian transforms as A1; therefore all integrals of the form
ψH ψ dτ vanish unless ψ
andψ belong to the same symmetry species It follows that the secular determinant factorizes into
four determinants
A1: H a1a1 = 1
6
(π1+ · · · + π6)H (π1+ · · · + π6) dτ = α + 2β
B1: H b1b1 = 1
6
(π1− π2+ · · ·)H (π1− π2+ · · ·) dτ = α − 2β
E1: H e1(a)e1(a) = α − β, H e1(b)e1(b) = α − β, H e1(a)e1(b)= 0
Hence α − β − ε 0
0 α − β − ε =0 solves toε = α − β (twice)
E2: H e2(a)e2(a) = α + β, H e2(b)e2(b) = α + β, H e2(a)e2(b)= 0
Hence α + β − ε 0
0 α + β − ε =0 solves toε = α + β (twice)
P15.17 Consider phenanthrene with carbon atoms as labeled in the figure below
b ⬘ c⬘
d ⬘
e ⬘
f ⬘
g ⬘ g f e d
(a) The 2p orbitals involved in the π system are the basis we are interested in To find the
irrepro-ducible representations spanned by this basis, consider how each basis is transformed under the symmetry operations of theC2vgroup To find the character of an operation in this basis, sum the coefficients of the basis terms that are unchanged by the operation
C2 −a −a −b −b −c −c −d −d −e −e −f −f −g −g 0
σ −a −a −b −b −c −c −d −d −e −e −f −f −g −g −14
Trang 8To 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 orderh of the group (as in Section 15.5(a)) The table below
illustrates the procedure, beginning at left with theC2vcharacter table
E C2 σv σ
v product E C2 σv σ
v sum/h
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 A1are 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 A1symmetry (No surprise here: the orbitals span only A2and B1.) An A2SALC is obtained by multiplying the characters 1, 1,−1, −1 by the first column:
1
4(a − a− a+ a) = 1
2(a − a).
The A2combination from the second column is the same There are seven distinct A2 combi-nations in all: 1/2(a − a), 1/2(b − b), , 1/2(g − g) The B1combination from the first column is:
1
4(a + a+ a+ a) = 1
2(a + a).
The B1 combination from the second column is the same There are seven distinct B1 com-binations in all: 12(a + a), 1
2(b + b), ,1
2(g + g) There are no B2combinations, 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:
This determinant has the same eigenvalues as as in exercise 14.16(b)b
Trang 9(c) The ground state of the molecule has A1symmetry 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 A1character 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/iincludes the totally symmetric species A1 Consider first transitions to another A1wavefunction, in which case we need the product A1µA1 Now A1A1= A1, and the only character that returns A1when multiplied by
A1is A1itself Thez component of the dipole operator belongs to species A1, soz-polarized
A1← A1transitions are allowed (Note: transitions from the A1ground state to an A1excited 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 (A2or
B1) to the other; in that case, the excited-state wavefunction will have symmetry of A1B1= B2
from the two singly occupied orbitals in the excited state The symmetry of the transition dipole, then, is A1µB2= µB2, and the only species that yields A1when multiplied by B2is B2itself Now they 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 oneC naxis (aC2), and there is aσh The point group, then,
is C2h
f g
h i
j k
k ⬘
j ⬘
i ⬘
h ⬘
g ⬘
f ⬘
e ⬘
d ⬘
b ⬘
(b) The 2p zorbitals are transformed under the symmetry operations of theC2hgroup as follows
i −a −a −b −b −c −c −j −j −k −k 0
σh −a −a −b −b −c −c . −j −j −k −k −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 theC2hcharacter table
E C2 i σh product E C2 i σh sum/h
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 Auare 1, 1, 1, 1, so the
Trang 10columns 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 Agsymmetry (No surprise: the orbitals span only Auand Bg) An AuSALC is obtained by multiplying the characters 1, 1,−1, −1 by the first column:
1
4(a + a+ a+ a) = 1
2(a + a).
The Aucombination from the second column is the same There are 11 distinct Aucombinations
in all: 1/2(a + a), 1/2(b + b), 1/2(k + k) The Bgcombination from the first column is:
1
4(a − a− a+ a) = 1
2(a − a).
The Bgcombination from the second column is the same There are 11 distinct Bgcombinations
in all: 1/2(a − a), 1/2(b − b), 1/2(k − k) There are no Bucombinations, 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:
The energies of the filled orbitals areα+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 Agsymmetry 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 Agcharacter 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 Ag Consider first transitions to another Agwavefunction, in which case we need the product AgµAg Now AgAg = 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 (Auor Bg) to the other; in that case, the excited-state wavefunction will have symmetry of AuBg= Bufrom the two singly occupied orbitals in the excited state The symmetry of the transition dipole, then, is AgµBu= µBu, and the only species that yields Agwhen multiplied by Buis Buitself Thex and y components of
the dipole operator belongs to species Bu, so these transitions are allowed