15 Molecular symmetry

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1 5 Molecular symmetry Solutions to exercises Discussion questions E5.(b) Symmetry operations Symmetry elements. Identity, E. 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 E5.2(b) E5.3(b) E5.4(b) E5.5(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 n axis guarantees that a molecule cannot exist in chiral forms. See Sections 5.4(a) and (b). The direct sum is the decomposition of the direct product. The procedure for the decomposition is the set of steps outlined in Section 5.5(a) on p. 47 and demonstrated in Illustration 5.. Numerical exercises CCl 4 has 4 C 3 axes (each C Cl axis), 3 C 2 axes (bisecting Cl C Cl angles), 3 S 4 axes (the same as the C 2 axes), and 6 dihedral mirror planes (each Cl C Cl plane). E5.6(b) Only molecules belonging to C s, C n, and C nv groups may be polar, so... (a) CH 3 Cl (C 3v ) may be polar along the C Cl bond; (b) HW 2 (CO) 0 (D 4h ) may not be polar (c) SnCl 4 (T d ) may not be polar E5.7(b) The factors of the integrand have the following characters under the operations of D 6h E 2C 6 2C 3 C 2 3C 2 3C 2 i 2S 3 2S 6 σ h 3σ d 3σ v p x z p z Integrand E5.8(b) The integrand has the same set of characters as species E u, so it does not include A g ; therefore the integral vanishes We need to evaluate the character sets for the product A g E 2u q, where q = x, y,orz E 2C 6 2C 3 C 2 3C 2 3C 2 i 2S 3 2S 6 σ h 3σ d 3σ v A g E 2u (x, y) Integrand

2 244 INSTRUCTOR S MANUAL To see whether the totally symmetric species A g is present, we form the sum over classes of the number of operations times the character of the integrand c(a g ) = (4) + 2( ) + 2() + ( 4) + 3(0) + 3(0) + (4) +2( ) + 2() + ( 4) + 3(0) + 3(0) = 0 E5.9(b) Since the species A g is absent, the transition is forbidden for x- ory-polarized light. A similar analysis leads to the conclusion that A g is absent from the product A g E 2u z; therefore the transition is forbidden. The classes of operations for D 2 are: E, C 2 (x), C 2 (y), and C 2 (z). How does the function xyz behave under each kind of operation? E leaves it unchanged. C 2 (x) leaves x unchanged and takes y to y and z to z, leaving the product xyz unchanged. C 2 (y) and C 2 (z) have similar effects, leaving one axis unchanged and taking the other two into their negatives. These observations are summarized as follows E C 2 (x) C 2 (y) C 2 (z) xyz A look at the character table shows that this set of characters belong to symmetry species A E5.0(b) A molecule cannot be chiral if it has an axis of improper rotation. The point group T d has S 4 axes and mirror planes (= S ), which preclude chirality. The T h group has, in addition, a centre of inversion (= S 2 ). E5.(b) The group multiplication table of group C 4v is E C + 4 C 2 σ v (x) σ v (y) σ d (xy) σ d ( xy) E E C + 4 C 2 σ v (x) σ v (y) σ d (xy) σ d ( xy) C + 4 C + 4 C 2 E σ d (xy) σ ( xy) σ v (y) σ v (x) E C 2 C + 4 σ d ( xy) σ (xy) σ v (x) σ v (y) C 2 C 2 C + 4 E σ v (y) σ v (x) σ d ( xy) σ d (xy) σ v (x) σ v (x) σ d ( xy) σ d (xy) σ v (y) E C 2 C + 4 σ v (y) σ v (y) σ d (xy) σ d ( xy) σ v (x) C 2 E C + 4 σ d (xy) σ d (xy) σ v (x) σ v (y) σ d ( xy) C + 4 E C 2 σ d ( xy) σ d ( xy) σ v (y) σ v (x) σ d (xy) C + 4 C 2 E E5.2(b) See ig. 5.. (a) Sharpened pencil: E,C,σ v ; therefore C v (b) Propellor: E,C 3, 3C 2 ; therefore D 3 (c) Square table: E,C 4, 4σ v ; therefore C 4v ; Rectangular table: E,C 2, 2σ v ; therefore C 2v (d) Person: E,σ v (approximately); therefore C s E5.3(b) We follow the flow chart in the text (ig. 5.4). The symmetry elements found in order as we proceed down the chart and the point groups are (a) Naphthalene: E, C 2,C 2,C 2, 3σ h,i; D 2h (b) Anthracene: E, C 2,C 2,C 2, 3σ h,i; D 2h

3 MOLECULAR SYMMETRY 245 (a) (b) (c) (d) igure 5. (c) Dichlorobenzenes: (i),2-dichlorobenzene: E, C 2,σ v,σ v ; C 2v (ii),3-dichlorobenzene: E, C 2,σ v,σ v ; C 2v E5.4(b) (a) (iii),4-dichlorobenzene: E, C 2,C 2,C 2, 3σ h,i; D 2h H (b) (e) I (c) O Xe O (f) T d (d) OC OC CO e CO OC CO e OC OC CO The following responses refer to the text flow chart (ig. 5.4) for assigning point groups. (a) H: linear, no i,so C v (b) I 7 : nonlinear, fewer than 2C n with n>2, C 5, 5C 2 perpendicular to C 5,σ h,so D 5h

4 246 INSTRUCTOR S MANUAL (c) XeO 2 2 : nonlinear, fewer than 2C n with n>2, C 2,noC 2 perpendicular to C 2,noσ h,2σ v, so C 2v (d) e 2 (CO) 9 : nonlinear, fewer than 2C n with n>2, C 3, 3C 2 perpendicular to C 3,σ h,so D 3h (e) cubane (C 8 H 8 ): nonlinear, more than 2C n with n>2, i, noc 5,so O h (f) tetrafluorocubane (23): nonlinear, more than 2C n with n>2, no i, so T d E5.5(b) (a) Only molecules belonging to C s, C n, and C nv groups may be polar. In Exercise 5.3b ortho-dichlorobenzene and meta-dichlorobenzene belong to C 2v and so may be polar; in Exercise 5.0b, H and XeO 2 2 belong to C nv 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 (S 2 ) or a mirror plane (S ). In Exercises 5.9b and 5.0b, all the molecules have mirror planes, so none can be chiral. E5.6(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 p x and p y orbitals available to a bonding N atom have the proper symmetry. If d orbitals are available (as in SO 3 ), all d orbitals except dz 2 could have nonzero overlap. E5.7(b) The product Ɣ f Ɣ(µ) Ɣ i must contain A (Example 5.7). Then, since Ɣ i = B, Ɣ(µ) = Ɣ(y) = B 2 (C 2v character table), we can draw up the following table of characters E C 2 σ v σ v B 2 B B B 2 = A 2 E5.8(b) Hence, the upper state is A 2, because A 2 A 2 = A. (a) Anthracene H H H H H D 2h H H H H H The components of µ span B 3u (x), B 2u (y), and B u (z). The totally symmetric ground state is A g. Since A g Ɣ = Ɣ in this group, the accessible upper terms are B 3u (x-polarized), B 2u (y-polarized), and B u (z-polarized). (b) Coronene, like benzene, belongs to the D 6h group. The integrand of the transition dipole moment must be or contain the A g symmetry species. That integrand for transitions from the ground state is A g qf, where q is x,y,orzand f is the symmetry species of the upper state. Since the ground state is already totally symmetric, the product qf must also have A g symmetry for the entire integrand to have A g symmetry. Since the different symmetry species are orthogonal, the only way qf can have A g symmetry is if q and f have the same symmetry. Such combinations include za 2u, xe u, and ye u. Therefore, we conclude that transitions are allowed to states with A 2u or E u symmetry.

5 MOLECULAR SYMMETRY 247 E5.9(b) E 2C 3 3σ v A A 2 E 2 0 sin θ cos θ Linear combinations of sin θ and cos θ Product The product does not contain A,so yes the integral vanishes. Solutions to problems P5.3 Consider ig The effect of σ h on a point P is to generate σ h P, and the effect of C 2 on σ h P is to generate the point C 2 σ h P. The same point is generated from P by the inversion i,soc 2 σ h P = ip for all points P. Hence, C 2 σ h = i, and i must be a member of the group. igure 5.2 P5.6 Representation D(C 3 )D(C 2 ) = = = D(C 6 ) and from the character table is either A or A 2. Hence, either D(σ v ) = D(σ d ) = respectively. Representation 2 + or D(C 3 )D(C 2 ) = ( ) = = D(C 6 ) and from the character table is either B or B 2. Hence, either D(σ v ) = D(σ d ) = or D(σ v ) = D(σ d ) = respectively. P5.8 A quick rule for determining the character without first having to set up the matrix representation is to count 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 occurs on the diagonal, and so count. The character of the identity is always equal to the dimension of the basis since each function contributes to the trace.

6 248 INSTRUCTOR S MANUAL E: all four orbitals are left unchanged; hence χ = 4 C 3 : One orbital is left unchanged; hence χ = C 2 : No orbitals are left unchanged; hence χ = 0 S 4 : No orbitals are left unchanged; hence χ = 0 σ d : Two orbitals are left unchanged; hence χ = 2 The character set 4,, 0, 0, 2 spans A + T 2. Inspection of the character table of the group T d shows that s spans A and that the three p orbitals on the C atom span T 2. Hence, the s and p orbitals of the C atom may form molecular orbitals with the four Hs orbitals. In T d, the d orbitals of the central atom span E + T 2 (character table, final column), and so only the T 2 set (d xy, d yz, d zx ) may contribute to molecular orbital formation with the H orbitals. P5.9 (a) In C 3v symmetry the Hs orbitals span the same irreducible representations as in NH 3, which is A + A + E. There is an additional A orbital because a fourth H atom lies on the C 3 axis. In C 3v, the d orbitals span A + E + E [see the final column of the C 3v character table]. Therefore, all five d orbitals may contribute to the bonding. (b) In C 2v symmetry the Hs orbitals span the same irreducible representations as in H 2 O, but one H 2 O fragment is rotated by 90 with respect to the other. Therefore, whereas in H 2 O the Hs orbitals span A + B 2 [H + H 2, H H 2 ], in the distorted CH 4 molecule they span A + B 2 + A + B [H + H 2, H H 2, H 3 + H 4, H 3 H 4 ]. In C 2v the d orbitals span 2A + B + B 2 + A 2 [C 2v character table]; therefore, all except A 2 (d xy ) may participate in bonding. P5.0 The most distinctive symmetry operation is the S 4 axis through the central atom and aromatic nitrogens on both ligands. That axis is also a C 2 axis. The group is S 4. P5.2 (a) Working through the flow diagram (ig. 5.4) in the text, we note that there are no C n axes with n>2 (for the C 3 axes present in a tetrahedron are not symmetry axes any longer), but it does have C 2 axes; in fact it has 2C 2 axes perpendicular to whichever C 2 we call principal; it has no σ h, but it has 2σ d. So the point group is D 2d. (b) Within this point group, the distortion belongs to the fully symmetric species A, for its motion is unchanged by the S 4 operation, either class of C 2,orσ d. (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 ig. 5.4, we note that there is only one C n axis with n>2, namely a C 4 axis; it has no C 2 axes perpendicular to the C 4, and it has no σ h, but it has 4σ v. So the point group is C 4v. (d) Within this point group, the distortion belongs to the fully symmetric species A. 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. P5.4 (a) xyz changes sign under the inversion operation (one of the symmetry elements of a cube); hence it does not span A g and its integral must be zero (b) xyz spans A in T d [Problem 5.3] and so its integral need not be zero (c) xyz xyz under z z (the σ h operation in D 6h ), and so its integral must be zero

7 MOLECULAR SYMMETRY 249 P5.6 We shall adapt the simpler subgroup C 6v of the full D 6h point group. The six π-orbitals span A + B + E + E 2, and are a = (π + π 2 + π 3 + π 4 + π 5 + π 6 ) 6 b = (π π 2 + π 3 π 4 + π 5 π 6 ) 6 (2π π 2 π 3 + 2π 4 π 5 π 6 ) e 2 = 2 2 (π 2 π 3 + π 5 π 6 ) (2π + π 2 π 3 2π 4 π 5 + π 6 ) e = 2 2 (π 2 + π 3 π 5 π 6 ) The hamiltonian transforms as A ; therefore all integrals of the form ψ Hψ dτ vanish unless ψ and ψ belong to the same symmetry species. It follows that the secular determinant factorizes into four determinants A : H a a = 6 (π + +π 6 )H (π + +π 6 ) dτ = α + 2β B : H b b = 6 (π π 2 + )H (π π 2 + ) dτ = α 2β E : H e (a)e (a) = α β, H e (b)e (b) = α β, H e (a)e (b) = 0 Hence α β ε 0 = 0 solves to ε = α β(twice) 0 α β ε E 2 : H e2 (a)e 2 (a) = α + β, H e2 (b)e 2 (b) = α + β, H e2 (a)e 2 (b) = 0 Hence α + β ε 0 = 0 solves to ε = α + β(twice) 0 α + β ε P5.7 Consider phenanthrene with carbon atoms as labeled in the figure below d c e b f a g a b g f e c d (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 C 2v group. To find the character of an operation in this basis, sum the coefficients of the basis terms that are unchanged by the operation. a a b b c c d d e e f f g g χ E a a b b c c d d e e f f g g 4 C 2 a a b b c c d d e e f f g g 0 σ v a a b b c c d d e e f f g g 0 σ v a a b b c c d d e e f f g g 4

8 250 INSTRUCTOR S MANUAL 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 5.5(a)). The table below illustrates the procedure, beginning at left with the C 2v character table. E C 2 σ v σ v product E C 2 σ v σ v sum/h A A B B The orbitals span 7A 2 + B 2. To find symmetry-adapted linear combinations (SALCs), follow the procedure described in Section 5.5(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. or example, the characters of species A are,,,, 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 A symmetry. (No surprise here: the orbitals span only A 2 and B.) An A 2 SALC is obtained by multiplying the characters,,, by the first column: 4 (a a a + a) = 2 (a a ). The A 2 combination from the second column is the same. There are seven distinct A 2 combinations in all: /2(a a ), /2(b b ),...,/2(g g ). The B combination from the first column is: 4 (a + a + a + a) = 2 (a + a ). The B combination from the second column is the same. There are seven distinct B combinations in all: 2 (a + a ), 2 (b + b ),..., 2 (g + g ). There are no B 2 combinations, as the columns sum to zero. (b) The structure is labeled to match the row and column numbers shown in the determinant. The Hückel secular determinant of phenanthrene is: a b c d e f g g f e d c b a a α E β β b β α E β β c 0 β α E β d 0 0 β α E β e β α E β f β α E β g 0 β β α E β g β α E β β 0 f β α E β e β α E β d β α E β 0 0 c β α E β 0 b β β α E β a β β α E This determinant has the same eigenvalues as as in exercise 4.6(b)b.

9 MOLECULAR SYMMETRY 25 (c) The ground state of the molecule has A 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 A 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 A. Consider first transitions to another A wavefunction, in which case we need the product A µa.nowa A = A, and the only character that returns A when multiplied by A is A itself. The z component of the dipole operator belongs to species A,soz-polarized A A transitions are allowed. (Note: transitions from the A ground state to an A 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 (A 2 or B ) to the other; in that case, the excited-state wavefunction will have symmetry of A B = B 2 from the two singly occupied orbitals in the excited state. The symmetry of the transition dipole, then, is A µb 2 = µb 2, and the only species that yields A when multiplied by B 2 is B 2 itself. Now the y component of the dipole operator belongs to species B 2, so these transitions are also allowed (y-polarized). P5.2 (a) ollowing the flow chart in ig. 5.4, not that the molecule is not linear (at least not in the mathematical sense); there is only one C n axis (a C 2 ), and there is a σ h. The point group, then, is C 2h. b d f h j k i g e c a a c e g i k j h f d b (b) The 2p z orbitals are transformed under the symmetry operations of the C 2h group as follows. a a b b c c... j j k k χ E a a b b c c... j j k k 22 C 2 a a b b c c... j j k k 0 i a a b b c c... j j k k 0 σ h 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 5.5(a)). The table below illustrates the procedure, beginning at left with the C 2h character table. E C 2 i σ h product E C 2 i σ h sum/h A g A u B g B u The orbitals span A u + B g. To find symmetry-adapted linear combinations (SALCs), follow the procedure described in Section 5.5(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. or example, the characters of species A u are,,,, so the

10 252 INSTRUCTOR S MANUAL 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 A g symmetry. (No surprise: the orbitals span only A u and B g ). An A u SALC is obtained by multiplying the characters,,, bythe first column: 4 (a + a + a + a) = 2 (a + a ). The A u combination from the second column is the same. There are distinct A u combinations in all: /2(a + a ), /2(b + b ),.../2(k + k ). The B g combination from the first column is: 4 (a a a + a) = 2 (a a ). The B g combination from the second column is the same. There are distinct B g combinations in all: /2(a a ), /2(b b ),.../2(k k ). There are no B u combinations, as the columns sum to zero. (c) The structure is labeled to match the row and column numbers shown in the determinant. The Hückel secular determinant is: a b c... i j k k j i... c b a a α E β b β α E β c 0 β α E i α E β j β α E β k β α E β k β α E β j β α E β i β α E c α E β 0 b β α E β a β α E The energies of the filled orbitals are α β, α β, α β, α β, α β, α +.365β, α β, α β, α β, α β, and α β. The π energy is β. (d) The ground state of the molecule has A g 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 A g 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 A g. Consider first transitions to another A g wavefunction, in which case we need the product A g µa g.nowa g A g = A g, and the only character that returns A g when multiplied by A g is A g itself. No component of the dipole operator belongs to species A g,sonoa g A g 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 (A u or B g ) to the other; in that case, the excited-state wavefunction will have symmetry of A u B g = B u from the two singly occupied orbitals in the excited state. The symmetry of the transition dipole, then, is A g µb u = µb u, and the only species that yields A g when multiplied by B u is B u itself. The x and y components of the dipole operator belongs to species B u, so these transitions are allowed.

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