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Chapter 7Answers to Problems

h g7.1 In I all d orbitals transform as the species H . Therefore they remain degenerate. Alldegenerate d orbitals will rise in energy due to electron-electron repulsions between the

60metal and “ligand” C .

7.2 See J. J. Zuckerman, J. Chem Educ., 1965, 42, 315 and R. Krishnamurthy and W. B.Schaap, J. Chem. Educ., 1969, 46, 799.

2 4h(a) ML - linear (D )

g z___ F d 2+

g xz yz___ ___ B d , d

g x -y xy___ ___ * d 2 2, d

3 3h(b) ML - trigonal planar (D )

x -y xy___ ___ e' d 2 2, d

1 z___ a ' d 2

xz yz___ ___ e" d , d

5 3h(c) ML - trigonal bipyramid (D )

1 z___ a ' d 2

x -y xy___ ___ e' d 2 2, d

xz yz___ ___ e" d , d

5 4v(d) ML - square pyramid (C )

1 x -y___ b d 2 2

1 z___ a d 2

2 xy___ b d

xz yz___ ___ e d , d

6 3d(e) ML - trigonal antiprism (D )

g xz yz___ ___ e d , d

g x -y xy___ ___ e d 2 2, d

1g z___ a d 2

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7 2v(f) ML - capped trigonal prism (C ) The order of the orbitals is not unambiguous in thiscase and will depend upon the details of the geometry. The ordering suggested below isbased on the following idealized model, in which the central atom lies in the plane of thefour ligand atoms from which the capped position emerges.

2 xy___ a d

1 z___ a d 2

1 xz___ b d

1 x -y___ a d 2 2

2 yz___ b d

7 5h(g) ML - pentagonal bipyramid (D )

1 z___ a ' d 2

2 x -y xy___ ___ e 'd 2 2, d

1 xz yz___ ___ e " d , d

8 4d(h) ML - square antiprism (D )

3 xz yz___ ___ e d , d

2 x -y xy___ ___ e d 2 2, d

1 z___ a d 2

8 h(i) ML - cubic (O ) Same splitting as an octahedron, but with reversed order.

2g xz yz xy___ ___ ___ t d , d , d

g x -y z___ ___ e d 2 2, d 2

9 3h(j) ML - tricapped trigonal prism (D )

xz yz___ ___ e" d , d

x -y xy___ ___ e' d 2 2, d

1 z___ a ' d 2

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7.3 One way of approaching this to consider an initial tetragonal distortion, involving stretchingalong z, followed by compression along x to give the orthorhombically distorted structure;

h 4h 2hi.e., O 6 D 6 D . This would lead to the following CFT orbital splitting diagram:

As a result of the orthorhombic distortion, orbitals with an x component will rise in energy,

4hwhile those with a z component will fall in energy. Relative to the D z-stretch case, the

x -y z xy xzseparation between d 2 2 and d 2 orbitals will increase, as will that between the d and d

xz yz 4h g 2h yzorbitals. The degeneracy between d and d in D (e ) will be lifted in D as the d orbitalbecomes more stable.

2h hThis is an allowed Jahn-Teller distortion, because D is centrosymmetric, as is O . Forexample, for d the proposed orthorhombic distortion would give a nondegenerate ground 1

state. This distortion would be more likely than a tetragonal z-stretch (which leads to adoubly degenerate ground state), but it is no more likely than a tetragonal z-compression(which leads to a nondegenerate state). As always, we cannot predict the nature of thedistortion that will occur on the basis of the Jahn-Teller theorem.

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37.4 Using the C axis shown below, clockwise rotations will result in the transformations listedin the following table.

3 3C C 2

x z y

y x z

z y x

xy zx yzd d d

yz zy yxd d d

yz xy zxd d d

x -y z -x y -zd 2 2 d 2 2 d 2 2

2z -x -y 2y -z -x 2x -y -zd 2 2 2 d 2 2 2 d 2 2 2

The first three rows of d orbital listings show the equivalence of the orbitals comprising the

2t set. The last two lines show the equivalence of the orbitals comprising the e set, when the

x -y 2z -x -ytransformed orbitals are seen to be related to d 2 2 and d 2 2 2 by Eqs. (7.1a) - (7.1d).

x -y z -x z -y 87.5 Taking the functions d 2 2, d 2 2, and d 2 2 as reference functions, C rotations about an axisperpendicular to the plane of the cloverleaf in each case (z, y, and x) will effect

xy xz yz x -y xytransformations to d , d , and d , respectively. The interconversion of d 2 2 and d isshown below.

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z -x z -yAs noted in the statement of the problem, d 2 2 and d 2 2 can be combined into the

z 8 8 8conventional function for d 2. As a result, the operations of C (z), C (y), and C (x) wouldinterconvert all five conventional d orbitals into each other or the components of the

z 8combination that makes up d 2, making them degenerate with each other. Of course, C is

hnot an operation in O , and the five d orbitals cannot be made completely degenerate by anyoperation in that point group.

3h7.6 (a) DP(E) = 2l + 1 = 5

3P(C ) = {sin (2.5)(120 )}/{sin 60 } = -0.866/0.866 = -1o o

2P(C ) = {sin (2.5)(180 )}/{sin 90 } = 1/1 = 1o o

P(F) = + sin {(2.5)(180 )} = 1o

3P(S ) = + {sin (2.5)(120 + 180 )}/{sin 150 } = 0.5/0.5 = 10 o o

3h 3 2 h 3 vD E 2C 3C F 2S 3F

d' 5 -1 1 1 1 1

d 1' = A ' + E' + E"

2h(b) DP(E) = 2l + 1 = 5

2P(C ) = {sin (2.5)(180 )}/{sin 90 } = 1/1 = 1o o

P(i) = + {2l + 1) = 5P(F) = + sin {(2.5)(180 )} = 1o

2h 2 2 2D E C (z) C (y) C (x) i F(xy) F(xz) F(yz)

d' 5 1 1 1 5 1 1 1

d g 1g 2g 3g' = 2A + B + B + B

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4v(c) CP(E) = 2l + 1 = 5

4P(C ) = {sin (2.5)(90 )}/{sin 45 } = -0.707/0.707 = -1o o

2P(C ) = {sin (2.5)(180 )}/{sin 90 } = 1/1 = 1o o

P(F) = + sin {(2.5)(180 )} = 1o

4v 4 2 v dC E 2C C 2F 2F

d' 5 -1 1 1 1

d 1 1 2' = A + B + B + E(d) I

P(E) = 2l + 1 = 5

5P(C ) = {sin (2.5)(72 )}/{sin 36 } = 0/0.588 =0o o

5P(C ) = {sin (2.5)(144 )}/{sin 72 } = 0/0.951 =02 o o

3P(C ) = {sin (2.5)(120 )}/{sin 60 } = -0.866/0.866 = -1o o

2P(C ) = {sin (2.5)(180 )}/{sin 90 } = 1/1 = 1o o

h 5 5 3 2I E 12C 12C 20C 15C2

d' 5 0 0 -1 1

d g h' = H Y H in I

7.7 With l = 3, f orbitals are inherently ungerade, requiring the use of the negative sign withEqs. (7.4) - (7.6).

d(a) TP(E) = 2l + 1 = 7

3P(C ) = {sin (3.5)(120 )}/{sin 60 } = 0.866/0.866 = 1o o

2P(C ) = {sin (3.5)(180 )}/{sin 90 } = -1/1 = -1o o

4P(S ) = – {sin (3.5)(270 )}/{sin 135 } = –(-0.707)/0.707 = 1o o

P(F) = – sin {(3.5)(180 )} = – (-1) = 1o

d 3 2 4 dT E 8C 3C 6S 6F

f' 7 1 -1 1 1

f 1 1 2' = A + T + T

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h d(b) O . By correlation of the T results obtained in part (a) to the corresponding ungerade

hspecies of O we obtain

d hT 6 O

1 2uA 6 A

1 2uT 6 T

2 1uT 6 T

Alternately, the reducible representation can be generated and decomposed in the usualway.P(E) = 2l + 1 = 7

3P(C ) = {sin (3.5)(120 )}/{sin 60 } = 0.866/0.866 = 1o o

2P(C ) = {sin (3.5)(180 )}/{sin 90 } = -1/1 = -1o o

4P(C ) = {sin (3.5)(90 )}/{sin 45 } = -0.707/0.707 = -1o o

P(i) = – (2l + 1) = -7

4P(S ) = – {sin (3.5)(270 )}/{sin 135 } = –(-0.707)/0.707 = 1o o

6P(S ) = – {sin (3.5)(240 )}/{sin 120 } = – (0.866)/0.866 = -1o o

P(F) = – sin {(3.5)(180 )} = – (-1) = 1o

h 3 2 4 2 4 6 h dO E 8C 6C 6C 3C i 6S 8S 3F 6F

f' 7 1 -1 -1 -1 -7 1 -1 1 1

f 2u 1u 2u' = A + T + T

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4h h(c) D . By correlation of the O results obtained in part (b) to the corresponding ungerade

4hspecies D of we obtain

h 4hO 6 D

2u 1uA 6 B

2u 2u uT 6 A + E

1u 2u uT 6 B + E

Alternately, the reducible representation can be generated and decomposed in the usualway.

P(E) = 2l + 1 = 7

4P(C ) = {sin (3.5)(90 )}/{sin 45 } = -0.707/0.707 = -1o o

2P(C ) = {sin (3.5)(180 )}/{sin 90 } = -1/1 = -1o o

P(i) = – (2l + 1) = -7

4P(S ) = – {sin (3.5)(270 )}/{sin 135 } = –(-0.707)/0.707 = 1o o

P(F) = – sin {(3.5)(180 )} = – (-1) = 1o

4h 4 2 2 2 4 h v dD E 2C C 2C ' 2C " i 2S F 2F 2F

f' 7 -1 -1 -1 -1 -7 1 1 1 1

f 2u 1u 2u u' = A + B + B + 2E

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3d h(d) D . By correlation of the O results obtained in part (b) to the corresponding ungerade

4hspecies D of we obtain

h 3dO 6 D

2u 2uA 6 A

2u 2u uT 6 A + E

1u 1u uT 6 A + E

Alternately, the reducible representation can be generated and decomposed in the usualway.

P(E) = 2l + 1 = 7

3P(C ) = {sin (3.5)(120 )}/{sin 60 } = 0.866/0.866 = 1o o

2P(C ) = {sin (3.5)(180 )}/{sin 90 } = -1/1 = -1o o

P(i) = – (2l + 1) = -7

6P(S ) = – {sin (3.5)(240 )}/{sin 120 } = – (0.866)/0.866 = -1o o

P(F) = – sin {(3.5)(180 )} = – (-1) = 1o

3d 3 2 6 dD E 2C 3C i 2S 3F

f' 7 1 -1 -7 -1 1

f 1u 2u u' = A + 2A + 2E

7.8 (a) Only degenerate states are subject to Jahn-Teller distortions. Assuming the ordering of dorbitals shown on the extreme right of Fig. 7.9, only the configurations d and d result in 1 3

g 1gdegenerate ground states and would be distorted. If the e orbitals are above the a orbital,the ground state configurations d and d would be degenerate and would be distorted. Of 3 5

gcourse, any excited states in which one or three electrons occupy the e orbitals would alsobe degenerate and result in Jahn-Teller distortions.

4h(b) D is a centrosymmetric point group, so any Jahn-Teller distortion would produce acentrosymmetric structure. Disregarding ligand structure, the only centrosymmetric

2h i 2hsubgroups are D and C . A D structure would result from shortening or lengthening atrans-related pair of bonds relative to the other pair of trans-related bonds. If the L–M–L'bond angle in such a structure were distorted from 90 , the symmetry would descend furthero

ito C . As always with the Jahn-Teller theorem, the precise distortion cannot be predicted.

4 4h g(c) Almost all ML D complexes are d , in which the e orbitals are filled, regardless of 7-8

1gtheir relative order with the a orbital. Consequently, square planar complexes havenondegenerate ground states and are immune to Jahn-Teller distortions.

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5 1 27.9 (a) As shown for PF (cf. problem 4.13), for the ligand F-SALCs, ' = 2A ' + E' + A ". The

3h zsymmetries of the d orbitals, taken from the direct product transformations in D , are d 2 =

1 x -y xy xz yzA ', (d 2 2, d ) = E', (d , d ) = E". Bonding and antibonding combinations can be formed

1 xz yzwith A ' and E' symmetries, but the E" symmetry (d , d ) pair must be nonbonding in thismodel. The following qualitative MO can be constructed.

(b) The upper three levels (boxed region, above) have identical splitting to that in the CFTtbp model (cf. problem 7.2 c). Electron pairs contributed by the ligands are sufficient to fillall lower levels. Therefore, any electrons contributed by the metal ion result in filling in theupper three levels, in the same manner as predicted in the CFT model.

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1 z(c) The a ' combination with the d 2 orbital would use opposite signs for the axial and

1 1 2 3 4 5equatorial ligand F AOs. The SALC would be M = 1//5{N + N + N – N – N }. The resulting LCAO would be:

x -y xyThe two e' SALCs, which combine with d 2 2 and d , involve only equatorial functions.

2 1 2 3 3 2 3They are M = 1//6{2N – N – N } and M = 1//2{N – N }, respectively. The resultingLCAOs, projected in the xy plane, would be:

1 2(d) The a ' and a " SALCs, which form nonbonding MOs in the d-only scheme, could form

z x ybonding and antibonding combinations with s and p AOs on the metal. The p and p AOscould form bonding and antibonding combinations with the e' SALCs, which in the d-only

x -y xyscheme form bonding and antibonding combinations with d 2 2 and d AOs. Thiscompetition would result in six mixed MOs of varying bonding, nonbonding, andantibonding character, depending upon the relative contributions of d and p AOs.

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7.10 (a) The representation of F-SALCs is based on the following model:

4v 4 2 v dC E 2C C 2F 2F

F' 5 1 1 3 1

F 1 1' = 2A + B + E

The AO symmetries of the central M atom are shown below.

z 1 x -y 1 xy 2 xz yz(n – 1)d: d 2 = A , d 2 2 = B , d = B , (d , d ) = E

1ns: s = A

z 1 x ynp: p = A , (p , p ) = E

xy 2The d orbital (b ) has no symmetry match with a SALC and must remain nonbonding.

(b) For simplicity, the following sketches show 2 = 90 . The three kinds of potentiallyo

bonding interactions between (n – 1)d orbitals and F-SALCs are the following:

The e (xz, yz) combinations are nonbonding at 2 = 90 and become only slightly bondingo

xwith 2 > 90 . A better bonding combination with the E SALCs can be formed with p ando

yp AOs (see below).

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The ns and np AOs can form the following bonding combinations:

1 1 zThe ns and np combinations for a are probably mixed. With a total of three a AOs (d 2, s,

z 1p ) and two a SALCs, there can be only five MOs. We will assume bonding and

z 1 1 z 1antibonding combinations for d 2 (a ) and s (a ), and treat p (a ) as essentially nonbonding.

x y xzThe two e combinations with p and p have better overlap with the E SALCs than d and

yzd . Therefore we will form bonding and antibonding combinations with these and assume

xz yzthat the d and d orbitals are essentially nonbonding.

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(c) The following MO scheme is based on the assumptions noted above.

(d) In the MO scheme shown for part (c), electrons from ligand would fill all levels belowthose in the box. The MOs in the box have the same symmetries and are in the samerelative energy order as the d orbitals in a square pyramidal crystal field [cf. answer to 7.2(d)]. Thus, the filling of electrons in MOs above the four lowest bonding levels isequivalent to the presumed filling of electrons in d orbitals of the CFT model.

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27.11 (a) The SALCs are identical to the oxygen F- and B-SALCs derived for CO in Chapter 4(pp. 118-126). Thus,

F g u' = E + E+ +

B u g' = A + A

(b) Assuming p orbitals on the ligands, the SALCs would have the following forms:

g uE E+ +

u A both xz and yz planes

gA both xz and yz planes

The normalized wave function expressions for these SALCS are as follows:

g z z u z zE = 1//2{N (1) + N (2)} E = 1//2{N (1) – N (2)}+ +

u x x u y yA (x) = 1//2{N (1) + N (2)} A (y) = 1//2{N (1) + N (2)}

g x x g y yA (x) = 1//2{N (1) – N (2)} A (y) = 1//2{N (1) – N (2)}

g g(c) s = E Y bonding and antibonding with E SALC+ +

z u up = E Y bonding and antibonding with E SALC+ +

x y u u(p , p ) = A Y bonding and antibonding with A SALCs

z g gd 2 = E Y bonding and antibonding with E SALC+ +

xz yz g g(d , d ) = A Y bonding and antibonding with A SALCs

x -y xy g(d 2 2, d ) = ) Y nonbonding.

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z(d) Both ns and (n - 1)d 2 AOs can form bonding and antibonding combinations with the

gE SALC, so s-d mixing is possible. Being higher in energy, the ns AO probably+

zcontributes more to antibonding, while the lower-energy (n - 1)d 2 AO contributes more tobonding. For simplicity, the following MO scheme assumes the ns AO on the metal forms

zan essentially nonbonding F level, leaving the (n - 1)d 2 AO to form bonding andn

antibonding MOs without showing possible s-d mixing.

The MOs in the box have the same symmetry and splitting as the presumed splitting of dorbitals in the CFT model (cf. problem 7.2 a). Electron pairs contributed by ligands aresufficient to fill all lower levels. Therefore, any electrons contributed by the metal ionresult in filling in the upper three levels in the same manner as predicted in the CFT model.

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7.12 (a) For F-SALCs, use the following four-vector basis set to obtain the reduciblerepresentation.

4h 4 2 2 2 4 h v dD E 2C C 2C ' 2C " i 2S F 2F 2F

F' 4 0 0 2 0 0 0 4 2 0

F 1g 1g u' = A + B + E

For B-SALCs, use the following eight-vector basis set to obtain the reduciblerepresentation.

4h 4 2 2 2 4 h v dD E 2C C 2C ' 2C " i 2S F 2F 2F

B' 8 0 0 -4 0 0 0 0 0 0

B 2g 2g g 2u 2u u' = A + B + E + A + B + E

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(b) The matches between SALCs and central-atom AOs is shown below. Among the B-

2g 2uSALCs, A and B have no matching AOs and must be nonbonding (n.b.).

1g 1g uF-SALCs A B E

z x -y x yAOs s, d 2 d 2 2 (p , p )

2g 2g g 2u 2u uB-SALCs A B E A B E

xy xz yz z x yAOs n.b. d (d , d ) p n.b. (p , p )

(c) F-SALCs:

1g 1 2 3 4G(A ) = ½(F + F + F + F )

1g 1 2 3 4G(B ) = ½(F – F + F – F )

u 1 3G (E ) = 1//2(F – F )"

u 2 4G (E ) = 1//2(F – F )$

B-SALCs:

2g 5 6 7 8A(A ) = ½(B + B – B – B ) (nonbonding)

2g 5 6 7 8A(B ) = ½(B – B + B + B )

g 2 4A (E ) = 1//2(B – B )"

g 1 3A (E ) = 1//2(B – B )$

2u 1 2 3 4A(A ) = ½(B + B + B + B )

2u 1 2 3 4A(B ) = ½(B – B + B – B ) (nonbonding)

u 6 8A (E ) = 1//2(B + B )"

u 5 7A (E ) = 1//2(B + B )$

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1g(d) G(A ) + s

1g zG(A ) + d 2

1g x -yG(B ) + d 2 2

u y u xG (E ) + p ; G (E ) + p" $

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2gA(A )

2g xyA(B ) + d

g xz g yzA (E ) + d ; A (E ) + d" $

2u zA(A ) + p

2uA(B )

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u x u yA (E ) + p ; A (E ) + p" $

7.13 We can save labor by doing the work in O and then correlating to the matching gerade

hspecies of O , simply adding g subscripts to the Mulliken symbols in O.

For G, L = 4.P(E) = 2l + 1 = 9

3P(C ) = {sin (4.5)(120 )}/{sin 60 } = 0/0.866 = 0o o

2P(C ) = {sin (4.5)(180 )}/{sin 90 } = 1/1 = 1o o

4P(C ) = {sin (4.5)(90 )}/{sin 45 } = 0.707/0.707 = 1o o

3 2 4 2O E 8C 3C 6C 6C '

G' 9 0 1 1 1

G 1g g 1g 2g' = A + E + T + T

For H, L = 5.P(E) = 2l + 1 = 11

3P(C ) = {sin (5.5)(120 )}/{sin 60 } = -0.866/0.866 = -1o o

2P(C ) = {sin (5.5)(180 )}/{sin 90 } = -1/1 = -1o o

4P(C ) = {sin (5.5)(90 )}/{sin 45 } = 0.707/0.707 = 1o o

3 2 4 2O E 8C 3C 6C 6C '

H' 11 -1 -1 1 -1

H g 1g 2g' = E + 2T + T

For I, L = 6.P(E) = 2l + 1 = 13

3P(C ) = {sin (6.5)(120 )}/{sin 60 } = 0.866/0.866 = 1o o

2P(C ) = {sin (6.5)(180 )}/{sin 90 } = 1/1 = 1o o

4P(C ) = {sin (6.5)(90 )}/{sin 45 } = -0.707/0.707 = -1o o

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3 2 4 2O E 8C 3C 6C 6C '

I' 13 1 1 -1 1

I 1g 2g g 1g 2g' = A + A + E + T + 2T

The splitting of g and i orbitals would be the same as the states, because both are gerade.

u 1u 2uThe h orbitals are ungerade, and contrary to the H state will split as e + 2t + t .

h 4h7.14 (a) Use the O 6 D correlation table.

Free-Ion

h 4hTerms Terms in O Terms in D

1g 1gS A A

1g 2g gP T A + E

g 2g 1g 1g 2g gD E + T A + B + B + E

2g 1g 2g 2g 1g 2g gF A + T + T A + B + B + 2E

1g g 1g 2g 1g 2g 1g 2g gG A + E + T + T 2A + A + B + B + 2E

g 1g 2g 1g 2g 1g 2g gH E + 2T + T A + 2A + B + B + 3E

1g 1g g 1g 2g 1g 2g 1g 2g gI A + A + E + T + 2T 2A + A + 2B + 2B + 3E

3(b) The most direct correlation for terms S through F is to use the R correlation table

3 3 h 3d 3d 3showing R 6 D . For the I terms, use O 6 D and then D 6 D .

Free-Ion

h 3Terms Terms in O Terms in D

1g 1S A A

1g 2P T A + E

g 2g 1D E + T A + 2E

2g 1g 2g 1 2F A + T + T A + 2A + 2E

1g g 1g 2g 1 2G A + E + T + T 2A + A + 3E

g 1g 2g 1 2H E + 2T + T A + 2A + 4E

1g 1g g 1g 2g 1 2I A + A + E + T + 2T 3A + 2A + 4E

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h 2d 4h(c) There is no correlation table for O 6 D . Use results in part (a) with the table for D

2d 2 2 2d 4h6 D (C ' 6 C '). Note that for D , the ligand-field terms are the same as for D

d 2dwithout the subscript g. Alternately, one could use the correlation T 6 D .

Free-Ion

4h 2dTerms Terms in D Terms in D

1g 1S A A

2g g 2P A + E A + E

1g 1g 2g g 1 1 2D A + B + B + E A + B + B + E

2g 1g 2g g 2 1 2F A + B + B + 2E A + B + B + 2E

1g 2g 1g 2g g 1 2 1 2G 2A + A + B + B + 2E 2A + A + B + B + 2E

1g 2g 1g 2g g 1 2 1 2H A + 2A + B + B + 3E A + 2A + B + B + 3E

1g 2g 1g 2g g 1 2 1 2I 2A + A + 2B + 2B + 3E 2A + A + 2B + 2B + 3E

4h 4h(d) One the basis of vector transformations, determine the correlation between D and D ,as follows:

4h 4hD D Basis

1g gA E z+ 2

2g g zA E R–

1g 2g gB + B ) (x – y , xy)2 2

g g x yE A (R , R ) (xz, yz)

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Free-Ion

4h 4hTerms Terms in D Terms in D

1g gS A E +

2g g g gP A + E E + A–

1g 1g 2g g g g gD A + B + B + E E + A + )+

2g 1g 2g g g g gF A + B + B + 2E E + 2 A + )–

1g 2g 1g 2g g g g g gG 2A + A + B + B + 2E 2E + E + 2A + )+ –

1g 2g 1g 2g g g g g gH A + 2A + B + B + 3E E + 2E + 3A + )+ –

1g 2g 1g 2g g g g g gI 2A + A + 2B + 2B + 3E 2E + E + 3A + 2)+ –

7.15 (a) There are four microstates:

ge ¼ ¼ ¿ ¿

2gt ¼¿ ¼¿ ¼¿ ¼¿ ¼¿ ¼¿ ¼¿ ¼¿ ¼¿ ¼¿ ¼¿ ¼¿

(b) Use Table 7.2, or apply Eqs. (7.2) - (7.6) for L = 4. Either way, the states emerging

1g g 1g 2gfrom G are A , E , T , T .2 2 2 2 2

t(c) For G, D = (2)(9) = 18.2

1g g 1g 2g tFor A + E + T + T , D = (2)(1) + (2)(2) + (2)(3) + (2)(3) = 18.2 2 2 2

7.16 The free-ion terms of d and d are the same because they arise from the same number of n 10-n

microstates, as shown by Eq. (7.14):

The free-ion terms are unique to a given number of microstates. Moreover, each free-ionterm splits into a characteristic set of ligand-field terms in any point group. Thus, theligand-field terms for d and d must be the same. n 10-n

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d7.17 The strong-field terms emerging from the d T configurations in the hypothetical extreme 2

hfield are the same as for the comparable d O configurations, except that the subscript g is 2

omitted for each. The extreme-field configurations, however, will be in the reverse energy

h 2 2order of the d O configurations; i.e., e < e t < t . Identifying the triplet terms versus the 2 2 1 1 2

hsinglet terms follows the same logic as shown in the d O case; i.e., each configuration 2

dgives rise to the same multiplicities for comparable T orbital terms. Making the

hcorrelations in the same way gives a diagram identical to that shown for d O (Fig. 7.26), 8

except for the gerade designations on the states.

h g7.18 Weak-Field Terms. In O , the free-ion D term of a d configuration becomes a E ground2 9 2

2gstate term and a T excited state terms. From the correlation tables, it can be seen that2

4hthese terms split on descent to D square planar geometry as follows:

g 1g 1g 2g 2g gE 6 A + B T 6 B + E2 2 2 2 2

4hExtreme Field Configurations and Strong-Field Terms. The configurations for the d D 9

square planar geometry in an extremely strong field are shown below in order of increasingenergy. The terms in a strong field are obtained by taking the orbital assignment of the

4hunpaired electron as a basis for a representation in D . All paired electrons can be ignored,because the representation for any pair must be the totally symmetric representation; e.g.,

1g 1g 1gB × B = A . With only one unpaired electron in all configurations, all terms must bedoublets.

1gb ¼ ¼¿ ¼¿ ¼¿

2gb ¼¿ ¼ ¼¿ ¼¿

1ga ¼¿ ¼¿ ¼ ¼¿

ge ¼¿ ¼¿ ¼¿ ¼¿ ¼¿ ¼¿ ¼¿ ¼

1g 2g 1g g B B A E2 2 2 2

The resulting qualitative correlation diagram is shown below.

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67.19 (a) FeF is a d high-spin complex, because F is a weak-field ligand. Therefore, there are3– 5 –

1g 6no spin-allowed transitions from the A ground state. Fe(CN) is a d low-spin complex,6 3– 5

2gbecause CN is a strong-field ligand. The ground state is T term. From the Tanabe-– 2

Sugano diagram for d low-spin cases (right side), it is apparent that there are at least three 5

2g 2g 1g 2g g 2g 2gpossible spin allowed transitions: T 6 [ A , T ], T 6 E , T 6 A . As the Tanabe-2 2 2 2 2 2 2

2g 1gSugano diagram indicates, the A and T states for the first transition are virtually2 2

degenerate and would give rise to a single broad band. These transitions account for the

6color of Fe(CN) .3–

(b) Tetrahedral complexes are not centrosymmetric, so the LaPorte Rule does not apply. Nonetheless, d orbitals have an inherent gerade character. Therefore, although molarabsorptivities are slightly higher for tetrahedral complexes compared to octahedralcomplexes, they are still relatively low by comparison to charge-transfer transitions.

(c) {The charge on the complex should be 3+, not 2+ as shown in the first printing.}

2 3[Cr(H O) ] is an octahedral d case. From either the Tanabe-Sugano diagram or Orgel3+ 3

2g 2g 2g 1g 2gdiagram, three bands are expected from the transitions A 6 T , A 6 T (F), A 64 4 4 4 4

1g 2g 2g gT (P). The last of these involves a change in configuration from t to t e ; i.e., a two-4 3 1 2

electron transition. This is a less probable event and therefore would have a lowerabsorptivity. Moreover, it is the highest energy transition and may be obscured by the tailof a stronger charge-transfer band in the ultraviolet.

2 6(d) A d high-spin complex such as [Fe(H O) ] is expected to show a single band from the 6 2+

2g gtransition T 6 E . However, both the ground state and excited state are degenerate and5 5

subject to Jahn-Teller distortions. The distortion is greater for the doubly degenerateexcited state. Therefore, the observed band splitting is primarily due to lifting of the

gdegeneracy of the upper E state.5

(e) With d configurations, both complex ions have no possible d-d transitions. The 0

observed intense colors arise from L6M charge transfer transitions, which are LaPorteallowed. Such transitions can have very high molar absorptivities.

7.20 (a) Use direct products for the configurations to determine the orbital term designations. Where two electrons are paired in the same orbital the product is totally symmetric. The

g 1g 2g 1gground state e a b consists of all paired electrons, so the term is A . For the first4 2 2 1

g 1g 2g 1g 2g 1gexcited singlet state e a b b , take the direct product B × B :4 2 1 1

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4h 4 2 2 2 4 h v vD E 2C C 2C ' 2C " i 2S F 2F 2F

2gB 1 -1 1 -1 1 1 -1 1 -1 1

1gB 1 -1 1 1 -1 1 -1 1 1 -1

2g' = A 1 1 1 -1 -1 1 1 1 -1 -1

2gTherefore, the first singlet excited state is A . For the second excited singlet state1

g 1g 2g 1g 1g 1g 1g 1ge a b b the direct product is A × B = B , and the resulting term is B . For the4 1 2 1 1

g 1g 2g 1g g 1gthird excited singlet state e a b b , the direct product is E × B :3 2 2 1

4h 4 2 2 2 4 h v vD E 2C C 2C ' 2C " i 2S F 2F 2F

gE 2 0 -2 0 0 2 0 -2 0 0

1gB 1 -1 1 1 -1 1 -1 1 1 -1

g' = E 2 0 -2 0 0 2 0 -2 0 0

gTherefore the term is E .1

(b) The three electronic transitions are

1g 2gA 6 A1 1

1g 1gA 6 B1 1

1g gA 6 E1 1

(c) For an electronic transition to be vibronically allowed, the transition moment

e v e v eIR R :R ' R ' dJ must be totally symmetric to be non-zero. In this specific case R is also

vtotally symmetric, and the ground vibrational state, R is totally symmetric for all moleculesexcept free radicals. Therefore the symmetry of the integral depends upon the symmetries

e vof the direct products : × R ' × R '. The symmetries of the dipole moment components,

4hobtained from the unit vector transformation properties listed in the D character table are

z 2u x y u: = A (: , : ) = E

2g 1g gThe symmetries of the excited electronic states, as shown in part (a), are A , B , and E . The symmetries of the components of : are ungerade, and the symmetries of the electronic

eexcited states are gerade, so the products : × R ' must all be ungerade. In order for the

vintegral to be totally symmetric and nonvanishing, the R ' must belong to an identical

eungerade species as : × R ' . The ungerade normal modes of vibration, which mightcouple with the electronic transitions, are

3 2u 5 2u 6 u 7 u< (A ) < (B ) < (E ) < (E )

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For the purpose of vibronic coupling, the gerade normal modes can be ignored. By

e vmatching symmetries of : × R ' with one or more of these R ' we obtain the followingresults, which show the normal modes that vibronically allow each electronic transition:

eTransition : × R ' Matching normal modes

1g 2g 2u 2g 1uA 6 A A × A = A none1 1

u 2g u 6 7 uE × A = E < , < (E )

1g 1g 2u 1g 2u 5 2uA 6 B A × B = B < (B )1 1

u 1g u 6 7 uE × B = E < , < (E )

1g g 2u g u 6 7 uA 6 E A × E = E < , < (E )1 1

u g 1u 2u 1u 2u 3 2u 5 2uE × E = A + A + B + B < (A ), < (B )

1g 2g z 2uIn the first case for A 6 A , the integral with the dipole moment component : (A )1 1

1uvanishes, because there is no matching normal mode with the resultant A symmetry.

1g 2gNonetheless, the transition A 6 A is vibronically allowed by coupling with the normal1 1

6 7 u emodes < and < (E ), because the direct product : × R ' with the degenerate pair of dipole

x y umoment components (: , : ) is E , making the full integral totally symmetric. In the last

1g g u gcase for A 6 E , the direct product E × E is a reducible representation, only two whose1 1

2u 2ucomponent irreducible representations (A and B ) match symmetries with normal modes.

1g gThose matches are sufficient to vibronically allow the A 6 E electronic transition.1 1