Download -108- Chapter 7 Answers to Problems 7.1 In I all d orbitals transform

Chapter 7
Answers to Problems
7.1 In Ih all d orbitals transform as the species Hg. Therefore they remain degenerate. All
degenerate d orbitals will rise in energy due to electron-electron repulsions between the
metal and “ligand” C60.
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.
(a) ML2 - linear (D4h)
___
___ ___
___ ___
Fg+ dz2
Bg dxz, dyz
*g dx2-y2, dxy
___ ___
___
___ ___
e' dx2-y2, dxy
a1' dz2
e" dxz, dyz
___
___ ___
___ ___
a1' dz2
e' dx2-y2, dxy
e" dxz, dyz
___
___
___
___ ___
b1 dx2-y2
a1 dz2
b2 dxy
e dxz, dyz
___ ___
___ ___
___
eg dxz, dyz
eg dx2-y2, dxy
a1g dz2
(b) ML3 - trigonal planar (D3h)
(c) ML5 - trigonal bipyramid (D3h)
(d) ML5 - square pyramid (C4v)
(e) ML6 - trigonal antiprism (D3d)
-108-
(f) ML7 - capped trigonal prism (C2v) The order of the orbitals is not unambiguous in this
case and will depend upon the details of the geometry. The ordering suggested below is
based on the following idealized model, in which the central atom lies in the plane of the
four ligand atoms from which the capped position emerges.
___
___
___
___
___
a2 dxy
a1 dz2
b1 dxz
a1 dx2-y2
b2 dyz
(g) ML7 - pentagonal bipyramid (D5h)
___
___ ___
___ ___
a1' dz2
e2'dx2-y2, dxy
e1" dxz, dyz
___ ___
___ ___
___
e3 dxz, dyz
e2 dx2-y2, dxy
a1 dz2
(h) ML8 - square antiprism (D4d)
(i) ML8 - cubic (Oh) Same splitting as an octahedron, but with reversed order.
___ ___ ___ t2g dxz, dyz, dxy
___ ___ eg dx2-y2, dz2
(j) ML9 - tricapped trigonal prism (D3h)
___ ___
___ ___
___
-109-
e" dxz, dyz
e' dx2-y2, dxy
a1' dz2
7.3 One way of approaching this to consider an initial tetragonal distortion, involving stretching
along z, followed by compression along x to give the orthorhombically distorted structure;
i.e., Oh 6 D4h 6 D2h. 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,
while those with a z component will fall in energy. Relative to the D4h z-stretch case, the
separation between dx2-y2 and dz2 orbitals will increase, as will that between the dxy and dxz
orbitals. The degeneracy between dxz and dyz in D4h (eg) will be lifted in D2h as the dyz orbital
becomes more stable.
This is an allowed Jahn-Teller distortion, because D2h is centrosymmetric, as is Oh. For
example, for d 1 the proposed orthorhombic distortion would give a nondegenerate ground
state. This distortion would be more likely than a tetragonal z-stretch (which leads to a
doubly 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 the
distortion that will occur on the basis of the Jahn-Teller theorem.
-110-
7.4 Using the C3 axis shown below, clockwise rotations will result in the transformations listed
in the following table.
C3
C3 2
x
z
y
y
x
z
z
y
x
dxy
dzx
dyz
dyz
dzy
dyx
dyz
dxy
dzx
dx2-y2
dz2-x2
dy2-z2
d2z2-x2-y2
d2y2-z2-x2
d2x2-y2-z2
The first three rows of d orbital listings show the equivalence of the orbitals comprising the
t2 set. The last two lines show the equivalence of the orbitals comprising the e set, when the
transformed orbitals are seen to be related to dx2-y2 and d2z2-x2-y2 by Eqs. (7.1a) - (7.1d).
7.5 Taking the functions dx2-y2, dz2-x2, and dz2-y2 as reference functions, C8 rotations about an axis
perpendicular to the plane of the cloverleaf in each case (z, y, and x) will effect
transformations to dxy, dxz, and dyz, respectively. The interconversion of dx2-y2 and dxy is
shown below.
-111-
As noted in the statement of the problem, dz2-x2 and dz2-y2 can be combined into the
conventional function for dz2. As a result, the operations of C8(z), C8(y), and C8(x) would
interconvert all five conventional d orbitals into each other or the components of the
combination that makes up dz2, making them degenerate with each other. Of course, C8 is
not an operation in Oh, and the five d orbitals cannot be made completely degenerate by any
operation in that point group.
7.6 (a) D3h
P(E) = 2l + 1 = 5
P(C3) = {sin (2.5)(120o)}/{sin 60o} = -0.866/0.866 = -1
P(C2) = {sin (2.5)(180o)}/{sin 90o} = 1/1 = 1
P(F) = + sin {(2.5)(180o)} = 1
P(S3) = + {sin (2.5)(1200 + 180o)}/{sin 150o} = 0.5/0.5 = 1
D3h
E
2C3
3C2
Fh
2S3
3Fv
'd
5
-1
1
1
1
1
'd = A1' + E' + E"
(b) D2h
P(E) = 2l + 1 = 5
P(C2) = {sin (2.5)(180o)}/{sin 90o} = 1/1 = 1
P(i) = + {2l + 1) = 5
P(F) = + sin {(2.5)(180o)} = 1
D2h
E
'd
5
C2(z) C2(y)
1
1
C2(x)
i
F(xy)
1
5
1
'd = 2Ag + B1g + B2g + B3g
-112-
F(xz) F(yz)
1
1
(c) C4v
P(E) = 2l + 1 = 5
P(C4) = {sin (2.5)(90o)}/{sin 45o} = -0.707/0.707 = -1
P(C2) = {sin (2.5)(180o)}/{sin 90o} = 1/1 = 1
P(F) = + sin {(2.5)(180o)} = 1
C4v
E
2C4
C2
2Fv
2Fd
'd
5
-1
1
1
1
'd = A1 + B1 + B2 + E
(d) I
P(E) = 2l + 1 = 5
P(C5) = {sin (2.5)(72o)}/{sin 36o} = 0/0.588 =0
P(C52 ) = {sin (2.5)(144o)}/{sin 72o} = 0/0.951 =0
P(C3) = {sin (2.5)(120o)}/{sin 60o} = -0.866/0.866 = -1
P(C2) = {sin (2.5)(180o)}/{sin 90o} = 1/1 = 1
Ih
E
12C5
12C52
20C3
15C2
'd
5
0
0
-1
1
'd = H Y Hg in Ih
7.7 With l = 3, f orbitals are inherently ungerade, requiring the use of the negative sign with
Eqs. (7.4) - (7.6).
(a) Td
P(E) = 2l + 1 = 7
P(C3) = {sin (3.5)(120o)}/{sin 60o} = 0.866/0.866 = 1
P(C2) = {sin (3.5)(180o)}/{sin 90o} = -1/1 = -1
P(S4) = – {sin (3.5)(270o)}/{sin 135o} = –(-0.707)/0.707 = 1
P(F) = – sin {(3.5)(180o)} = – (-1) = 1
Td
E
8C3
3C2
6S4
6Fd
'f
7
1
-1
1
1
'f = A1 + T1 + T2
-113-
(b) Oh. By correlation of the Td results obtained in part (a) to the corresponding ungerade
species of Oh we obtain
Td 6 Oh
A1 6 A2u
T1 6 T2u
T2 6 T1u
Alternately, the reducible representation can be generated and decomposed in the usual
way.
P(E) = 2l + 1 = 7
P(C3) = {sin (3.5)(120o)}/{sin 60o} = 0.866/0.866 = 1
P(C2) = {sin (3.5)(180o)}/{sin 90o} = -1/1 = -1
P(C4) = {sin (3.5)(90o)}/{sin 45o} = -0.707/0.707 = -1
P(i) = – (2l + 1) = -7
P(S4) = – {sin (3.5)(270o)}/{sin 135o} = –(-0.707)/0.707 = 1
P(S6) = – {sin (3.5)(240o)}/{sin 120o} = – (0.866)/0.866 = -1
P(F) = – sin {(3.5)(180o)} = – (-1) = 1
Oh
E
8C3
6C2
6C4
3C2
i
6S4
8S6
3Fh
6Fd
'f
7
1
-1
-1
-1
-7
1
-1
1
1
'f = A2u + T1u + T2u
-114-
(c) D4h. By correlation of the Oh results obtained in part (b) to the corresponding ungerade
species D4h of we obtain
Oh 6 D4h
A2u 6 B1u
T2u 6 A2u + Eu
T1u 6 B2u + Eu
Alternately, the reducible representation can be generated and decomposed in the usual
way.
P(E) = 2l + 1 = 7
P(C4) = {sin (3.5)(90o)}/{sin 45o} = -0.707/0.707 = -1
P(C2) = {sin (3.5)(180o)}/{sin 90o} = -1/1 = -1
P(i) = – (2l + 1) = -7
P(S4) = – {sin (3.5)(270o)}/{sin 135o} = –(-0.707)/0.707 = 1
P(F) = – sin {(3.5)(180o)} = – (-1) = 1
D4h
E
2C4
C2
2C2'
2C2"
i
2S4
Fh
2Fv
2Fd
'f
7
-1
-1
-1
-1
-7
1
1
1
1
'f = A2u + B1u + B2u + 2Eu
-115-
(d) D3d. By correlation of the Oh results obtained in part (b) to the corresponding ungerade
species D4h of we obtain
Oh 6 D3d
A2u 6 A2u
T2u 6 A2u + Eu
T1u 6 A1u + Eu
Alternately, the reducible representation can be generated and decomposed in the usual
way.
P(E) = 2l + 1 = 7
P(C3) = {sin (3.5)(120o)}/{sin 60o} = 0.866/0.866 = 1
P(C2) = {sin (3.5)(180o)}/{sin 90o} = -1/1 = -1
P(i) = – (2l + 1) = -7
P(S6) = – {sin (3.5)(240o)}/{sin 120o} = – (0.866)/0.866 = -1
P(F) = – sin {(3.5)(180o)} = – (-1) = 1
D3d
E
2C3
3C2
i
2S6
3Fd
'f
7
1
-1
-7
-1
1
'f = A1u + 2A2u + 2Eu
7.8 (a) Only degenerate states are subject to Jahn-Teller distortions. Assuming the ordering of d
orbitals shown on the extreme right of Fig. 7.9, only the configurations d 1 and d 3 result in
degenerate ground states and would be distorted. If the eg orbitals are above the a1g orbital,
the ground state configurations d 3 and d 5 would be degenerate and would be distorted. Of
course, any excited states in which one or three electrons occupy the eg orbitals would also
be degenerate and result in Jahn-Teller distortions.
(b) D4h is a centrosymmetric point group, so any Jahn-Teller distortion would produce a
centrosymmetric structure. Disregarding ligand structure, the only centrosymmetric
subgroups are D2h and Ci. A D2h structure would result from shortening or lengthening a
trans-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 90o, the symmetry would descend further
to Ci. As always with the Jahn-Teller theorem, the precise distortion cannot be predicted.
(c) Almost all ML4 D4h complexes are d 7-8, in which the eg orbitals are filled, regardless of
their relative order with the a1g orbital. Consequently, square planar complexes have
nondegenerate ground states and are immune to Jahn-Teller distortions.
-116-
7.9 (a) As shown for PF5 (cf. problem 4.13), for the ligand F-SALCs, ' = 2A1' + E' + A2". The
symmetries of the d orbitals, taken from the direct product transformations in D3h, are dz2 =
A1', (dx2-y2, dxy) = E', (dxz, dyz) = E". Bonding and antibonding combinations can be formed
with A1' and E' symmetries, but the E" symmetry (dxz, dyz) pair must be nonbonding in this
model. The following qualitative MO can be constructed.
(b) The upper three levels (boxed region, above) have identical splitting to that in the CFT
tbp model (cf. problem 7.2 c). Electron pairs contributed by the ligands are sufficient to fill
all lower levels. Therefore, any electrons contributed by the metal ion result in filling in the
upper three levels, in the same manner as predicted in the CFT model.
-117-
(c) The a1' combination with the dz2 orbital would use opposite signs for the axial and
equatorial ligand F AOs. The SALC would be M1 = 1//5{N1 + N2 + N3 – N4 – N5}. The
resulting LCAO would be:
The two e' SALCs, which combine with dx2-y2 and dxy, involve only equatorial functions.
They are M2 = 1//6{2N1 – N2 – N3} and M3 = 1//2{N2 – N3}, respectively. The resulting
LCAOs, projected in the xy plane, would be:
(d) The a1' and a2" SALCs, which form nonbonding MOs in the d-only scheme, could form
bonding and antibonding combinations with s and pz AOs on the metal. The px and py AOs
could form bonding and antibonding combinations with the e' SALCs, which in the d-only
scheme form bonding and antibonding combinations with dx2-y2 and dxy AOs. This
competition would result in six mixed MOs of varying bonding, nonbonding, and
antibonding character, depending upon the relative contributions of d and p AOs.
-118-
7.10 (a) The representation of F-SALCs is based on the following model:
C4v
E
2C4
C2
2Fv
2Fd
'F
5
1
1
3
1
'F = 2A1 + B1 + E
The AO symmetries of the central M atom are shown below.
(n – 1)d:
dz2 = A1, dx2-y2 = B1, dxy = B2, (dxz, dyz) = E
ns:
s = A1
np:
pz = A1, (px, py) = E
The dxy orbital (b2) has no symmetry match with a SALC and must remain nonbonding.
(b) For simplicity, the following sketches show 2 = 90o. The three kinds of potentially
bonding interactions between (n – 1)d orbitals and F-SALCs are the following:
The e (xz, yz) combinations are nonbonding at 2 = 90o and become only slightly bonding
with 2 > 90o. A better bonding combination with the E SALCs can be formed with px and
py AOs (see below).
-119-
The ns and np AOs can form the following bonding combinations:
The ns and np combinations for a1 are probably mixed. With a total of three a1 AOs (dz2, s,
pz) and two a1 SALCs, there can be only five MOs. We will assume bonding and
antibonding combinations for dz2 (a1) and s (a1), and treat pz (a1) as essentially nonbonding.
The two e combinations with px and py have better overlap with the E SALCs than dxz and
dyz. Therefore we will form bonding and antibonding combinations with these and assume
that the dxz and dyz orbitals are essentially nonbonding.
-120-
(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 below
those in the box. The MOs in the box have the same symmetries and are in the same
relative 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 is
equivalent to the presumed filling of electrons in d orbitals of the CFT model.
-121-
7.11 (a) The SALCs are identical to the oxygen F- and B-SALCs derived for CO2 in Chapter 4
(pp. 118-126). Thus,
' F = E g+ + E u+
'B = A u + A g
(b) Assuming p orbitals on the ligands, the SALCs would have the following forms:
Eg+
Eu +
Au
both xz and yz planes
Ag
both xz and yz planes
The normalized wave function expressions for these SALCS are as follows:
(c)
Eg+ = 1//2{Nz(1) + Nz(2)}
Eu+ = 1//2{Nz(1) – Nz(2)}
Au(x) = 1//2{Nx(1) + Nx(2)}
Au(y) = 1//2{Ny(1) + Ny(2)}
Ag(x) = 1//2{Nx(1) – Nx(2)}
Ag(y) = 1//2{Ny(1) – Ny(2)}
s = Eg+ Y bonding and antibonding with Eg+ SALC
pz = Eu+ Y bonding and antibonding with Eu+ SALC
(px, py) = Au Y bonding and antibonding with Au SALCs
dz2 = Eg+ Y bonding and antibonding with Eg+ SALC
(dxz, dyz) = Ag Y bonding and antibonding with Ag SALCs
(dx2-y2, dxy) = )g Y nonbonding.
-122-
(d) Both ns and (n - 1)dz2 AOs can form bonding and antibonding combinations with the
Eg+ SALC, so s-d mixing is possible. Being higher in energy, the ns AO probably
contributes more to antibonding, while the lower-energy (n - 1)dz2 AO contributes more to
bonding. For simplicity, the following MO scheme assumes the ns AO on the metal forms
an essentially nonbonding Fn level, leaving the (n - 1)dz2 AO to form bonding and
antibonding MOs without showing possible s-d mixing.
The MOs in the box have the same symmetry and splitting as the presumed splitting of d
orbitals in the CFT model (cf. problem 7.2 a). Electron pairs contributed by ligands are
sufficient to fill all lower levels. Therefore, any electrons contributed by the metal ion
result in filling in the upper three levels in the same manner as predicted in the CFT model.
-123-
7.12 (a) For F-SALCs, use the following four-vector basis set to obtain the reducible
representation.
D4h
E
2C4
C2
2C2'
2C2"
i
2S4
Fh
2Fv
2Fd
'F
4
0
0
2
0
0
0
4
2
0
'F = A1g + B1g + Eu
For B-SALCs, use the following eight-vector basis set to obtain the reducible
representation.
D4h
E
2C4
C2
2C2'
2C2"
i
2S4
Fh
2Fv
2Fd
'B
8
0
0
-4
0
0
0
0
0
0
'B = A2g + B2g + Eg + A2u + B2u + Eu
-124-
(b) The matches between SALCs and central-atom AOs is shown below. Among the BSALCs, A2g and B2u have no matching AOs and must be nonbonding (n.b.).
F-SALCs
A1g
B1g
Eu
AOs
s, dz2
dx2-y2
(px, py)
B-SALCs
A2g
B2g
Eg
A2u
B2u
Eu
AOs
n.b.
dxy
(dxz, dyz)
pz
n.b.
(px, py)
(c)
F-SALCs:
G(A1g) = ½(F1 + F2 + F3 + F4)
G(B1g) = ½(F1 – F2 + F3 – F4)
G"(Eu) = 1//2(F1 – F3)
G$(Eu) = 1//2(F2 – F4)
B-SALCs:
A(A2g) = ½(B5 + B6 – B7 – B8)
(nonbonding)
A(B2g) = ½(B5 – B6 + B7 + B8)
A"(Eg) = 1//2(B2 – B4)
A$(Eg) = 1//2(B1 – B3)
A(A2u) = ½(B1 + B2 + B3 + B4)
A(B2u) = ½(B1 – B2 + B3 – B4)
A"(Eu) = 1//2(B6 + B8)
A$(Eu) = 1//2(B5 + B7)
-125-
(nonbonding)
(d)
G(A1g) + s
G(A1g) + dz2
G(B1g) + dx2-y2
G"(Eu) + py; G $(Eu) + px
-126-
A(A2g)
A(B2g) + dxy
A"(Eg) + dxz; A$(Eg) + dyz
A(A2u) + pz
A(B2u)
-127-
A"(Eu) + px; A$(Eu) + py
7.13 We can save labor by doing the work in O and then correlating to the matching gerade
species of Oh, simply adding g subscripts to the Mulliken symbols in O.
For G, L = 4.
P(E) = 2l + 1 = 9
P(C3) = {sin (4.5)(120o)}/{sin 60o} = 0/0.866 = 0
P(C2) = {sin (4.5)(180o)}/{sin 90o} = 1/1 = 1
P(C4) = {sin (4.5)(90o)}/{sin 45o} = 0.707/0.707 = 1
O
E
8C3
3C2
6C4
6C2'
'G
9
0
1
1
1
'G = A1g + Eg + T1g + T2g
For H, L = 5.
P(E) = 2l + 1 = 11
P(C3) = {sin (5.5)(120o)}/{sin 60o} = -0.866/0.866 = -1
P(C2) = {sin (5.5)(180o)}/{sin 90o} = -1/1 = -1
P(C4) = {sin (5.5)(90o)}/{sin 45o} = 0.707/0.707 = 1
O
E
8C3
3C2
6C4
6C2'
'H
11
-1
-1
1
-1
'H = Eg + 2T1g + T2g
For I, L = 6.
P(E) = 2l + 1 = 13
P(C3) = {sin (6.5)(120o)}/{sin 60o} = 0.866/0.866 = 1
P(C2) = {sin (6.5)(180o)}/{sin 90o} = 1/1 = 1
P(C4) = {sin (6.5)(90o)}/{sin 45o} = -0.707/0.707 = -1
-128-
O
E
8C3
3C2
6C4
6C2'
'I
13
1
1
-1
1
'I = A1g + A2g + Eg + T1g + 2T2g
The splitting of g and i orbitals would be the same as the states, because both are gerade.
The h orbitals are ungerade, and contrary to the H state will split as eu + 2t1u + t2u.
7.14 (a) Use the Oh 6 D4h correlation table.
Free-Ion
Terms
Terms in Oh
Terms in D4h
S
A1g
A1g
P
T1g
A2g + Eg
D
Eg + T2g
A1g + B1g + B2g + Eg
F
A2g + T1g + T2g
A2g + B1g + B2g + 2Eg
G
A1g + Eg + T1g + T2g
2A1g + A2g + B1g + B2g + 2Eg
H
Eg + 2T1g + T2g
A1g + 2A2g + B1g + B2g + 3Eg
I
A1g + A1g + Eg + T1g + 2T2g
2A1g + A2g + 2B1g + 2B2g + 3Eg
(b) The most direct correlation for terms S through F is to use the R3 correlation table
showing R3 6 D3. For the I terms, use Oh 6 D3d and then D3d 6 D3.
Free-Ion
Terms
Terms in Oh
Terms in D3
S
A1g
A1
P
T1g
A2 + E
D
Eg + T2g
A1 + 2E
F
A2g + T1g + T2g
A1 + 2A2 + 2E
G
A1g + Eg + T1g + T2g
2A1 + A2 + 3E
H
Eg + 2T1g + T2g
A1 + 2A2 + 4E
I
A1g + A1g + Eg + T1g + 2T2g
3A1 + 2A2 + 4E
-129-
(c)
There is no correlation table for Oh 6 D2d. Use results in part (a) with the table for D4h
6 D2d (C2' 6 C2'). Note that for D2d, the ligand-field terms are the same as for D4h
without the subscript g. Alternately, one could use the correlation Td 6 D2d.
Free-Ion
Terms
Terms in D4h
Terms in D2d
S
A1g
A1
P
A2g + Eg
A2 + E
D
A1g + B1g + B2g + Eg
A1 + B1 + B2 + E
F
A2g + B1g + B2g + 2Eg
A2 + B1 + B2 + 2E
G
2A1g + A2g + B1g + B2g + 2Eg
2A1 + A2 + B1 + B2 + 2E
H
A1g + 2A2g + B1g + B2g + 3Eg
A1 + 2A2 + B1 + B2 + 3E
I
2A1g + A2g + 2B1g + 2B2g + 3Eg
2A1 + A2 + 2B1 + 2B2 + 3E
(d) One the basis of vector transformations, determine the correlation between D4h and D4h,
as follows:
D4h
D4h
Basis
A1g
Eg+
z2
A2g
Eg –
Rz
B1g + B2g
)g
(x2 – y2, xy)
Eg
Ag
(Rx, Ry) (xz, yz)
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Free-Ion
Terms
Terms in D4h
Terms in D4h
S
A1g
E g+
P
A2g + Eg
E g– + A g
D
A1g + B1g + B2g + Eg
E g+ + A g + ) g
F
A2g + B1g + B2g + 2Eg
Eg– + 2 A g + )g
G
2A1g + A2g + B1g + B2g + 2Eg
2Eg+ + Eg– + 2Ag + )g
H
A1g + 2A2g + B1g + B2g + 3Eg
Eg+ + 2Eg– + 3Ag + )g
I
2A1g + A2g + 2B1g + 2B2g + 3Eg
2Eg+ + Eg– + 3Ag + 2)g
7.15 (a) There are four microstates:
eg
t2g
¼
¼¿ 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
from 2G are 2A1g, 2Eg, 2T1g, 2T2g.
(c)
For 2G, Dt = (2)(9) = 18.
For 2A1g + 2Eg + 2T1g + 2T2g, Dt = (2)(1) + (2)(2) + (2)(3) + (2)(3) = 18.
7.16 The free-ion terms of d n and d 10-n are the same because they arise from the same number of
microstates, as shown by Eq. (7.14):
The free-ion terms are unique to a given number of microstates. Moreover, each free-ion
term splits into a characteristic set of ligand-field terms in any point group. Thus, the
ligand-field terms for d n and d 10-n must be the same.
-131-
7.17 The strong-field terms emerging from the d 2 Td configurations in the hypothetical extreme
field are the same as for the comparable d 2 Oh configurations, except that the subscript g is
omitted for each. The extreme-field configurations, however, will be in the reverse energy
order of the d 2 Oh configurations; i.e., e2 < e1t21 < t22. Identifying the triplet terms versus the
singlet terms follows the same logic as shown in the d 2 Oh case; i.e., each configuration
gives rise to the same multiplicities for comparable Td orbital terms. Making the
correlations in the same way gives a diagram identical to that shown for d 8 Oh (Fig. 7.26),
except for the gerade designations on the states.
7.18 Weak-Field Terms. In Oh, the free-ion 2D term of a d 9 configuration becomes a 2Eg ground
state term and a 2T2g excited state terms. From the correlation tables, it can be seen that
these terms split on descent to D4h square planar geometry as follows:
2
2
Eg 6 2A1g + 2B1g
T2g 6 2B2g + Eg
Extreme Field Configurations and Strong-Field Terms. The configurations for the d 9 D4h
square planar geometry in an extremely strong field are shown below in order of increasing
energy. The terms in a strong field are obtained by taking the orbital assignment of the
unpaired electron as a basis for a representation in D4h. All paired electrons can be ignored,
because the representation for any pair must be the totally symmetric representation; e.g.,
B1g × B1g = A1g. With only one unpaired electron in all configurations, all terms must be
doublets.
b1g
¼
¼¿
¼¿
¼¿
b2g
¼¿
¼
¼¿
¼¿
a1g
¼¿
¼¿
¼
¼¿
eg
¼¿ ¼¿
¼¿ ¼¿
¼¿ ¼¿
¼¿ ¼
2
2
2
2
B1g
B2g
A1g
Eg
The resulting qualitative correlation diagram is shown below.
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7.19 (a) FeF63– is a d 5 high-spin complex, because F– is a weak-field ligand. Therefore, there are
no spin-allowed transitions from the 6A1g ground state. Fe(CN)63– is a d 5 low-spin complex,
because CN– is a strong-field ligand. The ground state is 2T2g term. From the TanabeSugano diagram for d 5 low-spin cases (right side), it is apparent that there are at least three
possible spin allowed transitions: 2T2g 6 [2A2g, 2T1g], 2T2g 6 2Eg, 2T2g 6 2A2g. As the TanabeSugano diagram indicates, the 2A2g and 2T1g states for the first transition are virtually
degenerate and would give rise to a single broad band. These transitions account for the
color of Fe(CN)63–.
(b) Tetrahedral complexes are not centrosymmetric, so the LaPorte Rule does not apply.
Nonetheless, d orbitals have an inherent gerade character. Therefore, although molar
absorptivities are slightly higher for tetrahedral complexes compared to octahedral
complexes, 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.}
[Cr(H2O)3]3+ is an octahedral d 3 case. From either the Tanabe-Sugano diagram or Orgel
diagram, three bands are expected from the transitions 4A2g 6 4T2g, 4A2g 6 4T1g(F), 4A2g 6
4
T1g(P). The last of these involves a change in configuration from t2g3 to t2g1eg2; i.e., a twoelectron transition. This is a less probable event and therefore would have a lower
absorptivity. Moreover, it is the highest energy transition and may be obscured by the tail
of a stronger charge-transfer band in the ultraviolet.
(d) A d 6 high-spin complex such as [Fe(H2O)6]2+ is expected to show a single band from the
transition 5T2g 6 5Eg. However, both the ground state and excited state are degenerate and
subject to Jahn-Teller distortions. The distortion is greater for the doubly degenerate
excited state. Therefore, the observed band splitting is primarily due to lifting of the
degeneracy of the upper 5Eg state.
(e) With d 0 configurations, both complex ions have no possible d-d transitions. The
observed intense colors arise from L6M charge transfer transitions, which are LaPorte
allowed. 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
ground state eg4a1g2b2g2 consists of all paired electrons, so the term is 1A1g. For the first
excited singlet state eg4a1g2b2g1b1g1, take the direct product B2g × B1g:
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D4h
E
2C4
C2
2C2'
2C2"
i
2S4
Fh
2Fv
2Fv
B2g
1
-1
1
-1
1
1
-1
1
-1
1
B1g
1
-1
1
1
-1
1
-1
1
1
-1
' = A2g
1
1
1
-1
-1
1
1
1
-1
-1
Therefore, the first singlet excited state is 1A2g. For the second excited singlet state
eg4a1g1b2g2b1g1 the direct product is A1g × B1g = B1g , and the resulting term is 1B1g. For the
third excited singlet state eg3a1g2b2g2b1g1, the direct product is Eg × B1g:
D4h
E
2C4
C2
2C2'
2C2"
i
2S4
Fh
2Fv
2Fv
Eg
2
0
-2
0
0
2
0
-2
0
0
B1g
1
-1
1
1
-1
1
-1
1
1
-1
' = Eg
2
0
-2
0
0
2
0
-2
0
0
Therefore the term is 1Eg.
(b) The three electronic transitions are
1
A1g 6 1A2g
1
A1g 6 1B1g
1
A1g 6 1Eg
(c) For an electronic transition to be vibronically allowed, the transition moment
IReRv:Re' Rv' dJ must be totally symmetric to be non-zero. In this specific case Re is also
totally symmetric, and the ground vibrational state, Rv is totally symmetric for all molecules
except free radicals. Therefore the symmetry of the integral depends upon the symmetries
of the direct products : × Re' × Rv'. The symmetries of the dipole moment components,
obtained from the unit vector transformation properties listed in the D4h character table are
:z = A2u
(:x, :y) = Eu
The symmetries of the excited electronic states, as shown in part (a), are A2g, B1g, and Eg.
The symmetries of the components of : are ungerade, and the symmetries of the electronic
excited states are gerade, so the products : × Re' must all be ungerade. In order for the
integral to be totally symmetric and nonvanishing, the Rv' must belong to an identical
ungerade species as : × Re' . The ungerade normal modes of vibration, which might
couple with the electronic transitions, are
<3 (A2u)
<5 (B2u)
<6 (Eu)
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<7(Eu)
For the purpose of vibronic coupling, the gerade normal modes can be ignored. By
matching symmetries of : × Re' with one or more of these Rv' we obtain the following
results, which show the normal modes that vibronically allow each electronic transition:
Transition
: × Re'
Matching normal modes
1
A2u × A2g = A1u
none
Eu × A2g = Eu
<6, <7 (Eu)
A2u × B1g = B2u
<5 (B2u)
Eu × B1g = Eu
<6, <7 (Eu)
A2u × Eg = Eu
<6, <7 (Eu)
Eu × Eg = A1u + A2u + B1u + B2u
<3 (A2u), <5 (B2u)
1
1
A1g 6 1A2g
A1g 6 1B1g
A1g 6 1Eg
In the first case for 1A1g 6 1A2g, the integral with the dipole moment component :z (A2u)
vanishes, because there is no matching normal mode with the resultant A1u symmetry.
Nonetheless, the transition 1A1g 6 1A2g is vibronically allowed by coupling with the normal
modes <6 and <7 (Eu), because the direct product : × Re' with the degenerate pair of dipole
moment components (:x, :y) is Eu, making the full integral totally symmetric. In the last
case for 1A1g 6 1Eg, the direct product Eu × Eg is a reducible representation, only two whose
component irreducible representations (A2u and B2u) match symmetries with normal modes.
Those matches are sufficient to vibronically allow the 1A1g 6 1Eg electronic transition.
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