Assignment 8 Solutions

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Chem 370 - Spring, 2015
Assignment 8 - Solutions
10.1
a. All tetrahedral complexes are high spin. For Td d 6 the configuration is e3t23:
t2
Y 4 unpaired electrons
e
2+
7
b. Co(H2O)6 is d high-spin Oh because H2O is a weak-field ligand. (Do not confuse Co2+ with Co3+,
which tends to be low-spin, even with H2O.)
eg
Y 3 unpaired electrons
t2g
c. Cr(H2O)63+ is d3 Oh.
eg
Y 3 unpaired electrons
t2g
d. Square planar complexes have the following splitting scheme, filled in a low-spin manner. (The
middle two levels may be reversed in some cases, but that does not generally affect the overall
configuration or number of unpaired electrons.)
b1g (x2 - y2)
b2g (xy)
a1g (z2)
Y 1 unpaired electron
eg (xz, yz)
e.
10.2
Y n = 4.2 . 4
a. To have M3+ with 3d1, the neutral atom would need to be 3d24s2, which is Ti.
b. The most unpaired electrons is d5. [MBr4]– has a M3+ ion. These two facts make it Fe.
c. Diamagnetic [M(CN)6]3– would be a low-spin d6, because CN– is a high-field ligand. The M3+ ion
with a 3d6 configuration is Co.
d. If CFSE = –0.6Δo with no pairing terms, there are no paired electrons. The high-spin configuration
t2g3eg1 would have a CFSE = (–0.4)(3)Δo + (0.6)(1)Δo = –0.6Δo. This would be the configuration for
Cr2+. If pairing energy is ignored, the configuration d9 = t2g6eg3 would also fit this description, but
with pairing energy CFSE = (–0.4)(6)Δo + (0.6)(3)Δo + 4P = –0.6 + 4P. This would be the
configuration for Cu2+.
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10.7
This is {[Fe(H2O)6]2+}2{[Fe(CN)6]4–}. The cation is high-spin d 6 with μ = 4.89 B.M., and the anion is
low-spin d 6 with μ = 0. Thus, μ = (2/3)(4.89 B.M.) = 3.27. For 2.67 electrons,
, which is close to the expected average of two paramagnetic
cations and one diamagnetic anion.
10.8
Co2+ is d 7. Tetrahedral would be 3 unpaired electrons with μ = 3.87 B.M. Octahedral high-spin would
also be 3 unpaired electrons with μ = 3.87 B.M. Square planar would be 1 unpaired electron with μ =
1.73 B.M. A d 7 low-spin octahedral complex would also be 1 unpaired electron with μ = 1.73 B.M.
10.21 a. When compression occurs on the z axis, those orbitals with a z component (dxz, dyz, dz2) are
destabilized, and those with an xy component (dxy, dx2–y2) are stabilized. The degeneracies of Oh are
split on descent to D4h. The effect is the opposite of stretching along z (see below).
b. Stretching along z destabilizes those orbitals with an xy component (dxy, dx2–y2) and stabilizes those
orbitals with a z component (dxz, dyz, dz2). The splitting of the octahedral t2g and eg levels is shown
below.
Oh
D4h
b1g (dx2-y2)
+*1/2
eg
-*1/2
a1g (dz2)
)o
)o
b2g (dxy)
+2*2/3
t2g
-*2/3
increasing stretch along z
eg (dxz, dyz)
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10.23 a. The answers below also include the pairing energy (P) contributions, which the authors ignore in
their answers (as do many other sources). Each electron in a t2g orbital contributes –0.4Δo to CFSE,
and each electron in an eg orbital contributes +0.6 to CFSE.
*
Complex
Config
.
n
μ (B.M.)
CFSE
[Co(CO)4]–
d 10
0
0
0
[Cr(CN)6]4–
t2g4
2
2.83
–1.6Δo + P
[Fe(H2O)6]3+
t2g3eg2
5
5.92
0
[Co(NO2)6]4– *
t2g6eg1
1
1.73
–1.8Δo + 3P
[Co(NH3)6]3+
t2g6
0
0
–2.4Δo + 3P
MnO4–
d0
0
0
0
[Cu(H2O)6]2+
t2g6eg3
1
1.73
–0.6Δo + 4P
The authors’ answer book has [Co(NO2)6]4– as a high-spin d 7 case, but from the position of NO2– in
the spectrochemical series you would predict this to be low-spin with 1 unpaired electron, as shown
here.
b. Both [Co(CO)4]– and MnO4– have no CFSE. The octahedral ions (except [Fe(H2O)6]3+) have CFSE
values that favor octahedral coordination. The tetrahedral ions are stabilized by π bonding, with CO
as an acceptor and O2– as a donor. Fe3+ would have no CFSE, either as an octahedral or tetrahedral
complex.
c. Compare the CFSE values (ignoring pairing). Recall that Δt = (4/9)Δo.
Ion
Spin State
Oh CFSE
Td CFSE
Co(II)
d 7 high spin
–0.8Δo
–1.2Δt = (4/9)(–1.2)Δo = –0.53Δo
Ni(II)
d8
–1.2Δo
–0.8Δt = (4/9)(–0.8)Δo = –0.36Δo
As these results show, for Co(II) Oh CFSE is better than Td CFSE by only 0.27Δo. For Ni(II), Oh
CFSE is better by 0.84Δo. The smaller octahedral advantage with Co(II) makes it more likely that
tetrahedral complexes may be stable, whereas with Ni(II) the octahedral advantage is so great that
tetrahedral complexes are less likely.
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10.26 a. A square pyramid is C4v.
L
L
L
L
L
M
From the direct product transformation properties listed in the C4v character table, we can determine
that the five-fold degeneracy among d orbitals in the free ion is lifted as follows:
dz2 = a1
dx2–y2 = b1
dxy = b2
(dxz, dyz) = e
The dxy orbital is lowest, because its lobes lie between ligands, which are actually in a plane below.
The degenerate (dxz, dyz) orbitals with their lobes oriented between all ligands are next in order. The
dz2 orbital is next higher in energy, because it points directly at the axial ligand and its annular part is
in the xy plane. The dx2–y2 orbital is the highest in energy, because its lobes point directly at the four
basal ligands. The following CFT splitting scheme results:
_____ dx2–y2 (b1)
_____ dz2 (a1)
_____ _____ dxz, dyz (e)
_____ dxy (b2)
With the preceding CFT results, we can construct an MO scheme, using five σ-SALCs from the
ligands to interact with the d orbitals, whose symmetries we have already determined.
C4v
E
2C4
C2
2σv
2σd
Γσ
5
1
1
3
1
G
G/8
A1
5
2
1
6
2
16
2
A2
5
2
1
–6
–2
0
0
B1
5
–2
1
6
–2
8
1
B2
5
–2
1
–6
2
0
0
E
10
0
–2
0
0
8
1
Γσ = 2 A1 + B1 + E
All SALCs have same-symmetry matches with d orbital symmetries and will form bonding and
antibonding MOs. There are two A1 SALCs and only one A1 AO (dz2), so we can form only three
MOs of this symmetry. For simplicity, the scheme below assumes two bonding and one mixed
antibonding MO. We do not have a B2 symmetry SALC, so the dxy AO on the metal will be
nonbonding. Electron pairs from the five ligands will fill the bonding MOs. Electrons from the
metal will result in partial filling of the nonbonding b2 MO, then the antibonding e MO, the
antibonding a1 MO, and finally the antibonding b1 MO, depending on the metal ion’s d n
configuration. There is no simple way of predicting the order of the bonding MOs, so the order
shown below is just a reasonable guess. However, as seen with octahedral and tetrahedral cases, the
ordering of MOs filled from metal electrons is the same as predicted by CFT considerations.
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M
5L
ML 5
b1*
a1*
M d n filling starts here
e*
b2n
a1 + b1 + b2 + e
e
a1
2a 1 + b 1 + e
b1
a1
11.11 a. d 8 (Oh) is t2g6eg2. There is only one configuration for this, so it is 3A1g.
b. d 5 high spin is t2g3eg2, for which there is only one configuration, so it is 6A1g.
d 5 low spin is t2g5eg0. The unpaired electron can be in any of the three t2g orbitals, making this a
triply degenerate T terms. The Tanabe-Sugano diagram shows this as 2T2g.
c. d 4 (Td) is e2t22. Recall that all Td ML4 complexes are high spin. Here, the hole can be in any one of
the three t2 orbitals, so this is a triply degenerate state. The d 4 Td Orgel diagram is the same as the
d 1, d 6 Oh diagram, so this is a 5T2 state (no subscript g).
d. d 9 square planar (D4h). The highest orbital, b1g (dx2–y2), has one electron in it, so this is a nondegenerate state. In D4h this would be 2B1g.
11.13 [Ni(H2O)6]2+ is d 8 Oh. This case is covered by the left side of the second Orgel diagram, the diagram for
d 2, d 7 Oh and related cases. For d 8 Oh, the lowest energy transition, 3A2g ÷ 3T2g, is equivalent to Δo, so
from Figure 11-8 for [Ni(H2O)6]2+ Δo is ~8,700 cm–1. The other two bands arise from the transitions 3A2g
÷ 3T1g(F) and 3A2g ÷ 3T1g(P). The splitting of the bands is a result of Jahn-Teller distortion of the triply
degenerate excited states in each case. Note that the ground state (3A1g) is immune from Jahn-Teller
distortion.
11.14 a. [Cr(C2O4)3]3– has Cr3+ d3. Either from the Orgel diagram or the Tanabe-Sugano diagram, we expect
the following transitions: ν1 [4A2g ÷ 4T2g], ν2 [4A2g ÷ 4T1g(F)], ν3 [4A2g ÷ 4T1g(P)]. In d 3 Oh cases, Δo
corresponds to the energy of the first transition, ν1. For this complex, ν1 = 17,400 cm-1 = Δo.
b. [Ti(NCS)6]3– has Ti3+ d 1. The single band arises from the transition 2T2g ÷ 2Eg, as predicted from the
right side of the first Orgel diagram. Thus, Δo = 18,400 cm–1. Both the ground state and the excited
state are subject to Jahn-Teller distortions, but only the 2Eg state will have appreciable splitting,
because it arises from an imbalance in eg orbitals directed at ligands. This is the probable cause of
the band splitting.
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c. [Ni(en)3]2+ has Ni2+ d 8. The Orgel diagram for d 8 is the same as d 3 (see above), so we expect three
bands: ν1 [3A2g ÷ 3T2g], ν2 [3A2g ÷ 3T1g(F)], ν3 [3A2g ÷ 3T1g(P)]. As with the d 3 Oh case, for d 8 Oh Δo is
the energy of the single band, here ν = 11,200 cm–1 = Δo.
11.16 a.
b.
c.
d.
e.
t2g4eg2 – T
t2g6 – A
t2g3eg3 – E
t2g5 – T
eg1 – E
The pair in t2g orbitals can be in any of the three.
There is only one way of assigning the electrons.
An excited state; the pair in eg (or the unpaired electron) can be in either orbital.
The unpaired electron can be in any of the three t2g orbitals.
An excited state; the electron can be in either orbital.
11.17 Configurations corresponding to nondegenerate states will not have distortions. Those that are eg1 and
eg3, which lead to E states, will have large distortions, because eg orbitals are directed at ligands.
Configurations that correspond to triply degenerate states arise from an imbalance among t2g orbitals,
which lie between ligands. These have only small distortions.
11.23 a. At 80 K,
At 300 K,
State
Distortion
Distortion
t2gxegy
1
t2g 1
2
T2g
small
2
t2g 2
3
T2g
small
3
t2g 3
4
A2g
none
4
t2g 3eg1
5
Eg
large
t2g 4
3
small
5
t2g 3eg2
6
A1g
none
t2g 5
2
T2g
small
6
t2g 4eg2
5
T2g
small
t2g 6
1
A1g
none
7
t2g 5eg2
4
T1g
small
t2g 6eg1
2
large
8
t2g 6eg2
3
A2g
none
9
t2g 6eg3
2
large
Eg
t2gxegy
State
dn
T1g
Eg
, from which n = 0.19 . 0.
, from which n = 4.3 . 4.
b. The complex appears to be d 6 low-spin at 80K and d 6 high spin at 300 K. Looking at the TanabeSugano diagram, we would expect as many as five bands arising from the 1A1g ground state of the
low-spin complex at 80 K. More realistically, there are probably only two or three bands, because
the upper 1Eg and 1T2g states are nearly degenerate, and the transition to the highest state (1A2g)
probably falls in the uv. The transition to the accidentally degenerate 1Eg and 1T2g states may also
fall in the uv. By contrast, the d 6 high-spin complex should have only one band, 5T2g ÷ 5Eg. This
band may show some spitting, owing to strong Jahn-Teller distortion in the excited state.
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11.27 Adding NH3 causes replacement of the H2O ligands, and adding ethylendiamine (en) causes replacement
of NH3 ligands with en.
[Ni(H2O)6]2+
Color:
green
Complement:
red
NH3
[Ni(NH3)6]2+
blue
orange
en
[Ni(en)3]2+
violet
yellow
In the spectrochemical series, H2O is weaker than NH3, which is weaker than en. Thus, through the
series of substitutions, the value of Δo increases. Treating all of these as octahedral (the en complex is
actually D3), from the d8 Orgel diagram we predict three absorption bands, arising from the transitions
3
A2g÷3T2g, 3A2g÷3T1g(F), 3A2g÷3T1g(P). As Δo increases, the frequencies of all of these transitions
increase, which means the absorbed region of the visible spectrum shifts toward the blue end of the
visible spectrum. The absorbed regions are the complements of the colors we perceive. These
complements change from red to orange to yellow, a shift toward higher frequencies, consistent with
increasing Δo.