Chem 213: Advanced Ligand Field Theory

Chem 213: Advanced Ligand Field Theory

Todd Gingrich
May 6, 2008

D. Introduction to Electronic Spectra

1. (a) The following transitions could occur in an octahedral d3 complex. Which of these are formally
forbidden and why? Make an order of magnitude estimate of the extinction coefficient and oscilla-
tor strength of each transition. Propose mechanisms by which each transition could gain intensity.

i. 4A2g → 4T2u is formally allowed. It is clearly spin allowed, and it is also orbitally allowed
because the triple product A2g ⊗ T1u ⊗ T2u = A1g ⊕ Eg ⊕ T1g ⊕ T2g contains the totally
symmetric representation. Thus we expect ≈ 10, 000M −1cm−1 with an oscillator strength
of f ≈ 1.

ii. 4A2g → 4T2g is formally orbitally-forbidden because A2g ⊗T1u⊗T2g has u parity and therefore
does not contain a totally symmetric component. Thus we expect ≈ 100 with f ≈ 0.01.
This transistion could gain intensity by coupling to vibration with u parity.

iii. 4A2g → 4T1u is formally orbitally-forbidden since A2g ⊗ T1u ⊗ T1u = A2g ⊕ Eg ⊕ T1g ⊕ T2g,
which does not contain the A1g representation. Thus we expect ≈ 100 with f ≈ 0.01. This
transition could gain intensity by coupling to a A2g, Eg, T1g, or T2g vibration.

iv. 4A2g → 2Eg is spin-forbidden and orbitally-forbidden since A2g ⊗ T1u ⊗ Eg = T1u ⊕ T2u.
Therefore we expect ≈ 1 and f ≈ 10−4. This transition could gain intensity if spin-
orbit coupling were increased, which would occur in a heavier metal. The intensity would
also increase if the transition were to become orbitally-allowed by coupling to a T1g or T2g
vibration.

v. 4A2g → 2T2u is spin forbidden (but orbitally-allowed like (i)). We expect ≈ 1 and f ≈ 10−4,
but if comparing the intensities of (iv) and (v) we note that (iv) should be weaker as it is
doubly disallowed. This transition could be more intense if spin-orbit coupling is strengthened.

(b) Determine the symmetries of the normal modes of vibration of an octahedral complex. Use these to
justify intensity gains in the transitions of part (a) via vibronic interactions based on these modes.

First we seek the symmetries of the normal modes of vibration. We arbitrarily fix the loca-

tion of the metal atom (hereby removing the three degrees of translational freedom). Now we

consider the action of each of our symmetry elements on the basis set formed by local (x, y, z)

axis centered on each ligand atom. We record the traces for this representation of the group that

we call Γtot where by total we mean rotational plus vibrational. The rotations (Γrot) transform
as T1g based on a character table. Subtracting these traces out we can obtain the traces of the
vibrational representation, Γvib.

Oh E 8C3 6C2 6C4 3C2(=C42) i 6S4 8S6 3σh 6σd

Γtot 18 0 0 2 -2 0 0 0 4 2

Γrot 3 0 -1 1 -1 3 1 0 -1 -1

Γvib 15 0 1 1 -1 -3 -1 0 5 3

1

After applying the Great Orthogonality Theorem, we see that Γvib = A1g ⊕ Eg ⊕ T2g ⊕ 2T1u ⊕ T2u.
The T1u and T2u vibrations can couple with the 4A2g → 4T2g electronic transition to increase the
intensity of the band. The Eg and T2g vibrations could couple with the 4A2g → 4T2u electronic
transition to increase the intensity.

2

2. Absorption spectra of Cr(III) complexes are readily analyzed using the ligand field theory derived in
parts A-C.

(a) Summarize the atomic selection rules. The absorption of a free Cr(III) ion shows a weak band at
about 14,000 cm−1. Explain.

Atomic transitions are formally allowed iff ΔS = 0 and ΔL = 0, ±1 (excluding L = 0 → 0).
For a first row transition metal such as Chromium, the spin-orbit coupling is weak so the spin
orthogonality is close to rigorously true. Thus the weak band at 14,000 cm−1 must come from
an exitation from the ground 4F ground term to an exited quartet term. There is only one such
transition in a d3 free ion configuration, so we attribute the 14,000 cm−1 band to the 4F →4 P
transition. This transition is formally disallowed as ΔL = −2, but the dipole approximation of
light breaks down leading to the weak band.

(b) The electronic spectra of four Cr(III) complexes, CrLn6−, are summarized below. Assign the bands
(assuming they are all allowed transitions). Estimate Dq and B values from the Tanabe-Sugano

diagram. Use these values to calculate the positions of each band for each of the complexes. Draw

an energy level diagram showing the levels for all four compounds. On the basis of these results,

order the ligands in the table in the spectroschemical series and the nephelauxetic series.

L 17,400 ν(cm−1) 38,000
14,900 24,500 34,400
H2O 16,150 22,700 34,400
F− 26,600 22,600
Urea 32,500 ?
CN−

From low to high energy we assign the bands as: 4T2g ←4 A2g, 4T1g ←4 A2g, and 4T1g ←4 A2g.
Equations 30 and 31 in Problem Set C give analytic expressions for the transition energies with
respect to Dq and B. There is some ambiguity in our assignment of Dq and B since there are
two parameters and three data points. We use the low energy band to solve for Dq. We then use
that value to solve for B based on the high and low energy T1g bands. The two different values
of B are averaged to give:

L Dq(cm−1) B(cm−1)
1,740 686
H2O 1,490 843
F− 1,615 562
Urea 2,660 529
CN−

These parameters lead to bands which are consistent with the reported values:

L 4A2g → 4T2g 4A2g → 4T1g 4A2g → 4T1g
17,400 24,300 38,200
H2O 14,900 22,400 34,900
F− 16,150 21,950 34,950
Urea 26,600 32,500 55,250
CN−

In Figure 1 we can also plot (to scale) the energies of each state (neglecting the 3A term that all
states share).

From the calculated values of B and Dq, we can order the ligands with respect to the spectro-
chemical series (Dq) and the nephelauxetic series (B).

Spectrochemical Series: F− < Urea < H2O < CN−
Nephelauxetic Series: CN− < Urea < H2O < F−

3

20000 4T1g
10000

Energy (cm−1) 0 4T1g
−10000 4T2g
−20000

−30000

−40000 4A2g

F− Urea H2O CN−

Figure 1: Cr(III) Ligand Field Energies for Various Ligands.

(c) Read the paper by Krause, Trabjerg, and Ballhausen (Chem. Phys. Lett. 1969, 3, 297-299) and
comment critically on the significance of this two photon experiment.

The two-photon experiment of Krause et. al. provides insight into the structure of the doublet
manifold of a Cr(III) complex. A flash lamp is used to excite the complex into the 2Eg state.
Immediately after this flash, a spectrum is taken (the second photon) to record transitions from
the 2Eg state into higher energy doublet states. A spectrum taken without the initial flash is
used for background subtraction to eliminate the quartet excitations. There are two aspects of
the experiment that seem a little concerning. First, there is a shoulder at 25,000 cm−1, which
could not be explained by a doublet-doublet transition, but matches the 4A2g → 4T1g transition
exactly. This suggests that the background subtraction did not entirely eliminate the quartet-
quartet transitions. It is also unclear to me how the energy of initial flash excitation was controlled.
The authors assume that the flash excites into the 2Eg state and not any of the higher energy
doublet states. This assumption demands justification.

(d) Read the paper by Witzke (Theoret. Chim. Acta. 1971, 20, 171-184). This is typical of efforts
aimed at parameterizing d-d spectra. Summarize the conclusions and general methodology.

For practical use the most important conclusion is that the ratio C/B = 4 is a good approximation
for the free ion, but for an ion in the presence of a ligand field the ratio of C/B is not constant
and not necessarily ≈ 4. The constant ratio would only hold if the radial functions were not
complex-dependent. Full molecular orbital theory is really necessary to fully resolve this issue.
The paper also suggests that in order to better parameterize the d-d it requires a different B value
for the spin-allowed transitions compared to the spin-forbidden transitions. The paper starts by
discussing simple methodology (ie crude approximations). The method we used in part (b) of
this problem is employed as the simplest method for predicting spin-allowed transitions. For
spin-forbidden transitions the Tanabe-Sugano matrices (electrostatic matrices in the strong-field
basis set) were diagonalized numerically and the values of B and C which matched the observed
absorptions most closely were used. If B and C held rigorous in these situations then there should
be a value of B and C that will predict each possible transition, but in general the curves for C

4

as a funciton of B do not converge on a point. This is a consequence of the fact that B and C
can only make sense as a linear combination of F2 and F4 which are integrals of d orbitals. In the
bonded compound, those orbitals change shape, demanding that a full molecular orbital approach
be employed to actually predict the energy levels. Nevertheless, Witzke’s paper demonstrates
that the rough 3 parameter (Δ, B, C) approach works fairly well.
(e) The absorption spectra of Cr(en)33+ and of the cis and trans isomers of Cr(en)2(H2O)23+ are
shown below. Assign the observed bands and identify which spectra correspond to the cis and
trans isomers. Explain.

Figure 2: Absorption Spectra of Cr(en)33+ (—–), cis-Cr(en)2(H2O)32+ (- - - -), and trans-Cr(en)2(H2O)23+
(— —).

The assignment of the Cr(en)33+ spectra is straightforward. The low energy peak at ≈ 22, 000
cm−1 is the 4A2g → 4T2g transition while the higher energy peak at ≈ 27, 500 cm−1 is the
4A2g → 4T1g transition.
The Cr(en)2(H2O)23+ are more complicated. Section 5-c of Ballhausen treats the cis and trans
problem. The fields are expanded into the sum of cubic, tetragonal, and rhombic field compo-
nents. The tetragonal component causes the degeneracy of the cubic states to split most strongly,
so Ballhausen focuses on the effect of the cis/trans configuration on the strength of the tetragonal
component. It is shown that a trans complex will exhibit twice the tetragonal splitting as a cis
compound, so the broadened peak (— —) is taken to be the trans complex. In D4h symmetry we
must reassign these peaks with respect to the new point group. The Oh 4A2g → 4T1g transition
splits into: 4B1g → 4Eg (19, 500 cm−1), 4B1g → 4A2g (22, 500 cm−1). Splitting of the Oh
4A2g → 4T2g transition is not as pronounced in the spectrum, but some broadening is observed
as the transition splits into the two nondegenerate transitions: 4B1g → 4Eg and 4B1g → 4B2g.
Because the tetragonal splitting is so much weaker in the cis complex, the peaks do not re-
ally split, but some broadening is observed. As pointed out in Ballhausen, the ordering of the
excited states will switch (ie 4Eg will lie above 4A2g).

5

3. (a) Briefly discuss the stability of octahedral low- and high-spin d7 complexes.

d7 complexes with large ΔO will be most stable sacrificing the spin-pairing energy in order to
populate the (eg) set with as few electrons as possible. Hence low-spin d7 complexes are most
stable for large splitting. If the splitting is small, then the cost of spin-pairing may exceed the cost
of moving another electron from the (t2g) set into the (eg) set. In these situations the high-spin
(t2g)5(eg)2 configuration is favored.
(b) In both cases in (a), a distortion from Oh symmetry is predicted. Explain. What geometries
might you expect and how will the ground states be split? Comment on the relative magnitudes
of the splittings.

In the low spin case there is a single electron occupying the degenerate eg set so Jahn-Teller
splitting is expected. The splitting will be stong since the eg set points directly at the octahedral
ligands and therefore distortions can significantly affect the overlap between eg orbitals and the
molecular orbitals. In the high spin case we expect Jahn-Teller distortion because there is a single
vacancy in the triply degenerate t2g set. The splitting of this t2g set will be significantly smaller
than the low spin J-T splitting since distortions in geometry cannot strongly affect ligand-metal
orbital overlap integrals because the t2g orbitals do not point directly at the octahedral ligands.

The low spin distortion would distort into D4h geometry such that the dz2 level drops in en-
ergy and the dx2−y2 level rises. The d7 low spin ground state is 2Eg, which in D4h symmetry
splits into 2A1g ⊕2 B1g. Because dz2 is lower in energy than dx2−y2 it is clear that the 2A1g state
is lower in energy that the 2B1g state. An axial distortion still maintains zero overlap of the t2g
orbitals with the ligand orbitals, so a J-T distortion in the high spin case will not be along the
C4 axis. Rather, the elongation would be along the C3 axis to give D3d symmetry. In that case
the 4T1g ground state would be reduced to 4A2 ⊕4 E with 4A2 lying lower in energy.
(c) Shown below are absorption spectra for Co(H2O)62+ and CoCl42− in aqueous solution. Identify
which spectrum belongs to which compound. Explain.

Figure 3: Co(H2O)26+ and CoCl42− in aqueous solution

Co(H2O)62+ is the unstructured peak on the left because the peak is at higher energy. ΔO = 9 ΔT ,
4
so the lower energy peak is the tetrahedral complex: CoCl42−.

6

(d) The absorption spectra for two crystals of KCoF3 of different thickness at approximately 150 K
are shown below. What are the advantages of working at low-temperature and using crystalline

samples to study the spectroscopy of metal complexes?

Figure 4: Absorption spectra of two crystals of KCoF3 of different thicknesses at approximately 150 K

Assuming the two most intense features in this spectrum correspond to spin-allowed transitions,

assign these bands. Identify any other spin-allowed feature(s) and comment on their intensity.
Assuming the weak feature at ≈ 10, 000 cm−1 is the lowest energy Eg term, determine values for
Dq, B, and C. Comment on these results. Comment on any remaining features in the spectrum.
Why does the broad spin-allowed band around 19, 200 cm−1 appear structured?

Low temperatures and crystalline samples can allow one to resolve vibrational structure. These
samples are also adventageous since they eliminate coupling with the solvent and reorganization
effects.

We assume that the strongest three bands are the spin-allowed bands:

4T1g → 4T2g 7200 cm−1
4T1g → 4A2g 15000 cm−1
4T1g → 4T1g 19000 cm−1

The band at 15, 000 cm−1 is weaker than the other two because the 4A2g state has a smaller
degeneracy and the 4T1g and 4T2g states.

Because E(4A2g) = 12Dq − 43B + 14C and E(4T2g) = 2Dq − 43B + 14C, 10Dq can be obtained
as the difference between the position of the 4T1g → 4A2g band and the 4T1g → 4T2g band.
7800 cm−1. 4T1g states should
Hen√ce Dq = + 36 · Dq · B Also the the energy difference between the two solving for B gives
be 5 45B2 + 20Dq2. Setting this equal to 19, 000 cm−1 and

B = 880 cm−1. Finally, to find C we must use the 2Eg band at 10, 000 cm−1 and use:

√
3 5
E (2 Eg ) − E (4 T1g ) = −15Dq + B + 3C + 2 45B2 + 36 · Dq · B + 20Dq2

2

⇒ C = 3640 cm−1

The small value of Dq is a result of the π-donations of F−1 ions, which makes the ligand field
weaker. The values of B and C are quite typical. The structure in the peak at 19, 000 cm−1 is

due to different vibronically excited modes. I was puzzled by additional weak featers at 17, 500
cm−1 and 21, 500 cm−1. I consulted the original paper, which suggested that these peaks were

due to other doublet excitations.

7

4. The visible spectrum of VCl64− is strongly temperature dependent (see below). The positions of the
peak maxima at 77 K are 7, 200 cm−1, 11, 700 cm−1, and 19, 100 cm−1.

Figure 5: Visible spectrum of VCl64−
(a) Assign the bands in the spectrum.

VCl64− is a d3 compound with a 4A2g ground state. The band at 7, 200 cm−1 is due to the
spin-forbidden 4A2g → 2Eg transition. The band at 11, 700 cm−1 is due to the 4A2g → 4T2g
transition. The band at 19, 100 cm−1 is due to the 4A2g → 4T1g.
(b) Account for the temperature dependence of this spectrum.
The observed transitions are formally symmetry forbidden. The weak bands are possible through
vibronic coupling (see problem 1). As the vibronic coupling is weakened by lowering the temper-
ature, the bands weaken. Additional, the absorption maximums blue-shift because there is less
vibrational energy, so the electronic energy must increase in order to compensate.

8

5. The ability to predict the relative intensities of electronic transitions available to molecular systems
often proves to be a useful tool when assigning an absorption spectrum. Selection rules derived from
quantum mechanical arguments provide a means of doing just that. However, it is important to remem-
ber that selection rules for a particular molecule in question are only as reliable as the wavefunctions
and approximations used to derive them. This problem will deal with the derivation of selection rules
for a hydrogen atom in a beam of plane polarized light. The methodology used here can be extended
to molecular systems as well.

A development of the equations describing the interaction of an atomic or molecular system with

the time-dependent electric and magnetic fields of an incident light beam usually beings with Fermi’s

golden rule:

2π j Gˆ o 2 (1)
Poj = 2
φoj

By solving Maxwell’s equations along with the classical Hamiltonian for a charged particle moving in

an electric and magnetic field it may be shown that the time-dependent perturbation found in (1) can

be written as: i Qη
Mη c
Gˆ = Aη · ∇η (2)

η

where Aη is the vector potential produced by the time-dependent fields found in the incident radiation.

Aη = A(ω )uei ω k·r + A(ω )ue−i ω k·r (3)
c c

Here k is the wave vector pointing in the direction of the propoagation of the light wave and u is the

unit vector pointing in the direction of the oscillating electric field. If the wavefunctions j and o are

taken to be Born-Oppenheimer functions, then the sum over nuclear coordinates in (2) reduces to a

sum over electronic coordinates multiplied by some Frank-Condon factors. Using this idea and these

equations we have:

2π 2
Poj = 2
j Aω uei ω k·r · ∇η o φoj (4)
c

η

Here the index η runs over all electronic coordinates. Using (4) as a starting point, derive the selection
rules for electic dipole, magnetic dipole, and electric quadrupole allowed electronic transitions in a
hydrogen atom for light traveling in the z direction and polarized in the x direction.

Solution: Expanding out the exponential in (4) yields:

2π ω 2
c
Poj ≈ 2 j Aωu · ∇η o + j Aω u(i k · r) · ∇η o φoj (5)

ηη

but k = zˆ and u = xˆ, so the sums over the single hydrogen atom’s electronic coordinates collapses

2π ∂ ∂ω 2

Poj ≈ 2 j Aω ∂x o + j Aω ∂x (i c z) o φoj (6)

Solving for when the first matrix element is nonzero gives the electronic dipole selection rules. Lets
examine this matrix element more closely, making use of the fact that

[H, x] = − 2 ∂
m ∂x

9

∂ = − mAω nlm |[H, x]| nlm
nlm Aω ∂x n l m
2

= − mAω nlm |Hx − xH| n l m

2

= − mAω (En − En ) nlm |x| nlm

2

which is nonzero iff En = En ⇒ n = n since the energy of the nth eigenstate is only dependent on n
in the Hydrogen atom. The rest of the electronic dipole selection rules arise from solving for when

nlm |x| nl m = 0

We note that √ 4π (Y1−1 − Y11)
Hence, 2 3

x=r
2

√ 4π (
nlm |x| nl m = r 2 3
2 nlm Y1−1 nl m − nlm Y11 nl m ) (7)

We recall that |nlm = Rn,l(r)Ylm(θ, φ) = N Rn,l(r)Plm(cos θ)eimφ. From the exponential terms we
immediately see that:

Yml Y1±1Yml dτ = 0 unless m ± 1 = m

We also note that the character of an inversion is given by χ(i) = (−1)l and so the character of the
inversion for a direct prodcut is only 1 if l + l is odd. We also note that Yml belongs to the irrep of the
full rotation group, Γl while Y1±1 belongs to Γ1. We can take the direct product of these irreps to get

Γl ⊗ Γ1 = Γ(l+1) + Γl + Γ(l−1)

From this it is clear that the integral in (7) vanishes unless l ± 1 = l , giving the final dipole selection
rule for x-polarized light.

Now we return to (6) and look at the second matrix element to see when it is zero. This matrix element
will give us the electric quadrupole and the magnetic dipole selection rules. We note that

2 x ∂ +z ∂

[H, xz] = −
m ∂z ∂x

allowing us to write the second matrix element as:

∂ω = iωAω j ∂∂ + ∂ −x ∂ o
j Aω ∂x (i c z) o z +x z
2c ∂x ∂z ∂x ∂z

= iωAω j − m [H, xy] + ∂ −x ∂ o
2c z
2 ∂x ∂z

= iωAω j − m [H, xy ] o + iωAω j ∂ −x ∂ o (8)
2c 2c z
2 ∂x ∂z

The first matrix element of (8) is recognized as the electric quadruople operator after H is applied to
the bra and ket respectively as in the electric dipole case. The second matrix element of (8) is actually

10

the matrix element of an angular momentum Jy, which is linearly proportional to the magnetic dipole
operator. Hence the electric quadrupole selection rules arise from solving for when:

< nlm|[H, xy]|n l m >= 0

Observe that

< nlm|[H, xy]|n l m >= (En − En ) < nlm|xy|n l m >= (En − En ) < nlm|xy|n l m > (9)

which demands that n = n in order for the integral to not vanish. Additionally we can re-express y

with respect to the spherical harmonics:

√ 4π Y11 + Y1−1
y = −r 2 3

2

⇒ xy = − 2π (Y11)2 + (Y1−1)2
3

When we square Y1±1 there is a factor of e2imφ, which requires that m ± 2 = m for a nonvanishing inte-
gral. Also, the representation of the perturbation is now Γ1 ⊗ Γ1 which spans Γ2 ⊕ Γ1 ⊕ Γ0 ⊕ Γ−1 ⊕ Γ−2.
So due to rotation group considerations, |l − l | < 4. Also, due to the inversion considerations, l + l

must be even. This requires l = l or l ± 2 = l .

The magnetic dipole selection rules arise from solving for when:

< nlm|Jy|n l m >= 0

Jy is an axial pseudo-vector so it is g under inversion. Unlike the electric dipole selection rules this
requires that l = l . Re-expressing the matrix element with respect to the shift operators gives:

< nlm| 1 + J−)|nlm >
2 (J+

which is obviously nonzero iff m ± 1 = m . This completes the selection rules for magnetic dipoles.

11

6. The Fe atom in the protein rubredoxin is thought to reside in a roughly tetrahedral environment
consisting of four mercaptide S atoms. The UV/Vis spectra of ferric and ferrous rubredoxin are
dominated by charge transfer bands. The d − d bands were located in the near IR by Eaton and
Lovenberg (JACS 1970, 92, 7195-7198). Read this paper and answer the following questions:
(a) Briefly describe why circular dichroism is as useful as absorption spectroscopy in locating the
ligand field bands in this case. Explain the assignment of the band at 6250 cm−1.
Circular Dichroism (CD) provides a method to ensure that the feature at 6250 cm−1 is an elec-
tronic transition. The 5E → 5T transition is magnetic dipole allowed, so the fact that the band
shows up using CD confirms the assignment. Without the CD evidence, the feature could be
confused as a vibrational excitation (as was the case in ferric rubredoxin). Such vibrations, how-
ever, are not coupled together by the magnetic dipole moment, so the presence of the CD band
in ferrous rubredoxin indicates that there is actually an electronic transition at 6250 cm−1. This
band is assigned to the only spin-allowed transition: 5E → 5T .
(b) The authors state that the other spectral data indicate that the Fe site symmetry may be less
that Td. Assuming a distortion leading to D2d symmetry can occur, how will the d orbitals split
in this lower symmetry environment? How many excited states would you expect? A transition
has been detected in Mossbauer experiments that does not correspond to an iron nuclear spin flip
(850 cm−1). Which symmetry do you think is present in the protein? Explain.

In D2d symmetry, the e and t2 sets split out. The e set splits into the b1 symmetry x2 − y2 orbital
and the a1 symmetry z2 orbital. The t2 set splits into the b2 symmetry xy orbital and the e
symmetry (yz, xz) orbitals. The ground state is a 5A1 state with the b1, a1, b2, and e sets all
singly occupied with alligned spins. A spin allowed excitation requires a promotion of an electron
into the other spin-alligned E orbital. This electron could originate from the b1, a1, or b2 orbital,
so there are three spin-allowed excited states corresponding to the following transitions:

x2 − y2 → (yz, xz)
z2 → (yz, xz)
xy → (yz, xz)

The first of these two transitions are roughly equal in energy (both ≈ 6000cm−1). The second
transition is essential the energy of the splitting of the t2 set from tetrahedral splitting. It is
conceivable that if the distortion is strong enough, this energy splitting could be several hun-
dred wavenumbers. Hence, I suspect that the true symmetry is D2d and that the Mossbauer
experiments have revealed the splitting of the tetrahedral t2 set.

12

7. The first two spin-allowed d-d transitions in three octahedral d6 low-spin cobalt(III) complexes have
been reported at the following positions:

Complex ν1 (cm−1) ν2 (cm−1)
Co(H2O)63+ 17,000 25,000
Co(NH3)63+ 21,000 29,000
Co(CN)36− 32,100 38,500

Using the appropriate Tanabe-Sugano diagram in Figgis, predict the following transitions:

1A1g → 3A1g;1 A1g → 3T2g;1 A1g → 5T2g

Careful examination of a 3 mm thick single crystal of K3Co(CN)6 reveals a band at 26,000 cm−1,
= 0.25. Assign this band and obtain a value for C which fixes the C/B ratio for low-spin Co(III). Re-

calculate the positions of the three spin-forbidden transitions mentioned above for all three complexes.
Would you expect to be able to observe more than one of these transitions? Appropriate strong-field
energy expressions may be found in Table A29 of Griffith (ignore configuration interactions). Finally
draw an energy level diagram illustrating how the three configurations (t26g; t25geg; t24ge2g) split into their
ligand field states for hexacyanocobaltate and draw the transitions corresponding to the bands dis-
cussed above.

Solution: In a d6 complex the 1A1g, 1T1g, and 1T2g energies (neglecting confiuration interaction)
are:

E(1A1g) = 15A − 17B + 13C − 24Dq
E(1T1g) = 15A − 30B + 14C − 14Dq
E(1T2g) = 15A − 14B + 14C − 14Dq

So

ν1 = E(1T1g) − E(1A1g) = 10Dq − 13B + C
ν2 = E(1T2g) − E(1A1g) = 10Dq + 3B + C

So 16B = ν2 − ν1 ⇒ B = ν2 −ν1 . We assume C/B = 4 to obtain:
16

1 ν2 − ν1 = Dq
10 ν1 + 9 16

Hence we obtain

Complex ν1 (cm−1) ν2 (cm−1) Dq (cm−1) B (cm−1)
Co(H2O)36+ 17,000 25,000 2150 500
Co(NH3)63+ 21,000 29,000 2550 500
Co(CN)63− 32,100 38,500 3570 400

Now we must find the energy of the 3A1g, 3T2g, and 5T2g states. The 3A1g state can only arise from
the t23ge3g strong field configuration. Table A29 in Griffiths gives the energy of this state (including the
9A − 14B + 7C term from the hole formalism).

E(3A1g) = 15A − 26B + 11C + 6Dq

13

The 3T2g state can actually arise from several strong field configurations, but we take the diagonal
matrix element of the lowest energy strong field configuration

E(3T2g; t25geg1) = 15A − 22B + 12C − 14Dq

Similarly

E(5T2g; t42geg2) = 15A − 35B + 7C − 4Dq

From these energies we simply obtain the transitions energies:

E(1A1g → 3A1g) = 30Dq − 9B − 2C
E(1A1g → 3T2g) = 10Dq − 5B − C
E(1A1g → 5T2g) = 20Dq − 18B − 6C

So

Complex 1A1g → 3A1g 1A1g → 3T2g 1A1g → 5T2g
Co(H2O)36+ 34,500 17,000 22,000
Co(NH3)36+ 42,500 21,000 30,000
Co(CN)36− 64,600 32,100 54,600

The band at 26, 000 cm−1 from the single crystal of K3Co(CN)6 is due to the spin forbidden 1A1g →
3T1g transition. From this we no longer have to use our C/B = 4 estimate. We can now solve for Dq,
B, and C in K3Co(CN)6 and assume this C/B ratio for the other compounds.

Dq = 3425 cm−1, B = 400 cm−1, C = 3050 cm−1 and C/B = 7.6

Using this new value of C/B we get

Complex 1A1g → 1T1g 1A1g → 1T2g 1A1g → 3A1g 1A1g → 3T2g 1A1g → 5T2g 1A1g → 3T1g
Co(H2O)36+ 17,000 25,000 27,250 13,375 7,500 9375
Co(NH3)36+ 21,000 29,000 35,250 17,375 15,500 13,375
Co(CN)63− 32,100 38,500 58,800 29,200 43,000 26,000

It appears that some of the bands should be observable. For example, in Co(NH3)63+, the 1A1g → 3T2g
band is predicted to be at 17,375 cm−1. The for this band would be very small though, making it
exceedingly difficult to see.

The energy level diagram demonstrating how the strong field states split is given below. Only the
states we discussed are labeled.

14

t24g e2g 5T2g
t25g e1g 1T2g
t26g 1T1g
3T2g
3T1g

1A1g

Figure 6: Energy level diagram showing splitting of strong field states in a d6 configuration

15

8. Gillespite is a rare silicate mineral which has figured prominently in the development of mineral spec-
troscopy. Read the paper by Burns, Clark, and Stone (Inorg. Chem. 1966, 5, 1268-1272) and answer
the following questions:

(a) Derive all of the possible spin-allowed electronic transitions from the ground state.

We treat the problem as having D4h symmetry. In this point group the d orbitals split into
a1g, eg, b2g, and b1g sets corresponding to orbitals which are dominanted by dz2 , (dxz, dyz), dxy, and
dx2−y2 orbitals respectively. The paper shows that the ordering of these orbitals from low to high
energy are as listed. The paper also indicates that a high-spin ground state is observed. Hence the
ground state is the 5A1g state corresponding to the configuration (dz2 )2(dxz)1(dyz)1(dxy)1(dx2−y2 )1.
The spin accessible excited states will all be have a basic (dz2 )1(dxz)1(dyz)1(dxy)1(dx2−y2)1 config-
uration with the sixth electron in each of the possible orbitals. Because the half-filled configuration
behaves as A1g, the spin-allowed excited states are just given by

A1g ⊗ eg = Eg
A1g ⊗ b2g = B2g
A1g ⊗ b1g = B1g

Thus the spin allowed transitions are 5A1g → 5Eg, 5A1g → 5B2g, and 5A1g →5 B1g.
(b) Assume it is reasonable to consider only the local environment of the iron site and determine the

symmetries of the FeO4 vibrations. Do you think this assumption is reasonable?

We consider local x, y, and z coordinate systems on each of the 4 planar O atoms of the FeO4
unit. We neglect the axes on the Fe atom since these simply give translational motion. It is a
simple exercise to show that

D4h E 2C4 C2 2C2 2C2 i 2S4 σh 2σv 2σd
Γtot 12 0 0 -2 0 0 0 4 2 0
Γrot 3 1 -1 -1 -1 3 1 -1 -1 -1
Γvib 9 -1 1 -1 1 -3 -1 5 3 1

Applying the great orthogonality theorem yields Γvib = A1g ⊕ B1g ⊕ B2g ⊕ A2u ⊕ B2u ⊕ 2Eu
(c) Derive the assignments and polarizations which are reported in Table II of the paper.

I will only derive the transitions in the table from the 5A1g ground state since these are the tran-
sitions which were observed. In D4h the electric dipole vector can transform as Eu in the case of
(x, y) or as A2u in the case of z polarized light. For parallely polarized light, the electric dipole
transforms as A2u. In order for a transition to be allowed, it must couple to an ungerade vibra-
tion to overcome the Laporte selection rules. Since the ground state is A1g, a parallel transition
is allowed iff X ⊗ A2u ⊗ Y = A1g where X is the excited state and Y is the representation of
the vibration. First we consider the parallel 5A1g → 5Eg transition. Note that Eg ⊗ A2u = Eu
and Eu ⊗Eu contains A1g, so the 5A1g →5 Eg transtion is allowed if coupled with an Eu vibration.

Now we consdier 5A1g → 5B2g. Note that B2g ⊗ A2u = B1u. So in order to be allowed this would
have to couple to a B1u vibration, but there is no B1u vibration, making this transition forbidden.

Finally we consider 5A1g → 5B1g. This is very similar to the case before except now we require
a B2u vibration, which is present.

Next we consider the perpendicular transitions, which make use of x and y polarized light that
transforms as Eu. The 5A1g → 5Eg transition can actually be allowed through two different

16

vibrations since Eg ⊗ Eu = A1u ⊕ A2u ⊕ B1u ⊕ B2u. A A1u, A2u, B1u, or B2u vibration could
couple to these in order to give rise to an overall A1g matrix element. However only A2u and B2u
vibrations are present.
It is also straightforward to see that 5A1g → 5B2g and 5A1g → 5B1g are allowed if coupled to
Eu vibrations since:

Bng ⊗ Eu ⊗ Eu = Eu ⊗ Eu = A1g ⊕ A2g ⊕ B1g ⊕ B2g
This completes the derivation of Table II.

17

9. In addition to an absorption spectrum, the emission spectrum of a metal complex yields useful informa-
tion about transitions between ground states and low-lying excited states. Emission from a transition
metal copmlex with an unfilled d shell generally occurs from the lowest electronic excited state in the
molecule or from those states that can achieve a signficiant Boltzmann population relative to the lowest
excited state. In problem 2 of this set, the absorption spectra of Cr(III) complexes were discussed.
In this problem the absorption and emission behavior of octahedral Cr(III) will be linked together
by potential energy surface considerations. Background material may be found in N.J. Turro Modern
Molecular Photochemistry, pp. 52-96.

(a) Excitation from the 4A2g ground state into one of the spin-allowed 4Tng excited states is fol-
lowed by rapid internal conversion and vibrational relaxation inot the zeroth vibrational level of
the 4T2g term. From this point, two main photophysical processes occur, namely fluorescence
or intersystem crossing (ISC) into the 2Eg term from which phosphorescence is observed. This
emission is not Stokes shifted; it overlaps the very weak corresponding absorption band. This
leads to an unusual situation in that the fluorescence is found to the red of the phosphorescence.
The situation is illustrated in the following Jablonski diagram. A typical spectra is also shown.
Construct qualitative plots of energy vs. Q (a totally symmetric nuclear coordinate) in which

the intersections, energy gaps, and widths of the PE surfaces for the 4A2g, 4T2g, 2Eg, and 4T1g
terms are illustrated. Keeping in mind that each Dq/B value in a Tanabe-Sugano diagram cor-
responds to a different set of nuclear coordinates, provide plots for three cases: (a)Dq/B < 2.1;
(b) Dq/B = 2.1; (c) Dq/B > 2.1. Use the appropriate Tanabe-Sugano diagram in Figgis.

Solution: Note that at Dq/B = 2.1, E(4T2g) = E(2Eg). This equality is for a fixed value of
Q, so it is only for the vertical transition. We approximate the quartet terms as all having the
same width, and the doublet term is slightly wider.

4T1g 2Eg 4T1g 4T1g
4T2g 4T2g 4T2g

2Eg 2Eg
E
E
E

4A2g 4A2g 4A2g

Dq/B < 2.1 Dq/B = 2.1 Dq/B > 2.1
Q Q Q

18

(b) The luminescence behavior of some Oh Cr(III) complexes is tabulated below (energy in cm−1).

Discuss the data with regard to the spectrochemical series and your plots in part a.

CrCl63− νmax(4A2g → 4T2g) νmax(4A2g → 2Eg) ν0-0(phos) νmax(fluor)
CrF36−
Cr(urea)63+ 13,060 14,480 – 11,600
Cr(oxalate)33− 14,900 15,700 – 12,830
Cr(NH3)63+ 16,150 14,350 14,240 12,550
Cr(phen)33+ 20,800 13,420
Cr(CN)36− 17,500 ? 14,390 –
Cr(CH3)36− 21,550 14,350 15,120 –
23,800 15,300 13,720 –
26,600 13,700 12,430 –
12,470 –

Solution: As we read down the table we move through the spectrochemical series with increasing

Dq. As Dq increases, we move more into the strong-field case depicted by the third potential

energy diagram. In this situation the 2Eg surface drops in energy with increasing Dq as seen

by the second column of the table. Also, inner-system crossing becomes more favorable. The

first potential energy diagram reveals an energy barrier moving from the 4T2g to 2Eg surface, but
for Dq/B > 2.1 the spin cross-over is downhill from the excited quartet state. When the 2Eg
minima lies much lower than the 4T2g minimua, the Boltzman factor will heavily favor intersystem
crossing into the 2Eg state. This justifies why phosphorescence is seen rather than fluorescence

in the compounds with the largest values of Dq.

(c) Read the paper by Kenney, Clymire and Agnew (J. Am. Chem. Soc. 1995, 117, 1645-1646),
and summarize the interpretation of the spectrum reported in Figure 1. Draw an energy level
diagram as a function of pressure illustrating this situation.

Solution: In essence, Kenney et. al. report that they can take a weak field Cr(III) coupound

and make it strong field under high pressure since the ligands will be pushed closer to the Cr
center. In the strong field limit the 2Eg state is lower in energy than the 4T2g excited state, so
phosphorescence replaces the usual weak-field fluorescence. Figure 1 in their paper displays the

emission spectrum as a function of pressure, demonstrating the gradual onset of a sharp phos-

phorescence peak along with the gradual loss of the vibrationally structured fluorescence peak.

We can roughly say that Dq is proportional to pressure, allowing us to take the relative energy of
the 4A2g, 4T2g, and 2Eg states from a normal Tanabe-Sugano diagram. There is some unknown
scaling factor between Dq and P , but the general situation is captured by the spin-crossover of

the excited state in the following diagram. (Obviously the plot is not accurate near P = 0)

E/B 60

50

40 4T2g
30

2Eg
20

10
4A2g

0
0 2 4 6 8 10 12 14 16 18 20
Pressure (GPa)

19

10. Substitution reactions of low-spin Co(III) complexes generally proceed very slowly. Co(H2O)36+ is a
striking exception. Taube and coworkers suggested that the high-spin term 5T2g should be substitution

labile and thermally accessible. In order to ascertain whether this is reasonable, the energy gap between
the 1A1g and 5T2g must be determined (see the potential energy surfaces below).

Figure 7: Potential energy surfaces of Co(III)

(a) What is the problem with using straightforward ligand field analysis to find this energy difference?
What is the vertical transition energy in terms of ligand field parameters (EFC)?

Solution: The straightforward ligand field analysis only pertains to vertical transitions, in which
the ligand-metal distances are held static. The sextet term has a larger equilibrium bond length,
so the minima of that potential energy surface is actually lower than EFC would suggest. It is the
energy of this minima that determines the Boltzmann factor. In Problem 7 I already solved for
this vertical transition energy.

EFC = 20Dq − 18B − 6C

(b) Experimentally observed splittings in the electronic origin of the 3T1g term may be used to estimate
the position of the 5T2g potential energy surface (Wilson and Solomon JACS 1980, 102, 4084-
4095). The position of this surface may also be found using the following equation:

slope 5T2g 2 Sa1g (5T2g )
slope 3T1g Sa1g (3T1g )
=

where “slope” refers to the slope of the curve on the Tanabe-Sugano diagram and the quantity S
is the Huang-Rhys factor for the normal coordinate of a1g symmetry in the indicated state.

K (ΔQ)2
S=

2ω

20

The Tanabe-Sugano slope of the quintet term is twice that of the triplet term. It is then assumed
that the force constant for the 5T2g totally symmetric vibration is K∗ = αK where K is the

ground state value and α varies from 0 to 1. Locating the quintet surface now reduces to a

problem in analytical geometry.

E (1 A1g ) = 1 KQ2
2

E(5T2g = E0 + 1 K∗(Q − Q0)2
2

EFC for the 1A1g → 5T2g transition is just E(5T2g) evaluatied at Q = 0. Substituting into the ex-

pression for EFC the parameters Dq, B, and C, what is the separation of the minima of the ground
state and the first quintet state potential energy surfaces? Let ω = 357 cm−1. The reflectance
spectrum of CsCo(SO4)2 · 12H2O yields Dq = 2080 cm−1, B = 513 cm−1, and C = 4250 cm−1.
Sa1g (3T1g) can be assumed to be 2.5 from Solomon’s work. Using this experimental data, calcu-
late the energy gap E0. Estimate the activation energy for the spin crossover process in kcal/mole.

Solution: The lowest energy strong-field 3T1g state has a t25ge1g configuration, so the slope with
respect to Dq is roughly 10. From part (a) we see that the slop for the 5T2g state is 20. Hence

4 = Sa1g (5T2g)
Sa1g (3T1g)

We take the value of Sa1g (3T1g) from Solomon of 2.5, so we now have an estimate of Sa1g (5T2g).
The vertical transition energy is given by

E (5 T2g ) = E0 + 1 K ∗Q02 = 20Dq − 18B − 6C
2

⇒ E0 = 20Dq − 18B − 6C − ωSa1g (5T2g)

⇒ E0 = 20Dq − 18B − 6C − 4 ωSa1g (3T1g)

Plugging in the given values yields

E0 = 3296 cm−1

The activation energy is the energy at which the 1A1g surface intersects the 5T2g surface. For a
rough approximation at this energy we will set α = 1 and solve for the intersection of the parabolic

surfaces. The intersection occurs at

Q= EFC

2K(EFC − E0)

and the intersection energy is

Eact = EF2C E0) = 3300 cm−1 = 9.4kcal/mol
4(EFC −

21

11. From analysis of the emission spectrum of K3Co(CN)6, Hipps and Crosby concluded that the maximum
of the 1A1g → 3T1g transition in absorption to be at 20,300 cm−1. The emission peak at 14,000 cm−1
was attributed to the 3T1g → 1A1g transition.
(a) Is the predicted 20,300 cm−1 band consistent with the expected reduction in the Racah parameter
C from the free ion value of 5120 cm−1? Explain. If not, at approximately what energy would
you expect this transition?
For large Dq, the energy of the 3T1g state is approximately (neglecting configuration interac-
tion) given by
E(3T1g) = −14Dq + 15A − 30B + 12C
and from Problem 7
E(1T1g) = −14Dq + 15A − 30B + 14C = 32100 cm−1
so the transition 1A1g → 3T1g should have energy 32100 − 2C. With the assignment of Hipps
and Crosby this implies that C = 5900 cm−1. Hence the band position is not consistent with
the expected reduction in the Racah parameter C from its free ion value. I would expect C to
be reduced to about 4,000 cm−1, so the 1A1g → 3T1g band should be around 32100 − 8000 =
24100 cm−1.
(b) IR studies indicate that the a1(Co-C) stretch occurs at 414 cm−1. Discuss the following spectrum
in this context.

The fine structure on the spectra is spaced by 414 cm−1 corresponding to increasing quanta of
the a1 vibrations.
(c) Discuss the significance of the 1A1g → 3T1g transition with regard to molecular geometry and
ligand field parameters Dq, B, and C.
The excited triplet state adopts a differnt molecular geometry. We expect that the geometry
is still essentially octahedral, but the average bond lengths increase. At longer bond lengths the
overlap integrals are smaller, so Dq decreases. Also B and C tend to increase toward their free-
ion values. Essentially the excited electronic state acts like a distinct molecular species since its
molecular geometry changes. This means Dq, B, and C should be recalculated for the excited
state.

22

(d) The absorption spectrum of an excited state offers an interesting view of the energy level pattern
of a molecule. The hexacyanocobaltate ion gives rise to a structured d − d transient absorption
spectrum obtained directly after laser excitation of the ground state system. The spectrum is
shown below. From Griffith table A25, determine the energies of the relevant excited states.
Draw an energy level diagram illustrating these levels and the impact of configuration interaction
on their energies. Interpret the spectrum.

We are interested in the low-lying triplet states. We will take low-lying to mean all of the
triplet states arising from t52ge1g and t24geg2 configurations. From Table A29 of Griffiths we can
explicitly construct the electrostatic matrices including configuration interaction between the low-

lying states:

⎛ − 15B + 13C − 8Dq −10B 0√ √ ⎞
15A 15A − 23B +√11C − 8Dq 2B 3 −3B√ 2 ⎟⎟⎠
15A − 25B +√11C − 8Dq 3B√ 2
3T1g = ⎜⎝⎜ −10B 2B√3 B6
0√ 3B 2 B6

−3B 2 15A − 30B + 12C − 18Dq

⎛ − − 8Dq √ ⎞
15A 27B +√11C 2B 2 6B√ ⎠
3 T2g = ⎝ 2B 2 15A − 23B +√11C − 8Dq 3B 2
3B 2
6B 15A − 22B + 11C − 18Dq

3Eg = 15A − 25B + 11C − 8Dq

3A2g = 15A − 16B + 14C − 8Dq

The eigenvalues of these matrices give our energy levels, which are plotted using the values of
Dq = 3425 cm−1, B = 400 cm−1, C = 3050 cm−1.

23

x 104 x 104
5 3.5
4.5
4 3.45
3.5
3 3.4
2.5
2 3.35
1.5
1Energy (cm−1) 3.3
0.5 Energy (cm−1)
0 3T1g 3.25
3T
No Configuration Interaction 3.2 3T1g
2g 3.15 3T2g
3Eg
3E 3.1 3A
3.05
g 2g
3
3A2g No Configuration Interaction (b) With Configuration Interaction

With Configuration Interaction

(a)

Figure 8: Relevant excited triplet states with an without configuration interaction. Energies are given with
respect to the 3T1g state, which is assumed to be the intermediate. Dq = 3425 cm−1, B = 400 cm−1, C =
3050 cm−1

We note that the position of the bands does not seem to match the experiment by several thousand
wavenumbers. This is because the 3T1g state has a different value of Dq, B, and C. The change
in Dq is most drastic. By setting Dq = 2475 cm−1 the following plots were obtained, which have

much better agreement with the experiment.

x 104 x 104
4 2.65

3.5 2.6
2.55
3
2.5
2.5 2.45
Energy (cm−1)
2 Energy (cm−1) 2.4
2.35
1.5 3T1g 3T1g
3T2g 2.3 3T2g
1 3Eg 2.25 3Eg
3A2g 3A2g
0.5 2.2
With Configuration Interaction 2.15 With Configuration Interaction
0
No Configuration Interaction 2.1
No Configuration Interaction (b)
(a)

Figure 9: Relevant excited triplet states with an without configuration interaction. Energies are given with
respect to the 3T1g state, which is assumed to be the intermediate. Dq = 2475 cm−1, B = 400 cm−1, C =
3050 cm−1

Based on this calculation, the peak at 2.17 μm−1 is assigned to be both the 3T1g → 3T1g and
3T1g → 3T2g transition. The peak at 2.39 μm−1 is assigned to be the 3T1g → 3Eg transition.
And the shoulder at 2.6 μm−1 is due to both the 3T1g → 3T1g and 3T1g → 3T2g transitions. My
assignment differs slightly from the assignment I found in a JACS paper by Viaene et. al. Viaene
clumped the higher energy 3T1g → 3T1g transition in with the 3Eg transition to account for the
middle peak.

24

12. The previous problem dealt with the structured emission spectrum of hexacyanocobaltate. Lumines-
cence spectra are appropriate for investigation of vibrational progressions in electronic transitions,
since they can be measured with high sensitivity. The Frank-Condon principle may be used to eval-
uate the change in molecular geometry which accompanies an electronic transition. Such a vibronic
intensity analysis yields important information such as the excited-state geometry and excited-state
force constants. In this problem you will carry out a FC analysis of the hexacyanocobaltate emission
spectrum.

(a) Background material:

i. C.J. Ballhausen Molecular Electronic Structures of Transition Metal Complexes, sec. 4.7
ii. S.E. Schwartz J.Chem. Ed. 1973, 50, 608-610.
iii. G. Herzberg Spectra of Diatomic Molecules, chap. 4.
iv. M.A. Hitchman Trans. Met. Chem. 1985, 9, 1.

(b) The theory outlined below derives from work of Ansbacher (Z. Naturforsh. 1959, 14a, 889) and

Henderson and co-workers (J.Chem.Phys. 1964, 41, 580). The vibronic emission intensities are

given by:

Iν,ν∗ = 64 π4 cNν Eν4,ν Re2 ψν |ψν∗ 2
3 h4
∗

Here Eν,ν∗ is the energy of the electronic transition between vibrational states with the quantum
numbers ν and ν∗ for the excited and ground vibrational states respectively. Re is the average

value of the electronic transition dipole moment, and the integrals are the Frank-Condon overlap

integrals hereafter called Rν,ν∗ and N is the number of molecules in the initial state. In practice

the relative intensities can be related via the following equation.

I0,n = E0,n 4 R0,n 2
I0,0 E0,0 R0,0

The frequency dependence for emission is given by:

E0,n = E0,0 − nhν0

where E0,0 is the energy of the electronic transition between the lowest vibrational states of the
excited and ground electronic states, and ν0 is the eigenfrequency of the oscillator in the electronic
ground state. The most useful relations for the vibration overlap integrals are given by:

R0,0 = √ 2δ e− ρ2
2

1+δ
−2DδR0,n − √2n(δ2 − 1)R0,n−1
R0,n+1 = (1 + δ2) 2(n + 1)

δ = ν0
ν0∗

D = CΔS μν0∗

ρ= √ D
1 + δ2

When the displacement ΔS of the minimum of the PE surface of the excited state along the

stretching coordinate S is expressed in angstroms, the vibrational energy in wavenumbers and the

masses of the ligands in amu, the constant C takes on the value 0.1722.

The distortions calculated from FC analysis are distortions in the normal coordinate whose vibra-

tions comprise the progression in the spectrum. For the totally symmetric stretch in an octahedral

complex, the change in each metal ligand bond length is √1 ΔS . ΔS is normally determined by
6

calculating the vibronic band shape for different values of ΔS and seeing which value most satis-

factorily reproduces the observed spectrum.

25

(c) The energy of the vibronic peaks are given below:

E0,n(cm−1)
16,650
16,244
15,857
15,439
15,048
14,660
14,267
13,852
13,462

The origin for this progression is located at 17,020 cm−1. You may assume ν0 = 0.65. Calculate
ν0∗

the vibronic intensity distribution and plot the intensities for the following values of ΔS: (a)

0.05, (b) 0.3 and (c) 0.5 ˚A. It may be useful to write a simple computer program to do these

calculations. From a comparison of these distributions and the known spectrum determine the

distortion of the metal ligand bonds in the excited state.

We aim to develop an expression for I0,n.

I0,n = I0,0 E0,n 4 R0,n 2
E0,0 R0,0

I0,0 will be scaled to normalize the maximum intensity to 1.
E0,0 is the energy of the origin of the progression: 17,020 cm−1.

C = 0.1722

δ = 0.65

ν0 can be obtained by averaging the energy gaps between adjacent energy peaks.
μ is the reduced mass, which is well approximated by the reduced mass between a Cobalt atom and
a carbon atom of the ligand. Using these values and the equations given in the problem a script
was written to generate plots for three values of S. Comparing these plots to the luminescence

1 1 1

0.8 0.8 0.8
Relative Intensity
0.6 Relative Intensity0.6 0.6
Relative Intensity
0.4 0.4 0.4
S = 0.05 S = 0.3 S = 0.5

0.2 0.2 0.2

0 0 0
0.8 1 1.2 1.4 1.6 1.8 0.8 1 1.2 1.4 1.6 1.8 0.8 1 1.2 1.4 1.6 1.8

Energy (cm−1) x 104 Energy (cm−1) x 104 Energy (cm−1) x 104

spectrum from Problem 11 shows that S = 0.3 is the best fit of the three options. Tuning the

value of S a little more shows that S = 0.25 actually produces a slightly better fit. For the totally

symmetric stretch, ΔQ = √1 ΔS , so ΔQ ≈ 0.10 A˚.
6

(d) What does this distortion imply with respect to excited state substitution reactions of these kinds

of complexes? You have fit the spectrum using the symmetric stretching mode; what other mode

would you expect to be involved?

26

1

0.8 S = 0.25

Relative Intensity 0.6

0.4

0.2

0
0.9 1 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8

Energy (cm−1) x 104

This implies that the electronic excited state is very distorted, with a change in the ligand dis-
tance of 0.10 A˚. Hence the electronic excited state is expected to be more labile. The a1g vibration
does not alter the selection rules, so there is a progression of peaks due to different quanta of a1g
vibrations. The other involved mode is the mode that will allow the electronic transition from
A1g to T1g. From Problem 1 the octahedral vibrational modes are A1g ⊕ Eg ⊕ T2g ⊕ 2T1g ⊕ T2u.
The direct product of A1g, T1g, T1u, and the irrep of the vibration must contain A1g. This implies
that the allowing vibration is T1u.

(e) Recall the feature around 26,000 cm−1 observed in the absorption spectrum of a thick crystal
of K3Co(CN)6 (Problem 7). Estimate a value for the E0,0 of the emission band. How does this
change your result found in (c)? Comment on any changes in the quality of fit.

If the absorption peak is at 26,000 cm−1 rather than around 20,500 cm−1 then E0,0 ≈20,000
cm−1. This shift implies that the largest intensity emissions occur for even higher quanta of a1g
vibrations. This means ΔS is even larger than anticipated from the calculation in part (c). The
quality of the fit will be decreased since it will require essentially zero intenisty for the first several
quanta of a1g vibration, then a sharp peak occuring nearly 15 vibrational quanta below E0,0. The
slope of the peak intensity function cannot be very large for high number of quanta (ie there
cannot be a major difference between I0,15 and I0,16). Hence it will not be possible to fit the new
scenario to the luminescence spectra as closely.

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13. This problem deals with the absorption spectrum of the square planar complex Ni(CN)42−.

(a) The solution spectrum is composed mainly of charge transfer transitions and has been interpreted
by Gray and Ballhausen (JACS 1963, 83, 260-264) using an MO scheme that predicts the same
ordering of terms as does the ligand field theory for a strong tetragonal field.

What happens to this ordering if a weak ligand field treatment is used? Draw diagrams illus-
trating the effect of weak and strong tetragonal distortions on the energy levels and terms. The
d−d transitions and proposed assignments for a single crystal of BaNi(CN)4·4H2O are given below.

ν cm−1 Assignment

22,400 2 1A1g → 1A2g b2g → b1g
24,000 50 1A1g → 1B1g a1g → b1g
27,000 100 1A1g → 1Eg eg → b1g

Solution: We first consider the octahedral splitting then apply a weak tetragonal perturbation to

split the single electron t2g and eg levels. We know that in D4h, t2g → eg ⊕ t2g and eg → a1g ⊕ t1g.
Hence the ordering of the orbitals for a weak field treatment is eg < b2g < a1g < b1g since the
energy of the derivatives of the t2g set never cross the energy of the derivatives of the eg set. This
is illustrated in Figure 5-1 in Ballhausen. I have reproduced a similar figure below.

3d b1g
Free Ion b1g
eg
a1g

b2g

a1g
b2g
t2g
eg

eg

Octahedral Weak Tetragonal Strong Tetragonal

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(b) Determine the symmetries of the normal vibrations in the Ni(CN)24− unit. Determine the selection
rules for d − d transitions in the σ and π polarizations.

Solution: Conveniently we note that we already found the symmetries of vibration for an analo-
gous situation for Problem 8. We consider local x, y, and z coordinate systems on each of the C
and N atoms. The characters of interest are exactly twice he characters from Problem 8.

D4h E 2C4 C2 2C2 2C2 i 2S4 σh 2σv 2σd
Γtot 24 0 0 -4 0 0 0 8 4 0
Γrot 3 1 -1 -1 -1 3 1 -1 -1 -1
Γvib 21 -1 1 -3 1 -3 -1 9 5 1

Applying the great orthogonality theorem yields Γvib = 2A1g ⊕ A2g ⊕ 2B1g ⊕ 2B2g ⊕ Eg ⊕ 2A2u ⊕
2B2u ⊕ 4Eu.

We note that (x, y) transform as Eu while z transforms as A2u. The electronic states are all
g, so all transitions are Laporte forbidden. However, forbidden electronic transitions can couple
to the vibrational modes to become weakly allowed. This is summarized in the following tables.

σ Polarization (Eu):

Ground State Excited State Vibration
1 A1g 1 A2g
1 A1g 1B1g Eu
1 A1g 1 Eg Eu
A2u, B2u, Eu

π Polarization (A2u):

Ground State Excited State Vibration
1A1g 1A2g
1A1g 1 B1g Not Allowed (Requires A1u)
1A1g 1Eg B2u
Eu

(c) The square planar anions in crystals of tetracyanonickelate salts typically stack to form a nearly

linear chain of nickel atoms. This allows for the determination of the absorption spectrum of the
crystal perpendicular and parallel to the stacking axis. The band at 23,000 cm−1 is observed

only in the π polarization. This indicates that the D4h selection rules deduced in part (b) may
not be appropriate. Read the paper by Ballhausen, Bjerrum, Dingle, Eriks, and Hare (Inorg.

Chem. 1965, 4, 514-518). Here cogent arguments are advanced to the effect that the D4h excited
states 1B2g and 1Eg are unstable with respect to distortion to the D2d states 1B2 and 1E1 in
D4h symmetry. The B2u vibration will transform a D4h molecule into a D2d molecule. [Note:
There seems to be a discrepancy. The Ballhausen paper says that the B1u vibration transforms
D4h into D2d. I will use Ballhausen’s symmetry. Further Note: I believe the discrepancy comes
from the fact that Ballhausen’s ligands lie off the x, y axes while Gray’s lie on the axes. This
interchanges b2 with b1. Nevertheless, my method has been self consistent.] The 23,000 cm−1
band must therefore terminate in a 1B1g⊗ ? excited state. Fill in the “?” and explain the
answer. Work out selection rules for the 1A1g → 1B2 and 1A1g → 1E transitions. Observe the
symmetry change. Finally draw qualitative PE curves for the 1A1g, 1A2g, 1B2, and 1E terms on a
plot of energy vs. Q. Given the appearance of the 27,000 cm−1 band is this assignment convincing?

Solution: We want the excited state to be allowed by parallel light when coupled to a B1u
vibration but not allowed by perpendicular light when coupled to the same B1u vibration. We

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