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 formallyforbidden 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 allowedbecause the triple product A2g ⊗ T1u ⊗ T2u = A1g ⊕ Eg ⊕ T1g ⊕ T2g contains the totallysymmetric representation. Thus we expect ε ≈ 10, 000M−1cm−1 with an oscillator strengthof f ≈ 1.

ii. 4A2g → 4T2g is formally orbitally-forbidden because A2g⊗T1u⊗T2g has u parity and thereforedoes not contain a totally symmetric component. Thus we expect ε ≈ 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 ε ≈ 100 with f ≈ 0.01. Thistransition 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 ε ≈ 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 wouldalso 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 ε ≈ 1 and f ≈ 10−4,

but if comparing the intensities of (iv) and (v) we note that (iv) should be weaker as it isdoubly 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 tojustify 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 weconsider 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 thatwe call Γtot where by total we mean rotational plus vibrational. The rotations (Γrot) transformas T1g based on a character table. Subtracting these traces out we can obtain the traces of thevibrational representation, Γvib.

Oh E 8C3 6C2 6C4 3C2(=C24) i 6S4 8S6 3σh 6σd

Γtot 18 0 0 2 -2 0 0 0 4 2Γrot 3 0 -1 1 -1 3 1 0 -1 -1Γvib 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 theintensity of the band. The Eg and T2g vibrations could couple with the 4A2g → 4T2u electronictransition to increase the intensity.

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

(a) Summarize the atomic selection rules. The absorption of a free Cr(III) ion shows a weak band atabout 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 spinorthogonality is close to rigorously true. Thus the weak band at 14,000 cm−1 must come froman exitation from the ground 4F ground term to an exited quartet term. There is only one suchtransition in a d3 free ion configuration, so we attribute the 14,000 cm−1 band to the 4F →4 Ptransition. This transition is formally disallowed as ΔL = −2, but the dipole approximation oflight breaks down leading to the weak band.

(b) The electronic spectra of four Cr(III) complexes, CrLn−6 , are summarized below. Assign the bands

(assuming they are all allowed transitions). Estimate Dq and B values from the Tanabe-Suganodiagram. Use these values to calculate the positions of each band for each of the complexes. Drawan 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 ν(cm−1)H2O 17,400 24,500 38,000F− 14,900 22,700 34,400

Urea 16,150 22,600 34,400CN− 26,600 32,500 ?

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 withrespect to Dq and B. There is some ambiguity in our assignment of Dq and B since there aretwo parameters and three data points. We use the low energy band to solve for Dq. We then usethat value to solve for B based on the high and low energy T1g bands. The two different valuesof B are averaged to give:

L Dq(cm−1) B(cm−1)H2O 1,740 686F− 1,490 843

Urea 1,615 562CN− 2,660 529

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

L 4A2g → 4T2g4A2g → 4T1g

4A2g → 4T1g

H2O 17,400 24,300 38,200F− 14,900 22,400 34,900

Urea 16,150 21,950 34,950CN− 26,600 32,500 55,250

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

−40000

−30000

−20000

−10000

0

10000

20000

Ene

rgy

(cm

−1)

F− Urea H2O CN−

4A2g

4T2g

4T1g

4T1g

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) andcomment critically on the significance of this two photon experiment.

The two-photon experiment of Krause et. al. provides insight into the structure of the doubletmanifold 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 fromthe 2Eg state into higher energy doublet states. A spectrum taken without the initial flash isused for background subtraction to eliminate the quartet excitations. There are two aspects ofthe experiment that seem a little concerning. First, there is a shoulder at 25,000 cm−1, whichcould not be explained by a doublet-doublet transition, but matches the 4A2g → 4T1g transitionexactly. 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 energydoublet states. This assumption demands justification.

(d) Read the paper by Witzke (Theoret. Chim. Acta. 1971, 20, 171-184). This is typical of effortsaimed 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 approximationfor the free ion, but for an ion in the presence of a ligand field the ratio of C/B is not constantand not necessarily ≈ 4. The constant ratio would only hold if the radial functions were notcomplex-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 valuefor the spin-allowed transitions compared to the spin-forbidden transitions. The paper starts bydiscussing simple methodology (ie crude approximations). The method we used in part (b) ofthis problem is employed as the simplest method for predicting spin-allowed transitions. Forspin-forbidden transitions the Tanabe-Sugano matrices (electrostatic matrices in the strong-fieldbasis set) were diagonalized numerically and the values of B and C which matched the observedabsorptions most closely were used. If B and C held rigorous in these situations then there shouldbe a value of B and C that will predict each possible transition, but in general the curves for C

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as a funciton of B do not converge on a point. This is a consequence of the fact that B and Ccan only make sense as a linear combination of F2 and F4 which are integrals of d orbitals. In thebonded compound, those orbitals change shape, demanding that a full molecular orbital approachbe employed to actually predict the energy levels. Nevertheless, Witzke’s paper demonstratesthat the rough 3 parameter (Δ, B, C) approach works fairly well.

(e) The absorption spectra of Cr(en)3+3 and of the cis and trans isomers of Cr(en)2(H2O)3+2 areshown below. Assign the observed bands and identify which spectra correspond to the cis andtrans isomers. Explain.

Figure 2: Absorption Spectra of Cr(en)3+3 (—–), cis-Cr(en)2(H2O)3+2 (- - - -), and trans-Cr(en)2(H2O)3+2(— —).

The assignment of the Cr(en)3+3 spectra is straightforward. The low energy peak at ≈ 22, 000cm−1 is the 4A2g → 4T2g transition while the higher energy peak at ≈ 27, 500 cm−1 is the4A2g → 4T1g transition.

The Cr(en)2(H2O)3+2 are more complicated. Section 5-c of Ballhausen treats the cis and transproblem. 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 tetragonalcomponent. It is shown that a trans complex will exhibit twice the tetragonal splitting as a ciscompound, so the broadened peak (— —) is taken to be the trans complex. In D4h symmetry wemust reassign these peaks with respect to the new point group. The Oh

4A2g → 4T1g transitionsplits into: 4B1g → 4Eg (19, 500 cm−1), 4B1g → 4A2g (22, 500 cm−1). Splitting of the Oh4A2g → 4T2g transition is not as pronounced in the spectrum, but some broadening is observedas 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 theexcited states will switch (ie 4Eg will lie above 4A2g).

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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 topopulate the (eg) set with as few electrons as possible. Hence low-spin d7 complexes are moststable for large splitting. If the splitting is small, then the cost of spin-pairing may exceed the costof 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 geometriesmight you expect and how will the ground states be split? Comment on the relative magnitudesof the splittings.

In the low spin case there is a single electron occupying the degenerate eg set so Jahn-Tellersplitting is expected. The splitting will be stong since the eg set points directly at the octahedralligands and therefore distortions can significantly affect the overlap between eg orbitals and themolecular orbitals. In the high spin case we expect Jahn-Teller distortion because there is a singlevacancy in the triply degenerate t2g set. The splitting of this t2g set will be significantly smallerthan the low spin J-T splitting since distortions in geometry cannot strongly affect ligand-metalorbital 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 symmetrysplits into 2A1g ⊕2 B1g. Because dz2 is lower in energy than dx2−y2 it is clear that the 2A1g stateis 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 theC4 axis. Rather, the elongation would be along the C3 axis to give D3d symmetry. In that casethe 4T1g ground state would be reduced to 4A′

2 ⊕4 E′′ with 4A′2 lying lower in energy.

(c) Shown below are absorption spectra for Co(H2O)2+6 and CoCl2−4 in aqueous solution. Identifywhich spectrum belongs to which compound. Explain.

Figure 3: Co(H2O)2+6 and CoCl2−4 in aqueous solution

Co(H2O)2+6 is the unstructured peak on the left because the peak is at higher energy. ΔO = 94ΔT ,

so the lower energy peak is the tetrahedral complex: CoCl2−4 .

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(d) The absorption spectra for two crystals of KCoF3 of different thickness at approximately 150 Kare shown below. What are the advantages of working at low-temperature and using crystallinesamples 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 forDq, 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. Thesesamples are also adventageous since they eliminate coupling with the solvent and reorganizationeffects.

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 smallerdegeneracy and the 4T1g and 4T2g states.

Because E(4A2g) = 12Dq − 43B + 14C and E(4T2g) = 2Dq − 43B + 14C, 10Dq can be obtainedas the difference between the position of the 4T1g → 4A2g band and the 4T1g → 4T2g band.Hence Dq = 7800 cm−1. Also the the energy difference between the two 4T1g states shouldbe√

5√

45B2 + 36 ·Dq · B + 20Dq2. Setting this equal to 19, 000 cm−1 and solving for B givesB = 880 cm−1. Finally, to find C we must use the 2Eg band at 10, 000 cm−1 and use:

E(2Eg)− E(4T1g) = −15Dq +32B + 3C +

√5

2

√45B2 + 36 ·Dq · B + 20Dq2

⇒ C = 3640 cm−1

The small value of Dq is a result of the π-donations of F−1 ions, which makes the ligand fieldweaker. The values of B and C are quite typical. The structure in the peak at 19, 000 cm−1 isdue to different vibronically excited modes. I was puzzled by additional weak featers at 17, 500cm−1 and 21, 500 cm−1. I consulted the original paper, which suggested that these peaks weredue to other doublet excitations.

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4. The visible spectrum of VCl4−6 is strongly temperature dependent (see below). The positions of thepeak maxima at 77 K are 7, 200 cm−1, 11, 700 cm−1, and 19, 100 cm−1.

Figure 5: Visible spectrum of VCl4−6

(a) Assign the bands in the spectrum.

VCl4−6 is a d3 compound with a 4A2g ground state. The band at 7, 200 cm−1 is due to thespin-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 throughvibronic 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 lessvibrational energy, so the electronic energy must increase in order to compensate.

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5. The ability to predict the relative intensities of electronic transitions available to molecular systemsoften proves to be a useful tool when assigning an absorption spectrum. Selection rules derived fromquantum 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 wavefunctionsand approximations used to derive them. This problem will deal with the derivation of selection rulesfor a hydrogen atom in a beam of plane polarized light. The methodology used here can be extendedto molecular systems as well.

A development of the equations describing the interaction of an atomic or molecular system withthe time-dependent electric and magnetic fields of an incident light beam usually beings with Fermi’sgolden rule:

Poj =2π�2

∣∣∣⟨j ∣∣∣G∣∣∣ o⟩∣∣∣2 φoj (1)

By solving Maxwell’s equations along with the classical Hamiltonian for a charged particle moving inan electric and magnetic field it may be shown that the time-dependent perturbation found in (1) canbe written as:

G =∑

η

i�Qη

Mηc�Aη · ∇η (2)

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

�Aη = A(ω)�uei ωc

�k·�r +A(ω)�ue−i ωc

�k·�r (3)

Here �k is the wave vector pointing in the direction of the propoagation of the light wave and �u is theunit vector pointing in the direction of the oscillating electric field. If the wavefunctions j and o aretaken to be Born-Oppenheimer functions, then the sum over nuclear coordinates in (2) reduces to asum over electronic coordinates multiplied by some Frank-Condon factors. Using this idea and theseequations we have:

Poj =2π�2

∣∣∣∣∣⟨j

∣∣∣∣∣∑η

Aω�uei ω

c�k·�r · ∇η

∣∣∣∣∣ o⟩∣∣∣∣∣

2

φoj (4)

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

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

Poj ≈ 2π�2

∣∣∣∣∣⟨j

∣∣∣∣∣∑η

Aω�u · ∇η

∣∣∣∣∣ o⟩

+

⟨j

∣∣∣∣∣∑η

Aω�u(iω

c�k · �r) · ∇η

∣∣∣∣∣ o⟩∣∣∣∣∣

2

φoj (5)

but �k = z and �u = x, so the sums over the single hydrogen atom’s electronic coordinates collapses

Poj ≈ 2π�2

∣∣∣∣⟨j ∣∣∣∣Aω∂

∂x

∣∣∣∣ o⟩ +⟨j

∣∣∣∣Aω∂

∂x(iω

c�z)

∣∣∣∣ o⟩∣∣∣∣2 φoj (6)

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

[H,x] = −�2

m

∂

∂x

9

⟨nlm

∣∣∣∣Aω∂

∂x

∣∣∣∣n′l′m′⟩

= −mAω

�2〈nlm |[H,x]|nlm〉

= −mAω

�2〈nlm |Hx− xH |n′l′m′〉

= −mAω

�2(En − En′) 〈nlm |x|nlm〉

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

〈nlm |x|nl′m′〉 �= 0

We note that

x = r

√2

2

√4π3

(Y −11 − Y 1

1 )

Hence,

〈nlm |x|nl′m′〉 = r

√2

2

√4π3

(⟨nlm

∣∣Y −11

∣∣nl′m′⟩− ⟨nlm

∣∣Y 11

∣∣nl′m′⟩) (7)

We recall that |nlm〉 = Rn,l(r)Y ml (θ, φ) = NRn,l(r)Pm

l (cos θ)eimφ. From the exponential terms weimmediately see that: ∫

Y lmY

±11 Y l′

m′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 theinversion for a direct prodcut is only 1 if l+ l′ is odd. We also note that Y l

m belongs to the irrep of thefull rotation group, Γl while Y ±1

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 selectionrule for x-polarized light.

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

[H,xz] = −�2

m

(x∂

∂z+ z

∂

∂x

)allowing us to write the second matrix element as:⟨

j

∣∣∣∣Aω∂

∂x(iω

c�z)

∣∣∣∣ o⟩ =iωAω

2c

⟨j

∣∣∣∣(z ∂∂x + x∂

∂z

)+

(z∂

∂x− x ∂

∂z

)∣∣∣∣ o⟩=

iωAω

2c

⟨j

∣∣∣∣(−m�2[H,xy]

)+

(z∂

∂x− x ∂

∂z

)∣∣∣∣ o⟩=

iωAω

2c

⟨j∣∣∣(−m

�2[H,xy]

)∣∣∣ o⟩ +iωAω

2c

⟨j

∣∣∣∣(z ∂∂x − x ∂∂z)∣∣∣∣ o⟩ (8)

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

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the matrix element of an angular momentum Jy, which is linearly proportional to the magnetic dipoleoperator. 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 ywith respect to the spherical harmonics:

y = −r√

22

√4π3

(Y 1

1 + Y −11

)⇒ xy = −2π

3((Y 1

1 )2 + (Y −11 )2

)When we square Y ±1

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 thisrequires that l = l′. Re-expressing the matrix element with respect to the shift operators gives:

< nlm|12(J+ + J−)|nlm′ >

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

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6. The Fe atom in the protein rubredoxin is thought to reside in a roughly tetrahedral environmentconsisting of four mercaptide S atoms. The UV/Vis spectra of ferric and ferrous rubredoxin aredominated by charge transfer bands. The d − d bands were located in the near IR by Eaton andLovenberg (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 theligand 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 bandshows up using CD confirms the assignment. Without the CD evidence, the feature could beconfused 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 bandin ferrous rubredoxin indicates that there is actually an electronic transition at 6250 cm−1. Thisband 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 lessthat Td. Assuming a distortion leading to D2d symmetry can occur, how will the d orbitals splitin this lower symmetry environment? How many excited states would you expect? A transitionhas 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 orbitaland the a1 symmetry z2 orbital. The t2 set splits into the b2 symmetry xy orbital and the esymmetry (yz, xz) orbitals. The ground state is a 5A1 state with the b1, a1, b2, and e sets allsingly occupied with alligned spins. A spin allowed excitation requires a promotion of an electroninto 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 secondtransition is essential the energy of the splitting of the t2 set from tetrahedral splitting. It isconceivable 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 Mossbauerexperiments have revealed the splitting of the tetrahedral t2 set.

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

Complex ν1 (cm−1) ν2 (cm−1)Co(H2O)3+6 17,000 25,000Co(NH3)3+6 21,000 29,000Co(CN)3−6 32,100 38,500

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

1A1g → 3A1g;1A1g → 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,ε = 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-fieldenergy expressions may be found in Table A29 of Griffith (ignore configuration interactions). Finallydraw an energy level diagram illustrating how the three configurations (t62g; t

52geg; t42ge

2g) 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 − 24DqE(1T1g) = 15A− 30B + 14C − 14DqE(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−ν116 . We assume C/B = 4 to obtain:

110

[ν1 + 9

(ν2 − ν1

16

)]= Dq

Hence we obtain

Complex ν1 (cm−1) ν2 (cm−1) Dq (cm−1) B (cm−1)Co(H2O)3+6 17,000 25,000 2150 500Co(NH3)3+6 21,000 29,000 2550 500Co(CN)3−6 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 fromthe t32ge

3g 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 diagonalmatrix element of the lowest energy strong field configuration

E(3T2g; t52ge1g) = 15A− 22B + 12C − 14Dq

SimilarlyE(5T2g; t42ge

2g) = 15A− 35B + 7C − 4Dq

From these energies we simply obtain the transitions energies:

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

So

Complex 1A1g → 3A1g1A1g → 3T2g

1A1g → 5T2g

Co(H2O)3+6 34,500 17,000 22,000Co(NH3)3+6 42,500 21,000 30,000Co(CN)3−6 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 → 1T1g1A1g → 1T2g

1A1g → 3A1g1A1g → 3T2g

1A1g → 5T2g1A1g → 3T1g

Co(H2O)3+6 17,000 25,000 27,250 13,375 7,500 9375Co(NH3)3+6 21,000 29,000 35,250 17,375 15,500 13,375Co(CN)3−6 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)3+6 , the 1A1g → 3T2g

band is predicted to be at 17,375 cm−1. The ε for this band would be very small though, making itexceedingly difficult to see.

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

14

1A1g

3T1g

1T2g

1T1g3T2g

5T2g

t62g

t52ge1g

t42ge2g

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 answerthe 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 intoa1g, eg, b2g, and b1g sets corresponding to orbitals which are dominanted by dz2 , (dxz , dyz), dxy, anddx2−y2 orbitals respectively. The paper shows that the ordering of these orbitals from low to highenergy are as listed. The paper also indicates that a high-spin ground state is observed. Hence theground 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 configurationbehaves 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 asimple exercise to show that

D4h E 2C4 C2 2C′2 2C′′

2 i 2S4 σh 2σv 2σd

Γtot 12 0 0 -2 0 0 0 4 2 0Γrot 3 1 -1 -1 -1 3 1 -1 -1 -1Γvib 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 dipoletransforms 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 transitionis allowed iff X ⊗ A2u ⊗ Y = A1g where X is the excited state and Y is the representation ofthe 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 wouldhave 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 requirea B2u vibration, which is present.

Next we consider the perpendicular transitions, which make use of x and y polarized light thattransforms 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 couldcouple 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 toEu 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 transitionmetal copmlex with an unfilled d shell generally occurs from the lowest electronic excited state in themolecule or from those states that can achieve a signficiant Boltzmann population relative to the lowestexcited 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 togetherby potential energy surface considerations. Background material may be found in N.J. Turro ModernMolecular 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 ofthe 4T2g term. From this point, two main photophysical processes occur, namely fluorescenceor intersystem crossing (ISC) into the 2Eg term from which phosphorescence is observed. Thisemission is not Stokes shifted; it overlaps the very weak corresponding absorption band. Thisleads 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 ofQ, so it is only for the vertical transition. We approximate the quartet terms as all having thesame width, and the doublet term is slightly wider.

4A2g

2Eg

4T2g

4T1g

Dq/B < 2.1

Q

E

4A2g

2Eg

4T2g

4T1g

Dq/B = 2.1

Q

E

4A2g

2Eg

4T2g

4T1g

Dq/B > 2.1

Q

E

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.

νmax(4A2g → 4T2g) νmax(4A2g → 2Eg) ν0-0(phos) νmax(fluor)CrCl3−6 13,060 14,480 – 11,600CrF3−

6 14,900 15,700 – 12,830Cr(urea)3+6 16,150 14,350 14,240 12,550

Cr(oxalate)3−3 20,800 ? 13,420 –Cr(NH3)3+6 17,500 14,350 14,390 –Cr(phen)3+3 21,550 15,300 15,120 –Cr(CN)3−6 23,800 13,700 13,720 –Cr(CH3)3−6 26,600 12,470 12,430 –

Solution: As we read down the table we move through the spectrochemical series with increasingDq. As Dq increases, we move more into the strong-field case depicted by the third potentialenergy diagram. In this situation the 2Eg surface drops in energy with increasing Dq as seenby the second column of the table. Also, inner-system crossing becomes more favorable. Thefirst potential energy diagram reveals an energy barrier moving from the 4T2g to 2Eg surface, butfor 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 intersystemcrossing into the 2Eg state. This justifies why phosphorescence is seen rather than fluorescencein 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 leveldiagram as a function of pressure illustrating this situation.

Solution: In essence, Kenney et. al. report that they can take a weak field Cr(III) coupoundand make it strong field under high pressure since the ligands will be pushed closer to the Crcenter. In the strong field limit the 2Eg state is lower in energy than the 4T2g excited state, sophosphorescence replaces the usual weak-field fluorescence. Figure 1 in their paper displays theemission 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 ofthe 4A2g, 4T2g, and 2Eg states from a normal Tanabe-Sugano diagram. There is some unknownscaling factor between Dq and P , but the general situation is captured by the spin-crossover ofthe excited state in the following diagram. (Obviously the plot is not accurate near P = 0)

0 2 4 6 8 10 12 14 16 18 200

10

20

30

40

50

60

Pressure (GPa)

E/B

4A2g

2Eg

4T2g

19

10. Substitution reactions of low-spin Co(III) complexes generally proceed very slowly. Co(H2O)3+6 is astriking exception. Taube and coworkers suggested that the high-spin term 5T2g should be substitutionlabile and thermally accessible. In order to ascertain whether this is reasonable, the energy gap betweenthe 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 whichthe 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 theenergy of this minima that determines the Boltzmann factor. In Problem 7 I already solved forthis vertical transition energy.

EFC = 20Dq − 18B − 6C

(b) Experimentally observed splittings in the electronic origin of the 3T1g term may be used to estimatethe 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

slope 3T1g

]2

=Sa1g (5T2g)Sa1g (3T1g)

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

S =K(ΔQ)2

2�ω

20

The Tanabe-Sugano slope of the quintet term is twice that of the triplet term. It is then assumedthat the force constant for the 5T2g totally symmetric vibration is K∗ = αK where K is theground state value and α varies from 0 to 1. Locating the quintet surface now reduces to aproblem in analytical geometry.

E(1A1g) =12KQ2

E(5T2g = E0 +12K∗(Q−Q0)2

EFC for the 1A1g → 5T2g transition is just E(5T2g) evaluatied at Q = 0. Substituting into the ex-pression for EFC the parametersDq, B, and C, what is the separation of the minima of the groundstate and the first quintet state potential energy surfaces? Let �ω = 357 cm−1. The reflectancespectrum 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 t52ge1g 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(5T2g) = E0 +12K∗Q2

0 = 20Dq − 18B − 6C

⇒ E0 = 20Dq − 18B − 6C − �ωSa1g(5T2g)

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

Plugging in the given values yieldsE0 = 3296 cm−1

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

Q =EFC√

2K(EFC − E0)

and the intersection energy is

Eact =E2

FC

4(EFC − E0)= 3300 cm−1 = 9.4kcal/mol

21

11. From analysis of the emission spectrum of K3Co(CN)6, Hipps and Crosby concluded that the maximumof 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 parameterC from the free ion value of 5120 cm−1? Explain. If not, at approximately what energy wouldyou 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 Hippsand Crosby this implies that C = 5900 cm−1. Hence the band position is not consistent withthe expected reduction in the Racah parameter C from its free ion value. I would expect C tobe 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 spectrumin this context.

The fine structure on the spectra is spaced by 414 cm−1 corresponding to increasing quanta ofthe a1 vibrations.

(c) Discuss the significance of the 1A1g → 3T1g transition with regard to molecular geometry andligand field parameters Dq, B, and C.

The excited triplet state adopts a differnt molecular geometry. We expect that the geometryis still essentially octahedral, but the average bond lengths increase. At longer bond lengths theoverlap 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 itsmolecular geometry changes. This means Dq, B, and C should be recalculated for the excitedstate.

22

(d) The absorption spectrum of an excited state offers an interesting view of the energy level patternof a molecule. The hexacyanocobaltate ion gives rise to a structured d − d transient absorptionspectrum obtained directly after laser excitation of the ground state system. The spectrum isshown 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 interactionon their energies. Interpret the spectrum.

We are interested in the low-lying triplet states. We will take low-lying to mean all of thetriplet states arising from t52ge

1g and t42ge

2g configurations. From Table A29 of Griffiths we can

explicitly construct the electrostatic matrices including configuration interaction between the low-lying states:

3T1g =

⎛⎜⎜⎝15A− 15B + 13C − 8Dq −10B 0 −3B

√2

−10B 15A− 23B + 11C − 8Dq 2B√

3 3B√

20 2B

√3 15A− 25B + 11C − 8Dq B

√6

−3B√

2 3B√

2 B√

6 15A− 30B + 12C − 18Dq

⎞⎟⎟⎠

3T2g =

⎛⎝15A− 27B + 11C − 8Dq 2B√

2 6B2B√

2 15A− 23B + 11C − 8Dq 3B√

26B 3B

√2 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 ofDq = 3425 cm−1, B = 400 cm−1, C = 3050 cm−1.

23

0

0.5

1

1.5

2

2.5

3

3.5

4

4.5

5x 10

4

No Configuration Interaction With Configuration Interaction(a)

Ene

rgy

(cm

−1)

3T1g

3T2g

3Eg

3A2g

3

3.05

3.1

3.15

3.2

3.25

3.3

3.35

3.4

3.45

3.5x 10

4

No Configuration Interaction With Configuration Interaction(b)

Ene

rgy

(cm

−1)

3T1g

3T2g

3Eg

3A2g

Figure 8: Relevant excited triplet states with an without configuration interaction. Energies are given withrespect 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 thousandwavenumbers. This is because the 3T1g state has a different value of Dq, B, and C. The changein Dq is most drastic. By setting Dq = 2475 cm−1 the following plots were obtained, which havemuch better agreement with the experiment.

0

0.5

1

1.5

2

2.5

3

3.5

4x 10

4

No Configuration Interaction With Configuration Interaction(a)

Ene

rgy

(cm

−1)

3T1g

3T2g

3Eg

3A2g

2.1

2.15

2.2

2.25

2.3

2.35

2.4

2.45

2.5

2.55

2.6

2.65x 10

4

No Configuration Interaction With Configuration Interaction(b)

Ene

rgy

(cm

−1)

3T1g

3T2g

3Eg

3A2g

Figure 9: Relevant excited triplet states with an without configuration interaction. Energies are given withrespect 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 and3T1g → 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. Myassignment differs slightly from the assignment I found in a JACS paper by Viaene et. al. Viaeneclumped the higher energy 3T1g → 3T1g transition in with the 3Eg transition to account for themiddle 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 vibronicintensity analysis yields important information such as the excited-state geometry and excited-stateforce constants. In this problem you will carry out a FC analysis of the hexacyanocobaltate emissionspectrum.

(a) Background material:i. C.J. Ballhausen Molecular Electronic Structures of Transition Metal Complexes, sec. 4.7ii. 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) andHenderson and co-workers (J.Chem.Phys. 1964, 41, 580). The vibronic emission intensities aregiven by:

Iν,ν∗ =643π4

h4cNνE

4ν,ν∗R2

e 〈ψν |ψν∗〉2

Here Eν,ν∗ is the energy of the electronic transition between vibrational states with the quantumnumbers ν and ν∗ for the excited and ground vibrational states respectively. Re is the averagevalue of the electronic transition dipole moment, and the integrals are the Frank-Condon overlapintegrals hereafter called Rν,ν∗ and N is the number of molecules in the initial state. In practicethe relative intensities can be related via the following equation.

I0,n

I0,0=

[E0,n

E0,0

]4 [R0,n

R0,0

]2

The frequency dependence for emission is given by:

E0,n = E0,0 − nhν0where E0,0 is the energy of the electronic transition between the lowest vibrational states of theexcited and ground electronic states, and ν0 is the eigenfrequency of the oscillator in the electronicground state. The most useful relations for the vibration overlap integrals are given by:

R0,0 =2δ√1 + δ

e−ρ2

2

R0,n+1 =−2DδR0,n −

√2n(δ2 − 1)R0,n−1

(1 + δ2)√

2(n+ 1)

δ =ν0ν∗0

D = CΔS√μν∗0

ρ =D√

1 + δ2

When the displacement ΔS of the minimum of the PE surface of the excited state along thestretching coordinate S is expressed in angstroms, the vibrational energy in wavenumbers and themasses 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 octahedralcomplex, the change in each metal ligand bond length is 1√

6ΔS. ΔS is normally determined by

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)16,65016,24415,85715,43915,04814,66014,26713,85213,462

The origin for this progression is located at 17,020 cm−1. You may assume ν0ν∗0

= 0.65. Calculatethe 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 thesecalculations. From a comparison of these distributions and the known spectrum determine thedistortion of the metal ligand bonds in the excited state.

We aim to develop an expression for I0,n.

I0,n = I0,0

[E0,n

E0,0

]4 [R0,n

R0,0

]2

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 anda carbon atom of the ligand. Using these values and the equations given in the problem a scriptwas written to generate plots for three values of S. Comparing these plots to the luminescence

0.8 1 1.2 1.4 1.6 1.8

x 104

0

0.2

0.4

0.6

0.8

1

Energy (cm−1)

Rel

ativ

eIn

tens

ity

S = 0.05

0.8 1 1.2 1.4 1.6 1.8

x 104

0

0.2

0.4

0.6

0.8

1

Energy (cm−1)

Rel

ativ

eIn

tens

ity

S = 0.3

0.8 1 1.2 1.4 1.6 1.8

x 104

0

0.2

0.4

0.6

0.8

1

Energy (cm−1)

Rel

ativ

eIn

tens

ity

S = 0.5

spectrum from Problem 11 shows that S = 0.3 is the best fit of the three options. Tuning thevalue of S a little more shows that S = 0.25 actually produces a slightly better fit. For the totallysymmetric stretch, ΔQ = 1√

6ΔS, so ΔQ ≈ 0.10 A.

(d) What does this distortion imply with respect to excited state substitution reactions of these kindsof complexes? You have fit the spectrum using the symmetric stretching mode; what other modewould you expect to be involved?

26

0.9 1 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8

x 104

0

0.2

0.4

0.6

0.8

1

Energy (cm−1)

Rel

ativ

eIn

tens

ity

S = 0.25

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 vibrationdoes 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 fromA1g 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 impliesthat the allowing vibration is T1u.

(e) Recall the feature around 26,000 cm−1 observed in the absorption spectrum of a thick crystalof K3Co(CN)6 (Problem 7). Estimate a value for the E0,0 of the emission band. How does thischange 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 ≈20,000cm−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). Thequality of the fit will be decreased since it will require essentially zero intenisty for the first severalquanta of a1g vibration, then a sharp peak occuring nearly 15 vibrational quanta below E0,0. Theslope of the peak intensity function cannot be very large for high number of quanta (ie therecannot be a major difference between I0,15 and I0,16). Hence it will not be possible to fit the newscenario 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)2−4 .

(a) The solution spectrum is composed mainly of charge transfer transitions and has been interpretedby Gray and Ballhausen (JACS 1963, 83, 260-264) using an MO scheme that predicts the sameordering 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. Thed−d transitions and proposed assignments for a single crystal of BaNi(CN)4·4H2O are given below.

ν cm−1 ε Assignment22,400 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 tosplit 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 theenergy of the derivatives of the t2g set never cross the energy of the derivatives of the eg set. Thisis illustrated in Figure 5-1 in Ballhausen. I have reproduced a similar figure below.

3d

Free Ion

eg

t2g

Octahedral

eg

b2g

b1g

a1g

Weak Tetragonal

b1g

b2g

a1g

eg

Strong Tetragonal

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(b) Determine the symmetries of the normal vibrations in the Ni(CN)2−4 unit. Determine the selectionrules 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 Cand N atoms. The characters of interest are exactly twice he characters from Problem 8.

D4h E 2C4 C2 2C′2 2C′′

2 i 2S4 σh 2σv 2σd

Γtot 24 0 0 -4 0 0 0 8 4 0Γrot 3 1 -1 -1 -1 3 1 -1 -1 -1Γvib 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 allg, so all transitions are Laporte forbidden. However, forbidden electronic transitions can coupleto the vibrational modes to become weakly allowed. This is summarized in the following tables.

σ Polarization (Eu):

Ground State Excited State Vibration1A1g

1A2g Eu1A1g

1B1g Eu1A1g

1Eg A2u, B2u, Eu

π Polarization (A2u):

Ground State Excited State Vibration1A1g

1A2g Not Allowed (Requires A1u)1A1g

1B1g B2u1A1g

1Eg Eu

(c) The square planar anions in crystals of tetracyanonickelate salts typically stack to form a nearlylinear chain of nickel atoms. This allows for the determination of the absorption spectrum of thecrystal perpendicular and parallel to the stacking axis. The band at 23,000 cm−1 is observedonly in the π polarization. This indicates that the D4h selection rules deduced in part (b) maynot 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 excitedstates 1B2g and 1Eg are unstable with respect to distortion to the D2d states 1B2 and 1E1 inD4h 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 transformsD4h into D2d. I will use Ballhausen’s symmetry. Further Note: I believe the discrepancy comesfrom the fact that Ballhausen’s ligands lie off the x, y axes while Gray’s lie on the axes. Thisinterchanges 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 theanswer. Work out selection rules for the 1A1g → 1B2 and 1A1g → 1E transitions. Observe thesymmetry change. Finally draw qualitative PE curves for the 1A1g, 1A2g, 1B2, and 1E terms on aplot 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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assume that the final electronic state is of the form 1B1g⊗ ? and the ground state is 1A1g. Forparallel (z polarized) light we require

1A1g ⊗A2u ⊗B1u ⊗1 B1g ⊗ ? = A1g

For perpendicular light we require

1A1g ⊗ Eu ⊗B1u ⊗1 B1g ⊗ ? �= A1g

Equivalently these can be written as A2g ⊗ ? = A1g and Eg ⊗ ? �= A1g. The second equationimplies that “?” is a one-dimensional representation and the first equation tells us that A2g = ? .Hence the excited electronic state is 1B2g in D4h symmetry but 1B2 in D2d symmetry. From thiswe have the fact that 1A1g → 1B2 is allowed by parallel light by not allowed by perpendicularlight (assuming a coupling to the B1u vibration). The 1A1g → 1E transition is allowed byperpendicular light only. The qualitative potential energy surfaces are given below and are largelyinspired by Figure 6 of the Ballhausen paper. The square planar configuration corresponds toQ = 0. As Q moves away from 0, the configuration becomes more tetrahedral, passing throughthe D2d point group. I do not find the assignment all that convincing based on the peak at 27,000

1A1g

1B2

1E

1A2g

Q

E

cm−1. The peak appears in both the parallel and perpendicular spectra. In order to appear inthe parallel spectra, it requires that a Eg vibration be excited along with the B1u vibration thattakes the molecule from D4h to D2d symmetry. I would expect a broad peak from the vibrationalprogression, but this is not seen as it was in the peak at 23,000 cm−1.

(d) From the band width at half max. and the band extinction coefficient one finds an oscillatorstrength of 5 x 10−6 for the transition at 22,400 cm−1. Based on the assignment given in part(a), is this transition allowed by a magnetic dipole mechanism?

Solution: The magnetic dipole operator acts as Rx, Ry, and Rz , which transform as Eg andA2g. We note that 1A1g ⊗ A2g ⊗1 A2g = A1g, so the parallel magnetic dipole mechanism couldallow the transition.

(e) The band at 22,400 cm−1 is only present in the spectrum of the barium salt and absent in spectraof the strontium and calcium salts. It has been proposed that this feature may arise from a

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spin-forbidden d− d transition. Does this explanation seem plausible? The band at 22,400 cm−1

is temperature dependent, red-shifting ( 300 cm−1) and broadening with decreasing temperaturefrom 295 to 80 K. When crystals are doped with Pt(CN)2−4 anions, the band also red-shifts ( 500cm−1) and broadens. Below 90% nickel, the feature is no longer resolved. Given the structuraldata summarized below, propose an alternative assignment for this transition. (Hint: considerthe report by Miskowski and co-workers (Inorg. Chem. 1994, 2799).)

Crystal Ni· · ·Ni distance C-Ni· · ·Ni-C TorsionSrNi(CN)4 · 5H2O 3.64 A 0◦

CaNi(CN)4 · 5H2O 3.39 A 27◦

BaNi(CN)4 · 4H2O 3.36 A 45◦

Solution: It seems conceivable that the band at 22,400 cm−1 could be a spin-forbidden d − dtransition since the heavy metals in Ba would lead to a greater degree of spin-orbit coupling.However, one would expect these effects to also influence the spectrum of the strontium salt,at least slightly. It is more likely that the transition is due to metal-metal intervalence chargetransfer. In the barium salt, the stacks of (roughly) square planar molecules are rotated by 45degrees from eachother. This allows the barium metals to pack closer together, allowing for morebonding between the metal atoms.

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