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Chem 310 Fall 1999Solutions to Problem Set 4

1) How many microstates are possible for the following configurations:

(i) the ground configuration of Ti ?2+

(ii) the ground configuration of Co ?3+

Remember the meaning of C = (n!)/[r!(n-r)!] for the number of possible combinations. Inn r

the case of Ti we have a d configuration, and the number of microstates is (10!)/[(8!)( 2!)]2+ 2

= 45. Co is a d ion but is easier to visualize it in terms of the number of holes rather3+ 6

than the number of electrons, i.e., a d is equivalent to a d , and the possible combinations6 4

are (10!)/[(6!)(4!)] = 210.

2) Identify the ground state term symbols for Fe , Ni , Ti , Mn . 2+ 2+ 2+ 2+

OK, here we are using are knowledge of Hund’s Rules, i.e.,

1. The ground state will have the highest multiplicity (be high spin).2. For states of the same multiplicity, that with the higher L value will ne lower in

energy.3. For given L and S values, that with the lowest J value [J = *L+S*, ... , *L-S*] is the

lowest energy. For shells that are greater than half-filled, this latter ordering isreversed.

The idea then is to set out an array of the possible m values, slap the electrons into the boxesl

so as to generate the maximum M (= sum of the m ’s) with the maximum M (i.e., hold theL l s

electrons spin up). This creates a microstate which is the “wing tip” of the state that fulfilsrules 1 and 2. For the record the total degeneracy of the state (number of microstates withthe same energy) is (2S+1)(2L+1). Let’s do it...

Fe = d2+ 6

This is a greater than half-filled shell, we could treat it as a d system. Either way, the M4L

value is 2 and the M value also is 2. This is the “wing tip” microstate of a L = 2, S = 2S

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state, i.e., a D state, for which the multiplicity is 5 × 5 = 25. In other words the5

microstate identified is one of 25 twin siblings (energetically speaking). The possible Jstates, which arise from spin orbit coupling, are J = 4, 3,2,1,0. Because this is a greaterthan half-filled shell, the J = 4 state is the lowest in energy.

Ni = d2+ 8

This again is a greater than half-filled shell, we could treat it as a d system. Either way,2

the M value is 3 and the M value also is 1. This is the “wing tip” microstate of a L =L S

3, S = 1 state, i.e., a F state, for which the multiplicity is 3 × 7 = 21. In other words3

the microstate identified above is one 21 twin siblings (energetically speaking). The possibleJ states, which arise from spin orbit coupling, are J = 4, 3,2. Because this is a greater thanhalf-filled shell, the J = 4 state is the lowest in energy.

Ti = d2+ 2

Been there, done it. This is the same as the d ion above. The M value is 3 and the M8L S

value is 1. This is the “wing tip” microstate of a L = 3, S = 1 state, i.e., a F state, for3

which the multiplicity is 3 × 7 = 21. The possible J states, are J = 4, 3,2. Because thisis a less than half-filled shell, the J = 2 state is the lowest in energy.

Mn = d2+ 5

This is the classic high spin log jam. The M value is 0 and the M value also is 5/2. ThisL S

is the “wing tip” microstate of a L = 0, S = 5/2 state, i.e., a S state, for which the6

multiplicity is 6 × 1 = 6. As for the possible J states, which arise from spin orbit coupling,there is only one, with J = 5/2.

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3) Identify the ground term symbols from each of the following sets of terms.

(a) G, F, P, P1 3 3 1

(b) H, P, D, I, G3 5 1 1

(c) S, G, P, I6 4 4 2

Specify the spin and orbital angular momenta (L and S) for the ground state, and also the possible totalangular momentum values J. Specify lowest spin orbit component (assuming a > half-filled band), and alsothe degeneracy (number of microstates for the lowest spin orbit state).

Answer this question by inspection of the Term Symbol. Remember what the symbol istelling you. The prefix tells you the multiplicity (the total spin angular momentum S), themain letter the total orbital angular momentum L. The state with the lowest L and lowestS is the ground state. For a greater then half filled configuration the largest J component isthe lowest in energy. Hence:

(a) F ground state, with J = 4, 3, 2. J = 4 is the lowest state. The degeneracy of any3

J-state is 2J+1. In this case, the degeneracy is 9.

(b) D ground state, with J = 4, 3, 2, 1, 0. J = 4 is the lowest state. The degeneracy5

of any J-state is 2J+1. In this case, the degeneracy is 9.

(c) S ground state, with J = 5/2. J = 5/2 is the ground spin orbit state with a6

degeneracy of 6 microstates.

4) Close analysis of the electronic spectrum of Ni(H O) reveals absorption maxima at 8600, 13,500 and2 62+

25,300 cm . There are also two extremely weak bands at 15,400 and 18,400 cm . Consult the appropriate-1 -1

Tanabe Sugano diagram and assign all these transitions. Estimate 10Dq and B.

If you look at the Tanabe Sugano diagram for a d ion in the Appendix 5 of Schriver you’ll8

see the expected features - a A ground state and three spin allowed T excited states. All3 3

these we built in class. You’ll also see A and E terms trucking across almost parallel to1 1

the ground state, i.e., not dependent on 10Dq. The two weak bands correspond to these spinforbidden transitions, which amount to spin flips of electrons within the eg set. As you wouldexpect, the energies if these transitions should be largely independent of the value of 10Dq.

3A3T ,3T 3T 1A 1E

spin allowed spin forbiddenGroundState

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As for calculating 10Dq, remember that for a d ion the lowest frequency band gives 10Dq8

directly. Thus 10Dq = 8600 cm . You can get at B by numerical or graphical methods.-1

The ratio of < /< is 13.5/8.6, or about 1.57. Tracking on the TS diagram for the x-2 1

coordinate corresponding to this ratio in the two low energy T states yields a value of3

10Dq/B near 10. Hence, B is about 860 cm (compared to 1003 cm for gas phase Ni ).-1 -1 2+

5) Consult your class notes for the relationships between the excitation frequencies and the 10D and B valuesq

of a d ion. Use Jorgensen's f, g, h, k constants to predict the spin allowed transitions in the UV-visible3

spectrum of Cr(NH ) . The ruby laser contains Cr ions in a Oh environment surrounded by oxide ions.3 63+ 3+

Explain the excitation and emission processes on which the laser action hinges. Is the value of 10 Dqimportant?

The following data applies:

Cr g = 17.0, h = 0.21 NH f = 1.25, k = 1.43+3

B for Cr free ion = 1030 cm3+ -1

Thus 10Dq = (g × f) × 10 cm = 1.25 ×17.0 × 10 = 21,250 cm 3 -1 3 -1

and B = B (1- k.h) = 1030 (1 - 0.294) = 727 cmfree ion-1

4T1g

4T2g

4A2g

2Eg

2Eg

4T1g

4A2g

4T2g

Cr3+ (weak field) Cr3+ (medium field)

luminescence(with stokes shift)

phosphorescence

intersystem crossing

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You could calculate the three frequencies using the formulas in the class notes or, moresimply, read the frequencies off the Tanabe Sugano diagram. I took the latter route.

10Dq/B = 29.2

< = 28B = 20 ×10 cm (at best my eyes are good to 2 sig. figs)13 -1

< = 47B = 34 ×10 cm 13 -1

< = 61B = 24 ×10 cm 33 -1

The critical feature of the electronic structure of ruby crystals is the proximity of the doubletand quartet manifolds. The ligand field is such that they are right beside one another (about10Dq/B = 20). A spin-orbit driven intersystem crossing is most efficient under thesecircumstances, so that one can really build up a population inversion in the E state. 2

6S

B*

F*

B5T5E

Cr2+ (d to d)

Metal d-orbitals

ligand orbitals

CrO4- (LMCT)

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6) Salts of Mn(II) in aqueous solution are almost colourless. This we have ascribed to the fact that in a highspin d ion all transitions are spin-forbidden. That doesn't mean they don't occur, it just means that they will5

be very weak. What causes the apparent slight breakdown of the spin selection rule? The spectrum of atypical Mn(II) complex is provided in class notes. You will note that the lines are much sharper than thoseobserved for most other complexes. Why is this so?

The spin selection rules breaks down as spin-orbit couplingcauses an intersystem crossing - a change in multiplicity.Typically, these effects are much more important in the heaviermetals and lanthanides. The spectrum of Mn exhibits a series2+

of weak but sharp lines which arise because of spin turnovers,in which electrons are flipped inside the t and e sets rather2g g

than excited between them. These transitions result in very littlestructural change (they are not t to e ) during the excitation, so2g g

that the so-called Franck-Condon envelope (remember CHEM129?) is very narrow.

7) Solutions of [Cr(H O) ] ions are pale blue-green, but the chromate ion [CrO ] is a very intense yellow.2 62+

4-

Characterize the origins of the expected transitions and explain the relative intensities.

In Cr we are dealing with Laporte forbidden d-d transitions, while in the chromate ion the2+

transitions are ligand to metal charge transfer bands, which are very intense.

10

20

30

40

50

60

70

80

90

10 20 30

1

2

3

4

5

4F

E/B

10Dq/B

4T 4T ,4T 4A

Excited StatesHigh SpinGround State

Low SpinGround State

2E

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8) Shown below is a Tanabe-Sugano diagram.

(i) For what octahedral first row M ion is this figure2+

applicable?(ii) What is the physical significance of the dashed

vertical line?(iii) Indicate the term symbols of the states designated

by the numbers 1 to 5. (iv) Determine (using Jorgensen’s constants) whether

the M(NH ) complex will be high or low spin. 3 62+

This is a Sherlock Holmes type problem. A 2+ metal ion with a F ground state. That has4

to be d or a d . It has to be the latter, as there is a huge spin cross-over line (the dashed3 7

vertical line) in the middle. So that makes it Co . Cranking the handle as we have done2+

before the excited states for such a system (if high spin) in Oh symmetry are:

The excited states 2,3, and 4 above correspond to those shown at left. State 5 is the high spinground state, i.e., T, while state 1 is the low spin E ground state. 4 2

2Eg

4T1g

4A2g

4T2g

Cr3+ (very strong field)

phosphorescence

intersystem crossing

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Finally the synthesis of the electronic spectrum via Jorgensen parameters follows directlyfrom problem 5. With Co and ammonia as a ligand,2+

g = 9.3, k = 0.24, f =1.25, h = 1.4, B = 970 cmfree ion-1

This gives 10Dq = (g × f )× 10 cm = 11,625 cm and B = 970 (1 - 0.336) = 644 cm3 -1 -1 -1

Hence 10Dq/B = 18.0, i.e., an x-coordinate to the high spin side of the spin cross over line.

9) The complex [Cr(CN) ] has absorption bands at 264, 310 and 378 nm. Estimate ) and B using a Tanabe6 03-

Sugano diagram. First, decide how to assign the observed transitions, noting that the ligand has empty Borbitals. The complex also luminesces at 725 nm. Is this fluorescence or phosphorescence? (Hint. Checkthe full Tanabe Sugano diagram in Schriver for spin forbidden transitions.)

What you have to do here is haul out (again!) Appendix 5 of Schriver, in order to examinethe Tanabe Sugano diagram for a d metal ion. The lowest energy transition gives 10Dq3

directly (26,460 cm ), and from the ratio of the first and second bands (1.22) we can lock-1

in approximately where on the x-coordinate thesystem lies. I found it easier and quicker to estimateB from Jorgensen parameters k and h, and I got 597cm , a low value in keeping with cyanide’s good-1

ability to “cloud expand”.

One can then estimate 10Dq/B to be near 44 B,i.e., way out on the strong field side of the TanabeSugano diagram, as expected. The state diagramwill thus look like this:

The luminescence we are looking at is at very muchlower energy than the excitations, and results froman ISC (intersystem crossing) from the quartet todoublet manifolds. This is phosphorescence.

h<

B*

F*

B

M L

Metal d-orbitals

ligand orbitals

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10) The intense purple colour of the permanganate ion consists of two transitions at 18,500 and 32,200 cm .-1

Explain how you could estimate 10Dq for MnO from this information.4-

Mn(VII) is d complex so we are clearly dealing with LMCT bands from the oxygen lone0

pairs to the empty t and e sets. One can get a handle on 10Dq nonetheless simply by2g g

measuring the difference in the two lowest energy transitions, as shown below. For thissystem 10Dq = 32,000 - 18,500 = 13,500 cm .-1