Chm 363 Spring 2017, Exercise Set 3 Transition Metal Bonding and Spectra Mr. Linck Version 1.5 March 9, 2017 3.1 Transition Metal Bonding in Octahedral Compounds How do the metal 3d, 4s, and 4p orbitals transform under O h symmetry? 3.2 Transition Metal σ Bonding in Octahedral Compounds Build an mo diagram for bonding between a metal ion and six NH 3 ligands. HINTS: The ammonia orbital is an orbital pointing more or less toward the metal; that is, acts as an s or a p in. Also, the ammonia orbitals are below the metal orbitals. Finally, metal 3d orbitals are below 4s and 4p in building mo diagrams. 3.3 Transition Metal π Bonding in Octahedral Compounds For ligands such as F, there are additional orbitals on the ligand that we need to be concerned with. These are the p orbitals, two per fluorine. Find the irreducible representations generated by these twelve orbitals and reduce it. 3.4 Transition Metal π Bonding in Octahedral Compounds Make a mo diagram using only the metal orbitals and the p of the fluorine atoms. 3.5 Transition Metal Bonding in Octahedral Compounds Make the complete mo diagram for bonding of six fluoride ions with a metal. 3.6 Crude Spectral Analysis to Appreciate Bonding in TMC We will deal about spectra more later, but generally speaking one of the causes of color in transition metal ions involves transfer of electrons from the t 2g level to the e g level. In Figure 1 are the low energy spectra of CrF6 3, Cr(H 2 O) 6 3+, and Cr(NH 3 ) 6 3+, from right to left. Use the mo model to account for these data. Be complete. 3.7 Review Build an mo diagram for CN. HINTS: Assume the N 2s orbital is too stable to contribute. Assume sp hybridization on the C. 3.8 Review What is the HOMO of CN? What is the LUMO? 3.9 Review Where are the electrons that are in the π bond of CN? Where is the π* orbital of CN big?
3.17 2 Figure 1: Low Energy Spectra of CrL 6 with L = NH 3, H 2 O, and F, left to right. 3.10 Transition Metal Bonding in Octahedral Compounds Build an mo diagram for Cr(CN) 6 3. HINTS: (1) The last problem suggests the only orbitals of importance on the cyanide ion are the sp hybrid on C pointing away from the N and the π* orbitals of the CN. (2) The π* orbitals of the CN are rather high in energy. 3.11 Transition Metal Bonding in Octahedral Compounds What is the nature (bonding, antibonding, nonbonding) of the t 2g levels in Cr(NH 3 ) 3+ In CrF6 3? In Cr(CN) 6 3 3.12 Transition Metal Bonding in Octahedral Compounds What is the nature (bonding, antibonding, nonbonding) of the e g levels in Cr(NH 3 ) 6 3+? In CrF6 3? In Cr(CN) 6 3 3.13 Spectra in Transition Metal Compounds What should happen to the lowest energy transition as you move from Cr(NH 3 ) 6 3+ to CrF6 3 to Cr(CN) 6 3? 3.14 Spectra in Transition Metal Compounds The energies of the transition in Rh(en) 3 3+ is 34,600 cm 1 and that is Rh(CN) 6 3 is 44,900 cm 1. Do the relative positions make sense? Comment. 3.15 Spectra for One Electron Atoms Compounds with only one electron (or only one electron outside a filled core) have particularly simple spectra since the only energy term of concern is the orbital energy of the electron before and after the transition. Within the transition metals, what compounds fit this criterion? Give some examples. 6?
3.22 3 3.16 Multielectronic Ion Spectra What other energy term might be important in a multielectronic ion? 3.17 Electron-Electron Repulsion Do you know of any case we studied previously in this course where electron-electron repulsion was critical to the analysis? 3.18 Reduction of Spherical Irreducible Representations The RS state that we assigned to a d 1 configuration was 2 D. Verify that this is true. The angular momentum symbol, D is really just the name of an inrreducible representation in spherical symmetry. Spherical symmetry is an extreme example of an infinite group. Not only are there an infinite number of C 2 rotations (all in the same class), as there are in the linear groups, but there are also an infinite number of C 3 rotations (all in the same class), and, an infinite number of C 15.7 rotations. Nonetheless, a representation in spherical symmetry is reducible when a descent in symmetry occurs. Here are the characters of the D representation in spherical symmetry for certain rotations. What are the characters for the rotations of the O point group (O h with the consideration of only proper rotations) for this state? E C 2 C 3 C 4 C 5 C 6 C 7 C 8 D 5 1-1 -1 0 1 Cos(3π/14) Sin(π/7) Cos(π/8) Sin(π/8) Given our experience, are you surprised with this result? 3.19 Splitting of States The result from the last question means that the 2 D state of a d 1 configuration in octahedral geometry no longer is five fold degenerate (in a L sense), but splits into a set of three and a set of two, corresponding to the states 2 T 2g and 2 E g and configurations t 1 2g and e1 g. It is convenient to assume a center of gravity splitting, which means the sum of all orbital energies is equal to zero. For the d 1 system, this means three go down and two go up. If the total splitting between the two is said to be, find how far down in energy the three orbitals are and how far up in energy the two orbitals are such that the center of gravity rule holds. 3.20 Splitting of States If the result of the last paragraph holds, show that a plot of energy of state versus have slopes of -2/5 and +3/5. Make such a plot. 3.21 Reduction of Spherical Irreducible Representations There is a general formula for getting the characters of the irreducible representations in spherical symmetry for rotations (all we end up needing). It is (called by me the sin rule ): χ red (O) = Sin[L + 1 2 ]α Sin[α/2] (1) where O is a rotation of α radians and L is the value of the angular momentum quantum number of concern. Apply the sin rule to a P state in octahedral symmetry.
3.34 4 3.22 The g/u Issue If the P state of the last problem arises from d orbitals, is the answer g or u? 3.23 Reduction of Spherical Irreducible Representations Apply the sin rule to a P state in D 4h symmetry. 3.24 The g/u Issue If the P state of the last problem arises from p orbitals, is the answer g or u? 3.25 Reduction of Spherical Irreducible Representations Apply the sin rule to a F state from d orbitals in D 4h symmetry. 3.26 Review What are the states of a d 2 configuration? Why do they differ in energy? 3.27 Splitting of RS States How does the 3 P state of a d 2 configuration split in an octahedral field of ligands? 3.28 Splitting of RS States How does the 3 F state of a d 2 configuration split in an octahedral field of ligands? 3.29 Splitting of RS States List all the states of triplet nature that arise from a d 2 configuration split in an octahedral field of ligands. 3.30 The Strong Field Situation In a very strong field, our name for a large gap between the t 2g and e g electrons, where the splitting of the levels is much more important than electron-electron repulsion, we can identify that nature of the states. Then we can correlate the states between the strong field case and the weak field case. To start, put two electrons into e g levels, but maintain a triplet by making S z = 1. What is the energy of this state compared to any other possible configuration? 3.31 The Strong Field Situation How many ways can you put two identical electrons into the e g levels, but maintain a triplet by making S z = 1. Which of the states you listed in problem 29 is consistent with this level of degeneracy? 3.32 Strong Field Assignment Could we say that the configuration e 2 g, the least stable state, belongs to the 3 A 2g state? What is the L degeneracy of this state? 3.33 Possible Number of States How many ways can you put one electron in a t 2g orbital and the other in a e g one while maintaining S z = 1?
3.40 5 Figure 2: Weak Field and Strong Field Situations for d 2 Configuration. 3.34 Two Electron Wave Functions and Symmetry If we have an electron described by ψ 1 and a second described by ψ 2, then the two electron wave function, in order to satisfy the probablity requirements of wave functions, must be a product function: Ψ = ψ 1 ψ 2 If we know the symmetry of each of the one electron wave functions, can we get the symmetry of the product wave function? Have we ever done such a thing before? 3.35 Strong Field Assignment What is the symmetry of the product wave function for the configuration t 1 2g e1 g? 3.36 Strong Field Assignment Relative to the other possible configurations, what is the energy of the t 1 2g e1 g configuration? Could we say that there are two states belonging to the t 1 2g e1 g configuration, a 3 T 1g and a 3 T 2g? Does this match the number of ways you determined above? 3.37 Strong Field Assignment Which state from our list of states from problem 29 have we not accounted for? Is the degeneracy of this state consistent with the degeneracy expected from the t 2 2g configuration (providing, as usual, that you maintain S z = 1)? 3.38 Strong Field Assignment What is the relative energy of this last state? 3.39 The Correlation Using Symmetry Our situation is illustrated in Figure 2. Now we correlate the two sides. There are two steps in each correlation. First, we go to the strong field limit and ask what the slope of a plot of energy of level versus would be; this depends on the configuration. What would the slope be for the 3 A 2g state?
3.45 6 Figure 3: Weak Field and Strong Field Correlated for the 3 A 2g State. 3.40 The Correlation Using Symmetry The only place the 3 A 2g state can come from is the 3 F state. There is no other state of 3 A 2g symmetry and hence there is no mixing of this state with any other. The line is straight. Figure 3 shows this. Repeat this argument for the 3 T 2g state. What do you get? 3.41 The Correlation Using Symmetry The situation with the 3 T 1g states is slightly different as there are two of those that can mix. What is the slope of the most stable 3 T 1g state? Since it is most stable, it is likely to largely arise from the lower RS state, the 3 F state. Then the upper 3 T 1g state will correlate with the 3 P state. At any rate, these two states cannot cross as they have the same symmetry and would interact, The complete diagram is show in Figure 4. 3.42 Splitting of States, a Review Use the sin rule to show how a G state splits in octahedral symmetry. HINT: Consider only proper operations and use the fact that this G states comes from a d configuration to assess the gerade versus ungerade. 3.43 Use of Equation (1) Use the sin rule to determine how an F state from d orbitals will split in D 6h symmetry. HINT: See the last problem. 3.44 Building a Correlation Diagram Build a state correlation diagram (like Figure 4, also called a Tanabe-Sugano diagram, for an octahedral d 9 system. Make your slopes approximately correct. 3.45 Features of Correlation Diagrams In the spherical ion, a d 4 system has a 3 H excited state. Show that one component of this in an octahedral field has a slope sufficiently negative to eventually become more negative than the steepest declining state from the 5 D ground electronic state. What is the degeneracy
3.50 7 Figure 4: Weak Field and Strong Field Correlated for All states of the d 2 Configuration. of the steeply declining state? HINT: Remember to apply in the strong field limit the most stable state consistent with S z = 1; and that for the 5 D state the strong field limit must be applied consistent with S z = 2. 3.46 Interpretation of Spectral Data Increased pressure on a crystal of ruby (Cr(III) replacing some of the Al(III) in Al 2 O 3 ) causes various transitions (assuming, incorrectly, O h symmetry) to change in position. The 4 A 2g to 4 T 2g transition changes from 18020 cm 1 to 19880 cm 1 as the pressure goes from 1 to 350000 atm; The 4 A 2g to 2 E g transition changes from 14400 cm 1 to 13600 cm 1 as the pressure goes from 1 to 1Matm. Explain. 3.47 Magnetic Considerations When a compound has net spin, it also has a magnetic moment that is proportional to the length of the spin angular momentum vector, S(S + 1). The proportionality constant is about 2. Thus such a compound interacts with a magnetic field. In the simplest model, the strength of the magnetic moment (in units of the Bohr magneton) is given by the formula: µ = 2 S(S + 1) = 2 n/2(n/2 + 1) = n(n + 2) where we used the property that S = n/2 where n is the number of unpaired electrons (in states of maximum S z ). What happens to the magnetic moment of the compound in the last problem when the cross-over point is reached? 3.48 Magnetic Properties There is no difference between a high-spin and a low-spin d 8 system in octahedral symmetry. Is there a difference in D 4h symmetry? 3.49 Magnetic Moment Predict the magnetic moment for the following: FeF6 3, Fe(CN) 6 4, Mn(H 2 O) 6 2+, VF 3 Cu(NH 3 ) 6 2+. 6,
3.52 8 Figure 5: Very Approximate Graphical Representation of Molar Absorptivities as a Function of Various Restrictions. 3.50 The Intensity Issue Many of you have taken a colorless solution of Fe(H 2 O) 6 3+ and added NCS to produce an intense blood red solution. Maybe the wavelength of absorbance has changed. However, in this particular case, the major change is in the intensity of the absorbance, in how allowed the transition is. To get an absorbance two conditions are needed: (1) There must be a match between the energy of the photon, E = hν, and the energy gap between the two levels. (2) There must be a mechanism to get the energy from the photon into the molecule. The latter is the issue of allowedness. There are two main rules for determining allowedness. The first concerns whether or not the spin of the state changes. The second concerns a symmetry restriction that is most often used in terms of the symmetry parameter under the i operation, expressed as the g g (forbidden) or g u (allowed) rule. These two rules are usually called the spin forbidden (allowed) and the LaPorte forbidden (allowed) rules. The allowedness is proportional to the integral: ψ 1 (x, y, z)ψ 2 where the (x, y, z) means any of these (or all of them, depending on the symmetry). Show using symmetry (direct product) that if ψ 1 and ψ 2 are g, then the integral must vanish. 3.51 The Intensity Issue In Figure 5 is a representation of the magnitude of the molar absorptivity as a function of the restrictions. The only classification that needs explaining if the no sym, but past g g. This refers to a situation where you there is strictly no symmetry restriction because there is no g or u in the molecule, but where the atom remembers that its orbitals are mostly d, and hence, mostly g. An 0.025M solution of Cr(NH 3 ) 6 3+ has an absorbance of 0.76 at 464nm in a one cm cell. What are the intensity rules applied to this this compound?
3.61 9 3.52 The Intensity Issue An 0.0063M solution of Cr(en) 3 3+ has an absorbance of 0.50 at 464nm in a one cm cell. What is the molar absorptivity? Comment on the relative values of the wavelength and the molar absorptivity from this problem and the last one. 3.53 The Intensity Issue An 0.134M solution of Co(H 2 O) 6 2+ has an absorbance of 0.75 at 515nm. Find the molar absorptivity and comment on the intensity rules. 3.54 The Intensity Issue A 1.7 10 3 M solution of CoCl4 2 (tetrahedral) has an absorbance of 1.275 at 690 nm. Find the molar absorptivity and compare this result to the one in the last problem. 3.55 The Intensity Issue A 3.95 10 5 M solution of Fe(phen) 3 2+ has an absorbance of 0.40 at 520 nm. Find the molar absorptivity and comment on the intensity rules. 3.56 Charge Transfer Spectra Go back to problem 10. In that problem you had electrons in d orbitals (largely g ), and an empty orbital made from the π* orbitals of the cyanide, largely u. Would such a transition be allowed? How do you account for the data in the last problem using this argument? Such transitions are called charge transfer transitions since the charge moves from metal to ligand (or, in some cases, ligand to metal). 3.57 Charge Transfer Spectra The permanganate ion, MnO 4, has no d electrons. Yet it is a deep purple in color. What could be causing that color? Be specific. 3.58 Charge Transfer Specra Peaks are found in the spectrum of permanganate that are assigned to 1t 1 to 2e (at 15100 cm 1 ) and to 1t 1 to 3t 2 (at 29500 cm 1 ). Can you estimate a value for tetrahedral from these data? HINT: I certainly would draw an mo diagram first. Also, note the results have been modified to neglect the O 2s orbitals. 3.59 Charge Transfer Spectra Some data for the intense transition in Ru(NH 3 ) 5 pyx 2+, where pyx is a pyridine ligand substituted in the 4-position, are given in Table 1. Account for this data. HINT: Pay attention to your past knowledge about the relative rate of electrophilic substitution on substituted benzenes. 3.60 Intensities of Transitions In complexes of the type trans-ru(nh 3 ) 4 LL 2+, there are two peaks in absorption spectra (which is a t 2g to ligand transition) when L and L differ, but only one when L = L. Comment.
3.68 10 X Energy of Transition, cm 1 H 24600 CH 3 25100 NH 2 27900 Cl 23800 C(O)OH 20100 CN 19600 Table 1: Data for Spectra Peaks in Ru(NH 3 ) 5 pyx 2+ 3.61 Intensities of Polarized Transitions A crystal of [Et 4 N] 2 InCl 5 is doped with Mn(III) ion (i.e., some of the In(III) ions are replaced by Mn(III) ions) and studied with light both parallel and perpendicular to the z axis. In x,y polarization one peak is seen with two small shoulders at the low energy side. In z polarization, three weak bands are seen at similar positions. Account for this data given that the MnCl5 2 ion has C 4v symmetry. HINT: You can determine how a D state splits in C 4v symmetry because you know how d orbitals split in that symmetry. 3.62 Absorbance Band Width Think about an a 1g vibration of the ligands in an octahedral compound; sketch how the atoms would move. 3.63 Absorbance Band Width Remind yourself about the consequence of having an electron in a t 2g level. Depending on the nature of the ligand, this electron has what kind of bonding property? Would it be appropriate to put weakly in front of those words? 3.64 Absorbance Band Width Remind yourself about the consequnce of having an electron in an e g level. What kind of bonding? Relavtive to the t 2g, what kind of strength? 3.65 Absorbance Band Width Make a plot (on the same scale) for the energy of a t 2g level and the energy of an e g level as the molecule undergoes an a 1g vibration. You may have to think a little to do this ( Oh, no! ) but remember the last two problems. 3.66 Absorbance Band Width Becaues electrons move much more rapidly than nuclei, the transitions from the t 2g to the e g are strictly vertical in the plot from the last problem. At what energy do they occur? Can you predict that electronic transitions in transition metal compounds are broad? 3.67 Intensity The intensities of transition metal complex spectra usually increase at higher temperatures. Explain. HINT: Think about the last several problems with other vibrations, such as a t 1u one.
3.68 11 3.68 Using Properties to Determine Reality The mineral chalcopyrite has the formula CuFeS 2. You could formulate it as either a Cu(I) and Fe(III) compound or a Cu(II)/Fe(II) compound. What evidence could you use to distinguish these formulations? Could anything interfere with this analysis?