Chemistry 312 Problem Set 5 (due Friday, August 5)
1. I stated in lecture that, for octahedral octahe dral symmetry molecules, the ligand atom p orbitals oriented perpendicular to the metal-ligand bond would belong to the symmetry species T1g, T1u, T2g, and T2u. Using drawings or reducible representations, show that this is the case. You will have to choose your phases wisely. Using reducible representations: Oh "reducible
E 12
8 C3 0
6 C2 0
6 C4 0
3 C2 –4
i 0
6 S4 0
8 S6 0
3 !h 0
3 !d 0
which decomposes to T1g + T2g + T1u + T2u. In order to get the reducible representation characters, I used drawings like the following for determining the 3C2 character:
3+
3+
2. Diamagnetic complexes of cobalt (III) like [Co(NH3)6] and [Co(en)3] are orange3– yellow, while paramagnetic cobalt (III) complexes like [CoF6] and [Co(H2O)3F3] are blue. Explain, using abridged MO diagrams, this difference in color. All these complexes are octahedral. The orange-yellow complexes appear that color because they absorb (maximally) in the purple-violet region of the EM spectrum, whereas the blue complexes absorb in the orange. This means that orange-yellow complexes have a higher #O than the blue complexes, which makes sense because the diamagnetic complexes the energy separation between the t2g and eg* orbitals is great enough that the electrons will overcome repulsion and pair in the t2g orbitals before they populate the eg* level. See diagram below.
3. Predict which complex in each of the following pairs will have the lower energy d–d transition, and give a short reason for your choice: a. [Co(NH3)5Cl]2+ or [Co(NH3)5 NO2]2+ 2+
–
–
[Co(NH3)5Cl] because Cl is a lower field strength ion than NO2 , according to the spectrochemical series. This means that the crystal (ligand) field splitting energy, #O, is lower for Cl – than for NO2 – , making the d-d transition itself lower energy. 3–
3–
b. [Co(CN)6] or [Ir(CN) 6] 3–
[Co(CN)6] for very much the similar reason as part a; according to the second list on 3+ 3+ page 475 of the text, Co is a lower field strength ion than Ir , so the cobalt ion has the lower #O. 2+
4. a. The complex [Cu(H2O)6] ought to be octahedral; however, measurements of the bond lengths of the Cu–O bonds reveal that four of them, oriented along the x and y axes, have a bond length of 280 pm, and the other two, oriented along the z-axis, have a bond length of 318 pm. Use the octahedral MO diagram as a starting point, and show how that MO diagram changes to accommodate the new point group of the distorted molecule. Hint: there should be significant orbital changes. The new point group is D4h, which is going to be similar to the square planar complexes.
b. Why is this distortion (the relative lengthening of the z-axis oriented bonds) is required for this molecule? Hint: as always, it will be an energy-minimizing reason. The t2g orbitals in the Oh symmetry split into the eg and b1g orbitals under D4h symmetry; the eg* orbitals split into the a1g* and b1g* orbitals. When the orbitals split, they use the original degenerate energy level (t2g or eg*) as the barycenter of the energies of the separate orbitals. For instance, the energy that the a1g* orbital is lower than the original eg* orbital energy level by the same amount that the b1g* orbital is higher. Thus, for filled (octahedral) orbitals, there is no energy difference so there is no advantage for the molecule to distort into D4h symmetry. However, for the octahedral eg* orbitals (which were not completely filled), note that instead of having three electrons at the same energy level, there are now two lower and one higher than originally – even this small energy difference will make a particular structure more energetically stable. This is called the Jahn-Teller effect. c. Explain, using part b, why [Cu(en)3] a bidentate ligand.
2+
cannot be made. (en) = NH3CH2CH2 NH3 and is
Since (en) is a bidentate ligand, the complex is constrained; in other words, the z-axis Cu–O bonds cannot lengthen without untethering one of the (en) lone pairs and thus destroying the complex. d. In some copper complexes, it is the x and y axis-oriented metal-ligand bonds that are longer than the z axis-oriented metal-ligand bonds. Would the MO diagram for such a
distorted molecule be different than the one you created for part a? If so, show the new energy order of orbitals. See the abbreviated MO diagram below. The point is that the orbitals would still follow the rules of crystal field theory in avoiding electron-electron repulsion; for instance, 2 because the z-axis bonds are shorter relative to the x and y-axis bonds, the dz orbital would have more repulsion than the dx2 –y2 orbital and thus will have a higher energy. The point of this is that whether the z-axis bonds lengthen or the x and y-axis bonds lengthen will depend on the number of electrons populating these orbitals. The difference between this picture and the one with the z-axis bonds longer is the reversal of the e g and b1g orbitals. In the case below, the b1g orbital is twice as low (compared to the t2g energy level from where it came) as the eg orbitals are high, which means that one electron populating these orbitals will be lower energy than if it was populating these same levels in part a.
5. Consider a metal-ligand complex in which ethylene (C2H4) is the ligand; recall that the ethylene ligand is oriented such that the C=C bond is perpendicular to the line from the metal center to the center of the C=C bond. a. Draw the orbital(s) responsible for the ethylene ligand’s donation of electron density to a 3d metal center.
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!"#
#!"
b. Draw the orbital(s) responsible for the metal’s donation of electron density to the ethylene ligand.
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c. Using the Dewar-Chatt-Duncanson Model, explain whether you would expect (CO)5Cr(H2C=CH2) or (CO)3(PH3)2Cr(H2C=CH2) to have a longer C=C bond in the ethylene ligand. & ! &!
& !
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! ! '(! &
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The phosphine complex would be expected to most activate the ethylene bond. Carbonyl ligands are pi-acceptors and therefore reduce the electron density at the metal center available for back-donation to the side-on ethylene ligand. Phosphines, in contrast, are not significant pi-acceptors, and thereby increase electron density at the metal center that can be donated to the ethylene ligand’s anti-bonding orbital.
