CHM 501

Homework 4

Due May 2, 2016

1. Mo(pyr)2(CO)4 (pyr = pyridine) has two isomers. Predict the number of IR allowed CO stretching vibrations for each isomer. Based on this, would IR spectroscopy be a useful method to distinguish the two isomers? Why or why not?

The two isomers have the pyr cis or trans to each other:

For the D4h isomer the CO stretching vibrations transform as a1g + b1g + eu. Only the eu degenerate pair are allowed transitions.

For the C2v isomer the CO stretching vibrations transform as 2a1 + b1 + b2. All four are allowed transitions.

IR spectroscopy can easily distinguish the two isomers: the trans isomer will have a single peak in the 1700 - 2000 cm–1 region of the spectrum while the cis isomer will have 4 peaks in this region.

2. CoCl2 reacts with DMSO to give the compound [Co(DMSO)6][CoCl4], which is dark blue and has a magnetic moment of 4.6 μB. Account for the color and magnetic moment. Does the O or the S bond to the cobalt? Use a Lewis structure to justify your answer.

[Co(DMSO)6][CoCl4] has the Co2+ (d7) in two different environments: the cation is Oh and the anion is Td. The Td anion is high spin, accounting for 3 unpaired electrons, and has a low energy (in the red region of the spectrum) allowed transition accounting for the blue color. The observed magnetic moment suggests a total of four unpair electrons, so the cation must be a low spin complex (implying DMSO is a strong field ligand), adding one additional spin to the total. Since the octahedral complex will have only forbidden transitions (small molar absorption coefficients), it does not contribute significantly to the color of the molecular product.

There are two low energy resonance structures: Although both the O and S atoms have lone pairs, the resonance structure shown with a charge on the O atom will be the preferred site to bond to a cationic metal.

3. One synthetic method to prepare substituted chromium carbonyl complexes is photochemical:

Cr(CO)6 + L + hν → Cr(CO)5L + CO

Explain why this is an effective synthetic method using an appropriate bonding theory.

Cr(CO)6 is Cr0, d6. The low oxidation state suggests that a molecular orbital approach to understanding the bonding in this compound is more appropriate, so the ground state electron configuration is t2g6 (nb6). Upon excitation, the electron configuration becomes t2g5 (nb5)eg1 (σ*1). Population of the antibonding orbital weakens the Cr-CO bond so that one CO becomes a good leaving group and can be replaced by L.

4. [Cr(ox)3]3– has an absorption spectrum with the following bands and molar absorption coefficients:

λmax (cm–1)

ε (L·M–1cm–1)











Find 10Dq and B. Hint: Be sure your response is consistent with the Tanabe-Sugano diagram.

[Cr(ox)3]3– has Cr in the +3 oxidation state and is a d3 ion. The two lowest spin-allowed transitions are 4A2g4T2g and 4A2g4T1g. The tentative assignments for these two transitions are 17,500 cm–1 and 23,700 cm–1, respectively. Using the formulas provided in class, this should give 10Dq = 17,500 cm–1 and 3B = [(23,700 – 2×17,500)(23,700 – 17,500)]/(5×23,700 – 9×17,500) = 17,960 so B = 599 cm–1. This looks too small for B, since the free ion value for Cr3+ is 1030 cm–1. To compare this to the Tanabe-Sugano diagram, use Dq/B = 1750/599 = 2.9, E/B = 17500/599 = 29 and E/B = 23700/599 = 40. These are consistent with the Tanabe-Sugano diagram for both the spin-allowed and spin-forbidden transitions: this assignment predicts a closely spaced pair of spin-forbidden transitions at lower energy than the lowest spin-allowed transition and a third spin-forbidden transition at just slightly higer energy than the lowest energy spin-allowed transition.

So 10Dq = 17500 cm–1 and B = 599 cm–1 and the assignment of all of the transitions then become: 14,500 cm–1, 4A2g2Eg; 15,300 cm–1, 4A2g2T1g; 17,500 cm–1, 4A2g4T2g; 20,700 cm–1, 4A2g2T2g; and 23,700 cm–1, 4A2g4T1g.