Cume 2000 Feb Strass Rhco4 ANSWERS

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Inorganic Cume (Prof. Stanley) ANSWER KEYFeb 19, 2000

Please read the attached JACS communication and answer the following questions about the paper. You

have 2 hours for this closed book exam.

1. (10 pts) Sketch out and electron count the following complexes: [(6-C6H6)Rh(CO)2]+ and [Rh(CO)4]+.

RhCOOC

Rh(+1) d8C6H6 6e-2CO 4e-Total: 18e-

C

RhOC

C

CO

O

O

Rh(+1) d84CO 8e-Total: 16e-

2. (20 pts) Free CO has an IR stretching frequency of 2143 cm1. How can a metal carbonyl have aCO >2143 cm1 ? Clearly discuss and explain.

The HOMO (highest occupied molecular orbital) on CO is mainly lone pair in nature and localized on the carbon atom.But this orbital is slightly antibonding with respect to the C-O bond. So when the CO bonds to a metal using only thislone pair HOMO it is draining away some of the electron density in this orbital. This decreases the amount of -antibonding character between the C and O, thus increasing the bond strength between them. As the bond strengthincreases, the CO stretching frequency will also increase. Normally when a carbonyl ligand binds to a metal centerthere is also -backbonding occurring (due to delectron density on the metal center) and this weakens the C-O bondconsiderably more than the mild strengthening effect caused by donation of the HOMO electron density. So normally,one does not see metal carbonylCO > 2143 cm

1 except when there is little or no -backbonding occurring.

3. (10 pts) [(6-C6H6)Rh(CO)2]+ has the highest IRCO = 2115 cm1, while [Rh(CO)4]+ has its highest IRCO = 2138 cm1. Discuss and explain.

[Rh(CO)4]+ is more electron deficient (less electron rich) relative to [(6-C6H6)Rh(CO)2]

+. This means that there willbe less -backbonding to weaken the CO bond. This will give rise to the highest CO stretching frequency in the IR.The C6H6 group is more donating and less able to -backbond relative to two CO ligands. Also the [(

6-C6H6)Rh(CO)2]

+ is an 18e- complex compared to the 16e- [Rh(CO)4]+. While one cant always directly compare

electron counts to determine which complex is more electron-rich or deficient, the similarity of the two complexes(same metal, same oxidation state) does allow us to do a semi-direct comparison.


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4. (10 pts) The average Rh-CO bond distance in [(6-C6H6)Rh(CO)2]+ is 1.88 , while [Rh(CO)4]+ has an

average Rh-CO bond distance of 1.95 . Is the longer Rh-CO distance in [Rh(CO)4]+ consistent with whatthe IR stretching frequencies are telling you? Discuss.

Yes, the longer Rh-CO bond distance in [Rh(CO)4]+ indicates that it has weaker Rh-CO bonding (but stronger C-O

triple bonding). This does correspond to the higher CO stretching frequency observed that indicates strong C-O triplebonding, and by inference, weaker Rh-C bonding. The more electron-rich [(6-C6H6)Rh(CO)2]

+ complex has more -backbonding that strengthens the Rh-C bond at the expense of the C-O bond.

5. (10 pts) Why is the low basicity of the superweak anion (last paragraph) important for these studies?Discuss and explain. What would happen with a normal counter anion like BF4?

The minimal amount of -backbonding in [Rh(CO)4]+ is indicative that the metal center is quite electron deficient.

Coupling this with the cationic charge means that the Rh center is probably fairly electrophillic and will be looking forelectron donors to satisfy its desire for better donating ligands. The superweak counteranion is important to

minimize the amount of electron donation from the anion to the Rh center. The authors are looking for the mostelectron deficient metal center that will give them the least amount of -backbonding and the highest CO stretchingfrequency in the IR. The more localized the anionic charge on the anion, the stronger the electrostatic attractionbetween the cationic Rh center and the anion. The closer the anion gets to the metal, in large part driven by theelectrostatic attraction, the more likely it will start donating electron density to the electrophillic metal center. A normalanion like BF4

(which is considered to be a non-coordinating anion) is more likely to get closer to the Rh center anddonate a little more electron density to the Rh making it less electron deficient (more electron-rich). This will lead tomore -backbonding to the COs and a lower CO stretching frequency.

6. (10 pts) The authors consider [Rh(CO)4]+ to be a non-classical carbonyl because it hasCO(avg) > 2143cm1. They state that the highest IRCO for [Rh(CO)4]+ is 2138 cm1, but thatCO(avg) = 2167 cm1 (firstparagraph). How did they arrive atCO(avg) = 2167 cm1 and is this a fair comparison to the IRCO forfree CO of 2143 cm1?

The authors get the 2167 cm1 average value by combining the high wavenumber Raman values (2215, 2176 cm 1)with the IR value (2138 cm1). The 2138 cm1 value is a doubly degenerate vibration (Eu symmetry) and they usedthis twice in their averaging. This, I believe, is CHEATING!! While the 2143 cm1 value for the stretching frequencyof free CO represents both an IR and Raman value, one should only compare apples to apples. Therefore, I believethat one should only be comparing similar IR values. The totally symmetric Raman CO stretching frequencies arenaturally going to be at a considerably higher energy due to the concerted nature of the symmetrical stretching norma

mode(s). The less symmetrical Eu symmetry IR vibration is probably a closer and fairer comparison to the free COvalue. But if they used the IR value of 2138 cm1 they couldnt claim that [Rh(CO)4]+ is a non-classical metalcarbonyl.


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7. (10 pts) The reaction of [(6-C6H6)Rh(CO)2]+ with CO in the solid-state produces [Rh(CO)4]+. Theauthors state that neither CpRh(CO)2 nor Cp*Rh(CO)2 reacts in this fashion. Discuss why.

A benzene coordinates far more weakly than a Cp or Cp* anion, so it is far easier to displace coordinated benzenefrom the Rh than the strongly bonding Cp anions. It is a little surprising that COs can displace the somewhat betterdonating benzene ligand. But the authors would also probably find that [Rh(CO)4]

+ will react with benzene to make[(6-C6H6)Rh(CO)2]

+ under similarly mild conditions. So this isnt that big a surprise.

8. (10 pts) The isoelectronic [Pd(CO)4]2+ complex mentioned in the last column of the paper has asignificantly higher CO IR stretching frequency (2248 cm1) compared to [Rh(CO)4]+ (2138 cm1).Is this expected or not? Discuss and compare the two complexes.

[Pd(CO)4]2+ should have quite a bit higher CO than [Rh(CO)4]

+ for two reasons: Pd is more electronegative thanRh, and the Pd complex is dicationic versus the monocationic Rh complex. The higher positive charge on the Pd

complex will contract and lower the energy of the Pd d orbitals (in concert with the higher electronegativity) and makethem considerably harder to -backbond to the CO ligands.

9. (10 pts) A structure on the [Pd(CO)4]2+ complex has not be done to my knowledge. Given that the average

Rh-CO bond distance in [Rh(CO)4]+ is 1.95 (and other information in the paper), what would you predictfor the Pd-CO bond distance? Discuss, explain and justify.

[Pd(CO)4]2+ should have shorter Pd-CO bond distances relative to [Rh(CO)4]

+ for two reasons: Pd is moreelectronegative than Rh, and the Pd complex is dicationic versus the monocationic Rh complex. Both of these willwork to contract the Pd d orbitals. This contraction means that the COs have to get closer to the Pd center to havesimilar overlap between its lone pair and the appropriate empty metal orbital. This should be the case even thoughthe Pd-CO bonding should be weaker than that seen for the Rh complex. It is reasonable to make the argument thatthe considerably higher CO stretching frequency for the Pd complex implies that there should be weaker bondingbetween the Pd and the CO ligands. Weaker bonding usually translates as longer M-ligand bond distances. But thisreally only works well when one is comparing the same metal in similar oxidation states. But the orbital contractiondiscussed above is probably a major factor and should result in a shorter Pd-CO bond distance. I would estimate itaround 1.9 .


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Cume 2000 Feb Strass Rhco4 ANSWERS

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