Chemistry 324 Midterm 1 KEY Wednesday, October 19, 2011 Instructor: D. J. Berg

Transcription

1 Chem 324 Midterm 1 Fall 2011 Version 1 Page 1 of 9 Chemistry 324 Midterm 1 KEY Wednesday, October 19, 2011 Instructor: D. J. Berg Name: Answer all questions on the paper (use the back if necessary). There are 50 minutes and 50 marks so ration your time accordingly. A periodic table is included with this examination. There are 9 pages including a blank page for more workspace (p. 8) and a periodic table (p. 9). 1. Draw the structures and give the metal oxidation state and d n electron count for each of the following compounds. (3 pts each) (a) trans, mer-[cr(dien)cl 2 (OH 2 )] + Ox. t. = 3+ d n = 3 + NH 2 HN OH 2 Cl Cr Cl NH 2 (b) [Pd(µ-Cl) 2 ] n Ox. t. = 2+ d n = 8 (c) [(CO) 3 Co] 2 (µ-η 2 :η 2 -PhC CPh) Ox. t. = 0 d n = 9 Ph OC OC Co OC Co CO CO CO Ph

2 Chem 324 Midterm 1 Fall 2011 Version 1 Page 2 of 9 2. omewhat surprisingly, the complex Zr(CH 3 ) 4 (dppe) shows only ONE type of metal-bonded methyl resonance (of relative integration 12) in the 1 H NMR in addition to the resonances for dppe. Furthermore, the spectrum is not affected by cooling to -100 C, by changes in complex concentration OR by addition of excess dppe. What do these observations suggest about the structure of this complex in solution? [6 pts] The facts indicate that this is a 6-coordinate structure at all temperatures: dppe is a bidentate phosphine and there is no temperature or concentration effects making it unlikely that dppe is dissociating. Fluxional 6-coordinate structures are unusual and require high temperature when they do occur. However, if the structure is 6-coordinate, there is no octahedral structure that makes all 4 methyls equivalent. Thus we need to consider other possible structures and the most common option would be a trigonal prism. If we place the dppe ligand so that one P is in each trigonal face then all 4 methyls are equivalent because there are two mirror planes that inter-relate all four. This is the most likely structure that fits the facts (mirror planes shown by dashed lines):

3 Chem 324 Midterm 1 Fall 2011 Version 1 Page 3 of 9 3. Draw the ligands given by the following names or abbreviations [2 pts each]: (a) ox (b) acac O O (c) R or -BINAP

4 Chem 324 Midterm 1 Fall 2011 Version 1 Page 4 of 9 4. The ligand CH 2 CH 2 NH 2 forms a complex of the type [(κ 2 - N,-CH 2 CH 2 NH 2 )()(OH 2 )] that does not conduct electricity in solution. This complex is found to exist as two optical isomers. What does this observation tell you about the structure? Draw the isomers to illustrate your answer. [6 pts] This is 2+ and 4-coordinate so the options are square planar or tetrahedral structures (note it does not conduct electricity so it is presumably an intact neutral complex as written). The question is, what does the observation of optical isomers tell us about the structure adopted? A square planar structure might be the first choice but if the square plane is also a mirror plane no optical isomers are possible. That will be the case if the ligands have a plane of symmetry as is the case here. On the other hand, a tetrahedral structure has no mirror plane and is chiral at the metal so optical isomers are possible. Thus we conclude that this complex adopts a tetrahedral structure. superimposable H 2 N H 2 N H 2 O H 2 N OH 2 H 2 O geometric isomers (dif f erent ligands trans to NH 2 ) NH HN OH 2 H 2 O not superimposable 5. What is the relationship between the following pairs of isomers? [2 pts each] (a) [CrCl 3 (OH 2 ) 3 ] 3 H 2 O and [Cr(OH 2 ) 6 ] 3+ 3Cl - Hydration or ionization isomers (b) [Cr(NH 3 ) 6 ] 3+ [Fe(CN) 6 ] 3- and [Fe(NH 3 ) 6 ] 3+ [Cr(CN) 6 ] 3- Coordination isomers

5 Chem 324 Midterm 1 Fall 2011 Version 1 Page 5 of 9 6. Consider the complex [Cr(OH 2 ) 6 ] 3+. (a) how what happens to the d orbital crystal field splitting when the z axis water ligands are both replaced by NH 3 ligands. Draw a labeled splitting diagram to illustrate your answer. [6 pts] NH 3 is a stronger field ligand than water so replacement of the z axis waters with NH 3 will cause greater repulsion in the z direction than in the (still equivalent) x and y. This splits the e g set with d x2-y2 lower than d z2 and it also splits the t 2g set with d xy lower than either d xz or d yz : (b) Discuss what would happen to this structure if the metal oxidation state was reduced by one electron. [3 pts] Adding an electron gives [Cr(OH 2 ) 6 ] 2+ which is a d 4 ion and subject to a Jahn-Teller distortion (either an axial compression or elongation).

6 Chem 324 Midterm 1 Fall 2011 Version 1 Page 6 of 9 (c) What is the magnitude of the crystal field stabilization energy for this ion in terms of Δ o? [2 pts] Cr 3+ d 3 so CFE = 3(-0.4Δ o ) = -1.2Δ o (d) How does this compare with the crystal field stabilization energy of [Mo(OH 2 ) 6 ] 3+ in absolute energy terms (i.e. not in terms of Δ o )? [2 pts] This ion will also have three electrons in the t 2g level and a relative CFE of -1.2Δ o, however Δ o is much larger for the 2 nd row of the d block so the CFE in absolute energy terms will be much larger for Mo 3+ than for Cr 3+.

7 Chem 324 Midterm 1 Fall 2011 Version 1 Page 7 of 9 7. Illustrate, using a molecular orbital approach, why a π-donor like an alkoxide ( - OR) decreases Δ o in a d 6 complex. [6 pts] This question comes directly from section 6 of the notes. The key point is that the π levels of the ligand are filled, low in energy and of the correct symmetry to overlap with the t 2g metal orbitals which house the metal d electrons. ince the ligand π levels are filled, whatever electrons there are from the metal will now fill the t 2g * level and this is closer to the e g * level meaning that Δ o is less than in the case of pure σ-donor ligands.

8 Chem 324 Midterm 1 Fall 2011 Version 1 Page 8 of 9 (extra page for answers) END (Periodic Table follows)

9 Chem 324 Midterm 1 Fall 2011 Version 1 Page 9 of 9