100 points on 4 pages + a useful page 5

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Gerald Robinson
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1 /21 Name Chemistry 350 Spring 2012 Exam #3, April 27, minutes CCM 100 points on 4 pages a useful page 5 1. n discussing solid-state materials, we discussed band theory. Draw a depiction of filling of bands for the following types of compounds. Be sure to label the band gap (with approximate energy) and bands appropriately. (9 pts) a) an insulator vb = valence band cb=conduction band E = band gap > 4eV b) a conductor E = 0 ev c) semi-conductor E < 4eV 2. A little potpourri of transition metal questions like those on the problem set (12 pts) a) Determine the number of valence electrons on the central metal atom in the following complex using either of the methods we learned about in class. You must state which method you are using and show your work. So that there is less confusion, the answer for [Ni(OH 2 ) 6 ] 2 would be 20 electrons. onic: Cr(2) 4 Neutral: Cr 6 [Cr(PR 3 ) 4 (NO) 2 ] 4 (linear NO) 4 PR PR NO 2 2 NO b) Assuming the 18 electron rule applies, identify the second-row transition metal: [M(CO) 7 ] 18-(7x2) = 4, so Zirconium, Zr c) What charge, z, would be necessary for the following complex to obey the 18-electron rule? [Re(CO) 3 (SiMe 3 ) 3 ] z n: 18 7 (3x2) (3x1) = 2, so need 2 more electrons z=2- i: 18 (3x2) (3x2) = 6 electrons on Re. That is 1. So, 1 (3x1-) = 2-

Chem 350, Spring 2012 Exam 3 CCM, Page 2 3. Consider the following useful redox diagrams. (24 pts) a) Write a balanced equation for the VO 2 /V 3 couple in acid.

e) ill in the blanks appropriately Using the _Pourbaix diagram at right, it can easily be determined that the only stable oxidation state for vanadium in basic solution is V 5.

2 Chem 350, Spring 2012 Exam 3 CCM, Page 2 3. Consider the following useful redox diagrams. (24 pts) a) Write a balanced equation for the VO 2 /V 3 couple in acid. e- 2H VO 2 V 3 H 2 O b) Which vanadium species is the best oxidizing agent? c) Which vanadium species is the best reducing agent? d) Which vanadium species is most stable in acid? e) ill in the blanks appropriately Using the _Pourbaix diagram at right, it can easily be determined that the only stable oxidation state for vanadium in basic solution is V 5. t can also be seen that the species one would expect to find in bog water, which has ph=2 and a potential of 0.4 V, is VO 2. _1.10_ V a) Assuming standard conditions and based on the data on the useful page, label the cathode and anode in this cell and give the E 0 cell in the boxes provided. Cu 2 Zn = Cu Zn 2 E cell = E cat -E an = 0.34 (-0.76) = 1.10 V b) f water is added to the Zn/Zn 2 cell, the voltage of the cell will (circle one) increase decrease stay the same. c) iefly explain your reasoning for part g. The voltage will change with addition of water. This is mathematically obvious from the Nernst equation. Diluting the Zn 2 solution will decrease Q and, therefore, increase E cell. The system is attempting to create more Zn 2, basically. _cath ode an_ode /24

3 Chem 350, Spring 2012 Exam 3 CCM, Page 3 4. a)complete the following table and b) answer the question below. (29 pts) [e(co) 6 ] 0 [Co(en) 2 (NH 3 ) 2 ] 3 [Cr ] 4- correct name oxidation state of central metal atom number of valence electrons on central metal atom (e.g. [Ni(OH 2 ) 6 ] 2 has 20) number of d electrons (e.g. [Ni(OH 2 ) 6 ] 2 has 8) electron configuration of central metal ion likely molecular geometry around central atom spin, S = (at this point, answer as if all complexes are high spin octahedral ) below below below neutral: 8 2(6) =20 ionic: 8 2(6) = n: 9 2(4)2(2) -3 =18 i: 6 2(4) 2(2) = n:62(1)2(1)2(1)4=16 i: 42(2)2(2)2(2)= [Ar]4s 2 3d 6 [Ar]3d 6 [Ar]3d 4 octahedral octahedral octahedral Chiral? (circle) YES NO YES NO YES NO There is a little more room here so give the proper names for [e(co) 6 ] 0 :_hexacarbonyliron(0) [Co(en) 2 (NH 3 ) 2 ] 3 :_diamminebis(ethylenediamine)cobalt() [Cr ] 4- : dibromodichlorodifluorochromate() b) At least one of the above compounds is chiral. Draw two enantiomers here and label them appropriately. isomer of [Cr ] e.g. all cis Cr Cr /29

4 Chem 350, Spring 2012 Exam 3 CCM, Page 4 5. Shown is a portion of an engineering diagram for a common industrial process. (10 pts) a) Name the process Haber-Bosch Process b) Write a balanced chemical equation for the overall process. N 2 (g) 3 H 2 (g) 2 NH 3 (g) c) Comment on three portions of the engineering that help the reaction occur in a more commercially viable way (there are more than 3). Couple with other processes High temperature High pressure of N 2 and H 2 Use of catalyst Recycle unreacted gases Liquify NH 3 6. n the space provided, draw all of the possible stereoisomers of [] 0. Be sure to consider both possible coordination geometries. (5 points) 7. n class we same the demonstration of the reaction of [Ni(OH 2 ) 6 ] 2 with ethylenediamine, en. With the addition of en, the aqueous solution turned from green to blue to violet. (11 points) a) Based on these observations, comment on whether [Ni(OH 2 ) 6 ] 2 is inert or labile. Explain your reasoning. Labile. The substitution reaction occurs quickly, as observable with the color change. nert species do not react quickly if at all. b) Explain why you observed multiple color changes, not a single color change. The substitution of the hexacoordinated nickel does not occur in one step. One could imagine first substituting two waters with one en, then repeating that process, and then a third and final time. Only the initial change and the final color change are obvious to us on observation and the likelihood is that there is a statistical distribution of [Ni(en) n (OH 2 ) 6-2n ] 2 from the beginning until 3 equivalents of en are added. c) We also discussed the order of addition. f we had first mixed another reagent, NH 3 with the[ni(oh 2 ) 6 ] 2, we would see a different blue-colored solution. f we then added en, we would achieve the same violet solution above. f we add NH 3 to the violet solution it stays violet, i.e. there is no reaction. iefly explain why. The en ligand is bidentate, that is it has two lone pairs that can bind to the metal. Higher denticity ligands (here bidentate) are more likely to stay bound than lowe denticity (here monodentate) ligands; this is called the chelate effect. f you add the en first, it stays bound. Second, it replaces the NH 3. /26

Chem 350, Spring 2012 Exam 3 CCM, Page 5 Ligand onic Method Neutral Method

S 2, RN 2 ) 2 CR 3 (neutral) 3 N 6 (N 3 ) 3 5 -C 5 H 5 (Cp) 6 ( C 5 H 5 ) 5

3 H 5 (allyl) 4 (C 3 H 5 ) 3 RC CR 2 or 4 2 or 4 ief Spectrochemical Series:

5 Chem 350, Spring 2012 Exam 3 CCM, Page 5 Ligand onic Method Neutral Method NO (bent) 2 (NO ) 1 (O=N ) NO (linear) 2 (NO) 3 (O=N lp) =O, =S, =NR 4 (O 2, S 2, RN 2 ) 2 CR 3 (neutral) 3 N 6 (N 3 ) 3 5 -C 5 H 5 (Cp) 6 ( C 5 H 5 ) 5 6 -C 6 H 6 (benzene) C 7 H 7 6 ( C 7 H 7 ) 7 (cycloheptatrienyl) 3 -C 3 H 5 (allyl) 4 (C 3 H 5 ) 3 RC CR 2 or 4 2 or 4 ief Spectrochemical Series: - < H 2 O < NH 3 < CN -,CO en = ethylenediamine H 2 NCH 2 CH 2 NH 2 = bipy N N

π donor L L L π acceptor has empty π orbitals on ligand in to which d e- from M can be donated

π donor L L L π acceptor has empty π orbitals on ligand in to which d e- from M can be donated Name KEY D# Chemistry 350 Fall 2005 Exam #4, November 18, 2005 50 minutes CCM 100 points on 4 pages + a useful page 5 1. Consider the molecular orbital diagram shown for M N. (16 pts) a) Indicate the following: