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Name Chemistry 350 Spring 2012 Exam #3, April 27, 2012 50 minutes CCM 100 points on 4 pages + a useful page 5 1. In 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 b) a conductor E = 0 eV vb = valence band cb=conduction band E = band gap > 4eV 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(OH2)6]2+ would be 20 electrons. Ionic: Cr(+2) 4 4 PR3 8 2 NO 2 16 Neutral: Cr 4 PR3 2 NO 4+ 6 8 6 -4 16 [Cr(PR3)4(NO)2]4+ (linear 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(SiMe3)3]z n: 18 – 7 – (3x2) – (3x1) = 2, so need 2 more electrons… z=2i: 18 – (3x2) – (3x2) = 6 electrons on Re. That is 1+. So, 1+ + (3x1-) = 2- /21 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 VO2+/V3+ couple in acid. e- + 2H+ + VO2+ V3+ + H2O b) Which vanadium species is the best oxidizing agent? (circle one) VO2+ VO2+ V3+ V2+ c) Which vanadium species is the best reducing agent? (circle one) VO2+ VO2+ V3+ V2+ d) Which vanadium species is most stable in acid? (circle one) VO2+ VO2+ V3+ V2+ e) Fill 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 __V5+___ . It 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 __VO2+___. _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 E0cell in the boxes provided. Cu2+ + Zn = Cu + Zn2+ Ecell = Ecat-Ean = 0.34 – (-0.76) = 1.10 V b) If water is added to the Zn/Zn2+ cell, the voltage of the cell will (circle one) increase decrease stay the same. c) Briefly explain your reasoning for part g. The voltage will change with addition of water. This is mathematically obvious from the Nernst equation. Diluting the Zn2+ solution will decrease Q and, therefore, increase Ecell. The system is attempting to create more Zn2+, basically. _cath__ode __an_ode /24 Chem 350, Spring 2012 Exam 3 CCM, Page 3 4. a)Complete the following table and b) answer the question below. (29 pts) correct name oxidation state of central metal atom number of valence electrons on central metal atom [Fe(CO)6]0 [Co(en)2(NH3)2]3+ [CrF2Br2Cl2]4- below below below 0 3+ 2+ neutral: 8 + 2(6) =20 ionic: 8 + 2(6) = 20 n: 9 + 2(4)+2(2) -3 =18 i: 6 + 2(4) + 2(2) = 18 n:6+2(1)+2(1)+2(1)+4=16 i: 4+2(2)+2(2)+2(2)=16 20 18 16 8 6 4 [Ar]4s23d6 [Ar]3d6 [Ar]3d4 octahedral octahedral octahedral 1 2 2 (e.g. [Ni(OH2)6]2+ has 20) number of d electrons (e.g. [Ni(OH2)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 ) Chiral? (circle) YES NO YES NO YES NO There is a little more room here… so give the proper names for… [Fe(CO)6]0 :_hexacarbonyliron(0)__________________ [Co(en)2(NH3)2]3+:_diamminebis(ethylenediamine)cobalt(III) [CrF2Br2Cl2]4- :__dibromodichlorodifluorochromate(II) b) At least one of the above compounds is chiral. Draw two enantiomers here and label them appropriately. isomer of [CrF2Br2Cl2] 4- F 4- Cl F F Cl Cr Cr Cl 4- e.g. all cis F Br Br Cl Br Br /29 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. N2 (g) + 3 H2 (g) 2 NH3 (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 N2 and H2 Use of catalyst Recycle unreacted gases Liquify NH3 6. In the space provided, draw all of the possible stereoisomers of [TiBrClFI]0. Be sure to consider both possible coordination geometries. (5 points) Br Br F Ti I F Ti Ti Cl Cl F Br Br I I F F Ti Cl I Cl Ti Cl I Br 7. In class we same the demonstration of the reaction of [Ni(OH2)6]Cl2 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(OH2)6]2+ is inert or labile. Explain your reasoning. Labile. The substitution reaction occurs quickly, as observable with the color change. Inert 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(OH2)6-2n]2+ from the beginning until 3 equivalents of en are added. c) We also discussed the order of addition. If we had first mixed another reagent, NH3 with the[Ni(OH2)6]2+, we would see a different blue-colored solution. If we then added en, we would achieve the same violet solution above. If we add NH3 to the violet solution it stays violet, i.e. there is no reaction. Briefly 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. If you add the en first, it stays bound. Second, it replaces the NH3. /26 Chem 350, Spring 2012 Exam 3 CCM, Page 5 Ligand Ionic Method NO (bent) NO (linear) =O, =S, =NR CR N 5-C5H5 (Cp) 6-C6H6 (benzene) 7-C7H7 (cycloheptatrienyl) 3-C3H5 (allyl) RCCR 2 (NO–) 2 (NO+) 4 (O2–, S2–, RN2–) 3 (neutral) 6 (N3–) 6 ( C5H5 –) 6 6 ( C7H7+) Neutral Method 1 (O=N•) 3 (O=N• + lp) 2 3 3 5 6 7 4 (C3H5–) 2 or 4 3 2 or 4 Brief Spectrochemical Series: I- < H2O < NH3 < CN-,CO en = ethylenediamine H2NCH2CH2NH2 = bipy N N