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Chemistry 104B - Fall 2003 - JohnArnold - Midterm 2

Chemistry 104B - Fall 2003 - JohnArnold - Midterm 2 -...

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Unformatted text preview: Midterm Exam 2 Chemistry 104B October 30, 2003 Inorganic Chemistry 9.40 — 11.00 a.m. Name: k9! NOTE: THIS IS A CLOSED BOOK EXAM” 1. Looking at other materials or other students' work during the exam period will result in immediate confiscation of your exam and further disciplinary action. 2. Write answers in the spaces provided. Ask Sean for additional sheets of paper if needed, but be sure to secure these to your exam before you hand it in. 3. There are 8 pages to this exam (not including the Tanabe Sugano diagram). Take a few minutes to check that you have all pages and to read through the whole exam before you start writing. Good luck]. Grading: Problem Max Pts 1 . ____________ (6) (8) (6) (8) (10) (6) (4) (9) (12) _______ (8) TOTAL ____________ (77) pwflaweww p—l .0 1. (6 points). On the periodic table below, fill in the missing elements: Periodic Table 2. (8 points) In our discussion of the reactivity of transition metal complexes, we ofien refer to terms such as stability and inertness. Explain in words the difference between the two and provide an example of a complex that is stable, but not inert. 5+“bll’l1‘7 FCQG 1‘0 :1 1L4ermodynam',c 7uan+2¥/_ Ufa-GLIC mo/ECUIeS AAVE, Al‘jA Free energy) SVLAA/e Molecules AIM/C /ow free, envy} VAc/ef (amoeba/15 7L4Q7‘» genial: can)“; WJ unset/e mo/ew/es 74w 43 form J‘Lub/e ones. Iner+ne$5 repel“ 7‘» a lane’s}; jumHi-y, Requt/e, COM/ownls 711ml vLo {sit/”c c; «1414/ andJi “I eAer-qy VLO reac7l' “IA/X: )i/lel‘VLrom/oadnals fejmf‘c morc energy 7‘0 (‘8ch mm :3 Me- n one: no; mm m oxfihiion .44qu or 1:3qu 59% I‘A 1%0, Igvl fl; [‘5 (Padilla-w? A45 a “‘7‘ rule, mo “ruler eXcAaVe, 3. (6 points) Consult the attached Tanabe-Sugano diagrams to determine how many reasonably intense (i.e. spin-allowed) d—d transitions are expected in the spectra of strong-field octahedral complexes of ions having the following electronic configurations. (a) d“ 5 +ransi +I‘0W5 , 93 3 9 3 i * i7; E) 7/!) 7; 14/ 3 SO C 0“” ”‘ 9"“37 ““37 04’” dombaL / / ‘2’ (M Simnsii’iow-S . ' I |A’ "’3 (If) 61% 1/an A; j E deroioqioly :37"! 098ij 1 Y «Warm a 53319. peak 4. (8 points) Consider the species [Pt(py)(NH3)(CN)Cl] \{a) How many stereoisomers are possible for this formula? 9 (b) One isomer can be prepared Via the sequence: _ .tNHa - py NH3 CN py'u... - PtCI 2 __.. ____. Pt 4 cl/ \CN [M Explain Why this isomer results. ”2””; hon/8: weal Afwné e-GFeo‘l’ 50 ”may arrange ois RFYYiZ/SJ- fill—3) cits—[(Py‘HA/H‘gx Hug] baiweevx rue/Hue) P/ Acts #5 “(auger— +mns 9&1;ch \ 50. ., my [(PyX/v/g) P‘l’C/l] (0) Five other three reaction sequences are possible. Give the expected product for any two of these. Show your sequences. F91 6 3/4, CN ' a— A/ll % a [Pvl (6)0ij ~39 1Lf‘ans [PfchOCN/faq’z]- j” J P/ villus [FVL(P7)(CA/)L/ J. C} 1 M é’N'Pyl—cN 7‘ ,0, 72)] \fl‘CN / 5. (10 points) A researcher obtained the following data by measuring reaction rates in methanol for: trans-[PtCH3(PEt3)2C1]+ + pyridine —> trans-[PtCH3(PEt3)2pyridine] 2* + c1- kobs [pyridine] 14.0 x 10 "42.1 x 10 ‘2 25.1 x 10 ‘4 4.1 x 10 '2 , L75 6% 43.5x10"‘7.9x10'2 w” Vb ‘ 85.2x10'416.0x10'2 (a) What is the general rate law for reactions of this type? j): ; /<,£/)J + kJZF/JUU (b) Use the rate data to calculate k 1 and k 2 for substitution of pyridine for chloride in these square planar complexes. (Graph paper is available if you need it, otherwise say how you arrived at your answer). I‘ = 0 00036 5" ' MS» 57W“ 1 . ka= 0.05! M":"’ flop koLS Y5, [PX] (c) In substitution reactions of square-planar complexes where k1 is large, does this mean that the reaction must include a dissociative path? \ \ A0, #35 meqns Add“ flew: L5 a56h’\£/CW gait/chi" a£§/\S’VL€0( KmVMuJQ/ (For QXNV‘f/QJ [I VLLb 50(an I} more nucleoffillc Aux Hm ithmiVU Adda) (d) If the reaction did feature a dissociative path, what other pieces of experimental evidence would be most supportive of such a mechanism? . F0513 CLEPCMlamce, on tLe/ no. VLUr‘e MC VLAfl/ [Om/)3 fired/3 . More pas(\l}w [ii/F ' Clo/)qi'la" “pt?!“ Duaar bek‘“>”"lfl [(01.5 ‘VSeLinoowihj L/bawfl 6. (6 points) (a) For the series of complexes [C0X(NH3)5]2+ (X = F, Cl, Br, or I), the ligand field (d—d) absorption bands differ only slightly whereas the charge-transfer bands are shifted greatly. Explain why. Clea/«(3)23 omip/ one 0X“ é [Grier/i5 does noi' ohwje flo “\QA' mvob/ i>V+ H‘e/ LAiiJfiS kau& 7y>i¢ Cl; cmApL \‘onT-Lmiiml‘ emrjhfi Me owing “44" “1; .s J‘arlin each «car J .9/ Lam (barge XM $495615 (images Ci/07Z from Fez (b) [Cr(H20)6]3+ is pale blue-green whereas Cr042' is an intense yellow. Explain why this is so. 1 t wicker pica-r” 5 ‘l’V-e‘fifIW/‘IOV‘W‘ [gum/i5- lt‘3wls’ chi): Sim") L—‘> m “L‘Wfle’ ATJ i’f‘a’hsihové Kgr LfltuJ-j [\A QAC‘LI‘OMKC SfeCJ'I‘JM in e iflCi’ro"\l c sped?” (0) Two isomers of a Co(II) complex, believed to be cis— and trans— isomers of the type CoL4X2, both show two bands in the Visible region. Those for complex A show a = 60-80 whereas those for complex B show shoulders on the two bands, and both have in the range 3 = 30—50. Assign the isomers, explaining your reasoning. \ \ cowplex A15 as (oNPizx B is J'rans. IvLs banal: «UPC [955 /{n 7Lans€, AficavSe, A; campiex 1M; inversion S/MWGVL’Y aml deal ‘Lrwaliws are, quori—e, forLi/JJg/L 1A6 r War; 6ch CWchl, 4!, Tab“ ‘7943‘ J:\5v[arJ/\m break/i2} ”a in VCPSiow s/r’WZWLT/ ~ novargsgnvL [In A 7. (4 points) Octahedral complexes of high oxidation state metals, or of d metals of the second and third transition series are less labile than those with low oxidation numbers and those from the first row. Using your knowledge of substitution reactions in octahedral complexes, explain why this is so. \ fiuhsdlvuifin rear/41‘9”)“ 0A comp/eras vLeV‘A 4L0 58/ dl>5c>C;q Ii/L’N/HKCA Paid/ES Cleaves-5e, 05; M»; AoAJS, ”ij (’Lm’je W&4’@//(OAS’ mke "(J-“U m-L $.9an bin/664i on {ring (cu/0M£I‘C «Madam 2'42...) pea/5 mks, 5mg MIL sang, a; a resaH' ed; 467%0’ ML orbivlm/ OVeF/tp. S'L'Y’M‘ lawn wee mm. em” ,‘A/ML 4., Aren1~ mag/#23 ,4 leis Alix/6 @mpkxa 8. (9 points) Assign inner- or outer—sphere mechanisms for each of the following. (You must provide a reasonable explanation, otherwise points may be deducted.) (a) The main product of the reaction between [CrF(NCS)(H20)4]+ and Cr2+(aq) is [CrF(H20)5]2+. \ \ \lrme" chn Hm si'wv‘i'lltj Wmi‘emq' inelrxd {Edi/gel (1va (9394chch Pauli/c" wre/ inert and. & l/‘jani has Been )(S'efi‘eol, kaeckwnfisk (XS inner‘sW (b) The rates of reduction of [Co(py)(NH3)5]3+ by [F e(CN)6]4' are insensitive to substitution on the pyridine ligand; i.e all three of the following give similar rates: N \F/> ”Oil“ ”QM/'6 +0 lee, \IVchr Q‘fora 0V coMe‘I’ami’ Al‘ioljivt’ “SN”; (Onin/V- Acne} mugi‘ fiqciillml’c mm. tor C/Vo‘—9<30, Co int/57L (053 q “Semi-FY '5 (1‘5 “05"" AA”: ”Sakai. lxoweutf‘ di$oc§qiian of“ PX 050““ be’depeco'vmi' on {is e/CO‘LPO/Ukc r shit/é” +kere I: he [Gpeflllvfl¢ on If! eleclvron ctiono \ (0) Reduction of (A) by Cr2+ is much slower than re f (B). N CA "’55, 41‘ [\5 Feq c 47}, 1 2+ 2+ MUsqL [>8 00460 0— o_ o . (en)ZCo< 22 (en)2Co< :gfl Sflflcra o o H \ / C Amess, O O \A O/Cf“ 6’ 9. (12 points) Nitrogen monoxide NO, is a well-known, and important, molecule. In the context of transition metal chemistry, NO forms a wide range of complexes similar in some respects to those of CO. In contrast, the phosphorus analog, PO, is relatively unknown as a result of its highly reactive nature. In a recent paper in Organometallics (2000, 19, 4336) the electronic structure of P0 was investigated using calculational methods. In the two figures below, molecular orbital diagrams for two species, the free PO molecule and a (hypothetical) molybdenum complex (OP)M0(NH3)3 are drawn. By reference to the figures above, answer the following questions: a) Identify the HOMO and LUMO in both molecules using the symmetry labels provided in the figures (i.e. there’s no need to consult character tables to answer this). Po: 77’*l/omo (DP/“o W593 ‘. 7E #omo b) Reduction of PO to PO' results in a (calculated) decrease in the P-O stretching frequency of 205 cm'1 whereas corresponding reduction of NO to NO' decreases the N-O stretch by 514 cm]. What feature of the molecular orbital diagram explains the large difference between these numbers? In other words, why is the decrease much smaller for P0? r‘euLchl’Fovx FCSUH’S ln lncreasecl fofulayl';en 0‘; 7T “anthemfi Tr AWOL 14" ”/02 77* ls mac/L higher [\rt ene’i‘ly (W Lone) Ts s'l’mncjea because Highs Ls bell-9:" MAN“) O‘ml’f’ caulk toe‘l‘l'er— mulch Le—Fuoeen orElJ-al size M amfl)’ ”VS-0 COMFMJ’a 73113.0. Populal’lrfl ”Sher eMcV‘ay’ an” J?) and $65421); lZ>Qj *K éPr. Qd) Which orbitals on Mo are involved in n-bonding? Le) Is the latter interactio stabilizing o destabilizing? Why? \ a” [30/ Mo eleci'rons '“VOiueol OJ‘C. (Pimce‘l I\V\ i’o Lowoufij .7 M015 10. (8 points) Heme is a tetradentate ligand that binds strongly to divalent metal cations (see below). Myoglobin, an oxygen carrying protein, contains an Fez+ion bound within one heme group. You wish to determine whether this iron atom is in a high or low spin configuration within the protein. When you measure the molar magnetic susceptibility, however, you find that it is not consistent with the value predicted by spin-only magnetic moment for either a high- or low-spin configuration. You attribute this deviation to diamagnetic susceptibility. Why is that reasonable? Suggest an experiment that would allow you to make an estimate of diamagnetic susceptibility. After making this correction would you expect the outcome of your experiment to be consistent with high spin or low spin configuration? Why? &ldW\QjV\e+(‘p soSaepiiLilii/ “rises 912V“ Pam/J QZ‘MLN“ Ofi mimic-AN +ker‘e are, a. Jrecml' “my in ai/JFO‘LCI‘A (nucToMokoUle3. RQPiacirfl Fegkwii’i'x v-L SiMiiw‘ oiiqmmyne4’;\o )Kon (9.3. Cu“) (:1ch Mensuer'tg Masnoleic suscep'b‘él‘id‘l/ Woo/oi, ICQJ {3,1300% «mac tor +L¢ Fe,“ humans»; flagrantly, me+ be H5 Aer/awe L5 Fang/5 {33‘ ”A’s. (6:? 0x739“ Gaf‘f‘lfij reyuire; mm 05w gunman, Lé 532+le Le, .‘MH: 80 80 Willie?) 70 70 so 60 d‘ d“ . T I so 50 4O 40 ‘Tzuse'i in aEteJ in ' m m S 30 30 20 10 272(6) 20 10 20 30 4o AOI'B‘ £3.33 in in m m 70 “TM?“ 70 60 “T 60 50 d8 50 ”9 aTzugea) 40 _ 40 273(136‘) in Q: m m 30 30 20 20 3P 10 10 “1120:92) 2E[r§e3) 3F 10 20 so 40 ’D 10 20 30 4o Au/B' AclB' AO/B’ Tanabc—Sugano diagrams for d ' through d9 metals in octahedral ligand fields. (The g subscripts have been omitted for clarity.) [Adapted from K. F. Purcell and J. C. Kotz, Inorganic Chemistry, Saunders. Philadelphia. 1977: DD. 584 585.1 ...
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