Chem103fall2010Exam1KEY - Name SID Signature Problem 1...

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Unformatted text preview: Name SID Signature Problem 1: Problem 2: Problem 3: Problem 4: Problem 5: Problem 6: Total: Chem 103 Midterm 1 Prof. Chris Chang September 30, 2010 Version A (16 pts) (40 pts) (62 pts) (60 pts) (24 pts) (48 pts) (250 pts) 1. Metals in Biology. (16 pts) Match each of the following roles/substances in the left column With a relevant metal from the list. You may use each metal once and only once: Cu Mn Zn Fe Co Li Ba V Halo genation Photosynthesis C02 Fixation Structural support ‘ X-ray Contrast ck Detoxification of reactive oxygen and nitrogen species . Treatment for Manic Depression Vitamin B12 «C; 2) Electron Counting. (40 Pts) For the following bioinorganic complexes, provide aim; a) the formal oxidation state of each metal center b) the (1 count of each metal center c) the total electron count of the complex i) Catalyst for H2 production “gawk fl gfiéfik‘bfikfilfiw .. mem-“Ze: V M K910 E‘xjmy 2m ‘3}; @fit E2; fléwfimc it) as 53¢ 1:1,.» NZ if 43;” N2 \ NH Asp Glu Q—a O."\O \O\ / Glu\ OH2 A / H20“”‘Mn\0r~\O/Mn / \O‘ ’O/ \ HN/§N ‘— / Glu His His New? B ** iii) Precursor to heart imaging agent V‘fi‘wkv‘w‘)‘; C; c A 117;; M r ‘ VW‘AQ‘ Reece co OMe 321M“ EN? N 1 “co | “too )4 NEG—Tc“ Tc“ GEN OC i 0C i CO CO iii) Ribonucleatide Reductase, catalyst for deoxynucleotide formation 1 VVK{,AE&,)€XC:%Zm/€ 25% tom, (sextet > a “wk “ , Qfiq . QLWQ 5’? We» 9‘3 A» a “:1 5:: -é~ s "I "y ‘3 m 5/ 5;" RAW”. a; ‘tfl \, {EXP it.an "fit-sum.) M vii; (:9 t‘: M? 9E; '23, w?“ M 53$ X CM", (‘41;ka fif‘” 3) Molecular Orbital Theory. (62 Pts) Nitrite (N02) is an important nutrient found in the soil and taken up by plants, which is then incorporated into our diets. Research has shown nitrite is involved in metabolism, but its specific roles are not well-understood. Understanding its bonding may help elucidate its biological functions. V a) Draw the Lewis dot structure of N02". What is its molecular geometry? b) Provide an M0 diagram for this species with labels (6, 712, etc) and fill in the electrons. Draw the appropriate bonding and antibonding orbitals, showing orbital overlaps. *Hint: one way to do this problem is to use the s and p atomic orbitals of the two outer atoms to generate a set of molecular orbitals. This set can then interact with the s and p orbitals of the central atom to generate the complete MO diagram. c) What is the total bond order for N02"? \u. 4) Crystal Field Theory. (60 points) An interesting catalyst for N20 decomposition is a Ru porphyrin that evolves N2 and incorporates the O atom into olefin products. . a) The initial step involves loss of N2 to give a Ru species capable of reacting with N20. i. Draw the Crystal field splitting diagrams of I and II, showing how the crystal field splitting diagram of I changes with respect to the crystal field splitting diagram of II. g pig. ii. Give the d-electron count and metal oxidation state forI and II. 14+ ' An Ru-N bond lengths ‘ are rougth equivalent —» m (II) 1+ Q can , at 3'1 E 1: w 1““: 3"“? W x1e? E V “ V“? m “L ~ w- I I.» ._ - «m I,“ .2va My, organ, an“ a. 4% -- 7 M W *7 ME 3% I I I i 2 4t '3‘ a 9% b) Intermediate II can bind to two N20 to release to two N2 in a stepwise fashion to give 111 and IV. , i. Redraw the crystal field splitting diagram of II and show how the crystal field splitting diagrams of III and IV change with respect to II. ii. Give the d—electron count and metal oxidation state for III and IV. Ru:O bond length shorter than Flu—N 0 (II) (III) (IV) ,3? We“ $2" 22? " “ 2;? Mm- x’zw 7,1, 2, «‘2 ~ Wm”? M sh??? E 4 i 2 M g 2,, ., ' EZVV" ‘ lg. w" it? n “22;, E w “*3 . 9" 3— QM. 2% QM” } ‘4 ) 5g Adm 0) Intermediate IV reacts with olefins like propylene to give V. i. Redraw the crystal field splitting diagram of IV and Show how the crystal field splitting diagram of V changes with respect to IV. ii. Give d—electron count and metal oxidation state of V. V Ru—O bond length ', longer than Ru~N \ e x \ /\ \ N\ zN‘ —> \ N/RU\ / H 0 (IV) i * » k ' u" w ‘ - w?“ \\““> m 2‘ ’1 “L “W Z cflcirmj "nVV-«-l~-\«~~w§ %u a?» \ ‘ ‘5 k E m%mn W i,,.v,,.,-‘_v,.n..\) ‘ W7 - ’1 at“? (l M“, ,‘o d) For intermediate I, calculate the Crystal Field Stabilization Energy (CF SE). L 2 Z“ L $9 Jgficfiso + 7gb} now» 395; A§+3€E 5) Ligands. (16 pts) Sez‘ Glu- co a) hard/Soft/intermediate S b) G—dOIlOI‘, n-donor, 7" acceptor ’ n—donor n-donor o—donor n-acceptor (c—donor) (cs-donor) (o-donor) c) Prefer Ca + or Fe 6) Molecular Orbital (MO) Theory. (48 points) CF is a dangerous oxidant that can produce chlorofluorocarbons upon reaction with common laboratory solvents like chloroform and dichloromethane. a) Draw the Lewis dot structure of CF. b) Provide an M0 diagram for this species with labels (6, 7r, etc) and electrons. Draw the appropriate bonding and antibonding orbitals showing orbital overlaps. c) What is the bond order? an :O—F : “a, Bond order = (8 bonding e) - (6 antibondlng e) = 1 2 ...
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