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Chapter_3_homework_key1-1

Chapter_3_homework_key1-1 - Problem Solutions 3.1 Ethanol...

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Unformatted text preview: Problem Solutions 3.1. Ethanol and dimethyl ether are isomers because they have the same number and types of atoms but different properties. Furthermore, they are structural isomers because they have different numbers and types of chemical bonds. For example, ethanol has one C-O bond while dimethyl ether has two. 3.2. The three different forms of a chaise lounge are more like stereoisomers because they have the same number and types of "chemical bonds" but differ in the spatial arrangements of those "bonds." 3.3. The two chain theory formulations would be considered to be stereoisomers because they have the same numbers and types of chemical bonds (three each of Co-N, Nils-NH}, and NH3-Cl) but differ in the spatial arrangements of those bonds. 3.4. These are stereoisomers. Each has the same numbers and types of chemical bonds but differ in the spatial arrangements of those bonds. 3,5. A light beam of a given wavelength can be polarized such that only one plane of the electric or magnetic field is allowed. A molecule is optically active if it is capable of rotating that plane to the right or to the left when the light beam is passed through the substance. 3.6. Each of these sets of isomers (the 1,2; 1,3; and 1,4 for the MA4B; and the 1,2,3; 1,2,4; and 1,3,5 for the MA3B3) are geometric isomers because their different spatial arrangements result in different geometries. None of these geometric isomers are chiral because each contains at least one internal mirror plane, namely that of the hexagon itself. 3.7. For each case (MA,B; and MA3B3), these are geometric isomers because they have the same numbers and types of chemical bonds but the different spatial arrangements of these bonds result in different geometries. In the MA,B, set the 1,6 case does not contain an internal mirror plane and is chiral. The 1,5 isomer is actually the nonsuperimposable mirror image of the 1,6 isomer. Similarly, in the MAsBa the 1,2,4 isomer is chiral. The 1,2,5 isomer is its nonsuperimposabie mirror image. 3. 8 . The two mirror images of bromodichloroiodozincate(II), shown below, can be superimposed upon each other by rotating one isonier 180° about the Zn-Br axis. Therefore, this molecule is not chiral. The same conclusion is arrived at by considering that the molecule has at least one internal mirror plane. Br : Br I Rotate about the ‘ : Zn-Br axis by 180° ’1 I "'5 Cl ffl/ \ ' / Xx C1 I l 1 C! : c1 3. 9 . These two mirror images are nonsuperimposable. No matter how one is rotated in space, it cannot be made equivalent to the other. Therefore, this molecule is chiral. The same conclusion is arrived at by considering the fact that this molecule does not contain any internal mirror planes. Cl Br 0 7 7c \ I, '. F 3.10 In a molecule of formula Mabcd, only a tetrahedral structure will be chiral. It will have no internal mirror planes. In the square planar molecule, the plane of the square is an internal mirror plane. Therefore the square planar Mabcd is not chiral. 3.11. (a) If the mirror image of a molecule cannot be rotated in space so as to be equivalent to the original molecule, it is said to have a nonsuperimposable mirror image. Such a molecule is chiral. (b) If a molecule does not possess an internal mirror plane (a plane that passes through the molecule such that every atom in the molecule can be reflected through the plane into another equivalent atom) it is chiral. [There are a few exceptions to this rule but they are beyond the scope of this text] 20 3,12. Only PCIBrI is chiral because it has a ' bl ' ' C1 1 c1 nonsupenmposa e mirror Image. 1 i . . . . _ _ 1 The mirror image 18 superlmposable on the magma] A] 1 Al molecule. To see this, picture the mirror image rotated / \ i \ 180' about the Al-Cl axis. . I Br l ' ’df} P l P Itsx ' ’ ' c1 Here, the mirror image cannot be superimposed on I . 1 the original molecule. Br ! Br 3,13 . Only AlClBrI has an internal mirror plane, the one that the entire molecule sits in. Therefore, this molecule is not chiral whereas the corresponding phosphorus compound is. 3.14. There are three possible isomers of PEG]; as shown below. Each of these possesses at least one internal mirror plane. Therefore, none of these isomers is chiral. ' 1: F Cl ’ fl,- CI ....r’ Cl “we: Cl P K F P R F P R C1 C1 F F Cl (:1 [Note: For the purposes of considering the chirality of the above three forms, we are assuming that the above molecules are nonfluxional. ] 3.15. The structural formulas are as follows: EFor (b) and (CL-the structure of only the cations are shown; the overall +1 positive charge of these two cations is also omitted for clarity.] NH3 (3) (b) (c) _ .PPh3 H3N NH3 ONO ' ' PPh3 H3N NH3 py ON 'NH3 21 3.16. There are two additional internal mirror planes in (a) and one additional in (b), (c), and (d)- Thcse are shown below. 22 3.17. Neither of these compounds is chiral because each contains at least one internal minm' plane. (a) (b) 3.19.. (a) [010204) 313‘ -- chiral due to propeller shape with no internal mirror planes (b) Not chiral due to several internal mirror planes including the plane of the molecule. . NH:2 \ CH x 2 '3 {CH2 3.20,. (a) [C0(EDTA)]‘; (b) [CI‘C12(NTA)] 3' The EDTA complex does not contain an internal mirror plane and is chiral. The NTA complex always contains such a plane and is not chiral. Charges on the structures are omitted for clarity. 24 3,21. Prior to Werner’s synthesis and resolution of coordination compounds containing chelating gusts, the evidence for his coordination theory had been " negative." That is to say, his theory often y two isomers for a given MAB: or MA333 complex (where A and B are monodentate “dialed on] lip-Dds) if the geometry of the coordination sphere was octahedral. Planar hexagonal and trigonal 'c geometries, on the other hand, yielded three isomers in each case. His and other research w'smatl groups indeed could synthesize only two isomers of these compounds. But what if the correct geometry was hexagonal or trigonal prismatic and the third isomer was just particularly difficult to synthesize? Werner had only "negative" evidence for his octahedral coordination theory. It was the We of a given isomer that Werner claimed supported his ideas. However, octahedral coordination spba'cs for a complex of general formula M(AA)2B2 (where AA = a chelating bidentate ligand) should have one chiral and one nonchiral form. Werner and his group were able to confirm the existence of this number of isomers and therefore provide "positive" evidence for his theory and the octahedral configuration of his secondary valence. 3.22. cis- and trans- dichlorobis(ethyl- C +1 H N +1 cnediamine)cobalt ' 2 “C113 (11]) chloride. The ‘fHZ cation is shown in H2C/ NH “‘13 NH2 each case. H (3/ ,CH2 2 ‘- 3/ CH2 NH2 NHZ Cl trans 3-24: For the complex cation of this compound (shown without its charge), there would be four Tillble geometric isomers as shown below. Only the last, ((1), lacks an internal mirror plane and is c 1 . . (a) (b) Hzri Nn2\ H2C \ NH2 NH: Cl 25 3.25. The +1 charge on the structures of the cations have been omitted for clarity. \ (sz NH ' / ”HZ ‘ NH H2}: . \CH2 2 H C ’ 2 \ . /CH2 N02 trans 3.26. Since both the ethylenediamine and oxalate ligands can only span the cis positions of an octahedron, there is only one possible geometric isomer for this complex. This cation (shown below without its +1 charge) lacks a plane of symmetry and is chiral. HZN H CH2 3.27. No, the structure would then have a plane of symmetry and would no longerbe chrial. The plane of symmetry would contain the C-C bond of the methyl substituted ethylenedlamme, pass through the platinum cation, and then bisect the C-C bond of the phenyl (CaHs) - SUbSfitUth ethylenediamine. 3.29. (a) potassium chloro(nitrilotriacetato)thiocyanatocobaltateflll) 26 3.29. (b) Since the four binding cites of the NTA must be adjacent to each other, the two monodentate ligands (SCN— and Cl‘) also must always be cis to each other. However, either the chloride or the miocyanatc could be trans to the nitrogen of the NTA as shown below. In either case, the complex anion contains an internal mirror plane and is therefore not chiral. ‘ SCN 3.30. If the coordination sphere were trigonal prismatic instead of . . . , o octahedral, the resultmg complex anion would have only one possrble ‘ q geometric isomer, This structure, however, would not possess an 0 C"‘0 internal plane of symmetry and would be chiral. ‘ C1 SCN o’C\0 3.31. Both of the square planar structures possess a plane of symmetry (the plane of the molecule), but the one tetrahedral structure does not. Therefore the square planar versions are nonchiral while the tetrahedral is chiral. I 27 3.32. (a) Since this ligand can only bind to a metal center through an oxygen atom, it cannot be ambidentate. - (b) Since compounds containing six Coh-N interactions were always yellow whereas those containing five CoH—N and one Cob-0 interactions were red, Werner and J orgensen were able to confirm the structure of this red compound. Today, we know that this compound must have one Coy-0 bond because the nitrate ion does not have a lone pair of electrons on a nitrogen atom. ‘ 3.33. Since the nitrate ligand is capable only of 0-bonding to a metal center, both the bis(ethylene— - .diamine)dinitratocobalt(III) chloride and the bis(ethylenediamine)dinitritocobaltflll) chloride possess four Cob-N and two Coy-O interactions: Therefore it follows that both of these compounds should be the same color while the bis(ethylenediamine)dinitrocobalt(lll) chloride, in which the complex cation contains six COM—N interactions, should be a different color. 3.34. (a) [Co(en)2(N02)1][AuCl4], bis(ethylenediamine)dinitrocobalt(III) tetrachloroaurateflll) (b) [Co(en)2(ONO)3][AuCl4], bis(ethylenediamine)dinitritocobalt(HI) tetrachloroaurateflll) 3.35. The Lewis and VSEPR structures of thiocyanate are shown in (i) and (ii) below. The sulfur has three lone pairs available to bind with a metal center while the nitrogen has only one. Assuming the four electron pairs around the sulfur (three lone pairs and one bonding pair) (i) _. are tetrahedrally dispersed, S-bonded _ \ SCN- complexes should possess a i .. _ : ~11 nonlinear M-S-C interaction. Given the i: : S :C 2::Nz] - J linear arrangement of the triple bond .. and the one lone pair about the nitrogen atom,on the other hand (or is it on the " other tooth?) N-bonded thiocyanate \ _ ' complexes should possess a linear M« : ' ‘ N-C interaction. These two cases are M0 0 "J U shown in structures (iii) and (iv), (109 ' 1300 respectively. (ii) ' —C‘ENI 3.36. Given the fact that both the carbon and nitrogen ends of the cyanide ion have a linear arrange- ment of the triple bond and the lone pair, we would expect both the cyano and isocyano forms to form linear bonds with a metal center. , _ _ _ __ _ [13:14.] @N. :CEN—M 180° \goojv 28 3.37. S—bonded thiocyanate will take up more space around a metal atom or ionbecause the -SCN is able to rotate about the M-S bond and sweep out a fairly large cone-shaped volume. N—bonded thiocyanate , on the other hand (tooth?) rotates about the M-N bond and sweeps out a much smaller, needle-like cylindrical volume. In the [Co(NI-I3)5NCS]2+cation, the five ammonia ligands sweep out _ cone-like volumes and take up a fair amount of volume. They do not leave much volume available for the thiocyanate so it is more stable in the isothiocyanate "needle" form that does not require as much volume. In the [C0(CN)SSCN]3~ anion, the linear CN‘ ligands take very little volume and so the S« bonded SCN', that takes up more space, is favored. 3.38. The three possible linkage isomers are as follows: (a) trans-bis(ethylenediamine)dithiocyanatocopperfil) (b) trans-bis(ethylenediamine)diisothiocyanatocopperfll) (c) trans-bis(ethylenediamine)isothiocyanatothiocyanatocopperfll) (a) SCN Cb) NCS NH NH NH NH HZC/ Q HZC / ~Z , CH CH Hzc Cu / 2 H C/ CH I 2 CH 2 \ --“" 2 \ ~— ....-- CH N 2 NH2 N 2 ”Hz 2 SCN NCS SCN NH / AJ‘NHz (c) H2; (2 \CH2 HZC “ x \ /CH2 NHZ NHZ NCS 3.42. (a) [Co(NH3)5(N03)]SO4: An ionization isomer would be [Co(NH;),SO4]N03, pentaammine- sulfatocobaltflII) nitrate. (b) [Cr(en)3][Cr(C204)3]: A coordination isomer would be .{Cr(en)2(C;O4)][Cr(C204)2(en)], bis(ethylenediamine)oxalatochromiumflll) (ethylenediamine)dioxalatochromate(III). 3.43. (a) [Pt(NH3)4Clz]Br2: An ionization isomer would be [Pt(NH3)4BrgClz, dibromotetraammine- platinum(IV) chloride. 29 (b) [Cu(NH3)..][PtC14]: A coordination isomer would be [CUCNH3),CI][PI(NH3)C131= triamminechlorocoppedll) amminetrichloropiatinatefil). [PKNH3)4(C204)]C12 tetraammi neoxalatoplatinumfl'V) chloride ionization isomer [Pt(NH3)4C12]C204 cis-tetraamminedichloroplatinumflV) oxalate geometric iSOmer [Pl(NH3)4C12]C204 cans-tetraamminedi chloroplatinum(IV) oxalate 3.43. The following ionization and geometric isomers are possible. No optical, coordination, or linkage isomers are possible. 3.45. At least one of each type of isomer except optical is possible as shown below. [PdeBhHPdWOzML tetraamminepailadiumfll) tetranitropalladate(H) coordination isomer [Pd(NH3)2(N02)2]. ois-diamminedinitropalladiumfll) geometric isomer {PdtNH3)2(N02)2]. trans—diamminedinitropalladiumfll) linkage isomer [Pd(NH3)2(0N0)21. trans—diamminedinin'itopalladiumfll) 30 346. All five types of isomers can be illustrated as shown below. 31 [VC12(en)2]N02 Lrans-dichlorobis(ethylenediamine)vanadium(III) nitrite geometric isomer [VC12(en)2]N02 R~cis-dichlorobis(ethylenediamine)vanadiumfl]]) nitrite ionization isomer [VCl(en)2N02]Cl R-cis—chlorobis(clhylenediaminehitrovanadinmflfl) chloride linkage isomer [VCl(en)20N0]Cl R-cis-chlorobis(ethylenediamine)nilritovanadiumflm ' chloride optical isomer [VCl(en)20NO]Cl S-cis-chlbrobis(ed1ylenediamine)nitritovanadiumflII) ' chloride ...
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