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PS5 Solutions Problem 7.1. Co(II)CO3 + NH4Cl(aq) + air (O2) ----> [Co(NH3)4(CO3)]Cl the carbonate product is pink and contains bidentate carbonate. [Co(NH3)4(CO3)]Cl + 2HCl(aq) ----> CO2(g) + H2O + [Co(NH3)4Cl2]+(aq) The cobalt-chloride product is violet and could exist as a cis or trans isomer. Since the analogous chiral en complex, which must contain cis chlorides, is also violet, it is likely to be cis. [Co(NH3)4Cl2]+(aq) + conc. HCl ---> [Co(NH3)3Cl3]•NH4Cl Here the product is green and could exist as the fac or mer isomer. Later we'll see why the mer isomer is formed. Comment on the reaction that generates gas (carbon dioxide): While Co(III) in acidic conditions can oxidize water to generate O2(g), (a) with the ammine ligands its oxidation potential is lowered so it can't oxidize water, and (b) all the subsequent compounds are Co(III) complexes, suggesting another gas-forming reaction: 2H+ + CO32- H2CO3 H2O + CO2(g) Problem 7.4. The starting Co complex must be neutral to explain the lack of conductivity. Assuming the ammonia ligands didn't migrate during the reaction, the original compound must have the structure shown below (mer isomer). Cl H 3N H 3N Co Cl product NH3 OH2 NO2 H 3N NH3 Co H 3N NO2 O2N starting material [Co(NH3)3(NO2) 3](aq) + HCl ---> trans-[Co(NH3)3(H2O)(Cl)2] 1 Problem 7.5. [ZrCl4(dppe)] + 2MgMe2 ----> [ZrMe4(dppe)] Me P Me Me Zr Me P Me "octahedral" Me P Zr Me P "trigonal prism" Me In the "octahedral" geometry there are 2 types of different Me groups (that is, they cannot be interconverted by a symmetry operation); these are cis and trans to a P, respectively. However in the "trigonal prism" structure all the Me groups are equivalent, so this is consistent with the NMR data. In both cases it is useful to consider a C2 axis that includes Zr and bisects the dppe CH2CH2 backbone. 2. (a) [Cr(H2O)4Cl2]Cl cis and trans isomers for the cation; 6-coordinate, octahedral Cr(III) (b) K2[OsCl5N] octahedral Os(VI) (c) (Ph4As)2[PtCl2(Me)(H)] square planar Pt(II); cis and trans isomers (d) Ru(PEt3)3H3 mer isomer; octahedral Ru(III) (e) [(NH3)5Co-NH2-Co(NH3)4(H2O]Cl5 Two octahedral Co(III)s with a single bridging NH2 (f) [(NH3)2Pt(µ-Cl)2-Pt(NH3)2]Cl2 Two square planar Pt(II)s with 2 bridging chlorides. 3. For this formula, the octahedral geometry can yield only 2 isomers (cis and trans). However, the trigonal prism can have 3 isomers: Cl Cl Cl Cl Cl Cl 4. In addition to the usual isomers (cis and trans) of [M(AA)2e2] molecules we also have linkage isomers of SCN–. In the figures below en is abbreviated as N~N. Note that the pairs D and E, F+G, and H+I are enantiomers. 2 A B C SCN N N Co N N NCS NCS N N Co N N SCN SCN N Co N N SCN E D N N N N Co N SCN NCS N NCS N SCN N SCN N G N N Co N SCN N NCS N SCN Co N N NCS N N Co N F I H N Co Note: all are cations N N N SCN NCS Co N N NCS 5. (a) The trans isomer should have no dipole moment (all metal-ligand bond dipoles cancel) while the cis isomer will have a dipole bisecting the B-M-B bond (b) Consider the 2 structures: mer A fac A A A Co B B A B B Co A B B Each will have a dipole moment along the axis indicated; how do they compare? -for mer the dipole will be (µM-A + µM-B) since the other bond dipoles cancel -for fac a little trigonometry and vector addition shows it will have a somewhat larger dipole: (1+21/2)/(2)1/2µM-A + (1+21/2)/(2)1/2µM-B If the molecular dipole can be quantified, these should be distinguishable. [Note: make sure you understand part a. The quantitative aspects of part b, which are more difficult, would not appear on an exam, but you do need to know the definitions of fac and mer isomers.] 3 6. Here are the isomers: Note: all are cations Cl N S Co Cl trans cis N N S S S Co N Λ Cl Cl Cl Cl Co cis N N S S S N N Co S N Cl Cl Cl Cl Co N S Λ Δ S Δ 7. (a) See pictures below. (i) ML2. For a linear arrangement of 2 ligands, we define the internuclear axis to be the z-axis. Then the electrons in the dz2 orbital will suffer the most repulsions (the 2 ligands = negative point charges are on the z-axis), followed by the other orbitals with a z-component, and, at lowest energy, those with the least repulsions. (ii) trigonal planar ML3: let xy be the molecular plane. The ligands no longer line up nicely on the axes, but it should be clear the greatest repulsions are to orbitals in the xy plane. The dxz and dyz orbitals should be degenerate, but their energy relative to the dz2 orbital is not obvious (how important is the repulsion from the donut in the xy-plane?) (iii) pyramidal ML3: Let the plane of the 3 ligands be parallel to the xy plane, and let the z-axis pass through the metal and through the center of the L3 triangle. energy dxy, dx2-y2 dxz, dyz dz2 dxz, dyz dxy, dx2-y2 (i) dxz, dyz dxy, dx2-y2 dz2 (ii) dz2 (iii) 8. (a) [Ni(CN)4]2- could be square planar (D4h) or tetrahedral (Td). (b) [Ni(CN)5]3- could be square pyramidal (C4v) or trigonal bipyramidal (D3h). (c) [Ni(CN)6]4- will be octahedral (Oh). Note that all of these are Ni(II) d8. The electron configurations and resultant number of unpaired electrons for each geometry are shown below. Note: the results for octahedral, tetrahedral and square planar geometries are straightforward. However, it is difficult to predict the separations between the energy levels (and hence the electron configurations) for the other 2 geometries. 4 b1g Oh(2) D4h(0) Td(2) eg t2 t2g b2g eg e D3h(0 or 2) a1' b1 a1' e' e'' or a1g e' a1 e'' b2 b1 or a1 b2 e e C4v (0 or 2) 9. (a) Fe(III) is d5 and CN– is a strong field ligand, so the complex is low-spin, with 1 unpaired electron. LFSE= 5x2/5Δo = 2 Δo. (b) [Co(NH3)6]3+ is low spin d6; LFSE = 6x2/5 Δo = 12/5 Δo and all the electrons are paired, so it is diamagnetic. (c) [Fe(H2O)6]2+ is high-spin d6, with 4 unpaired electrons. LFSE = 8/5 Δo – 6/5 Δo = 2/5 Δo. (d) [CoF6]3– is high-spin d6; same results as in (c). (e) [Ru(NH3)6]2+ Since Ru is a 2nd-row metal, we expect a large Δo and a low P, so it should be low spin. LFSE = 6x2/5 Δo = 12/5 Δo ; diamagnetic. (f) [CoCl4]2– is tetrahedral (so high spin) and d7. LFSE = 12/5 Δt – 6/5 Δt = 6/5 Δt and there are 3 unpaired electrons. (g) [Mn(H2O)6]2+ is high spin d5; LFSE = 0; 5 unpaired electrons. (h) [Cr(H2O)6]2+ is high spin d4; LFSE = 6/5 Δo - 3/5 Δo = 3/5 Δo; 4 unpaired electrons. 10. The true value for ΔH hydration should be larger (more negative) than the spherical ion value by an amount equal to the LFSE. For Ni(II) (d8), the LFSE = 1.2 Δo. Since Δo = 8600 cm-1 the LFSE = 10,320 cm-1 . From the back of the textbook, 1 cm-1 = 11.96 J/mol, so LFSE = 123.45 kJ/mol. Then Corrected ΔHhydration = 3123 kJ/mol. 5