Exam 3 Key - Chem 10181 Fall 2007(1 EXAM 3 KEY November 8...

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Unformatted text preview: Chem 10181, Fall 2007 (1) EXAM 3 KEY November 8, 2007 1 Both [Co(NO2)6]3- and [CoF6]3- are octahedral complexes of cobalt, but they differ in a number of ways. One is orange, the other blue; one is paramagnetic, the other diamagnetic. Their UV-visible spectra are shown below. (Note that the scales are not the same; "m" is synonymous with nm. The spectrum on the right is from Cotton, F. A.; Meyers, M. D. J. Am. Chem. Soc. 1960, 82, 5023-5026.) In the complex containing nitrite, NO2-, it is the nitrogen atom that is attached to the cobalt. 1000 , M-1cm-1 800 600 , M-1cm-1 400 200 0 400 450 500 550 600 , nm 2 6 650 700 750 800 Optical spectrum, Na [Co(NO ) ], aqueous solution 3 (a) (12 pts.) On the same energy scale, draw the d-orbital splitting diagrams for both octahedral complex ions [Co(NO2)6]3- and [CoF6]3-. Fill each diagram with the appropriate number of electrons, and SPECIFY which complex is orange and which is blue, as well as which is paramagnetic and which is diamagnetic. Both compounds are Co(III), therefore both have six d electrons. From the spectra, Co(NO2)63- has a much larger energy gap than does CoF63- (d-d band at = 475 nm vs. 800 nm or so). Therefore, if one is paramagnetic, it must be CoF63(i.e., that must be high-spin while the nitrite complex is lowspin). Co(NO2)63- absorbs in the blue, so it will appear orange, while CoF63- absorbs in the redorange region, so it will appear blue. Co(NO2)63diamagnetic orange CoF63paramagnetic blue Chem 10181, Fall 2007 EXAM 3 KEY November 8, 2007 2 (b) (c) (10 pts.) Which ligand exerts a stronger ligand field, NO2- or F-? Explain the difference in terms of the bonding of the two ligands to the transition metal. (Note that the pKa of HNO2 is 3.14, while the pKa of HF is 3.19.) The ligands are very similar in basicity, so it is unlikely that the strength of the bonding is very different. However, F- has filled orbitals (it is isoelectronic with Cl- which was discussed in class), which will push the t2g orbitals up in energy and therefore weaken the ligand field. NO2- has a bond to nitrogen; it is isoelectronic to allyl, so the empty * orbital has significant N character and will therefore lower the energy of the t2g orbitals (analogous to CO that was discussed in class). (8 pts.) K3CoF6 shows an additional absorption at = 252 nm that is about 50 times more intense than its absorptions shown on the previous page (Allen, G. C.; El-Sharkawy, G. A. M.; Warren, K. D. Inorg. Chem. 1971, 10, 2538-2546). Suggest an explanation for this intense absorption band, being as specific as possible. The high intensity indicates that this band is not a d-d transition. Since F- does not have any empty orbitals, it must be a ligand-to-metal charge transfer transition. Chem 10181, Fall 2007 EXAM 3 KEY November 8, 2007 3 A few years ago, introductory chemistry students at Notre Dame prepared an isomer of [Co(en)2(NO2)2]+ (isomer A) from Na3[Co(NO2)6] as described below (en = ethylenediamine, NH2CH2CH2NH2). They also prepared a second isomer of [Co(en)2(NO2)2]+ (isomer B) starting from [Co(H2O)6]2+: Na3[Co(NO2)6] + 2 en 70 C [Co(en)2(NO2)2](NO2) HNO3 H2O2 HNO 3 [Co(en)2(NO2)2](NO3) Isomer A [Co(en)2(NO2)2](NO3) Isomer B room [Co(H2O)6](NO3)2 + 2 en + 2 NO2 temperature [Co(en)2(NO2)2] (d) (8 pts.) The reaction involving Na3[Co(NO2)6] requires heating to 70 C, while the reaction of [Co(H2O)6]2+ takes place readily at room temperature. Rationalize this difference in reaction conditions. [Co(NO2)6]3- is kinetically inert (see part (a)), since it has no eg electrons and no empty t2g orbitals. Co(H2O)62+ is d7, so it will certainly have at least one eg electron (in fact, since it is high-spin, it has two), which makes it kinetically labile. The reactions shown require replacing the original ligands with en (and with nitrite, in the second reaction), so since [Co(NO2)6]3- will exchange ligands much more slowly than Co(H2O)62+, the former requires heating to go at a reasonable rate while the latter reacts speedily even at room temperature. Chem 10181, Fall 2007 EXAM 3 KEY November 8, 2007 4 (e) (10 pts.) The 13C NMR spectra of isomer A and isomer B (obtained in D2O solution) are shown below. Draw CLEAR 3-D REPRESENTATIONS of these two octahedral complexes. Be sure to specify how you assigned which compound was A and which was B on the basis of the 13C NMR spectra. CH2 CH2 NH2 NO2 NH2 H2 N Co NO2 NH2 CH2 CH2 + NO2 H2N CH2 CH2 NH2 NH2 Co NO2 NH2 CH2 CH2 + Isomer A (or its mirror image) Isomer B In isomer B, all four carbon atoms are in the same environment. In isomer A, the two carbons on each en are different (one is attached to the N opposite an NO2-, while the other is attached to a nitrogen opposite the other en). Therefore A will have two peaks in the 13C NMR while B has only one. Chem 10181, Fall 2007 (2) (a) EXAM 3 KEY November 8, 2007 5 The hexachloroplatinate ion, PtCl62-, forms a relatively insoluble potassium salt. The solubility of K2[PtCl6] in water at room temperature is 4.81 g/L. (10 pts.) Calculate the Ksp of K2[PtCl6]. FW of K2[PtCl6] = 2(39.10) + 195.08 + 6(35.45) = 485.98 4.81 g/L K2[PtCl6]/(485.98 g/mol) = 0.00990 mol/L K2[PtCl6] 2 K+ + PtCl62Ksp = [K+]2[PtCl62-] = [0.0198]2[0.00990] Ksp = 3.88 10-6 (12 pts.) In order to recover more of the precious metal platinum, a chemist wishes to precipitate more of the K2[PtCl6] by adding KNO3. How many grams of KNO3 would need to be added to 1.00 L of a saturated solution of K2[PtCl6] to cause 3.50 g K2[PtCl6] to precipitate? (If you were UNABLE to do part (a), you may assume that the Ksp of K2[PtCl6] is 1.00 10-5.) If 3.50 g K2[PtCl6] have precipitated, then 4.81g - 3.50g = 1.31 g K2[PtCl6] (= 0.00270 mol) remain dissolved. So [PtCl62-] = 0.00270 M Ksp = [K+]2[PtCl62-] 3.88 10-6 = [K+]2[0.00270] [K+] = 0.0379 M Of the total K+, 2(0.00270) = 0.00540 mol come from the dissolved K2[PtCl6], so 0.0379 - 0.0054 = 0.0325 mol KNO3 must have been added. 0.0325 mol KNO3 101.10 g KNO3/mol = 3.29 g KNO3 (b) Chem 10181, Fall 2007 (3) EXAM 3 KEY November 8, 2007 6 Zinc carbonate, ZnCO3, is rather insoluble in pure water (Ksp = 1.4 10-11). However, zinc carbonate is much more soluble in the presence of ammonia, due its strong complexation of zinc to form Zn(NH3)42+ (Kf = 2.9 109). (6 pts.) Write a chemical equation that represents the major reaction that takes place when ZnCO3 dissolves in the presence of ammonia, and give its Keq. ZnCO3 (s) + 4 NH3 Zn(NH3)42+ (aq) + CO32- (aq) Keq = KfKsp = 0.0406 (a) (b) (12 pts.) Calculate the minimum number of moles of NH3 that would need to be added to 1.00 L of solution to just dissolve 10.00 g ZnCO3. 10.00 g ZnCO3/(125.40 g ZnCO3/mol) = 0.0797 mol ZnCO3 ZnCO3 (s) + 4 NH3 Zn(NH3)42+ (aq) + CO32- (aq) Since all the ZnCO3 dissolves according to this major equation, [Zn(NH3)42+] = [CO32-] = 0.0797 M. [Zn(NH3 )42+][CO32- ] = Keq [NH3 ]4 [0.0797][0.0797] = 0.0406 [NH3]4 [NH3] = 0.629 M This is the concentration of free ammonia. There is also 4(0.0797 mol) = 0.319 mol NH3 bound to the Zn. So the total NH3 that would need to be added is 0.629 + 0.319 = 0.948 mol NH3. Chem 10181, Fall 2007 (c) EXAM 3 KEY November 8, 2007 7 (12 pts.) To the rather basic solution in part (b) (pH ~ 11.5) is added a solution of concentrated aqueous HCl dropwise until the pH is lowered to 9.50. (The volume change involved in this addition is negligible.) Will ZnCO3 precipitate from the solution after the addition of the HCl? Justify your answer. The Ka of NH4+ is 5.6 10-10 and the Ka of HCO3 is 4.8 10-11. Making the solution more acidic decreases the [NH3], which causes the [Zn2+] to increase. But making the solution more acidic also decreases the [CO32-], which allows more [Zn2+] to be present in solution before precipitation begins! So which effect is more significant? Start with the speciation of the two bases: [H3O+] = 10-9.5 = 3.16 10-10 [CO32-][H3O+] = 4.8 x 10-11 HCO3- + H2O CO32- + H3O+ [HCO3-] [CO32- ] = 0.15 [HCO3-] [NH3 ][H3O+] = 5.6 x 10-10 [NH4+] NH4+ + H2O NH3 + H3O+ [NH3] = 1.77 [NH4+] We will initially assume that the solution remains homogeneous, and then test that assumption by comparing [Zn2+][CO32-] to Ksp. If the Zn2+ remains predominantly complexed by NH3 (we will check this assumption presently), then the total amounts of carbonate and ammonia remain 0.0797 and 0.629 M, respectively, so: [CO32-] + [HCO3-] = [CO32-] + [CO32-]/0.15 = 0.0797 [CO32-] = 0.0104 [NH3] + [NH4+] = [NH3] + [NH3]/1.77 = 0.629 [NH3] = 0.402 From the complex ion formation equilibrium, one can calculate [Zn2+]: [Zn(NH3)42+] Kf = [Zn2+][NH3]4 [0.0797] 2.9 x 109 = [Zn2+][0.402]4 [Zn2+] = 1.05 10-9 M {so it is indeed mostly complexed} [Zn2+][CO32-] = [1.05 10-9][0.0104] = 1.09 10-11 Since [Zn2+][CO32-] < Ksp (1.09 10-11 < 1.4 10-11) then ZnCO3 will not precipitate. Chem 10181, Fall 2007 EXAM 3 KEY November 8, 2007 8 Note that it is not enough to note that the [CO32-] decreases much more than the [NH3], since the solubility depends on the first power of [CO32-], while complexation depends on the fourth power of [NH3]. So while at pH 9.5, the change in the CO32- concentration is more important, and ZnCO3 is soluble, as the pH is lowered still further, the change in [NH3]4 becomes more significant, and ZnCO3 will precipitate! ...
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This homework help was uploaded on 04/17/2008 for the course CHEM 10181 taught by Professor Brown during the Fall '08 term at Notre Dame.

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