Chapter (a) White pigment; (b) green pigment; (c) starting material for making other silver compounds (and as a laboratory reagent).

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1 Chapter 20 PROPERTIES OF THE d TRANSITION METALS Exercises 20.2 (a) 2 VO 2+ (aq) + Zn(s) + 4 H + (aq) 2 V + (aq) + Zn 2+ (aq) + 2 H 2 O(l) 2 V + (aq) + Zn(s) 2 V 2+ (aq) + Zn 2+ (aq) (b) 2 Cr + (aq) + 7 H 2 O(l) Cr 2 O 2 7 (aq) + 4 H + (aq) + 6 e FeO 2 4 (aq) + 8 H + (aq) + e Fe + (aq) + 4 H 2 O(l) 2 Cr + (aq) + 2 FeO 2 4 (aq) + 2 H + (aq) Cr 2 O 2 7 (aq) + 2 Fe + (aq) + H 2 O(l) (c) Cu(OH) 2 (s) CuO(s) + H 2 O(l) (d) 2 Cu 2+ (aq) + 4 I (aq) 2 CuI(s) + I 2 (aq) (e) 2 Au(s) + Cl 2 (g) 2 AuCl (s) 20.4 (a) White pigment; (b) green pigment; (c) starting material for making other silver compounds (and as a laboratory reagent) Chlorine is being reduced (from 0 to ), and carbon is being oxidized (from 0 to +2) Although aluminum is plentiful in the Earth s crust, it is usually found in such minerals as clays, from which extraction is very difficult. It is only from the rare bauxite ore that aluminum is commercially obtainable. Even then, high-cost electrolytic methods must be used for the extraction process. Iron can be obtained from the plentiful deposits of iron oxides by a low-cost coke reduction process The chemical equation is Ni(s) + 4 CO(g) Ni(CO) 4 (g) Formation of this compound requires a net decrease of three moles of gas, resulting in a decrease in entropy. Hence the forward reaction must be enthalpy driven (that is, exothermic). The decomposition reaction is favored at higher temperatures. This is logical when the relationship G = H T S is considered. For the reverse reaction, both H and S are positive, thus as the temperature increases, ( T S ) will become 5

2 6 Chapter 20 increasing negative, to the point where the term exceeds H and the reverse reaction becomes spontaneous (negative G ) (a) Cobalt: [Co(OH 2 ) 6 ] 2+ (aq) + 4 Cl (aq) [CoCl 4 ] 2 (aq) + 6 H 2 O(l) (b) Copper: 2 Cu(s) + 2 H + (aq) + 4 Cl (aq) 2 [CuCl 2 ] (aq) + H 2 (g) [CuCl 2 ] (aq) CuCl(s) + Cl (aq) (c) Chromium: 2 CrO 2 4 (aq) + 2 H + (aq) Cr 2 O 2 7 (aq) + H 2 O(l) 20.4 A halide ion is an obvious choice because a large negative weak field ligand will favor a tetrahedral arrangement. Iodide would be the specific choice because it is the largest of the halides. Also, vanadium(ii) is in a very low oxidation state for the metal. To stabilize it, a reducing anion is preferable and iodide is strongly reducing The aqueous iron(iii) ion hydrolyses in aqueous solution: [Fe(OH 2 ) 6 ] + (aq) + H 2 O(l) [Fe(OH 2 ) 5 (OH)] 2+ (aq) + H O + (aq) thus the yellow color of the hydroxy-species predominates unless acid is added to drive the equilibrium to the left to give the pale purple color of the hexaaquairon(iii) ion Oxygen. Iron(III) is a hard acid, while oxygen is a hard base (and sulfur is a soft base) Chromium(VI) oxide should be acidic. It is the metal in the higher oxidation state (that is, with the more oxygens) that will exhibit the more acidic properties Fluoride ion; [CoF 6 ] Three.

3 Properties of the d Transition Metals (a) Cobalt(II): [Co(OH 2 ) 6 ] 2+ (aq) + 4 Cl (aq) [CoCl 4 ] 2 (aq) + 6 H 2 O(l) Co 2+ (aq) + 2 OH (aq) Co(OH) 2 (s) (b) Chromate: CrO 2 4 (aq) + Ba 2+ (aq) BaCrO 4 (s) 2 CrO 2 4 (aq) + 2 H + (aq) Cr 2 O 2 7 (aq) + H 2 O(l) CrO 2 4 (aq) + 8 H + (aq) + e Cr + (aq) + 4 H 2 O(l) SO 2 (aq) + 2 H 2 O(l) SO 2 4 (aq) + 4 H + (aq) + 2 e (c) Copper(II): Cu 2+ (aq) + Zn(s) Cu(s) + Zn 2+ (aq) [Cu(OH 2 ) 6 ] 2+ (aq) + 4 NH (aq) [Cu(NH ) 4 ] 2+ (aq) + 6 H 2 O(l) (a) Copper; (b) iron; (c) iron and molybdenum; (d) cobalt. Beyond the Basics 20.0 Ni(s) + 4 CO(g) Ni(CO) 4 (g) Hº = [( 602.9) 4( 0.5)] kj mol = 60.9 kj mol Sº = [(+40.6) 4(+97.7) (+29.9)] J mol K = 40. J mol K Gº = Hº T Sº = ( 60.9 kj mol ) (298 K)( 0.40 kj mol K ) = 8.7 kj mol ln K = Gº/RT = ( J mol )/(8. J mol K )(298 K) = 5.6 K = When K =, Gº = 0. T = Hº/ Sº = ( 60.9 kj mol )/( 0.40 kj mol K ) = 92 K = 9ºC At room temperature the formation of the tetracarbonylnickel(0) is favored, but when warmed the nickel carbonyl complex decomposes again. This is the route used for nickel purification.

4 8 Chapter First we must calculate the two lattice energies: 2 ( mol ) ( C J 9 2 ( C) m )( m) U CuF 0 = 956 kj mol 2 ( mol ) ( C 2 J 9 2 ( C) m )( m) U CuF 2 0 = 2868 kj mol Then we set up Born-Haber cycles: 8 8

5 Properties of the d Transition Metals 9 H f º(CuF(s)) = [(+7) + (55) + (752) + ( 28) + ( 956)] kj mol = 8 kj mol H f º(CuF 2 (s)) = [(+7) + (55) + (752) + (+964) + 2( 28) + ( 2868)] kj mol = 6 kj mol Cu(s) + F 2 (g) CuF(s) Cu(s) + F 2 (g) CuF 2 (s) In each case, the entropy change will be negative, so the enthalpy decrease needs to be sufficent to overcome the entropy factor [Co(OS(CH ) 2 ) 6 ] 2+ 2 ClO 4 and [Co(OS(CH ) 2 ) 6 ] 2+ [CoCl 4 ] From the Appendix, VO 2+ (aq) + 2 H + (aq) + e V + (aq) + H 2 O(l) Eº = +0.7 V Gº = (F)(+0.7 V) = 0.7 F V + (aq) + e V 2+ (aq) Eº = V Gº = (F)( V) = F VO 2+ (aq) + 2 H + (aq) + 2 e V 2+ (aq) + H 2 O(l) Gº = 0.2 F = 2 FEº, Eº = V 2 Cr 2+ (aq) 2 Cr + (aq) + 2 e Eº = V VO 2+ (aq) + 2 H + (aq) + 2 e V 2+ (aq) + H 2 O(l) Eº = V VO 2+ (aq) + 2 Cr 2+ (aq) + 2 H + (aq) V 2+ (aq) + 2 Cr + (aq) + H 2 O(l) Eº = V Gº = nfeº = RTlnK 4 nfe ( C mol ) ( ln K RT 8. V C mol K K = V) 298 K 8.7 Probable two-step mechanism: VO 2+ (aq) + Cr 2+ (aq) + 2 H + (aq) V + (aq) + Cr + (aq) + H 2 O(l) V + (aq) + Cr 2+ (aq) V 2+ (aq) + Cr + (aq) 20.8

6 20 Chapter (a) The equations for K w and K sp can be combined to provide the required expression. Fe + (aq) + H 2 O(l) Fe(OH) (s) + H + (aq) Fe(OH) (s) Fe + (aq) + OH (aq) 2 H 2 O(l) H O + (aq) + OH (aq) K = [H + ] /[Fe + ] K sp = [Fe + ][OH ] = K w = [H + ][OH ] = K (b) K K w sp [H ] [OH ] [Fe ][OH ] [H ] [Fe ] Substituting: [Fe + ] = ( ) /( ) = mol L (c) Mol Fe(OH) = ( mol L )( 0 6 L) = mol Mass Fe(OH) = (06.9 g mol - )(2 0 9 mol) = g Very little iron(iii) hydroxide is lost! [A] Manganese(II) nitrate; [B] manganese(iv) oxide; [C] nitrogen dioxide; [D] manganese(ii) chloride; [E] dichlorine; [F] dibromine; [G] permanganate ion; [H] dioxygen; [I] dinitrogen tetraoxide. Mn(NO ) 2 (s) MnO 2 (s) + 2 NO 2 (g) MnO 2 (s) + 4 HCl(aq) MnCl 2 (aq) + Cl 2 (g) + 2 H 2 O(l) Cl 2 (aq) + 2 Br (aq) 2 Cl (aq) + Br 2 (aq) MnO 4 (aq) + 2 H 2 O(l) + e MnO 2 (s) + 4 OH (aq) H 2 O 2 (aq) + 2 OH (aq) O 2 (g) + 2 H 2 O(l) + 2 e 2 NO 2 (g) N 2 O 4 (g) The crystal must contain some chromium(iii) ions to provide the charge balance in fact, it is easy to calculate that, of every 92 chromium ions, 76 must be chromium(ii) and 6 chromium(iii) FeI (s) + 2 e 2 FeI 2 (aq) + 2 I (aq) 2 H 2 O(l) O 2 (g) + 4 H + (aq) + 4 e Net reaction: 4 FeI (aq) + 2 H 2 O(l) 4 FeI 2 (aq) + 4 HI (aq) + O 2 (g)

7 Properties of the d Transition Metals Under normal acid conditions, manganese is reduced from +7 to +2 oxidation state: MnO 4 (aq) + 8 H + (aq) + 5 e Mn 2+ (aq) + 4 H 2 O(l) Fluoride stabilizes high oxidation states, thus it would not be unreasonable to propose that reduction occurs only down to the + oxidation state: MnO 4 (aq) + 6 F (aq) + 8 H + (aq) + 4 e [MnF 6 ] (aq) + 4 H 2 O(l) This would require (20 ml) ( 5 / 4 ) = 25 ml of titrant to oxidize the iron(ii) ion CuCN(s) + CN (aq) [Cu(CN) 2 ] (aq) NiCo 2 O 4 contains Ni 2+ and Co +. Cobalt is readily oxidized to the lowspin d6 configuration, making this formulation CFSE energetically favourable. Theoretical CoNi 2 O 4 would require Ni + and Co 2+. Nickel is hard to oxidize to the + state and there would not be the CFSE advantage of the other permutation (in fact the Co 2+ would be oxidized by the Ni + ) (a) MnO: [Mn] + [ 2] = 0, Mn = +2 Mn O 4 : [Mn] + 4[ 2] = 0, Mn = + 8 / Mn 2 O : 2[Mn] + [ 2] = 0, Mn = + MnO 2 : [Mn] + 2[ 2] = 0, Mn = +4 Mn 2 O 7 : 2[Mn] + 7[ 2] = 0, Mn = +7 (b) Mn O 4 is a mixed oxide containing (Mn 2+ )(Mn + ) 2 (O 2 ) 4. (c) MnO should be basic and Mn 2 O 7 acidic. Low oxidation state metal oxides are basic, and high-oxidation-state metal oxides are acidic. (d) MnO 2 is the analog of ClO 2 (Mn 2 O 7 is the analog of Cl 2 O 7, but neither of these is a common oxide) For the first reaction: H = [2 Hº f (CO)] [ Hº f (CO 2 ) + Hº f (CH 4 )] = [2 ( )] [( 94) + ( 75)] kj mol

8 22 Chapter 20 = +247 kj mol For the second reaction: H = [ Hº f (CO)] [ Hº f (H 2 O) + Hº f (CH 4 )] = [( )] [( 242) + ( 75)] kj mol = +206 kj mol Both reaction steps are endothermic; thus they are driven by the entropy factor (increasing moles of gas).

Reactions in aqueous solutions Redox reactions

Reactions in aqueous solutions Redox reactions Redox reactions In precipitation reactions, cations and anions come together to form an insoluble ionic compound. In neutralization reactions, H + ions and