Metals - Homework solutions

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1 Metals - Homework solutions Q Ex 1,2,3,6,8,10,12,13,14,17,20 and Prob a) Melting point decreases down the groups Melting point increases on going left to right b) The radii increase on going down a group The radii decrease going left to right c) The peroxides / superoxides are least stable for the smaller ions Usually, anions are stabilized by cations of similar size. This can be rationalized using lattice energy arguments. 8.2 a) Mg 2+ is likely to form a stable complex. Binding in complexes is strongly influenced by the charge to size ratio of the cation. Small high charge cations form the most tightly bound complexes. b) Sr is most likely to dissolve in liquid ammonia. Dissolution depends upon how easy it is to break up the metal lattice, how readily the metal ionizes and how well solvated the resulting cation will be. Sr wins over Be due to it s lower ionization and lattice energies. c) K + is best bound by cryptand. K + is the correct size to fit inside the ligand and Li + is too small. 8.3 a) Ca 2+ is eight coordinate (cubic) in CaF2 Mo 2+ is six coordinated (trigonal prismatic) in MoS2 For ionic compounds (like CaF2) large coordination numbers are often found for large cations. For more covalent compounds like MoS2 lower coordination numbers may be favored due tot he covalent bonding. b) CdI2 has a layer structure based upon hexagonal close packed I - with half of the octahedral holes filled with Cd 2+ ions. MoCl2 has an octahedral metal cluster structure. Metal-metal bonding occurs as there are partly filled d-orbitals with the correct size for overlap.

2 c) CaO has a rock salt structure (6 coordinate). BeO has a Wurtzite structure (4 coordinate). The smaller size of Be2+ favor the lower coordination number.

3 d)molybdenum (II) acetate is a metal-metal bonded dimer. Basic beryllium acetate does not contain any metal--metal bonds (no partially filled orbitals in Be 2+ ) and it has a structure based on a tetrahedral arrangement of Be around a central oxide ion. 8.6 SEE LATIMER DIAGRAMS IN BACK OF BOOK a) Cr 2+ (aq) + Fe 3+ (aq) -----> Cr 3+ (aq) + Fe 2 +(aq) M(II) cation become less reducing on going from left to right b) CrO4 2- (aq) + MoO2(s) + 2H2O -----> MoO4 2- (aq) + Cr(OH)3(s) + OH - (aq) High oxidation states become increasingly stable on going down a group. c) MnO4 - (aq) + 4H + (aq) + 3e > MnO2(s) + 2H2O Cr2O7 2- (aq) + 14H + (aq) + 6e > 2Cr 3+ (aq) + 7H2O 2MnO4 - (aq) + 2Cr 3+ (aq) + 5H2O > 2MnO2(s) + 6H + (aq) + Cr2O7 2- (aq) 8.8 b) First row metals Ti ---> Ni. The heavier early metals will not form stable difluorides. The dihalides of metals like Mo or W contain metalmetal bonds or disproportionate. The fluorides are not good candidates form metal-metal bonds and are likely to be unstable with respect to disproportionation. c) The early heavy transition metals Mo, W, Re, Nb, form metal- metal bonded halides. ReCl3, MoCl2 etc a) No reaction as molybdate is not oxidizing b) 6K2MoO4(s) + 10H3O + (aq) > [Mo6O19] 2- (aq) + 12K + (aq) + 15H2O To get molybdate to condense we need to remove oxide. We can do this by adding acid and producing water.

4 2K + (aq) c) 5ReCl5(s) + 2KMnO4(aq) + 12H2O > 5ReO4 - (aq) + 2Mn 2+ (aq) H + (aq) + 25Cl - (aq) Re(VII) is not very oxidizing when compared to Mn(VII)

5 d) 6MoCl2(s) +2 HCl(aq) -----> 2H + (aq) + Mo6Cl14 2- (aq) MoCl2 contains metal-metal bonded clusters. These are not oxidized by aqueous acid. The compound dissolves producing discrete cluster anions a) FeF2 predominantly ionic - rutile structure. Fe and F differ markedly in electronegativity and Fe 2+ is not a very high oxidation state. b) NiI2, PtS and NbCl4 are significantly covalent NiI2 - layered structure, soft anions PtS - Square planar Pt(II) in a 3-D net, soft anions NbCl4 - chain structure, high formal oxidation state metal c) WCl2 metal-metal bonded - partly filled d-orbitals, d-orbitals are the right size for overlap and bond formation a) 2TiO(s) + 6HCl(aq) > H2(g) + 2Ti 3+ (aq) + 2H2O + 6Cl - (aq) Ti(II) is a reducing oxidation state. However, in acid Ti 3+ is stable. b) Ce 4+ (aq) + Fe 2+ (aq) -----> Ce 3+ (aq) + Fe 3+ (aq) Ce 4+ is oxidzing and Fe 3+ is not oxidzing c) Rb9O2 + 7H2O > 9RbOH + 5/2H2 Rb9O2 is a suboxide. It behaves like a mixture of Rb metal and Rb2O. d) Na(am) + MeOH -----> NaOMe + Na + + 1/2H2 Na(am) is a good reducing agent and is capable of reducing H + from the methanol a) Mo2(Ac)4 σ 2 π 4 δ 2 δ* 0 π* 0 σ* 0 Bond order = 4 b) Cr2(O2CC2H5)4 σ 2 π 4 δ 2 δ* 0 π* 0 σ* 0 Bond order = 4

6 c) Cu2(Ac)4 σ 2 π 4 δ 2 δ* 2 π* 4 σ* 2 Bond order = a) On going down the groups species with an oxidation state two less than the group oxidation state become increasingly stable. Pb, Sn and Tl commonly display the inert pair effect. b) i) Sn 2+ + PbO > Sn 4+ + Pb 2+ Sn 2+ is more reducing than Pb 2+ ii) Tl 3+ + Al > Tl + Al 3+ Al is more electropositive than Tl iii) In + (aq) > In + In 3+ (aq) The chemistry of In + is solvent dependent. In MeCN it is stable but in water it disproportionates. In + is reducing and is capable of reducing itself. iv) Sn /2O2 + H2O -----> Sn OH - Sn(II) is a reducing species v) Tl + + O2 ----> NO reaction Tl 3+ is a powerful oxidizing agent and can not be prepared by the aerial oxidation of aqueous Tl Most of the lanthanide elements only have stable trivalent cations in aqueous solution. However, Ce and Eu can form Ce 4+ and Eu 2+ cations in solution. These cations have very different size/charge ratios form the other lanthanide M(III) cations and can be easily separated. Prob 8.8

7 a) is correct b) is incorrect. Hydrogen is evolved. c) is correct d) is incorrect. WBr2 has a metal-metal bonded structure.

M10/4/CHEMI/SPM/ENG/TZ2/XX+ CHEMISTRY. Wednesday 12 May 2010 (afternoon) 45 minutes INSTRUCTIONS TO CANDIDATES

M10/4/CHEMI/SPM/ENG/TZ2/XX+ CHEMISTRY. Wednesday 12 May 2010 (afternoon) 45 minutes INSTRUCTIONS TO CANDIDATES M10/4/CHEMI/SPM/ENG/TZ/XX+ 106116 CHEMISTRY standard level Paper 1 Wednesday 1 May 010 (afternoon) 45 minutes INSTRUCTIONS TO CANDIDATES Do not open this examination paper until instructed to do so. Answer