Chemistry Labs

International Chemistry Olympiad · 2012

Problems

  1. Problem 1Boron hydrides BXxHXy\ce{B_{x}H_{y}} were developed by Alfred Stock; William Lipscomb won the 1976 Nobel Prize for elucidating their bonding. (a) Using mass percent of boron and molar mass, derive the molecular formulae of two boron hydrides: A is a liquid at 25 °C, 83.1 % B, M=65.1M = 65.1 g mol−1^{-1}; B is a solid, 88.5 % B, M=122.2M = 122.2 g mol−1^{-1}. (b) The styx code counts B–H–B bridges (s), three-centre BBB bonds (t), two-centre B–B bonds (y) and BHX2\ce{BH2} groups (x); BX2HX6\ce{B2H6} is 2002. Propose a structure for tetraborane BX4HX10\ce{B4H10} with styx = 4012. (c) A compound BX4CClX6O\ce{B4CCl6O} has two types of B atoms (tetrahedral : trigonal planar = 1:3) and a C≡O\ce{C#O} triple bond; suggest a structure. (d) From D(B−Cl)=443D(\ce{B-Cl}) = 443, D(Cl−Cl)=242D(\ce{Cl-Cl}) = 242 kJ mol−1^{-1}, ΔfH∘(BClX3(g))=−403\Delta_fH^{\circ}(\ce{BCl3(g)}) = -403 and ΔfH∘(BX2ClX4(g))=−489\Delta_fH^{\circ}(\ce{B2Cl4(g)}) = -489 kJ mol−1^{-1}, estimate the B–B bond dissociation enthalpy in BX2ClX4\ce{B2Cl4}. (e) In a reaction scheme starting from BX2HX6\ce{B2H6}, give the boron-containing products: 1 = BX2HX6\ce{B2H6} + NaOH\ce{NaOH}/then CHX3OH\ce{CH3OH}, HX+\ce{H+}; 2 = product of BX2HX6\ce{B2H6} + CX6HX6\ce{C6H6}, OX2\ce{O2} chemistry identified by cryoscopy (ΔTf=0.205\Delta T_f = 0.205 °C for 0.312 g in 25.0 g benzene, Kf=5.12K_f = 5.12 kg K mol−1^{-1}); 3 = BX2HX6\ce{B2H6} + ClX2\ce{Cl2}; 4 = BX2HX6\ce{B2H6} + NHX3\ce{NH3} (1:1 adduct); 5 = thermal product of 4, b.p. 55 °C.Solutions: 1
  2. Problem 2Square-planar Pt(II) complexes substitute with retention of stereochemistry, and the rate of substitution of X by Y depends on the ligand trans to X (trans effect): CNX−>HX−>NOX2X−,IX−>BrX−,ClX−>pyridine,NHX3,OHX−,HX2O\ce{CN-} > \ce{H-} > \ce{NO2-}, \ce{I-} > \ce{Br-}, \ce{Cl-} > \text{pyridine}, \ce{NH3}, \ce{OH-}, \ce{H2O}. (a) Draw all stereoisomers of square-planar Pt(py)(NHX3)BrCl\ce{Pt(py)(NH3)BrCl} (py = pyridine). (b) Write reaction schemes, via intermediates, to prepare each stereoisomer of [Pt(NHX3)(NOX2)ClX2]X−\ce{[Pt(NH3)(NO2)Cl2]-} in water starting from PtClX4X2−\ce{PtCl4^{2-}}, NHX3\ce{NH3}, NOX2X−\ce{NO2-}, exploiting the trans effect. (c) Substitution follows Rate=kS[MLX3X]+kY[Y][MLX3X]\mathrm{Rate} = k_S[\ce{ML3X}] + k_Y[\ce{Y}][\ce{ML3X}]; for [py]≫[Pt][\ce{py}] \gg [\ce{Pt}], Rate=kobs[Pt]\mathrm{Rate} = k_{obs}[\ce{Pt}]. For the displacement of ClX−\ce{Cl-} by pyridine in methanol at 25 °C: [py][\text{py}] = 0.122, 0.061, 0.030 mol dm−3^{-3} give kobsk_{obs} = 7.20×10−47.20 \times 10^{-4}, 3.45×10−43.45 \times 10^{-4}, 1.75×10−41.75 \times 10^{-4} s−1^{-1}. Find kSk_S and kYk_Y and decide which pathway dominates at [py]=0.10[\text{py}] = 0.10 mol dm−3^{-3}. (d) A gold nanoparticle of diameter 13 nm carries 90 oligonucleotide groups, 98 % bound to a Pt(IV) complex; a 1.0 cm3^3 vessel contains 1.0×10−61.0 \times 10^{-6} mol dm−3^{-3} in Pt (density of Au = 19.3 g cm−3^{-3}). Calculate the masses of Pt and Au used.Solutions: 1