Chemistry Labs

Problem 1

Boron 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.
Step 1 of 4: Formulae of A and B
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n(B)A=0.831×65.110.81=5.0⇒BX5HX11;n(B)B=0.885×122.210.81=10.0⇒BX10HX14n(\mathrm{B})_A = \dfrac{0.831 \times 65.1}{10.81} = 5.0 \Rightarrow \ce{B5H11};\qquad n(\mathrm{B})_B = \dfrac{0.885 \times 122.2}{10.81} = 10.0 \Rightarrow \ce{B10H14}
Analysis

Dividing the boron mass per mole by the atomic mass gives 5.0 B atoms for A (leaving ≈11\approx 11 H) and 10.0 for B (leaving ≈14\approx 14 H): BX5HX11\ce{B5H11} and BX10HX14\ce{B10H14}, both classic Stock boranes.