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 3 of 4: B–B bond enthalpy by Hess cycle
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ΔrH(BX2ClX4(g)+ClX2(g)→2 BClX3(g))=2(−403)−(−489)=−317 kJ;−317=[4(443)+242+D(B−B)]−6(443)⇒D(B−B)=327 kJ mol−1\Delta_rH(\ce{B2Cl4(g) + Cl2(g) -> 2BCl3(g)}) = 2(-403) - (-489) = -317\ \text{kJ};\quad -317 = [4(443) + 242 + D(\ce{B-B})] - 6(443) \Rightarrow D(\ce{B-B}) = 327\ \text{kJ mol}^{-1}
Analysis

The atomisation route breaks 4 B–Cl bonds and 1 B–B bond in BX2ClX4\ce{B2Cl4} plus 1 Cl–Cl bond, and reforms 6 B–Cl bonds in 2 BClX32\,\ce{BCl3}: ΔrH=4(443)+D(B−B)+242−6(443)=−317\Delta_rH = 4(443) + D(\ce{B-B}) + 242 - 6(443) = -317 kJ, giving D(B−B)=327D(\ce{B-B}) = 327 kJ mol−1^{-1}.

Common pitfall. Use half a mole of ClX2\ce{Cl2} carefully: the cycle compares BX2ClX4+ClX2\ce{B2Cl4 + Cl2} to 2BClX32\ce{BCl3}, so only one extra Cl–Cl bond is broken.