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Electron-Pair Sharing Reactions · Paper 2 — Free Chemistry HL Practice Questions

1ChallengeSAQ-LCoordinate covalent bonding7 marksPaper 2~11 min

Data

- Electronegativity values: N=3.0\text{N} = 3.0, H=2.2\text{H} = 2.2, B=2.0\text{B} = 2.0 - Bond dissociation energies: N–H=391 kJ mol−1\text{N–H} = 391\,\text{kJ mol}^{-1}; B–H=389 kJ mol−1\text{B–H} = 389\,\text{kJ mol}^{-1}; N–B (coordinate)=310 kJ mol−1\text{N–B (coordinate)} = 310\,\text{kJ mol}^{-1} - Enthalpy of sublimation of NH3BH3(s)\text{NH}_3\text{BH}_3\text{(s)}: +85 kJ mol−1+85\,\text{kJ mol}^{-1}
Ammonium borane (NH3BH3\text{NH}_3\text{BH}_3) is a stable white crystalline solid at room temperature. It decomposes at around 110 ∘C110\,^\circ\text{C} rather than melting. The molecule contains a coordinate covalent (dative) bond between the nitrogen and boron atoms.
(a)
Calculate the enthalpy change, ΔH\Delta H, for the gas-phase reaction: NH3(g)+BH3(g)→NH3BH3(g)\text{NH}_3\text{(g)} + \text{BH}_3\text{(g)} \rightarrow \text{NH}_3\text{BH}_3\text{(g)} State the type of bond formed and identify which atom donates both electrons. [3 marks]
(b)
Using the electronegativity data provided, explain why the N–B coordinate bond has a lower bond dissociation energy than the N–H covalent bond. [2 marks]
(c)
Evaluate the claim: "The existence of a stable solid such as NH3BH3\text{NH}_3\text{BH}_3 shows that coordinate covalent bonds are stronger than ordinary covalent bonds." In your answer, use both the bond energy data and the enthalpy of sublimation to support your evaluation. [2 marks]
diagram

Solutions

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