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Electron Sharing Reactions · Paper 2 — Free Chemistry SL Practice Questions

1MasterySAQ-SCovalent bonds and electron sharing5 marksPaper 2~8 min
Hydrogen peroxide, H2O2\text{H}_2\text{O}_2, is used as a disinfectant and bleaching agent. Its structural formula is H–O–O–H, containing one O–O bond and two O–H bonds. Bond enthalpies: O–O=+146 kJ mol−1\text{O–O} = +146\,\text{kJ mol}^{-1}; O–H=+463 kJ mol−1\text{O–H} = +463\,\text{kJ mol}^{-1}
(a)
State the number of electrons shared in the O–O bond of H2O2\text{H}_2\text{O}_2. [1 mark]
(b)
Calculate the total energy, in kJ mol−1\text{kJ mol}^{-1}, required to break all bonds in one mole of H2O2\text{H}_2\text{O}_2 molecules. [2 marks]
(c)
Deduce, with reference to atomic orbital overlap, why the O–O bond enthalpy is significantly lower than the O–H bond enthalpy, and predict whether the O–O bond in H2O2\text{H}_2\text{O}_2 would be stronger or weaker than a C–C single bond (bond enthalpy +347 kJ mol−1+347\,\text{kJ mol}^{-1}), justifying your answer. [2 marks]
diagram

Solutions

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