CCEA A-Level Chemistry 1110
Formulae, Equations and Amounts of Substance
Atomic Structure
Bonding
Intermolecular Forces and Structure
Shapes of Molecules and Ions
Redox
The Halogens
Acid-Base Titrations
Qualitative Tests
Formulae and Amounts of a Substance
Nomenclature and Isomerism in Organic Chemistry
Alkanes
Alkenes
Halogenoalkanes
Alcohols
Infrared Spectroscopy
Energetics
Kinetics
Equilibrium
Group II Elements and Their Compounds
Lattice Enthalpy, Entropy and Free Energy
Rates of Reaction
Equilibrium and Acid-Base Equilibria
Isomerism
Aldehydes and Ketones
Carboxylic Acids
Derivatives of Carboxylic Acids
Aromatic Chemistry
Spectroscopy and Chromatography
Volumetric Analysis
Transition Metals
Electrode Potentials
Amines
Amides and Polymers
Amino Acids and Proteins
Lattice Enthalpy, Entropy and Free Energy
Born–Haber cycles for Group II halides and oxides
- 3 things to know
- Unit A2 1
Overview
Recall questions on how a Group II cycle differs from a Group I one, worked calculations for magnesium chloride, magnesium oxide and calcium oxide, why the halogen terms are doubled, the second electron affinity of oxygen and why it is endothermic, why the second ionisation energy is much larger, and comparing MgO with CaO and with MgCl₂.
What you need to know
Answer each one before you reveal it.
Why must the atomisation and electron affinity terms for chlorine be doubled in the cycle for magnesium chloride?
Explain why the second electron affinity of oxygen is endothermic.
Questions from this subtopic
Each one with the model answer underneath.
- 1
How does a Born–Haber cycle for a Group II halide differ from one for a Group I halide?
Model answer
Two ionisation energies are needed for the metal because the ion carries a 2+ charge, and two moles of halogen atoms must be formed and each given an electron.
- 2
Using ΔHf(MgCl₂) = −641 kJ mol⁻¹, ΔHat(Mg) = +148 kJ mol⁻¹, the first and second ionisation energies of magnesium = +738 and +1451 kJ mol⁻¹, ΔHat(Cl) = +121 kJ mol⁻¹ and the first electron affinity of chlorine = −349 kJ mol⁻¹, calculate the lattice enthalpy of magnesium chloride.
Model answer
The route to the gaseous ions totals 148 + 738 + 1451 + 2(121) + 2(−349) = +1881 kJ mol⁻¹, so the lattice enthalpy is −641 − 1881 = −2522 kJ mol⁻¹.
- 3
Why must the atomisation and electron affinity terms for chlorine be doubled in the cycle for magnesium chloride?
Model answer
One mole of MgCl₂ contains two moles of chloride ions, so two moles of chlorine atoms must be formed and each must gain an electron.
- 4
Define the second electron affinity of oxygen.
Model answer
It is the enthalpy change when one electron is added to each ion in one mole of gaseous O⁻ ions to form one mole of gaseous O²⁻ ions.
- 5
Explain why the second electron affinity of oxygen is endothermic.
Model answer
An electron is being forced onto an ion that already carries a negative charge, so energy must be supplied to overcome the repulsion between them.
- 6
Using ΔHf(MgO) = −602 kJ mol⁻¹, ΔHat(Mg) = +148 kJ mol⁻¹, ionisation energies +738 and +1451 kJ mol⁻¹, ΔHat(O) = +249 kJ mol⁻¹, the first electron affinity of oxygen = −141 kJ mol⁻¹ and the second = +798 kJ mol⁻¹, calculate the lattice enthalpy of magnesium oxide.
Model answer
The steps to the gaseous ions total 148 + 738 + 1451 + 249 − 141 + 798 = +3243 kJ mol⁻¹, so the lattice enthalpy is −602 − 3243 = −3845 kJ mol⁻¹.
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