Questions on lattice enthalpy and Born–Haber cycles

Question 1

Explain what the data in the table show about the bonding in calcium fluoride and in calcium iodide, and why the two compounds differ. The table gives each compound's experimental (Born–Haber) and theoretical (perfect ionic model) lattice enthalpy of formation, and the radius of its anion.

Accepted Answer

Indicative content

for calcium fluoride the experimental value is only 21 kJ mol⁻¹, under 1%, more exothermic than the theoretical one, so the perfect ionic model fits it closely and the bonding is almost purely ionic
for calcium iodide the experimental value is 169 kJ mol⁻¹, about 8%, more exothermic than the theoretical one, a far larger departure from the ionic model
the perfect ionic model treats every ion as a perfect sphere with its charge spread evenly, so it leaves out any distortion of the electron clouds
the cation is Ca²⁺ in both compounds, so its polarising power is the same and the difference must come from the anion
I⁻ is much larger than F⁻, 220 pm against 133 pm, with its outer electrons further from the nucleus and more shielded, so its electron cloud is far more easily polarised
Ca²⁺ draws the iodide electron cloud towards itself, putting electron density between the nuclei, so CaI₂ has covalent character and its real lattice is held more strongly than the model predicts

Levels

Level 3 (5–6 marks) both comparisons are read from the table with the sizes of the departures, the perfect ionic model is described, the common cation is used to locate the difference in the anion, and the polarisation of the larger iodide ion is explained as the cause of the covalent character and the stronger real lattice
Level 2 (3–4 marks) the closer fit for calcium fluoride is noted and the covalent character of calcium iodide is explained by polarisation of the larger anion; the sizes of the departures, what the perfect ionic model assumes, or the role of the shared cation is not covered
Level 1 (1–2 marks) one compound is called more ionic, typically with the iodide ion said to polarise the calcium ion, and the data are not used

Question 2

Explain why the second ionisation energy of magnesium is larger than its first, and why the second electron affinity of oxygen is endothermic although its first electron affinity is exothermic.

Accepted Answer

B1 the second electron is removed from an Mg⁺ ion, which is already positively charged B1 the same nuclear charge acts on one fewer electron, so the remaining electrons are held more strongly and more energy is needed to remove one B1 the first electron affinity is exothermic because the incoming electron is attracted by the oxygen nucleus B1 the second electron is added to an O⁻ ion, which is already negative, so energy is needed to overcome the repulsion; reject 'oxygen has a full outer shell'

Question 3

Calculate the lattice enthalpy of formation of calcium chloride from the data in the table.

Accepted Answer

M1 the atomisation and the electron affinity of chlorine are each doubled: 2 × 122 = +244 and 2 × (−349) = −698 M1 the steps to the gaseous ions sum to 178 + 590 + 1145 + 244 − 698 = +1459 kJ mol⁻¹ M1 ΔLEH = ΔfH − 1459, from rearranging the cycle; error carried forward from M2 where the chlorine terms were not doubled A1 ΔLEH = −796 − 1459 = −2255 kJ mol⁻¹, four significant figures, negative sign required

Question 4

Calculate the second electron affinity of oxygen from the data in the table for calcium oxide.

Accepted Answer

M1 ΔfH = ΔatH(Ca) + IE₁ + IE₂ + ΔatH(O) + EA₁ + EA₂ + ΔLEH, from Hess's law applied to the cycle M1 the known steps before the lattice sum to 178 + 590 + 1145 + 249 − 141 = +2021 kJ mol⁻¹ M1 EA₂ = −635 − 2021 − (−3454), error carried forward from M2 A1 EA₂ = +798 kJ mol⁻¹, three significant figures, positive sign shown; ignore a missing + sign only if the answer is described as endothermic

Question 5

Complete the Born–Haber cycle for potassium chloride in the diagram: give the species, with state symbols, at levels A, B and C, and name the single step marked with a question mark.

Accepted Answer

B1 level A is K(g) + ½Cl₂(g), with state symbols B1 level B is K⁺(g) + e⁻ + Cl(g); the electron must be shown B1 level C is K⁺(g) + Cl⁻(g), the electron now on the chloride ion B1 the enthalpy of formation of KCl(s), from the elements in their standard states

Question 6

Explain why the difference between the experimental and the theoretical lattice enthalpies of the silver halides increases from silver fluoride to silver iodide.

Accepted Answer

B1 the experimental value is more exothermic than the theoretical value because the bonding has some covalent character B1 the Ag⁺ ion polarises the halide ion, distorting its electron cloud towards the cation B1 the halide ions get larger from F⁻ to I⁻, with outer electrons further from the nucleus and more shielded B1 so the larger ions are more easily polarised, the covalent character increases and the difference grows; reject 'the bonding becomes covalent'

Question 7

Compare the lattice enthalpies of formation of potassium chloride, −718 kJ mol⁻¹, and calcium oxide, −3454 kJ mol⁻¹, and account for the difference.

Accepted Answer

B1 the lattice enthalpy of CaO is almost five times as exothermic as that of KCl B1 Ca²⁺ and O²⁻ carry twice the charges of K⁺ and Cl⁻, so the attraction between a pair of ions is four times as strong at the same distance B1 Ca²⁺ is smaller than K⁺ and O²⁻ smaller than Cl⁻, so the ions in CaO sit closer together and attract more strongly B1 both effects make the attraction stronger, and together they account for a factor larger than four; ignore any reference to the enthalpies of formation

Question 8

Evaluate this statement: 'The experimental lattice enthalpy of silver chloride is 72 kJ mol⁻¹ more exothermic than the theoretical value, so the Born–Haber cycle for silver chloride must contain an experimental error.'

Accepted Answer

B1 the same cycle method gives sodium chloride within 3% of the theoretical value, so the method itself is sound B1 the silver halide differences are systematic: always in the same direction and growing from fluoride to iodide, which random error would not produce B1 the difference arises because the perfect ionic model ignores polarisation of the anion by Ag⁺, so it underestimates the attraction in a lattice with covalent character B1 the statement is therefore wrong: the model, not the measurement, is the source of the difference

Question 9

Explain why magnesium carbonate decomposes on heating at a lower temperature than barium carbonate.

Accepted Answer

B1 both decompose to the metal oxide and carbon dioxide, MCO₃(s) → MO(s) + CO₂(g) B1 Mg²⁺ is smaller than Ba²⁺ with the same charge, so it has a higher charge density and a greater polarising power B1 Mg²⁺ distorts the electron cloud of the large carbonate ion more, drawing electron density towards itself B1 this weakens a C–O bond in the carbonate ion, so less energy is needed to break it and release CO₂; reject 'the Mg–O bond is weaker'

Question 10

State what is meant by the first electron affinity of an element, and write an equation, with state symbols, for the first electron affinity of oxygen.

Accepted Answer

B1 the enthalpy change when one mole of gaseous atoms each gain one electron B1 to form one mole of gaseous 1− ions; accept 'negative ions' for '1− ions' B1 O(g) + e⁻ → O⁻(g), with every state symbol shown; allow e for e⁻

Question 11

Calculate the lattice enthalpy of formation of sodium bromide from the data in the table.

Accepted Answer

M1 ΔfH = ΔatH(Na) + ΔatH(Br) + IE₁(Na) + EA₁(Br) + ΔLEH, from Hess's law applied to the cycle M1 the steps to the gaseous ions sum to 107 + 112 + 496 − 325 = +390 kJ mol⁻¹ A1 ΔLEH = −361 − 390 = −751 kJ mol⁻¹, three significant figures; the negative sign is required, and +751 is awarded the two method marks only

Question 12

Calculate the first electron affinity of chlorine from the data in the table for potassium chloride.

Accepted Answer

M1 ΔfH = ΔatH(K) + IE₁(K) + ΔatH(Cl) + EA₁(Cl) + ΔLEH, so EA₁(Cl) = ΔfH − ΔatH(K) − IE₁(K) − ΔatH(Cl) − ΔLEH M1 EA₁(Cl) = −437 − 89 − 419 − 122 + 718 A1 EA₁(Cl) = −349 kJ mol⁻¹, three significant figures, negative sign required

Question 13

Describe the perfect ionic model used to calculate a theoretical lattice enthalpy.

Accepted Answer

B1 each ion is treated as a perfect sphere B1 with its charge evenly distributed, so there is no polarisation and no covalent character B1 the lattice enthalpy is calculated from the electrostatic attractions and repulsions between the ions, using their charges and the distances between their centres

Question 14

Suggest why the experimental and theoretical lattice enthalpies of caesium fluoride agree to within 1%.

Accepted Answer

B1 Cs⁺ is a large ion with a single charge, so it has a low charge density and is only weakly polarising B1 F⁻ is a small anion whose electron cloud is held tightly and is hard to distort B1 there is very little polarisation, so the bonding is almost purely ionic and the perfect ionic model fits

Question 15

Suggest why magnesium forms MgCl₂ with chlorine rather than MgCl, even though forming Mg²⁺ needs the second ionisation energy of magnesium, +1451 kJ mol⁻¹.

Accepted Answer

B1 forming Mg²⁺ costs an extra 1451 kJ mol⁻¹, and a second chlorine atom must also be atomised and ionised B1 the lattice of Mg²⁺ and Cl⁻ ions has a far more exothermic lattice enthalpy, because the cation has twice the charge and is smaller, and there are more ions per formula unit B1 the extra lattice enthalpy more than repays the second ionisation energy, so the enthalpy of formation of MgCl₂ is much more negative than that of MgCl would be

Question 16

Explain why the lattice enthalpy of an ionic compound cannot be measured directly, and how a Born–Haber cycle is used to find it.

Accepted Answer

B1 a gas of separate ions cannot be prepared and reacted in a calorimeter, so the lattice-forming step cannot be carried out and measured B1 by Hess's law the enthalpy change from the elements to the solid is the same by any route B1 the other steps in the cycle, atomisation, ionisation energy, electron affinity and formation, are measured, so the lattice enthalpy is the only unknown and is found by subtraction

Question 17

Explain why the first electron affinity of bromine, −325 kJ mol⁻¹, is less exothermic than that of chlorine, −349 kJ mol⁻¹.

Accepted Answer

B1 a bromine atom is larger, so the incoming electron enters a shell further from the nucleus, 4p rather than 3p B1 the incoming electron is more shielded from the nucleus by more inner shells of electrons B1 the attraction between the nucleus and the incoming electron is weaker, so less energy is released; ignore any reference to nuclear charge alone

Question 18

Suggest why the first electron affinity of fluorine, −328 kJ mol⁻¹, is less exothermic than that of chlorine, −349 kJ mol⁻¹, although a fluorine atom is smaller.

Accepted Answer

B1 the incoming electron enters the 2p subshell of fluorine, which already holds five electrons in a small volume B1 the repulsion between the incoming electron and those electrons is large B1 the repulsion cancels part of the stronger attraction to the nucleus, so less energy is released than for chlorine

Question 19

Suggest why lithium carbonate decomposes when heated with a Bunsen burner but sodium carbonate does not.

Accepted Answer

B1 Li⁺ is much smaller than Na⁺ with the same charge, so its charge density is higher B1 Li⁺ polarises the carbonate ion more, distorting its electron cloud and weakening a C–O bond B1 lithium carbonate therefore decomposes to Li₂O and CO₂ at a temperature a Bunsen burner reaches, while sodium carbonate needs a far higher one

Question 20

Justify the inclusion of two enthalpies of atomisation of chlorine, and of both the first and the second ionisation energies of magnesium, in the Born–Haber cycle for magnesium chloride, MgCl₂.

Accepted Answer

B1 the enthalpy of atomisation of chlorine, ½Cl₂(g) → Cl(g), forms one mole of atoms, and two moles of gaseous chlorine atoms are needed for one mole of MgCl₂, so the value is doubled; allow one Cl–Cl bond enthalpy instead B1 the lattice holds Mg²⁺ ions, so two electrons are removed from each gaseous magnesium atom, Mg(g) → Mg⁺(g) → Mg²⁺(g), which needs the first and then the second ionisation energy B1 the first electron affinity of chlorine is doubled for the same reason, one for each Cl⁻ ion; every step must match the one mole of MgCl₂(s) that the lattice enthalpy and the enthalpy of formation refer to

Question 21

State what is meant by the term lattice enthalpy of formation.

Accepted Answer

B1 the enthalpy change when one mole of a solid ionic compound is formed B1 from its gaseous ions, under standard conditions; reject 'from its elements', which describes the enthalpy of formation

Question 22

Give equations, with state symbols, for the enthalpy of atomisation of chlorine and for the second ionisation energy of calcium.

Accepted Answer

B1 ½Cl₂(g) → Cl(g), one mole of chlorine atoms formed; reject Cl₂(g) → 2Cl(g), which forms two moles B1 Ca⁺(g) → Ca²⁺(g) + e⁻, starting from the gaseous 1+ ion; reject Ca(g) → Ca²⁺(g) + 2e⁻

Question 23

Give equations, with state symbols, for the thermal decomposition of calcium nitrate and of potassium nitrate.

Accepted Answer

B1 2Ca(NO₃)₂(s) → 2CaO(s) + 4NO₂(g) + O₂(g), balanced; allow multiples B1 2KNO₃(s) → 2KNO₂(s) + O₂(g), the nitrite and oxygen only; reject any equation giving NO₂

Compound	Experimental / kJ mol⁻¹	Theoretical / kJ mol⁻¹	Anion radius / pm
CaF₂	−2630	−2609	133
CaI₂	−2074	−1905	220

Enthalpy change	Value / kJ mol⁻¹
Enthalpy of formation of CaCl₂(s)	−796
Enthalpy of atomisation of calcium	+178
First ionisation energy of calcium	+590
Second ionisation energy of calcium	+1145
Enthalpy of atomisation of chlorine	+122
First electron affinity of chlorine	−349

Enthalpy change	Value / kJ mol⁻¹
Enthalpy of formation of CaO(s)	−635
Enthalpy of atomisation of calcium	+178
First ionisation energy of calcium	+590
Second ionisation energy of calcium	+1145
Enthalpy of atomisation of oxygen	+249
First electron affinity of oxygen	−141
Lattice enthalpy of formation of CaO	−3454

Enthalpy change	Value / kJ mol⁻¹
Enthalpy of formation of NaBr(s)	−361
Enthalpy of atomisation of sodium	+107
Enthalpy of atomisation of bromine	+112
First ionisation energy of sodium	+496
First electron affinity of bromine	−325

Enthalpy change	Value / kJ mol⁻¹
Enthalpy of formation of KCl(s)	−437
Enthalpy of atomisation of potassium	+89
First ionisation energy of potassium	+419
Enthalpy of atomisation of chlorine	+122
Lattice enthalpy of formation of KCl	−718