Lattice Energy & Born-Haber Cycles (Cambridge (CIE) A Level Chemistry): Flashcards

False.

The second electron affinity is endothermic because the incoming electron must overcome repulsion from an already negatively charged ion. Energy is absorbed and the value is positive.

Cl (g) + e^- → Cl^- (g)

Going down Group 17, the first electron affinity generally becomes less exothermic because there are more electron shells increasing shielding.

Fluorine has a very small atomic radius, causing high electron density around the nucleus. This leads to increased repulsion between the incoming electron and existing electrons, reducing the overall attraction and making EA₁ less exothermic.

True.

Increasing nuclear charge across a period strengthens attraction for the incoming electron, making electron affinity more exothermic up to Group 17.

1. Nuclear charge: higher nuclear charge makes EA more exothermic.

2. Atomic radius: larger radius makes EA less exothermic.

3. Shielding: more inner shells weaken nuclear attraction, making EA less exothermic.

The second electron affinity is represented by:

X^- (g) + e^- → X^2- (g)

and is endothermic.

Chlorine has the most exothermic first electron affinity in Group 17, with a value of -345 kJ mol^-1.

A Born-Haber cycle is a specific application of Hess's law for ionic compounds, used to calculate lattice enthalpy indirectly from other measurable enthalpy changes.

Energy increases going up the diagram. Endothermic steps are shown with arrows pointing upwards and exothermic steps with arrows pointing downwards.

False.

Atomisation is endothermic (bond breaking), so it is shown with an upward arrow in the Born-Haber cycle.

1. Atomisation of K (s) → K (g)

2. First ionisation energy of K (g) → K⁺ (g)

3. Atomisation of ½Cl₂ (g) → Cl (g)

4. First electron affinity of Cl (g) → Cl^- (g)

5. Enthalpy of formation of KCl (s)

6. Apply Hess's law to find lattice energy

In a Born-Haber cycle for NaCl, the equation Na (g) → Na⁺ (g) + e^- represents the ionisation energy of sodium, which is an endothermic process.

False.

A Born-Haber cycle can be used to calculate any unknown enthalpy change in the cycle, such as enthalpy of formation, electron affinity or ionisation energy, as long as all other values are known.

Mg forms a 2+ ion, requiring removal of two electrons (IE₁ and IE₂). Oxygen forms a 2- ion, requiring addition of two electrons (EA₁ and EA₂).

The enthalpy change for the direct route in a Born-Haber cycle is the enthalpy of formation and this must equal the sum of all steps in the indirect route.

Na (s) + ½Cl₂ (g) → NaCl (s) ΔH_f^θ = -411 kJ mol^-1

ΔH_latt^θ = ΔH_f^θ - ΔH₁^θ

where ΔH₁^θ is the sum of all enthalpy changes needed to convert elements in their standard states to gaseous ions.

True.

MgCl₂ contains two moles of Cl^- ions, so two moles of electrons are added to two moles of chlorine atoms. The electron affinity value must be doubled to account for this.

To calculate ΔH_latt^θ for KCl:

ΔH_latt^θ = ΔH_f^θ - [(ΔH_at^θ K) + (ΔH_at^θ Cl) + (IE₁ K) + (EA₁ Cl)]

ΔH_latt^θ = (-437) - [(+90) + (+122) + (+418) + (-349)]

ΔH_latt^θ = (-437) - (281) = -718 kJ mol^-1

False.

The direct route is the enthalpy change of formation. The lattice energy is part of the indirect route. The stage being calculated is always the direct route.

MgO requires IE₂ (second ionisation energy of Mg) and EA₂ (second electron affinity of O), because Mg forms Mg²⁺ and O forms O^2-.

The simplified Born-Haber equation is:

ΔH_f^θ = ΔH₁^θ + ΔH_latt^θ

Rearranging to find lattice energy:

ΔH_latt^θ = ΔH_f^θ - ΔH₁^θ

Hess's law: the total enthalpy change is independent of the route taken, so the sum of enthalpy changes along the indirect route equals the enthalpy change of the direct route.

True.

Brackets help keep track of positive and negative values when summing multiple enthalpy terms, reducing the chance of arithmetic sign errors.

1. Ionic charge: higher charge makes lattice energy more exothermic.

2. Ionic radius: smaller radius makes lattice energy more exothermic.

False.

As ionic radius increases, charge density decreases and ions are further apart, so electrostatic attraction is weaker and the lattice energy becomes less exothermic.

Both CsF and KF contain F^- ions. Because Cs⁺ is larger than K⁺, the lattice energy of CsF is less exothermic than that of KF.

Ca²⁺ and O^2- have higher ionic charges than K⁺ and Cl^-, leading to greater charge density and stronger electrostatic attraction between ions. Ca²⁺ and O^2- are also smaller ions, further increasing the exothermic value.

True.

Higher ionic charge means a stronger attraction between oppositely charged ions, so more energy is released when the lattice forms, giving a more exothermic lattice energy.

Charge density is a measure of the ionic charge relative to the ionic volume. It is higher for smaller ions or ions with greater charge.

Larger ions have lower charge density. The charge is spread over a greater volume and the ions are further apart, weakening the electrostatic attraction between oppositely charged ions and reducing the energy released.

For ions of the same charge, going down a group increases ionic radius and decreases charge density, making lattice energy less exothermic.

MgO has the more exothermic lattice energy. Mg²⁺ is smaller than Ca²⁺, so it has a higher charge density. This leads to stronger electrostatic attraction between Mg²⁺ and O^2- and more energy released when the lattice forms.