Lattice Energy and Energetics Study Notes

Core Definitions and Learning Outcomes

  • Standard Enthalpy Change of Atomisation (ΔatH\Delta_{at}H): The enthalpy change when one mole of gaseous atoms is formed from an element in its standard state, measured at standard conditions (298K298\,K and 100kPa100\,kPa).

  • Electron Affinity (EA): The energy change when each atom in one mole of atoms in the gaseous state gains an electron to form a 1-1 ion.

  • Lattice Energy (ΔLEH\Delta_{LE}H): The energy change when one mole of an ionic compound is formed from its ions in the gaseous state (measured at a stated temperature, usually 298K298\,K, and pressure, usually 100kPa100\,kPa).

  • Enthalpy Change of Solution (ΔsolH\Delta_{sol}H): The enthalpy change when one mole of an ionic solid dissolves in water to form an infinitely dilute solution.

  • Enthalpy Change of Hydration (ΔhydH\Delta_{hyd}H): The enthalpy change when one mole of an ion in its gaseous state is completely hydrated by water.

Lattice Energy Fundamentals

  • In covalent compounds, bond enthalpies measure the bond strength. In ionic compounds, the equivalent measurement is lattice energy.

  • Example Equations:

    • Na+(g)+Cl(g)NaCl(s)Na^+(g) + Cl^-(g) \rightarrow NaCl(s)

    • ΔLEH=780kJmol1\Delta_{LE}H = -780\,kJ\,mol^{-1}

    • Mg2+(g)+2Cl(g)MgCl2(s)Mg^{2+}(g) + 2Cl^-(g) \rightarrow MgCl_2(s)

    • ΔLEH=2526kJmol1\Delta_{LE}H = -2526\,kJ\,mol^{-1}

  • Lattice Enthalpy vs. Lattice Energy: At A Level, these terms are treated as synonymous.

Factors Affecting Lattice Energy Magnitude

Lattice energies are determined by four primary factors:

  1. Magnitude of Charges on the Ions: Higher charges result in stronger electrostatic attractions.

  2. Sum of the Ionic Radii: Smaller ions can get closer together, increasing the force of attraction.

  3. Type of Lattice Structure.

  4. Extent of Covalent Interactions: The presence of covalency increases the magnitude (makes it more negative).

Case Study: Magnesium Chloride vs. Sodium Chloride The lattice energy of MgCl2MgCl_2 is significantly higher than NaClNaCl because:

  • Magnesium (Mg2+Mg^{2+}) has twice the charge of sodium (Na+Na^+).

  • There are twice as many chloride ions, leading to more cation-to-anion interactions.

  • The ionic radius of Mg2+Mg^{2+} is smaller than Na+Na^+, resulting in a shorter distance between ion centers.

Data Table: Lattice Energies of Various Compounds

Compound

Inter-ionic Distance (nm)

Charges on Ions

Lattice Energy (kJmol1kJ\,mol^{-1})

LiFLiF

0.2070.207

+1,1+1, -1

1031-1031

NaFNaF

0.2350.235

+1,1+1, -1

918-918

CaF2CaF_2

0.2330.233

+2,1+2, -1

2630-2630

Li2OLi_2O

0.2140.214

+1,2+1, -2

2814-2814

MgOMgO

0.2120.212

+2,2+2, -2

3791-3791

Al2O3Al_2O_3

0.1930.193

+3,2+3, -2

15504-15\,504

Atomisation and Electron Affinity Details

Standard Enthalpy Change of Atomisation Examples:

  • C(s)C(g)C(s) \rightarrow C(g) (ΔatH=+717kJmol1\Delta_{at}H = +717\,kJ\,mol^{-1})

  • Na(s)Na(g)Na(s) \rightarrow Na(g) (ΔatH=+107kJmol1\Delta_{at}H = +107\,kJ\,mol^{-1})

  • 12H2(g)H(g)\frac{1}{2}H_2(g) \rightarrow H(g) (ΔatH=+218kJmol1\Delta_{at}H = +218\,kJ\,mol^{-1})

Electron Affinity Trends:

  • 1st Electron Affinity: Usually negative (exothermic). Exceptions include noble gases where the new electron must enter a new quantum shell, experiencing repulsion from existing valence electrons.

    • Cl(g)+eCl(g)Cl(g) + e^- \rightarrow Cl^-(g) (1stEA=349kJmol11^{st} EA = -349\,kJ\,mol^{-1})

    • O(g)+eO(g)O(g) + e^- \rightarrow O^-(g) (1stEA=325kJmol11^{st} EA = -325\,kJ\,mol^{-1})

  • 2nd Electron Affinity: Always positive (endothermic). Energy is required to overcome the electrostatic repulsion between the incoming electron and the already negative ion.

Born-Haber Cycles

Born-Haber cycles apply Hess's Law to calculate lattice energy, which cannot be measured directly. They relate the enthalpy of formation (ΔfH\Delta_f H) to the enthalpy of atomisation (ΔatH\Delta_{at} H), ionisation energies (IE), electron affinities (EA), and lattice energy (ΔLEH\Delta_{LE} H).

Example: Calculating Lattice Energy for Sodium Chloride (NaClNaCl) Data:

  • ΔfH[NaCl(s)]=411kJmol1\Delta_f H[NaCl(s)] = -411\,kJ\,mol^{-1}

  • ΔatH[Na(s)]=+107kJmol1\Delta_{at} H[Na(s)] = +107\,kJ\,mol^{-1}

  • 1stIE[Na(g)]=+496kJmol11^{st} IE[Na(g)] = +496\,kJ\,mol^{-1}

  • ΔatH[Cl2(g)]=+122kJmol1\Delta_{at} H[Cl_2(g)] = +122\,kJ\,mol^{-1}

  • EA[Cl(g)]=349kJmol1EA[Cl(g)] = -349\,kJ\,mol^{-1}

Applying Hess's Law: +107+496+122+(349)+ΔLEH=411+107 + 496 + 122 + (-349) + \Delta_{LE}H = -411 ΔLEH=411107496122+349=787kJmol1\Delta_{LE}H = -411 - 107 - 496 - 122 + 349 = -787\,kJ\,mol^{-1}

Experimental vs. Theoretical Lattice Energy

  • Experimental Lattice Energy: Calculated indirectly via Born-Haber cycles based on experimental data.

  • Theoretical Lattice Energy: Calculated using electrostatic theory equations (e.g., Born-Landé, Born-Mayer, or Kapustinskii).

Theoretical Assumptions:

  1. Ions are in direct contact.

  2. Ions are perfectly spherical.

  3. Charge is evenly distributed (ions act as point charges).

Interpreting Differences:

  • If values match (e.g., Sodium halides), the compound is close to "100% ionic."

  • Large differences (e.g., Silver halides) indicate the ionic model is insufficient because the bonding has significant covalent character.

  • Covalency makes the experimental lattice energy more negative (stronger bonding) than the theoretical value.

Polarisation and Fajan's Rules

Covalent character in ionic substances arises from the polarisation of the anion by the cation. The cation distorts the electron cloud of the anion, causing it to move closer or overlap.

Fajan’s Rules for Increased Polarisation:

  1. Cation: High charge and small size (high charge density or "polarising power").

  2. Anion: High charge and large size (more easily "polarisable").

Examples:

  • Mg2+ClMg^{2+}-Cl^- has more covalent character than Na+ClNa^+-Cl^- because Mg2+Mg^{2+} has a higher charge density.

  • AgIAgI has more covalent character than AgFAgF because the large iodide ion (II^-) is more easily polarised than the small fluoride ion (FF^-).

Enthalpy of Solution and Hydration

Enthalpy Change of Solution (ΔsolH\Delta_{sol}H):

  • Involves breaking the lattice (endothermic) and hydrating gaseous ions (exothermic).

  • ΔsolH=ΔLEH+ΔhydH\Delta_{sol}H = -\Delta_{LE}H + \sum \Delta_{hyd}H

  • Measured at infinite dilution (where further dilution releases no more heat). This is found via extrapolation because it cannot be measured directly.

Interactions in Hydration:

  • Cations: Ion-dipole interactions with the δ\delta- Oxygen of water. Transition metals may form dative covalent bonds.

  • Anions: Ion-dipole interactions with δ+\delta+ Hydrogen and hydrogen bonding using anion lone pairs.

Trends in ΔhydH\Delta_{hyd}H:

  • Increases (becomes more negative) with increasing ionic charge.

  • Increases (becomes more negative) with decreasing ionic radius (higher charge density correlates to more negative ΔhydH\Delta_{hyd}H).

Entropy and Solubility

Substances are more likely to be soluble if ΔsolH\Delta_{sol}H is exothermic. However, many compounds with endothermic ΔsolH\Delta_{sol}H still dissolve due to entropy.

The Entropy Equation: ΔStotal=ΔHT+ΔSsystem\Delta S_{total} = -\frac{\Delta H}{T} + \Delta S_{system}

  • Transition to Solution: Disordered solution state and an increase in the number of particles result in a positive ΔSsystem\Delta S_{system}.

  • For endothermic salts, solubility depends on whether T×ΔSsystemT \times \Delta S_{system} overcomes the endothermic ΔH\Delta H. Increasing temperature makes ΔStotal\Delta S_{total} more positive, which makes the reaction feasible (\Delta G < 0).

Questions & Discussion

Checkpoint 12B.1

  • Q: Why is ΔLEH\Delta_{LE}H for NaFNaF larger (more negative) than KClKCl?

  • A: Na+Na^+ is smaller than K+K^+ and FF^- is smaller than ClCl^-. The smaller inter-ionic distance in NaFNaF leads to stronger electrostatic attractions.

  • Q: Why is ΔLEH\Delta_{LE}H for CaOCaO four times larger than KFKF?

  • A: Inter-ionic distances are similar (0.240nm0.240\,nm vs 0.271nm0.271\,nm), but the product of charges for Ca2+O2Ca^{2+}O^{2-} is 2×2=42 \times 2 = 4, compared to 1×1=11 \times 1 = 1 for KFKF.

Checkpoint 12B.2

  • Q: Why do CaF2CaF_2 and AgFAgF differ in agreement between theoretical and experimental values?

  • A: CaF2CaF_2 is close to 100% ionic. AgFAgF has significant covalent bonding, which electrostatic theory does not account for.

  • Q: Calculate ordering of polarising power for Mg2+Mg^{2+}, Al3+Al^{3+}, Li2+Li^{2+}, Na+Na^+, Ca2+Ca^{2+}, K+K^+.

  • A: Order (based on charge/radius2radius^2): K^+ < Na^+ < Ca^{2+} < Li^+ < Mg^{2+} < Al^{3+}.

Checkpoint 12B.3

  • Q: Explain what happens to the thermometer when 1g1\,g of NaFNaF dissolves in 250cm3250\,cm^3 of water (ΔsolH=+0.3kJmol1\Delta_{sol}H = +0.3\,kJ\,mol^{-1}).

  • A: No measurable change. The temperature decrease is too small for a standard thermometer (graduated in 1C1\,^{\circ}C) to register.