🟣 This is Higher Level (HL) content.
Lattice Enthalpy
📘 IB Definition
Lattice enthalpy is the enthalpy change when one mole of an ionic compound is separated into its constituent gaseous ions under standard conditions. It is always endothermic (positive ΔH) because energy must be input to overcome the electrostatic attractions.
\(\text{NaCl(s)} \rightarrow \text{Na}^+(g) + \text{Cl}^-(g) \quad \Delta H_{lat} > 0\)
Coulomb's Law
The electrostatic force F between two ions is governed by Coulomb's Law. While you don't need to calculate F numerically, you must understand qualitatively how the two key factors affect lattice enthalpy:
\[ F \propto \frac{q^+ \times q^-}{r^2} \]
↑ Ionic Charge (q)
Higher charge = stronger attraction = greater lattice enthalpy.
MgO (2+ / 2−) has a much higher lattice enthalpy than NaCl (1+ / 1−).
↓ Ionic Radius (r)
Smaller ions = closer together = greater lattice enthalpy.
NaF has a higher lattice enthalpy than NaI because F⁻ is much smaller than I⁻.
Charge Density
Charge density combines both factors into one concept: it is the ratio of ionic charge to ionic volume. Ions with high charge density (small radius, high charge) create the strongest electrostatic fields and thus the highest lattice enthalpies.
Comparison Example
Why does MgO have a much higher melting point than NaCl?
1. Mg²⁺ has a higher charge than Na⁺ → greater electrostatic attraction
2. Mg²⁺ has a smaller ionic radius than Na⁺ → ions closer together
3. O²⁻ has a higher charge than Cl⁻ → greater electrostatic attraction
∴ MgO has a much greater lattice enthalpy → higher melting point
Formal Charge & the Sulfate Ion
HL candidates must apply formal charge to justify the preferred structure of polyatomic ions — specifically the sulfate ion \(\text{SO}_4^{2-}\).
Formal Charge Formula
\[\text{FC} = V - L - \frac{B}{2}\]
Structure 1: Octet-Compliant
4 × single bonds (S obeys octet)
FC on S: 6 − 0 − 4 = +2 ❌
FC on each O: 6 − 6 − 1 = −1
High charge separation — unfavourable
Structure 2: Expanded Octet ✓
2 × double bonds + 2 × single bonds
FC on S: 6 − 0 − 6 = 0 ✅
FC on double-bonded O: 6 − 4 − 2 = 0
FC on single-bonded O: 6 − 6 − 1 = −1
Minimal charge separation — preferred
🔑 Why the Expanded Octet Wins
Period 3 elements like sulfur have accessible, empty d-orbitals that allow expansion beyond the octet. The expanded structure minimises formal charge to zero on three of five atoms, with the necessary negative charges placed on the most electronegative element (oxygen). Experimental bond lengths confirm a fractional bond order of ~1.5, consistent with resonance hybrids of this structure.