Ionic Bonding

An ionic bond is not really a "bond" in the shared-pair sense at all — it is pure electrostatics. One atom, usually a metal with a low ionisation enthalpy, gives away an electron and becomes a cation; a second atom, usually a non-metal with a highly negative electron gain enthalpy, accepts that electron and becomes an anion. The two ions then attract each other, and in the solid state that attraction extends in three dimensions, packing enormous numbers of ions into an orderly crystal lattice rather than a single isolated ion pair. Three factors decide whether this route is favourable:

• Low ionisation enthalpy of the metal — electrons must be easy to remove.
• High (more negative) electron gain enthalpy of the non-metal — accepting the electron must release energy.
• High lattice enthalpy of the resulting solid — this is usually the largest energy term in the whole process, and its size is what ultimately makes ionic-compound formation exothermic overall.

Lattice enthalpy ($\Delta_{lattice}H$) is the enthalpy change when one mole of an ionic solid forms from its constituent gaseous ions. It is a huge, favourable (energy-releasing) step, but it cannot be measured directly in a calorimeter — you cannot isolate a beaker of gaseous $\text{Na}^+$ and $\text{Cl}^-$ ions and watch them condense. Instead it is calculated indirectly using a Born–Haber cycle, an application of Hess's law that stitches lattice enthalpy together with enthalpies that can be measured:

where: $\Delta_f H^\circ$ is the standard enthalpy of formation of the solid from its elements, $\Delta_{sub}H$ is the sublimation enthalpy of the metal (solid to gas), $\Delta_{diss}H(\text{X}_2)$ is the bond dissociation enthalpy of the diatomic non-metal, $\Delta_i H(\text{M})$ is the metal's ionisation enthalpy, $\Delta_{eg}H(\text{X})$ is the non-metal's electron gain enthalpy, and $\Delta_{lattice}H$ is the term we usually want. Every term on the right is a step in one path from elements to gaseous ions to solid; the left side is the direct path. Because enthalpy is a state function, both paths give the same total change, so one unknown term can always be solved once the rest are known. The half-factor on $\Delta_{diss}H(\text{X}_2)$ deserves a moment: forming one mole of $\text{NaCl}$ needs only one mole of Cl atoms, but dissociation enthalpy is quoted per mole of $\text{Cl}_2$ molecules, so only half of it is "spent" per formula unit.

Ionic solids share a family of properties that follow directly from this picture: they are hard and brittle (a strong, rigid lattice that fractures rather than bends), have high melting and boiling points (a lot of energy is needed to pull ions apart), and conduct electricity only when molten or in aqueous solution, because only then are the ions free to move — in the solid, they are locked in fixed lattice positions.

Explain why $\text{MgO}$ has a far higher melting point (about $2852\ ^\circ\text{C}$) than $\text{NaCl}$ (about $801\ ^\circ\text{C}$).

Step 1. Identify the ions and their charges: $\text{NaCl}$ is built from $\text{Na}^+$ and $\text{Cl}^-$, each of charge magnitude $1$, while $\text{MgO}$ is built from $\text{Mg}^{2+}$ and $\text{O}^{2-}$, each of charge magnitude $2$.

Step 2. Recall the qualitative Coulombic rule that lattice enthalpy grows with the product of the ionic charges and shrinks as the ions get bigger (larger inter-ionic distance): $\Delta_{lattice}H \propto \dfrac{|q_+ q_-|}{r_+ + r_-}$.

Step 3. Compare the charge products: for $\text{NaCl}$, $|q_+ q_-| = 1\times1 = 1$; for $\text{MgO}$, $|q_+ q_-| = 2\times2 = 4$ — four times larger. The ionic radii of $\text{Mg}^{2+}$/$\text{O}^{2-}$ are not dramatically different from $\text{Na}^+$/$\text{Cl}^-$, so this quadrupled charge product dominates the comparison.

Step 4. Since melting an ionic solid means overcoming exactly the electrostatic forces holding the lattice together, a much larger lattice enthalpy means a much higher melting point. This is why $\text{MgO}$'s melting point ($2852\ ^\circ\text{C}$) so far exceeds $\text{NaCl}$'s ($801\ ^\circ\text{C}$): a fourfold greater charge product gives $\text{MgO}$ a far stronger lattice.

• Mixing up the sign convention for lattice enthalpy — it can be quoted as the (negative) energy released on forming the lattice from gaseous ions, or as the (positive) energy needed to break it apart. Check which convention a question uses before plugging numbers into the cycle.
• Forgetting to halve the dissociation enthalpy of a diatomic non-metal like $\text{Cl}_2$ or $\text{O}_2$ when only one atom of it is needed per formula unit.
• Treating ionic and covalent bonding as strictly separate boxes — real bonds usually have partial covalent character, and the Born–Haber numbers describe an idealised, purely ionic model.
• Assuming ionic solids conduct electricity in the solid state — they do not, because the ions are frozen in fixed lattice positions; conduction needs mobile ions (molten or dissolved).

Ask the tutor: Stuck on which sign to use for a Born–Haber step, or want to compare lattice enthalpies for a different pair of ionic compounds — ask the AI tutor to walk through the cycle with you.