Ionic bonding

Atoms form ions to reach a stable, full outer shell - usually the electron configuration of the nearest noble gas (a stable octet, or a duplet for the lightest elements).

• Metals lose electrons to form positive cations. A Group 1 metal loses 1 electron ($\text{Na} \rightarrow \text{Na}^+ + e^-$), Group 2 loses 2 ($\text{Mg} \rightarrow \text{Mg}^{2+} + 2e^-$), Group 13 loses 3 ($\text{Al}^{3+}$).
• Non-metals gain electrons to form negative anions. Group 17 gains 1 ($\text{Cl} + e^- \rightarrow \text{Cl}^-$), Group 16 gains 2 ($\text{O}^{2-}$), Group 15 can gain 3 ($\text{N}^{3-}$).

The charge equals the number of electrons transferred. Notice the ion has the same electron configuration as a noble gas but a different nuclear charge, so a cation is much smaller than its parent atom and an anion is larger.

For the main groups the ionic charge follows the group number. Transition metals can form more than one ion (e.g. $\text{Fe}^{2+}$ and $\text{Fe}^{3+}$), so a Roman numeral names the charge: iron(II) and iron(III).

Many ions are polyatomic - a covalently bonded group carrying an overall charge, such as $\text{NO}_3^-$, $\text{SO}_4^{2-}$, $\text{CO}_3^{2-}$, $\text{OH}^-$ and $\text{NH}_4^+$ - and these stay together as a unit when forming compounds.

An ionic bond is the electrostatic attraction between oppositely charged ions. Because this attraction acts equally in all directions (it is non-directional), each ion surrounds itself with as many oppositely charged neighbours as possible.

To write an ionic formula the total positive charge must equal the total negative charge so the compound is electrically neutral. A quick method is to cross over the magnitudes of the charges:

• $\text{Na}^+$ and $\text{Cl}^-$ give $\text{NaCl}$ (1:1).
• $\text{Mg}^{2+}$ and $\text{Cl}^-$ give $\text{MgCl}_2$.
• $\text{Al}^{3+}$ and $\text{O}^{2-}$ give $\text{Al}_2\text{O}_3$.

With polyatomic ions, place the ion in brackets before adding a subscript: $\text{Mg}^{2+}$ with $\text{NO}_3^-$ gives $\text{Mg(NO}_3)_2$; $\text{Ca}^{2+}$ with $\text{OH}^-$ gives $\text{Ca(OH)}_2$.

An ionic formula is always the simplest whole-number ratio - an empirical formula - because the lattice extends indefinitely and contains no discrete molecules. The bond is strongest when charges are high and ions are small, following $F \propto \dfrac{q_+ q_-}{r^2}$, so $\text{MgO}$ (2+ / 2-) is held far more strongly than $\text{NaCl}$ (1+ / 1-).

Ionic compounds form a giant ionic lattice: a regular 3-D array of alternating cations and anions, each surrounded by oppositely charged neighbours (in $\text{NaCl}$, each ion has 6 nearest neighbours). Their properties all follow from the strong, non-directional electrostatic forces throughout this lattice.

• High melting and boiling points - large amounts of energy are needed to overcome the many strong ionic attractions; higher charge and smaller ions raise the melting point further (e.g. $\text{MgO} \gg \text{NaCl}$).
• Brittle - a blow shifts the layers so that like charges align; the resulting repulsion splits the crystal.
• Electrical conductivity - they do not conduct when solid (ions are locked in place) but do conduct when molten or dissolved, because the ions are then free to move and carry charge.
• Solubility - many dissolve in water because polar water molecules surround and stabilise (hydrate) the separated ions, though some ionic compounds are insoluble.

Ionic compounds are also non-volatile and tend to have characteristic regular crystal shapes that reflect the underlying lattice.

Lattice enthalpy ($\Delta H_{lat}^{\ominus}$) is the enthalpy change when one mole of a solid ionic compound is separated into its gaseous ions, e.g.

$$\text{NaCl(s)} \rightarrow \text{Na}^+(g) + \text{Cl}^-(g)$$

Defined this way it is endothermic (positive); a larger value means a stronger ionic lattice. (Some sources define lattice enthalpy as the reverse, exothermic, formation process - always check the direction given.)

Lattice enthalpy depends on the same two factors as bond strength, summarised by $\Delta H_{lat}^{\ominus} \propto \dfrac{q_+ q_-}{r_+ + r_-}$:

• Ionic charge - the biggest effect; $\text{MgO}$ (2+/2-) has a far larger lattice enthalpy than $\text{NaCl}$ (1+/1-).
• Ionic radius - as ions get larger the ionic centres sit further apart, so lattice enthalpy decreases. Keeping the cation fixed and going down a group of anions shows this with a single variable: $\text{LiF} > \text{LiCl} > \text{LiBr}$, since the halide ion grows from $\text{F}^-$ to $\text{Cl}^-$ to $\text{Br}^-$.

Lattice enthalpy cannot be measured directly, so it is found indirectly using a Born-Haber cycle. This applies Hess's law: the enthalpy of formation of the solid equals the sum of the steps that build gaseous ions from the elements (atomisation, ionisation energy, electron affinity) plus lattice formation. The data booklet provides ionic radii and lattice enthalpy values for comparison.