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S2.1

The ionic model

Structure 2 · Models of bonding and structure · SL and HL

🎯What you need to be able to do

  • Predict the charge of an ion from the electron configuration of its atom, including the different charges formed by a transition element.
  • Describe the ionic bond as the electrostatic attraction between oppositely charged ions.
  • Deduce the formula and name of an ionic compound from its component ions, including the polyatomic ions listed in the syllabus, and interconvert names and formulas of binary ionic compounds.
  • Explain why ionic compounds exist as three-dimensional lattices represented by empirical formulas.
  • Explain the volatility, electrical conductivity and solubility of ionic compounds.
  • Use lattice enthalpy as a measure of ionic bond strength, and explain how ionic radius and ionic charge affect it.

📚The chemistry

Forming ions

Metal atoms lose electrons to form positive ions, cations; non-metal atoms gain electrons to form negative ions, anions. In each case the driving pattern is the achievement of a stable noble-gas electron configuration, and the charge follows directly from the electron configuration you learned in S1.3.

Group 1 \( ns^1 \) → 1+
Group 2 \( ns^2 \) → 2+
Group 13 → 3+
Group 15 → 3−
Group 16 → 2−
Group 17 → 1−

Transition elements form ions of more than one charge, and the syllabus requires this. Iron gives \( \mathrm{Fe^{2+}} \) and \( \mathrm{Fe^{3+}} \); copper gives \( \mathrm{Cu^{+}} \) and \( \mathrm{Cu^{2+}} \). The reason is that the 4s and 3d sublevels are close in energy, so removing a third or fourth electron does not cost dramatically more than removing the second — the full explanation is in S3.1. In names and formulas the charge is therefore stated in Roman numerals: iron(II) chloride is \( \mathrm{FeCl_2} \), iron(III) chloride is \( \mathrm{FeCl_3} \).

The ionic bond

An ionic bond is the electrostatic attraction between oppositely charged ions. Two features distinguish it from a covalent bond and both matter later:

  • It is non-directional. The attraction acts equally in all directions, so an ion attracts every oppositely charged ion around it, not one partner.
  • It only forms between different elements — one must lose electrons and the other gain them — whereas a covalent bond can form between identical atoms.

Formulas and names

An ionic compound is electrically neutral overall, so the formula is whatever ratio makes the charges cancel. Balance the charges and then reduce to the simplest ratio.

Binary ionic compounds are named cation first, then anion with the suffix “-ide”: sodium chloride, magnesium oxide, calcium nitride. Compounds containing a polyatomic ion keep that ion’s name unchanged.

Seven polyatomic ions must be known by name and formula. There is no way round learning these:

ammonium \( \mathrm{NH_4^{+}} \)
hydroxide \( \mathrm{OH^{-}} \)
nitrate \( \mathrm{NO_3^{-}} \)
hydrogencarbonate \( \mathrm{HCO_3^{-}} \)
carbonate \( \mathrm{CO_3^{2-}} \)
sulfate \( \mathrm{SO_4^{2-}} \)
phosphate \( \mathrm{PO_4^{3-}} \)

When more than one polyatomic ion is needed, bracket it: calcium hydroxide is \( \mathrm{Ca(OH)_2} \), not \( \mathrm{CaOH_2} \), and aluminium sulfate is \( \mathrm{Al_2(SO_4)_3} \). Without the brackets the subscript applies only to the last atom, which is a different compound entirely.

Failing to reduce, or reducing when you should not. Magnesium oxide is \( \mathrm{MgO} \), not \( \mathrm{Mg_2O_2} \) — the 2+ and 2− already cancel one to one. But note the reverse trap in covalent chemistry: hydrogen peroxide really is \( \mathrm{H_2O_2} \) and not \( \mathrm{HO} \), because molecular formulas are not reduced. Ionic formulas are empirical; molecular formulas are not.

The lattice, and why the formula is empirical

Because the ionic bond is non-directional, ions pack into a giant three-dimensional lattice in which every cation is surrounded by anions and vice versa. In sodium chloride each \( \mathrm{Na^{+}} \) is surrounded by six \( \mathrm{Cl^{-}} \) and each \( \mathrm{Cl^{-}} \) by six \( \mathrm{Na^{+}} \).

So there is no such thing as a molecule of sodium chloride. The formula \( \mathrm{NaCl} \) does not describe a particle; it states the ratio in which the ions are present. Ionic formulas are therefore always empirical formulas, and the correct term for \( \mathrm{NaCl} \) is a formula unit.

Explaining the properties

Every property of an ionic compound comes back to the same two facts: the lattice is held by strong electrostatic forces acting in all directions, and it is built from charged particles.

  • Low volatility, high melting and boiling points. Melting requires the electrostatic attractions between ions throughout the lattice to be overcome, and there are a very large number of them. That takes a great deal of energy, so ionic solids have high melting points and evaporate hardly at all — they are non-volatile.
  • Electrical conductivity. A substance conducts if it contains charged particles that are free to move. In a solid ionic lattice the ions are held in fixed positions, so it does not conduct. Molten or dissolved in water, the ions become mobile and it does conduct — and in doing so is decomposed, which is electrolysis (R3.2).
  • Solubility. Many ionic compounds dissolve in water and few dissolve in non-polar solvents. Water molecules are polar: the \( \delta- \) oxygen is attracted to cations and the \( \delta+ \) hydrogens to anions. These ion–dipole attractions release energy (hydration) which can compensate for the energy needed to break up the lattice. A non-polar solvent cannot form such interactions, so it cannot pay for the lattice, and the solid does not dissolve.
  • Brittleness. Not on the syllabus list but worth knowing: strike an ionic crystal and the layers shift, bringing like charges into contact; the resulting repulsion splits the crystal along a clean plane. Contrast metals in S2.3.

Lattice enthalpy

Lattice enthalpy is the measure of the strength of the ionic bonding in a compound. It is the energy change when one mole of a solid ionic compound is separated into its gaseous ions, and it is governed by Coulomb’s law: the electrostatic force is proportional to the product of the charges and inversely proportional to the distance between the centres of the ions.

Larger ionic charge
→ stronger attraction
→ larger lattice enthalpy
→ higher melting point
Smaller ionic radius
→ ions closer together
→ stronger attraction
→ larger lattice enthalpy

Charge is much the more powerful of the two. Magnesium oxide (2+ with 2−) melts at about 2850 °C while sodium chloride (1+ with 1−) melts at 801 °C, even though the ions are of comparable size — because the charge product is four times greater. Radius explains the trends within a group: down group 1 the chlorides have progressively smaller lattice enthalpies as the cation grows.

The full thermodynamic treatment of lattice enthalpy, via the Born–Haber cycle, is AHL material in R1.2. Here you only need it as a measure of bond strength and the two factors that control it.

✏️Worked example

(a) Deduce the formula of the compound formed between aluminium and oxygen, and name it.
(b) Write the formula of iron(III) sulfate and of ammonium phosphate.
(c) Sodium chloride melts at 801 °C; magnesium oxide melts at 2852 °C; potassium chloride melts at 770 °C. Explain both differences.
(d) Explain why solid magnesium chloride does not conduct electricity but molten magnesium chloride does, and state what happens at each electrode when the melt is electrolysed.

(a) Aluminium is in group 13, so it forms \( \mathrm{Al^{3+}} \); oxygen is in group 16, so it forms \( \mathrm{O^{2-}} \). For neutrality the charges must cancel: the lowest common multiple of 3 and 2 is 6, so two \( \mathrm{Al^{3+}} \) (total 6+) balance three \( \mathrm{O^{2-}} \) (total 6−). The formula is \( \mathrm{Al_2O_3} \), aluminium oxide.

(b) Iron(III) is \( \mathrm{Fe^{3+}} \) and sulfate is \( \mathrm{SO_4^{2-}} \); balancing 3 against 2 gives two iron ions and three sulfate ions, so \( \mathrm{Fe_2(SO_4)_3} \). Ammonium is \( \mathrm{NH_4^{+}} \) and phosphate is \( \mathrm{PO_4^{3-}} \), so three ammonium ions are needed: \( \mathrm{(NH_4)_3PO_4} \). Both need brackets, because the subscript applies to the whole polyatomic ion.

(c) Two separate comparisons, each changing one variable.

MgO against NaCl — the charges differ. \( \mathrm{Mg^{2+}} \) and \( \mathrm{O^{2-}} \) carry double the charge of \( \mathrm{Na^{+}} \) and \( \mathrm{Cl^{-}} \), so the electrostatic attraction is far stronger, the lattice enthalpy far larger, and much more energy is needed to separate the ions. (The ions are also smaller, which adds to the effect, but charge is the dominant reason.)

KCl against NaCl — the charges are the same, so the difference must be size. \( \mathrm{K^{+}} \) has an extra occupied main energy level and so is larger than \( \mathrm{Na^{+}} \). The centres of the ions are further apart, the electrostatic attraction is weaker, the lattice enthalpy smaller, and the melting point lower.

(d) Conduction requires mobile charge carriers. In the solid, the \( \mathrm{Mg^{2+}} \) and \( \mathrm{Cl^{-}} \) ions are held in fixed positions in the lattice by strong electrostatic attractions, so although charged particles are present they cannot move, and no current flows. On melting, the lattice breaks down and the ions are free to move towards the electrodes, so the liquid conducts.

At the cathode (negative), magnesium ions are reduced: \( \mathrm{Mg^{2+}} + 2\mathrm{e^{-}} \rightarrow \mathrm{Mg}(l) \). At the anode (positive), chloride ions are oxidised: \( 2\mathrm{Cl^{-}} \rightarrow \mathrm{Cl_2}(g) + 2\mathrm{e^{-}} \).

Check it. For any formula you write, add up the charges and confirm they come to zero: \( \mathrm{Fe_2(SO_4)_3} \) gives \( 2(+3) + 3(-2) = 0 \). For the melting-point argument in (c), check you have changed only one variable per comparison — if you find yourself explaining MgO against KCl, you are varying charge and radius at once and the answer becomes unfalsifiable. For (d), check the electrode assignment against the direction of charge: positive ions must travel to the negative electrode, and reduction is gain of electrons, so the cathode is always where reduction happens.
“NaCl conducts when dissolved because the bonds break and release electrons.” Nothing releases electrons. The charge carriers in a molten or aqueous ionic compound are the ions themselves, which already existed in the solid — what changes is only that they become mobile. Writing about “free electrons” describes a metal (S2.3), not an ionic compound, and it is one of the quickest ways to lose a whole explanation mark.

📝Practise

Work through these on paper, then reveal the answer.

1. Write formulas for: calcium nitrate, sodium hydrogencarbonate, ammonium sulfate, chromium(III) oxide, and lithium phosphate.
Calcium nitrate: \( \mathrm{Ca^{2+}} \) with \( \mathrm{NO_3^{-}} \) → \( \mathrm{Ca(NO_3)_2} \). Sodium hydrogencarbonate: \( \mathrm{Na^{+}} \) with \( \mathrm{HCO_3^{-}} \) → \( \mathrm{NaHCO_3} \) (no brackets needed for a single polyatomic ion). Ammonium sulfate: \( \mathrm{NH_4^{+}} \) with \( \mathrm{SO_4^{2-}} \) → \( \mathrm{(NH_4)_2SO_4} \). Chromium(III) oxide: \( \mathrm{Cr^{3+}} \) with \( \mathrm{O^{2-}} \) → \( \mathrm{Cr_2O_3} \). Lithium phosphate: \( \mathrm{Li^{+}} \) with \( \mathrm{PO_4^{3-}} \) → \( \mathrm{Li_3PO_4} \). Check each by confirming the charges sum to zero.
2. Predict, using electron configurations, the charge of the ion formed by each of: K, Ba, S, N and Br.
K is \( [\mathrm{Ar}]\,4s^1 \): losing the single 4s electron gives the argon configuration, so \( \mathrm{K^{+}} \). Ba is \( [\mathrm{Xe}]\,6s^2 \): losing both gives \( \mathrm{Ba^{2+}} \). S is \( [\mathrm{Ne}]\,3s^2\,3p^4 \): gaining two electrons fills 3p and gives the argon configuration, so \( \mathrm{S^{2-}} \). N is \( 1s^2\,2s^2\,2p^3 \): gaining three gives the neon configuration, so \( \mathrm{N^{3-}} \). Br is \( [\mathrm{Ar}]\,4s^2\,3d^{10}\,4p^5 \): gaining one gives the krypton configuration, so \( \mathrm{Br^{-}} \). The general rule is that the ion adopts the configuration of the nearest noble gas, since that requires the fewest electrons to move.
3. Explain why sodium chloride dissolves readily in water but not in hexane.
Dissolving requires the energy cost of breaking down the lattice to be compensated by energy released when the ions interact with solvent molecules. Water is polar, with \( \delta- \) on oxygen and \( \delta+ \) on the hydrogens. The \( \delta- \) oxygen atoms are attracted to \( \mathrm{Na^{+}} \) and the \( \delta+ \) hydrogens to \( \mathrm{Cl^{-}} \), forming strong ion–dipole interactions and surrounding each ion with a shell of water molecules — hydration. The energy released on hydration is comparable to the lattice enthalpy, so the solid dissolves. Hexane is non-polar: its molecules have no permanent dipole, so they can only offer weak London (dispersion) interactions to an ion. Far too little energy is released to compensate for breaking a lattice held by full ionic charges, so sodium chloride is insoluble in hexane. The general principle is “like dissolves like”, and the reasoning is energetic, not magical.
4. Place the following in order of increasing melting point and justify the order: NaF, MgO, KBr, CaO.
Sort by charge product first, then ionic radius. NaF and KBr are both 1+ / 1−; MgO and CaO are both 2+ / 2−. So the two 2 : 2 compounds come above the two 1 : 1 compounds. Within each pair, the compound with the larger ions melts lower, because the ionic centres are further apart and the electrostatic attraction is weaker. \( \mathrm{K^{+}} \) and \( \mathrm{Br^{-}} \) are larger than \( \mathrm{Na^{+}} \) and \( \mathrm{F^{-}} \); \( \mathrm{Ca^{2+}} \) is larger than \( \mathrm{Mg^{2+}} \). Order: KBr < NaF < CaO < MgO. Actual values: 734, 993, 2572 and 2852 °C — note how much bigger the charge effect is than the size effect, exactly as the Coulomb argument predicts.
5. Explain why the formula of an ionic compound is described as an empirical formula, and why it is wrong to speak of "a molecule of magnesium chloride".
The ionic bond is non-directional: each ion attracts every oppositely charged ion around it equally in all directions. Ions therefore pack into a giant three-dimensional lattice which extends throughout the crystal, and there is no point at which one can identify a discrete group of atoms bonded to each other and not to their neighbours. Since there are no discrete particles, there are no molecules, and \( \mathrm{MgCl_2} \) cannot be a molecular formula. What the formula does state is the simplest whole-number ratio of ions present — one \( \mathrm{Mg^{2+}} \) for every two \( \mathrm{Cl^{-}} \), as required for electrical neutrality — which is precisely the definition of an empirical formula. The correct name for the entity \( \mathrm{MgCl_2} \) represents is a formula unit.
6. Aluminium oxide has a much higher melting point than sodium oxide. Identify the two factors responsible and state which is more important, then predict which of MgCl2 and NaCl has the greater lattice enthalpy.
The ions in \( \mathrm{Al_2O_3} \) are \( \mathrm{Al^{3+}} \) and \( \mathrm{O^{2-}} \); in \( \mathrm{Na_2O} \) they are \( \mathrm{Na^{+}} \) and \( \mathrm{O^{2-}} \). Two factors differ: (i) charge — the aluminium ion carries 3+ against 1+, so the product of the charges is three times greater and the electrostatic attraction correspondingly stronger; (ii) radius — \( \mathrm{Al^{3+}} \) is considerably smaller than \( \mathrm{Na^{+}} \) (higher nuclear charge pulling in the same number of electrons, and one fewer occupied energy level than the sodium atom it came from), so the ionic centres are closer and the attraction stronger again. Both push the same way, and charge is the more important, since it enters Coulomb’s law as a product while distance enters as an inverse. For the prediction: \( \mathrm{MgCl_2} \) has \( \mathrm{Mg^{2+}} \) against \( \mathrm{Na^{+}} \), a doubled cation charge, and \( \mathrm{Mg^{2+}} \) is also smaller than \( \mathrm{Na^{+}} \). So magnesium chloride has the greater lattice enthalpy, and correspondingly the higher melting point (714 °C against 801 °C — and the fact that this particular prediction goes the “wrong” way experimentally is a good reminder that \( \mathrm{MgCl_2} \) has significant covalent character, which is the subject of S2.4).

🔗Go deeper — other people’s work

These are external resources, not mine. If one stops working, tell me and everything above it on this page still stands.

  • PhET — Salts and Solubility and Sugar and Salt Solutions, which show the hydration shell forming around each ion as the lattice dissolves.
  • The Royal Society of Chemistry periodic table — ionic radii for every common ion, if you want to test the radius argument against real numbers rather than assertions.
  • Any crystal-structure viewer (for example the VESTA or CrystalMaker demonstrations) — rotating a rock-salt lattice makes the “no molecules” point immediate in a way no diagram does.