Ionic bonding

1. Overview

When a reactive metal combines with a reactive non-metal, electrons move from the metal atoms to the non-metal atoms, and the compound is built from charged particles called ions. The force that holds these ions together is ionic bonding. This section covers the definition you must learn, how the ions form and what charges they carry, and the three examples you must be able to describe: sodium chloride, magnesium oxide and calcium fluoride. These ideas underpin dot-and-cross diagrams, giant ionic lattices and the properties of salts.

Key Definitions

• Ionic bonding: the electrostatic attraction between oppositely charged ions, that is, between positively charged cations and negatively charged anions.
• Ion: an atom, or a group of covalently bonded atoms, that carries an overall electric charge because it has an unequal number of protons and electrons.
• Cation: a positively charged ion, which has fewer electrons than protons; metal atoms form cations by losing electrons, for example Na⁺, Mg²⁺ and Ca²⁺.
• Anion: a negatively charged ion, which has more electrons than protons; non-metal atoms form anions by gaining electrons, for example Cl⁻, O²⁻ and F⁻.
• Electrostatic attraction: the force that pulls opposite electric charges towards each other; between two ions it is stronger when their charges are greater in size and when their centres are closer together.
• Giant ionic lattice: a regular three-dimensional arrangement of a very large number of cations and anions, in which each ion is surrounded by ions of the opposite charge and is attracted to all of them.
• Formula unit: the group of ions shown by the formula of an ionic compound, in the simplest whole-number ratio; one formula unit of CaF₂ is one Ca²⁺ ion and two F⁻ ions. Ionic compounds are not made of molecules.
• Isoelectronic: having the same number of electrons; for example Na⁺, Mg²⁺, O²⁻ and F⁻ each have 10 electrons, arranged as 1s² 2s² 2p⁶.

Content

What an ionic bond is

Learn this definition word for word:

Ionic bonding is the electrostatic attraction between oppositely charged ions (positively charged cations and negatively charged anions).

Metals form cations by losing electrons; non-metals form anions by gaining them.

The bond is the attraction, not the movement of electrons. Electron transfer is how the ions are made when elements react. Sodium chloride crystallised from sea water has exactly the same ionic bonding as sodium chloride made from sodium and chlorine.

Electrostatic attraction acts in all directions. A Na⁺ ion does not belong to one particular Cl⁻ ion; it attracts every Cl⁻ ion around it. That is why ionic compounds form giant lattices rather than molecules.

Which elements form ionic bonds?

Ionic bonding forms between elements with a large difference in electronegativity, usually a metal from Group 1 or 2 and a non-metal from Group 16 or 17. With the Data Booklet's Pauling values:

Compound	Electronegativities	Difference	Bonding
NaCl	Na 0.9, Cl 3.0	2.1	ionic
MgO	Mg 1.3, O 3.5	2.2	ionic
CaF₂	Ca 1.0, F 4.0	3.0	ionic
AlCl₃	Al 1.5, Cl 3.0	1.5	covalent

"Metal plus non-metal" is only a guide. Aluminium chloride contains a metal, yet its bonding is covalent: it turns into a vapour of Al₂Cl₆ molecules at only about 180 °C.

One compound can contain both kinds of bond. In sodium hydroxide, ionic bonding holds Na⁺ and OH⁻ together, while the O-H bond inside each hydroxide ion is covalent.

How the ions form

You can predict an ion's charge from the element's group:

Group	1	2	16	17
Electrons lost or gained	lose 1	lose 2	gain 2	gain 1
Ion charge	1+	2+	2−	1−
Examples	Li⁺, Na⁺, K⁺	Mg²⁺, Ca²⁺	O²⁻, S²⁻	F⁻, Cl⁻

A Group 1 or 2 metal atom loses its outer s electrons. A Group 16 or 17 atom gains electrons to fill its outer p sub-shell. In both cases the ion ends up with the electronic configuration of a noble gas.

The formula of the compound balances the charges so that the total is zero. Calcium forms Ca²⁺ and fluorine forms F⁻, so two F⁻ ions balance each Ca²⁺: 1 × (+2) + 2 × (−1) = 0, giving CaF₂.

Why do the ions form? Not because atoms "want" a full outer shell. Removing the outer electron from one mole of gaseous sodium atoms takes in 494 kJ (Data Booklet). That is more than the energy released when one mole of gaseous chlorine atoms each gain an electron, so making the separate ions takes in energy overall. The compound forms because the oppositely charged ions then pull together into a solid, which releases far more energy than the ions cost to make.

The three examples: NaCl, MgO and CaF₂

	Sodium chloride	Magnesium oxide	Calcium fluoride
Metal atom	Na 1s² 2s² 2p⁶ 3s¹	Mg 1s² 2s² 2p⁶ 3s²	Ca 1s² 2s² 2p⁶ 3s² 3p⁶ 4s²
Cation	Na⁺ 1s² 2s² 2p⁶ (as Ne)	Mg²⁺ 1s² 2s² 2p⁶ (as Ne)	Ca²⁺ 1s² 2s² 2p⁶ 3s² 3p⁶ (as Ar)
Non-metal atom	Cl 1s² 2s² 2p⁶ 3s² 3p⁵	O 1s² 2s² 2p⁴	F 1s² 2s² 2p⁵
Anion	Cl⁻ 1s² 2s² 2p⁶ 3s² 3p⁶ (as Ar)	O²⁻ 1s² 2s² 2p⁶ (as Ne)	F⁻ 1s² 2s² 2p⁶ (as Ne)
Electrons moved per formula unit	1	2	2 (one to each F)
Ratio of ions	1 Na⁺ : 1 Cl⁻	1 Mg²⁺ : 1 O²⁻	1 Ca²⁺ : 2 F⁻
Equation	2Na(s) + Cl₂(g) → 2NaCl(s)	2Mg(s) + O₂(g) → 2MgO(s)	Ca(s) + F₂(g) → CaF₂(s)

Sodium chloride. Each sodium atom transfers its 3s electron to a chlorine atom. Na⁺ has 11 protons and 10 electrons; Cl⁻ has 17 protons and 18 electrons.

Magnesium oxide. Each magnesium atom transfers both 3s electrons to one oxygen atom. Mg²⁺ and O²⁻ both have 10 electrons (they are isoelectronic), but Mg²⁺ is much smaller because its 12 protons pull those electrons in more strongly than the 8 protons of O²⁻.

Calcium fluoride. A calcium atom has two 4s electrons to lose, but a fluorine atom can take only one. So each calcium atom gives one electron to each of two fluorine atoms, and there are twice as many F⁻ ions as Ca²⁺ ions.

Showing the bonding: dot-and-cross diagrams

A dot-and-cross diagram for an ionic compound shows separate ions, not shared electrons:

1. Draw each ion on its own inside square brackets, with its charge at the top right, outside the bracket.
2. Show the outer shell of each anion: for the anions on this page it is full, with eight electrons. Use crosses for electrons that came from the metal and dots for the non-metal's own electrons, or the other way round, but keep to one key.
3. For each cation, either show no outer electrons, or show its new outer shell, the full shell below the one it emptied (eight electrons for Na⁺, Mg²⁺ and Ca²⁺, but two for Li⁺). Both are common practice; the charge must be shown either way.
4. Show the right number of each ion. For CaF₂, draw one [Ca]²⁺ and two separate [F]⁻ ions, each F⁻ with seven dots and one cross.
5. Never draw a line or overlapping shells between the ions: those mean shared electrons, which is covalent bonding.

Ionic bonding in the solid: a giant lattice

In the solid, the ions pack into a giant ionic lattice, a regular three-dimensional arrangement repeated in every direction. Each ion's nearest neighbours carry the opposite charge, so the attractions outweigh the repulsions between ions of the same charge.

• In sodium chloride, each Na⁺ is surrounded by six Cl⁻ ions and each Cl⁻ by six Na⁺ ions.
• Magnesium oxide has the same arrangement, with Mg²⁺ and O²⁻.
• In calcium fluoride, the lattice holds two F⁻ ions for every Ca²⁺. A Ca²⁺ ion is not bonded to just two F⁻ ions: like every ion in the lattice, it attracts all the F⁻ ions around it.

The formula NaCl gives the ratio of ions, not a molecule, so we speak of a formula unit.

The bonding explains two key properties. Melting points are high, because many strong attractions acting in all directions must be overcome. The solid does not conduct electricity, because its ions are fixed in place; when molten or dissolved in water, the ions are free to move and carry charge.

How strong is ionic bonding?

In general, ionic bonds are much stronger than intermolecular forces. Between two ions, the attraction is stronger when:

• the charges on the ions are greater in size, and
• the ions are smaller, so their centres are closer together.

Compound	Sizes of the charges	Distance between neighbouring ion centres / nm	Melting point / °C
NaCl	1 and 1	0.282	801
CaF₂	2 and 1	0.237	1418
MgO	2 and 2	0.211	above 2800

(You do not need to learn the distances; they are here so you can compare.)

MgO against NaCl. Both factors favour MgO. At the same distance apart, ions with charges of size 2 and 2 attract each other four times as strongly as ions with charges of size 1 and 1. The smaller ions in MgO also bring the centres about 25% closer. The charge effect is the larger of the two, so the 2+ and 2− charges are the main reason magnesium oxide melts at a far higher temperature.

CaF₂ in between. Ca²⁺ carries a 2+ charge but F⁻ only a 1− charge, and the ion centres are closer than in NaCl but further apart than in MgO. So the attraction is stronger than in NaCl and weaker than in MgO. This simple charge-and-size picture puts the three melting points in the right order.

Worked example 1: describing the bonding in lithium oxide

Question: Lithium burns in oxygen to form lithium oxide, Li₂O. Describe the bonding in lithium oxide. Include the electronic configuration of each ion and the number of electrons transferred per formula unit. (Pauling electronegativities: Li 1.0, O 3.5)

Step 1: decide the type of bonding. Lithium is a Group 1 metal and oxygen a Group 16 non-metal. The electronegativity difference is 3.5 − 1.0 = 2.5, which is large, so the bonding is ionic.

Step 2: work out the ions.

• Li, 1s² 2s¹, loses its 2s electron to form Li⁺, 1s², the configuration of helium. Its outer shell is now the full first shell, with two electrons, not eight.
• O, 1s² 2s² 2p⁴, gains two electrons to form O²⁻, 1s² 2s² 2p⁶, the configuration of neon.

Step 3: balance the charges. 2 × (+1) + 1 × (−2) = 0, so there are two Li⁺ ions for each O²⁻ ion. Each oxygen atom gains one electron from each of two lithium atoms: 2 electrons per formula unit.

$$4\text{Li(s)} + \text{O}_2\text{(g)} \rightarrow 2\text{Li}_2\text{O(s)}$$

Answer: Lithium oxide has ionic bonding: the electrostatic attraction between oppositely charged Li⁺ ions (1s²) and O²⁻ ions (1s² 2s² 2p⁶), in a 2 : 1 ratio, acting in all directions through a giant lattice. Two electrons are transferred per formula unit, one from each lithium atom.

Worked example 2: counting the ions in calcium fluoride

Question: A crystal of calcium fluoride, CaF₂, has a mass of 2.50 g. Calculate (a) the total number of ions in the crystal and (b) the number of fluoride ions. ($A_{\text{r}}$: Ca 40.1, F 19.0; Avogadro constant $L = 6.022 \times 10^{23}$ mol⁻¹)

Step 1: molar mass. $$M(\text{CaF}_2) = 40.1 + (2 \times 19.0) = 78.1\ \text{g mol}^{-1}$$

Step 2: amount of formula units. $$n(\text{CaF}_2) = \frac{m}{M} = \frac{2.50}{78.1} = 0.03201\ \text{mol}$$

Step 3: ions per formula unit. Each formula unit is one Ca²⁺ and two F⁻, so three ions. $$n(\text{ions}) = 3 \times 0.03201 = 0.09603\ \text{mol}$$

Step 4: number of ions. $$N = n \times L = 0.09603 \times 6.022 \times 10^{23} = 5.783 \times 10^{22}$$

Step 5: fluoride ions only. Two of the three ions in each formula unit are F⁻. $$N(\text{F}^-) = 2 \times 0.03201 \times 6.022 \times 10^{23} = 3.8553 \times 10^{22}$$

Answer: (a) $5.78 \times 10^{22}$ ions; (b) $3.86 \times 10^{22}$ fluoride ions (3 s.f., matching the 2.50 g given).

Sense check. The crystal holds $1.93 \times 10^{22}$ Ca²⁺ ions, half as many as F⁻. The total charge is $(1.93 \times 10^{22} \times 2) - (3.86 \times 10^{22} \times 1) = 0$, as it must be.

Worked example 3: electrons transferred in making magnesium oxide

Question: Magnesium burns in oxygen to form 5.00 g of magnesium oxide, MgO. Calculate the number of electrons transferred from magnesium atoms to oxygen atoms. ($A_{\text{r}}$: Mg 24.3, O 16.0; $L = 6.022 \times 10^{23}$ mol⁻¹)

Step 1: amount of MgO. $$M(\text{MgO}) = 24.3 + 16.0 = 40.3\ \text{g mol}^{-1} \qquad n(\text{MgO}) = \frac{5.00}{40.3} = 0.12407\ \text{mol}$$

Step 2: electrons per formula unit. Each Mg atom loses two electrons to become Mg²⁺, and each O atom gains them to become O²⁻. $$n(\text{e}^-) = 2 \times 0.12407 = 0.24814\ \text{mol}$$

Step 3: number of electrons. $$N(\text{e}^-) = 0.24814 \times 6.022 \times 10^{23} = 1.494 \times 10^{23}$$

Answer: $1.49 \times 10^{23}$ electrons are transferred (3 s.f.).

Making the same amount, 0.124 mol, of NaCl would move only half as many electrons, because each sodium atom loses one electron, not two.

Key Equations

Charge balance in an ionic formula (the total charge of a formula unit is zero; for CaF₂, 1 × (+2) + 2 × (−1) = 0): $$(\text{number of cations} \times \text{cation charge}) + (\text{number of anions} \times \text{anion charge}) = 0$$

Forming the ions ($\text{e}^-$ is an electron; these half-equations only show where the electrons go, not separate steps that happen on their own, so state symbols are left out here; the overall equations, with state symbols, are in the table of the three examples): $$\text{Na} \rightarrow \text{Na}^{+} + \text{e}^{-} \qquad \text{Cl}_2 + 2\text{e}^{-} \rightarrow 2\text{Cl}^{-}$$ $$\text{Mg} \rightarrow \text{Mg}^{2+} + 2\text{e}^{-} \qquad \text{O}_2 + 4\text{e}^{-} \rightarrow 2\text{O}^{2-}$$ $$\text{Ca} \rightarrow \text{Ca}^{2+} + 2\text{e}^{-} \qquad \text{F}_2 + 2\text{e}^{-} \rightarrow 2\text{F}^{-}$$

Amount and number of particles ($m$ = mass in g; $M$ = molar mass in g mol⁻¹; $n$ = amount in mol; $N$ = number of particles; $L$ = Avogadro constant, $6.022 \times 10^{23}$ mol⁻¹): $$n = \frac{m}{M} \qquad N = n \times L$$

Ions and electrons from formula units: $$n(\text{ions}) = n(\text{formula units}) \times \text{ions per formula unit}$$ $$n(\text{electrons transferred}) = n(\text{formula units}) \times \text{electrons moved per formula unit}$$

Common Mistakes to Avoid

• Mistake: defining ionic bonding as "the transfer of electrons from a metal to a non-metal". Fix: the bond is the attraction: in sodium chloride, "the electrostatic attraction between oppositely charged Na⁺ and Cl⁻ ions". Electron transfer only explains how the ions formed.
• Mistake: wrong ion charges, and so wrong formulas, such as F²⁻, CaF or MgO₂. Fix: take the charge from the group (Group 2 gives 2+, Group 16 gives 2−, Group 17 gives 1−) and balance: Ca²⁺ needs two F⁻, so the formula is CaF₂.
• Mistake: drawing an ionic compound as a molecule, with a shared pair or overlapping shells between the ions. Fix: draw separate ions in square brackets with their charges. MgO is [Mg]²⁺ and [O]²⁻, with eight electrons in the oxide ion's outer shell, two of them shown as crosses.
• Mistake: explaining the melting point of an ionic solid with "intermolecular forces". Fix: there are no molecules. Say "strong electrostatic attraction between oppositely charged ions throughout a giant lattice, which takes a lot of energy to overcome", for example to melt NaCl at 801 °C.
• Mistake: comparing MgO with NaCl using only one factor, or just writing "stronger bonds". Fix: give both reasons: the charges are twice the size (2+ and 2− against 1+ and 1−), and Mg²⁺ and O²⁻ are smaller than Na⁺ and Cl⁻, so their centres are closer.
• Mistake: using the moles of formula units as the moles of ions. Fix: multiply by the ions per formula unit. In Worked example 2, 0.0320 mol of CaF₂ contains 0.0960 mol of ions; skipping the × 3 gives $1.93 \times 10^{22}$ ions instead of $5.78 \times 10^{22}$.
• Mistake: wrong configurations for the ions, such as writing Ca²⁺ as 1s² 2s² 2p⁶ 3s² 3p⁶ 4s², or saying Na⁺ has the configuration of argon. Fix: a metal loses its outer s electrons first. Ca²⁺ is 1s² 2s² 2p⁶ 3s² 3p⁶ (as Ar); Na⁺ is 1s² 2s² 2p⁶ (as Ne).
• Mistake: assuming every metal and non-metal compound is ionic. Fix: check the electronegativity difference. Aluminium chloride (difference 1.5) is covalent and forms Al₂Cl₆ molecules, while NaCl (difference 2.1) is ionic.

Exam Tips

• Define ionic bonding with all three key ideas: "electrostatic attraction", "oppositely charged" and "ions". "Attraction between ions" alone is incomplete, and "transfer of electrons" describes how the ions form, not the bond.
• Describe the bonding in a named compound in four steps: the electrons each atom loses or gains, the ions formed with their charges (and configurations if asked), the ratio of ions, and the electrostatic attraction between them in a giant lattice.
• Draw dot-and-cross diagrams with every ion in square brackets, its charge outside, full outer shells on the anions, a clear key for dots and crosses, and the right number of each ion.
• Explain a difference in melting point by naming the bonding and the particles, saying which attraction is stronger and why (size of charge and size of ion), and linking that to the energy needed to overcome it.
• Deduce the bonding from data by linking each piece of evidence to a conclusion, for example "it conducts when molten but not as a solid, so it is ionic: its ions are fixed in the solid and free to move in the liquid". Do not mix terms such as "giant molecular with ionic bonding".
• When asked for the full electronic configuration of an ion, write out every sub-shell, for example 1s² 2s² 2p⁶ 3s² 3p⁶ for Cl⁻, rather than the [Ar] shorthand.