Unit 2 Compound Structure and Properties

AP Chemistry Ionic Bonds

How ions form and attract, how to predict their charges, and how to write and use ionic formulas, in four worked examples.

An ionic bond is the attraction between oppositely charged ions, and it is the reason a metal and a nonmetal combine in fixed ratios. Questions on it ask you to say how the ions form, predict the charge of each, write the formula that results, and count the ions in a sample or a solution. How to tell an ionic bond from a covalent or metallic one is in ionic vs covalent bonds, and how strong the attraction is in lattice energy; this page covers formation and formulas.

How Ionic Bonds Form and How to Predict Ion Charges

Metals have low ionization energy and lose their valence electrons to become cations. Nonmetals have a strong attraction for electrons and gain them to become anions. The electrons move from the metal to the nonmetal, and the resulting ions, each with a noble-gas configuration, attract each other in a repeating lattice. The usual charges follow the group number:

Metals lose electrons: group 1 gives +1, group 2 gives +2, and group 13 gives +3 (aluminum).
Nonmetals gain electrons: group 17 gives −1, group 16 gives −2, and group 15 gives −3.
Transition metals can form more than one charge, so the charge is given in the name, such as iron(III).

Fig. 1 · Usual ion charge by group

Worked Example 1: Showing the Electron Transfer in NaCl, MgO, and MgCl₂

Show how electrons transfer to form sodium chloride, magnesium oxide, and magnesium chloride, and give the electron count of each ion.

NaCl: Na (11 electrons, 1 valence electron) gives up 1 electron to become Na⁺ (10 electrons, like neon). Cl (17 electrons, 7 valence) gains 1 to become Cl⁻ (18 electrons, like argon). One electron moves, so the ratio is 1:1.
MgO: Mg (12 electrons) loses 2 to become Mg²⁺ (10 electrons). O (8 electrons) gains 2 to become O²⁻ (10 electrons). Two electrons move, so the ratio is 1:1.
MgCl₂: Mg still loses 2 electrons, but each Cl can accept only 1, so two Cl atoms are needed, giving Mg²⁺ and 2 Cl⁻.

In each case the total electrons lost equals the total electrons gained, so the compound is neutral. The ions in a pair, such as Na⁺ and Ne or O²⁻ and Mg²⁺, have the same number of electrons, which is why the periodic table tells you the charges.

Worked Example 2: Writing Ionic Formulas by Balancing Charge

Write the formula for aluminum sulfide, calcium nitride, magnesium nitrate, and ammonium phosphate.

Aluminum sulfide: Al³⁺ and S²⁻. The least common multiple of 3 and 2 is 6, so 2 Al (+6) and 3 S (−6): Al₂S₃.
Calcium nitride: Ca²⁺ and N³⁻. 3 Ca (+6) and 2 N (−6): Ca₃N₂.
Magnesium nitrate: Mg²⁺ and NO₃⁻. One Mg (+2) and 2 nitrate (−2): Mg(NO₃)₂.
Ammonium phosphate: NH₄⁺ and PO₄³⁻. 3 ammonium (+3) and 1 phosphate (−3): (NH₄)₃PO₄.

Check by adding the charges, which must total zero, and put parentheses around a polyatomic ion only when its subscript is greater than 1. A quick shortcut is to use the numerical value of each charge as the subscript of the other ion, then reduce. The names and charges of the polyatomic ions are on the polyatomic ions page, and the naming rules are in naming compounds.

Worked Example 3: Comparing the Size of an Atom With Its Ion

Which is larger, a sodium atom or a Na⁺ ion, and a chlorine atom or a Cl⁻ ion? Explain.

Sodium: the atom (about 190 pm) is much larger than Na⁺ (about 102 pm). The ion has lost its outer 3s electron, so the outer shell is the smaller n = 2 shell, and 11 protons now pull on only 10 electrons.
Chlorine: Cl⁻ (about 181 pm) is larger than the Cl atom (about 99 pm). The ion has 18 electrons held by 17 protons, so electron repulsion increases and the effective pull per electron decreases.

Cations are smaller than their atoms, and anions are larger. These ion sizes are the distances used when comparing the strength of ionic attractions in lattice energy, and the reasoning follows the nuclear-charge argument in periodic trends.

Worked Example 4: Counting the Ions Released When an Ionic Compound Dissolves

A solution is made by dissolving aluminum sulfate to a concentration of 0.050 M Al₂(SO₄)₃. Find the concentration of each ion and of all ions together.

Dissociation: Al₂(SO₄)₃(s) → 2Al³⁺(aq) + 3SO₄²⁻(aq).
[Al³⁺]: 2 × 0.050 = 0.10 M.
[SO₄²⁻]: 3 × 0.050 = 0.15 M.
Total ions: 0.10 + 0.15 = 0.25 M, five times the formula-unit concentration.

The subscripts in the formula are the mole ratios for dissociation, which is the same bookkeeping used in molarity and in net ionic equations. A solution with more ions per liter holds more charge carriers and tends to conduct better, so 0.050 M Al₂(SO₄)₃ is expected to conduct more than 0.050 M NaCl.

Five Ionic Bond Practice Questions With Answers

Write the formula for barium nitride.
Show answer
Ba²⁺ and N³⁻, so 3 Ba (+6) and 2 N (−6): Ba₃N₂.
Write the formula for ammonium sulfate.
Show answer
NH₄⁺ and SO₄²⁻: (NH₄)₂SO₄.
Write the formula for calcium phosphate.
Show answer
Ca²⁺ and PO₄³⁻: 3 Ca (+6) and 2 phosphate (−6), so Ca₃(PO₄)₂.
How many moles of each ion are produced by dissolving 0.25 mol of Na₃PO₄?
Show answer
0.75 mol Na⁺ and 0.25 mol PO₄³⁻, or 1.00 mol of ions in total.
Why is the formula of sodium chloride NaCl but the formula of magnesium chloride MgCl₂?
Show answer
Na⁺ and Cl⁻ have charges that cancel one to one, but Mg²⁺ needs two Cl⁻ ions to balance its +2 charge.

Common Ionic Bond Mistakes

Writing a molecule for an ionic compound. The formula is the simplest ratio of ions in a lattice, called a formula unit.
Putting the ion charges in the formula. Charges are used to find the subscripts and are not written in the final formula.
Forgetting parentheses around a polyatomic ion that needs a subscript, as in Mg(NO₃)₂.
Unbalanced charge. Always add the charges at the end and confirm they total zero.
Changing a polyatomic ion's subscripts to balance charge. Only the number of ions can change.
Calling the dissolved ions molecules. Dissolving separates the ions from each other, so count them from the formula.

Writing formulas and counting ions are short skills that feed into stoichiometry, equilibrium, and acid-base problems. To see how a stronger Unit 2 would shift your overall estimate, the AP Chem Score Calculator turns section scores into a 1–5 range.

Frequently Asked Questions

What is an ionic bond?

An ionic bond is the electrostatic attraction between positively charged cations and negatively charged anions. It forms when one or more electrons transfer from a metal atom to a nonmetal atom, typically giving both ions a noble-gas electron configuration.

How do you predict the charge of an ion?

For main-group elements, find the group. Metals in groups 1, 2, and 13 lose 1, 2, and 3 electrons to form +1, +2, and +3 ions, and nonmetals in groups 15, 16, and 17 gain 3, 2, and 1 electrons to form −3, −2, and −1 ions. Transition metals vary and their charge is usually given in the name, as in iron(III).

How do you write the formula of an ionic compound?

Choose the smallest numbers of each ion that make the total positive charge equal the total negative charge, and write them as subscripts. Put a polyatomic ion in parentheses if its subscript is greater than 1, as in Mg(NO₃)₂.

Why do ionic compounds have formula units instead of molecules?

An ionic solid is a lattice of many cations and anions, not a set of separate molecules. The formula gives only the simplest ratio of ions, called a formula unit, such as NaCl or MgCl₂.

How many ions are produced when an ionic compound dissolves?

Count them from the formula. One formula unit of Al₂(SO₄)₃ gives 2 Al³⁺ ions and 3 SO₄²⁻ ions, so 5 ions in total. This matters for ion concentrations and for how well a solution conducts electricity.

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