Structure of ionic solids — AP Chemistry Study Guide

For: AP Chemistry candidates sitting AP Chemistry.

Covers: Coordination number, unit cell structure of common ionic lattices (NaCl, CsCl, ZnS), ionic radius ratio rules, density calculation of ionic unit cells, and the relationship between ionic structure and bulk physical properties.

You should already know: Ionic bonding basics. Unit cell contribution rules for crystalline solids. Coulomb's law for electrostatic attraction.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Chemistry style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Structure of ionic solids?

Ionic solids are crystalline solids composed of oppositely charged ions arranged in a continuous, repeating lattice structure, rather than discrete molecules. The study of ionic solid structure focuses on how cations and anions pack to maximize electrostatic attraction between opposite charges and minimize repulsion between like charges, resulting in the lowest potential energy state for the crystal. This topic falls within Unit 2 (Molecular and Ionic Compound Structure and Properties) of the AP Chemistry CED, contributing approximately 2-3% of the total exam score, and appears in both multiple-choice (MCQ) and free-response (FRQ) sections. Questions typically ask to count ions in a unit cell, calculate density, predict structure from radius ratio, or connect nanoscale structure to observable bulk properties like melting point and conductivity. Unlike molecular solids, the repeating lattice extends throughout the entire crystal, so the formula of an ionic compound reflects the whole-number ratio of ions, not a molecular formula.

2. Counting Ions in Ionic Unit Cells

Ionic unit cells contain two distinct types of ions (cations and anions) arranged on a lattice, so counting ions follows the same contribution rules as metallic unit cells, but requires separate counting for each ion and confirmation of charge neutrality. The contribution rules are unchanged from other cubic unit cells: ions on corners are shared between 8 unit cells, so they contribute $\frac{1}{8}$ to the unit cell; ions on edges are shared between 4 unit cells, contributing $\frac{1}{4}$; ions on faces are shared between 2 unit cells, contributing $\frac{1}{2}$; and ions fully inside the unit cell contribute $1$. For any neutral ionic compound, the total positive charge from cations must equal the total negative charge from anions, so you can use this rule to confirm your count matches the compound's empirical formula. Common 1:1 ionic lattices have well-known counts: NaCl has 4 cations and 4 anions per unit cell, CsCl has 1 of each, and ZnS has 4 of each.

Worked Example

Calcium fluoride (${\text{CaF}}_2$) has ${\text{Ca}}^{2+}$ ions occupying all corner and face positions of a cubic unit cell, and all ${\text{F}}^{-}$ ions are fully contained inside the unit cell. How many total ions are in one unit cell of ${\text{CaF}}_2$?

1. Count the number of ${\text{Ca}}^{2+}$ ions first: 8 ${\text{Ca}}^{2+}$ at corners × $\frac{1}{8}$ contribution = 1, plus 6 ${\text{Ca}}^{2+}$ on faces × $\frac{1}{2}$ contribution = 3. Total ${\text{Ca}}^{2+}$ = 1 + 3 = 4.
2. Calculate total positive charge: 4 × (+2) = +8. For neutrality, total negative charge must equal -8.
3. Each ${\text{F}}^{-}$ has a charge of -1, so we need 8 ${\text{F}}^{-}$ ions. All ${\text{F}}^{-}$ are inside the unit cell, so each contributes 1, giving 8 total ${\text{F}}^{-}$.
4. Total ions per unit cell = 4 ${\text{Ca}}^{2+}$ + 8 ${\text{F}}^{-}$ = 12.

Exam tip: Always confirm your ion count matches the empirical formula and charge neutrality rule—if your ratio does not match the compound formula, you used the wrong contribution factor for one of the ions.

3. Ionic Radius Ratio Rule

In ionic lattices, anions pack in a close-packed arrangement, and cations fit into the interstitial voids (empty spaces) between anions. The type of void (and thus the coordination number, or number of oppositely charged ions surrounding a given ion) that a cation can stably fit into is determined by the radius ratio $\frac{r^{+}}{r^{-}}$, where $r^{+}$ is the cation radius and $r^{-}$ is the anion radius. The rule arises because if the cation is too small for the void, it will not touch the surrounding anions, leaving repulsive interactions between anions that make the structure unstable. If the cation is too large for the void, it pushes the anions apart, also creating an unstable arrangement. The accepted radius ratio ranges and corresponding coordination numbers are:

• < 0.225: Coordination number 2 (linear)
• 0.225 – 0.414: Coordination number 4 (tetrahedral)
• 0.414 – 0.732: Coordination number 6 (octahedral)

• 0.732: Coordination number 8 (cubic)

Worked Example

The ionic radius of ${\text{Na}}^{+}$ is 102 pm, and the ionic radius of ${\text{Cl}}^{-}$ is 181 pm. Predict the coordination number of ${\text{Na}}^{+}$ in NaCl.

1. Calculate the radius ratio: $\frac{r^{+}}{r^{-}}=\frac{102\text{pm}}{181\text{pm}}\approx 0.56$.
2. Compare the ratio to the established ranges: 0.414 < 0.56 < 0.732.
3. This range corresponds to coordination number 6 (octahedral voids), which matches the known structure of NaCl, where each ${\text{Na}}^{+}$ is surrounded by 6 ${\text{Cl}}^{-}$ ions.
4. Confirm: A coordination number of 8 would require a ratio greater than 0.732, which 0.56 is not, so the octahedral arrangement is the stable structure.

Exam tip: Always put the cation radius in the numerator and anion radius in the denominator—swapping the ratio is the most common mistake on this type of question.

4. Density Calculation for Ionic Unit Cells

Density of an ionic solid can be calculated directly from its unit cell parameters, using the definition of density as mass divided by volume ($d=\frac{m}{V}$). For a unit cell, the total mass is equal to the mass of all formula units contained in the unit cell: if $Z$ is the number of formula units per unit cell, $M$ is the molar mass of the compound, and $N_A$ is Avogadro's number, the total mass of the unit cell is $\frac{Z\times M}{N_A}$. For a cubic unit cell, volume is equal to the edge length $a$ cubed, so $V=a^3$. Combining these gives the density formula: $$d=\frac{Z\times M}{N_A\times a^3}$$ To get density in the standard unit of ${\text{g/cm}}^3$, edge length must be converted from picometers (pm, the common unit for ionic radii) to centimeters: $1\text{pm}=10^{-10}\text{cm}$.

Worked Example

CsCl has Z = 1 formula unit per unit cell, a molar mass of 168.36 g/mol, and an edge length of 412 pm. Calculate the density of CsCl.

1. Convert edge length to cm: $a=412\text{pm}\times 10^{-10}\text{cm/pm}=4.12\times 10^{-8}\text{cm}$.
2. Calculate unit cell volume: $a^3=(4.12\times 10^{-8}\text{cm}{)}^3\approx 7.00\times 10^{-23}{\text{cm}}^3$.
3. Plug into the density formula: $d=\frac{(1\times 168.36\text{g/mol})}{(6.022\times 10^{23}{\text{mol}}^{-1}\times 7.00\times 10^{-23}{\text{cm}}^3)}\approx \frac{168.36}{42.15}\approx 4.0{\text{g/cm}}^3$.
4. This matches the expected density range for ionic solids (1–5 g/cm³), so the calculation is reasonable.

Exam tip: If your final density is several orders of magnitude outside 1–5 g/cm³, you almost certainly forgot to convert edge length to centimeters—check your unit conversions first.

5. Structure and Physical Properties of Ionic Solids

The arrangement of ions in a crystalline ionic lattice, held together by strong electrostatic ionic bonds, gives ionic solids their characteristic bulk properties. First, ionic solids have high melting and boiling points, because a large amount of energy is required to overcome the strong ionic bonds. Melting point strength follows Coulomb's law: attraction is proportional to $\frac{q_1{q}_2}{r^2}$, where $q_1,q_2$ are ion charges and $r$ is the distance between ions. Higher ion charge leads to much stronger attraction and higher melting point, while larger ionic radii increase $r$ and weaken attraction, lowering melting point. Second, ionic solids are brittle: applying force shifts layers of ions, aligning like charges next to each other, and the resulting repulsion splits the crystal along cleavage planes. Third, ionic solids do not conduct electricity in the solid state, because ions are fixed in the lattice and cannot move to carry charge. When molten or dissolved in water, the lattice breaks apart and ions become mobile, so they conduct electricity.

Worked Example

Magnesium oxide (MgO) and sodium chloride (NaCl) both have the same crystal structure (face-centered cubic, 1:1 ion ratio). MgO melts at 2800°C, while NaCl melts at 801°C. Explain the large difference in melting point.

1. Melting point depends on the strength of electrostatic attraction between ions in the lattice, described by Coulomb's law.
2. NaCl has ${\text{Na}}^{+}$ (+1) and ${\text{Cl}}^{-}$ (-1), so the charge product $q_1{q}_2=(1)(1)=1$. MgO has ${\text{Mg}}^{2+}$ (+2) and ${\text{O}}^{2-}$ (-2), so the charge product is $(2)(2)=4$.
3. The electrostatic attraction in MgO is 4 times stronger than in NaCl, even though the internuclear distance $r$ is similar for both compounds.
4. Much more thermal energy is required to overcome the stronger attraction in MgO, leading to a much higher melting point than NaCl.

Exam tip: Always explicitly reference Coulomb's law when explaining melting point differences in FRQ answers—AP readers require this explicit connection to earn full credit.

6. Common Pitfalls (and how to avoid them)

• Wrong move: Counting only one ion and assuming the number of the second ion is equal, regardless of the compound's formula. Why: Most common examples are 1:1, so students forget to use charge neutrality for compounds with other ratios. Correct move: After counting the first ion, use the compound's empirical formula and charge neutrality to calculate the number of the second ion, then confirm the ratio matches.
• Wrong move: Calculating radius ratio as $\frac{r^{-}}{r^{+}}$ (anion over cation) instead of $\frac{r^{+}}{r^{-}}$. Why: Questions often list anions first in the problem statement, leading students to swap the values. Correct move: Write the ratio formula $\frac{r^{+}}{r^{-}}$ on your paper before plugging in any numerical values.
• Wrong move: Claiming solid ionic solids conduct electricity because they contain charged ions. Why: Students confuse the presence of ions with mobility of ions. Correct move: Remember that ions are fixed in the solid lattice, so conductivity only occurs when ions are mobile (molten or dissolved).
• Wrong move: Using edge length in picometers directly in the density formula, leading to incorrect order of magnitude. Why: Students forget the unit conversion step when working quickly. Correct move: Convert edge length from pm to cm immediately after writing it down, before any other calculations.
• Wrong move: Assuming higher coordination number always leads to higher density, regardless of ionic radii. Why: Students associate higher coordination with tighter packing, but ignore the effect of ion size on volume. Correct move: Always calculate density from unit cell mass and volume, do not guess based on coordination number alone.

7. Practice Questions (AP Chemistry Style)

Question 1 (Multiple Choice)

Sodium sulfide has the formula ${\text{Na}}_2\text{S}$. ${\text{S}}^{2-}$ ions form a face-centered cubic unit cell, with 4 ${\text{S}}^{2-}$ ions per unit cell. How many ${\text{Na}}^{+}$ ions are present in one unit cell of ${\text{Na}}_2\text{S}$? A) 2 B) 4 C) 6 D) 8

Worked Solution: First, calculate total negative charge from the 4 ${\text{S}}^{2-}$ ions: $4\times (-2)=-8$. For charge neutrality, total positive charge must equal +8. Each ${\text{Na}}^{+}$ has a +1 charge, so 8 ${\text{Na}}^{+}$ ions are required to balance the charge. This also matches the 2:1 ${\text{Na}}^{+}$:${\text{S}}^{2-}$ ratio from the formula ${\text{Na}}_2\text{S}$ (8:4 = 2:1). The correct answer is D.

Question 2 (Free Response)

Silver bromide (AgBr) has a cubic unit cell with edge length 5.77 Å (1 Å = 100 pm), and Z = 4 formula units per unit cell. (a) Calculate the density of AgBr in ${\text{g/cm}}^3$. (b) The ionic radius of ${\text{Ag}}^{+}$ is 114 pm, and the ionic radius of ${\text{Br}}^{-}$ is 196 pm. Predict the coordination number of ${\text{Ag}}^{+}$ in AgBr. (c) Explain why solid AgBr is brittle, while metals are malleable.

Worked Solution: (a) Convert edge length to cm: $a=5.77\mathring{A}\times 100pm/\mathring{A}\times 10^{-10}cm/pm=5.77\times 10^{-8}cm$. Molar mass of AgBr = $107.87+79.90=187.77g/mol$. Use the density formula: $$d=\frac{ZM}{N_A{a}^3}=\frac{4\times 187.77}{(6.022\times 10^{23})(5.77\times 10^{-8}{)}^3}\approx \frac{751.08}{(6.022\times 10^{23})(1.92\times 10^{-22})}\approx 6.48{\text{g/cm}}^3$$

(b) Calculate radius ratio: $\frac{r^{+}}{r^{-}}=\frac{114}{196}\approx 0.58$. This falls between 0.414 and 0.732, so the coordination number of ${\text{Ag}}^{+}$ is 6.

(c) In ionic solids like AgBr, applying force shifts layers of ions, aligning like charges next to each other. The electrostatic repulsion between aligned like charges causes the crystal to split along cleavage planes, making it brittle. In metals, the delocalized metallic bonding allows layers to slide past each other without repulsion between like charges, so metals are malleable instead of brittle.

Question 3 (Application / Real-World Style)

A geologist measures the density of a pure lithium fluoride (LiF) mineral sample from a pegmatite deposit and gets 2.59 g/cm³. LiF has the same structure as NaCl, with Z = 4 formula units per unit cell, and an edge length of 402 pm. The theoretical molar mass of LiF is 25.94 g/mol. Calculate the experimental molar mass from the density measurement, and explain why the experimental value differs from theory.

Worked Solution: Rearrange the density formula to solve for molar mass: $M=\frac{d{N}_A{a}^3}{Z}$. Convert edge length: $a=402pm=4.02\times 10^{-8}cm$, so $a^3=6.497\times 10^{-23}c{m}^3$. Plug in values: $$M=\frac{(2.59g/c{m}^3)(6.022\times 10^{23}mo{l}^{-1})(6.497\times 10^{-23}c{m}^3)}{4}\approx 25.3g/mol$$ The experimental molar mass is slightly lower than the theoretical value. This occurs because natural mineral crystals have small lattice defects, where some ion positions are vacant. Vacancies reduce the total mass per unit cell, leading to a lower measured density and lower calculated molar mass than the perfect lattice theoretical value.

8. Quick Reference Cheatsheet

9. What's Next

This topic lays the foundational framework for connecting nanoscale crystal structure to macroscale bulk properties, a core skill across AP Chemistry. Next you will study properties of molecular solids, where you will compare weak intermolecular attractions in molecular solids to the strong electrostatic ionic bonds in ionic solids, explaining large differences in melting point and conductivity between the two compound classes. This chapter is also a prerequisite for understanding lattice energy calculations, solid-state defects, and colligative properties of ionic solutions in later units. Without mastering unit cell counting and density calculation for ionic solids, you will struggle with quantitative crystalline solid problems on the FRQ section.

Intermolecular forces Lattice energy Colligative properties Solid state thermodynamics