Structure of metals and alloys — AP Chemistry Study Guide

For: AP Chemistry candidates sitting AP Chemistry.

Covers: metallic bonding and the electron sea model, crystalline unit cell edge length-atomic radius relationships for pure metals, classification of substitutional and interstitial alloys, and structure-property relationships for metals and alloys for the AP Chemistry exam.

You should already know: Basics of ionic and covalent bonding, Fundamentals of crystalline lattice structure, Relationship between bonding and physical properties.

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 metals and alloys?

This topic falls under Unit 2 (Molecular and Ionic Compound Structure and Properties) of the AP Chemistry Course and Exam Description (CED), accounting for approximately 7-9% of the total AP exam weight, and appears regularly on both multiple choice (MCQ) and free response (FRQ) sections. The structure of metals and alloys describes the arrangement of atoms and the nature of bonding in metallic substances, which explains their characteristic physical properties (electrical and thermal conductivity, malleability, ductility, melting point) that differ dramatically from ionic, molecular, or covalent network solids. Alloys are homogeneous mixtures of metals (or metals and small nonmetals) engineered to have improved functional properties over pure metals, making them one of the most widely used classes of materials in industrial and consumer applications. On the AP exam, questions on this topic typically ask you to relate bonding and atomic-scale structure to macroscopic properties, classify alloys by type, compare properties of pure metals and alloys, and derive and calculate atomic radius from unit cell parameters for pure metallic crystals.

2. Metallic Bonding and the Electron Sea Model

Metallic bonding is defined as the electrostatic attraction between positively charged metal cations (nuclei plus core electrons) and a delocalized "sea" of mobile valence electrons that extends uniformly throughout the entire crystalline solid. Unlike covalent bonding, where electrons are localized between two specific atoms, or ionic bonding, where electrons are fully transferred to form fixed charged ions in a lattice, the delocalization of valence electrons in metals directly explains all their signature properties.

Electrical conductivity arises because mobile delocalized electrons can flow through the lattice when a voltage is applied, carrying electric current. Thermal conductivity occurs because delocalized electrons rapidly transfer kinetic energy through the solid when one region is heated. Malleability (ability to be hammered into sheets) and ductility (ability to be drawn into wires) occur because when an external force is applied, layers of metal cations can slide past one another without breaking the metallic bonding: the delocalized electron sea simply rearranges to maintain electrostatic attraction around the shifted cations. By contrast, ionic solids fracture when struck because shifting layers align like charges, leading to repulsive forces that break the solid. Melting point of metals correlates with the strength of the electrostatic attraction: higher cation charge and more delocalized electrons lead to stronger attraction and higher melting points.

Worked Example

Problem: A student claims that sodium metal is malleable because its metallic bonds are weak. Evaluate this claim and justify your answer using the electron sea model.

1. The student’s claim is incorrect. Weak metallic bonds would result in a low melting point, not malleability.
2. The electron sea model describes metallic bonding as strong electrostatic attraction between delocalized valence electrons and fixed sodium cations in a crystalline lattice.
3. When an external force is applied to sodium, layers of sodium cations can slide past one another. The delocalized electron sea rapidly rearranges to maintain electrostatic attraction between cations and electrons, so the solid does not fracture.
4. Malleability is a result of delocalized bonding, not weak bonding.

Exam tip: On AP exam FRQs, always explicitly connect the property of the metal to the delocalization of electrons. Examiners award points for this specific link, not just a generic reference to "metallic bonding".

3. Crystalline Unit Cell Geometry for Pure Metals

Nearly all pure metals form crystalline solids with close-packed atomic arrangements, because close packing maximizes attractive forces between cations and the electron sea, lowering the overall energy of the solid. The three most common cubic unit cell structures for metals are primitive cubic (low 52% packing efficiency, rare), body-centered cubic (BCC, 68% packing efficiency, found in iron and sodium), and face-centered cubic (FCC/CCP, 74% packing efficiency, found in copper and aluminum).

A common AP exam question asks you to relate the edge length of the unit cell ($a$) to the atomic radius ($r$) of the metal atom, based on the assumption that atoms are hard spheres that touch along the direction of packing. These relationships can always be derived from the Pythagorean theorem, so you do not need to rely solely on memorization:

• Primitive cubic (atoms touch along edge): $a=2r$
• BCC (atoms touch along body diagonal of the cube): $\sqrt{3}a=4r⟹r=\frac{a\sqrt{3}}{4}$
• FCC (atoms touch along face diagonal of the cube): $\sqrt{2}a=4r⟹r=\frac{a\sqrt{2}}{4}$

Worked Example

Problem: Silver crystallizes in a face-centered cubic unit cell with an edge length of 408 pm. Calculate the atomic radius of a silver atom, in picometers.

1. In FCC, atoms are located at each cube corner and the center of each face. Atoms touch along the face diagonal of the cube, so the total length of the face diagonal equals four atomic radii (corner radius + face center radius + opposite corner radius = $r+2r+r=4r$).
2. For a square face of the cube with edge length $a$, the Pythagorean theorem gives: $(facediagonal{)}^2=a^2+a^2=2a^2$, so $facediagonal=\sqrt{2}a$.
3. Equate the two expressions for face diagonal and rearrange to solve for $r$: $\sqrt{2}a=4r⟹r=\frac{\sqrt{2}a}{4}$.
4. Substitute $a=408$ pm: $r=\frac{(1.414)(408)}{4}\approx 144$ pm.

Exam tip: Always show your derivation of the $a$-$r$ relationship on FRQs, even if you have memorized the formula. AP graders award points for reasoning, not just the final numerical answer.

4. Structure and Classification of Alloys

Alloys are homogeneous mixtures of two or more elements, at least one of which is a metal, that retain bulk metallic properties. They are classified into two main types based on the relative atomic size of the added component, which determines its position in the original metal’s crystalline lattice:

1. Substitutional alloys: Form when the added element has an atomic radius within ~15% of the original metal’s atomic radius. The added atom replaces the original metal atom in the crystalline lattice. Common examples include brass (copper + zinc) and 14-karat gold (gold + copper).
2. Interstitial alloys: Form when the added element has an atomic radius more than ~30% smaller than the original metal’s atomic radius. The small added atom fits into the empty interstitial gaps between the original metal atoms in the lattice. The most common example is carbon steel (iron + carbon).

All alloys almost always have higher hardness and strength, and lower electrical conductivity, than pure metals. The added atoms disrupt the regular crystalline lattice of the pure metal: this makes it harder for layers of atoms to slide past one another (increasing hardness) and disrupts the delocalized electron sea (reducing conductivity).

Worked Example

Problem: Sterling silver is an alloy of silver (atomic radius 144 pm) and copper (atomic radius 128 pm). (a) Classify this alloy. (b) Compare the hardness of sterling silver to pure silver, and justify your answer.

1. Calculate the percent difference in atomic radii: $\frac{|144-128|}{144}\times 100\%\approx 11\%$, which is less than the 15% threshold for substitutional alloys.
2. Because copper atoms are similar in size to silver atoms, copper atoms replace silver atoms in the crystalline lattice of silver, so sterling silver is a substitutional alloy.
3. Pure silver has a uniform, regular lattice, so layers of silver atoms can slide past one another easily, making it soft.
4. The slightly different-sized copper atoms disrupt the regular lattice structure, creating obstructions that prevent layers of silver atoms from sliding easily. This makes sterling silver harder than pure silver.

Exam tip: Do not assume all alloys with a nonmetal are interstitial. Always base classification on relative atomic size, not whether the added element is metal or nonmetal.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Claiming metals are malleable because metallic bonds are weak. Why: Students confuse the ability of layers to slide with bond weakness. Weak bonds lead to low melting points, not malleability. Correct move: Connect malleability to the delocalized electron sea that rearranges to maintain bonding when layers slide.
• Wrong move: For BCC unit cells, using the face diagonal instead of the body diagonal to relate $a$ and $r$. Why: Students mix up atom positions in BCC vs FCC. BCC has an atom at the cube center, not face centers, so atoms touch along the body diagonal. Correct move: Draw a quick sketch of the unit cell before solving, marking where atoms touch.
• Wrong move: Classifying any alloy with a nonmetal as automatically interstitial. Why: Students associate nonmetals with small size, but some nonmetals (e.g., silicon, 111 pm) are similar in size to many metals. Correct move: Always calculate the percent difference in atomic radii to classify, regardless of element type.
• Wrong move: Claiming alloys are harder than pure metals because they have stronger metallic bonds. Why: Students confuse lattice disruption with bond strength. Increased hardness does not come from stronger bonds. Correct move: Explain increased hardness as a result of a disrupted lattice that prevents sliding of atomic layers.
• Wrong move: For FCC, writing $a=2r$ instead of $a=2\sqrt{2}r$. Why: Students forget that two radii come from each end of the face diagonal, leading to 4r total, not 2r. Correct move: Derive the relationship from the Pythagorean theorem every time, rather than relying on memory.

6. Practice Questions (AP Chemistry Style)

Question 1 (Multiple Choice)

Which of the following correctly explains why pure aluminum conducts electricity while solid aluminum oxide (an ionic solid) does not? A) Aluminum has covalent bonds, while aluminum oxide has ionic bonds. B) Aluminum has delocalized mobile electrons that can flow through the solid, while all electrons in aluminum oxide are localized on fixed ions. C) Metallic bonds are weaker than ionic bonds, so electrons can escape more easily from aluminum. D) Aluminum has a lower melting point than aluminum oxide, so electrons move faster in solid aluminum.

Worked Solution: First, eliminate incorrect options. Option A is false: aluminum has metallic bonding, not covalent bonding. Option C is false: conductivity does not depend on bond weakness, and weak bonds do not explain conductivity. Option D is false: melting point does not determine conductivity in solids. Option B correctly matches the electron sea model: delocalized mobile electrons in metals carry current, while all electrons in ionic solids are bound to fixed ions, so no current flow is possible. Correct answer: B

Question 2 (Free Response)

Iron crystallizes in a body-centered cubic unit cell with an edge length of 287 pm. (a) Derive the relationship between edge length $a$ and atomic radius $r$ for BCC. Show your reasoning. (b) Calculate the atomic radius of iron, in pm. (c) Carbon steel is an alloy of iron (atomic radius 126 pm) and carbon (atomic radius 77 pm). Classify this alloy, and explain why it is stronger than pure iron.

Worked Solution: (a) In BCC, atoms are located at all 8 cube corners and one atom at the center of the cube. Atoms touch along the body diagonal that runs from the center of one corner atom through the central atom to the center of the opposite corner atom, so total body diagonal length = $r+2r+r=4r$. For a cube, the 3D Pythagorean theorem gives $(bodydiagonal{)}^2=a^2+a^2+a^2=3a^2$, so $bodydiagonal=\sqrt{3}a$. Equating the two expressions gives $\sqrt{3}a=4r$, the required relationship. (b) Rearrange to solve for $r$: $r=\frac{\sqrt{3}a}{4}=\frac{(1.732)(287)}{4}\approx 124$ pm, which matches the expected value for iron. (c) The percent difference in atomic radii is $\frac{126-77}{126}\times 100\%\approx 39\%$, so carbon is much smaller than iron. Carbon fits into the interstitial gaps between iron atoms, so this is an interstitial alloy. The added carbon atoms disrupt the regular lattice of pure iron, preventing layers of iron atoms from sliding past one another easily when force is applied. This makes carbon steel stronger than pure iron.

Question 3 (Application / Real-World Style)

Cast iron used for cookware is an alloy containing 95% iron (atomic radius 126 pm) and 5% silicon (atomic radius 111 pm) by mass. Classify this alloy and explain why it is more brittle than pure iron, making it suitable for shaping into cast cookware.

Worked Solution: First calculate the percent difference in atomic radii: $\frac{|126-111|}{126}\times 100\%\approx 12\%$, which is less than 15%, so this is a substitutional alloy. Pure iron has a regular lattice that allows layers to slide, making it ductile and hard to shape into fixed intricate forms like cast cookware. The silicon atoms disrupt the regular iron lattice, increasing hardness and making the alloy more brittle: it breaks cleanly when stressed rather than bending, which makes it easier to cast into fixed shapes during manufacturing and holds its shape well during use.

7. Quick Reference Cheatsheet

8. What's Next

This topic establishes the core atomic-scale reasoning for linking structure to properties in crystalline solids, a key skill for the entire AP Chemistry course. Next, you will extend the unit cell geometry skills you learned here to ionic crystalline solids, where you will calculate density and ionic radii from unit cell parameters, and connect ionic structure to solubility and melting point. Without mastering the derivation of edge length-radius relationships for metallic unit cells, you will struggle to apply that same reasoning to more complex ionic unit cells, a common FRQ topic. This topic also provides the foundation for materials chemistry, which appears across later units of the course.

• Ionic Solid Structure and Unit Cells
• Bonding and Physical Properties of Solids
• Materials Chemistry of Alloys