Structure of Metals and Alloys - AP Chemistry

SECTION 1

Historical Context & Motivation

Humans have shaped metals for millennia—copper, bronze, and iron each defined entire ages of civilization—yet a coherent atomic-level picture of why metals conduct electricity, bend without shattering, and gleam with characteristic luster arrived only in the twentieth century. Early metallurgists understood empirically that mixing tin into copper produced a harder alloy (bronze), but they could not explain the mechanism. The development of X-ray crystallography and quantum mechanics finally revealed that the macroscopic properties of metals arise from two intertwined features: a regular crystal lattice of cations and a delocalized sea of electrons that permeates the structure.

~3300 BCE • Bronze Age Begins

Artisans in the Near East discover that alloying copper with tin produces bronze—harder and more durable than either pure metal—launching millennia of empirical metallurgy.

1912 • X-ray Diffraction of Crystals

Max von Laue demonstrates that X-rays diffract through crystal lattices, providing the first tool to determine the arrangement of atoms inside metals.

1913 • Bragg's Law & Metallic Structures

W. H. and W. L. Bragg use X-ray diffraction to solve the crystal structures of simple metals, identifying face-centered cubic (FCC), body-centered cubic (BCC), and hexagonal close-packed (HCP) arrangements.

1927–1928 • Free Electron Model

Arnold Sommerfeld applies quantum mechanics to Drude's classical electron gas, creating the free-electron model that explains metallic conductivity and heat capacity quantitatively.

1930s–Present • Band Theory & Modern Alloy Design

Band theory refines the electron-sea picture into continuous energy bands. Computational alloy design now predicts new superalloys for jet engines and biomedical implants.

The central question this lesson addresses is: How does the arrangement of atoms and the nature of metallic bonding account for the physical properties of pure metals and their alloys? Understanding the answer is essential not only for AP Chemistry but for appreciating materials science broadly.

SECTION 2

Core Principles of Metallic Bonding

Metallic bonding differs fundamentally from ionic and covalent bonding. In a metallic solid, each atom releases one or more valence electrons into a communal pool that is not associated with any single atom. The resulting structure consists of an ordered array of metal cations immersed in a delocalized electron sea. Because these valence electrons are shared by all cations simultaneously—rather than being transferred (ionic) or shared between two atoms (covalent)—the bonding is nondirectional and extends uniformly throughout the solid.

Electron Sea Model

Valence electrons are delocalized across the entire lattice. The electrostatic attraction between the mobile electron cloud and the stationary cation cores provides the cohesive force that holds the metal together.

Close-Packed Lattice

Metal atoms tend to adopt arrangements that maximize the number of nearest neighbors (coordination number up to 12), minimizing empty space and maximizing bonding contacts.

Nondirectional Bonding

Unlike covalent bonds that lock atoms at specific angles, metallic bonds are isotropic. Layers of atoms can slide past one another without breaking the overall bonding network, producing malleability and ductility.

Electrical & Thermal Conductivity

The freely mobile electrons carry charge under an applied voltage (electrical conductivity) and transport kinetic energy through the lattice (thermal conductivity).

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KEY TAKEAWAY

Think of a metallic solid as a crowded stadium where every spectator (metal cation) has tossed coins into a giant shared fountain (the electron sea). No single spectator 'owns' any particular coin, yet the fountain's presence keeps everyone in their seats—bonded together. When you push one row of spectators sideways, the fountain simply reshapes itself, which is why metals dent instead of shatter.

SECTION 3

Visualizing the Metallic Lattice

The violet spheres represent metal cations arranged in a regular lattice. The small cyan circles represent delocalized valence electrons that move freely throughout the structure, creating the 'electron sea' responsible for electrical conductivity, luster, and malleability.

In the diagram above, the regularity of the cation positions reflects the crystalline order found in real metals. Notice that the electrons are not localized between specific pairs of cations the way a covalent bond would be; instead, the cyan dots are scattered throughout the lattice. This delocalization is what makes metallic bonding fundamentally different from covalent or ionic interactions. When an electric field is applied across the solid, these mobile electrons drift in one direction, constituting an electrical current. Similarly, when one face of the metal is illuminated, the free electrons oscillate in response to the electromagnetic wave and re-emit light, producing the characteristic metallic luster.

Inside one copper conductor Mobile electrons in copper allow charge to move through the conductor.

SECTION 4

Crystal Packing & Unit Cells

The arrangement of atoms in a metallic crystal is described by its unit cell—the smallest repeating three-dimensional pattern that, when translated in all directions, regenerates the entire lattice. Three unit-cell types dominate metallic elements: body-centered cubic (BCC), face-centered cubic (FCC), and hexagonal close-packed (HCP). The packing efficiency and coordination number differ among these, directly influencing density and mechanical behavior.

Property	BCC	FCC	HCP
Atoms per unit cell	2	4	6 (hexagonal cell)
Coordination number	8	12	12
Packing efficiency	68%	74%	74%
Edge–radius relation	a = 4r / √3	a = 2√2 · r	a = 2r
Example metals	Fe, W, Cr, Na	Cu, Au, Ag, Al	Mg, Zn, Ti

Comparison of the three common metallic crystal structures

DENSITY FROM UNIT CELL

ρ = (n × M) / (Nₐ × a³)

ρ = density (g/cm³); n = atoms per unit cell; M = molar mass (g/mol); Nₐ = Avogadro's number (6.022 × 10²³ mol⁻¹); a = edge length (cm)

FCC EDGE–RADIUS RELATIONSHIP

a = 2√2 · r

In an FCC cell, atoms touch along the face diagonal. The face diagonal equals 4r, and because the face diagonal of a cube is a√2, solving gives a = 4r / √2 = 2√2 · r.

Both FCC and HCP achieve the theoretical maximum packing fraction of 74%, meaning only 26% of the crystal volume is void space. BCC structures are slightly less efficient at 68%. These differences matter: FCC metals like copper and gold are generally more ductile because their close-packed planes allow more slip systems, while BCC metals like iron exhibit greater hardness at room temperature. When the AP exam asks you to relate crystal structure to macroscopic properties, the coordination number and packing efficiency are the key links.

SECTION 5

Alloys — Substitutional and Interstitial

An alloy is a mixture of a metal with one or more other elements (metallic or nonmetallic) that retains metallic bonding and properties. Alloys are not compounds with fixed stoichiometric ratios; rather, they are solid solutions whose composition can vary continuously over some range. The two principal types—substitutional alloys and interstitial alloys—differ in how the solute atoms fit into the host lattice.

Left: In a substitutional alloy (e.g., brass), solute atoms of similar radius replace host atoms at regular lattice sites. Right: In an interstitial alloy (e.g., steel), small solute atoms such as carbon occupy the holes between larger host atoms without displacing them.

For a substitutional alloy to form readily, the Hume-Rothery rules provide useful guidelines: the atomic radii of the two metals should differ by no more than about 15%, the metals should have similar electronegativities and crystal structures, and they should share the same valence. Brass (Cu–Zn) is the classic example—copper and zinc have comparable radii and both adopt FCC or HCP structures under the right compositions. In contrast, interstitial alloys form when the solute atom is much smaller than the host. Carbon (atomic radius ≈ 77 pm) fits into the interstitial holes of the iron lattice (atomic radius ≈ 126 pm), producing steel. The small interstitial atoms impede the motion of dislocations—line defects in the lattice—thereby increasing hardness and tensile strength relative to the pure metal.

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AP Exam Tip

The College Board frequently tests whether students can identify an alloy as substitutional or interstitial based on the relative sizes of the component atoms. If the atoms are similar in size, predict substitutional; if one atom is much smaller (H, C, N, B), predict interstitial. Also be prepared to explain why alloys are generally harder than the pure host metal: the differently sized solute atoms disrupt the regular lattice, making it more difficult for planes of atoms to slide.

One alloy microstructure reference An alloy’s microstructure contains crystalline grains separated by grain boundaries.

SECTION 6

Worked Example — Calculating Density from a Unit Cell

A common AP-level problem asks you to calculate the density of a metal from its unit-cell type and atomic radius. The following example walks through the complete calculation for gold, which crystallizes in an FCC lattice.

Density of Gold from its FCC Unit Cell

1
Step 1 — Identify Given Information

Gold crystallizes in an FCC structure. Its atomic radius is r = 144 pm = 1.44 × 10⁻⁸ cm. Molar mass M = 196.97 g/mol. FCC cells contain n = 4 atoms per unit cell.
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Step 2 — Calculate Edge Length (a)

For FCC: a = 2√2 · r = 2(1.414)(1.44 × 10⁻⁸ cm) = 4.073 × 10⁻⁸ cm.

a = 4.073 × 10⁻⁸ cm
3
Step 3 — Calculate Unit Cell Volume

V = a³ = (4.073 × 10⁻⁸ cm)³ = 6.756 × 10⁻²³ cm³.

V = 6.756 × 10⁻²³ cm³
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Step 4 — Apply the Density Formula

ρ = (n × M) / (Nₐ × a³) = (4 × 196.97 g/mol) / (6.022 × 10²³ mol⁻¹ × 6.756 × 10⁻²³ cm³). Numerator = 787.88 g/mol. Denominator = 40.67 cm³/mol.

ρ ≈ 19.4 g/cm³
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Step 5 — Verify Reasonableness

The accepted density of gold is 19.3 g/cm³. Our calculated value of 19.4 g/cm³ matches within rounding, confirming both the FCC model and our arithmetic.

SECTION 7

Properties of Metals vs. Other Solid Types

AP Chemistry expects you to compare metallic solids with ionic, covalent-network, and molecular solids. The electron-sea model predicts metallic properties; contrasting it with the bonding in other solid types clarifies why each class behaves differently.

Property	Metallic Solid	Ionic Solid	Covalent Network	Molecular Solid
Particles	Cations + e⁻ sea	Cations + anions	Atoms (covalent bonds)	Molecules (IMFs)
Conductivity (solid)	High (mobile e⁻)	None (ions fixed)	None (usually)	None
Malleability	High	Brittle	Very hard, brittle	Soft
Melting point	Variable (Hg to W)	High	Very high	Low
Luster	Shiny (e⁻ re-emit light)	No	No	No

Comparison of the four classes of crystalline solids

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KEY TAKEAWAY

The unique combination of high electrical and thermal conductivity, malleability, ductility, and luster in metals all trace back to a single feature: delocalized valence electrons. No other bonding model provides freely mobile charge carriers in the solid state. When an ionic crystal is struck, like charges are forced into alignment and the crystal shatters; when a metal is struck, the electron sea simply redistributes, allowing layers to glide smoothly.

Cleavage in a salt crystal An ionic crystal can cleave when displacement brings like charges into neighboring positions.

SECTION 8

Connection to Band Theory

The electron-sea model is a powerful qualitative tool, but a more rigorous quantum-mechanical treatment leads to band theory. When N atoms come together in a metallic solid, their N discrete atomic orbitals overlap and split into a near-continuous band of N closely spaced molecular orbitals. The lower-energy portion of this band is filled with electrons while the upper portion remains empty, and because the energy gaps between adjacent levels are negligibly small, electrons can be promoted with essentially zero energy input—this is the quantum explanation for metallic conductivity.

Feature	Electron-Sea Model	Band Theory
Electron description	Free, delocalized gas of e⁻	Electrons fill a continuum of energy levels (bands)
Conductivity	Mobile e⁻ carry current	Partially filled band allows electron promotion with negligible energy
Insulators explained?	Not directly	Yes—large band gap prevents electron promotion
Semiconductors?	Not explained	Small band gap; conductivity increases with temperature
AP relevance	Sufficient for qualitative questions	Beyond AP scope; useful for advanced understanding

Electron-sea model vs. band theory

For the AP Chemistry exam, you are not required to invoke band theory, but understanding that the electron-sea model is a simplified version of a deeper quantum picture helps you appreciate its limitations. When a question asks why metals conduct but ionic solids do not in the solid state, the electron-sea answer—mobile delocalized electrons versus fixed ions in a lattice—is perfectly sufficient and expected.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

Which of the following best explains why metals are malleable while ionic compounds are brittle?

A. Metallic bonds are weaker than ionic bonds, so metal atoms separate more easily under stress.
B. In metals, delocalized electrons allow layers of cations to slide without disrupting the bonding, whereas shifting ions in an ionic solid aligns like charges, causing repulsion and fracture.
C. Metal atoms are smaller than ions, so they can rearrange more freely in the lattice.
D. Metallic solids lack a regular lattice structure, so there are no defined planes to fracture along.

Show answer (B)

B is correct. In metallic solids, the nondirectional electron sea adjusts as cation layers slide, maintaining the bonding interaction. In ionic solids, displacing a layer brings like-charged ions adjacent to one another, generating strong repulsive forces that shatter the crystal. A is wrong because malleability is about bond type, not bond strength—many metals have very strong metallic bonds. C is wrong because atomic size alone does not explain malleability; the nature of the bonding does. D is wrong because metals do have regular crystalline lattices.

Question 2: BASIC CALCULATION

Silver crystallizes in an FCC unit cell with an atomic radius of 144 pm. What is the edge length of the unit cell?

A. 288 pm
B. 333 pm
C. 407 pm
D. 576 pm

Show answer (C)

C is correct. For FCC: a = 2√2 × r = 2(1.414)(144 pm) = 407 pm. A (288 pm) would be 2r, which is the relationship for a simple cubic cell where atoms touch along an edge. B (333 pm) incorrectly uses a = 4r/√3, which is the BCC relationship. D (576 pm) is 4r, the face diagonal, not the edge length.

Question 3: INTERMEDIATE

Steel is an alloy of iron and carbon, and brass is an alloy of copper and zinc. Which statement correctly classifies these alloys?

A. Steel is substitutional; brass is interstitial.
B. Steel is interstitial; brass is substitutional.
C. Both are substitutional because both involve metallic elements.
D. Both are interstitial because the solute atoms are always smaller than the solvent atoms.

Show answer (B)

B is correct. Carbon atoms (radius ≈ 77 pm) are much smaller than iron atoms (radius ≈ 126 pm), so carbon fits into the interstitial holes of the iron FCC/BCC lattice—making steel an interstitial alloy. Copper (radius ≈ 128 pm) and zinc (radius ≈ 134 pm) have similar atomic radii, so zinc replaces copper atoms at lattice sites—making brass substitutional. A reverses the classifications. C is wrong because carbon is a nonmetal and its small size dictates interstitial placement regardless. D is wrong because while Zn is slightly smaller than Cu, the size difference is within the 15% Hume-Rothery threshold, leading to substitutional, not interstitial, behavior.

Question 4: APPLIED

4 pts

Aluminum crystallizes in a face-centered cubic (FCC) unit cell. The molar mass of aluminum is 26.98 g/mol and its density is 2.70 g/cm³. (a) Determine the edge length of the aluminum unit cell in cm. Show your work. (2 pts) (b) Calculate the atomic radius of aluminum in pm. (1 pt) (c) Explain why aluminum is a good electrical conductor, referencing its bonding and structure. (1 pt)

Show answer

(a) [2 pts] Using ρ = nM/(Nₐa³), solve for a³ = nM/(ρNₐ). For FCC, n = 4. a³ = (4 × 26.98 g/mol) / (2.70 g/cm³ × 6.022 × 10²³ mol⁻¹) = 107.92 / (1.626 × 10²⁴) = 6.637 × 10⁻²³ cm³. a =

(6.637 × 10⁻²³)^{1/3}

= 4.049 × 10⁻⁸ cm = 404.9 pm. [1 pt for correct setup, 1 pt for correct answer with units.] (b) [1 pt] For FCC: a = 2√2·r → r = a/(2√2) = 404.9 pm / 2.828 = 143.2 pm. [1 pt for correct calculation.] (c) [1 pt] Aluminum is metallic, so its valence electrons are delocalized throughout the crystal lattice in an electron sea. These mobile electrons are free to move in response to an applied electric field, enabling high electrical conductivity. [1 pt for connecting delocalized electrons to conductivity.]

Question 5: CRITICAL THINKING

4 pts

A student measures the densities and hardness values (on a relative scale) of four solid samples and records the following data. Sample A: Density 8.96 g/cm³, Hardness 3.0 (pure copper, FCC) Sample B: Density 8.52 g/cm³, Hardness 3.5 (brass, Cu–Zn alloy) Sample C: Density 7.87 g/cm³, Hardness 4.0 (pure iron, BCC) Sample D: Density 7.85 g/cm³, Hardness 6.5 (steel, Fe–C alloy) (a) Explain why brass (Sample B) has a lower density than pure copper (Sample A) despite having a similar crystal structure. (1 pt) (b) Explain, at the atomic level, why steel (Sample D) is significantly harder than pure iron (Sample C), even though their densities are nearly identical. (1 pt) (c) Classify brass and steel as substitutional or interstitial alloys, and justify each classification using the data and atomic radii (Cu: 128 pm, Zn: 134 pm, Fe: 126 pm, C: 77 pm). (1 pt) (d) A student claims that alloying always decreases density. Using the data, evaluate this claim. (1 pt)

Show answer

(a) [1 pt] Zinc has a slightly higher molar mass (65.38 g/mol) than copper (63.55 g/mol) but a significantly larger atomic radius (134 pm vs 128 pm). When zinc substitutes for copper in the FCC lattice, the unit cell expands. The volume increase (proportional to r³) outpaces the small mass increase per atom, resulting in a net decrease in density. (b) [1 pt] In steel, small carbon atoms occupy interstitial holes in the iron lattice. These interstitial atoms impede the movement of dislocations (planes of atoms that slip under stress), making it much harder for layers of iron atoms to slide past one another. This dislocation-pinning mechanism dramatically increases hardness with only a small change in composition. (c) [1 pt] Brass is substitutional: Cu (128 pm) and Zn (134 pm) differ in radius by only about 4.7%, well within the ~15% threshold for substitutional alloy formation. Steel is interstitial: C (77 pm) is only about 61% the size of Fe (126 pm), so carbon atoms are small enough to fit in the interstitial voids of the iron lattice. (d) [1 pt] The claim is not universally true. While brass (8.52 g/cm³) is less dense than pure copper (8.96 g/cm³), steel (7.85 g/cm³) is essentially the same density as pure iron (7.87 g/cm³). In interstitial alloys, the small solute atoms fill existing voids, adding mass without significantly expanding the unit cell, so density changes are minimal. The effect of alloying on density depends on the type of alloy and the properties of the solute.

SUMMARY

Lesson Summary

Summary

Metallic solids consist of metal cations arranged in a crystal lattice held together by a delocalized sea of valence electrons. This nondirectional bonding explains the hallmark properties of metals: electrical and thermal conductivity (mobile electrons), malleability and ductility (layers slide without breaking bonds), and metallic luster (electrons absorb and re-emit visible light). The three dominant unit-cell types—BCC, FCC, and HCP—differ in coordination number and packing efficiency, and the density formula ρ = nM/(Nₐa³) connects the atomic-scale unit cell to a measurable macroscopic property.

Alloys are solid solutions that modify metallic properties. In substitutional alloys (e.g., brass), solute atoms of similar size replace host atoms at lattice sites. In interstitial alloys (e.g., steel), much smaller atoms such as carbon occupy the holes between host atoms. Both types disrupt the regular lattice, increasing hardness by impeding dislocation motion. For the AP exam, connect the electron-sea model to conductivity and malleability, distinguish substitutional from interstitial alloys by relative atomic size, and apply unit-cell geometry to density calculations.