College Chemistry • Bonding & Molecular Structure

Structure of Metals and Alloys

How the arrangement of atoms in metallic crystals and alloys governs the physical properties of the materials we engineer.

Laboratory test tubes with an atom and science notebook.
Inside this lesson 10 sections 5 questions 9 figures

SECTION 1

Historical Context & Motivation

The study of metals stretches back thousands of years, from the earliest copper smelting in Anatolia to the sophisticated steel alloys of the Industrial Revolution. Yet understanding why metals behave as they do—why they conduct electricity, deform without shattering, and gain strength when mixed with other elements—required a revolution in how scientists conceived of atomic arrangement. The concept of the crystal lattice and the metallic bond only emerged in the twentieth century, when X-ray diffraction and quantum mechanics provided the tools to probe atomic-scale structure. Before these breakthroughs, metallurgists relied on empirical trial and error—heating, hammering, and quenching—guided by centuries of craft knowledge but lacking a coherent theoretical framework.

1912 • Laue's X-ray Diffraction

Max von Laue demonstrated that crystals diffract X-rays, proving atoms are arranged in periodic three-dimensional lattices. This technique opened the door to determining the exact positions of atoms in metals.

1913 • Bragg's Law

William Henry Bragg and William Lawrence Bragg derived Bragg's Law (nλ = 2d sin θ), enabling precise measurement of interplanar spacings in metallic crystals. Their work on copper and other metals revealed the face-centered cubic and body-centered cubic structures.

1928 • Sommerfeld's Free Electron Model

Arnold Sommerfeld applied quantum mechanics to the Drude free-electron model, explaining metallic conductivity and heat capacity through the concept of delocalized valence electrons obeying Fermi–Dirac statistics.

1930s • Hume-Rothery Rules

William Hume-Rothery formulated empirical rules predicting the solubility of one metal in another, connecting atomic size, electronegativity, and valence to alloy phase stability—bridging chemistry and metallurgy.

1984 • Discovery of Quasicrystals

Dan Shechtman observed an Al–Mn alloy with fivefold rotational symmetry, violating classical crystallographic rules. Quasicrystals expanded the definition of ordered solids and earned Shechtman the 2011 Nobel Prize in Chemistry.

Together, these milestones frame the central question of this lesson: how does the spatial arrangement of metal atoms—and the nature of the bonds holding them together—explain the remarkable and tunable properties of metallic materials? To answer this, we must first establish the principles of metallic bonding, then examine the crystal structures metals adopt, and finally explore how alloy formation modifies these structures at the atomic level.

SECTION 2

Core Principles of Metallic Structure

Metallic structure rests on several interconnected principles that distinguish metals from ionic and covalent solids. In a metallic crystal, atoms pack together as closely as geometry permits, surrendering their valence electrons to a communal electron sea that pervades the lattice. This metallic bond is non-directional: each cation core interacts equally with delocalized electrons in every direction, which contrasts sharply with the highly directional covalent bond. The result is a set of characteristic physical properties—high electrical and thermal conductivity, luster, malleability, and ductility—that can be rationalized directly from the bonding model and the crystal packing.

1

Metallic Bonding

Valence electrons are delocalized across the entire crystal, forming an electron sea that binds positively charged metal cation cores together through electrostatic attraction. The bond is non-directional, allowing atoms to slide past one another without breaking the overall bonding framework.
2

Close-Packed Structures

Metal atoms tend to adopt arrangements that maximize the number of nearest neighbors (coordination number) and minimize void space. The two most efficient packings—FCC and HCP—achieve a packing efficiency of 74%, the theoretical maximum for identical spheres.
3

Unit Cell & Lattice

The unit cell is the smallest repeating structural unit that, when translated in three dimensions, reconstructs the entire crystal. Identifying the unit cell allows calculation of density, interatomic distances, and packing fraction from a handful of parameters.
4

Interstitial & Substitutional Sites

Voids (holes) within the lattice can accommodate smaller atoms (interstitial alloys), while atoms of similar size can replace host atoms at lattice points (substitutional alloys). Both mechanisms modify properties systematically.
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KEY TAKEAWAY

Think of a metallic crystal as a stadium full of fans doing 'the wave.' Each fan (atom) stays roughly in place, but the wave (electron density) moves freely through the crowd. This communal sharing of electrons is what makes metals electrical conductors—charge carriers are not locked between two atoms as in a covalent bond, but free to drift through the entire lattice when a voltage is applied.

SECTION 3

Visualizing Crystal Packing

The three most common crystal structures adopted by metals are face-centered cubic (FCC), body-centered cubic (BCC), and hexagonal close-packed (HCP). Each structure can be described by its unit cell—a parallelepiped whose edges are defined by the lattice parameter a (and c in the hexagonal case). The following diagram compares the three unit cells side by side, showing atom positions, coordination numbers, and packing efficiencies.

Common Metallic Crystal StructuresFCCFace-Centered CubicCoordination #: 12Atoms/cell: 4Packing: 74%a = 2√2 rCu, Al, Au, Ag, NiBCCBody-Centered CubicCoordination #: 8Atoms/cell: 2Packing: 68%a = 4r / √3Fe, W, Cr, Na, KHCPHexagonal Close-PackedCoordination #: 12Atoms/cell: 6Packing: 74%c/a = 1.633 (ideal)Mg, Zn, Ti, Co
Comparison of the three most common metallic crystal structures. Corner atoms (shown at full saturation) define the unit cell geometry, while translucent spheres mark face-center or body-center positions. FCC and HCP both achieve the maximum packing efficiency of 74%, but differ in their stacking sequence (ABCABC vs. ABABAB). BCC, with its lower coordination number of 8, packs less efficiently at 68% but is adopted by many transition metals (Fe, W, Cr) at standard conditions.

The distinction between FCC and HCP lies in the stacking sequence of close-packed layers. Both structures begin with a layer of atoms arranged in a hexagonal pattern (layer A), followed by a second layer (B) nestled into the hollows of the first. The critical choice occurs at the third layer: if it sits directly above layer A, the sequence repeats as ABABAB and the structure is HCP; if it occupies the remaining set of hollows not aligned with either A or B, the sequence is ABCABC and the structure is FCC. Despite having the same packing efficiency, these two arrangements yield different slip systems and therefore different mechanical behaviors—FCC metals like copper and aluminum are generally more ductile than HCP metals like zinc or magnesium.

💡

Counting Atoms in a Unit Cell

Atoms shared among adjacent unit cells contribute only a fraction of themselves to any single cell. A corner atom is shared by 8 cells (contributes ⅛), an edge atom by 4 cells (contributes ¼), a face atom by 2 cells (contributes ½), and a body-center atom belongs entirely to one cell (contributes 1). For FCC: 8 corners × ⅛ + 6 faces × ½ = 1 + 3 = 4 atoms per unit cell.

SECTION 4

Mathematical Framework

Quantitative analysis of metallic crystal structures revolves around relating the lattice parameter a to the atomic radius r, and then using the unit cell geometry to predict macroscopic density. These relationships arise from the geometry of close contact—atoms touch along the face diagonal in FCC, along the body diagonal in BCC, and along the basal plane edge in HCP.

Lattice Parameter–Radius Relations

FCC RELATIONSHIP

a=2√2⋅ra = 2√2 · r

In FCC, atoms touch along the face diagonal (length = a√2), which spans 4 atomic radii: a√2 = 4r.

BCC RELATIONSHIP

a=(4r)/√3a = (4r) / √3

In BCC, atoms touch along the body diagonal (length = a√3), which spans 4 atomic radii: a√3 = 4r.

Density from the Unit Cell

CRYSTALLOGRAPHIC DENSITY

ρ=(n×M)/(Na×a3)ρ = (n × M) / (Nₐ × a³)

where n = number of atoms per unit cell, M = molar mass (g/mol), Nₐ = Avogadro's number (6.022 × 10²³ mol⁻¹), and a = lattice parameter in cm.

Packing Efficiency

ATOMIC PACKING FACTOR (APF)

APF=(n×(4/3)πr3)/a3APF = (n × (4/3)πr³) / a³

The atomic packing factor is the fraction of the unit cell volume actually occupied by atoms. Substituting the appropriate a–r relationship yields APF = 0.74 for FCC and HCP, and APF = 0.68 for BCC.

These equations connect atomic-scale parameters (radius, unit cell geometry) to the measurable bulk property of density. Experimentally, one can determine a via X-ray diffraction using Bragg's Law (nλ = 2d sin θ), then compare the calculated density with the measured value to confirm the crystal structure assignment.

SECTION 5

Types of Alloys & Structural Modifications

An alloy is a mixture of two or more elements in which the primary component is a metal. Alloying alters the crystal structure at the atomic level, disrupting the regularity of the lattice in ways that change mechanical, thermal, and electrical properties. Two broad categories of solid-solution alloys are distinguished by how the solute atoms are incorporated into the host lattice.

Alloy Types: Substitutional vs. InterstitialSubstitutional Alloy(e.g., Brass: Cu + Zn)Host atom (Cu)Solute atom (Zn)Solute replaces host atlattice positionsRequires similar atomic radii (±15%)Interstitial Alloy(e.g., Steel: Fe + C)Host atom (Fe)Solute atom (C)Small solute atoms fit inholes between host atomsSolute must be much smaller (H, C, N, B)
Left: In a substitutional alloy (e.g., brass), solute atoms (Zn, gold spheres) replace host atoms (Cu, blue spheres) at regular lattice sites. This requires the two elements to have similar atomic radii. Right: In an interstitial alloy (e.g., steel), much smaller solute atoms (C, red spheres) occupy the interstices—holes between the larger host atoms (Fe). Both mechanisms disrupt the regularity of the lattice, impeding dislocation motion and increasing hardness.

Hume-Rothery Rules for Substitutional Alloys

Not every pair of metals forms a continuous substitutional solid solution. The Hume-Rothery rules provide four empirical criteria that must all be satisfied for extensive mutual solubility.

  1. Atomic size factor: The atomic radii of the two elements must differ by no more than about 15%. Larger mismatches generate excessive lattice strain.
  2. Crystal structure: Both elements should adopt the same crystal structure in their pure forms to enable unlimited solubility.
  3. Electronegativity: The two elements should have similar electronegativities; otherwise, the interaction becomes partially ionic and an intermetallic compound forms instead of a solid solution.
  4. Valence: Metals of lower valence tend to dissolve more of a higher-valence metal than vice versa, owing to favorable electron-to-atom ratios.

The Cu–Ni system is the textbook example of a system that satisfies all four rules: both metals are FCC, their atomic radii differ by only 2.7%, their electronegativities are nearly identical (1.90 vs. 1.91 on the Pauling scale), and both commonly exhibit +2 valence states. As a result, Cu and Ni are completely miscible in the solid state over the entire composition range. Conversely, the Cu–Ag system violates the size criterion (radii differ by ~12%), limiting mutual solubility and producing a eutectic phase diagram.

SECTION 6

Worked Example: Density of Aluminum

Aluminum crystallizes in the FCC structure with an atomic radius of 143 pm. Using the crystallographic density equation, we can predict its bulk density and compare it with the measured value of 2.70 g/cm³.

Calculating the Density of Aluminum from its Crystal Structure

  1. 1

    Step 1 — Identify Given Values

    Crystal structure: FCC → n = 4 atoms per unit cell. Atomic radius: r = 143 pm = 143 × 10⁻¹⁰ cm. Molar mass of Al: M = 26.98 g/mol. Avogadro's number: Nₐ = 6.022 × 10²³ mol⁻¹.
  2. 2

    Step 2 — Calculate the Lattice Parameter

    For FCC, a = 2√2 × r = 2 × 1.4142 × 143 pm = 404.5 pm = 4.045 × 10⁻⁸ cm.
    a = 4.045 × 10⁻⁸ cm
  3. 3

    Step 3 — Calculate the Unit Cell Volume

    V = a³ = (4.045 × 10⁻⁸ cm)³ = 6.621 × 10⁻²³ cm³.
    V = 6.621 × 10⁻²³ cm³
  4. 4

    Step 4 — Apply the Density Equation

    ρ = (n × M) / (Nₐ × a³) = (4 × 26.98 g/mol) / (6.022 × 10²³ mol⁻¹ × 6.621 × 10⁻²³ cm³) = 107.92 / 39.87 = 2.71 g/cm³.
    ρ = 2.71 g/cm³
  5. 5

    Step 5 — Compare with Experiment

    The calculated density of 2.71 g/cm³ matches the experimentally measured value of 2.70 g/cm³ to within 0.4%, confirming that aluminum indeed crystallizes in the FCC structure and that the reported atomic radius is accurate. Small deviations arise from lattice defects, thermal expansion, and impurities in real samples.
    Excellent agreement: 0.4% deviation

SECTION 7

Comparing Crystal Structures & Alloy Types

Understanding the relative advantages and limitations of different crystal structures and alloy types is essential for materials selection in engineering contexts. The following table summarizes key properties of the three main metallic structures.

PropertyFCCBCCHCP
Coordination Number12812
Atoms per Unit Cell426
APF0.740.680.74
Slip Systems12 (4 planes × 3 directions)48 (but pencil glide)3 (1 basal plane × 3 directions)
DuctilityHighModerateLow
Example MetalsCu, Al, Au, Ni, PbFe (α), W, Cr, Mo, NaMg, Zn, Ti (α), Co (α)

Comparison of FCC, BCC, and HCP crystal structures

Substitutional vs. Interstitial Alloys

FeatureSubstitutionalInterstitial
Solute SizeSimilar to host (within ±15%)Much smaller than host
Solute PositionReplaces host at lattice siteOccupies tetrahedral or octahedral holes
Composition RangeCan be continuous (0–100%)Limited (typically < 10 at%)
Effect on PropertiesSolid-solution strengthening, modified conductivityDramatic hardening, reduced ductility
Classic ExampleBrass (Cu–Zn), Sterling silver (Ag–Cu)Steel (Fe–C), Nitinol (Ni–Ti with N)

Comparison of substitutional and interstitial alloy types

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KEY TAKEAWAY

Alloying is essentially atomic-level sabotage of the perfect crystal lattice—and that is precisely the point. In a pure metal, planes of atoms slide easily over one another (ductile but soft). Introducing solute atoms of different size or bonding character is like scattering rocks onto an ice rink: the irregularities impede the gliding motion of dislocations, making the material harder and stronger. This is why a small amount of carbon (< 2%) transforms soft iron into steel.

SECTION 8

Connections to Advanced Theory

The introductory treatment of metallic bonding as an 'electron sea' provides powerful qualitative predictions, but a more complete picture emerges from band theory. In a crystalline solid containing N atoms, each atomic orbital splits into N closely spaced molecular orbitals, forming a quasi-continuous energy band. In metals, the valence band is only partially filled or overlaps with the conduction band, which accounts for electrical conductivity. In semiconductors and insulators, the valence and conduction bands are separated by a band gap whose magnitude determines conductivity.

ConceptElectron Sea ModelBand Theory
Electron DescriptionDelocalized cloud of electrons shared by all cationsElectrons occupy delocalized Bloch states in continuous energy bands
ConductivityQualitative: free electrons carry currentQuantitative: conductivity depends on density of states at the Fermi level
Insulator DistinctionNot explainedBand gap > 3 eV prevents thermal excitation across gap
Alloy EffectsLattice distortion impedes electron flowImpurity scattering modifies mean free path and resistivity
ScopeMetals onlyMetals, semiconductors, and insulators unified

Electron sea model vs. band theory

Beyond electronic structure, advanced topics include intermetallic compounds (ordered alloy phases with fixed stoichiometry and distinct crystal structures, such as Ni₃Al in superalloys), phase diagrams (graphical maps showing which phases are thermodynamically stable as a function of composition and temperature), and quasicrystals (aperiodic structures that possess long-range order without translational periodicity). These topics build directly on the structural foundations laid in this lesson, and you will encounter them in materials science, solid-state chemistry, and condensed matter physics courses.

ℹ

Looking Ahead

If you continue into materials science, you will learn to read binary phase diagrams—plots of temperature vs. composition that predict whether a given alloy will be a single solid solution, a two-phase mixture, or an intermetallic compound at any temperature. The lever rule, tie lines, and the Gibbs phase rule all extend the structural concepts introduced here into the thermodynamic domain.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

Explain why FCC metals like copper and aluminum are generally more ductile than HCP metals like magnesium. Your answer should reference the concept of slip systems.
Show answer
Ductility depends on the ability of planes of atoms to slide past one another under stress—a process mediated by dislocation motion along slip systems. A slip system consists of a close-packed plane and a close-packed direction within that plane. FCC structures have 12 independent slip systems (4 {111} planes × 3 ⟨110⟩ directions), providing many orientations for plastic deformation. In contrast, HCP structures have only 3 primary basal slip systems (1 {0001} plane × 3 ⟨112̄0⟩ directions). With fewer slip pathways available, HCP metals resist plastic deformation and fracture more readily, appearing brittle compared to FCC metals.

Question 2: BASIC CALCULATION

Iron crystallizes in a BCC structure with an atomic radius of 126 pm. Calculate the lattice parameter a and the number of atoms per unit cell.
Show answer
For BCC, atoms touch along the body diagonal: a√3 = 4r. Therefore a = 4r / √3 = 4 × 126 pm / 1.732 = 291 pm. The number of atoms per BCC unit cell is calculated as: 8 corner atoms × (1/8) + 1 body-center atom × 1 = 1 + 1 = 2 atoms per unit cell. This lattice parameter agrees well with the experimentally measured value of approximately 287 pm for α-iron; the slight discrepancy arises from the imprecision in the tabulated radius value.

Question 3: INTERMEDIATE

Gold (FCC, M = 196.97 g/mol) has a measured density of 19.30 g/cm³. Using the crystallographic density equation, determine the atomic radius of gold in picometers.
Show answer
Starting from ρ = nM / (Nₐa³), solve for a: a³ = nM / (Nₐρ) = (4 × 196.97) / (6.022 × 10²³ × 19.30) = 787.88 / (1.162 × 10²⁵) = 6.781 × 10⁻²³ cm³. Taking the cube root: a = (6.781×10−23)1/3(6.781 × 10⁻²³)^{1/3} = 4.077 × 10⁻⁸ cm = 407.7 pm. For FCC, a = 2√2r, so r = a / (2√2) = 407.7 / 2.828 = 144.2 pm. This is in excellent agreement with the tabulated metallic radius of gold (144 pm), validating the FCC assignment.

Question 4: APPLIED

A metallurgist wishes to create a substitutional alloy of copper (r = 128 pm, FCC, EN = 1.90, common valence = +1 or +2). Using the Hume-Rothery rules, evaluate whether each of the following elements would form an extensive solid solution with copper: (a) nickel (r = 125 pm, FCC, EN = 1.91, valence = +2), (b) zinc (r = 137 pm, HCP, EN = 1.65, valence = +2), (c) lead (r = 175 pm, FCC, EN = 2.33, valence = +2 or +4).
Show answer
(a) Nickel: Size difference = |128 − 125| / 128 × 100% = 2.3% (< 15% ✓). Same crystal structure (FCC ✓). Nearly identical electronegativity (1.90 vs. 1.91 ✓). Same common valence (+2 ✓). All four rules are satisfied, so Cu and Ni form a complete solid solution—which is experimentally confirmed. (b) Zinc: Size difference = |137 − 128| / 128 × 100% = 7.0% (< 15% ✓). Different crystal structures (HCP vs. FCC ✗). Electronegativity difference is moderate (0.25). Zinc is partially soluble in copper (up to ~35 at% Zn, as in brass), but the crystal structure mismatch prevents unlimited solubility. (c) Lead: Size difference = |175 − 128| / 128 × 100% = 36.7% (>> 15% ✗). This alone disqualifies lead. Additionally, the electronegativity difference is significant (0.43). Lead is virtually insoluble in solid copper.

Question 5: CRITICAL THINKING

Carbon dissolves interstitially in BCC iron (α-Fe) up to only about 0.02 wt% at 727 °C, but in FCC iron (γ-Fe, stable above 912 °C) carbon dissolves up to about 2.14 wt%. Given that BCC has a lower packing efficiency (68%) than FCC (74%) and therefore more total void space, explain the apparent paradox: why does the less efficiently packed BCC structure accommodate less interstitial carbon than the more tightly packed FCC structure?
Show answer
This is one of the most instructive paradoxes in metallurgy. Although BCC has more total void volume (32% vs. 26%), what matters for interstitial solubility is the size of the individual interstitial sites, not the aggregate void fraction. In BCC, the void space is distributed among many small, irregularly shaped tetrahedral and octahedral holes. The largest octahedral hole in BCC has a radius ratio r/R ≈ 0.155, meaning it can accommodate atoms only up to about 0.155 × 126 pm ≈ 19.5 pm without severe lattice distortion. In FCC, the octahedral holes are larger and more symmetric, with r/R ≈ 0.414, allowing atoms up to 0.414 × 128 pm ≈ 53 pm. Since carbon has a radius of approximately 77 pm, it distorts both lattices, but the distortion in BCC is far more severe because it must stretch the smaller holes asymmetrically. This greater strain energy severely limits how much carbon the BCC lattice can accommodate before it becomes thermodynamically unfavorable.

SUMMARY

Summary

Summary

Metals are held together by the metallic bond, a non-directional electrostatic interaction between positively charged cation cores and a delocalized electron sea. This bonding model explains metallic luster, conductivity, malleability, and ductility. Metallic atoms arrange themselves into repeating crystal lattices described by unit cells. The three most common structures are FCC (coordination 12, APF 0.74), BCC (coordination 8, APF 0.68), and HCP (coordination 12, APF 0.74). The crystallographic density equation ρ = nM / (Nₐa³) connects atomic-scale geometry to measurable bulk density.

An alloy modifies the host lattice by introducing solute atoms either at lattice sites (substitutional alloys, governed by the Hume-Rothery rules) or within interstitial voids (interstitial alloys). Both mechanisms impede dislocation motion, increasing hardness and strength. The simple electron-sea model is refined by band theory, which explains conductivity differences among metals, semiconductors, and insulators through the concept of partially filled vs. gapped energy bands.

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