Structure of Metals and Alloys

AP Chemistry· AP Chemistry CED — Molecular and Ionic Compound Structure and Properties· 14 min read

1. Metallic Bonding and the Electron Sea Model★★☆☆☆⏱ 4 min

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 (electrons localized between two atoms) or ionic bonding (electrons fully transferred to fixed charged ions), delocalization of valence electrons in metals directly explains all their signature properties.

📘 Definition

Metallic Bonding

Electrostatic attraction between delocalized mobile valence electrons and positively charged metal cations arranged in a crystalline lattice.

Example:

Solid sodium has metallic bonding between Na⁺ cations and a sea of delocalized 3s valence electrons.

Key properties of metals are directly explained by the electron sea model:

Electrical/thermal conductivity: Mobile delocalized electrons flow under an applied voltage to carry current, and rapidly transfer kinetic energy through heated regions of the solid.
Malleability and ductility: When external force is applied, layers of cations slide past each other, and the electron sea rearranges to maintain electrostatic attraction, so the solid does not fracture.
Melting point correlation: Stronger electrostatic attraction (higher cation charge, more delocalized electrons) leads to higher melting points.

📐 Worked Example

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 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".

2. Crystalline Unit Cell Geometry for Pure Metals★★★☆☆⏱ 5 min

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 (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): $a = 2 r$
BCC (atoms touch along body diagonal): $3 a = 4 r ⟹ r = \frac{a 3}{4}$
FCC (atoms touch along face diagonal): $2 a = 4 r ⟹ r = \frac{a 2}{4}$

📐 Worked Example

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: $r + 2 r + r = 4 r$ .
2
For a square face of the cube with edge length $a$ , the Pythagorean theorem gives:
3
$(face\ diagonal)^2 = a^2 + a^2 = 2a^2 \implies face\ diagonal = \sqrt{2}a$
4
Equate the two expressions for face diagonal and rearrange to solve for $r$ :
5
$\sqrt{2}a = 4r \implies r = \frac{\sqrt{2}a}{4}$
6
Substitute $a = 408$ pm:
7
$r = \frac{(1.414)(408)}{4} \approx 144\text{ pm}$

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.

3. Structure and Classification of Alloys★★☆☆☆⏱ 3 min

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:

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: brass (copper + zinc) and 14-karat gold (gold + copper).
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: 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

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:
2
$\frac{|144 - 128|}{144} \times 100\% \approx 11\%$
3
The difference is less than the 15% threshold for substitutional alloys. 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.
4
Pure silver has a uniform, regular lattice, so layers of silver atoms can slide past one another easily, making it soft. 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.

4. AP-Style Concept Check★★★☆☆⏱ 2 min

✓ Quick check

Test your understanding with this AP-style multiple choice question:

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.
Reveal answer
B —
A is incorrect: aluminum has metallic bonding, not covalent bonding. C is incorrect: conductivity does not depend on bond weakness. D is incorrect: melting point does not determine conductivity in solids. B correctly matches the electron sea model explanation for conductivity.

📐 Worked Example

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. (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.

1
(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, so total body diagonal length = $r + 2 r + r = 4 r$ .
2
For a cube, the 3D Pythagorean theorem gives:
3
$(body\ diagonal)^2 = a^2 + a^2 + a^2 = 3a^2 \implies body\ diagonal = \sqrt{3}a$
4
Equating the two expressions gives $3 a = 4 r$ , the required relationship.
5
(b) Rearrange to solve for $r$ :
6
$r = \frac{\sqrt{3}a}{4} = \frac{(1.732)(287)}{4} \approx 124\text{ pm}$
7
(c) The percent difference in atomic radii is ~39%, so carbon is much smaller than iron. Carbon fits into interstitial gaps, so this is an interstitial alloy. Added carbon atoms disrupt the regular iron lattice, preventing layers of atoms from sliding easily, making carbon steel stronger than pure iron.

5. Common Pitfalls

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 $a = 2 r$ instead of $a = 22 r$ .

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. Quick Reference Cheatsheet

Category	Formula / Rule	Notes
Metallic Bonding Model	Delocalized electron sea	Electrostatic attraction between mobile valence electrons and metal cations; explains conductivity, malleability, ductility
Primitive Cubic $a$ - $r$ relation	$a = 2 r$	Atoms touch along edge; 52% packing efficiency, rare for pure metals
Body-Centered Cubic (BCC) $a$ - $r$ relation	$r = \frac{a 3}{4}$	Atoms touch along body diagonal; 68% packing efficiency, found in Na, Fe
Face-Centered Cubic (FCC/CCP) $a$ - $r$ relation	$r = \frac{a 2}{4}$	Atoms touch along face diagonal; 74% packing efficiency, most common close-packed metal structure
Substitutional Alloy	Size difference < 15%	Added atom replaces original metal atom in lattice; examples: brass, 14k gold
Interstitial Alloy	Size difference > 30%	Added atom fits into gaps between original metal atoms; examples: carbon steel
Alloy Property Trend (vs pure metal)	Hardness/strength ↑, Conductivity ↓	Change caused by disrupted lattice, not change in metallic bond strength

What's Next

This topic establishes the core atomic-scale reasoning for linking structure to properties in crystalline solids, a key skill tested repeatedly across both multiple choice and free response sections of the AP Chemistry exam. 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 high-weight FRQ topic. This topic also provides the foundation for materials chemistry, which appears across later units of the course.