Structure of Metals and Alloys: Study Guide (Chemistry 2.4)

A different kind of solid

In Topic 2.3 you modelled an ionic solid as an array of cations and anions. A metal also has positive ions in a regular array, but there are no anions. So what holds the positive ions together?

Metal atoms have few valence electrons and hold them weakly (low ionization energies, Topic 1.7). In a solid metal, every atom releases its valence electrons into a shared pool. These electrons are delocalized: they are not tied to any one atom or any one bond. They move through the whole structure.

• What remains of each atom is a positive ion core: the nucleus plus the inner (core) electrons.
• The cores sit in a regular, closely packed array.
• The delocalized valence electrons form a “sea of electrons” around and between the cores.
• Metallic bonding is the attraction between the positive cores and the negative sea. It acts in every direction, so no core has a fixed “partner”.

Drawing the electron-sea model

A good particulate model of a metal has three features.

1. Identical cores in a regular array. All the circles for one metal are the same size and are packed closely in neat rows.
2. Electrons spread through the whole structure. Draw the valence electrons as dots (or a shaded region) between and around the cores, not attached to any one core.
3. Charges that balance. Each core carries a charge equal to the number of valence electrons its atom gave to the sea. Sodium gives 1 electron, so the model shows Na⁺ cores and one dot per core. Magnesium gives 2, so 16 Mg atoms become 16 Mg²⁺ cores and 32 electrons. Aluminium gives 3, so 12 Al atoms become 12 Al³⁺ cores and 36 electrons. In every case the total charge is zero.

What the model explains

You will study the properties of solids fully in Topic 3.2. The model already explains three of them.

• Electrical and thermal conductivity. The delocalized electrons are free to move, so they carry charge (and energy) through the metal.
• Malleability and ductility. When a force pushes one layer of cores past another, the cores simply move into new positions inside the same electron sea. The sea adjusts around them, and no like charges are forced together. Compare this with an ionic crystal, where sliding one layer lines up like charges and the crystal shatters (Topic 2.3, Figure 2).
• Lustre (background): the free electrons interact with light, which is why clean metal surfaces are shiny. You will not be assessed on this.

Alloys: adding a second element

An alloy is a mixture of a metal with one or more other elements, usually other metals or small nonmetal atoms such as carbon. The mixture keeps metallic bonding: the cores still sit in an array, and the electron sea is still there. Unlike a compound, an alloy has no fixed formula; its composition can vary.

Where the second kind of atom goes depends on its size compared with the host metal atoms.

Type	Size of added atoms	Where they go	Example
Interstitial	Much smaller than the host atoms	In the gaps (interstices) between host atoms; no host atom is removed	Steel: carbon atoms (radius 75 pm) in iron (radius 126 pm)
Substitutional	Similar to the host atoms	In place of some host atoms, at normal lattice positions	Brass: zinc atoms (134 pm) replace some copper atoms (128 pm)

The radii in the table come from standard data: metallic radii for Fe, Cu and Zn, and the covalent radius for carbon. They are different kinds of radius, but the conclusion is clear either way: a carbon atom is far smaller than an iron atom, while zinc and copper atoms are nearly the same size.

A quick way to compare sizes is the percent difference relative to the host:

percent difference = |r(host) − r(added)| ÷ r(host) × 100

• Cu and Zn: |128 − 134| ÷ 128 × 100 = 4.7 %, so similar sizes: substitutional.
• Fe and C: |126 − 75| ÷ 126 × 100 = 40.5 %, so very different sizes: interstitial (the carbon radius is only 0.60 of the iron radius).

Background, not required. Metallurgists use a rough guide that substitution works well when the radii differ by less than about 15 %. You do not need this number; “comparable radius” and “significantly different radius” are enough for the exam.

How alloying changes properties

• Interstitial alloys. The small atoms in the gaps get in the way when layers of host atoms try to slide, and they help hold neighbouring layers in place. The lattice becomes more rigid, so the alloy is harder and stronger but less malleable and less ductile than the pure metal. Steel is harder and less malleable than pure iron for this reason.
• Substitutional alloys. The added atoms are slightly different in size, so the rows of atoms are no longer perfectly even. Layers slide less easily, so the alloy is usually harder than the pure host metal; brass is harder than pure copper.
• Conductivity. Both kinds of alloy keep the sea of mobile electrons, so they still conduct electricity, although usually not quite as well as the pure metal.

Worked example 1: classifying alloys from atomic radii

Question. Use the radii below to decide whether each pair is more likely to form an interstitial or a substitutional alloy. Radii: Au 144 pm, Ag 144 pm, Fe 126 pm (metallic radii); N 71 pm (covalent radius).

(a) gold with silver (b) iron with nitrogen

1. Gold and silver. Percent difference = |144 − 144| ÷ 144 × 100 = 0 %. The atoms are the same size, so silver atoms can take the place of gold atoms in the lattice. Substitutional.
2. Iron and nitrogen. Percent difference = |126 − 71| ÷ 126 × 100 = 43.7 %. The nitrogen atom is little more than half the size of an iron atom (71 ÷ 126 = 0.56). It cannot replace an iron atom in the array, but it can fit into the gaps. Interstitial.

Answer. (a) Substitutional. (b) Interstitial.

Check. The deciding question is always the same: are the atoms of comparable size or very different size? The type of element (metal or nonmetal) is a hint, because the small atoms in interstitial alloys are usually nonmetals such as C, N, H or B, but it is the size that matters.

Worked example 2: from mass percent to a particulate model of brass

Question. A sample of brass contains 70 % copper and 30 % zinc by mass. (a) What fraction of the atoms are zinc atoms? (b) How should a particulate model with 10 atoms in the lattice be drawn?

1. Take 100 g. It contains 70 g of Cu and 30 g of Zn.
2. Moles of each. n(Cu) = 70 g ÷ 63.55 g mol⁻¹ = 1.1015 mol. n(Zn) = 30 g ÷ 65.38 g mol⁻¹ = 0.4589 mol.
3. Atom (mole) fraction of Zn. 0.4589 ÷ (1.1015 + 0.4589) = 0.4589 ÷ 1.5604 = 0.294.
4. Scale to 10 atoms. 10 × 0.294 = 2.94 zinc atoms and 7.06 copper atoms.

Answer. (a) About 0.29 (29 %) of the atoms are zinc. (b) Draw 7 copper atoms and 3 zinc atoms, all at normal lattice positions in one regular array, with the zinc atoms drawn very slightly larger and labelled. That is about 2.4 copper atoms for every zinc atom (1.1015 ÷ 0.4589 = 2.40).

Check. Cu and Zn have similar molar masses (63.55 and 65.38), so the atom fraction (29 %) is close to the mass fraction (30 %). That is not true when the atoms have very different masses, as the next example shows.

Worked example 3: how many carbon atoms are in steel?

Question. A steel contains 0.80 % carbon by mass; the rest is iron. How many carbon atoms are there for every 100 iron atoms?

1. Take 100 g: 0.80 g C and 99.20 g Fe.
2. n(C) = 0.80 ÷ 12.01 = 0.06661 mol. n(Fe) = 99.20 ÷ 55.85 = 1.7762 mol.
3. Carbon atoms per 100 iron atoms = 100 × 0.06661 ÷ 1.7762 = 3.75.

Answer. About 3.8 carbon atoms for every 100 iron atoms.

Interpretation. A carbon atom has less than a quarter of the mass of an iron atom (an iron atom is 55.85 ÷ 12.01 = 4.65 times heavier), so a small mass percent still means a noticeable number of atoms. A model of this steel should show many iron atoms with an occasional small carbon atom in a gap, not a carbon atom in every gap.

Common misconceptions

• “A metal is held together by ionic bonds because it contains positive ions.” No. There are no anions. The cores are held by their shared attraction to the delocalized electrons.
• “Each electron in the sea belongs to one atom and stays near it.” No. The valence electrons are delocalized across the whole sample.
• “The electron sea makes the metal negatively charged.” No. The sea’s charge exactly balances the charge of the cores.
• “In an interstitial alloy, the small atoms replace some of the host atoms.” No. That describes a substitutional alloy. Interstitial atoms fill the gaps; every host atom stays in place.
• “Steel is hard because carbon is hard.” The hardness comes from the carbon atoms disrupting the sliding of iron layers, not from carbon being hard on its own.
• “Alloys do not conduct because they are mixtures.” Alloys keep the electron sea and still conduct.
• “An alloy is a compound with a formula.” An alloy is a mixture with a variable composition.

Where this leads

You now have models for ionic and metallic solids. Next, Topic 2.5, Lewis Diagrams, turns to covalent bonding and how to represent the electrons in molecules. In Topic 3.2 you will compare the properties of ionic, metallic, covalent network and molecular solids using these models. Try the practice questions now, then use the revision notes and the checklist to consolidate.