CHEM 548: Materials Chemistry

Lecture 9: Solid Solutions and Vegard's Law

Wed 09/23/2026

Solid solutions, Vegard's law and its deviations, and band-gap engineering

Reading: Woodward, Karen, Evans, and Vogt, Solid State Materials Chemistry
Ch. 2 §2.5 · Ch. 4 §§4.1–4.2

Solid Solutions and Vegard's Law

Learning Objectives

Define a solid solution and compare to liquid solutions.
State Vegard's law: unit cell dimensions (and some properties) vary linearly with composition.
Apply Vegard's law to calculate lattice parameters and band gaps of intermediate compositions.
Recognize and explain deviations from Vegard's law:
- Charge transfer (e.g., Mn³⁺/Ru⁵⁺ in Ca-based perovskites).
- Ordering of different-size ions → more efficient packing.
Appreciate the importance of solid solutions in semiconductors (epitaxy, heterojunction lasers, LEDs).
Use Vegard's law to estimate composition from property measurements (e.g., lattice constant, band gap).

Vegard's Law

$a(x) = x\,a_{BX} + (1-x)\,a_{AX}$

A solid solution is the solid-state analog of a liquid solution: just as water and ethanol mix homogeneously in any ratio, solids can form homogeneous crystals across a full composition range.

AlAs–GaAs forms a continuous solid solution
Vegard’s law: the unit cell size varies linearly between the two end members
Both cubic (zinc blende) structures — AlAs has a slightly smaller lattice constant (~5.653 Å), GaAs a larger one
By adjusting composition, we can obtain any lattice parameter between these limits; conversely, by measuring the lattice parameter, we can determine the composition
Mathematically: the lattice constant is the weighted sum of each end member’s constant — x the AlAs fraction, 1-x the GaAs fraction

That linear baseline also makes deviations from Vegard's law visible and chemically meaningful.

Deviations from Vegard's Law

A positive deviation from the linear Vegard's-law line in the calcium ruthenate and calcium manganate solid solution, with causes listed.

Not all solid solutions obey Vegard’s law, and the author’s thesis advisor noted that Vegard himself doubted its validity—ironically, the linear case now bears his name.

Example: solid solutions between CaRuO₃ (Ru⁴⁺) and CaMnO₃ (Mn⁴⁺)
Middle compositions: the lattice parameter curve shows a positive deviation (bowing upward)
Spectroscopy reveals charge transfer: Mn³⁺ and Ru⁵⁺ appear
Their average ionic size is larger than Mn⁴⁺ and Ru⁴⁺ — explaining the expansion
Deviations can also be negative: random distributions of differently sized ions reorder into a more efficient packing → the lattice contracts below the linear Vegard prediction

The same linear interpolation can describe an electronic property, not only a lattice dimension.

Band Gap Engineering

A linear band-gap relation between CdS and CdSe with equations for predicting band gap and composition.

Semiconductor Solid Solutions

n semiconductor devices, solid solutions are critical because many involve epitaxial thin films—a crystal layer grown on a substrate with bonds matched at the interface.

Lattice constants of adjacent layers must be closely matched to minimize strain; band gaps must vary to achieve desired functions (e.g., in heterojunction lasers)
GaAs–AlAs especially useful: lattice parameters nearly identical → epitaxial growth across the full range
Band gap tunable from ~2.0 eV (AlAs) to ~1.4 eV (GaAs)
Such tunability underpins many LEDs and lasers, where compositions with direct band gaps are essential
Semiconductors highlight the importance of Vegard’s law, but the principle applies broadly to other solid solutions as well

One familiar non-device application makes this continuous tuning visible as a continuous change in color.

CdS-CdSe Solid Solutions

Cadmium sulfide to cadmium selenide pigments ranging from yellow through orange and red to nearly black.

Example: Orange CdS-CdSe

Composition calculation

Band-gap graph

So I would say the ideal band gap for making orange is something pretty close to 2.25.

So composition do we need if we wanted to make the band gap exactly 2.25?

Okay, to get the value of x here all we really have to do are to plug in the band gaps for the end members.

Let's call a cadmium sulfide with the band gap of 2.4 and b cadmium cellanide with the band gap of 1.7 and the band gap of the intermediate composition we want that to be 2.25.

[advance: show the composition calculation]

So when you plug that all into the formula you get 0.21.

So the way we've set this up is that this is going to be the selenium content in our solid solution.

So crystal that had composition cadmium, sulfur 0.79, selenium 0.21 should have a band gap very close to 2.25 provided vagards laws followed.

[advance: show the band-gap graph]

A graph assuming perfect Vegard’s law shows that x = 0.21 corresponds to a band gap of 2.25 eV.

It’s easy to confuse whether 0.21 refers to Se or S, but a reality check helps: since 2.25 eV is much closer to CdS (2.4 eV) than CdSe (1.7 eV), the composition must be S-rich.

This confirms the solution as CdS₀.79Se₀.21.

That calculation brings the main ideas together before we summarize what solid solutions let us predict.

Summary

Solid solutions: homogeneous crystals spanning compositions between two end members.
Vegard's law: lattice constant (and sometimes other properties) varies linearly with composition.
Deviations occur due to charge transfer or cation/anion ordering, leading to positive or negative curvature.
Many semiconductor properties (e.g., band gap) also approximate Vegard's law, enabling band gap engineering.
Applications:
- GaAs–AlAs: nearly constant lattice parameter but tunable band gap → key for epitaxial devices (lasers, LEDs).
- CdS–CdSe: tunable band gap controls pigment color (yellow → red → black).
Practical use: Vegard's law lets us predict composition from structure or property measurements.

To summarize this lecture: solid solutions are homogeneous crystals that span compositions between two end members.

Vegard's law: the lattice constant, and sometimes other properties, varies linearly with composition
Deviations: charge transfer or ordering of cations and anions — they show up as positive or negative curvature
Many semiconductor properties, like the band gap, also approximate Vegard's law → band gap engineering
GaAs and AlAs: nearly constant lattice parameter while the band gap is tuned — key for epitaxial devices such as lasers and LEDs
CdS and CdSe: a tunable band gap controls pigment color from yellow through red to black

The practical use to take away is that Vegard's law lets us predict composition from structure or from property measurements.

Why phase diagrams

Opening page for phase diagrams, listing the learning objectives: the Gibbs phase rule, reading a eutectic diagram, the liquidus, solidus, eutectic point and miscibility gap, the lever rule, and real examples.

Next we turn to phase diagrams, which are the tools we use to plan the synthesis of a solid-state material.

Once we know the phase diagram of a system, we can pick a temperature that gives us the phase we want, and when we need a single crystal we can pick conditions that grow one
By the end of this lecture you should be able to state the Gibbs phase rule and apply it to one-component and two-component systems
Read a binary phase diagram for a system that forms no compounds
Tell the liquidus, the solidus, the eutectic point and the miscibility gap apart
Use the lever rule to calculate phase fractions in a two-phase region
Recognize real systems: water, diopside and anorthite, magnesium oxide and calcium oxide
Today: the simplest diagrams, no compound formation — the next lecture adds compounds at compositions between the end members

First we need the rule that tells us how many phases a system can hold at equilibrium, the Gibbs phase rule.

The Gibbs phase rule

One component: water

Pressure against temperature phase diagram of water, with the solid, liquid and gas fields, the triple point and the critical point marked.

Water is a one-component system, so the phase rule gives us C plus two minus P, which is three minus P.

The degrees of freedom can never be negative, so no point on this diagram holds more than three phases at equilibrium
At the triple point three phases meet, P is three, and the degrees of freedom are zero — any change in pressure or temperature takes us off that point
Inside a single-phase region, solid ice for instance, P is one and we have two degrees of freedom: vary pressure and temperature independently and still have ice
Along a phase boundary, like the melting line between ice and liquid water, P is two and we have one degree of freedom — change the pressure and the temperature has to follow in a defined way

Now let's take that framework to a condensed system with more than one component.

A binary diagram: the axes

$A binary temperature against mole-fraction diagram for a system with no compounds, unlabelled except for the liquid, A plus liquid, B plus liquid and A plus B fields.$

Binary system with no compounds

The binary eutectic diagram carrying the phase rule in its two-component form, F equals three minus P; advancing labels each field with the number of phases it holds.

Phase counts in each field

With two components the phase rule gives us degrees of freedom equal to three minus P.

The degrees of freedom cannot be negative, so no region of a binary phase diagram holds more than two phases, and no point in it holds more than three
Let's count the phases field by field

[advance: label each field with the number of phases it holds]

The gray region at the top is liquid, a single phase
Below it on the left, solid A with liquid, two phases; on the right, solid B with liquid, again two phases
At the bottom, below the horizontal line, solid A with solid B, two phases
Every field holds one phase or two, exactly as the phase rule requires

Next we name those fields and the one special point where three phases meet.

Liquid, two-phase fields, eutectic point

The binary eutectic diagram with the one-phase liquid field, the three two-phase fields and the three-phase eutectic point circled at the bottom of the liquid valley.

The solidus

The binary eutectic diagram with the horizontal solidus line drawn in blue across the diagram and labelled.

The liquidus, then cooling from point a

The binary eutectic diagram with the liquidus curve drawn and labelled; advancing marks point a in the liquid field and then cools it down to point b on the liquidus.

Point a: a homogeneous liquid

Cooling to point b on the liquidus

Then we have this curved line, which separates the liquid region from the rest of the phase diagram, and we call it the liquidus.

[advance: mark point a in the liquid field]

Let's take a random point: point a, at zero point six on the x-axis
So the composition there is forty percent A and sixty percent B, and at a high enough temperature it is a homogeneous liquid
Let's see what happens as we cool it down from that temperature

[advance: cool from point a down to point b on the liquidus]

When we reach the liquidus, at point b, the liquid is in equilibrium with solid B, and the first crystals of B appear
What happens if we cool a little further?

A little further down is point c, inside the two-phase field.

Tie line at point c, then the lever rule

The binary eutectic diagram with a horizontal tie line drawn through point c from the liquidus at d to pure B at e; advancing adds the lever-rule ratio and then the worked numbers.

The lever-rule ratio

Working the numbers

At point c the system is in a two-phase region, solid B together with liquid, and the liquid no longer has the composition we started with.

To find the compositions of the two coexisting phases, we draw a horizontal tie line from point c out to the single-phase regions on either side
On the left the tie line meets the liquidus at point d, where the liquid is forty-three percent B and fifty-seven percent A
On the right it runs out to point e, at pure B
So at point c: solid B in equilibrium with a liquid of A zero point five seven, B zero point four three, while the overall composition of the sample is still sixty percent B
Now we want to know how much of each

[advance: the lever rule as a ratio of segments]

And to do that we use something called the lever rule: we look at the lengths of the segments on this line from d to e
The mole fraction of solid B is the length of the segment opposite to B, that is d to c, divided by the length of the whole line from d to e

[advance: work the numbers]

Point c is at zero point six, so d to c is zero point six minus zero point four three, which is zero point one seven
The full distance d to e is one minus zero point four three, which is zero point five seven
The lever rule gives us zero point one seven over zero point five seven, about thirty percent — so at point c the sample is thirty percent solid B and seventy percent liquid
Using the opposite side of the lever makes sense as soon as we slide c toward d, because there the sample has to be almost all liquid
As a check: take thirty percent pure B and seventy percent of a liquid that is A zero point five seven B zero point four three, combine them, and you get back forty percent A and sixty percent B, the composition we started with — try that one yourself
The overall composition is conserved; only the way it is divided between the two phases changes

Let's see this kind of diagram in a real pair of materials.

A real eutectic: diopside and anorthite

The diopside and anorthite temperature against weight-percent phase diagram at one bar, with its liquidus branches, eutectic and solidus, beside polyhedral structure drawings of the two minerals.

Melting a solid solution: silver and gold

A binary diagram with complete solid solution formation: a lens between the liquidus and solidus with points f, e, d and g marked; advancing annotates the four temperatures of the melting sequence.

Melting from T₁ to T₄

Now let's look at a different type of phase diagram, one where the solids form a solid solution.

A solution is a homogeneous mixture: liquids often mix that way, gases always do, and some solids do too
Silver and gold both take the cubic close-packed structure, so any ratio gives a homogeneous solid with silver and gold atoms distributed at random over the lattice
Suppose a sample that is sixty-five percent silver and thirty-five percent gold
You might expect it to melt at a single temperature when we heat it, but the behavior is richer than that

[advance: walk the melting sequence from T one to T four]

Point g, T one: the solid is homogeneous, sixty-five percent B and thirty-five percent A
Point f, T two: the sample begins to melt, but not all at once
Point e, T three: a two-phase equilibrium — a liquid of about forty-seven percent B and fifty-three percent A, read off the liquidus, in equilibrium with a solid of about eighty-three percent B and seventeen percent A, read off the solidus
As we keep heating, the fraction of liquid grows and the solid that is left gets richer in B
Point d, T four: the last of the solid melts and we have a homogeneous liquid again, back at the overall composition, sixty-five percent B and thirty-five percent A

There is a real mineral pair that behaves exactly this way.

A real solid solution: forsterite and fayalite

The forsterite to fayalite phase diagram, a narrow liquid-plus-solid lens above the olivine solid-solution field, beside a polyhedral drawing of the olivine structure.

Partial solid solutions, lever rule at point G

A binary diagram with partial solid solution formation, showing the A-ss and B-ss fields and points d, g and h, and asking which phases are present at g; advancing draws the tie line and then works the lever rule.

Tie line from d to h

The fraction of each phase

Solid solutions can also form incompletely, and then below the solidus we get two solids coexisting.

We label them A s s and B s s, solid solutions rather than pure A and pure B, and we read their exact compositions straight off the phase diagram
So take point g. The coordinates we are given are d at x B of zero point one, g at zero point three, and h at zero point nine five
Which phases are present at point g, and in what ratio?

[advance: draw the tie line from d through g to h]

Point g lies in a two-phase region, so we find the coexisting phases by drawing a horizontal tie line out to the single-phase regions on either side
To the left the tie line meets a solid solution at x equals zero point one, which is ninety percent A and ten percent B
To the right it meets a solid solution at x equals zero point nine five, which is ninety-five percent B and five percent A
Those two solid solutions are the phases at point g, and the lever rule gives us their relative amounts

[advance: work the lever rule between d and h]

The mole fraction of the B-rich phase, the phase we meet at point h, is the length of the segment from g to d divided by the length of the whole segment from d to h
Plugging in the numbers, zero point two over zero point eight five, we come up with a phase fraction of twenty-four percent
So the sample is twenty-four percent of the phase that is ninety-five percent B and five percent A, and seventy-six percent of the phase that is rich in A

A pair of real oxides shows this same partial mixing.

A real miscibility gap: MgO and CaO

The magnesium oxide and calcium oxide phase diagram with its limited MgO-rich and CaO-rich solid-solution fields and the two-phase miscibility gap between them, beside a rock-salt structure drawing.

As another real-life example, consider the phase diagram of magnesium oxide and calcium oxide.

Both adopt the rock salt structure, so we might expect them to form a complete solid solution
In reality a fifty-fifty mixture gives us two crystalline phases: one rich in magnesium oxide with a little calcium dissolved in it, and one rich in calcium oxide with a little magnesium dissolved in it
That two-phase region separating the two limited solid-solution fields is called a miscibility gap
Just as oil and water are immiscible liquids, magnesium oxide and calcium oxide mix only slightly in the solid state — partial miscibility rather than a full solid solution

Let's pull the phase-diagram half of the lecture together.

Summary: phase diagrams

Summary page for phase diagrams, listing their role in synthesis, the Gibbs phase rule, one-component and binary diagrams, the lever rule, and complete and incomplete solid solutions with miscibility gaps.

To summarize the phase-diagram half of this lecture: phase diagrams are the tools we use to plan a synthesis, to grow crystals, and to understand what is at equilibrium in a solid-state system.

The Gibbs phase rule ties the number of components, the number of phases and the degrees of freedom together, and it simplifies when we hold the pressure constant
A one-component diagram like water shows us triple points, phase boundaries, and the limit the phase rule puts on the degrees of freedom
Binary diagrams give us the liquidus and the solidus, eutectic behavior, and solid solutions
The lever rule turns a tie line into the proportion of each phase in a two-phase region
Real materials, minerals, oxides and alloys, give us complete solid solutions, incomplete ones, and the miscibility gaps between them

The practice questions for today cover both halves of the lecture.

Practice questions

Work the Lecture 9 practice questions before the next class. They cover solid solutions, Vegard's law, deviations from it, band-gap engineering in semiconductors and pigments, the Gibbs phase rule, reading a binary phase diagram, the liquidus, solidus and eutectic point, the lever rule, and solid solutions and miscibility gaps in real materials.

Open the Lecture 9 practice questions

Every question carries a worked explanation, so you can check your reasoning as you go.