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

Solid solutionDefine a solid solution and compare it to a liquid solution.

Vegard's lawUnit cell dimensions (and some properties) vary linearly with composition.

Apply itCalculate lattice parameters and band gaps of intermediate compositions.

DeviationsCharge transfer (Mn³⁺/Ru⁵⁺ in Ca-based perovskites), and ordering of different-size ions into more efficient packing.

SemiconductorsSolid solutions matter in semiconductors: epitaxy, heterojunction lasers, LEDs.

Composition from measurementEstimate composition from a property measurement such as a lattice constant or a band gap.

Vegard's Law

The unit cell parameters (and unit cell volume) vary linearly from one end member to the other as a function of composition.

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

$a(x)$ for solid solution · $a_{BX}$ for AlAs · $a_{AX}$ for GaAs

Cell parameter in angstroms plotted against x in Al(x)Ga(1-x)As: the measured points fall on a straight line between the GaAs and AlAs end members.

Per Solid State Materials Chemistry, Figure 2.5a, section 2.5

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) — AlAs the slightly larger lattice constant (~5.66 Å), GaAs the smaller (~5.65 Å)
Composition → lattice parameter: any value between the limits; and back again, measure the parameter and get 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

Deviations from Vegard's Law can be either positive (as shown below) or negative.

Cell volume in cubic angstroms across the CaRuO3 to CaMnO3 solid solution: the measured points bow above the straight line joining the end members.

Causes for deviations from Vegard's law

Site orderingDifferent-size ions order onto distinct sites.

InhomogeneitiesThe sample is not a single homogeneous phase.

Changes in oxidation stateCharge transfer between the two cations.

Per Solid State Materials Chemistry, Figure 2.5b, section 2.5

Band Gap Engineering

For semiconductors it is not unusual that the band gap will follow Vegard's law.

$E_g(x) = x E_g(\mathrm{B}) + (1-x) E_g(\mathrm{A})$

$x = \dfrac{E_g(x) - E_g(\mathrm{A})}{E_g(\mathrm{B}) - E_g(\mathrm{A})}$

Semiconductor Solid Solutions

Heterojunction laser

The solid solution between GaAs ($E_g$ = 1.4 eV, $a$ = 5.65 Å) and AlAs ($E_g$ = 2.1 eV, $a$ = 5.66 Å) is among the most important for optoelectronic devices.

Adapted from P. Bhattacharya, Semiconductor Optoelectronic Devices; alloy curves after Vurgaftman et al. (2001)

In 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 closely matched to minimize strain; band gaps must differ to achieve the function, as in heterojunction lasers
GaAs–AlAs especially useful: lattice parameters nearly identical, so epitaxial growth works across the full range
Band gap tunable from ~2.1 eV (AlAs) to ~1.4 eV (GaAs)
That tunability underpins many LEDs and lasers, where compositions with direct band gaps are essential
Semiconductors highlight Vegard's law, but the principle applies broadly to other solid solutions

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

CdS-CdSe Solid Solutions

CdS_1-xSe_x solid solutions are excellent pigments (cadmium yellow, cadmium orange). By controlling the composition, we can control the band gap and hence the color.

A lump of bright yellow cadmium sulfide powder on a white dish. — CdS ($E_g$ = 2.4 eV)

Pressed discs of CdS-CdSe solid solutions graded from yellow through orange and red to near black as the selenium content rises. — CdS_1-xSe_x compositions

A pile of dark grey-black cadmium selenide powder on a white dish. — CdSe ($E_g$ = 1.7 eV)

A fanned set of cadmium pigment paint swatches running from yellow through orange to deep red. — Cadmium pigments

Example: Orange CdS-CdSe

What composition in the CdS_1-xSe_x solid solution will have a band gap of 2.25 eV?

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

Which composition of CdS_1-xSe_x has a band gap of 2.25 eV?

Let's write the equation that gives us x.

[advance: show the inverse equation]

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.

a = CdS, band gap 2.4; b = CdSe, band gap 1.7
Target: the intermediate composition at 2.25

Now let's put those three numbers into the equation.

[advance: plug in the two band gaps]

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

The way we've set this up, that 0.21 is the selenium content
CdS_0.79Se_0.21 should land very close to 2.25 — provided Vegard's law is followed

Let's write that composition down and find the point on the band-gap graph.

[advance: name the composition, and mark it on 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.

Easy to confuse whether 0.21 is Se or S — the reality check settles it
2.25 eV is much closer to CdS (2.4) than CdSe (1.7), so the composition must be S-rich
Confirms CdS_0.79Se_0.21

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

Summary

Vegard's law lets us predict composition from structure or property measurements.

Solid solutions	Homogeneous crystals spanning the compositions between two end members
Vegard's law	The lattice constant, and sometimes other properties, varies linearly with composition
Deviations	Charge transfer or cation/anion ordering, giving positive or negative curvature
Semiconductor properties	The band gap also approximates Vegard's law, which is what makes band gap engineering possible
GaAs–AlAs	Nearly constant lattice parameter but tunable band gap, the key to epitaxial devices such as lasers and LEDs
CdS–CdSe	Tunable band gap controls the pigment color, yellow to red to black

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 — positive or negative curvature
Many semiconductor properties, the band gap above all, also approximate Vegard's law — that is band gap engineering
GaAs and AlAs: nearly constant lattice parameter while the band gap is tuned, key for epitaxial lasers and LEDs
CdS and CdSe: a tunable band gap controls pigment color from yellow through red to black
Take away: Vegard's law lets us predict composition from structure or from property measurements

The second half of today is phase diagrams.

Why phase diagrams

Gibbs phase ruleExplain it and apply it to one- and two-component systems.

Phase diagramsInterpret diagrams with no compound formation, the eutectic-type systems.

Key regionsDistinguish the liquidus, the solidus, the eutectic point, and the miscibility gap.

Lever ruleCalculate phase fractions in the two-phase regions.

Real systemsRecognize eutectic and solid-solution examples: diopside–anorthite, forsterite–fayalite, MgO–CaO.

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

Know the phase diagram and we can pick a temperature that gives the phase we want, or conditions that grow a single crystal
State the Gibbs phase rule, apply it to one-component and two-component systems
Read a binary phase diagram for a system that forms no compounds
Tell the liquidus, solidus, eutectic point and miscibility gap apart
Use the lever rule for phase fractions in a two-phase region
Real systems: diopside and anorthite, forsterite and fayalite, magnesium oxide and calcium oxide

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

$F = C + 2 - P$

$F$degrees of freedom: variables we can change without a phase appearing or disappearing (temperature, pressure, composition of each phase)

$C$minimum number of independent species (elements or compounds) needed to express every phase’s composition. Water: H₂O (C = 1). MgO–CaO: MgO, CaO (C = 2). Cu/Cu₂O/O₂(g): Cu, O₂ (C = 2).

$P$number of phases

For condensed phase systems where the pressure is constant this reduces to

$F = C + 1 - P$

For more details: Gibbs Phase Rule for Material Science (LearnChemE)

One component: water

Pressure against temperature phase diagram for water, pressure in atmospheres on a logarithmic axis and temperature in degrees Celsius from minus 200 to 600: solid, liquid, gas and supercritical fluid fields separated by boundary curves that meet at the triple point (0.0098 degrees Celsius, 0.00603 atmospheres), the liquid-gas curve ending at the critical point (374.4 degrees Celsius, 217.7 atmospheres).

Per Solid State Materials Chemistry, Figure 4.1, section 4.1

A binary diagram: the axes

Per Solid State Materials Chemistry, Figure 4.2, section 4.2.1

Binary system with no compounds or solid solutions

$F = C + 1 - P$

$F = 3 - P$

Liquid, two-phase fields, eutectic point

$F = C + 1 - P$

$F = 3 - P$

At high temperature the gray region is a homogeneous liquid, where we can vary composition and temperature and still have a single phase, so this region has two degrees of freedom.

One phase, so the liquid's composition is the overall composition
The three other regions: two phases coexist — solid A with solid B below the horizontal line, or a solid with liquid above it
Two phases: F = 1 — choose the temperature and each phase's composition is fixed
One special point where liquid, solid A and solid B all coexist — the eutectic point
Three phases: F = 0 — the temperature and all three compositions are fixed
In a eutectic diagram it is also the lowest temperature at which any liquid exists, the minimum melting temperature

Let's name the two lines that bound these fields, starting with the horizontal one.

The solidus

The liquidus, then cooling from point a

Point ahomogeneous liquid, A_0.4B_0.6

Point bliquid A_0.4B_0.6 in equilibrium with solid B

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

Liquidus: the curve that walls the liquid field off from everything below it

Let's pick a point up in the liquid field.

[advance: mark point a in the liquid field]

Let's take a random point on this phase diagram.

Point a, at zero point six on the x-axis — composition forty percent A, sixty percent B
High enough temperature, so 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.

Point b on the liquidus: liquid plus the first crystals of B
Straight down from a, still x = 0.6 — composition does not change on cooling

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

Tie line at point c, then the lever rule

Point cliquid A_0.57B_0.43 in equilibrium with solid B

$F = 1$fix T and both phase compositions are set; moving c changes only the amounts

Fraction of solid B ($f$)

$f = \dfrac{l_{dc}}{l_{de}}$

$f = \dfrac{0.6 - 0.43}{1.0 - 0.43} = 0.30$

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.

Draw a horizontal tie line from c out to the single-phase regions on either side
Left: it meets the liquidus at point d, liquid at forty-three percent B, fifty-seven percent A
Right: out to point e, at pure B
So at c, solid B with a liquid of A 0.57, B 0.43, while the overall composition is still sixty percent B
F = 3 − 2 = 1: pick the temperature and both compositions are set, liquid at d, solid B at e
Overall composition is not a variable: move c along the tie line and only the amounts change

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.

Look at the lengths of the segments on the line from d to e
Fraction of solid B = the segment opposite to B, d to c, over the whole line d to e

Now let's read those two lengths off the x-axis and put the numbers in.

[advance: work the numbers]

Reading off the x-axis, point c is at zero point six, so the distance from d to c is zero point six minus zero point four three, which is zero point one seven, and the full distance from d to e is one minus zero point four three, which is zero point five seven.

0.17 / 0.57 ≈ thirty percent — at c the sample is thirty percent solid B, seventy percent liquid
The opposite side of the lever makes sense at once: slide c toward d and the sample has to be almost all liquid
Check: thirty percent pure B plus seventy percent of a liquid at A 0.57 B 0.43 gives back forty percent A, sixty percent B — try that one yourself
Overall composition is conserved; only its division between the two phases changes

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

A real eutectic: diopside and anorthite

Polyhedral drawing of the diopside structure inside its unit-cell box: blue silicate tetrahedra and orange magnesium octahedra sharing red oxygen corners, with two large pale blue calcium spheres between them. — Diopside CaMgSi₂O₆

Polyhedral drawing of the anorthite structure inside its unit-cell box: a framework of blue and grey aluminosilicate tetrahedra linked at red oxygen corners, with pale blue calcium spheres in the cavities of the framework. — Anorthite CaAl₂Si₂O₈

Binary system with complete solid solution formation

T₄sample completely melts, composition A_0.35B_0.65

T₃equilibrium mixture of liquid rich in A and solid rich in B

T₂sample begins to melt

T₁homogeneous solid, composition A_0.35B_0.65

Per Solid State Materials Chemistry, Figure 4.4, section 4.2.3

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 cubic close packing, so any ratio gives a homogeneous solid, the atoms distributed at random over the lattice
Take a sample at sixty-five percent gold and thirty-five percent silver
You might expect a single melting temperature — the behavior is richer than that

Let's heat the sample and follow it from T one up to T four.

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

We start at point g, at temperature T one, where the solid is homogeneous, sixty-five percent B and thirty-five percent A.

T two, point f: the sample begins to melt, but not all at once
T three, point e: two-phase equilibrium — liquid about forty-seven percent B off the liquidus, solid about eighty-three percent B off the solidus
Keep heating and the liquid fraction grows while the remaining solid gets richer in B
T four, point d: the last solid melts, homogeneous liquid again, back at 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

Forsterite-fayalite phase diagram: a lens between the liquidus and solidus running from the fayalite melting point up to the forsterite melting point.

Polyhedral drawing of the olivine structure: silicate tetrahedra sharing corners with octahedrally coordinated metal sites. — olivine structure

Partial solid solutions, lever rule at point G

Considering the coordinates of points d ($x_B$ = 0.1), g ($x_B$ = 0.3) and h ($x_B$ = 0.95) what phases are present at g? In what ratio?

Phases at gA_0.9B_0.1 + A_0.05B_0.95

Fraction of A_0.05B_0.95

$f = \dfrac{l_{gd}}{l_{dh}} = \dfrac{0.30 - 0.10}{0.95 - 0.10} = 0.24$

Per Solid State Materials Chemistry, Figure 4.5, section 4.2.3

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

Label them A s s and B s s, solid solutions rather than pure A and pure B, and read their compositions straight off the diagram
Given: d at x_B 0.1, g at 0.3, h at 0.95
Which phases at g, and in what ratio?

Let's draw the tie line through point g and read the two phases off its ends.

[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.

Left: a solid solution at x = 0.1 — ninety percent A, ten percent B
Right: a solid solution at x = 0.95 — ninety-five percent B, 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]

So let's work out the fraction of the sample in the phase that is rich in B, the phase we meet at point h.

That fraction = segment g to d over the whole segment d to h
0.2 / 0.85 = twenty-four percent
So twenty-four percent of the ninety-five percent B phase, seventy-six percent of the A-rich phase

A pair of real oxides shows this same partial mixing.

A real miscibility gap: MgO and CaO

The rock salt structure: two interpenetrating face-centred cubic arrays with every ion octahedrally coordinated. — Rock salt structure

Taken from: https://ocw.mit.edu/courses/materials-science-and-engineering/3-012-fundamentals-of-materials-science-fall-2005/lecture-notes/lec19t.pdf

Summary: phase diagrams

Phase diagrams are the tools for planning synthesis, growing crystals, and reading what is at equilibrium.

Gibbs phase rule	Links components, phases and degrees of freedom, and simplifies under constant pressure
Single-component diagrams	Water shows the triple point, the phase boundaries, and the constraints on the degrees of freedom
Binary diagrams	The liquidus and the solidus, eutectic behavior, and solid solutions
Lever rule	Turns a tie line into the proportion of each phase in a two-phase region
Real materials	Minerals, oxides and alloys give complete and incomplete solid solutions, and the miscibility gaps between them

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 components, phases and degrees of freedom together, and simplifies at constant pressure
A one-component diagram like water: triple points, phase boundaries, and the limit the rule puts on the degrees of freedom
Binary diagrams: liquidus and solidus, eutectic behavior, 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: 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.

Open the Lecture 9 practice questions