Solid Solutions

Introduction

A binary alloy need not form separate pure-A and pure-B solids. If its components substitute for one another in one crystal structure over the entire composition range, the solid state is a continuous solid solution. Its temperature–composition diagram usually has a liquidus and a solidus enclosing a two-phase liquid-plus-solid region. The difference between equilibrium solid and liquid compositions also allows controlled purification by moving a molten zone through a solid.

Core explanation

At fixed pressure, a common idealised isomorphous diagram has a liquidus curve above a solidus curve. Above the liquidus the sample is all liquid; below the solidus it is one solid solution; between them solid and liquid coexist. A horizontal tie line through the two-phase region gives liquid composition x L at the liquidus and solid composition x S at the solidus. The overall composition z lies between them, and the lever rule determines the amounts of solid and liquid.

“Complete miscibility in the solid state” means one equilibrium solid-solution phase is possible throughout the binary composition range at the diagrammed conditions. It does not mean liquid and solid have the same composition at a given temperature. Generally x S differs from x L because each component partitions differently between structures. On cooling an initially uniform melt, the first solid has its own tie-line composition. Continued freezing changes the residual liquid composition and therefore the composition of later solid. If diffusion in the solid is slow, the final crystal can retain a nonuniform composition called coring or segregation even though the equilibrium diagram predicts uniformity after sufficient annealing.

The distribution or segregation coefficient k is often defined approximately as concentration of a dilute impurity in solid divided by that in liquid at a local solid–liquid equilibrium interface. When k<1, the impurity prefers the liquid, so newly formed solid contains less impurity than the adjacent melt. As the solidification front advances, rejected impurity accumulates ahead of it and can concentrate in the last-to-freeze region. The actual concentration profile depends on melt mixing, diffusion, growth speed and whether local equilibrium holds. A single k value is a model parameter, not a complete purification guarantee.

Zone refining exploits this partitioning. A narrow molten region moves slowly through a solid rod. Material freezes at the trailing edge and melts at the leading edge. For k<1, impurity tends to remain in the molten zone and is swept toward one end. Repeating passes can improve purity; the contaminated end is then removed. If k>1 under the chosen convention, impurity prefers solid and the simple sweep direction differs. Purification depends on the particular impurity–host pair and process conditions.

The word “solid solution” need not imply that the solid is molecularly disordered in exactly the way an ideal liquid is. Solute atoms may replace host atoms, occupy interstitial sites or develop short-range order. Complete solid miscibility is thermodynamic within the chosen phase region; lattice structure and enthalpy of mixing determine whether that condition is met. Some alloys instead show limited solid solubility and eutectic reactions, so the appropriate diagram must be selected before using an isomorphous tie line.

For a two-component, two-phase solid–liquid equilibrium, the full phase rule gives F=2. At fixed pressure, one intensive freedom remains. Choosing temperature determines the solid and liquid endpoint compositions along a tie line; the overall z chooses the phase amounts. This matches the VLE and liquid–liquid reasoning from earlier pages, demonstrating that tie-line balances are general.

Step-by-step reasoning

Locate the overall composition and a temperature in the liquid-plus-solid region. Draw a horizontal tie line to the solidus and liquidus. Label solid and liquid endpoints separately, then use a consistent material basis in the lever rule. For refining, determine whether k is below or above one and predict which phase retains the impurity at the moving interface.

Visual explanation

Sketch two lens-like curves between the melting points of pure A and B. The upper line is liquidus and the lower line solidus. A horizontal tie across the lens connects a solid composition on one boundary to a different liquid composition on the other. A moving narrow molten band on a rod carries rejected impurities along its travel direction when k<1.

Real-world analogy

Imagine a travelling crowd-control gate that lets most people settle behind it while a particular group tends to remain at the moving gate. Repeated sweeps move that group toward one end. This captures zone refining's selective partitioning, although real segregation follows chemical-potential equilibrium and transport rather than personal preference.

Real-world example

High-purity semiconductor material can be processed by a travelling melt zone to reduce selected impurities. Engineers measure how each impurity partitions and choose zone speed, number of passes and contaminated-end removal accordingly. The method is useful only when the impurity's distribution and material's melting behaviour support it.

Why?

Why can a homogeneous melt freeze into a compositionally uneven rod? At each advancing interface, equilibrium solid and liquid generally have different compositions. The residual melt changes as one component is preferentially incorporated or rejected. If solid-state diffusion is too slow to homogenise earlier layers, their initial compositions remain recorded along the rod.

Common misconception

Complete solid miscibility does not mean x S=x L at every temperature, nor does it guarantee that a rapidly frozen sample is uniform. Another error is assuming zone refining removes every impurity equally. Its success depends on impurity-specific k, interfacial equilibrium, diffusion and practical removal of the enriched end.

Worked example

At one T, a binary liquid-plus-solid tie line has solid x B^S=0.30 and liquid x B^L=0.70. An alloy has overall z B=0.50. The liquid fraction is (0.50−0.30)/(0.70−0.30)=0.50 and solid fraction is 0.50. The first solid's composition is not the overall alloy's 0.50. For a dilute impurity with k=0.20 and adjacent liquid concentration 100 ppm, local equilibrium predicts newly frozen solid near 20 ppm under that coefficient convention.

Quick check

1. If impurity distribution coefficient k<1, where is it enriched at a solidifying interface? Answer: It is enriched in the liquid relative to the newly formed solid, allowing a travelling molten zone to carry it toward one end.

Exam focus

Read solidus versus liquidus from their definitions rather than swapping curves. Label x S, x L and overall z. Define the convention for k before using it, since some texts define a reciprocal coefficient. Explain how finite diffusion and cooling rate can make observed segregation differ from equilibrium phase amounts.

Advanced insight

Scheil-type solidification models approximate complete mixing in the liquid and negligible diffusion in the solid, leading to strong enrichment of the last liquid when k<1. Full equilibrium instead allows diffusion throughout the solid and predicts a different path. Zone-refining design lies between these limiting descriptions and requires coupled phase equilibrium and transport.

Summary

A completely miscible binary solid solution can have distinct solidus and liquidus curves with different solid and liquid equilibrium compositions. Tie lines and the lever rule give phase compositions and amounts. Moving-zone purification uses impurity partitioning, especially k<1, but its outcome depends on transport and growth conditions.

Practice questions

1. What region lies between solidus and liquidus in a simple isomorphous binary diagram? Answer: A two-phase region containing solid solution and liquid in equilibrium. 2. If x B^S=0.25, x B^L=0.65 and z B=0.45, find liquid fraction. Answer: (0.45−0.25)/(0.65−0.25)=0.50 on the stated composition basis. 3. Why may a quickly frozen alloy show coring despite complete equilibrium solid miscibility? Answer: Successive solid layers form at changing interfacial compositions, and slow solid diffusion may not homogenise them during rapid cooling. 4. For k=0.5, does impurity prefer new solid or adjacent liquid? Answer: Adjacent liquid, because the solid concentration is half the liquid concentration under k=C s/C l.