CHEM 548: Materials Chemistry

Lecture 11: Free Energy, Phase Diagrams, and Ionic Bonding

Wed 09/30/2026 · Meeting 11

Reading: Woodward, Karen, Evans, and Vogt, Solid State Materials Chemistry
Ch. 5, §5.1

Cartoon title page asking where phase diagrams come from, with a stork delivering solid, liquid, and triple point.

Cartoon of entropy and enthalpy balancing on a seesaw, with temperature deciding which one wins.

Title text: matter seeks to minimize the Helmholtz free energy F equals U minus TS at constant temperature and volume.

$F=U-TS$

$Entropy of mixing equation and its curve, always positive and peaking at mole fraction one half.$

Internal energy of mixing equation with curves for negative, zero, and positive epsilon.

Free energy of mixing curves at several temperatures relative to the consolute temperature.

At high temperature entropy dominates and the free-energy curve is smooth with one minimum so the mixture is completely miscible, but at lower temperature if unlike bonds cost energy (ε > 0) the curve develops two dips near the pure ends with a barrier in the middle so the alloy separates into two phases, and the special point where the single dip just splits into two is called the consolute temperature given by T_c = (p × ε) / (2 × k_B), for example with p = 12 and ε = 0.010 eV (10 meV) the consolute temperature is about 696 K while with ε = 0.015 eV (15 meV) it is about 1040 K, which means that the larger the energy penalty the higher the consolute temperature because more thermal disorder is needed to keep the system mixed

High T: one minimum → completely miscible; lower T with ε > 0: two dips, a barrier, two phases
Single dip just splits in two at the consolute temperature, T_c = (p × ε) / (2 × k_B)
Numbers: p = 12, ε = 10 meV → about 696 K; ε = 15 meV → about 1040 K
Larger energy penalty → higher consolute temperature

So to read the coexisting compositions off a curve with two dips, we use what is called the method of tangents.

Free-energy curve with a common tangent touching at the two equilibrium compositions.

Method of tangents lets us find the compositions of the two phases that can coexist at a given temperature by drawing a straight line that just touches the free-energy curve at two points, and these tangent points give the equilibrium compositions directly, so in this case the line touches at x = 0.132 (13.2%) for the alpha phase and at x = 0.859 (85.9%) for the beta phase, meaning if the overall composition lies anywhere between these values the system will split into a mixture of alpha and beta with those compositions.

Straight line just touches the curve at two points → the equilibrium compositions
Here: x = 0.132 alpha, x = 0.859 beta
Overall composition between them → splits into alpha + beta at those compositions

If we repeat that tangent construction at temperature after temperature, we can generate the whole phase diagram.

Free-energy curves at several temperatures projected down to build the miscibility-gap phase diagram.

By applying the method of tangents to the free-energy curves at different temperatures we can trace out the pairs of compositions that coexist, and when we project those tangent points down onto the temperature–composition plane we generate the phase diagram, so at high temperature the curve has only one minimum and the system is fully mixed, while at lower temperatures two minima appear and the tangent gives two coexisting compositions, and by repeating this at many temperatures we build the binodal curve that outlines the miscibility gap with the consolute temperature at its peak.

Tangent points projected down onto the temperature–composition plane
High T: one minimum, fully mixed; lower T: two coexisting compositions
Many temperatures → the binodal curve outlining the miscibility gap, consolute temperature at its peak

Let's see how this works in a real system with no heat of mixing, silicon-germanium.

Free-energy curves and the lens-shaped phase diagram of the ideal silicon-germanium system.

In the Si–Ge system there is no heat of mixing, so the solid free energy is F_s(x,T) = RT[x ln(x) + (1–x) ln(1–x)] and the liquid free energy is F_l(x,T) = RT[x ln(x) + (1–x) ln(1–x)] + ΔS[x(T_B – T) + (1–x)(T_A – T)], where ΔS is the entropy of fusion per mole and T_A and T_B are the melting points of pure Si and pure Ge, so the entropy of mixing is the same in both phases but the liquid curve is shifted by how far the temperature is below each melting point, which means at low temperature the solid has lower free energy and is stable, at high temperature the liquid has lower free energy and is stable, and in between the curves cross so a common tangent defines the liquidus and solidus, giving the smooth phase diagram with complete miscibility that we see for Si–Ge.

Same entropy of mixing in both phases; liquid curve shifted by distance below each melting point
Low T: solid stable; high T: liquid stable; in between the curves cross
Common tangent defines liquidus and solidus → the smooth diagram with complete miscibility

Now let's give the solid a strongly negative heat of mixing and see what changes.

Free-energy curves and phase diagram for a solid with a negative heat of mixing, showing a congruent melting maximum.

Free-energy curves at falling temperatures generating a eutectic phase diagram.

From high T (1500 K, single-phase liquid L), through 1100–1000 K where the liquid free energy rises and mutual tangents to the solid minima define α+L and β+L fields with tie-lines giving equilibrium compositions (lever rule), down to the eutectic at ~940 K where the liquid minimum just touches the common tangent between α and β (an invariant three-phase point, L ⇌ α+β, depressed below the pure melting points by positive solid heat of mixing), and finally at lower T (e.g. 700 K) the liquid lies above the α–β tangent leaving only two solids (α+β), with phase compositions from tangent points and fractions from the lever rule—illustrating how positive enthalpy of mixing produces the V-shaped liquidus, eutectic reaction, and α+β solid region [oai_citation:0‡Nauman_ch12_phase_equilibria.pdf](file-service://file-Q1ewSHDHZyMzxiqeDyNy2m).

1500 K: single-phase liquid L
1100–1000 K: tangents to the solid minima define α+L and β+L fields, tie-lines + lever rule
~940 K eutectic: liquid minimum just touches the α–β tangent — invariant three-phase point, L ⇌ α+β
700 K: liquid above the tangent → only two solids, α+β
Positive enthalpy of mixing → V-shaped liquidus, eutectic reaction, α+β region

Sometimes a new phase appears in the middle of the diagram; let's look at an intermediate solid solution.

Rounded free-energy nose producing an intermediate solid-solution phase, beside the real aluminum-nickel diagram.

Generic line-compound topology beside the real cadmium-tellurium phase diagram with its vertical CdTe line.

What is a “line compound,” and why is it a vertical line?Some compounds only form at one exact composition.

Cd–Te: that’s CdTe (50 at% Te) — off-stoichiometric alloys split into Cd + CdTe (left) or Te + CdTe (right)
Why vertical: the solid-CdTe curve has a sharp, wedge-like minimum at stoichiometry
Common tangent from the liquid touches at a single composition → projects to a vertical CdTe line, no composition range
A rounded minimum would touch at two compositions → a finite γ + L band, a solid-solution instead
Right panel, real data: congruent melting of CdTe at ≈ 1098 °C, far above Cd ≈ 321 °C and Te ≈ 450 °C, vertical α-CdTe field
Left sketch: the generic A–γ–B topology for such cases

That completes our free-energy picture of phase diagrams; here are the homework questions on it.

Homework Questions

Define the parameter ε in the mean-field model of mixing. Physically, what does it mean if ε < 0? What about ε > 0?
Contrast the phase diagram topologies of an isomorphous alloy (Si–Ge) and an endothermically mixing alloy (with a eutectic). What underlying free-energy features produce the differences?
Calculate the entropy of mixing (per mole, in J/mol·K) for a binary solution with mole fraction x_A = 0.25, x_B = 0.75. Use R = 8.314 J/molK.
Using the regular solution model with p = 12, ε = 12 meV, compute the critical temperature T_c. Express your answer in Kelvin.
(Constants: 1 eV = 1.602 × 10⁻¹⁹ J, k_B = 8.617 × 10⁻⁵ eV/K).

Here are the homework questions on this material.

One: define the parameter epsilon in the mean-field model of mixing; physically, what does epsilon less than zero mean? Greater than zero?
Two: contrast the phase diagram topologies of an isomorphous alloy, silicon-germanium, and an endothermically mixing alloy with a eutectic; what free-energy features produce the differences?
Three: calculate the entropy of mixing, per mole, for x A equal to 0.25 and x B equal to 0.75, using R equal to 8.314 joules per mole kelvin
Four: regular solution model, p equal to 12, epsilon equal to 12 milli electron volts → compute T c in kelvin

Those are yours to work through; next we leave phase diagrams behind and start chemical bonding.

Ionic Bonding

By the end of this lecture, you should be able to:

Explain why chemical bonding connects structure to properties of materials.
Define ionic bonding as electrostatic interactions between point charges.
Apply Coulomb’s law to estimate attractive vs. repulsive interactions between ions.
Describe how coordination number and structure type affect the Madelung constant.
Explain what the Madelung constant represents and why it differs for NaCl, CsCl, ZnS, etc.
Define lattice energy and distinguish between lattice energy and lattice enthalpy.
Use the Born–Mayer equation to estimate lattice energies, recognizing the role of repulsive terms.

Coulomb interaction energy between point charges: cation-anion attractive, like charges repulsive.

Ionic bonding can be described in a simple, though not fully realistic, way by treating a crystal as a collection of point charges.

Potassium fluoride: K ions as +1 charges, fluoride as –1; calculate the electrostatic, or Coulomb, interactions
Energy: multiply the charges, divide by four pi times the permittivity of free space, divide by the distance
+1 charge means +1 times the elementary charge, 1.602 × 10⁻¹⁹ coulombs
Opposite charges → negative energy, attraction; like charges → positive, repulsion — two cations push apart until the energy tends toward zero
Magnitude grows with larger charges or shorter distances
Coulomb energy falls only as one over distance → significant even over relatively long ranges

A crystal has far more than two ions, so the next step is to sum these interactions over the whole lattice.

Madelung sum over successive neighbor shells of rock salt converging to 1.7476.

Once we have the energy expression for two interacting charges, the next step is to sum all such interactions in the crystal.

Sodium chloride, rock salt structure; start from a central chloride ion, look at successive shells
Nearest: six sodium ions, octahedral, at the interatomic spacing → six attractive interactions
Second: twelve chlorides along the cell edges, edge times the square root of two → repulsive
Third: eight sodiums at the cube corners, edge times the square root of three → attractive
Shell by shell: a series that alternates in sign and decreases in magnitude
Summed to infinity it converges to 1.7476 for sodium chloride — the Madelung constant

That converged sum is the Madelung constant, and its value depends on the structure; let's compare structure types.

Madelung Constants and Structure Type

Structure Type	Coordination	A	Structure Type	Coordination	A
Cesium chloride	Cs^[8]Cl^[8]	1.763	Fluorite	Ca^[8]F₂^[4]	2.519
Rock salt	Na^[6]Cl^[6]	1.748	Rutile	Ti^[6]O₂^[3]	2.408
Wurtzite	Zn^[4]S^[4]	1.641	Cadmium chloride	Cd^[6]Cl₂^[3]	2.244
Zinc blende	Zn^[4]S^[4]	1.638	Cadmium iodide	Cd^[6]I₂^[3]	2.192

Electrostatic interactions favor symmetric structures with close packing of anions and cations and high coordination numbers.

Different crystal structures give different Madelung constants because each has distinct neighbor shells.

1:1 compounds: constant decreases as coordination drops — eight-coordinate CsCl → six-coordinate NaCl → four-coordinate zinc blende and wurtzite
General rule: higher symmetry and higher coordination maximize ionic bonding
1:2 compounds: larger overall — more ions per formula unit — same coordination trend
Eight-coordinate CaF₂ → six-coordinate rutile TiO₂ → layered CdCl₂ and CdI₂: constant decreases
Rutile vs layered, all six-coordinate cations, three-coordinate anions: rutile spreads its octahedra in three dimensions, minimizing cation–cation repulsion
Layered types concentrate cations in certain planes → more repulsion, lower Madelung constant

Now that we can handle the electrostatics, let's define the quantity we are really after: the lattice energy.

Definition of lattice energy as the energy released forming a solid from gas-phase ions.

Coulomb attraction and exponential repulsion terms that combine into the Born-Mayer equation.

Energy versus interatomic distance, with the total-energy minimum at the equilibrium separation d zero.

Born-Mayer equation; clicks add its derivative set to zero and then the solved equilibrium form.

If we go back to the lattice energy expression, we can take its derivative with respect to the interatomic distance d and set it equal to zero.

Setting the derivative to zero identifies the minimum of the function, which corresponds to the equilibrium separation between ions.

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The Coulomb term varies as 1 over d, so its derivative is –1 over d², a negative contribution.

The repulsive term stays an exponential upon differentiation but gains a factor of 1 over rho.

Setting the total derivative equal to zero and evaluating at d₀ gives the condition for the minimum of the lattice energy.

[advance: reveal source state 24]

Solving for the constant B and substituting it back gives the Born–Mayer equation, which allows calculation of lattice energy.

Inputs: the ionic charges, Avogadro’s number, the elementary charge, and the permittivity of free space, all physical constants; the Madelung constant, which depends on structure type; and the equilibrium distance d₀, which can be estimated by adding the ionic radii of the two ions.

With these values, the Born–Mayer equation provides the lattice energy of an ionic compound.

Before we go further, it is worth remembering where this energy U fits in the bigger picture, so let me bring back the free-energy expression we used for phase diagrams.

Regular-solution free energy F of x and T, shown again to connect lattice energy back to phase diagrams.

Summary – Chapter 5: Ionic Bonding

Bonding refresher bridges earlier crystal structure topics to upcoming band structure.
Ionic bonding model: treat ions as point charges; interaction energy from Coulomb’s law.
Attractive vs. repulsive forces depend on charge sign and distance.
Long-range nature of Coulomb interactions requires summing over all shells of neighbors.
Summation leads to the Madelung constant (e.g., 1.748 for NaCl rock salt).
- Higher coordination → larger Madelung constant → stronger bonding.
- Layered structures (e.g., CdCl₂, CdI₂) show reduced Madelung constants due to added cation–cation repulsions.
Lattice energy = energy released when ions form a solid from infinite separation (exothermic).
Born–Mayer model combines:
- Coulombic attraction (favors collapse),
- Exponential repulsion (prevents collapse),
- Minimum → equilibrium distance d₀.
Born–Mayer equation allows quantitative lattice energy estimates using charges, Madelung constant, and ionic radii.

Let's read back through the summary of chapter five so far.

The bonding refresher bridges our earlier crystal structure topics to the upcoming band structure
Ions as point charges, interaction energy from Coulomb's law; attraction versus repulsion by charge sign and distance
Long-range Coulomb → sum over all shells → the Madelung constant, 1.748 for NaCl in the rock salt structure
Higher coordination → larger constant, stronger bonding; layered CdCl₂/CdI₂ reduced by added cation-cation repulsions
Lattice energy: released when ions form a solid from infinite separation, exothermic
Born-Mayer: Coulomb attraction favors collapse, exponential repulsion prevents it → minimum at d zero; the equation gives quantitative estimates from charges, Madelung constant, and ionic radii

Here is your homework on this material.

Part 2

By the end of this lecture, you should be able to:

Recall the form and physical meaning of the Born–Mayer equation.
Use ionic radii (Shannon radii) to estimate the equilibrium distance d₀.
Calculate the lattice energy of an ionic compound (example: KCl, rock salt structure).
Compare calculated lattice energies to experimental values obtained from the Born–Haber cycle.
Explain why the Born–Mayer equation underestimates lattice energies and identify additional corrections (dispersion, zero-point energy).
Relate cation/anion size ratios to coordination number and stable crystal structures (fluorite, rutile, layered).
Describe how anion polarizability and induced dipoles stabilize layered structures in heavier halides.

Born-Mayer lattice energy worked example for KCl; a click reveals the calculated result of minus 679 kilojoules.

In the In the last lecture, we derived the Born–Mayer equation for calculating the lattice energy of ionic compounds.

Two parts: the electrostatic potential energy of a lattice of point charges, and a repulsive correction, –ρ/d₀, for short-range ion repulsion
Example: KCl, rock salt structure
Needed: charges, Avogadro’s number, elementary charge 1.602 × 10⁻¹⁹ C, permittivity of free space, Madelung constant 1.748, and d₀
Hardest is d₀, from Shannon (Shannon–Prewitt) radii: six-coordinate K⁺ 138 pm + Cl⁻ 181 pm → d₀ = 319 pm
Charges of 1e (absolute values, the negative sign accounting for attraction), multiply by Avogadro’s number, divide by 4πϵ₀, substitute d₀, add the repulsion correction with ρ = 34.5 pm

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Put it in the calculator: minus 679 kilojoules
Without the repulsive term it would be minus 761 — the repulsive term reduces the lattice energy by a ballpark of about 10%
Now we might ask ourselves: how close is that to the true lattice energy?

The experimental answer comes from the Born-Haber cycle.

Born-Haber cycle for KCl determining the experimental lattice energy.

The Born–Haber cycle is a method of determining the experimental lattice energy of an ionic compound.

KCl: measure the direct enthalpy of formation from K metal and ½Cl₂(g), releasing 436 kJ·mol⁻¹, or reconstruct step by step
Vaporize K, +90; dissociate Cl₂, +122
Ionize K to K⁺, +419; electron gain enthalpy of Cl, –349, the only exothermic step up to this point
Gas-phase K⁺ and Cl⁻ combine to KCl: by definition the lattice energy, 718 kJ·mol⁻¹ released
This lattice enthalpy, nearly equal to lattice energy, provides the driving force: although forming ions from neutral atoms is endothermic, the large exothermic lattice energy stabilizes the compound
Theory about –680 vs experiment –718: the equation slightly underestimates

So let's compare the calculated and experimental values across the alkali halides.

Calculated versus experimental lattice energies for the alkali metal halides.

Complete Lattice Energy Expression

$U_L = -\frac{z_1 z_2 e^2 N_\mathrm{A}}{4πε_0 d_0}\,A\left(1-\frac{ρ}{d_0}\right)-\frac{N_\mathrm{A}C}{d_0^{\,6}}+2.25\,N_\mathrm{A}hυ_\mathrm{max}$

[A] Coulomb energy
[B] Repulsive interaction (9-17%)
[C] Dispersion forces (0.1-5%)
[D] Zero point energy (<1.5%)

Substance	[A] kJ	[B] kJ	[C] kJ	[D] kJ
NaCl (U=−766)	−859	99	−12	7
MgO (U= −3921)	−4631	698	−6	18

Table of observed MX2 halide structures by cation and anion, with fluorite, rutile, and cadmium iodide drawings.

To predict the most stable structures in MX₂ compounds, we compare possible cation–anion arrangements.

Fluorite: cation 8 anion neighbors, anion 4; rutile: 6 and 3; layered CdI₂/CdCl₂: half the octahedral holes filled, alternating filled and empty layers differing only in stacking
Fluorite has the largest Madelung constant → naively expected to dominate; observed for large cations with relatively electronegative anions
Smaller cations → rutile; larger, less electronegative halides → layered; PbCl₂ or SrI₂ variants have less symmetry, coordination number 7
Stability depends not just on Madelung constant but on ion sizes and electronegativity
What explains these trends? Pause the video, discuss or jot down ideas: what causes these structures to emerge?
The size argument first: in the fluoride compositions, small cations give rutile, large cations the fluorite structure

Wilson and Madden ran exactly this kind of calculation; let's see what a rigid ionic model predicts.

Rigid-ion calculated lattice energies of MBr2 versus cation radius for fluorite, rutile, and layered CdI2.

Shannon-Prewitt ionic radii versus crystal radii for strontium-fluorine, differing by a 14 picometer shift.

Before continuing, we need to discuss ionic radii.

Shannon and Prewitt, and Shannon’s 1976 revision: two tabulated sets, ionic radii and crystal radii
Directly measurable: the cation–anion center distance, e.g. Sr–F; the split into radii fixed the oxide and fluoride values to Pauling’s earlier values
Electron-density studies suggested Pauling had overestimated anion sizes → anion radii reduced by 14 pm, cation radii increased by the same
The sum is unchanged; ionic radii follow Pauling’s convention, crystal radii shift 14 pm from anion to cation
Crystal radii arguably better reflect electron-density boundaries, but most of the community uses ionic radii, and I now generally follow that convention

With that convention settled, let's do some hard-sphere geometry.

Cation inside a cube of anions; clicks add the Mg-F distance, the F-F distance, and their comparison.

But in what follows, we're going to talk about these ions as though they really were hard spheres.

We want to look at where the anions start to overlap, and for that it's best to use the crystal radii; we use them for the rest of this lecture.

Take a cation and put it into a cube of anions, as in the fluoride structure; there's a couple of distances we can get from simple trigonometry.

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The distance from the center of the cation to the center of the anion, D1, with magnesium and fluoride, is just 205 picometers, and we can get it simply by adding the radii of the two species.

Because it's a cube, that also fixes the distance from one fluoride center to the neighboring fluoride center.

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Two times D1, the cation anion distance, is the body diagonal of the cube; divide by the square root of three to get the edge length, the fluorine-fluorine distance.

That gives an estimate of 237 picometers from the center of one fluoride ion to the next.

[advance: reveal source state 40]

Compare the actual radius of a fluoride ion: 119 picometers, times two, is 238 picometers.

So the size of the magnesium cation is such that to get the optimal cation anion distance, we practically have to bring the fluoride ions together so they touch, and that is not favorable: as the fluorides get close we get repulsions between the core electrons.

So it starts to become unfavorable to put eight anions around the cation when the cation gets too small.

So let's lower the coordination number to six and redo the same geometry.

Cation in the equatorial plane of an octahedron of fluoride ions; a click adds the distance geometry.

If we were to reduce the coordination number to six, like it is in Rutile, here I'm just looking at, let's say, the equatorial plane of an octahedron.

The full octahedron would have one anion above and below; I'm not drawing those so we can see the geometry. Now what is the anion anion distance?

[advance: reveal source state 42]

A bit of trigonometry: from a 45° triangle, the side length is d₁ times sin 45°, and doubling it gives an F–F separation of 290 pm.

That is much larger than 238 pm, so the anions are not in contact and core–core repulsions between fluoride ions are avoided.

So smaller cations or larger anions favor lower coordination numbers: otherwise the anions are forced closer together, with unfavorable repulsions.

Eventually even six coordination gets too crowded; let's run the same numbers for zinc sulfide.

Zinc cation among sulfide ions; a click adds the S-S distance and its comparison to two sulfide radii.

Pauling radius ratio rule with coordination number cutoffs at 0.7 and 0.4.

MX2 halide structure table with ionic radii; a click replaces the entries with radius ratios.

All right, let's go back to our table here once again.

Remember: green entries coordination number eight, blue and red six, the white entries seven
So what kind of radius ratio do we get for these compounds?

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Updated crystal radii give more accurate values than Pauling had, but the cutoffs are not absolute: ions are not hard spheres
The trend: ratio decreases → coordination number decreases
Fluorides: eightfold to sixfold crossover below about 0.96; chlorides, softer, between 0.79 and 0.68
Useful guidelines, not exact rules
They do not explain every case: zinc chloride, zinc bromide, manganese iodide adopt layered CdI₂-type structures rather than rutile, despite both being six-coordinate and rutile's higher Madelung constant — why?

To answer that, let's go back to the Wilson and Madden calculations we saw a few slides ago.

Rigid-ion MBr2 lattice energies shown again before polarization is added.

MBr2 lattice energies; a click adds the polarization result that stabilizes the layered structure.

In the same study, an additional term was introduced to make the modeling more realistic by allowing the anion’s electron cloud to deform, essentially accounting for polarization.

A large iodide ion: negative charge spread over a wide electron cloud
A small cation such as magnesium approaches → the cloud shifts toward it, an induced dipole
Electron distribution no longer symmetric → polarization must be included in the lattice energy

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With polarization, the CdI₂-type layered structure emerges as the most stable for small cation sizes, at least for the bromides in the middle of the series
Induced dipole lowers the layered structure's energy significantly; fluorite essentially unchanged, rutile decreases only slightly
Stronger for layered CdI₂ and CdCl₂: cations concentrated in planes → stronger polarization interactions, larger stabilization

To see why the layered structure benefits so much, let's look at the anion's environment in each of the three structures.

Anion coordination environments in fluorite, rutile, and CdI2, with their polarization consequences.

Layered CdI2-type structure showing the induced dipole pulling anion charge toward the cation layers.

MX2 halide structure table shown again after the polarization argument.

Summary

Born–Mayer equation combines Coulomb attraction with a short-range repulsive term to approximate lattice energy.
Example: For KCl, calculated lattice energy ≈ −679 kJ/mol vs. experimental −718 kJ/mol (Born–Haber).
Shannon radii provide coordination-specific values to estimate d₀.
The model systematically underestimates lattice energies by ~3–6%.
Adding dispersion and zero-point energy terms improves accuracy slightly, but the main trend is well captured.
Radius ratio rule: smaller cations (or larger anions) lower coordination numbers → 8 → 6 → 4.
Structure stability trends:
- Large cations + small anions → fluorite (CN = 8).
- Small cations → rutile (CN = 6).
- Large, polarizable anions (Br⁻, I⁻) → layered CdI₂-type structures, stabilized by anion polarization.
Symmetry of cation arrangement dictates whether polarization cancels (fluorite, rutile) or enhances stability (layered).

Let's read back through the summary.

Born-Mayer: Coulomb attraction plus short-range repulsion; KCl calculated about minus 679 versus experimental minus 718 from the Born-Haber cycle
Shannon radii: coordination-specific values to estimate d zero
Systematic underestimate of about three to six percent; dispersion and zero-point terms improve accuracy slightly, the main trend well captured
Radius ratio rule: smaller cations, or larger anions, lower the coordination number, eight to six to four
Stability trends: large cations, small anions → fluorite, CN eight; small cations → rutile, CN six; polarizable bromide and iodide → layered CdI₂-type, stabilized by anion polarization
Cation-arrangement symmetry dictates whether polarization cancels, fluorite and rutile, or enhances stability, layered

And here is the last homework problem.

Homework

5.6 With the exception of helium, all noble gases solidify at low temperature. The lack of ionic or covalent bonding means that atoms are held together by dispersion forces alone. Given the melting points of the noble gases: Ne = 24 K, Ar = 84 K, Kr = 116 K, Xe = 161 K, what can you say about the strength of the London dispersion forces as the principal quantum number of the outermost shell increases? What is the explanation for this trend?

Practice questions

Work the Lecture 11 practice questions before the next class. They cover the free energy of mixing and the method of tangents, Coulomb interactions and the Madelung constant, the Born–Mayer lattice energy and the Born–Haber cycle, and radius ratios, anion polarization, and MX₂ structure choice.

Open the Lecture 11 practice questions

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