CHEM 548: Materials Chemistry
Lecture 11: Ionic Bonding
Wed 09/30/2026 · Meeting 11
Reading: Woodward, Karen, Evans, and Vogt, Solid State Materials Chemistry
Welcome back to CHEM 548.
Chapter five: ionic bonding
Coulomb's law and the Madelung constant
Lattice energy and the Born-Mayer equationIonic radii and why MX₂ halides adopt different structuresReading : Chapter 5, §5.1, Woodward, Karen, Evans, and Vogt
Let's start with what you should be able to do after this lecture.
Ionic Bonding
By the end of this lecture, you should be able to:
Bonding → properties explain why chemical bonding connects structure to properties of materials.
Ionic bonding define it as electrostatic interactions between point charges .
Coulomb’s law apply it to estimate attractive vs. repulsive interactions between ions.
Madelung constant describe how coordination number and structure type affect it.
Structure dependence explain what the Madelung constant represents and why it differs for NaCl, CsCl, ZnS , etc.
Lattice energy define it, and distinguish lattice energy from lattice enthalpy .
Born–Mayer equation use it to estimate lattice energies, recognizing the role of repulsive terms .
Chapter five is all about chemical bonding.
Bonding links structure to properties ; properties need band structure
First taught band structure cold; students needed a bonding refresher first
Order: ionic , then atomic orbitals, then covalent
Goals: ionic bonding as point-charge electrostatics ; Coulomb attraction vs. repulsion
Madelung constant : what it is, why NaCl, CsCl, ZnS differLattice energy vs. lattice enthalpy; estimate with Born-Mayer
So let's start ionic bonding with the simplest possible picture, point charges and Coulomb's law.
Coulomb Interaction
$U_C = \frac{\textcolor{#ff5a5f}{(z_1 e)\cdot(z_2 e)}}{4\pi\varepsilon_0 \textcolor{#8d90e8}{d}}$
magnitude of the charges
distance between charges
Cation–Anion Interaction – Attractive
Anion–Anion Interaction – Repulsive
Cation–Cation Interaction – Repulsive
Per Solid State Materials Chemistry , §5.1.1, Eq. (5.1).
The simplest way to describe ionic bonding is to treat the crystal as a collection of point charges.
Not fully realistic, but goes a long way
KF : K⁺ as +1, F⁻ as −1, sum the Coulomb interactionsE = q₁q₂ / 4πε₀d ; ε₀ = permittivity of free space+1 = 1.602 × 10⁻¹⁹ C , the elementary charge
Opposite charges negative, attract ; like charges positive, repel
Larger charges, shorter d → stronger; 1/d falls slowly, long range
A crystal holds far more than two ions, so the next step is to sum these interactions over the whole lattice.
Madelung Sum
shell
running sum
6 Na⁺ at d +6/√1 → 6.000 12 Cl⁻ at √2 d −12/√2 → −2.485 8 Na⁺ at √3 d +8/√3 → 2.134
$V_\mathrm{C} = \frac{(z_1 e)\cdot(z_2 e)}{4\pi\varepsilon_0 d}\left(\frac{6}{\sqrt{1}} - \frac{12}{\sqrt{2}} + \frac{8}{\sqrt{3}} - \frac{6}{\sqrt{4}} + \frac{24}{\sqrt{5}} - \frac{24}{\sqrt{6}} - \frac{12}{\sqrt{8}} \cdots\right)$
If the shells are chosen in the right way (charge neutral shells) this sum converges to a value of $1.7476$. This is the Madelung constant, $A$ for the NaCl structure type
Per Solid State Materials Chemistry , §5.1.1, Eq. (5.2).
With the energy of two charges in hand, we sum every interaction in the crystal, shell by shell.
Rock salt , central Cl⁻Add shells one at a time , keep a running sum
First come the nearest neighbors.
[advance: add the first shell, six Na⁺ at d]
The nearest neighbors are six sodium ions in an octahedron at the interatomic distance d.
6 Na⁺ at d , attractiveRunning sum starts at 6
Next come the second neighbors.
[advance: add the second shell, twelve Cl⁻ at √2 d]
The second neighbors are twelve chlorides along the cell edges at d times √2.
12 Cl⁻ at √2 d , like charges, repulsive 12/√2: sum down to −2.485
Then come the third neighbors.
[advance: add the third shell, eight Na⁺ at √3 d]
The third neighbors are eight sodiums at the cube corners at d times √3, attractive again.
8/√3: sum back up to 2.134
Then 6 at 2d, 24 at √5 d, 24 at √6 d, 12 at √8 d
Alternating, slow: charge-neutral shells converge to A = 1.7476
The Madelung constant depends on the structure, so let's compare structure types.
Madelung Constants and Structure Type
1:1 compounds, MX
1:2 compounds, MX₂
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 3D network Ti[6] O₂[3] 2.408
Wurtzite Zn[4] S[4] 1.641
Cadmium chloride layered Cd[6] Cl₂[3] 2.244
Zinc blende Zn[4] S[4] 1.638
Cadmium iodide layered Cd[6] I₂[3] 2.192
Electrostatic interactions favor symmetric structures with close packing of anions and cations and high coordination numbers.
Values per Solid State Materials Chemistry , §5.1.1, Table 5.1.
Each structure type has its own set of neighbor shells, and so its own Madelung constant.
1:1 , A falls with CN: CsCl 1.763, rock salt 1.748, wurtzite/zinc blende ~1.641:2 larger, more ions per formula unitFluorite 2.519, rutile 2.408, CdCl₂ 2.244, CdI₂ 2.192
Rutile vs. layered : all 6:3 coordinationRutile spreads octahedra in 3D; layers crowd cations, more repulsion, lower A
Electrostatics favors symmetric, high-coordination structures
Now that we can handle the electrostatics, let's define the quantity we are really after, the lattice energy.
Lattice Energy
Lattice Energy ($U_\mathrm{L}$):
The amount of energy that is released when one mole of a crystal is formed from its constituent ions at infinite separation in the gas phase at $T$ = 0 K
$a\,\mathrm{M}^{y+}(g) \;+\; b\,\mathrm{X}^{z-}(g) \rightarrow \mathrm{M}_a\mathrm{X}_b(s)$
Lattice energy is exothermic
Lattice energy is approximately equal to lattice enthalpy, $U_\mathrm{L} = H_\mathrm{L}$
Per Solid State Materials Chemistry , §5.1.2, Eq. (5.4).
Textbooks define lattice energy in different ways, so let me fix ours.
U_L : energy released forming one mole of crystal from gas-phase ionsIons from infinite separation, at 0 K
a M^y+ + b X^z− (gas) → one formula unit of solid
Energy released, so U_L is negative
Experiment gives lattice enthalpy ; differs by a few kJ/mol, treat as equal
To calculate it, we combine the Coulomb attraction with a short-range repulsion.
Lattice Energy
$U_\mathrm{C} = \frac{z_1 z_2 e^2 N_\mathrm{A}}{4\pi\varepsilon_0 d}A$
$U_\mathrm{r} = Be^{-\frac{\textcolor{#8d90e8}{d}}{\textcolor{#e0b060}{\rho}}}$
distance
0.345 Å
The lattice energy in the absence of repulsive forces
Repulsive interactions, caused by overlap of core orbitals
Combining these two gives the Born-Mayer equation
$U_L = U_C + U_r = \frac{z_1 z_2 e^2 N_\mathrm{A}}{4\pi\varepsilon_0 d}A + Be^{-d/\rho}$
Per Solid State Materials Chemistry , §5.1.2, Eqs. (5.5) and (5.6).
We start from the Coulomb energy of the whole lattice of point charges.
Left: charge product × N_A / 4πε₀d × A
Negative, more favorable closer; alone the ions would collapse
So we need a repulsive term to hold them apart.
[advance: the repulsive term]
What stops them is a short-range repulsion from the overlap of the core orbitals.
Exponential B e^(−d/ρ) , B empirical, ρ = 0.345 Å
Now we add the two terms together.
[advance: the Born-Mayer equation]
The repulsion is positive and the Coulomb term is negative, and adding the two gives the Born-Mayer equation.
Every parameter known except B : charges, A, ρ, d
Let's plot what those two terms do as the ions approach each other.
Lattice Energy
200
300
400
500
600
700
800
−1000
−500
0
500
1000
interatomic distance, d (pm)
energy (kJ/mol)
d 0 = 319U 0 = −679repulsive potential energy, U r
coulomb potential energy, U C
U C + U r dU L /dd = 0
Per Solid State Materials Chemistry , §5.1.2, Figure 5.2.
Here are the two terms plotted against the distance between the ions.
Coulomb : more and more negative as the ions close inRepulsion : negligible far out, exponential, steep at short range Sum has a minimum : position d₀, depth U₀, slope zero
We can find that minimum exactly by taking the derivative of the lattice energy.
Born-Mayer Equation
$U_L = \frac{z_1 z_2 e^2 N_\mathrm{A}}{4\pi\varepsilon_0 d}A + Be^{-d/\rho}$
$\frac{\mathrm{d}U_L}{\mathrm{d}d} = -\frac{z_1 z_2 e^2 N_\mathrm{A}}{4\pi\varepsilon_0 d_0^2}A - \frac{Be^{-d_0/\rho}}{\rho} = 0$
$\frac{\mathrm{d}U_L}{\mathrm{d}d} = -\frac{z_1 z_2 e^2 N_\mathrm{A}}{4\pi\varepsilon_0 d_0^2}A - \frac{Be^{-d_0/\rho}}{\rho} = 0$
$U_L = -\frac{|z_1 z_2| e^2 N_\mathrm{A}}{4\pi\varepsilon_0 d_0}A\left(1 - \frac{\textcolor{#8d90e8}{\rho}}{\textcolor{#ff5a5f}{d_0}}\right)$
0.345 Å
Equilibrium distance
Per Solid State Materials Chemistry , §5.1.2, Eqs. (5.6)–(5.8).
Let's go back to the lattice energy expression and find its minimum.
Minimum: dU/dd = 0 , at the equilibrium separation
Setting it to zero also pins down B
So let's take the derivative term by term.
[advance: add the derivative set to zero at d₀]
The derivative has one term from the Coulomb energy and one from the repulsion.
Coulomb : 1/d → −1/d²Repulsion : stays exponential, picks up 1/ρ from the chain ruleSum = 0, evaluated at d₀ : the minimum condition
Now we solve that condition for B and put it back into the lattice energy.
[advance: add the solved Born-Mayer equation]
Solving for B and substituting it back gives the Born-Mayer equation.
U_L = Coulomb energy at d₀ × (1 − ρ/d₀), ρ = 0.345 Å; parenthesis = repulsion correction
What do we need to know to calculate a lattice energy? Charges of the ions; N_A, e, ε₀ are physical constantsA from the structure typed₀ = cation radius + anion radiusWith those, the lattice energy of an ionic compound
The U in the free energy of mixing from last lecture is this same bond energy, so the energies a bonding model gives us are what place the phase boundaries.
Summary – Chapter 5: Ionic Bonding
From point charges to a quantitative lattice energy, in eight steps.
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 . Attraction vs. repulsion depends on charge sign and distance . Long-range nature Coulomb interactions require summing over all shells of neighbors . Madelung constant that summation converges to A (1.748 for NaCl rock salt). Higher coordination → larger A → stronger bonding; layered CdCl₂ and CdI₂ are reduced by added cation–cation repulsions. Lattice energy energy released when ions form a solid from infinite separation — exothermic . Born–Mayer model Coulombic attraction favors collapse, exponential repulsion prevents it; the minimum sets the equilibrium distance d₀ .Born–Mayer equation quantitative lattice-energy estimates from charges , the Madelung constant , and ionic radii .
Let me summarize what we have built so far, from point charges to a lattice energy.
Bonding refresher bridges crystal structures and band structures
Point charges , Coulomb pairs: sign sets attraction, shorter distance stronger1/d falls slowly: sum every shell → Madelung constant , 1.748 rock salt
Higher CN, larger A; CdCl₂, CdI₂ lose some to cation-cation repulsion
Lattice energy : released from infinite separation, exothermicBorn-Mayer : repulsion stops collapse, minimum sets d₀; charges, A, radii give a number
In the second part we put that equation to work on a real compound.
Part 2
By the end of this lecture, you should be able to:
Born–Mayer equation recall its form and its physical meaning .
Shannon radii use ionic radii to estimate the equilibrium distance d₀ .
Lattice energy calculate it for an ionic compound — example: KCl, rock salt structure .
Born–Haber cycle compare calculated lattice energies to the experimental values it gives.
Corrections explain why Born–Mayer underestimates , and identify dispersion and zero-point energy .
Size ratios relate cation/anion size to coordination number and stable structures (fluorite, rutile, layered).
Anion polarizability describe how induced dipoles stabilize layered structures in heavier halides.
In the first part we derived the Born-Mayer equation, and the first goal is to recall its form and what each term means.
Shannon radii : cation radius + anion radius = d₀ Lattice energy : worked through for KCl, rock salt Born–Haber cycle : the experimental value we check againstCorrections : Born–Mayer comes out low; add dispersion and zero-point energy Size ratios : cation/anion size → CN → fluorite, rutile, or layeredAnion polarizability : induced dipoles stabilize the layered heavier halides
Let's start with the calculation for potassium chloride.
Lattice Energy – Example KCl
$U_L = -\frac{|z_1 z_2| e^2 N_\mathrm{A}}{4\pi\varepsilon_0 d_0}A\left(1 - \frac{\rho}{d_0}\right)$
Shannon radii
http://www.knowledgedoor.com/2/elements_handbook
$d_0 = r(\mathrm{K}^+) + r(\mathrm{Cl}^-) = 138\ \mathrm{pm} + 181\ \mathrm{pm} = 319\ \mathrm{pm}$
$U_L = -\frac{(1)(1)\left(1.602\times10^{-19}\,\mathrm{C}\right)^2\left(6.022\times10^{23}\,\mathrm{mol^{-1}}\right)}{4\pi\left(8.8542\times10^{-12}\,\frac{\mathrm{C}}{\mathrm{V\,m}}\right)\left(319\times10^{-12}\,\mathrm{m}\right)}\,1.748\left(1 - \frac{34.5\ \mathrm{pm}}{319\ \mathrm{pm}}\right)$
$U_L = -\left(7.61\times10^{5}\ \mathrm{J\,mol^{-1}}\right)\left(1 - 0.108\right) = -6.79\times10^{5}\ \mathrm{J\,mol^{-1}} = -679\ \mathrm{kJ\,mol^{-1}}$
Per Solid State Materials Chemistry , §5.1.2, Eq. (5.8); radii from the Shannon–Prewitt table linked above.
Let's calculate the lattice energy of potassium chloride with the Born-Mayer equation.
Prefactor: point-charge electrostatics
(1 − ρ/d₀): repulsion when ions get too close
Our example is potassium chloride, which has the rock salt structure.
[advance: show the KCl structure]
KCl has the rock salt structure, so the Madelung constant is 1.748.
Easy inputs: charges, e = 1.602 × 10⁻¹⁹ C , N_A, ε₀
Harder: d₀ , cation center to anion center
We estimate d₀ from ionic radii.
[advance: add the Shannon radii and d₀]
The best set of ionic radii are the Shannon radii, also called Shannon-Prewitt radii.
Radii depend on CN : rock salt is 6:6
K⁺ 138 + Cl⁻ 181 = 319 pm Cell edge 629 pm → true K-Cl ~315 pm , close
Now we can put every number into the equation.
[advance: add the substituted Born-Mayer equation]
We use the charges as absolute values, 1 and 1, and the minus sign in front accounts for the attraction.
e² × N_A / (4πε₀ × d₀ in m)
× A = 1.748 × (1 − 34.5/319)
Let's see what that comes out to.
[advance: add the evaluated lattice energy]
Put all of that into the calculator and you get minus 679 kilojoules per mole.
Coulomb part alone −761 kJ
Factor 1 − 0.108: repulsion takes ~10% off
How close to the true lattice energy ?
The experimental answer comes from the Born-Haber cycle.
Born-Haber Cycle
Experimental determination of Lattice Energy
K(s) + ½Cl₂(g)
KCl(s)
ΔH°f
−436
all energies kJ·mol⁻¹
K(g) + ½Cl₂(g)
+90
K(s) → K(g)
K(g) + Cl(g)
+122
½Cl₂(g) → Cl(g)
K⁺(g) + e⁻ + Cl(g)
+419
IE, K(g) → K⁺(g) + e⁻
K⁺(g) + Cl⁻(g)
−349
Heg , Cl(g) + e⁻ → Cl⁻(g)
−718
HL , K⁺(g) + Cl⁻(g) → KCl(s)
HL ≈ U L
K(g) + Cl(g) → K⁺(g) + Cl⁻(g): +419 − 349 = +70, uphill
The lattice energy pays for the electron transfer.
Born–Mayer −679 vs. experiment −718: about 5% too small
Per Solid State Materials Chemistry , §5.1.3, Figure 5.3.
The Born-Haber cycle gives us the experimental lattice energy.
Direct: K(s) + ½Cl₂(g) → KCl(s), −436 kJ/mol
KCl more stable than the elements by that much
Or we can go around the cycle one step at a time.
[advance: vaporize potassium]
First we vaporize the potassium, which costs 90 kJ per mole.
Then we split the chlorine molecules into atoms.
[advance: split the chlorine molecule]
Breaking half a Cl₂ molecule into a chlorine atom costs 122.
Next we ionize the potassium atoms.
[advance: ionize potassium]
Ionizing potassium costs 419, the most expensive step in the cycle.
K(g) → K⁺(g) + e⁻: +419 , ionization energy
Now the electron goes onto the chlorine atom.
[advance: add the electron to chlorine]
Adding the electron to chlorine releases 349, the electron gain enthalpy.
Cl(g) + e⁻ → Cl⁻(g): −349
Negative of the electron affinity; first exothermic step
Now the gaseous ions come together into the solid.
[advance: form solid KCl from the gaseous ions]
Letting K⁺ and Cl⁻ in the gas phase come together into solid KCl is, by definition, the lattice energy.
Cycle closes on −436 → −718 kJ/mol
Lattice enthalpy, H_L ≈ U_L
One thing about this cycle is worth a second look.
[advance: show the uphill electron transfer]
As chemists we usually say that potassium readily gives its electron to chlorine.
K(g) + Cl(g) → ions: 419 − 349 = +70, uphill
Reaction still very exothermic: lattice energy drives it
Now back to how accurate the Born-Mayer equation was.
[advance: compare with Born-Mayer]
The Born-Mayer equation gave minus 679 kJ per mole, and the experimental value is minus 718.
−679 vs. −718 : about 5% too small Not bad for point charges
Let's compare calculated and experimental values across the alkali halides.
Calculated vs. Experimental Lattice Energy
1.9
2.2
2.5
2.8
3.1
3.4
3.7
−550
−650
−750
−850
−950
−1050
cation–anion distance (Å)
lattice energy (kJ/mol)
Calculated with
Born-Mayer equation
fluorides
chlorides
bromides
iodides
Per Solid State Materials Chemistry , §5.1.3.
Across the alkali metal halides, the Born-Mayer equation underestimates the lattice energy by only about three to six percent.
Reasonably accurate for a point-charge model
Right: the shared rock salt structure
Same charges, same A: energy depends only on d → smooth curve
Points: Born-Haber values , fluorides, chlorides, bromides, iodides
Larger ions, larger d, lattice energy drops
If we wanted to do better than a few percent, we would need every term in the lattice energy.
Complete Lattice Energy Expression
$U_\mathrm{L} = \underbrace{\textcolor{#8fbde8}{-\frac{|z_1 z_2| e^2 N_\mathrm{A}}{4\pi\varepsilon_0 d_0}A}}_{\textcolor{#8fbde8}{[\mathrm{A}]}}\underbrace{\textcolor{#e0b060}{\left(1-\frac{\rho}{d_0}\right)}}_{\textcolor{#e0b060}{[\mathrm{B}]}}$$\;\underbrace{\textcolor{#8fe3b4}{-\frac{N_\mathrm{A}C}{d_0^{\,6}}}}_{\textcolor{#8fe3b4}{[\mathrm{C}]}}$ $\;\underbrace{\textcolor{#ff8b8f}{+\,2.25\,N_\mathrm{A}h\nu_\mathrm{max}}}_{\textcolor{#ff8b8f}{[\mathrm{D}]}}$
[A] Coulomb ion–ion attraction, negative, the dominant term
[B] Repulsion core-orbital overlap, positive, 9–17% of [A]
[C] Dispersion van der Waals, 1/d ⁶, stabilizes by 0.1–5%
[D] Zero point quantum, opposite sign to [C], <1.5%
kJ/mol [A] [B] [C] [D] $U_\mathrm{L}$
NaCl −859 +99 −12 +7 −766
MgO −4631 +698 −6 +18 −3921
Per Solid State Materials Chemistry , §5.1.3.
Up to this point we have used only the first two terms of the lattice energy.
[A] Coulomb : ion-ion attraction, negative, dominant [B] repulsion : ~9–17% of [A], positive
To be more accurate we can add two more terms.
[advance: add the dispersion term]
The first is a term that goes as one over d to the sixth, the van der Waals or dispersion attraction between the ions.
[C] dispersion , van der Waals, 1/d⁶More negative, more stable , up to ~5%
The last term is the zero-point energy.
[advance: add the zero-point term]
The zero-point energy comes from quantum mechanics, and it has the opposite sign to the dispersion term.
[D] zero-point : quantum, opposite sign to [C], < 1.5%[C] and [D] small, nearly cancel : often neglected
Let's see how big each term is for two real compounds.
[advance: add the NaCl and MgO values]
For sodium chloride the Coulomb term is minus 859 kJ per mole and the repulsion takes back 99.
NaCl : −859 + 99 − 12 + 7 ≈ −766 kJ/mol , terms roundedMgO , 2+/2−: −4631, +698; C and D only −6 and +18
With lattice energies in hand, we can ask which structure an MX₂ halide should adopt.
Which Structure Does an MX₂ Halide Adopt?
Observed structure by cation and anion size; ionic radii in pm.
Cationionic radius
F⁻133
Cl⁻181
Br⁻196
I⁻220
Mg²⁺72 rutile CdI₂/CdCl₂ CdI₂/CdCl₂ CdI₂/CdCl₂ Zn²⁺74 rutile tetrahedral tetrahedral tetrahedral Mn²⁺83 rutile CdI₂/CdCl₂ CdI₂/CdCl₂ CdI₂/CdCl₂ Ca²⁺100 fluorite rutile rutile CdI₂/CdCl₂ Sr²⁺118 fluorite fluorite PbCl₂ SrI₂ Ba²⁺135 fluorite PbCl₂ PbCl₂ PbCl₂
fluorite, Ca[8] F₂[4] A = 2.519 rutile, Mg[6] F₂[3] A = 2.408 Cd[6] I₂[3] A = 2.192
fluorite — CN 8
rutile — CN 6
CdI₂/CdCl₂ — layered, CN 6
PbCl₂ and SrI₂ — CN 7
tetrahedral — CN 4
Why would a structure with a smaller Madelung constant win, and why should that depend on ion size and electronegativity?
Madelung constants per Solid State Materials Chemistry , §5.1.1, Table 5.1. Zinc halides: ZnCl₂, J. Brynestad, H. L. Yakel, Inorg. Chem. 1978 , 17 , 1376; ZnBr₂, C. Chieh, M. A. White, Z. Kristallogr. 1984 , 166 , 189; ZnI₂, P.-H. Fourcroy et al., Acta Cryst. B 1978 , 34 , 3160. Other structure assignments and radii are the lecture’s own.
Let's take everything we have learned about ionic bonding and ask which structure an MX₂ compound should adopt.
Table: six M²⁺ down the side, four halides across, ionic radii in pm
Candidates: fluorite 8:4 , rutile 6:3 , layered CdI₂/CdCl₂ 6:3
Layered: half the octahedral holes, filled layer / empty layer ; the two differ only in stacking
A: fluorite 2.519 > rutile 2.408 > CdI₂ 2.192; smaller A → less negative Coulomb energy
Naive guess: fluorite everywhere
Now let's fill in the structures that are actually observed.
[advance: observed structures fill the table]
In the lower left of the table, large cations with the more electronegative halides, fluorite is indeed what we observe.
More than A: fluorides, smaller cation → rutile
Larger, less electronegative halide → layered
PbCl₂, SrI₂: less symmetric, cation CN 7
Zinc odd one out: Cl, Br, I tetrahedral, CN 4 , though Mg²⁺ is the same size; back to it at radius ratios
Stability: A plus ion size plus anion electronegativity
So what explains these trends?
[advance: the discussion question]
Why would a structure with a smaller Madelung constant win, and why should that depend on ion size and electronegativity?
A minute with a neighbor : what makes each structure win its region
Let's look first at the size argument: among the fluorides, small cations give rutile and large cations give fluorite.
Calculated Lattice Energy
60
80
100
120
140
−2500
−2300
−2100
−1900
−1700
cation radius (pm)
lattice energy, U L (kJ/mol)
MBr2
CdI2
(layered)
rutile
fluorite
Calculations with a rigid ionic model (similar to the Born‑Mayer equation)
Rutile structure has the lowest E when the cation is small
Fluorite structure has the lowest E when the cation is large
CdI₂ layered structure not stable for any cation size
M. Wilson, P. A. Madden, J. Phys. Cond. Matt. 1994 , 6 , 159–170.
Wilson and Madden calculated the lattice energies of the three structures as the cation radius changes.
Rigid ion , like Born-Mayer, better repulsion plus dispersionPlot: MBr₂ , U_L against cation radius
Start with the small cations on the left.
[advance: rutile lowest for small cations]
For small cations the rutile structure has the most negative lattice energy, so it is the most stable.
Small cation: rutile lowest
Matches the small-cation fluorides in the table
Now follow the curves as the cation gets larger.
[advance: fluorite lowest for large cations]
As the cations get larger there is a crossover, and the fluorite structure becomes the most stable.
Large cation: crossover to fluorite
A simple model predicts the crossover
Now look at the layered structure.
[advance: layered never most stable]
The layered CdI₂ structure is significantly higher in energy at every cation size.
CdI₂ higher at every size Cations crowded into layers → shorter M-M, more repulsion
Before we push the size argument further, we need a short detour on how ionic radii are defined.
Ionic Radii vs Crystal Radii
Shannon-Prewitt Ionic Radii
(O²⁻, F⁻ chosen to match Pauling radii)
Shannon-Prewitt Crystal Radii
(O²⁻, F⁻ 14 pm smaller)
F⁻
r = 133 pm
r = 1.33 Å
Sr²⁺
r = 118 pm
r = 1.18 Å
F⁻
r = 119 pm
r = 1.19 Å
Sr²⁺
r = 132 pm
r = 1.32 Å
d(Sr–F) = 133 + 118 pm
d(Sr–F) = 251 pm
d(Sr–F) = 119 + 132 pm
d(Sr–F) = 251 pm
Same distance either way: the anion loses 14 pm and the cation gains the same 14 pm.
Shannon and Prewitt, and Shannon's 1976 revision, tabulated two sets of radii: ionic radii and crystal radii.
Measurable: center-to-center distance , Sr to F here
Split fixed by Pauling's oxide and fluoride radii ; electron density: anions too big
Crystal radii : anion −14 pm, cation +14 pmSum unchanged: 133 + 118 = 119 + 132 = 251 pm
Community mostly uses ionic radii , as I generally do
Hard-sphere anion overlap: crystal radii for the rest of today
Let's start by asking how big a cation fits in a cube of eight anions.
How big a cation fits a cube of 8?
Anion radius = 1. All lengths run between ion centers.
The highlighted square is the cube's top face.
Cut vertically through the gold face diagonal.
2 2 d = 2√2 2√2 2 1 √2 1 + r d₁ d₂ One square face, viewed straight on.
The rectangular slice, viewed straight on.
From the center, take half of each side.
At contact, the center-to-corner distance is 1 + r.
Touching anions: edge = $1 + 1 = 2$
Face diagonal: $d = \sqrt{2^2 + 2^2} = 2\sqrt2$
Slice dimensions: $2 \times 2\sqrt2$
Half an edge: $2/2 = 1$ Half a face diagonal: $2\sqrt2/2 = \sqrt2$
$1+r = \sqrt{1^2+(\sqrt2)^2} = \sqrt3$$r = \sqrt3 - 1 = 0.732$
cube of 8 octahedron of 6 tetrahedron of 4
Now the real ions: Mg2+ in a cube of eight F− , hard spheres with crystal radii r(Mg2+ ) = 86 pm and r(F− ) = 119 pm.
Now the real ions: Mg2+ in a cube of eight F− , hard spheres with crystal radii r(Mg2+ ) = 86 pm and r(F− ) = 119 pm.
$\textcolor{#ff8b8f}{d_1}$, Mg–F center to corner: $86 + 119 = 205$ pm
Now the real ions: Mg2+ in a cube of eight F− , hard spheres with crystal radii r(Mg2+ ) = 86 pm and r(F− ) = 119 pm.
$\textcolor{#ff8b8f}{d_1}$, Mg–F center to corner: $86 + 119 = 205$ pm
86/119 = 0.72, just under the 0.732 limit derived aboveMg2+ is slightly too small for the cube, so the fluoride ions are already touching, with no room to close further.
Now we treat the ions as hard spheres and put a cation in a cube of eight anions, as in fluorite.
General answer first, then Mg²⁺ and F⁻
Anion radius = 1 , lengths between centersHighlighted top face; right, face on
Anions touch along the edge: edge = 2
First we need the diagonal of that face.
[advance: draw the face diagonal]
The face diagonal runs corner to corner across the square.
Pythagoras: √(2² + 2²) = 2√2
Now we cut the cube vertically through that diagonal.
[advance: cut the vertical slice through the face diagonal]
The vertical cut through the gold face diagonal gives a rectangle.
Sides 2√2 by 2
Cube center = rectangle center ; anion at each corner
From the center, we take half of each side.
[advance: mark the center and the two half sides]
From the center of the rectangle to a corner anion is a right triangle.
Half an edge: 2/2 = 1
Half the face diagonal: 2√2/2 = √2
Now we put the cation at the center and let it touch the corners.
[advance: add the cation touching the corner anions]
When the cation touches the anions, the center-to-corner distance is 1 plus r.
Hypotenuse √(1² + √2²) = √3
1 + r = √3 → r = 0.732
Ratio 0.732: touches all eight while anions touch
Smaller cation: anions touch first
Now let's put in real ions, magnesium and fluoride.
[advance: add the magnesium-fluoride distances]
Here are the magnesium-fluoride distances, from the crystal radii already on the slide, 86 pm and 119 pm.
d₁ , pink, Mg-F center to corner: 86 + 119 = 205 pm d₂ , green, F-F along the cube edgeUnit cube: edge 2 ↔ center-to-corner √3, so d₂ = 205 × 2/√3 = 237 pm
Two fluoride radii: 238 pm
Next is the radius ratio, and what it says about magnesium in a cube.
[advance: add the radius ratio for magnesium fluoride]
The radius ratio is 86 over 119, or 0.72, just under the 0.732 we derived.
Mg slightly too small : fluorides already touching
Eight in contact would push fluorides together, core-electron repulsion
Too-small cation: eight neighbors unfavorable
So let's lower the coordination number to six and redo the geometry.
How big a cation fits an octahedron of 6?
Six anions at the corners of an octahedron with the cation at the center.
1 1 √2 d₁ d₂ The square of four anions at true size. The other two sit above and below the plane.
The four anions in the plane touch, so their centers form a square with sides of 2. The cation sits at the center, 1 across and 1 up from a corner, so the cation–anion distance is $\sqrt{1^2 + 1^2} = \sqrt2$.
When the cation just touches the anions, that distance is one anion radius plus one cation radius: $1 + r = \sqrt2$.
$r = \sqrt2 - 1 = 0.414$sodium in the NaCl structure
cube of 8 · 0.732 octahedron of 6 · 0.414 tetrahedron of 4
Now the real ions: Mg2+ in an octahedron of six F− , same crystal radii: r(Mg2+ ) = 86 pm, r(F− ) = 119 pm.
Now the real ions: Mg2+ in an octahedron of six F− , same crystal radii: r(Mg2+ ) = 86 pm, r(F− ) = 119 pm.
$\textcolor{#ff8b8f}{d_1}$, Mg–F center to corner: $86 + 119 = 205$ pm
Now the real ions: Mg2+ in an octahedron of six F− , same crystal radii: r(Mg2+ ) = 86 pm, r(F− ) = 119 pm.
$\textcolor{#ff8b8f}{d_1}$, Mg–F center to corner: $86 + 119 = 205$ pm
86/119 = 0.72, well above the 0.414 limit derived aboveMg2+ is more than big enough for six, so the fluoride ions are not in contact and avoid core–core repulsion.
Now we lower the coordination number to six, an octahedron, as in rutile.
Left: octahedron , cation at center
Right: equatorial square of four anions, true size
Axial pair above and below left out
Anion radius still 1; Mg and F again after
First we need the distance from the cation to a corner of that square.
[advance: show the center-to-corner distance]
The four anions in the plane touch, so their centers form a square with sides of 2.
Cation at the center ; to a corner: 1 across, 1 up
√(1² + 1²) = √2
Now we write that √2 in terms of the two radii.
[advance: split the distance between anion and cation]
When the cation just touches the anions, the √2 is one anion radius plus one cation radius.
Anion 1, cation r: 1 + r = √2
That gives the radius of the cation that just fills the octahedron.
[advance: show the octahedral radius ratio]
The cation radius is √2 minus 1, which is 0.414.
Six-coordination limit ; classic case Na⁺ in NaClTwo limits: 0.732 for eight, 0.414 for six
Now back to magnesium and fluoride, this time in an octahedron.
[advance: add the magnesium-fluoride distances]
Here are the distances for magnesium in the octahedron, from the same crystal radii, 86 pm and 119 pm.
d₁ , pink, Mg-F center to corner: 205 pm againd₂ , green, F-F square edge: 45° triangle, 2 d₁ sin 45° = 290 pm Two fluoride radii only 238 pm
Next is the radius ratio for magnesium with six neighbors.
[advance: add the radius ratio for six-coordinate magnesium]
The ratio is still 0.72, and now that is well above the 0.414 limit.
Mg more than big enough for six
290 vs. 238 pm: fluorides don't touch , no core-core repulsion
Smaller cations, larger anions → lower CN , or anions forced together
MgF₂ is rutile , six-coordinate Mg
Eventually even six gets too crowded, so let's do the tetrahedron and then beryllium sulfide.
How big a cation fits a tetrahedron of 4?
Four anions on alternate corners of a cube. The gold plane is the diagonal section shown at the right.
√2/2 1 √(3/2) √2 2 the other two anions, edge on The gold diagonal section at true size, √2 by 2. Two anions and the cation lie in it. The other two anions touch it at the dot.
Touching anions: face diagonal $= 2$, so cube edge $e = \sqrt2$
Center to anion: $\sqrt{(\sqrt2/2)^2 + 1^2} = \sqrt{3/2}$
The anion takes 1, the cation the rest.
$r = \sqrt{3/2} - 1 = 0.225$zinc in sphalerite
cube of 8 · 0.732 octahedron of 6 · 0.414 tetrahedron of 4 · 0.225
Now the real ions: Be2+ in a tetrahedron of four S2− , crystal radii r(Be2+ ) = 41 pm, r(S2− ) = 170 pm.
Now the real ions: Be2+ in a tetrahedron of four S2− , crystal radii r(Be2+ ) = 41 pm, r(S2− ) = 170 pm.
Be–S center to corner: $41 + 170 = 211$ pm
Now the real ions: Be2+ in a tetrahedron of four S2− , crystal radii r(Be2+ ) = 41 pm, r(S2− ) = 170 pm.
Be–S center to corner: $41 + 170 = 211$ pm
41/170 = 0.24: below 0.414 for six, just above the 0.225 limit derived aboveFour sulfides just clear each other around Be2+ , and BeS takes the zinc blende structure.
The last case is four-coordination, a tetrahedron.
Anions on alternate cube corners
Gold plane: diagonal section ; right, true size, √2 by 2
Two anions + cation in the section; other two edge on, at the dot
Real ions this time: Be²⁺ and S²⁻
First we need the size of that cube.
[advance: find the cube edge]
The anions touch along the face diagonals of the cube.
Face diagonal = 2
e² + e² = 2² → e = √2
Now we find the distance from the center to an anion.
[advance: find the center-to-anion distance]
In the diagonal section, we go from the cube center to an anion center.
Across: half an edge, √2/2 ; up: half a face diagonal, 1
Square, add, root: √(3/2)
Again the anion takes 1 of that distance and the cation takes the rest.
[advance: split the distance and give the tetrahedral radius ratio]
The anion takes 1, and the cation takes the rest.
r = √(3/2) − 1 = 0.225
Example: Zn in sphalerite , zinc blende
All three limits: 0.732, 0.414, 0.225 for eight, six, four
Now let's place the real ions, beryllium in a tetrahedron of four sulfides.
[advance: add the beryllium-sulfide distances]
Here are the distances with four sulfides around the beryllium, from crystal radii of 41 pm for beryllium and 170 pm for sulfide.
Be-S: 41 + 170 = 211 pm
Tetrahedron edge: S-S = √(8/3) × 211 = 345 pm
Two sulfide radii: 340 pm , so sulfides just clear
Next is the radius ratio for beryllium sulfide.
[advance: add the radius ratio for beryllium sulfide]
The radius ratio is 41 over 170, or 0.24.
Below 0.414 for six, just above 0.225 for four
Four sulfides just clear : BeS is zinc blende , four-coordinate Be
Pauling turned this geometric argument into a simple rule, the radius ratio rule.
Radius Ratio, $\rho$
$\rho = \dfrac{r_\mathrm{cation}}{r_\mathrm{anion}}$
$\rho > 0.7$ — CN tends to be 8
$0.7 > \rho > 0.4$ — CN tends to be 6
$\rho < 0.4$ — CN tends to be 4
Cutoffs as the lecture teaches them, 0.7 and 0.4. Solid State Materials Chemistry , §1.4.2, note 17, gives the exact hard-sphere minima instead: √2 − 1 ≈ 0.414 for an octahedral hole and (√3/√2) − 1 ≈ 0.225 for a tetrahedral hole.
Early in the twentieth century, Linus Pauling proposed the radius ratio rule.
Sustainable CN depends on ρ = r_cation / r_anion
This ρ is a ratio, not the Born-Mayer ρ
ρ > 0.7 : eight fit, CN 8Below 0.7 : CN 6; below 0.4 : CN 4Rounded limits; textbook exact: 0.414 octahedral, 0.225 tetrahedral
Let's apply those radius ratios to our table of MX₂ halides.
Structures of MX₂ Halides
Cationcrystal radius
F⁻119
Cl⁻167
Br⁻182
I⁻206
Mg²⁺86 rutile ρ = 0.72CdI₂/CdCl₂ ρ = 0.51CdI₂/CdCl₂ ρ = 0.47CdI₂/CdCl₂ ρ = 0.42Zn²⁺88 rutile ρ = 0.74tetrahedral ρ = 0.53tetrahedral ρ = 0.48tetrahedral ρ = 0.43Mn²⁺97 rutile ρ = 0.82CdI₂/CdCl₂ ρ = 0.58CdI₂/CdCl₂ ρ = 0.53CdI₂/CdCl₂ ρ = 0.47Ca²⁺114 fluorite ρ = 0.96rutile ρ = 0.68rutile ρ = 0.63CdI₂/CdCl₂ ρ = 0.55Sr²⁺132 fluorite ρ = 1.11fluorite ρ = 0.79PbCl₂ ρ = 0.73SrI₂ ρ = 0.64Ba²⁺149 fluorite ρ = 1.25PbCl₂ ρ = 0.89PbCl₂ ρ = 0.82PbCl₂ ρ = 0.72
fluorite, Ca[8] F₂[4] A = 2.519 rutile, Mg[6] F₂[3] A = 2.408 Cd[6] I₂[3] A = 2.192
Green is CN 8, blue and pink CN 6, white CN 7, gold CN 4 — so what radius ratio does each compound have?
Structure assignments and radii are the lecture’s own; each ratio is the header row’s crystal radii divided cell by cell.
Let's go back to our table once again, now with crystal radii along the edges.
Structures and colors unchanged
Crystal radii from the geometry: cation +14 pm, anion −14 pm, so every ratio up
So what radius ratio does each of these compounds have?
[advance: add the radius-ratio grid]
Now we replace each structure with its radius ratio, the cation radius divided by the anion radius.
Ratio falls, CN falls ; where 8 → 6 depends on the anion Fluorides: between Ca 0.96 and Mn 0.82 , so MgF₂, ZnF₂, MnF₂ rutile above 0.7
Softer chlorides: lower, Sr 0.79 to Ca 0.68 , near 0.7; guides, not exact rules
Zn Cl, Br, I ≈ 0.5, expect 6, are CN 4 ; Mg same size, 86 vs 88 pm, all CN 6
Filled 3d, χ 1.65 vs 1.31 → partly covalent , four directed bonds, as in zinc blende
Rutile vs layered, both CN 6: MgBr₂, MnI₂ at 0.47, rutile higher A, yet CdI₂-type
Let's go back to the Wilson and Madden calculations, which also predict rutile over the layered structure.
Calculated Lattice Energy
60
80
100
120
140
−2500
−2300
−2100
−1900
−1700
cation radius (pm)
lattice energy, U L (kJ/mol)
MBr2
CdI2
(layered)
rutile
fluorite
Calculations with a rigid ionic model (similar to the Born‑Mayer equation)
Rutile structure has the lowest E when the cation is small
Fluorite structure has the lowest E when the cation is large
CdI₂ layered structure not stable for any cation size
M. Wilson, P. A. Madden, J. Phys. Cond. Matt. 1994 , 6 , 159–170.
Here again are Wilson and Madden's rigid ion lattice energies for the bromides.
Rigid ion , like Born-MayerLarge cation fluorite , small cation rutileLayered CdI₂ never most stable : model says never seen
But we see it; layered compounds a very active research area
So what stabilizes these layered structures?
Calculated Lattice Energy, MBr₂ Halides
60
80
100
120
140
−2600
−2400
−2200
−2000
cation radius (pm)
Lattice energy, U L (kJ/mol)
fluorite
rutile
CdI2 (layered)
Introduce
polarization of the
electron cloud
surrounding the
anion (creating an
induced dipole)
Now the layered CdI₂/CdCl₂ structures emerge as the lowest energy structure for small cations.
M. Wilson, P. A. Madden, J. Phys. Cond. Matt. 1994 , 6 , 159–170.
In the same study, Wilson and Madden added a term that lets the anion's electron cloud deform.
Term accounts for polarization
Large iodide: charge spread over a wide electron cloud
Small Mg²⁺ close: cloud shifts, induced dipole
Asymmetric density → polarization term in the lattice energy
Here is what that term does to the three structures.
[advance: add the result with polarization included]
With polarization included, the layered CdI₂ and CdCl₂ structures become the lowest in energy for small cations.
At least for the bromides , mid-series
Layered lowered a lot ; fluorite unchanged; rutile slightly
Cations concentrated in planes → stronger polarization
To see why the layered structure benefits so much, let's look at the anion's environment in each structure.
Anion Environment
F
Fluorite
Anion environment:
tetrahedral
Site symmetry:
43m (Td )
No anion
polarization
F
Rutile
Anion environment:
~trigonal planar
Site symmetry:
2mm (C2v )
Minimal anion
polarization
I
CdI2
Anion environment:
trigonal pyramidal
Site symmetry:
3m (C3v )
Significant anion
polarization
Whether the anion polarizes depends on how the cations are arranged around it.
Each cation pulls the anion's cloud toward itself
Fluorite : perfect tetrahedron, pulls cancel, no dipole Rutile : near trigonal planar, angles off 120°, very small dipole CdI₂ : trigonal pyramidal, all three cations one side, large dipole Layered gains real stabilization ; fluorite almost none
Here is what that induced dipole does inside the layered structure.
Anion Polarization
empty layer
B
A
B
A
I⁻ anion
Cd²⁺ cation
shifted electron cloud weaker I⁻–I⁻ repulsion
Each Cd²⁺ sits symmetrically among its I⁻, but each I⁻ has all its Cd²⁺ neighbours on one side.
The induced dipole pulls each anion’s negative charge toward the layer that contains the cations.
That reduces the repulsion between anions across the empty layer.
Here are two slabs of the CdI₂ structure, seen edge on.
Key: large blue I⁻ , small red Cd²⁺
Slab: I⁻ / Cd²⁺ / I⁻; between slabs an empty layer , no cations
Now look at one iodide next to the empty layer.
[advance: highlight one iodide and its cadmium neighbors]
Each cadmium sits symmetrically, with iodides above and below it, but each iodide has all its cadmium neighbors on one side.
Cd²⁺ symmetric ; I⁻ one-sided
Other side faces the empty layer : nothing pulls back
So the cations pull its electron cloud one way.
[advance: shift the iodide clouds toward the cadmium layers]
The induced dipole pulls the negative charge of each anion toward the layer that contains the cations.
Cloud moves toward its own Cd²⁺ layer ; nucleus stays
Drawn larger than life
That shift helps the layered structure in a second way.
[advance: mark the anion-anion repulsion across the gap]
We also reduce the repulsive interaction between anions across this empty layer.
Less charge faces the gap: weaker I⁻–I⁻ repulsion
Both effects stabilize the layered structure
With polarization in hand, let's come back one last time to our table of MX₂ structures.
Structures of MX₂ Halides
The same map once more, now read down the halide columns for polarizability; radii in pm.
Cationcrystal radius
F⁻119
Cl⁻167
Br⁻182
I⁻206
Mg²⁺86 rutile CdI₂/CdCl₂ CdI₂/CdCl₂ CdI₂/CdCl₂ Zn²⁺88 rutile tetrahedral tetrahedral tetrahedral Mn²⁺97 rutile CdI₂/CdCl₂ CdI₂/CdCl₂ CdI₂/CdCl₂ Ca²⁺114 fluorite rutile rutile CdI₂/CdCl₂ Sr²⁺132 fluorite fluorite PbCl₂ SrI₂ Ba²⁺149 fluorite PbCl₂ PbCl₂ PbCl₂
fluorite, Ca[8] F₂[4] A = 2.519 rutile, Mg[6] F₂[3] A = 2.408 Cd[6] I₂[3] A = 2.192
fluorite — CN 8
rutile — CN 6
CdI₂/CdCl₂ — layered, CN 6
PbCl₂ and SrI₂ — CN 7
tetrahedral — CN 4
Structure assignments and radii are the lecture’s own.
This time we read the table down the halide columns, for polarizability.
Larger, softer anion → stronger polarization
F⁻ hardly; Cl⁻ somewhat; Br⁻, I⁻ highly polarizable
Polarizable anion favors asymmetric cation arrangement
Heavy halides CdI₂/CdCl₂ ; fluorides of same cations stay rutile
Bottom render: anion on the outside of a sheet , cations one side
That completes the ionic bonding story, so let me summarize the second part.
Summary
From the Born–Mayer equation to anion polarization: what the second half of the lecture settled.
Born–Mayer equation Combines Coulomb attraction with a short-range repulsive term to approximate lattice energy. Worked 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 . Accuracy 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 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 . Polarization symmetry Symmetry of cation arrangement dictates whether polarization cancels (fluorite, rutile) or enhances stability (layered).
Let me summarize the second part, from the Born-Mayer equation to anion polarization.
Born-Mayer : Coulomb + short-range repulsion; KCl −679 vs. −718 Born-HaberShannon radii at the right CN give d₀Underestimates 3–6% ; dispersion, zero-point help slightly
Radius ratio : smaller cation, larger anion, CN 8 → 6 → 4MX₂: fluorite large cation; rutile smaller; CdI₂-type for polarizable Br⁻, I⁻
Polarization cancels in fluorite , nearly in rutile, adds up in layered
Before next class, work through the practice questions.
Practice questions
Work the Lecture 11 practice questions before the next class.
Open the Lecture 11 practice questions
Your homework is the Lecture 11 practice questions.
Course website, Practice ; slide link goes straight there
Madelung constant Born-Mayer and Born-Haber lattice energiesRadius ratios , anion polarizationWorked explanation on every question
Next time we review the first half of the course for the first examination, and after that we pick up bonding again with atomic orbitals and covalent bonding.