University Physics IV · Molecular and Solid-State Physics · 12.2

Ionic, Metallic & van der Waals Bonding

Covalency is not the only way to build a solid. Price the electron transfer, sum a conditionally convergent lattice series against a repulsive core, and put metallic delocalisation and R⁻⁶ dispersion on the same electronvolt scale — then say which approximation each number rests on.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Once the nuclei are held fixed, a crystal's cohesive energy becomes a bookkeeping problem: what does it cost to prepare the charges, and what does the lattice pay back? In an ionic solid the preparation is never free — stripping sodium costs 5.14 eV and chlorine returns only 3.61 eV, a net debt of 1.53 eV — so all of the binding comes from the electrostatics afterwards, summed over the whole lattice rather than over one pair. That sum is the Madelung constant, a pure number fixed by geometry, and it is only conditionally convergent: taken shell by shell it oscillates and settles nowhere, which is why neutral groupings or Ewald summation are needed to reach 1.7476 for rocksalt.

Balanced against a Born core B/Rⁿ whose exponent is fitted rather than derived, it returns 7.81 eV per NaCl pair against 8.2 eV measured. Run the same accounting on argon, where nothing transfers and the only attraction is the instantaneous-dipole R⁻⁶ term, and you get 0.0895 eV per atom — some ninety times weaker. Metallic binding refuses the accounting altogether: delocalised valence electrons make the energy per atom scale as √Z rather than Z, so no sum over pairs reproduces it.

The whole scheme assumes point ions and pairwise additivity, and each place those fail — polarisable ions, residual covalency, three-body dispersion, genuinely many-body metallic cohesion — is where the last few percent lives.

Simple definition: Ionic, metallic and van der Waals bonding are the three non-covalent ways a solid holds together: transferred charge summed over a whole lattice, valence electrons delocalised across the crystal, and fluctuating dipoles attracting as R⁻⁶.
Example: NaCl binds at 8.2 eV per ion pair, sodium metal at 1.11 eV per atom, and solid argon at 0.080 eV per atom — the same three mechanisms spanning a factor of a hundred in strength.

Cost of making the ionsΔE = I − AR_c = e²/(4πε₀ ΔE)

Na→Cl costs +1.53 eV, repaid only inside R_c = 9.4 Å — charge transfer is a debt the lattice pays.

I ionisation energy, A electron affinity, both in eV; e²/4πε₀ = 14.400 eV⋅Å

Madelung constantα = Σⱼ (±1)/pⱼ, with Rⱼ = pⱼ R

1.7476 rocksalt · 1.7627 CsCl · 1.6381 zincblende · 1.3863 for a one-dimensional alternating chain.

pⱼ is the jth neighbour's distance in units of the nearest-neighbour distance R; the sign follows the charge product

Born–Landé lattice energyU(R₀) = −(α e²/4πε₀R₀)(1 − 1/n)

NaCl: 1.7476 × 14.400/2.82 = 8.92 eV, times 0.875, gives 7.81 eV per pair against 8.2 eV measured.

R₀ nearest-neighbour distance in Å; n is the Born exponent, 6 to 12, fitted to the compressibility

Dispersion in a rare-gas crystalu/N = 2ε[A₁₂(σ/R)¹² − A₆(σ/R)⁶]

Minimum at R₀ = 1.09σ = 3.71 Å, depth 89.5 meV per atom — about one ninetieth of NaCl's binding.

ε well depth (Ar 10.4 meV), σ the zero-crossing (Ar 3.40 Å); fcc lattice sums A₁₂ = 12.132, A₆ = 14.454

Metallic cohesion is not pairwiseE_coh ∝ −√Z |β|, not −Z |β|

A pair potential fitted to the dimer over-binds an fcc metal by √12 ≈ 3.5 — no sum over pairs will do.

Z the coordination number (12 in fcc and hcp, 8 in bcc); β the nearest-neighbour hopping integral, in eV

Where R₀ comes fromnλ = 2d sinθdₕₖₗ = a/√(h² + k² + l²)

Cu Kα on NaCl (200): d = 2.82 Å puts the peak at 2θ = 31.7°, and that is the R₀ every card above needs.

a the conventional cube edge (NaCl 5.64 Å), hkl the Miller indices, λ the X-ray wavelength in Å

Charge transfer is a debt, not a payment

Before any lattice exists, ask what an ion pair costs. Removing sodium's 3s electron takes its first ionisation energy, I = 5.14 eV; attaching it to chlorine returns the electron affinity, A = 3.61 eV. The net is I − A = +1.53 eV — a debt. Only the Coulomb attraction repays it, and with e²/4πε₀ = 14.400 eV⋅Å the repayment is 14.40/R eV, which reaches 1.53 eV at R_c = 9.4 Å. Outside that radius the neutral pair Na + Cl is the lower state, which is exactly why gas-phase NaCl dissociates into neutral atoms rather than ions. The same arithmetic sets the halide trend: caesium's I = 3.89 eV against fluorine's A = 3.40 eV leaves a debt of only 0.49 eV, while LiI must find 5.39 − 3.06 = 2.33 eV. And the transfer is never complete — overlap of the two charge clouds leaves residual covalency, which Pauling's ionicity scale puts at 0.94 for NaCl and only 0.62 for ZnS.

The Madelung sum is pure geometry, and it barely converges

Write every neighbour's distance as Rⱼ = pⱼR, with R the nearest-neighbour separation. The electrostatic energy of one ion is then −(e²/4πε₀R)⋅Σⱼ(±1)/pⱼ, and that dimensionless sum is the Madelung constant α: geometry alone, identical for every rocksalt whatever the chemistry. Summing it in spherical shells is a trap. Rocksalt gives 6 − 12/√2 + 8/√3 − 6/2 + 24/√5 …, that is 6 − 8.485 + 4.619 − 3.000 + 10.733, whose partial sums run 6, −2.49, 2.13, −0.87, 9.87 — oscillating with growing amplitude, because a shell at pⱼ holds of order p² ions against a 1/p term. The series is only conditionally convergent, so rearranged it can be made to give anything at all. Evjen's fix is to sum over neutral cubes carrying no net charge and no dipole; Ewald's splits the sum into a fast real-space part and a fast reciprocal-space part. Both give 1.7476 for rocksalt, 1.7627 for caesium chloride, 1.6381 for zincblende — within 7% of one another, so α is not what selects a structure. The radius ratio and the repulsive core are.

Balance against the Born core and read off a number

Point charges alone collapse: −1/R has no minimum. The stop comes from Pauli exclusion as the closed shells overlap, modelled either as a power law B/Rⁿ or as a Born–Mayer exponential λe^(−R/ρ). Take the power law: U(R) = −αe²/4πε₀R + B/Rⁿ, and set dU/dR = 0 to eliminate the unknown B. What survives is U(R₀) = −(αe²/4πε₀R₀)(1 − 1/n): the core returns exactly 1/n of the Coulomb term and nothing more, whatever B happened to be. For NaCl with α = 1.7476, R₀ = 2.82 Å and n = 8, the Coulomb term is 1.7476 × 14.400/2.82 = 8.92 eV, the core hands back 1.12 eV, and the lattice energy is 7.81 eV per ion pair against 8.2 eV measured. That 4% is not rounding: n is fitted to the measured compressibility rather than derived, dispersion between the two polarisable ions is missing, and zero-point vibration is ignored. Against neutral atoms, subtract the 1.53 eV debt to get 6.28 eV — above H₂'s covalent 4.75 eV, which is what it means to say ionic cohesion rivals covalent.

Metallic binding is non-directional and irreducibly many-body

Take the same lattice and let the valence electrons leave their parent ions entirely. The binding now points nowhere: there is no bond axis to break, so metals adopt the highest coordination available — 12 in fcc and hcp, 8 in bcc — and slip planes glide without severing directed bonds, which is why metals are ductile where ionic and covalent crystals cleave. The energy scale runs from 1.11 eV per atom in sodium to 8.90 eV in tungsten, tracking the number of unfilled d states rather than any transferred charge. And the sum will not factorise. In a second-moment tight-binding estimate the cohesive energy per atom goes as −√Z rather than −Z, because the band width grows as the square root of the number of hopping paths. A pair potential fitted to the diatomic molecule therefore over-binds an fcc metal by √12 ≈ 3.5. Ionic and dispersion energies are sums over pairs; metallic energy is not one, and that is precisely why embedded-atom and tight-binding potentials had to be invented.

Why dispersion goes as R⁻⁶, and how weak that makes it

Neutral closed-shell atoms with no permanent moment still attract. An instantaneous fluctuation gives atom 1 a dipole p₁; the field it makes at distance R falls as p₁/R³; that field induces p₂ ∝ αp₁/R³ in atom 2; and the interaction −p₁⋅E₂ then goes as −αp₁²/R⁶. Two powers of R⁻³ multiplied together — that is the whole origin of the London R⁻⁶ law, and because the fluctuations are quantum it survives at T = 0. Wrap it in the Lennard-Jones form U(R) = 4ε[(σ/R)¹² − (σ/R)⁶], where the R⁻¹² is chosen for arithmetic convenience and has no physical basis; real overlap repulsion is exponential. For argon, ε = 10.4 meV and σ = 3.40 Å. Summing over the fcc lattice with A₁₂ = 12.132 and A₆ = 14.454 puts the minimum at R₀ = 1.09σ = 3.71 Å with 89.5 meV per atom, against 3.76 Å and 80 meV measured — zero-point motion of these light, weakly bound atoms accounts for most of that gap. Set 0.0895 eV beside NaCl's 8.2 eV per pair and dispersion is about ninety times weaker.

Lattice, basis, and where R₀ actually comes from

Every number above needed R₀, and R₀ is measured, not assumed. A crystal is a Bravais lattice — the set of translations leaving it invariant — plus a basis, the atoms attached to each lattice point. NaCl is not a simple cubic array of alternating ions; it is an fcc lattice with a two-atom basis, Na⁺ at (0,0,0) and Cl⁻ at (½,0,0)a, so the primitive cell holds exactly one ion pair while the conventional cube of side a = 5.64 Å holds four. The nearest-neighbour distance is a/2 = 2.82 Å. Bragg's law nλ = 2d sinθ with dₕₖₗ = a/√(h²+k²+l²) then turns angles into that spacing: Cu Kα at λ = 1.5418 Å reflects from the (200) planes, d = 2.82 Å, at sinθ = 1.5418/5.64 = 0.2734, so θ = 15.9° and the peak stands at 2θ = 31.7°. The division of labour is worth naming: the lattice fixes where the peaks sit, the basis fixes how bright they are. NaCl's (111) reflection is weak because its structure factor goes as f_Na − f_Cl, and in KCl, where K⁺ and Cl⁻ are isoelectronic, it very nearly vanishes.

Change one variable at a time

Make the relationship visible.

Interactive model

Madelung constant α1.75

Nearest-neighbour distance R₀2.82 Å

Born exponent n8

Set α = 1.75, R₀ = 2.82 Å and n = 8 for rocksalt NaCl and read 7.82 eV. Now drag n from 6 to 12: the well deepens by only 0.75 eV, because the core ever only returns 1/n. Shrink R₀ to 2.20 Å instead and it deepens by 2.2 eV — bond length, not core hardness, buys cohesion.

COULOMB αe²/4πε₀R₀8.94 eV

CORE RETURNS 1/n OF IT1.12 eV

LATTICE ENERGY PER PAIR7.82 eV

LESS THE 1.53 eV DEBT6.29 eV

Live interpretationCOULOMB αe²/4πε₀R₀: 8.94 eV. CORE RETURNS 1/n OF IT: 1.12 eV. LATTICE ENERGY PER PAIR: 7.82 eV. LESS THE 1.53 eV DEBT: 6.29 eV

Catch the common trap

Explain before calculating.

Rocksalt NaCl has six nearest neighbours at R₀ = 2.82 Å. A student estimates the lattice energy from those six ions alone, obtaining 6 × 14.400/2.82 = 30.6 eV per ion pair, and is startled that the measured value is 8.2 eV. What is the correct diagnosis?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasySodium's first ionisation energy is 5.14 eV and chlorine's electron affinity is 3.61 eV. Taking e²/4πε₀ = 14.40 eV⋅Å, find the separation at which an Na⁺Cl⁻ pair first becomes lower in energy than the neutral atoms, then estimate the pair's binding at the observed gas-phase bond length of 2.36 Å.

At infinite separation the transfer costs ΔE = I − A = 5.14 − 3.61 = 1.53 eV. Making the ions takes energy in; nothing has been gained yet.
Bringing the two ions to separation R returns Coulomb energy −e²/4πε₀R = −14.40/R eV, with R in ångström.
The ionic curve drops below the neutral one when 14.40/R = 1.53, so R_c = 14.40/1.53 = 9.41 Å. Outside that radius Na + Cl is the lower state, which is why gas-phase NaCl dissociates into neutral atoms rather than ions.
At the observed bond length, Coulomb gives 14.40/2.36 = 6.10 eV, leaving 6.10 − 1.53 = 4.57 eV once the debt is paid — above the measured 4.3 eV, because the overlapping closed shells have not yet been charged for.

AnswerR_c = 9.4 Å; inside it the ionic pair wins. The point-charge estimate of 4.57 eV overshoots the measured 4.3 eV by roughly the cost of the Born repulsive core.

MediumRocksalt NaCl has α = 1.7476, nearest-neighbour distance R₀ = 2.82 Å and Born exponent n = 8. With e²/4πε₀ = 14.400 eV⋅Å, find the lattice energy per ion pair relative to free ions and relative to free neutral atoms, and say why it falls short of the measured 8.2 eV.

Coulomb term: αe²/4πε₀R₀ = 1.7476 × 14.400/2.82 = 1.7476 × 5.1064 = 8.92 eV per ion pair.
Setting dU/dR = 0 eliminates the unknown B and leaves the core returning exactly 1/n of that: 8.92/8 = 1.12 eV.
Lattice energy from free ions: U₀ = 8.92 × (1 − 1/8) = 8.92 × 0.875 = 7.81 eV per pair.
Against the measured 8.2 eV that is about 4% low. The exponent n is fitted to the compressibility rather than derived, dispersion between the two polarisable ions is missing, and zero-point vibration is ignored.
Relative to neutral atoms, subtract the 1.53 eV transfer debt: 7.81 − 1.53 = 6.28 eV per pair — still well above the 4.75 eV covalent bond of H₂.

Answer7.81 eV per ion pair measured from free ions, 6.28 eV from free neutral atoms; the 4% shortfall against 8.2 eV is dispersion, zero-point motion, and a Born exponent that was fitted, not derived.

HardSolid argon is fcc with Lennard-Jones parameters ε = 10.4 meV and σ = 3.40 Å, and lattice sums A₁₂ = 12.132 and A₆ = 14.454. Find the equilibrium nearest-neighbour distance and the cohesive energy per atom, compare with the measured 3.76 Å and 80 meV, and set the result beside NaCl's 8.2 eV per ion pair.

Energy per atom is u(R) = 2ε[A₁₂(σ/R)¹² − A₆(σ/R)⁶]; the factor 2 rather than 4 stops every pair being counted twice.
Set du/dR = 0: 12A₁₂σ¹²/R¹³ = 6A₆σ⁶/R⁷, so (R₀/σ)⁶ = 2A₁₂/A₆ = 24.264/14.454 = 1.6787, giving R₀/σ = 1.6787¹⁄⁶ = 1.090 and R₀ = 1.090 × 3.40 = 3.71 Å.
At that minimum (σ/R₀)⁶ = 1/1.6787 = 0.5957, so u₀ = −εA₆²/(2A₁₂) = −10.4 × 208.92/24.264 = −10.4 × 8.610 = −89.5 meV per atom.
Both are close but not exact: 3.71 Å against 3.76 Å, and 89.5 meV against 80 meV. Argon is light and weakly bound, so zero-point motion pushes the lattice apart and shallows the well — a purely classical minimum cannot see it.
On the same scale NaCl gives 8.2 eV per ion pair, so dispersion binds argon 8.2/0.0895 = 92 times more weakly. Hence argon is solid only below 84 K while rocksalt melts at 1074 K.

AnswerR₀ = 3.71 Å and 89.5 meV per atom, against 3.76 Å and 80 meV measured; dispersion is about 92 times weaker than the ionic Coulomb lattice.