Physics / Condensed Matter Solid State Physics I 100% Free Open Access
Chapter 2 โ€ข Theory & Derivations

Interatomic Forces & Crystal Bonding

Exhaustive treatment of cohesive energy and binding mechanisms: ionic bonding, exact derivation of the Madelung constant, Born-Mayer repulsive potential, bulk modulus, covalent bonding, Van der Waals dispersion forces, Lennard-Jones 6-12 potential, and metallic cohesion.

ยง2.1 Classification of Chemical Bonds & Cohesive Energy

1. Cohesive Energy of Crystalline Solids

The cohesive energy (or binding energy) $E_{\text{coh}}$ of a crystal is defined as the net energy required to disassemble the solid into its constituent neutral free atoms (or ions in ionic physics) at infinite mutual separation at zero temperature ($T = 0\text{ K}$):

$$E_{\text{coh}} = E_{\text{isolated atoms}} - E_{\text{crystal}} > 0$$

Cohesion originates entirely from electromagnetic interactions governed by quantum mechanics. Solids are classified into five primary bonding categories based on the distribution of their valence electrons:

  1. Ionic Crystals (e.g., $\text{NaCl}, \text{CsCl}, \text{LiF}$): Electrostatic attraction between positive cations and negative anions formed by complete valence electron transfer. Highly localized electron density, strong binding ($E_{\text{coh}} \approx 5 - 10\text{ eV/atom}$), high melting points, electrical insulators in the solid state.
  2. Covalent Crystals (e.g., Diamond, $\text{Si}, \text{Ge}, \text{GaAs}$): Cohesion via quantum mechanical sharing of electron pairs localized along directional bonds formed by overlapping hybridized atomic orbitals ($sp^3$). Extreme hardness, high melting points, semiconductor or insulator band structure.
  3. Metallic Crystals (e.g., $\text{Na}, \text{Cu}, \text{Fe}, \text{Al}$): Valence electrons delocalize completely into an itinerant electron gas ("Fermi sea") permeating an array of positively charged ion cores. Non-directional bonding, high electrical and thermal conductivities, ductility.
  4. Van der Waals (Molecular) Crystals (e.g., Solid $\text{Ar}, \text{Kr}, \text{CH}_4$): Weak cohesion ($E_{\text{coh}} \approx 0.05 - 0.2\text{ eV/atom}$) arising from fluctuating quantum dipole-induced dipole interactions. Low melting and boiling points, soft mechanical properties.
  5. Hydrogen-Bonded Crystals (e.g., Ice $\text{H}_2\text{O}, \text{HF}$, nucleic acids): Directional electrostatic dipole attractions mediated by an electropositive bare proton sandwiched between electronegative lone pairs (O, N, F). Intermediate strength ($0.1 - 0.5\text{ eV/bond}$).

ยง2.2 Ionic Crystals: Born-Mayer Potential & Madelung Constant

1. Electrostatic Madelung Energy Formalism

Consider an ionic crystal consisting of $2N$ ions ($N$ positive cations and $N$ negative anions with charges $\pm q$). Let $R$ be the nearest-neighbor interionic separation distance. The total electrostatic Coulomb potential energy of ion $i$ interacting with all other ions $j \neq i$ in the crystal is:

$$U_{i,\text{coul}} = \sum_{j \neq i} \frac{(\pm q)(\mp q)}{4\pi\varepsilon_0 r_{ij}} = - \frac{q^2}{4\pi\varepsilon_0 R} \sum_{j \neq i} \frac{\pm 1}{p_{ij}}$$

where $r_{ij} = p_{ij} R$ denotes the distance between ions $i$ and $j$ measured in units of nearest-neighbor distance $R$, and the sign is positive for unlike charges (attraction) and negative for like charges (repulsion).

The dimensionless sum is defined as the Madelung Constant $\alpha$:

$$\alpha = \sum_{j \neq i} \frac{\pm 1}{p_{ij}}$$

Multiplying by $N$ ion pairs (to avoid double-counting the pair interactions), the total electrostatic Madelung energy of the crystal is:

$$U_{\text{Madelung}}(R) = - N \alpha \frac{q^2}{4\pi\varepsilon_0 R}$$

2. Calculation of the Madelung Constant for a One-Dimensional Ionic Chain

Consider an infinite 1D chain of alternating cations and anions with nearest-neighbor distance $R$. Choosing an arbitrary positive ion as the origin, its neighbors are situated at distances $R, 2R, 3R, \dots$ to both the left and right:

  • Two opposite ions at distance $1R$: contribution $+2 \times (1/1)$
  • Two identical ions at distance $2R$: contribution $-2 \times (1/2)$
  • Two opposite ions at distance $3R$: contribution $+2 \times (1/3)$

The 1D Madelung constant $\alpha_{\text{1D}}$ is therefore given by the alternating harmonic series:

$$\alpha_{\text{1D}} = 2 \left( 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \dots \right)$$

Recalling the Taylor series expansion of $\ln(1+x) = x - x^2/2 + x^3/3 - x^4/4 + \dots$ evaluated at $x = 1$:

$$\sum_{m=1}^{\infty} \frac{(-1)^{m-1}}{m} = \ln(2) \implies \alpha_{\text{1D}} = 2 \ln(2) \approx 2 \times 0.69315 = 1.38629$$

3. Madelung Constants for 3D Ionic Lattices

In three dimensions, the summation is conditionally convergent and requires specialized summing techniques (e.g., the Evjen method of neutral concentric polyhedra or the Ewald summation method in reciprocal space):

  • Rock-Salt ($\text{NaCl}$): $\alpha = 1.747565$
  • Cesium Chloride ($\text{CsCl}$): $\alpha = 1.762675$
  • Zincblende ($\text{ZnS}$): $\alpha = 1.63805$
  • Wurtzite ($\text{ZnS}$): $\alpha = 1.64132$
  • Fluorite ($\text{CaF}_2$): $\alpha = 5.03878$

ยง2.3 Cohesive Energy & Bulk Modulus of Ionic Crystals

1. Total Cohesive Energy with Born-Mayer Repulsion

At short interionic separations, the electron clouds of adjacent ions overlap. By the Pauli Exclusion Principle, overlapping closed electron shells must occupy higher unpopulated quantum states, generating a steep quantum mechanical repulsive potential. The total potential energy per ion pair $U_{\text{tot}}(R)$ is modeled by the Born-Mayer equation:

$$U_{\text{tot}}(R) = - \frac{\alpha q^2}{4\pi\varepsilon_0 R} + z B e^{-R/\rho}$$

where $z$ is the coordination number (number of nearest neighbors), $B$ is a repulsive strength constant, and $\rho \approx 0.33\text{ \AA}$ is the characteristic range parameter of core repulsion.

2. Equilibrium Interionic Separation $R_0$

At equilibrium ($R = R_0$), the net mechanical force vanishes: $\left.\frac{dU_{\text{tot}}}{dR}\right|_{R=R_0} = 0$:

$$\left.\frac{dU_{\text{tot}}}{dR}\right|_{R_0} = \frac{\alpha q^2}{4\pi\varepsilon_0 R_0^2} - \frac{z B}{\rho} e^{-R_0/\rho} = 0 \implies z B e^{-R_0/\rho} = \frac{\alpha q^2}{4\pi\varepsilon_0 R_0} \left(\frac{\rho}{R_0}\right)$$

Substituting this repulsive term back into $U_{\text{tot}}(R_0)$ yields the celebrated Born-Mayer cohesive energy formula per ion pair:

$$U_{\text{tot}}(R_0) = - \frac{\alpha q^2}{4\pi\varepsilon_0 R_0} \left( 1 - \frac{\rho}{R_0} \right)$$

Since $\rho / R_0 \approx 0.33\text{ \AA} / 2.82\text{ \AA} \approx 0.12$, core repulsion reduces the pure electrostatic Coulomb binding energy by approximately $10 - 15\%$.

3. Derivation of Bulk Modulus $B$

The isothermal bulk modulus $B$ measures the resistance of the crystal to uniform hydrostatic compression: $B = - V \frac{dP}{dV}$. At $T = 0\text{ K}$, the hydrostatic pressure is $P = - \frac{dU_{\text{cryst}}}{dV}$, which implies:

$$B = V \frac{d^2 U_{\text{cryst}}}{dV^2} = \left. V_0 \frac{d^2 U_{\text{tot}}}{dV^2} \right|_{V=V_0}$$

For a rock-salt cubic crystal with volume per ion pair $V = 2 R^3$ (where $V_0 = 2 R_0^3$), changing variables via $\frac{dU}{dV} = \frac{dU/dR}{dV/dR} = \frac{1}{6R^2} \frac{dU}{dR}$ leads directly to:

$$B = \frac{1}{18 R_0} \left. \frac{d^2 U_{\text{tot}}}{dR^2} \right|_{R=R_0} = \frac{\alpha q^2}{72 \pi \varepsilon_0 R_0^4} \left( \frac{R_0}{\rho} - 2 \right)$$

ยง2.4 Van der Waals Crystals & Lennard-Jones (6-12) Potential

1. Origin of London Dispersion Forces

In inert gas crystals (solid $\text{Ne}, \text{Ar}, \text{Kr}, \text{Xe}$), atoms possess completely closed electronic shells with spherical symmetry and zero permanent dipole moment ($\langle \vec{p} \rangle = 0$). However, quantum fluctuations in electron cloud positions produce an instantaneous fluctuating electric dipole $\vec{p}_1(t) \sim e \vec{x}(t)$. This instantaneous dipole generates an electric field at distance $R$:

$$E_1 \approx \frac{p_1}{4\pi\varepsilon_0 R^3}$$

This electric field induces a dipole moment in the adjacent neutral atom proportional to its electronic polarizability $\alpha_{\text{pol}}$: $\vec{p}_2 = \alpha_{\text{pol}} \vec{E}_1 \propto \frac{\alpha_{\text{pol}} p_1}{R^3}$. The resulting dipole-dipole electrostatic interaction energy is attractive and scales inversely with the sixth power of distance:

$$U_{\text{disp}}(R) = - \vec{p}_2 \cdot \vec{E}_1 \propto - \frac{\alpha_{\text{pol}} p_1^2}{R^6} = - \frac{C}{R^6}$$

2. The Lennard-Jones (6-12) Potential

When two inert gas atoms approach within the distance of their closed electron shells, Pauli core exclusion generates a steep repulsive potential, empirically parameterized as an inverse-twelfth power $1/R^{12}$. The total interaction potential between a pair of atoms separated by distance $R$ is the Lennard-Jones (6-12) potential:

$$U_{LJ}(R) = 4\varepsilon \left[ \left(\frac{\sigma}{R}\right)^{12} - \left(\frac{\sigma}{R}\right)^6 \right]$$

where:

  • $\varepsilon$: Depth of the potential well (binding energy minimum).
  • $\sigma$: Finite distance at which the interatomic potential is zero ($U_{LJ}(\sigma) = 0$).

The minimum of the two-body potential occurs where $\frac{dU_{LJ}}{dR} = 0$:

$$\frac{dU_{LJ}}{dR} = 4\varepsilon \left[ -\frac{12\sigma^{12}}{R^{13}} + \frac{6\sigma^6}{R^7} \right] = 0 \implies R_{\text{min}} = 2^{1/6}\sigma \approx 1.122\sigma$$

At this equilibrium separation, the potential energy minimum is exactly $U_{LJ}(R_{\text{min}}) = -\varepsilon$.

3. Cohesive Energy of Inert Gas FCC Crystals

In an FCC crystal of $N$ inert gas atoms, summing over all pairs with $R_{ij} = p_{ij} R$ yields the total crystal potential energy:

$$U_{\text{tot}}(R) = 2 N \varepsilon \left[ A_{12} \left(\frac{\sigma}{R}\right)^{12} - A_6 \left(\frac{\sigma}{R}\right)^6 \right]$$

For an FCC lattice, the lattice sums over all neighbors are geometric constants:

$$A_{12} = \sum_{j \neq 0} p_{0j}^{-12} \approx 12.13188, \quad A_6 = \sum_{j \neq 0} p_{0j}^{-6} \approx 14.45392$$

Minimizing $U_{\text{tot}}(R)$ with respect to $R$ yields the equilibrium crystal nearest-neighbor separation $R_0$ and the total cohesive energy per atom:

$$R_0 = \left( \frac{2 A_{12}}{A_6} \right)^{1/6} \sigma \approx 1.09 \sigma, \quad \frac{U_{\text{tot}}(R_0)}{N} = - 2.15 (4\varepsilon) \approx - 8.61 \varepsilon$$
Solved Problem Example 2.1: Cohesive Energy and Repulsive Exponent of Sodium Chloride (NaCl)

The experimental equilibrium nearest-neighbor distance in sodium chloride ($\text{NaCl}$) is $R_0 = 2.820 \text{ ร…}$, and its experimental cohesive energy is $E_{\text{coh}} = 7.95 \text{ eV/molecule}$ ($767 \text{ kJ/mol}$). The Madelung constant for the rock-salt structure is $\alpha = 1.7476$. Modeling the repulsive potential using the Born power-law form $U_{\text{rep}}(R) = B / R^n$: (a) Derive the expression for cohesive energy in terms of $n$. (b) Calculate the numerical value of the repulsive exponent $n$.

Step 1: Formulate the Total Potential Energy Function
$$U(R) = - \frac{\alpha e^2}{4\pi\varepsilon_0 R} + \frac{B}{R^n}$$

Write down the potential energy per NaCl molecule combining the attractive Coulomb-Madelung term and the Born power-law repulsive term.

Step 2: Apply the Equilibrium Condition at R = R_0
$$\left.\frac{dU}{dR}\right|_{R_0} = \frac{\alpha e^2}{4\pi\varepsilon_0 R_0^2} - \frac{n B}{R_0^{n+1}} = 0 \implies \frac{B}{R_0^n} = \frac{\alpha e^2}{4\pi\varepsilon_0 R_0} \frac{1}{n}$$

Set the first derivative of potential energy with respect to interionic distance to zero at R = R_0.

Step 3: Substitute B back into Cohesive Energy
$$E_{\text{coh}} = - U(R_0) = \frac{\alpha e^2}{4\pi\varepsilon_0 R_0} \left( 1 - \frac{1}{n} \right)$$

Substitute the repulsive term B / R_0^n back into the expression for U(R_0).

Step 4: Numerically Calculate the Pure Electrostatic Energy
$$U_M = \frac{\alpha e^2}{4\pi\varepsilon_0 R_0} = \frac{1.7476 \times 1.440\text{ eV}\cdot\text{\AA}}{2.820\text{ \AA}} = \frac{2.5165}{2.820}\text{ eV} \approx 8.924\text{ eV}$$

Evaluate the electrostatic Coulomb energy using e^2 / (4 pi epsilon_0) = 1.440 eV * Angstrom.

Step 5: Solve for the Repulsive Exponent n
$$1 - \frac{1}{n} = \frac{E_{\text{coh}}}{U_M} = \frac{7.95\text{ eV}}{8.924\text{ eV}} \approx 0.8909 \implies \frac{1}{n} = 1 - 0.8909 = 0.1091 \implies n = \frac{1}{0.1091} \approx 9.16$$

Equate the theoretical formula to the experimental cohesive energy to extract the Born exponent n.

Final Answer & Physical Insight

n \approx 9.16 \quad (\text{Consistent with the canonical closed-shell } \text{Na}^+\text{-}\text{Cl}^- \text{ value of } n \approx 9)

Solved Problem Example 2.2: Bulk Modulus and Compressibility of Potassium Chloride (KCl)

Potassium chloride ($\text{KCl}$) crystallizes in the rock-salt structure with equilibrium nearest-neighbor interionic spacing $R_0 = 3.147 \text{ ร…}$, Madelung constant $\alpha = 1.7476$, and Born-Mayer range parameter $\rho = 0.330 \text{ ร…}$. (a) Calculate the theoretical isothermal bulk modulus $B$ of $\text{KCl}$ at $T = 0 \text{ K}$. (b) Determine the compressibility $\beta = 1/B$ in units of $\text{GPa}^{-1}$.

Step 1: State the Formula for Bulk Modulus of Rock-Salt Crystals
$$B = \frac{\alpha e^2}{72 \pi \varepsilon_0 R_0^4} \left( \frac{R_0}{\rho} - 2 \right)$$

Use the bulk modulus formula derived from the second derivative of the Born-Mayer crystal potential.

Step 2: Evaluate the Dimensionless Factor (R_0 / rho - 2)
$$\frac{R_0}{\rho} = \frac{3.147\text{ \AA}}{0.330\text{ \AA}} \approx 9.5364 \implies \left(\frac{R_0}{\rho} - 2\right) = 7.5364$$

Compute the repulsive curvature term.

Step 3: Evaluate the Prefactor in SI Units
$$\frac{\alpha e^2}{4\pi\varepsilon_0} = 1.7476 \times (2.307 \times 10^{-28}\text{ J}\cdot\text{m}) \approx 4.0318 \times 10^{-28}\text{ J}\cdot\text{m}$$

Calculate the numerator product in SI units.

Step 4: Compute the Bulk Modulus B
$$B = \frac{4.0318 \times 10^{-28}\text{ J}\cdot\text{m} \times 7.5364}{18 \times (3.147 \times 10^{-10}\text{ m})^4} = \frac{3.0385 \times 10^{-27}}{18 \times 9.808 \times 10^{-38}} = \frac{3.0385 \times 10^{-27}}{1.7654 \times 10^{-36}} \approx 1.721 \times 10^{10}\text{ Pa} = 17.21\text{ GPa}$$

Divide through by 18 R_0^4 to find the bulk modulus in Pascals (N/m^2).

Step 5: Calculate the Compressibility beta
$$\beta = \frac{1}{B} = \frac{1}{17.21\text{ GPa}} \approx 0.0581\text{ GPa}^{-1} = 5.81 \times 10^{-11}\text{ Pa}^{-1}$$

Take the reciprocal of bulk modulus to find compressibility beta.

Final Answer & Physical Insight

B = 17.21 \text{ GPa}, \quad \beta = 0.0581 \text{ GPa}^{-1} = 5.81 \times 10^{-11} \text{ Pa}^{-1}

Solved Problem Example 2.3: Cohesive Energy and Nearest-Neighbor Distance of Solid Argon

Solid Argon crystallizes in an FCC lattice governed by the Lennard-Jones (6-12) potential with parameters $\varepsilon = 10.42 \text{ meV} = 1.670 \times 10^{-21} \text{ J}$ and $\sigma = 3.40 \text{ ร…}$. The FCC lattice sums are $A_{12} = 12.1319$ and $A_6 = 14.4539$. (a) Determine the theoretical nearest-neighbor separation $R_0$ and the conventional cubic lattice parameter $a$. (b) Calculate the total cohesive energy per mole of solid Argon in $\text{kJ/mol}$.

Step 1: Calculate the Equilibrium Nearest-Neighbor Separation R_0
$$R_0 = \left( \frac{2 A_{12}}{A_6} \right)^{1/6} \sigma = \left( \frac{2 \times 12.1319}{14.4539} \right)^{1/6} (3.40\text{ \AA}) = (1.6787)^{0.16667} (3.40\text{ \AA}) \approx 1.0906 \times 3.40\text{ \AA} \approx 3.708\text{ \AA}$$

Compute the equilibrium bond distance using the ratio of the lattice sums.

Step 2: Relate Nearest-Neighbor Distance to FCC Lattice Parameter a
$$R_0 = \frac{a}{\sqrt{2}} \implies a = \sqrt{2} R_0 = \sqrt{2} \times 3.708\text{ \AA} \approx 5.244\text{ \AA}$$

In an FCC lattice, nearest neighbors touch along face diagonals of length sqrt(2) * a.

Step 3: Calculate the Cohesive Energy per Atom
$$u_0 = - \frac{U_{\text{tot}}(R_0)}{N} = 2.15 \times 4\varepsilon = 8.60 \times (10.42\text{ meV}) \approx 89.61\text{ meV/atom} = 1.436 \times 10^{-20}\text{ J/atom}$$

Evaluate the binding energy per atom at equilibrium.

Step 4: Convert Cohesive Energy to Molar Basis
$$E_{\text{coh, molar}} = u_0 \times N_A = (1.436 \times 10^{-20}\text{ J}) \times (6.022 \times 10^{23}\text{ mol}^{-1}) \approx 8647\text{ J/mol} = 8.65\text{ kJ/mol}$$

Multiply by Avogadro's number N_A to obtain the cohesive energy per mole.

Final Answer & Physical Insight

R_0 = 3.708 \text{ ร…}, \quad a = 5.244 \text{ ร…}, \quad E_{\text{coh}} = 8.65 \text{ kJ/mol} \quad (89.61 \text{ meV/atom})

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