Here are the formulas for the energy of each interaction.
London Dispersion force
V = −3/4(α^2I)/r^6
where α is the polarizability, r is the distance, and I is the first ionization energy. The negative sign indicates the attractive interaction.
Dipole-Dipole Interaction
V = −(2/3)(μ_A^2μ_B^2)/((4πϵ_0)^2r^6)1/(k_BT)
where mu_i are the dipole moments, epsilon is the permittivity of the medium. k_B is Boltzmann's constant, and T is the Kelvin temperature.
Ion-Dipole interaction
V = −(qμ)/((4πϵ_0)r^2)
where q is the charge on the ion.
EXAMPLE
Calculate the dipole-dipole interaction energy at 298 K between two HF molecules. Assume that the positive and negative ends are separated by 400 pm. ϵ₀ = 8.854 × 10⁻¹² C²N⁻¹m⁻²; µ = 1.92 D; k_B = 1.381 × 10⁻²³ J•K⁻¹
Solution
µ = 1.92 D × (3.336 × 10⁻³⁰" C·m")/(1" D") = 6.41 × 10⁻³⁰ C•m
V = −(2/3)((μ_A^2μ_B^2)/(4πϵ_0)^2r^6)1/(k_BT)
V = −(2/3)((6.41 × 10^-30" C·m")^4)/((4π × 8.854 × 10^-12" C²N⁻¹m⁻²")^2(400 × 10⁻¹⁰" m")^6) × 1/(1.381 × 10⁻²³ J·K⁻¹ × 298" K") = 5.37 × 10⁻²¹ J
On a molar basis,
V = 5.37 × 10⁻²¹ J × 6.022 × 10²³ mol⁻¹ = 3240 J/mol = 3.24 kJ/mol
