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