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# Intermolecular Forces of Attraction

## Key Questions

• 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