Is there a difference in the dipole moment of a gas molecule in isolation compared to being dissolved in a solvent?
1 Answer
Yes, there is a difference, but only for the determination, i.e. the electric field generated by the solvent's dipole imparts induced dipole moments on nearby nuclei, making the total electric dipole moment deviate away from the pure gas electric dipole moment.
The true dipole moment (i.e. without external deviations) will not change.
This brief reading explains this fairly well, albeit being a bit old.
This influence is explicitly seen in the equations for the molar polarizability,
Isolated gas molecule (dilute gas)
#M/rho (epsilon - 1)/3 = (4piN_0)/3(alpha + mu^2/(3k_BT))# where:
#M# is the molar mass of the gas.#rho# is the density of the gas.#epsilon# is the dielectric constant. Some values can be seen here.#N_0# is Avogadro's number.#alpha# is the polarizability.#mu# is the dipole moment.#k_B# is the Boltzmann constant.#T# is the temperature.
Or, use
Molecule dissolved in a nonpolar condensed phase (dilute solution)
#M/rho (epsilon - 1)/(epsilon + 2) = (4piN_0)/3(alpha + mu^2/(3k_BT))#
Molecule dissolved in a polar condensed phase (dilute solution)
#(4piN)/3 (alpha + mu^2/(3k_B T)) = ((2epsilon + epsilon_1)(epsilon - 1))/(9epsilon)# where
#epsilon_1# is the low dielectric constant of the dissolved molecule, regarded as a sphere.#N# is the number of solute particles per unit volume.
At least locally, one can say that the electric field generated by the dipoles of the solvent molecules generates induced dipole moments in the solute gas, contributing to the total electric dipole moment measured.
As a further note, the less polar the solvent in which a dilute gas is dissolved, the more similar the total dipole moment will be to that of a pure gas molecule.
In addition, if the solvent dielectric constant is close to
