Molecular Simulation/Charge-Dipole Interactions

A cation (+) has an attractive interaction with the negative pole of a dipolar molecule, but a repulsive interaction with the positive pole of a dipolar molecule.

Charge-Dipole Interactions

Vector geometry can be used to derive an expression for the potential energy of an ion interacting with a dipolar molecule in terms of the distance and angle of orientation between the ion and the center of the dipole.

Charge-Dipole interactions occur in the presence of a atom with a formal net charge such as Na⁺ (q_ion = +1) or Cl^- (q_ion=-1) and a dipole. A dipole is a vector ($\mu$) which connects two charged species of different signs i.e (q_ion=+1 with q_ion=-1 NaCl) over a distance $r$

The dipole moment of a molecule depends on a few factors.

1. Differences in electronegativity of the atoms in the molecule. The larger the difference, the greater the dipole moment.
2. The distance between the positive and negative ends of the dipole. A larger distance corresponds to a larger dipole moment.
3. Multiple polar bonds pointing in the same direction increase the molecular dipole moment were as if they are pointing in opposite directions the dipoles will cancel out each other.

For diatomic molecules, the dipole moment can be calculated by the formula $\mu =qr$ were $\mu$ is the dipole moment q is the atomic charges and r is the distance.

polar bonds point in the same direction increase the molecular dipole moment

polar bonds in opposite directions will cancel out

from the potential energy of a Charge-Dipole interaction equation

$𝒱(r,\theta )=-\frac{q_{ion}\mu \cos \theta }{4\pi {ϵ}_0{r}^2}$

you can see that there are 4 factors that affect the potential energy

• the distance $(1/r^2)$ from the charge (q_ion) to the dipole ($\mu$)
• the magnitude of the dipole ($\mu$)
• the magnitude of the charge (q_ion)
• the angle of the interaction $(\theta )$

Depending on the angle the interactions can be repulsive when the charge is close to the like side of the dipole (i.e, + +) attractive if the charge is close to the opposite charged end of the dipole (i.e., + -) and zero if the positive and negative ends of the dipole are equal distances from the charge (e.g., perpendicular).

Derivation of Charge-Dipole Interactions Energy

Express interaction as the sum of Coulombic interactions.

$𝒱(r)=-\frac{q_{ion}{q}_d}{4\pi {ϵ}_0|\overrightarrow{AB}|}+\frac{q_{ion}{q}_d}{4\pi {ϵ}_0|\overrightarrow{AC}|}$

$=-\frac{q_{ion}{q}_d}{4\pi {ϵ}_0}[\frac{1}{|\overrightarrow{AB}|}-\frac{1}{|\overrightarrow{AC}|}]$

Distance from charges to poles

$|\overrightarrow{AB}|=\sqrt{(r-0.5l\cos \theta {)}^2+(0.5l\sin \theta {)}^2}\approx \sqrt{(r-0.5l\cos \theta {)}^2}=r-0.5l\cos \theta$

This approximation is valid of the distance between the charge and dipole (r) is much greater than the length of the dipole, making the $0.5l\sin {\theta }^2$ term negligible.

$|\overrightarrow{AC}|=\sqrt{(r+0.5l\cos \theta {)}^2+(0.5l\sin \theta {)}^2}\approx \sqrt{(r+0.5l\cos \theta {)}^2}=r+0.5l\cos \theta$

Substitute lengths into the potential energy equation

$𝒱(r,\theta )=-\frac{q_{ion}{q}_d}{4\pi {ϵ}_0}[\frac{1}{r+0.5l\cos \theta }-\frac{1}{r-0.5l\cos \theta }]$

$𝒱(r,\theta )=-\frac{q_{ion}{q}_d}{4\pi {ϵ}_0}[\frac{r-0.5l\cos \theta }{(r+0.5l\cos \theta )(r-0.5l\cos \theta )}-\frac{r+0.5l\cos \theta }{(r+0.5l\cos \theta )(r-0.5l\cos \theta )}]$

$𝒱(r,\theta )=-\frac{q_{ion}{q}_d}{4\pi {ϵ}_0}\left[\frac{(r-0.5l\cos \theta )-(r+0.5l\cos \theta )}{r^2-0.25l^2{\cos }^2\theta }\right]$

$=-\frac{q_{ion}{q}_d}{4\pi {ϵ}_0}\left[\frac{(l\cos \theta )}{(r^2-0.25l^2{\cos }^2\theta )}\right]$

$r^2>>0.25l^2{\cos }^2\theta$

$𝒱(r,\theta )=-\frac{q_{ion}{q}_d}{4\pi {ϵ}_0}\left[\frac{l\cos \theta }{r^2}\right]$

$\mu =q_{d}l$

$𝒱(r,\theta )=-\frac{q_{ion}\mu \cos \theta }{4\pi {ϵ}_0{r}^2}$

Example: Charge-Dipole Interaction

What is the interaction energy of a ammonia molecule (μ=1.42 D) with a Cl^- ion that is 2.15 Å away when the dipole moment of ammonia is aligned with the Cl^- ion at (θ=0)?

$𝒱(r,\theta )=-\frac{q_{ion}\mu \cos (\theta )}{4\pi {ϵ}_0{r}^2}$

$𝒱(r,\theta )=-\frac{(-1e)(1.42D)\cos (0)}{4\pi {ϵ}_0(2.15{)}^2}$

$𝒱(r,\theta )=44\text{kJ/mol}$