I came across two formulae for dipole-dipole interaction energies on Chemistry LibreTexts — Dipole-Dipole Interactions:

$$V=-\frac{2\mu_1 \mu_2}{4\pi\epsilon_0r^3}\tag{3}$$



Which of the above two is the formula for the interaction energy of dipole-dipole interactions? Also, what is the difference between the two formulae?

  • 1
    $\begingroup$ Where did you get the equations from? The first one hides a lot of details about the orientational dependence of the dipoles and the second one has a temperature dependence which is a bit unusual. $\endgroup$
    – Unskilled
    May 23 '21 at 18:43
  • 2
    $\begingroup$ What's unusual about that? As temperature grows, the dipoles oscillate faster and interact less - that is, if we consider their oscillations at all, which the second formula does and the first one doesn't. $\endgroup$ May 23 '21 at 18:51
  • $\begingroup$ In short, there is a great deal of difference between "dipoles" and "induced dipoles". $\endgroup$ May 23 '21 at 19:00
  • $\begingroup$ @IvanNeretin ok I mostly had seen dipole energies given in the case when they have their positions and orientations fixed in space. So the second one is an internal energy of two point dipole free particles? $\endgroup$
    – Unskilled
    May 23 '21 at 19:06
  • 1
    $\begingroup$ @PV. Then what is your question about? Your link says it all, and in more detail than can fit in an answer. $\endgroup$ May 23 '21 at 19:28

They are both dipole-dipole energies as from the link but their contexts are different. Eq. 3 (as numbered in the LibreTexts link),

$$V(r) = - \frac{\mu_{1}\mu_{2}}{4\pi\epsilon_{0}r^{3}} \tag{3}$$

is specifically for the case when the dipoles are aligned. The more general Eq. 4

$$V(r) = - \frac{\mu_{1}\mu_{2}}{4\pi\epsilon_{0}r^{3}_{12}} (\cos\theta_{12} - 3\cos\theta_{1}\cos\theta_{2}) \tag{4}$$

is preferable because it includes the dependence on the orientation of the dipoles in space. And I believe Eq. 7

$$V = - \frac{2\mu_{\mathrm{A}}^{2}\mu_{\mathrm{B}}^{2}}{3(4\pi\epsilon_{0})^{2}r^{6}} \frac{1}{k_{\mathrm{B}}T} \tag{7}$$

is the energy derived by thermally averaging the dipole–dipole energies when the dipoles are fixed in space but are free to rotate.

So Eq. 3 is the dipole–dipole energy when the dipoles are both in a specific orientation, Eq. 4 reveals the orientational dependence of the dipole-dipole energy, and Eq. 7 is the thermally averaged energy for two dipoles that have a fixed separation $r$ but are free to rotate.

  • $\begingroup$ What is thermal averaging of the dipole-dipole? $\endgroup$
    – PV.
    May 23 '21 at 20:19
  • 2
    $\begingroup$ A thermally averaged energy is the energy you get by summing up all the energies of all the different configurations (dipole orientations) but each energy you add into your sum is weighted differently depending on the energy of that particular configuration and the temperature. $\endgroup$
    – Unskilled
    May 23 '21 at 20:26
  • $\begingroup$ Thankyou so much! $\endgroup$
    – PV.
    May 23 '21 at 20:29
  • $\begingroup$ You're welcome, the thermally averaged energy is also known as the internal energy. If you want to know more about how this averaging is done exactly I would recommend looking into a topic called statistical mechanics, the subject can be quite heavy though. $\endgroup$
    – Unskilled
    May 23 '21 at 20:44
  • $\begingroup$ I did do statistical mechanics recently, but looks like it will take some more time to fully grasp the concepts. $\endgroup$
    – PV.
    May 23 '21 at 20:48

An equivalent and easier formula when a molecule's coordinates are known is to use vectors. The energy is then

$$V=\frac{1}{4\pi\epsilon_0}\left(\frac{\vec\mu_1\cdot\vec\mu_2}{r^3}-3\frac{(\vec\mu_1\cdot\vec R)(\vec\mu_2\cdot\vec R)}{r^5}\right)$$

where $\vec\mu_1\cdot\vec\mu_2=|\vec\mu_1||\vec\mu_2|\cos(\theta)$ with angle $\theta$ between vectors $\vec\mu_{1,2}$. The vector $\vec R$ is the vector separating the mid point of the dipoles and assumed to be large compared to the molecules dipoles themselves. The value of $r$ is the length of vector $R$ and is the separation of the two species.

The second formula you give is that averaged over angles and Boltzmann distribution of energies. It is also called the Keesom energy and should have the relative permittivity ($\epsilon$) of the medium added in the denominator. Change $\epsilon_0\to \epsilon_0\epsilon$. This equation applies when $k_\mathrm{B}T\gt \mu_1\mu_2/(4\pi\epsilon_0\epsilon r^3)$.


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