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14

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} - ...


11

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_{...


4

Think about the probability of finding an electron is a specific region and it will be clearer To understand where partial charges come from remember that electrons are not located in one specific place (Heisenberg!) but spread around in a cloud. In molecules, those clouds are called molecular orbitals which are more complex in shape and distribution than ...


2

Very good question but keep in mind that partial charge concept is a gross oversimplification. The very first sentence on Wikipedia on partial charge Partial charge is not fully correct either. Consider the partial charge as a book-keeping "number". Just like the oxidation state concept, a "partial" charge can be estimated for a diatomic ...


2

In aqueous medium, the proton ($\ce{H+}$) is thought to exist as hydronium, $\ce{H3O+}$. However, in reality, probably bigger complexes of water may also be associated with one proton. In liquid water, everything is interconnected with strong hydrogen bonds anyways, and it is difficult to distinguish the free species. All of this means that it would be ...


1

In H2O, as O is more electronegative than hydrogen, the resultant bond dipole is towards O, which means both the lone pair and bond pair dipole are acting in the same direction and the dipole moment of H2O is high. (1.84 D) On the other hand, In the case of F2O, the bond dipole is acting towards fluorine, so in F2O the lone pair and bond pair dipole are ...


1

After going through a standard book on inorganic chemistry , I found the answer. The direction of dipole moment in $F_2O$ is just reversed as in the case of $NF_3$ i.e from oxygen atom towards fluorine atoms. Reference: Concise Inorganic Chemistry 5th edition by J.D.LEE; Chapter "CHEMICAL BONDING" topic "APPLICATION OF DIPOLE MOMENT".


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