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High symmetry molecules fit into crystal lattices especially well (higher m.p.), but are volatile for having fewer van der Waals interactions (lower b.p.). $$\begin{array}{lrr} \hline \text{Compound} & \text{m.p.}/\pu{°C} & \text{b.p.}/\pu{°C} \\ \hline \text{pentane} & −130 & 36.1 \\ \text{isopentane} & −160 & 27.2 \\ \text{... 19 This is something I spent a lot of time thinking about during my PhD, so let me see if I can explain it. What you are talking about is the relationship between macroscopically observable properties (like viscosity), the relaxation time spectrum (which is a way of looking at the time scales involved in molecular movements), and the structure of the molecules ... 18 Short answer: there are many electrostatic interactions between two non-polar molecules. Beyond monopole (full charges) and permanent dipole moments (polar molecules), there is a full multipole expansion for the electrostatic potential around any molecule. (This is technically true for atoms and ions too, but higher-order terms are really only useful for ... 17 The normal distinction between "steric" and "electronic" is based on whether the effect is transmitted through space or through bonds All the normal physical interactions we experience are arguably electronic. When you touch your desk, you feel force because of interactions between the molecules of the desk and the molecules of your hand ... 14 Draw the structures in 3D and then you will see why one is polar and the other not. \ce{CF4}: As you can see this molecules adopts a tetrahedral geometry which is perfectly symmetrical in every direction and so the dipoles of the four \ce{C-F} bonds cancel out, leaving no overall dipole. \ce{CHF3}: Although the molecule has some symmetries, it is ... 14 The enthalpy of vaporization of \ce{HCN} is higher than for \ce{NH3}, which suggests that \ce{HCN} molecules interact more strongly than \ce{NH3} molecules. \ce{C-H} bonds are not usually considered good hydrogen bond donors, but \ce{HCN} is unusual. For example \ce{HCN} has a \mathrm pK_\mathrm a value of 9.2, indicating that the \ce{CN} ... 14 Well, it turns out that this is a very active area of research. I will only summarize what I understand to be true about the covalent nature of the hydrogen bond, so I'm sure the explanation could be more detailed and potentially more accurate in some places (I hope someone gives a more detailed answer), but here's what I've got. As you said, it has been ... 14 Yes, this is technically possible. A basic tutorial for this is in the excellent Psi4Numpy project, which I'll reproduce here with minor modifications. Their example fits the counterpoise-corrected MP2/aug-cc-pVDZ total interaction energy of the helium dimer. from __future__ import print_function import psi4 import numpy as np import matplotlib as mpl ... 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. 4V(r) = - \frac{\mu_{1}\mu_{2}}{4\pi\epsilon_{0}r^{3}_{12}} (\cos\theta_{12} - ...

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The definition of the van der Waals force that I like (and apparently so too do the folks who have contributed to the Wikipedia article on the topic) is much more *specific about what is and is not a van der Waals force. This definition originates in the International Union of Pure & Applied Chemistry Compendium of Chemical Terminology The van der Waals ...

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It is safe to say that there will always be intermolecular forces at play. At the time where you will consider these you should already have a good idea about the molecules involved in your system. Based on the composition and molecular structures you can make certain assumptions. In a molecule it is straight forward to estimate (bond) polarities based on ...

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You have a possible answer to your question in proteins, an example which includes some long polymer chains. Intramolecular interactions - while not necessarily the driving force for formation of a collapsed protein globule (usually argued to be due to the hydrophobic effect, requiring intermolecular interactions) - are the basis for higher order structure ...

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TL;DR: Lewis $\to$ Non-Lewis $\mathbf{E(2)}$ values have no direct physical significance, are intrinsically un-measurable, and serve only to quantify the extent to which the "real" wavefunction for a system deviates from the fictional idealized Lewis-structure wavefunction. $E(2)$ values do, however, correlate with a variety of trends in chemical bonding ...

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The strength of the ionic bond depends on Coulomb's law for the force acting between two charged particles where larger force translates to a stronger ionic bond. The equation is $$F = \frac{-k\cdot q_1 \cdot q_2}{r^2}$$ $k$ is a constant; $q_1$ and $q_2$ are the charges on the ions and $r$ is the distance between the ions. So the larger the charge the ...

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An ionic bond could maybe be described as an inter-ionic force. All electron interactions are most accurately described by wavefunctions and quantum mechanics, but in practice we use successively more detailed approximations for convenience, stopping at the lowest level of detail that suits our needs at the time. At the lowest-detail end of the spectrum, ...

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You have identified a fairly common but counter-intuitive result. At least it's counter-intuitive based on the way the properties of fluorine are described in undergraduate chemistry classes. Specifically, we're told it's very electronegative and hence forms very strong hydrogen bonds, and that it is very reactive. This is, however, a poor description of ...

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Graphite has got a structure similar to books stacked on top of each other. Multiple layers on top of each other and each layer going by the name graphene. Atoms in each individual layer is covalently bonded, which is quite strong. Remember covalent bond is the one that holds diamond together, which is one of the hardest substances. Atoms in the individual ...

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It is a reasonable rule of thumb, but certainly not always true. Compare for example the viscosities of Dodecane ($\ce{C12H26}$, $\mu=1.374 \ce{\; mPa\cdot s}$, no $\ce{OH}$ groups) and Ethanol ($\ce{C2H5OH}$, $\mu=1.095 \ce{\; mPa\cdot s}$, 1 $\ce{OH}$ group) (source). There you can see that ethanol has a higher number of $\ce{OH}$-groups, but a lower ...

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You know $\ce{CO_2}$ is gaseous at room temperature, so let's put that at the bottom. Methanol forms hydrogen bonds, so that will be above bromomethane which does not. At last we have rubidium fluoride which is a salt. Salts generally have a very high boiling point (> 1000 °C, much higher than molecular structures) because of the ionic (electrostatic) ...

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This is known as the chelate effect. The main reason why you observe this is that cations in solution have an ordered solvent shell around them, especially in polar solvents where there will be defined solvent geometry around the shell; often octahedral for metal ions in water. The formation of a complex with a bidentate ligand such as malonate is more ...

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Single point energy arises in the framework of the Born–Oppenheimer approximation and corresponds to just one point on the potential energy surface. Physically it is the total energy of the molecular system with its nuclei beeing fixed (or clamped) at some particular locations in space. In other words, it is total energy of the molecular system within the so-...

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Although Danny Rodriguez has already excellently exposed what the dispersion force is in simple terms, the word dispersion still demands a better explanation in my opinion. According to Wikipedia: The London theory has much similarity to the quantum mechanical theory of light dispersion, which is why London coined the phrase "dispersion effect." In ...

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I will quote the paper quite a bit, but I'll try and summarize a bit after the quotes and equations. You might want to start at the bottom and work backward, a lot of this is just for later reference. First, some background. Equation 3: $$E^{(2)} = \sum_{AB} \sum_{n=6,8,10,\dots} s_{n} \frac{C_{n}^{AB}}{r_{AB}^{n}} f_{d,n}(r_{AB})$$ Here, the first ...

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You are right, in a sense. A more accurate statement would be that as the temperature decreases, the average lifetime of a hydrogen bond increases. Thus, on average the liquid/solid/molecule will be more stable. This is because, as you point out, hydrogen bonds are weak enough to be broken by collisions due to thermal motions (translation and rotation). ...

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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 $\vec\mu_{... 10 There are a lot of reasons for why some molecules have stronger IMF's than others, but the trend of increasing IMF for increasing relative molecular mass ($M_\mathrm r\$) is due to an increase in London Dispersion Forces, part of the greater set of van der Waals forces. As you may or may not know, these alkanes are non-polar molecules (that is, they don't ...

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