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Disclaimer: All of this "jazz" will be about reaching a mere rule-of-thumb. You can't just compare whole families of organic compounds with each other. There are more factors to consider than below, mostly based on isomerism notions. However, as most of the A grade exams emphasize on the lighter aliphatic compounds, we can understand each other here. :) It'...

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I think @Ben Norris's answer is great, but I wanted to add one other reason that non-polar molecules will dissolve in non-polar solvents, even in the absence of interactions: entropy. If there are no intermolecular interactions at all (ie, the solvent molecules ignore each other and the solute molecules, and the solute molecules ignore each other and the ...

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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{... 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 ... 16 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 ... 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 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 ... 13 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 ... 13 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 ... 13 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 ... 13 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 ... 12 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 ... 12 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, ... 12 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 ... 11 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 ... 11 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) ... 11 It refers to the electrons that are distributed (or dispersed) in the electron cloud which then ensues an attraction between atoms or molecules. Electrons can be dispersed in any number of different ways in an electron cloud. One way, however, is where all the electrons appear in one side of the electron cloud. When this polarization happens, the atom has ... 11 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 ... 11 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 ... 11 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). ... 11 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 ... 10 Van der Waals forces are the attractions/repulsions - the forces - between molecules or atoms, other than attractions like ionic attractions, and covalent attractions. These forces are: Keesom Effect - This is an effect caused by two polar atoms interacting with each other. Two permanent dipoles are involved, meaning the molecules/atoms involved are polar. ... 10 When an ionic compound dissolves in water, the water molecules surround the individual ions. Since water is very polar, when water surrounds an ion there is a decrease in overall energy. However, to pull the ions apart in the first place, there has to be an energy increase to overcome the lattice energy. If the lattice energy (cost) is larger than the energy ... 10 Is the following statement always true? The more −OH functional groups in the molecule, the more is its viscosity? It's hard to find anything that is "always" true in chemistry, but I'd bet that within a series of molecules where the only variable is the number of \ce{OH} groups, your statement is generally true, so I basically agree with your ... 10 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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