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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 ...


9

In Principles of Nucleic Acid Structure, W. Saenger argues that hydrogen-bonded bases contain at least two hydrogen bonds (forming a "cyclic" pattern). Often, there is a tautomeric form possible that also makes two hydrogen bonds, with two covalent bonds turning into hydrogen bonds, two hydrogen bonds turning into covalent bonds, and double bonds moving ...


6

The parameter $a$ from the van der Waals equation $$\left(p+\frac{an^2}{V^2}\right)\left(V-nb\right)=nRT$$ is not to be taken only as the measure of strength of intermolecular bonding. Rather, it is overall measure of significance and frequency of intermolecular interaction. It may be based on purely elastic electrostatic interactions, where only the ...


5

London dispersion forces are instantaneous dipole-induced dipole interactions. All compounds can have an instantaneous dipole moment, including ions, so London dispersion forces exist for all compounds. Unlike molecules, ions are charged. The electrostatic interactions between charges are far stronger than the London dispersion forces between ions, so ...


3

I found a nice figure and the relevant statement in a paper by Frenking and Krapp (Unicorns in the world of chemical bonding models, 2006, https://doi.org/10.1002/jcc.20543): The crucial term which is responsible for repulsive interactions in chemical bonds except in two‐electron systems such as H2 is the Pauli repulsion. The three terms (a) ...


3

In milk, casein exists as a colloidal aggregate (casein micelle) with the properties you state, but purified casein is different. It is rich in proline (shown below), has no disulfide bridges, and hence little tertiary structure to hide the hydrophobic residues in the core of the protein with the hydrophilic residues on the surface to bind to water. In ...


3

Q1 and Q2 have the same approach. One notices that they are homologous alkanes and picks the biggest and the smallest, having the most London dispersion/van der Waals force and least, respectively. Other forces are known not to be present between alkanes.$^1$ (There is a small difference between branched and linear alkanes, but that is negligible compared to ...


3

Can anyone explain why the exchange contribution to the total energy is negative? I find it misleading that exchange interaction is treated as something that changes total energy of the system. This lowering of energy is actually due to the Hartree-Fock scheme being in principle inexact, and is not really specific to indistinguishable particles. Let me ...


3

I'll try to give my interpretation of the "physical" explanation of why exchange would lower the energy. For the true wavefunction, the motion of all the electrons should be correlated, with the classical view being that the electrons are avoiding each other to minimize repulsion. With Hartree-Fock, we find an approximate wavefunction by solving for 1-...


3

I have been watching this question with keen interest, but wanted to let someone else go for the bounty -- However it expires tomorrow and no one has answered, so I will give my perspective. Why does the exchange interaction in Hartree-Fock theory lower the total energy? The answer is given in the question: This integral is always positive, and so ...


3

The effect of intramolecular forces versus intermolecular forces on boiling and melting points is sometimes seen when we compare the boiling points of ortho- and para-nitrophenol (or similar compounds). Para-nitrophenol shows intermolecular hydrogen bonding, which causes an increase in boiling point as different molecules bond better with each other, as ...


3

Besides the computational methods clearly described by @Martin, you can identify hydrogen bonds using a number of experimental methods, many are indirect such as melting and boiling points, more specific are changes in absorption/fluorescence and changes in ir and raman spectra. Microwave spectra can be used in the gas phase. The best methods are ...


2

Slight correction on @airhuff's answer. "if a force is not an ionic or covalent chemical bond or electrostatic interaction, then it is a van der Waals force." The IUPAC Gold Book's definition precludes "electrostatic interaction of ions or of ionic groups" from being VdW forces but all VdW forces are electrostatic in nature: forces between electrostatic ...


2

A solution that has a maximum or minimum vapour pressure (vs mole fraction) is called an azeotrope. The liquid is in equilibrium with the vapour and mole fractions in the liquid are the same as in the vapour at a given temperature. As the composition of liquid and vapour are the same the $dp/dx_A=0$ where $p$ is pressure and $x_A$ mole fraction of species A....


2

Can't there ever be hydrogen bonds between 2 purines or between 2 pyrimidines? Yes, it is possible to have hydrogen-bonded base pairs (and triplets and quadruplexes) beyond the canonical Watson-Crick base pairs A:T and G:C. You can explore these in 3D on the DSSR-Jmol site for example. If you look at the 1ehz structure (a tRNA from yeast), which loads ...


2

This is a tough question to answer because the intermolecular distances are similar in the solid to liquid transition unlike those in the liquid to gas phase transition. In the case of the elements there is a correlation between the Debye temperature and the melting temperature. The Debye temperature is that temperature at which the atoms gain their full ...


2

Olive oil is a triglyceride, and the main fatty acid is oleic acid below. Now the bromine does not add to the double bond between oxygen and carbon; instead the bromine reacts with carbon-carbon double bond and add itself to it. If you use e.g. coconut oil which has minimal unsaturated (meaning having carbon-carbon double bond) fatty acids (and assume you ...


2

Let me crash the party here. TL;DR: Dispersion interactions are not due to fluctuating induced-dipole attractions, even if everyone tries to explain it that way. (I know that this must seem like an outrageous statement, and surely downvote fingers are itching now. Bear with me.) Based on the Hellmann-Feynman theorem, it is known that the forces acting on ...


2

You are correct that simply stating that the Schrodinger equation is time dependent does not really answer the question. In fact, time-dependence is not a component of the derivation of the dispersion forces. At a surface level, they just represent how much the ground state energy of a system of polarizable molecules is lowered by their interactions. It's ...


1

You may consider the vdW equation as the deviation from the ideal gas state equation $pV_\mathrm{m} = RT$. If you look at the vdW equation $$\left(P+\frac a{V_m^2}\right)(V_m-b)=R T$$ you can easily see why the corrections have their signs. The pressure term correction has the positive sign, as the real gas has due molecule cohesion smaller volume ...


1

The Hamiltonian for the collection of charged particles (electrons and nuclei) comprising a pair of molecules or ions has terms describing the kinetic energy of individual particles and pair-potential terms describing Coulombic interactions, that is, all interactions between particles are Coulombic. Where a Pauli repulsion term can crop up is when ...


1

Reason 1: For example if you have an porous ore and you melt it, the surface area will decrease dramatically. If you just roast it, the surface area will stay about the same. A higher surface area will aid calcination proceses as more oxygen gets into contact with the surface and will eventually react. Reason 2: A higher temperature means more energy to ...


1

Remember the energy of the bonds never increase or decrease. They remain the same. When things are heated what actually happens is the molecules begin to move around (in solids they just oscillate while in liquids and gases the movement becomes much more prominent) and their kinetic energy increases and the molecules bump into each other harder and quite ...


1

Permanent Dipoles occur when two atoms bonded to each other have significantly different electronegativities. When the atomic number of an element increases the number of electrons present in them also increases. These electrons are constantly moving around in the atom. At some point, there are more electrons on one side of the atom than the other. This ...


1

Although individual dispersion forces are weak, they are cumulative, and increase with molar mass. As a general rule, boiling point increases with molar mass. Polar molecules will have higher boiling points when compared to molecules with similar molar masses. For example, ethanol($\ce{CH3CH2OH}$) has a higher boiling point than dimethyl ether ($\ce{...


1

The relatively high boiling points for such low molecular weight compounds suggests a primary effect from hydrogen bonding. Using ethanamide as a reference point, propanamide has a higher molecular weight, which should raise the boiling point a little, but three relatively nonpolar carbons (plus hydrogens) vs two in ethanamide. A lower boiling point would ...


1

I haven't done it in vmd, but a rdf is really kind of like a histogram. For a particle of interest, you count how many particles are between r and $\Delta r$ from it and record that number in a bin, then move outwards, recording numbers in each shell into a bin for that shell. That tells you the rdf of that particle interacting with everything else. If there ...


1

It is true that Q1 and Q2 can be addressed having the same approach since all of them are hydrocarbons, which do not have any other forces acting on them other than the London dispersion and van der Waals forces. Also, they are homologous alkanes, which increase those forces according to their size. Yet, these forces also depend on how branched their ...


1

This is because when these liquids are mixed, H-bonding type interactions are formed between hydrogen atom of chloroform (partial positive charge due to 3 Cl atoms) and 'pi' electron cloud of benzene ring. Thus, chloroform-benzene interactions are stronger than chloroform-chloroform and benzene-benzene interactions.


1

Before getting to the azeotropes, it's worth reminding ourselves of the relevant assumption that we make about ideal mixtures: that the intermolecular interactions have constant energy regardless of the mixture. Given the important role that intermolecular interactions have in determining the boiling point of a pure liquid, it's surprising that so many ...


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