Deducing compounds from spectroscopic data for isomers of formula C4H8O Here's the problem:

I already know the $\pu{3600 cm-1}$ line indicates the $\ce{-OH}$ group and the $\pu{1640 cm-1}$ line indicates the $\ce{C}$-double-bond-$\ce{C}$ alkene.

From the $\ce{^13C}$ NMR data: the two carbons with chemical shifts of 100+ are $\mathrm{sp^2}$ carbons and the remaining two are $\mathrm{sp^3}$ carbons.

I have tried to work out compound D, and suggest this structure:

I am stuck on figuring out compound C. I suggest the structure below, but am unsure whether if it is correct.

Your two proposed structures are correct, although your rationalisation of the 13C spectrum for Cpd D is not quite correct around the double bond. In a nutshell, carbon chemical shifts can be calculated as a function of their α, β and γ contributions. α for what is directly bonded to that carbon, β for substituents one carbon away, and γ for substituents two carbons away. Almost without exception, alkene carbons exhibit a downfield β shift and upfield γ shift for substituted alkenes. So, for both compounds C and D, the carbon closest to the substituents will be shifted furthest downfield. The β and γ contributions for the following substituents are shown (someone more tech savvy than me might tidy this up) :

[Substituent; β ;γ]
[-H; 0 ; 0]
[-CH2O-; +14.2 ; -8.2]
[-CH3 ; +9.0 ; -7.0]

The electronics contributing to β and γ shifts is not well understood, but is more complex than a consideration of just electronegativities/electron withdrawing potential.

Below are the spectra for the two compounds:

Your confusion with the description of the quintet label arises from how people report splitting patterns in the literature. I have discussed this on another question, but essentially the two methods of reporting splittings are to (a) report the observed splitting pattern (here a quintet) and (b) report the calculated/expected splitting pattern (here a doublet of quartets). As you can see, (or least I can), reporting the observed pattern can lead to some confusion, and gives no real information about how this splitting pattern arises.

Further, reporting couplings to the nearest Hz (here 6Hz) might be fine for some occasions where the difference in coupling for the two different partners is less or approximate to the observed linewidth, however, with a good sample and a good operator on a good magnet, what is reported as a quintet may very well appear different, especially once some apodization is applied. For example, below is a expansion of a simulated spectrum for compound C with Jbc=5.8 and Jbe=6.2 (smaller couplings from d and f ignored), and observed linewidth of 0.5Hz. The top spectrum is the normal spectrum, and the bottom spectrum is what it looks like with a gaussian linewidth function applied (gb 0.1, lb -1). They look quite different, and the bottom spectrum is very hard to rationalise as a quintet, but easily recognizable as a doublet of quartets.


Separation of Liquids based on Density I was doing some practice problems when I came across this one

The answer key says the answer is B but I don't know why. Isn't the difference in the molar masses of the two substances in D greater than that in B? Or am I supposed to look for how many times greater the molar mass of one is than the other?
Thanks.

The ability of liquid mixtures to separate is not based on density, but on polarity of each liquid. $\ce{CCl_4}$ and $\ce{C_6H_{14}}$ are both non-polar; they will mix well in solution and not separate easily. $\ce{CH_3OH}$ is polar due to presence of a polar alcohol group. Mixtures of non-polar and polar compounds will separate easily.

Density comes into play when you need to determine which of the compounds settle on top of the other. So, for (B), you would use the fact that the density of $\ce{CH_3OH}$ (0.79 g/mL) is less than $\ce{CCl_4}$ (1.59 g/mL) to say that the methanol layer would float on top of the $\ce{CCl_4}$ layer.


Why do some chemical reactions require many steps? I posted the following question in Physics SE and was advised to transfer it to Chemistry SE.

I studied physics in college ten years ago and I recently started to learn biochemistry. I enjoy finding out that some familiar concepts in physics play important roles in biochemistry such as entropy and the Gibbs free energy.

For example, as a (ex-)student of physics, I am happy to know that the Gibbs free energy determines the directions of chemical reactions. I feel this is a good example where a sort of fundamental law of physics determines how a phenomenon looks like.

However, I still can not understand why the chemical reactions in a body need to be so complex. Many chemical systems consist of more than several steps to achieve their purposes. According to Wikipedia, glycolysis takes ten steps through its process. Why are so many steps necessary?

I tried to find out a physical law that prohibit the glycolysis process from being achieved in one or two steps, but I could not find an answer. I would like to know (or discuss) whether there is a physical law that makes chemical systems so complex (many steps required).

My assumption is that some physical law prohibit the existence of an enzyme that realizes a one-step process of glycolysis.

There is no fundamental law preventing simple chemical reactions: things are complex because of the combinatorial complexity of chemical compounds

The complexity of many chemical reactions is a byproduct of the fact that there is a very, very large variety of possible chemicals. Much of that complexity happens because of the almost infinite way even some simple elements can be combined together to give complicated structures (carbon being the archetypal example). Theoretically, for example (theoretical because not all of the examples can exist in 3D space) there are 366,319 ways to build different alkane compounds from just 20 carbon atoms and hydrogen atoms (see this question here and this entry in the Encyclopaedia of integer sequences). And this number drastically understates the real complexity as it ignores mirror images and more complicated ways of joining the carbon atoms together (like in rings, for example). The complexity just gets more mind boggling if you start adding other elements to the mix.

No physical law prevent us making any possible compound in one step. But the sheer complexity of the end products makes simple ways to reach many of them extraordinarily unlikely from the laws of probability alone, never mind the specific ways chemical components can be easily joined up to make more complex things.

Here is a simple analogy. Let's say you want to assemble a Lego model of the Star Wars Death Star weapon. There are 4,016 pieces of lego that have to be assembled in the right combination and the right order. There is no physical law that says you couldn't somehow do that in a single step. But no sane person's intuition would assume that this was easy or likely. It isn't physical law that prevents one step assembly: it is combinatorial complexity. Chemistry is, do I really need to say this, more complicated than Lego: not least because atoms can be joined up in many more complex ways than the simple, standard-sized physical pins that join Lego bricks together.

Both nature and synthetic chemists have explored many ways to achieve particular end products from simpler building blocks. Sometimes new chemical Death Star equivalents (like the geometrically beautiful hydrocarbon dodecahedrane, which, incidentally, has 20 carbons but isn't counted in the list of 20 carbon alkanes) are made only after long sequences of reactions. The original synthesis of dodecahedrane took 29 steps but others found better, higher yielding, routes that took only 20. Many important drugs are first synthesised in long sequences of reactions but are later found to be available via much shorter routes (there is nothing like the economics of manufacturing cost to encourage creativity).

So the reason many chemical reactions take multiple steps isn't physical laws but probability theory. There are just too many possible chemicals and too many ways to combine things for single step routes to most given products to be likely to work. Doing one thing at a time (just like you would if building the Lego Death Star) is the way to get what you want.


A student is provided with 500 mL of 600 ppm solutions of fructose. What volume of this solution in millilitres contains 0.15 fructose? A student is provided with 500 mL of 600 ppm solutions of fructose.  What volume of this solution in millilitres contains 0.15 fructose?

Does this mean that I just have to use 600 ppm as the concentration and the 0.15 as the weight? Like this:

\begin{align*}
{600 \space ppm}
&=\frac {0.15 g}{x}\cdot 10^{6} \\
x&=\frac {0.15 }{600}\cdot 10^{6}  \\
x&=250 \space mL\\
&
\end{align*}

And when can 500 mL be used?

Assuming it's $0.15gm$ of fructose,$$ppm=\frac{Weight\;of\;solute_{(in\;gm)}}{Weight\;of\;solution_{(in\;gm)}}10^6$$
Since density of water is $1gm/ml$, $500ml$ of water would weigh $500gm$
$$\therefore600=\frac{W_b}{500}10^6$$
$$\therefore W_b=0.3gm$$

Thus, $500ml$ of water contains $0.3gm$ of fructose. You can find out what volume will contain $0.15gm$ of it.


What happens if hydrogen peroxide is added to silver oxides? I have some powders that contain $\ce{Ag2O}$ and $\ce{AgO}$.

What happens if I add $\ce{H2O2}$ and some water to these silver oxides?

AgO (silver(I, III) oxide) is unstable and decomposes to produce $\ce{O2}$ in aqueous solutions. Hydrogen peroxide is thermodynamically unstable too and slowly decomposes to form water and oxygen. Decomposition of hydrogen peroxide can be catalyzed by different compounds, including transition metals (such as Ag) and their compounds. Probably, the silver (I) oxide will directly/inderectly catalyze the decomposition of hydrogen peroxide according to the reaction:$$\ce{Ag2O + H2O2 → 2Ag + H2O + O2}$$

Given that the majority of silver compounds/salts are photosensitive, they will be decomposed when exposed to regular light, especially to the ultraviolet light. Even if you have other substances/compounds as impurities in hydrogen peroxide solution which can react with silver oxide you still will probably end up with elemental silver ($\ce{Ag}$).

  and some water?

Sounds like a joke. The available hydrogen peroxide solutions are approx. 3% (drug store) or 30% (most chemical labs), thus, the solutions already have a plenty of water, at least 70%. Do you really need more water?


Finding change in enthalpy for a reaction with an atom in elemental state using bond energy For example, consider $\ce{C(s) + O2(g) -> CO2(g)}$.

If we know the bond energies for $\ce{O=O}$ and $\ce{C=O}$, can we determine $\Delta H$, even though there is carbon in the reaction? Can we ignore it, or is there a certain way to compute the energy?

There are two things we need to note with this. For one, you need to know the bond enthalpies if you are going to obtain $\Delta H$, not the bond energies. Second, we can't just use the bond enthalpies in this case because that would ignore that we have $C(s)$ when we need $C(g)$ to make this reaction occur. We have to explicitly include the atomization energy of carbon (the energy needed to break graphite into single carbon atoms).


Hazards of Keeping Freshwater Fish in a Container that was Used for Adipic Acid I have an Intermediate Bulk Container (IBC) which is a large 3' x 3'x 3' plastic cube. The label on the container described it as storing Adipic Acid.

Is it safe to use this container to grow plant life and harvest fish life for aquaponics? Or is there a possibility that this acid has leached into the plastic?

My quick research has shown that adipic acid is a mild skin irritant.

Adipic acid occurs naturally in beets and sugarcane, common foods, so it is not likely to be very toxic to fish if sufficiently diluted by washing. As @DGS and @ghost state, normal cleaning of the container followed by adjusting the pH for the fish should be OK. If you need to use an electric in-tank heater, though, check that it is safe to use in a plastic container.


Baking soda + oil solubility I have a question regarding the solubility of baking soda (bicarb soda) in oil. I am pretty sure that it has poor solubility, but I have a chemistry dilemma and cannot seem to solve it.

I make all-natural deodorant, and the main ingredients are coconut oil, shea butter and baking soda (amongst a few other ingredients). We have been making it for years, and it was always soft and smooth, but this time the baking soda is not dissolving and leaves a gritty texture (even after sifting baking soda to remove lumps).

Does anyone have any idea why this might be? We heat the ingredients to melting point on the stove, then let cool to room temperature and re-solidify in the fridge for a day, before using.

The only thing I can think of is that we have recently changed to a new brand of baking soda; however, with adequate mixing, surely that would not change the final texture so much.

Thank you!

The baking soda grain size might be bigger in this new brand. When you sift, you break up individual grains that have clumped together due to moisture or being pressed together tightly but there is a fundamental particle size from how the baking soda was manufactured that would require further processing to reduce (e.g. milling). The larger grain size would adversely affect the solubility. It could also be that the baking soda you buy now is either missing some additives or has some additives that are detrimental to your final product.


How does the composition of a mixed solvent affect solubility? I am trying to find out how varying the proportion of each solvent in a binary (water/ethanol) solvent solution affects the solubility of an organic acid. I know the solubilities of the acid in each solution. Is it as simple as a linear relationship between solvent composition, or is there something else going on?

This question deals with the same topic, and the conclusion there is that it is immensely difficult to predict solubility in mixed-solvent systems. My question is different for two reasons - I only want a qualitative answer, and I know the solubilities for the compound in both pure solvents.

In general, solubility is determined by the intermolecular interaction between a solvent and a solute. Electrostatic forces, hydrogen bonding, electron-pair donor/electron-pair acceptor interactions and solvophobic interactions are examples. The sum of the interaction forces can be related to the polarity of the molecule.

The dissolution of a substance requires that the interaction energy of the solute molecules and the interaction energy of the solvent molecules be overcome. In other words, for the solvation of A in a solvent B, we expect a stronger interaction A-B than A-A or B-B. Considering this, you will have the trend of solubility of polar and non-polar solutes in solvents: the old "like dissolves like" rule.

Now, if you have a solvent mixture, the interaction between unlike solvent molecules is important and leads to deviations from the ideal behaviour from Raoult's law, for example the water/alcohol mixture. Solvent mixtures can exhibit different physical properties compared with their pure components. Coming back to intermolecular forces, it has been found that the ratio of the solvent in the solvent shell of solvation can be different from that in the bulk solution, leading to a preferential solvation. This is caused by the variation of free energy of solvation of the solute in different solvents.

The acid-base equilibria is affected by a change in the 1) acidity or basicity of the solvent, 2) the relative permittivity of the solvent, $\epsilon_{r}$, and 3) its ability to solvate the species in the equilibria. There exists some formulas that relates the equilibrium constant and the relative permittivity (Born, Onsager, etc). In the case of the acetic acid, for example, the relative permittivity will have a strong influence in the solvation of the anionic and cationic species of the equilibria
$$\text{CH}_{3}\text{CO}_{2}\text{H}\rightleftarrows\text{CH}_{3}\text{CO}_{2}^{-}+\text{H}^{+}$$
since the two charges attract each other. The good solvation of the ions in a high relative permittivity solvent shift to the right the ionization. Consider that for water ($\epsilon_{r}\approx80$) and ethanol ($\epsilon_{r}\approx33$) the miscibility is similar, but also have distinct pKa's. Therefore, taking into account only the electrical interaction, between solute and solvent, is a poor approximation. The other problem is that it is practically impossible to find solvents that differ only in their relative permittivities and not in their acidity or basicity.

So, for instance, data on solubility in mixed solvents may be very difficult to obtain, but qualitatively the whole thing works in the form of intermolecular interactions.


Stability of pyridine derivatives Arrange the following pyridine derivatives in the order of increasing stability
of adduct formed with $\ce{BBr3}.$ Explain briefly.

I'm pretty confused about which structure is the most stable. I'd say structure III is the most stable as it has two methyl groups to donate electrons to nitrogen and increase its Lewis basicity.
However, I think III may not be the most stable because of steric hindrance. In fact, structure II could be the most stable as the methyl group is closer to nitrogen and there's no steric hindrance. Therefore, the answer should be III < I < II.
Is my reasoning correct? Are there any factors to consider for stability?

I believe that the order of stability is II > III > I. The adduct should be formed between the lone pair on nitrogen and the Lewis acid. From there, I would apply the M+, I+ effects of the methyl groups as are known from benzene derivatives (although the substitution chemistry of pyridine is different from benzene).


Clean silver with aluminum foil and * I already have two questions touching on this subject and there is a third question as well.

The thing is that all the other question talk about cleaning silver with aluminum foil and carbonated water. The recipe I use though is to soak the silverware and the aluminium foil in salt water (simply mix table salt and cold water). I got the idea many years ago watching a commercial for a product called "SilverStar" or something like that on TV-Shop which was essentially a 0.5 cm thick aluminium plats which you were supposed to but in a bowl with saltwater and all your jewellery and it would magically become polished.

The explanations for the "recipes" using carbonated water are all along the lines that the acid dissolves part of the aluminum and then the aluminum ions in solution somehow react with the silveroxide.

The question then is if I only use table salt and (still) water will the chemistry be more or less the same but only much slower (relying on CO2 dissolving in the water from the air) or is there something else going on?

My understanding of what was happening with the salt water was that the salt water acted as a "wire" and the aluminium acted as the "sacrificial metal" in a Cathodic Protection. I always thought that you could essentially put the aluminium in one bowl and the silver in another and connect them with a copper wire.

So the question is twofold

What is the purpose of the table salt in my recipe?
Is there any (chemical) difference between "my" recipe with table salt and the more mainstream one with different flavours of carbonated water?

Since you want to remove the tarnished silver ($\ce {Ag2O}$), you would convert it into silver metal (or reduce it). Therefore, you would oxidize the aluminum ($\ce {Al -> Al^3+}$). The silver and aluminum need to be in physical contact and the sodium chloride or the carbonated water is just providing the medium to transfer the ions; the aluminum does not dissolve first.


Can diamond undergo a self sustaining combustion reaction all the way to carbon dioxide? I know charcoal can smoulder in an incomplete combustion to produce carbon monoxide so I assume diamond also can. If it's heated to a sufficiently high temperature, will the reaction produce enough heat to sustain the full combustion reaction all the way to carbon dioxide even in air that's only 21 °C. If so, does the reaction of carbon monoxide with oxygen occur fast enough to be perceived as smouldering or does it occur slowly enough to be perceived as flaming combustion?

Yes, diamond will combust in air. Regardless of the ambient air temperature, e.g. your example of $21\ \mathrm{^\circ C}$, you of course have to heat it to it's ignition temperature somehow, whether in a furnace, a flame, etc.  The autoignition temperature for diamond is around $900\ \mathrm{^\circ C}$ (source 1, source 2), compared to about $730\ \mathrm{^\circ C}$ for graphite.  The autoignition temperature is described in this Wikipedia article as:

  "...the lowest temperature at which it spontaneously ignites in normal
  atmosphere without an external source of ignition, such as a flame or
  spark. This temperature is required to supply the activation energy
  needed for combustion."

So, once you have initiated combustion by some means heating it to about $900\ \mathrm{^\circ C}$, it will continue to burn, primarily to $\ce{CO2}$.  Note that combustion does not require that a flame be produced or maintained.

How efficiently diamond combusts in air, in other words how much of the carbon is converted to $\ce{CO2}$ vs. $\ce{CO}$ and other partially oxidized carbon species is a function of a few different things. The same is true when burning graphite, wood, or most anything else. Is it a single smooth "rock" or a finely ground powder? How much of what sort of impurities does it contain? How efficiently ventilated is the system? These issues and other conditions of the combustion will determine the degree to which $\ce{CO}$ and other partially combusted carbon species are produced.


What are the meanings of dotted and wavy lines in structural formulas? So I came across this diagram,the other day:

As  a chemistry student, I am well-versed with the dashed and wedged lines, but I was wondering what the wavy and the dotted line represent?

I was wondering what the wavy and the dotted line represent?

A dashed line indicates that the bond is extending behind the plane of the drawing surface
A bold-wedged line indicates that the bond is protruding out from the plane of the drawing surface
A solid line indicates that the bond exists in the plane of the drawing surface.
A wavy line indicates that the stereochemistry of the bond is unknown.
A dotted line indicates that the bond is not a full bond, it is only a partial bond as in a hydrogen bond or a partially formed or broken bond in a transition state. Also, as Mrigank Pawagi notes in the comments, the dotted line can be used along with the solid line to denote bonds that have partial double bond character due to resonance.
An arrow indicates a dative bond in which both electrons in the bond originate from one atom (not common)


Can chiral atoms have same substituents but with different orientations? So consider this molecule. We have to determine whether nitrogen atom is chiral or not.

MY IDEAS:
Now going by all I have read, for chiral atoms only criteria is substituents to be different. Here two substituents are same but as we can see orientations of the substituents is different.
So aren't the two substituents technically different because after all they aren't going to react the same way to plane polarized light?

I tried to break it into possibilities:

Now in first case I think we could rotate around one of the sigma bonds attached to nitrogen and form a conformation with a plane of symmetry.
So is in this case nitrogen not a chiral atom?

Now in the second case I do think we can't prove the two substituents to be same, no matter how much we rotate. So chiral?

CONCLUSION: I am a beginner in this topic and not really confident about my thought process

Can we really use conformers of molecules to support their symmetry because after all conformers are ever changing, not stuck in one orientation?

Also what's the logic behind checking the plane and centre of symmetry of molecules? How does that relate to rotation of PPL light?

->If two substituents of a sp3 hybridized atom are identical, we say not optically active.
 But both being identical will rotate PPL light in same direction, so a net rotation will be produced.
->So isn't it optically active?

No, it is not an asymmetric centre. To test this, you can draw the two potential stereoisomers:

Note that isomers A and B are identical (by C2 rotation), so the nitrogen cannot be an asymmetric centre. If it were one, then these isomers would by definition be distinct.

The attached groups are identical, from the point of view of the nitrogen - both have (R) stereochemistry. If one were (R) and the other (S), the nitrogen would constitute a pseudoasymmetric centre - https://goldbook.iupac.org/html/P/P04921.html

Hope this helps.

Also, don't get confused about the difference between the nitrogen being a sterogenic centre, and the molecule as a whole being chiral. The molecule is chiral, since it is non-superimposable on its mirror image. Chirality is a molecular property, and although frequently people will refer to a particular atom in a molecule as "chiral", this is a misnomer. A better description of an atom with 4 different groups attached is "sterogenic". It is entirely possible to have a molecule containing stereogenic centres but the molecule is achiral (e.g. meso compounds).

The logic behind checking the plane and centre of symmetry is that planes and centres of symmetry are conserved on reflection in a mirror, so checking for these symmetry elements is a shortcut to checking whether a molecule satisfies the definition of chirality.


Do electrons of opposite spin attract each other? Electrons of opposite spin create magnetic fields that are opposite to each other. So, is it possible that, 'such electron pairs attract each other' or ' they distract each other less than electron of same spin'? Thanks.

Electrons can sort of attract. Let's review magnetic dipoles. First, suppose we have a current loop $I$ rendered in blue at radius $a$ around the $z$-axis in the $xy$-plane and we want to compute the magnetic field at some point on the $z$-axis marked in red. See Figure $1$.

The source point vector is
$$\vec{r}_s=\langle a\cos\phi,a\sin\phi,0\rangle$$
So the element of arc length along the current path is
$$d\vec\ell=d\vec{r}_s=\langle-a\sin\phi,a\cos\phi,0\rangle\,d\phi$$
The field point vector is
$$\vec{r}_f=\langle0,0,z\rangle$$
So the vector from the source point to the field point is
$$\vec{r}=\vec{r}_f-\vec{r}_s=\langle-a\cos\phi,-a\sin\phi,z\rangle$$
And its magnitude
$$r=\left\lVert\vec{r}\right\rVert=\sqrt{a^2+z^2}$$
And a unit vector in its direction
$$\hat{r}=\frac{\vec{r}}{r}=\frac{\langle-a\cos\phi,-a\sin\phi,z\rangle}{\sqrt{a^2+z^2}}$$
So we can get the magnetic field at $\vec{r}_f$ via the Biot-Savart law
$$\begin{align}\vec B&=\frac{\mu_0}{4\pi}\int\frac{Id\vec\ell\times\hat{r}}{r^2}\\
&=\frac{\mu_0I}{4\pi}\int_0^{2\pi}\frac{\langle-a\sin\phi,a\cos\phi,0\rangle\,d\phi\times\langle-a\cos\phi,-a\sin\phi,z\rangle}{(a^2+z^2)^{3/2}}\\
&=\frac{\mu_0I}{4\pi(a^2+z^2)^{3/2}}\int_0^{2\pi}\langle az\cos\phi,az\sin\phi,a^2\rangle\,d\phi\\
&=\frac{\mu_0}{4\pi}\frac{2\pi a^2I}{(a^2+z^2)^{3/2}}\hat k\approx\frac{\mu_0}{4\pi}\frac{2m}{r^3}\hat k\end{align}$$
Where the magnetic dipole moment of the loop is $m=\pi a^2I$ and we are looking at the far field of the dipole so $a^2+z^2\approx r^2$. The magnetic scalar potential of a dipole, valid away from its source current, is
$$\Phi_m=\frac{\mu_0}{4\pi}\frac{mz}{r^3}$$
so
$$\begin{align}\vec B&=-\vec{\nabla}\Phi_m=-\frac{\mu_0m}{4\pi}\left\langle-\frac32\frac{2xz}{r^5},-\frac32\frac{2yz}{r^5},-\frac32\frac{2z^2}{r^5}+\frac{r^2}{r^5}\right\rangle\\
&=\frac{\mu_0m}{4\pi r^5}\langle3xz,3yz,3z^2-r^2\rangle\end{align}$$
Along the $z$-axis, $x=y=0$, $r=z$ so
$$\vec B=\frac{\mu_0}{4\pi}\frac{2m}{r^3}\hat k$$
So the scalar potential checks. Along the $x$-axis, $y=z=0$, $r=x$ and
$$\vec B=-\frac{\mu_0m}{4\pi x^3}\hat k$$
so
$$\frac{\partial B_z}{\partial x}=\frac{3\mu_0m}{4\pi r^4}$$
This is the same as for a scalar potential of
$$\Phi_m=-\frac{\partial B_z}{\partial x}\,xz$$
because then
$$\vec B=-\vec{\nabla}\Phi_m=\frac{\partial B_z}{\partial x}\langle z,0,x\rangle=\frac{3\mu_0m}{4\pi r^4}\langle0,0,a\cos\phi\rangle$$
For the field at our loop due to an identical loop at $\langle-r,0,0\rangle$. So then the force on our loop is
$$\begin{align}\vec F&=\int Id\vec\ell\times\vec B\\
&=\int_0^{2\pi}I\langle-a\sin\phi,a\cos\phi,0\rangle\,d\phi\times\frac{3\mu_0m}{4\pi r^4}\langle0,0,a\cos\phi\rangle\\
&=\frac{3\mu_0mI}{4\pi r^4}\int_0^{2\pi}\langle a^2\cos^2\phi,a^2\sin\phi\cos\phi,0\rangle\,d\phi\\
&=\frac{3\mu_0mI\pi a^2}{4\pi r^4}\hat i=\frac{3\mu_0m^2}{4\pi r^4}\hat i\end{align}$$
So in this configuration the force is repulsive if the dipoles are aligned but attractive if pointed opposite one another. You can see this if you have a neocube where oppositely aligned strings of magnet spheres attract at the equators.
For electrons things are more complicated. If we have little bits of charge $\Delta q$ and mass $\Delta m$ orbiting around an axis at distance $a$ and angular frequency $\omega$ then the charge passing a point in the orbit per unit time is $I=f\Delta q=\frac{\omega\Delta q}{2\pi}$ so the magnetic dipole moment is $m=I\pi a^2=\frac12\omega a^2\Delta q$. The angular momentum is $L=rp=a\Delta ma\omega$ so we would get
$$\frac mL=\frac12\frac{\Delta q}{\Delta m}$$
But life isn't so simple for a couple of reasons: the ratio is actually a factor of $g_e\approx2.002$. Also there is a question of whether we want to consider the total angular momentum of the electrons or the component at right angles to the vector from one to another. We choose the latter and this component is $L_z=\frac12\hbar$ so we get
$$m=\frac{g_eq_e\hbar}{4m_e}$$
So the magnetic force is
$$F=\frac{3\mu_0}{4\pi r^4}\left(\frac{g_eq_e\hbar}{4m_e}\right)^2$$
Comparing with the electrostatic force,
$$F=\frac{q_e^2}{4\pi\epsilon_0r^2}=\frac{\mu_0c^2q_e^2}{4\pi r^2}$$
We see that the magnetostatic force is greater when
$$r\lt\frac{\sqrt3g_e\hbar}{4m_ec}=\frac{\sqrt3(2.002)(1.055\times10^{-34}\text{J}\cdot\text{s})}{4(9.11\times10^{-31}\text{kg})(2.998\times10^8\text{m}/\text{s})}=3.35\times10^{-13}\text{m}$$
This is so close that it kind of negates the whole calculation though: the electrons are going to be moving pretty fast so there will be a magnetic force due to their relative motion and at these distances you really need quantum electrodynamics to analyze the problem. Still, it gives you an idea of how close you have to get so that the net force has at least a chance of being attractive.


Structure of the periodic table The periodic table I understand is arranged by order of increasing atomic number after Henry Moseley's X-ray experiment.

But I also learned that the periodic table also has its shape due to the types of subshells present in certain sections called blocks of the table.

However, I still do not understand who shaped the modern periodic table so it mimics the filling of lattermost subshells of atoms of each element (in order of increasing number of electrons- from left to right as well as top to bottom for the elements.)

My Question is : was the periodic table Mendeleev and Moseley devised not in a shape that can be used to predict the filling order of each elements, and that the periodic table was later adapted to fit this concept of orbitals in?

Mendeleev was the first to realize that properties recur among the elements. To highlight this, he devised the early periodic table based on atomic masses. Moseley came along and stated that the periodic properties of elements was a function of their atomic numbers rather than their atomic masses.

It was Niels Bohr who came up with the idea of shells and subshells. He developed his own take on the periodic table, which looked somewhat like this:

Finally, coming to your question, A lot of derivatives of the periodic table were made by various scientists during the 1920's - 1940's. The version we use today was made by Glenn T. Seaborg and his colleagues.

  Seaborg and collaborators had synthetically produced several new elements with atomic numbers beyond uranium, the last naturally occurring element in the table. Seaborg saw that these elements, the transuranics (plus the three elements preceding uranium) demanded a new row in the table, something Mendeleev had not foreseen. Seaborg’s table added the row for those elements beneath a similar row for the rare earth elements, whose proper place had never been quite clear, either.

Further reading:

How the periodic table went from a sketch to an enduring masterpiece
Further development of the Periodic Table - Wikipedia


Electrolytic refining of copper In refining copper, an aqueous solution of copper(II) sulphate is electrolyzed, using impure copper as the anode and pure copper as the cathode.

At the anode, metal impurities higher than copper in the electrochemical series such as zinc, iron and lead are oxidized first, then copper loses its electrons to form copper(II) ions. Metal impurities lower than copper in the electrochemical series are hardly oxidized and deposit at the bottom of the tank as sludge. So, I have two questions:

If ions from metals more reactive than copper exist in the solution, why aren't they preferentially discharged first to form zinc or iron metal on the other copper electrode first rather than copper?
Why can the pure copper cathode be thickened with pure copper only?

To your first question: Consider what happens if you put an iron nail or a zinc pellet into a copper(II) solution: the less noble metals will oxidize and reduce the copper ions to copper(0). So even if some of the less noble metal ions are reduced and deposited at the cathode, they would react with the copper ions (that are the most numerous in this scenario).

To your second question: if ions of more noble metals than copper were in the solution, they would be deposited on the cathode. However, by carefully tuning the voltage and current, no more noble metal ions are produced at the anode. As stated by Ivan Neretin in the comments, the cathode will reduce what is available to be reduced: copper ions (which are most numerous and the potential fits).


Studying chemical engineering My question is pretty simple. Does studying chemical engineering require memorizing manufacturing processes and other complex chemical mechanisms, or does it rely more on the practical side of things focusing on how stuff actually works.

I can speak based on my experience studying chemical engineering as an undergraduate at an ABET accredited university in the United States and working in industry.

A typical curriculum would include a focus on fundamentals which can then be applied in specific industries or in different cases rather than an in depth study of specific manufacturing processes or specific chemical reactions. This process is intended to develop a core of knowledge to approach specific processes or reactions and also as a preparation for learning more

So for example an undergraduate might start off taking calculus, and general chemistry and proceed to additional chemistry like organic and possibly phsyical chemistry. Starting in chemical engineering a student will encounter fundamental fields of chemical engineering which are fundamental tools such as mass and energy balance, thermodynamics, fluid dynamics, heat transfer, mass transfer, and chemical kinetics. These are all tools which can be used as a chemical engineer to evaluate processes. Also a review and study of types of what's called unit operations would be included.

Unit Operations are like the building blocks of many processes. The include operations like heat exchangers, distillation or absorbion columns, cooling towers, reactors, tanks, and more.

An understanding of chemical engineering fields such as kinetics or thermodynamics or heat transfer combined with an understanding of unit operations can be combined to learn more about a specific process.

I am sure there are some folks out there with more insight so I'll be very interested to hear from them as well!


Correlation between the equilibrium constant and delta G I understand this up until the last two steps. How does $\ln(K_\mathrm{eq})$ become $2{,}303\log(K_\mathrm{eq})$ and how does $-RT\cdot2{,}303$ become $-1{,}42$?

The first step is simple:

$$\frac{{\ln x}}{{\log x}} = \ln 10 \approx 2.303$$

thus

$$ - RT\ln {K_{{\text{eq}}}} \approx  - RT \times 2.303 \times \log {K_{{\text{eq}}}}$$

The second step is terrible because $\text{kcal}$ is used and the units are missing:

$R = 1.9872041 \times {10^{ - 3}}{\text{ kcal K}^{ - 1}}{\text{ mo}}{{\text{l}}^{ - 1}}$

$T = 310{\text{ K}}$ (my guess)

$$\begin{aligned}
  \Delta G =  & \; {-} RT \times 2.303 \times \log {K_{{\text{eq}}}} \\
   =  & \; {-}1.9872041 \times {10^{ - 3}}{\text{ kcal K}^{ - 1}}{\text{ mol}^{ - 1}} \times 310{\text{ K}} \times {\text{2}}{\text{.303}} \times \log {K_{{\text{eq}}}} \\
   =  & \; {-}1.42{\text{ kcal mol}^{ - 1}} \times \log {K_{{\text{eq}}}} \\
\end{aligned} $$


α-Halo ketone and strong base: Favorskii vs SN2 Let us consider an α-halo ketone, with at least one α-H on the α carbon not containing the halogen. If I add a strong base, what is likely to happen: $\mathrm{S_N2}$ or Favorskii rearrangement?

I have learnt that α-halo carbonyls undergo excellent $\mathrm{S_N2}$ due to the lowered energy of its LUMO. And the Favorskii is a characteristic of the α-halo carbonyls. How to decide between the two?

I would comment however I am new and have yet to unlock that feature.

Do you have an exact substrate of interest? I have studied the Favorskii rearrangement a tiny bit in a graduate mechanisms course and if your seeking a very general answer to a general question; The favorskii seems to be favorable when your substrate is containing rings that are conformationally rigid where our alpha-halo-ketone moiety is found.

Check out Eaton's 1964 synthesis of Cubane to find two really elegant Favorskii reactions in the synthesis of Cubane. Notice for almost all of our examples of favorskii reactivity we notice a common pattern in the substrate scope?

Rings! My theory is conformational rigidity of a ring is partially responsible and we can confidently point our fingers to Favorskii where cyclopropanone intermediates are possible AND energetically favorable (conformation argument).

If our substrate scope is including non-ring alpha-halo-ketones.. We may want to consider alternatives like SN2 to predict reactivity.


What is the proper term for lines of constant volume and constant enthalpy on a thermodynamic diagram? I know an isotherm is a line of constant temperature.
Isobar is a constant pressure line.
Isentrope is a line of constant entropy.

But what do you call a line of constant volume? An Isochore?
And what do you call a line of constant enthalpy? An "isenthalp"?

But what do you call a line of constant volume? An Isochore?

Correct.

  And what do you call a line of constant enthalpy?

The process is isenthalpic. More than one engineering/thermodynamic text I found by Googling indicates they are simply referred to as "isenthalpic lines." See this for example.


What can I use to make evaporating fake snow? I want to create the effect that it's snowing in my living room. Ideally, there would be a chemical or substance that would be able to be sprayed out of a snow machine or pressurized canister etc. This substance would start off solid and would have a chemical reaction with the air that would cause it to evaporate into thin air, a là Harry Potter. Or even if the reaction took a couple hours it would be ok.

The phase transition you are talking about (solid to gas) is called sublimation. There are a few compounds that sublime at room temperature and atmospheric pressure, but the one I think would best suit your purpose is dry ice. If spread into the air as a fine powder, it could sublime before it even touches the ground, especially if the air is warm.

It's important to consider the safety aspects of this, though. $\ce{CO2}$ sublimes at -$78º\mathrm{C}$, so if too much of this powder lands on people, you could end up with some cases of frostbite. Additionally, you would need to keep this space well-ventilated since you will be significantly increasing the concentration of $\ce{CO2}$. I wouldn't imagine it would be too hard to meet these standards, though.


Is NaCl the only precipitate when you add HCl to concentrated NaCl When aqueous HCl is added to a saturated salt water solution, is NaCl the only precipitate or does some HCl crystalize as well?

Sources online say that NaCl crystallizes but they don't say what happens to the HCl

https://bouman.chem.georgetown.edu/S02/lect8/lect8.htm
http://www.csun.edu/~ml727939/coursework/695/common%20ion%20effect/recrystallization%20of%20NaCl%20solution%20with%20HCl.htm

$\ce{HCl}$ is at ambient conditions a gas, very well soluble in water, dissociating as a strong acid.

When there is excess of it in solution, that cannot be kept solved, it does not precipitate, but evaporates or even boils out, depending on scenario.


Compound class naming convention: heteropoly-acid vs isopoly-acid There are heteropoly-acids and isopoly-acids, two classes of polynuclear compounds with and without heteroatoms within its molecular structures, respectively. These are the names widely used in the literature for a long time[1].
Why two non-homogeneous prefixes ("iso-" vs "hetero-") are used? Wouldn't it make more sense to denote these names with consistent prefixes showing parity ("hetero-" vs "homo-"; "aniso-" vs "iso")? For instance, IUPAC Red Book utilized the terms "homopolyatomic" vs "heteropolyatomic" in chapter IR-5 Compositional Nomenclature, and Overview of Names of Ions and Radicals [2, p. 68].
Wouldn't heteropoly-acid and homopoly-acids pair be morphologically more accurate, since it's a heteroatom presence that allows that distinction, just like in case of homocyclic vs heterocyclic or homopolyatomic vs heteropolyatomic species?
References

Middleton, A. R. J. Chem. Educ. 1933, 10 (12), 726 DOI: 10.1021/ed010p726.
IUPAC “Red Book” Nomenclature of Inorganic Chemistry; Connelly, N. G., Royal Society of Chemistry (Great Britain), International Union of Pure and Applied Chemistry, Eds.; IUPAC Recommendations; Royal Society of Chemistry Publishing/IUPAC: Cambridge, UK, 2005.

Wouldn't heteropoly-acid and homopoly-acids pair be morphologically more accurate (than heteropoly-acid and isopoly-acids)?

I completely agree with andselisk of this suggestion because the former makes more sense than latter in this regards. For instance, as correctly pointed out in the question, the IUPAC Red Book has utilized the terms "homopolyatomic" vs "heteropolyatomic" in the chapter, IR-5: Compositional Nomenclature, and Overview of Names of Ions and Radicals (p. 68-82, Ref.1). A few examples given for the homopolyatomic cations in the section IR-5.3.2.3 are $\ce{O2^+}$, $\ce{S4^2+}$, $\ce{Hg2^2+}$, and $\ce{Bi5^4+}$ (p.71). Likewise, few examples of heteropolyatomic cations are also given in the following section IR-5.3.2.4, which are $\ce{NH4^+}$, $\ce{H3O^+}$, $\ce{PH4^+}$, and $\ce{SbF4^+}$ (p.71). Thus, it make perfect sense of using the terms "homopolyatomic" and "heteropolyatomic" in the context, since it's the presence of heteroatom in the latter case ($\ce{N, O, P,}$ and $\ce{Sb}$, respectively) that allows the distinction.
Nonetheless, in another chapter in Ref.1 (IR-8: Inorganic Acids and Derivatives, p. 124-141) has utilized the terms "homopolyoxy acids" and "heteropolyoxy acids" to described polyoxy acids:

Examples 1, 3–6 and 14 above demonstrate that homo- and heteropolyoxoacids and their partially dehydronated forms […] (p.136, ref.1)

For instance, the example 1 and 4 mentioned in the excerpt are $\ce{H2PO7^2-}$ and $\ce{H4[SiW_{12}O_{40}]}$, respectively (Section: IR-8.4, p. 135, Ref.1). As we all know that $\ce{H4PO7}$ is an example of isopoly-acids and $\ce{H4[SiW_{12}O_{40}]}$ (Ref.2) is an example of heteropoly-acids. Thus, even if these unconventional names are frequently use in inorganic chemistry literature to date, I think IUPAC is already recognized the necessity to keep consistency in the naming.

References:

Nomenclature of Inorganic Chemistry: IUPAC recommendations 2005 (Issued by the Division of Chemical Nomenclature and Structure Representation in collaboration with the Division of Inorganic Chemistry); Prepared for publication by N. G. Connelly, et al.; RSC Publishing (for the International Union of Pure and Applied Chemistry): Cambridge, United Kingdom, 2005 (ISBN: 0-85404-438-8).
Li Min Yang, Zhi Kui Yin, and Li Qiang Wu, "$\ce{H4[SiW_{12}O_{40}]}$: An efficient catalyst for the synthesis of new spiro-[dibenzo[a,i]xanthene-14,3′-indoline]-2′,8,13-triones," Chinese Chemical Letters 2012, 23(3), 265-268 (DOI: https://doi.org/10.1016/j.cclet.2012.01.002).


Can plastic leak into water from a HDPE "2" plastic container at room temperature? Here in USA supermarkets often sell water in plastic 1 gallon jugs. These jugs seem to all be stamped with the text "HDPE" (meaning that the material is High-density polyethylene) and a resin number of "2" on the bottom.

I had one of these jugs, containing distilled water, on a counter in the kitchen, and another one in the refrigerator. For some reason the water on the counter had a light plastic taste, whilst the water in the fridge did not.

Could this taste be because of plastic leaking into the water? Or what else could cause the plastic taste?

HDPE
has a molecular weight more than several hundreds of thousands,
"leaking" is plainly impossible.
This taste is the tase of stale water, nobody knows
what that taste is actually.
Maybe its just the taste of not cold enough water,
lacking some dissolved air, or some bacteriae
develop after contact with air and light.
(google for pseudomonas)
The only substance which ever got into the jug prior
to the water is the air (or some other gas?) used to
inflate the jug in production.
But that cause would impair the taste of the water in
the fridge as well.

BTW why do You drink such botteled water?
What about the tap water in your town/city/village?


How does hybridisation affect an otherwise chiral centre? In basic theory, a carbon atom with four nonidentical substituents attached, makes a chiral centre. Thus any molecule is chiral as long as it has a chiral centre (except meso compounds). I thought identifying a chiral centre was straight forward until I saw the following video.

Is Carbon (with the ability to bond 4 covalent bonds) the only atom qualified as a chiral centre (stereogenic centre)? Wikipedia link states, other atoms like Nitrogen or Sulfur could also be a chiral centre.

The above video shows that other atoms too can be chiral as well... However a comment in the same video, replies that a hybridised Nitrogen atom is not qualified to be a chiral centre. Original comment : That Nitrogen is sp2 hybridized, the lone pair of electron are in a unhybridized p-orbital; it would not be considered a chiral center.

With reference to this question 1-, it has number of comments, but not quite the answer I am after. On the other hand this question 2- has more emphasis on the lone pair for chirality - an atom with lone pairs can't be a chiral centre.

So given these, why can't a hybridised atom be a chiral centre?

I really wouldn't put too much stock into YouTube comments. Often, it's just the blind leading the blind. The comments on this video have all sorts of incorrect statements, it's really not where you want to be learning from...

Firstly, chiral centres are not restricted to only carbon. Source: IUPAC Gold Book. With that out of the way, we can enumerate the three main possibilities for nitrogen.

$\mathrm{sp^3}$ nitrogen with a lone pair is usually not considered to be a chiral centre, as the rate of inversion is very rapid (google "nitrogen inversion").

There are exceptions, such as if the amine geometry is constrained by a bicyclic system (something similar to quinuclidine), but that's a discussion for another day.
$\mathrm{sp^3}$ nitrogen without a lone pair e.g. in ammonium cations, can be chiral if there are four different groups attached. So these two molecules are enantiomers of each other.

In fact, it is this context in which Wikipedia says that nitrogen can be a chiral centre.

  Racemization by Walden inversion may be restricted (such as ammonium cations), or slow, which allows the existence of chirality.

The implication is that nitrogen inversion is fast in the case where there is a lone pair. However, it is very unclear, so I don't blame you.
$\mathrm{sp^2}$ nitrogen, e.g. amide nitrogen, is planar so it cannot be chiral. So, this video you watched is incorrect in stating that the amide nitrogen is a chiral centre.

And, that also means the comment you quoted is correct. But, I would caution that it is not a direct cause and effect relationship between "$\mathrm{sp^2}$ hybridisation" and "not a chiral centre". It is more like: because it is $\mathrm{sp^2}$, therefore the geometry around it is planar; and because it is planar, it cannot be chiral.

As was mentioned in the comments, phosphorus and sulfur behave differently. This has been covered multiple times here; here are some links. 1 2 3 4


What household substances could be distilled I'm looking to experiment with distillation to have a basis for understanding the theory slightly better, and get a better feel for it.

I'm hoping for some suggestions as to what I could try distilling, such as purifying drinking alcohol. Any single idea would be appreciated, best answer would be awarded to an idea where:

The resulting purity would be straight forward to approximately test
No part of the distillation presents a significant hazard (e.g. toxic fumes, high volatility)
No particularly specialist or expensive equipment is required (e.g. extreme temperatures)

I think that pretty much covers it. I'm just looking for one or two ideas which I can experiment with besides alcohol.

Thanks

This is tricky. Besides water, nearly everything has some hazard associated with it.  And even water distillation involves high temperature (100C).  Be careful distilling alcohol: the azeotrope of distilled alcohol is 95% ethanol which is VERY flammable! Plus, technically speaking, distilling alcohol may be illegal where you live.

One thing you could do is distill water with dissolved substances, most likely table salt (NaCl).  This would help you test the principle that distillation is a technique for separating substances. In this case, distilling salt water should yield pure water. A more visual example would be brewed coffee or tea, which is a suspension and should separate into water and crud!

And after that, I'd be careful! For example, concentrating acetic acid (white vinegar) is dangerous, as concentrated acetic acid is both corrosive and has choking fumes.  I would not even consider trying something that is not food grade, given your requirements.


How to calculate the bond energy of C-Cl? $$\ce{C(s) + 2 Cl2 (g)  -> CCl4 (g)}$$
\begin{align}
\Delta H^\circ_\mathrm{f} (\ce{C(g)}) &= 121~\mathrm{kJ/mol}\\
\Delta H^\circ_\mathrm{f} (\ce{Cl(g)}) &= 716~\mathrm{kJ/mol}\\
\Delta H^\circ_\mathrm{f} (\ce{CCl4(g)}) &= -106.48~\mathrm{kJ/mol}\\
\end{align}

Basically I already answered this, but I was a bit confused about the result, cause I don't know if there is a bond that is negative, cause am familiar with the energy given to "dissociation" of a bond or the "formation" of it , but here it's given the energy to the formation of chlorine and carbon, I don't know if that's the same as the formation of the bond or not!

\begin{align}
\Delta H^\circ_\mathrm{f}(\ce{C-Cl}) &= \frac{\Delta H^\circ_\mathrm{f}(\ce{C(g)}) + 4 \Delta H^\circ_\mathrm{f}(\ce{Cl(g)}) - \Delta H^\circ_\mathrm{f}(\ce{CCl_4(g))}}{4} \\
\Delta H^\circ_\mathrm{f}(\ce{C-Cl}) &= \frac{121 + 4 \cdot 716 + 106.48}{4} \\
\Delta H^\circ_\mathrm{f}(\ce{C-Cl}) &= 772.87~\mathrm{kJ/mol}
\end{align}

Should I take the $\triangle H^\circ_\mathrm{f}(\ce{Cl(g)})$ same as $\triangle H^\circ_f(\ce{Cl-Cl})$? In that case I'm going to multiply it by $2$, not $4$.
Should the  $\Delta H^\circ_\mathrm{f}(\ce{C-Cl})$ be prefixed by a minus $-$ or a plus $+$, cause I don't know if it's an energy that is needed to formulate $\ce{CCl4}$? I'm not saying that the enthalpy of the bond should be negative, but the direction.

The bond dissociation enthalpy* $D$ is the enthalpy change for conversion of bonded into dissociated molecules following homolytic cleavage of the bond in question:
$$\ce{A-B (g)->A(g) + B(g)},\qquad D(\ce{A-B})=\Delta H^\circ_f (\ce{A(g)})+\Delta H^\circ_f (\ce{B(g)})-\Delta H^\circ_f (\ce{A-B(g)})$$
This is the property usually discussed as representing the "strength" of a bond, and is positive because the process of breaking a bond requires an input of heat (is endothermic). Breaking stronger bonds requires a greater positive enthalpy change.
The enthalpy change for formation of the bond by the dissociated molecular constituents is negative the dissociation enthalpy:
$$\Delta H^\circ (\ce{A-B})=-D (\ce{A-B})$$
and is a negative quantity because heat is released when bonds are formed (the process is exothermic).
The enthalpy of formation of a $\ce{Cl-Cl}$ bond is the enthalpy change for the following process:
$$\ce{2Cl(g)-> Cl2(g)}$$
and therefore can be expressed as
$$\Delta H^\circ (\ce{Cl-Cl})=\Delta H^\circ_f (\ce{Cl-Cl(g)})-2\Delta H^\circ_f (\ce{Cl(g)})$$
and since $\Delta H^\circ_f (\ce{Cl2(g)})=0$,
$$\Delta H^\circ (\ce{Cl-Cl})=-2\Delta H^\circ_f (\ce{Cl(g)})=-D(\ce{Cl-Cl})$$
In the process
$$\ce{C(g) + 4 Cl (g)  -> CCl4 (g)}$$
four $\ce{C-Cl}$ bonds are formed, so the average heat of formation of one bond is
$$\Delta H^\circ(\ce{C-Cl}) = \frac{1}{4}\left(\Delta H^\circ_\mathrm{f}(\ce{CCl_4(g))}-\Delta H^\circ_\mathrm{f}(\ce{C(g)}) - 4 \Delta H^\circ_\mathrm{f}(\ce{Cl(g)}) \right) $$
which is a negative quantity (exothermic).
Note again that the corresponding bond dissociation enthalpy is then a positive quantity.

*Sources (from IUPAC Gold Book website):

PAC, 1990, 62, 2167. (Glossary of atmospheric chemistry terms (Recommendations 1990)) on page 2176.
PAC, 1994, 66, 1077. (Glossary of terms used in physical organic chemistry (IUPAC Recommendations 1994)) on page 1089.


Effect of concentration on molar conductance It has been stated that the molar conductance($\Lambda_m$) of strong electrolytes is not affected to a greater extent on dilution and so to find the limiting value of molar concentration($\Lambda_m^0$) we can extrapolate the graph of molar conductance vs the concentration ($\sqrt c$). But as the concentration would near zero shouldn't the molar conductance decrease sharply to the range of say $10^{-8}$ which is the conductance of distilled water as the amount of solute that is the strong electrolyte would be negligible to consider?Also in this case extrapolating the graph should not be possible?

The molar conductivity of a strong electrolyte $(\Lambda_m)$ is given as the ratio of measured conductivity $(\kappa)$ to the molar concentration $(c)$:
$$\Lambda_m=\frac{\kappa}{c}$$

However, as you note, the value of $\Lambda_m$ is not independent of concentration. The reason for this is that $\kappa$ does not scale linearly with concentration.

$$\Lambda_m = \Lambda_m^\circ -K\sqrt{c}$$

In this equation, $\Lambda_m^\circ$ is the limiting or intrinsic conductivity of the electrolyte and $K$ is an empirical constant. The interpretation of $\Lambda_m^\circ$ is the molar conductivity at infinite dilution as $c\rightarrow0$.

$$\lim_{c\rightarrow 0}{\Lambda_m}=\Lambda_m^\circ$$

The infinite dilution scenario is different from the scenario when there were no ions in solution to begin with, just as mathematically $\lim_{x \rightarrow 0}{f(x)}$ does not always equal $f(0)$ (often because $f(0)$ is undefined).

In the limiting case of $c\rightarrow 0$, the ion concentration never reaches zero, but it does keep getting close. You can never dilute a solution containing a solute to zero concentration (i.e. pure solvent).

If you graph $\Lambda_m$ as a function of $\sqrt{c}$, you should get a straight line with a non-zero intercept $(\Lambda_m^\circ)$. This value is an empirical parameter used to model data. It can also tell you something about the intrinsic conductivity of ions (see this Wikipedia article). However, as you say, at exactly $c=0$, he conductivity should be negligible. At any other value of $c$, however the conductivity should follow from the equation given. Thus, the conductivity is not necessary linear in the infinitesimal range around $c=0$:

$$\Lambda_m=\begin{cases}
10^{-8}\approx0, &c=0\\
\Lambda_m^\circ - K\sqrt{c}, &c>0 \end{cases}$$


when Al2O3 melts which bonds are broken? Previously, I was trying to understand why $\ce{Al2O3}$ has a lower melting point than MgO. An answer on quora said

  Do not confuse the character of the bond between the metal and oxygen with the bonds between molecules in the compound. The polar character of the bonds between aluminum and oxygen is indeed less than between magnesium and oxygen but this describes the energy needed to separate the metal from oxygen. When a solid melts it doesn’t break down the bonds between oxygen and aluminum or magnesium. The intermolecular bonds (that is bonds between one $\ce{Al2O3}$ and another $\ce{Al2O3}$) are in fact weaker than the bonds between two magnesium oxide molecules. This results in the greater melting point of magnesium oxide in spite of the less covalent character of the bonds within the molecule.

But if $\ce{Al2O3}$ is ionic and forms a giant lattice, which bonds that are "between one $\ce{Al2O3}$ and another $\ce{Al2O3}$" are broken?

Right, the text you are mentioning is not clear enough. It is not correct  to say that the molecule $\ce{Al_2O_3}$ exists and is separated from the next $\ce{Al_2O_3}$. Such a molecule does not exist. A sample of $\ce{Al_2O_3}$ is made of a superposition of $\ce{Al^{3+}}$ ions and $\ce{O^{2-}}$ ions. It is not made of a superposition of molecules $\ce{Al_2O_3}$. Now the fact that MgO melts at a higher temperature than $\ce{Al_2O_3}$ is due to the stronger attraction between $\ce{Mg^{2+}}$ and $\ce{O^{2-}}$ ions. For some reason, the attraction between $\ce{Al^{3+}}$ ions and $\ce{O^{2-}}$ is weaker.


Reaction between magnesium metal and sodium hydroxide: Does this even happen? Came across this question in a book:

  Among aluminum, zinc, tin and magnesium, which metals (if any) will react with an aqueous solution of sodium hydroxide to displace dihydrogen?

The answer to the question is : Aluminum, zinc and tin.

Their reasoning?

  Under normal conditions, magnesium metal does not react with an aqueous solution of sodium hydroxide.

Given that the book does not go on to explain the rationale behind this one-liner, naturally my question would be: "Why not?"

QUESTION:

Why doesn't magnesium react with an aqueous solution of sodium hydroxide and release dihydrogen?

[@paracetamol thinks]

Initially, I was under the impression that they were talking about the oxide layer that forms over magnesium metal (the very reason we buff the surface of magnesium ribbons with sandpaper before igniting it), which would give the reaction a slow start. Then, the statement could've been: "Magnesium metal does not immediately react with aqueous sodium hydroxide (owing to the presence of an oxide layer)".

But earlier on in the chapter, there is explicit mention of the reaction between aluminum metal ( which also naturally come with an oxide coat) and aqueous sodium hydroxide, to form sodium aluminate and dihydrogen. There was no mention of an oxide layer hampering the reaction.

[@paracetamol thinks harder]

I then (cursorily) looked at the Electrochemistry aspect of the reaction; because whenever we're asked if a metal reacts with a mineral acid (to liberate dihydrogen), we're expected to flip open our table of standard electrode reduction potentials and say:

  Metals having a higher electrode reduction potential than that for the reaction $$\ce{2H^{+} -> H_{2}}$$ (which is arbitrarily taken as zero volts) will be able to liberate dihydrogen from the solution. However, metals such as copper, silver, mercury, etc which have a lower standard electrode reduction potential than that for $H^+$ will not liberate dihydrogen.

When I compared the standard electrode reduction potentials of aluminum, zinc, tin and magnesium, I've observed that all of them have electrode reduction potentials that are higher than the electrode reduction potential for $H^+$. Meaning, that all those metals (magnesium included) should be able to displace dihydrogen from the solution, at least, from the high-school electrochemistry perspective.
 (Yes, I know the textbook mentions an alkali in the question, but what I'm trying to point out is, that since the four metals have higher electrode reduction potentials that $H^+$, then they ought to behave the same way. So if aluminum, zinc and tin can liberate dihydrogen (as the textbook claims) then why not magnesium too?)

[@paracetamol Googles]

I Googled my question, and arrived at this Quora post. Same question as mine, but it's generated such lackluster answers, that they would only be worthy of...well...Quora. O:)

[@paracetamol bamboozled]

I can't for the world figure out why magnesium doesn't react with aqueous sodium hydroxide (if it doesn't). Anyone?

For a metal to react with sodium hydroxide solution it's oxide must be amphoteric, not just basic.  Alkaline earth metals are called that for a reason; except for beryllium the oxides are purely basic (and in fact moderate to strong in basic strength) in water, so sodium hydroxide won't make those metals react.

Magnesium, as a typical alkaline earth metal, can be made to react with acids displacing hydrogen, even when the "acid" is a transition metal salt that hydrolyzes slightly like cupric sulfate.


How to calculate the concentration of NO at equilibrium? $\ce{N2(g) + O2(g) <=> 2NO(g)}$, $K_c = 4.1 \cdot 10^{-4} (T=2000~^\circ\mathrm{C})$
  What is $\ce{[NO]}$ when a mixture of $0.20~\mathrm{mol}$ $\ce{N2(g)}$ and $0.15~\mathrm{mol}$ $\ce{O2(g)}$ reach equilibrium in a $1.0~\mathrm{L}$ container at $T=2000~^\circ\mathrm{C}$?

When K is $\cdot10^{-4}$, you can assume the change is negligible, and just using the concentrations you can get from the given mole amount and the container volume, I get the right answer. When plugging back into, to see if it matches the give $K_c$, it's right:
$$((0.20~\mathrm{M}~\ce{N2})(0.15~\mathrm{M}~\ce{O2}) \cdot 4.1\cdot10^{-4} K_c) = x^2.$$ Ans: $\ce{[NO]} = 0.0035~\mathrm{M}$.

From what I've read and heard you should be able to do it 'the hard' way, using the quadratic formula, like you would if $K_c$ wasn't to $10^{-4}$, and small enough. But I can't seem to get the right answer trying to do it that way.

Rearranging everything, the equation I get to plug into the quadratic formula is
$$x^2 - 0.35x + 0.03,$$ then I get $0.2$ and $0.15$ as the answers. $0.2$ is too high, subtracting it from the $0.15~\mathrm{M}$ $\ce{O2}$ initial wouldn't be right. So using the $4.15$ as $x$, the change in $\mathrm{M}$, and subtraction that from the initial concentrations I have, then plugging that into the equation for $K_c$, even tried rearranging it in case I have got something mixed up, but I don't get anything close to the given $K_c$.


Is it absolutely necessary to use centrifuge to separate ethylene glycol from precipitate? I've been trying to synthesize intermetallics of $\ce{XY}$ type using transition metal salts and main group element salts. For my system, EG, DEG, and TEG are good solvents. However, because of them being highly viscous, I've experienced normal drying doesn't do the trick and that the precipitate remains puffy even after drying (which shouldn't be the case) before sintering but by using centrifugation, I've been successful at removing the puffy texture of the precipitate.

Recently, I came upon this paper where they haven't used centrifugation for the main sample. Could anyone please explain how does this work?

What they do in the paper is what i was going to suggest even before reading it: Addition of ethanol in the ethylene glycol suspension lowers the viscosity of the solution making it more filterable. Since their nanoparticles are insoluble in ethanol they can add as much is necessary. After filtering the solid out they wash with plenty of ethanol to remove the leftover ethylene glycol (which since it has a high boiling point cant be dried easily). Hence they end up with their product contaminated with ethanol left over from the washings. However, ethanol is very easy to get rid off by applying vacuum to the powder due to being reasonably volatile. So the ethanol contaminated solid can be dried in a schlenk line or even a rotary evaporator overnight to remove any ethanol left. Note that you can do this not only with ethanol but any solvet that is volatile (b.p up to 110 deg C, you can do more but is a pain), miscible with the solvent you are trying to get rid of, and wont react with the compound that is being isolated. Therefore the answer to your question is: no, in general it is not necessary. However, if your precipitate is too fine for the filter you have available, and you cant get it to crash out as larger particles or if it is a very small amount and you cant scale it up you might actually need to use one.


Confusion regarding $\delta$ charge compared with electronegativity I heard on a lecture of an online site that, let us say, we have two orbitals (1 and 2). Now, it will be like a spherical cloud. Now, what they say is that

Now, we can notice that for orbital number 1. There is a dark region and a light region (right side). Now, what they say is that the darker region means the electron finding probability is more there. Similarly, for the number 2, electron finging probability is more on left side.
Therefore, we say that 1 develops $\delta +$ charge and 2 develops $\delta -$ charge. Therefore, the two will attract each other.
Now, my teacher on the other hand told it a bit differently:

The more electronegativity means more $\delta -$ charge. But, I have a confusion in it. Between $\ce{H}$ and $\ce{F}$. I have drawn a line. Fluorine has 7 electrons. My teacher says that $\ce{F}$ and $\ce{H}$ share that one electron where $\ce{H}$ has one and fluorine 7th electron. Now, since $\ce{F}$ is more electronegative. The electron (by hydrogen) is more pushed towards fluorine. My confusion is that if $\ce{F}$ is more electronegative than $\ce{H}$. Then, the electron should be present more towards the $\ce{H}$ side.

Although, OP indicated the confusion regarding $\delta^\pm $ charge compared with electronegativity, the text body indicated whole lot of confutations over interatomic/intermolecular forces. Thus, I start with the electronegativity.
According to IUPAC Gold Book, electronegativity is the concept introduced by Linus Pauling as the power of an atom to attract electrons to itself. There are several definitions of this quantity. However, Wikipedia gives a simple to understand version better suited for OP:

Electronegativity is the measurement of the tendency of an atom to attract a shared pair of electrons (or electron density).

The relative electronegativity of $\ce{H}$ atom has given a value of $\pu{2.1 eV^{-\frac{1}{2}}}$ and the electronegativity for each of other atoms in the periodic table have been calculated relative to this value for $\ce{H}$ atom (IUPAC Gold Book). These calculations are beyond OP’s field of study so should be disregarded. The important point for OP to consider is the higher the associated electronegativity, the more an atom or a substituent group attracts electrons. For instance, the electronegativity of $\ce{F}$ is $\pu{4.0 eV^{-\frac{1}{2}}}$ and hence, shared electron fair in $\ce{H-F}$ bond stays more closed to $\ce{F}$ than to $\ce{H}$. As a result, $\ce{F}$ carries permeant $\delta^- $ charge while $\ce{H}$ carries   permeant $\delta^+ $ charge in the $\ce{HF}$ molecule, creating a permanent dipole. Hence, $\ce{HF}$ is a polar molecule. Therefore, among $\ce{HF}$ molecules, there are always dipole-dipole interactions (the image 2 of OP's text). Since the positive end of these dipole is $\ce{H}$ atom, the relevant dipole-dipole interaction is called $\ce{H}$-bonding, the strongest intermolecular force among dipole-dipole interactions. These dipole-dipole interactions occur only in molecules.
Now, I think I have addressed the OP's main confusion:

My confusion is that if $\ce{F}$ is more electronegative than $\ce{H}$, then, the electron should be present more towards the $\ce{H}$ side.

According to the above explanation, PO's understanding of electronegativity is not correct. In reality, if an atom is relatively more electronegative than that of the other atom it attaches to, that means it attracts the shared pair of electrons, increasing its electron density compared to its protons (gaining $\delta^- $ charge). Meanwhile, the other atom feels decreased electron density compared to its protons (gaining $\delta^+ $ charge). This is completely opposite of what OP's indicated understanding.
On the other hand, the London dispersion force is also one of the intermolecular forces, but is the weakest among them and occurs between atoms as well. For example, why do you think gaseous $\ce{He}$ become a liquid when the temperature is lowered sufficiently? You can find the answer in chem.purdue.edu:

The London dispersion force is a temporary attractive force that results when the electrons in two adjacent atoms/molecules occupy positions that make the atoms/molecules form temporary dipoles. Thus, this force is sometimes called an induced dipole-induced dipole attraction. London forces are the attractive forces that cause nonpolar substances to condense to liquids and to freeze into solids when the temperature is lowered sufficiently.

An atom or a molecule can develop a temporary (instantaneous) dipole when its electrons are distributed unsymmetrically about the nucleus because of the constant motion of the electrons (e.g., the image 1 of OP's text). The explanation can be depicted in an image:

For instance, assume two helium atoms approaching each other. When they get closed to each other, the electrostatic attraction force between the nucleus of first $\ce{He}$ atom with two positive charges and the two electrons in the second atom causes each atom to polarize  (creating an induced dipole in each atom). Note that at the same time, there is  an electrostatic repulsion acing as well between electrons of each other as depicted in the image above.
Dispersion forces are present between all atoms or molecules. In molecules, these forces are present regardless of whether they are polar or nonpolar. London dispersion forces depends on few factors:

Larger and heavier atoms and molecules exhibit stronger dispersion forces than smaller and lighter ones. In a larger atom or molecule, the valence electrons are farther from the nuclei than in a smaller atom or molecule. They are less tightly held and can more easily form temporary dipoles (e.g., $\ce{Cl2}$ is a gas at room temperature and $\pu{1 atm}$ pressure, while $\ce{Br2}$ is a liquid and $\ce{I2}$ is a solid under the same conditions).
The strength of London dispersion forces varies according to molecular shapes. For example, $n$-$\ce{C5H12}$ and $neo$-$\ce{C5H12}$ are both pentanes, but former is more cylindrical shape while the latter is more spherical shape. Although, both have the same molar mass, $n$-pentane is a liquid (average boiling point of pentanes: $\pu{36.1 ^\circ C}$) at room temperature and $\pu{1 atm}$ pressure, while $neo$-pentane is a gas (boiling point: $\pu{9.5 ^\circ C}$) under the identical conditions.

Also, London dispersion forces are depend of the molecule's polarizability (The ease with which the electron distribution around an atom or molecule can be distorted is called the polarizability). London dispersion forces tend to be stronger between molecules that are easily polarized, but weaker between molecules that are not easily polarized.


Bromination of cyclohexene Cyclohexene is reacted with bromine in carbon tetrachloride in the dark. The product of the reaction is:

I know that reaction with bromine in $\ce{CCl4}$ will result in vicinal dibromocyclohexane. So the answer is either A or B. But how to decide between A and B? Since its an SN2 reaction it would lead to inversion of the reacting compound. Does this have to do anything with the answer?

To answer this question you have to look at the reaction mechanism. Since the reaction is conducted in the dark, you don't have any radicals to worry about. Under these condition bromine reacts with an electron-rich alkene in a electrophilic addition reaction in the course of which a cyclic bromonium ion is formed. This bromonium ion is highly reactive and in the absence of other nucleophiles reacts with the leftover $\ce{Br-}$ ion in an $\text{S}_{\text{N}}2$ reaction which leads to the opening of the 3-ring and formation of the trans product.

Update:

As @BenNorris pointed out, the ring structure in the question are drawn wrongly. So, I will add the correct version: The dibrominated product will initially be formed in a diaxial conformation (this is always the case due to the ring geometry and the mechanism of the $\text{S}_{\text{N}}2$ reaction). This is usually a very unfavourable comformation so the ring will subsequently flip (if possible) to the diequatorial conformation.


Examples of non-volatile, but flammable liquids? As far as I know, most flammable liquids are volatile. Are there flammable liquids that are non-volatile?

The element caesium melts at 28.4°C which is slightly above "room temperature" but below human body temperature, so I think we can consider it a liquid in "normal" conditions. It is so flammable it readily self-ignites in the air, but if you put it in inert atmosphere you will see that it isn't particularly volatile.


Omission of locants for functional groups in IUPAC names I'm having a bit of conflict while naming molecules as per IUPAC the nomenclature rules. If we take the following compound for example,

IUPAC would say that that correct name is 2-phenylacetaldehyde. But then for this molecule,

IUPAC designate it the name is chloroacetaldehyde.

Why does the aldehyde with a benzene ring as a functional group receive a '2' while the aldehyde with a halide as a functional group does not? I figured the molecules mentioned above wouldn't require a number to designate their functional groups given they're both substituents to an ethane parent chain.

Also, would it be acceptable to drop 'aldehyde'  and replace it with 'ethanal'. I've always been taught to add '-al' with a '-CHO' functional group?

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There is only one possible phenylacetaldehyde, so its IUPAC name is in fact phenylacetaldehyde. 2-Phenylacetaldehyde is its systematic IUPAC name, but many chemicals' official and systematic IUPAC names differ, like propan-2-one for example. The reasoning for this is often historical and to reflect what is most commonly used in the scientific community.

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Chloroacetaldehyde is quite obviously an aldehyde, but  switching the position of the hydrogen and the chlorine as you suggested results in a completely different class of molecule known as an acyl chloride. There is therefore only one possible chloroacetaldehyde and no need to specify the chlorine's location, though its systematic IUPAC name is indeed 2-chloroethanal.


Efficiency of various Ewald-type schemes for the summation of electrostatic interactions I am looking into implementation of Ewald summation techniques for a number of point charges in periodic 3D space (i.e. a molecular simulation box with periodic boundary conditions). The “mainstream” Ewald-type techniques I know are:

original Ewald scheme
particle-mesh Ewald (PME)
smooth particle-mesh Ewald (SPME)

However, while I can find papers describing each of those, I have not been able to find a comprehensive article/white paper/technique note comparing their implementations and their CPU timings on a typical dense system (liquid, protein in solvent, whatever…). What resources do you know of on this topic?

I also unsuccessfully tried to look for such a paper in the past but below are some useful resources I've found. The only definitive answer I can give you is that traditional Ewald summation will be slower than a particle-mesh technique for any relevant system sizes (this is backed up in several of the references below).

Chapter 4.9 of the GROMACS user manual has some very nice discussion on their various implemenations of long-range electrostatics and appropriate references.

The LAMMPS documentation also has quite a few references and some bits of helpful discussion.

You can also try digging through the AMBER manual which uses a PME implementation but I haven't read any of this myself.

Also here you can find implementation details for the PPPM method in HOOMD (helpful if you're interested in GPGPU implementations).

I think the most relevant bit of discussion is probably this excerpt in the GROMACS manual:

  The Particle-Particle Particle-Mesh methods of Hockney & Eastwood can
  also be applied in GROMACS for the treatment of long range
  electrostatic interactions [99]. Although the P3M method was the ﬁrst
  efﬁcient long-range electrostatics method for molecular simulation,
  the smooth PME (SPME) method has largely replaced P3M as the method of
  choice in atomistic simulations. One performance disadvantage of the
  original P3M method was that it required 3 3D-FFT back transforms to
  obtain the forces on the particles. But this is not required for P3M
  and the forces can be derived through analytical differentiation of
  the potential, as done in PME. The resulting method is termed P3M-AD.
  The only remaining difference between P3M-AD and PME is the
  optimization of the lattice Green inﬂuence function for error
  minimization that P3M uses. However, in 2012 it has been shown that
  the SPME inﬂuence function can be modiﬁed to obtain P3M [100]. This
  means that the advantage of error minimization in P3M-AD can be used
  at the same computational cost and with the same code as PME, just by
  adding a few lines to modify the inﬂuence function. However, at
  optimal parameter setting the effect of error minimization in P3M-AD
  is less than 10%. P3M-AD does show large accuracy gains with
  interlaced (also known as staggered) grids, but that is not supported
  in GROMACS (yet).

If you're really looking to dive in, you can look at the source code for these packages (I believe you need a license for AMBER) to see the actual implementations.


Why does the binding of DMSO to metal via the S atom (k-S) strengthen the S=O bond? So, DMSO (dimethyl sulfoxide) can form metal complexes in both $\ce{\kappa-O}$ and $\ce{\kappa-S}$ mode i.e. binding with the oxygen or the sulfur respectively. The general explanation given is that softer metal ions bind to the soft centre $\ce{S}$ and hard metal ions bind to the hard centre $\ce{O}$.
From the IR data it can be inferred that the binding of DMSO to a metal ion via $\ce{O}$ causes the $\ce{S=O}$ bond stretching frequency to go down. For example the $\ce{S=O}$ stretch in free DMSO (liquid) is $\pu{1055 cm-1}$, whereas the stretch in $\ce{[Co(dmso)6]+}$ appears at $\pu{950 cm-1}$. This makes sense to me because the oxygen can only act as $\pi$ donor (apart from the usual $\sigma$ donation), and donate electrons to the metal $\mathrm{t_{2g}}$ from bonding $\pi$ orbitals, which will weaken the bond between $\ce{S}$ and $\ce{O}$.
However, when DMSO binds to a metal ion via $\ce{S}$, the $\ce{S=O}$ bond stretching frequency goes up. For example, in $\ce{PdCl2(dmso)2}$ shows the $\ce{S=O}$ stretch at $\pu{1116 cm-1}$.
But how is this possible? The sulfur is already electron deficient due to the strongly electronegative $\ce{O}$, and should be a $\pi$-acceptor, and accept electrons from metal into the $\pi^*$ orbitals. This means that the $\ce{S=O}$ bond should also be weakened in this case. But this does not happen. Why? Does it have something to do with the molecular orbitals of DMSO?
[Note: According to this post, the $\ce{O}$ is mainly $\pi$-donor and the $\ce{S}$ $\pi$-acceptor for DMSO, and I agree with the argument in that answer. But it does not say anything about the strengthening of the $\ce{S=O}$ bond in $\ce{S}$ bonded complexes.]
The pictures of HOMO and LUMO of DMSO respectively:

The confusion seems to arise from the incorrect statement that the S-O bond has pi character and that there exist $\pi$ and $\pi^*$ molecular orbitals in DMSO. Instead, DMSO (despite the way it is often drawn) is most accurately depicted as having three single bonds and one lone pair on the the S atom (clearly visible as a dominant element in the HOMO shown in the question). The trigonal pyramidal geometry (also shown in the question) is consistent with this description.
The S-O bond is a dative bond, meaning that the S atom provides both electrons rather than having one electron come from each atom. That results in a formal positive charge on S and negative charge on the O. The O therefore is best represented as having one bond (to S) and three lone pairs, similar to the oxygen atom in an alkoxide ion. Because of the charge separation, the S-O bond also has a strong ionic component.
[Aside: so-called hypervalent sulfur compounds are typically drawn with double bonds that are actually combined sigma + ionic bonds rather than sigma + pi bonds. DMSO is not the only one. Keep this in mind in the future when trying to understand properties of sulfur-containing molecules.]
We now consider what happens when DMSO interacts with a transition metal. Similar to the O of an alkoxide ion, the O atom in DMSO can interact with metals as both a $\sigma$ and $\pi$ donor because of the multiple lone pairs. Both the $\sigma$ and $\pi$ donations result in decreased electron density on O, reducing its charge and weakening the S+/O- ionic interaction. The weaker ionic interaction means that the bond overall weakens, consistent with the decrease in the observed stretching frequency. There is no need for an S-O $\pi$ interaction to exist in order for this to happen.
If instead the S atom interacts with the metal, it does so primarily as a $\sigma$ donor via its one lone pair. There may be capacity for a $\pi$ acceptor interaction with $\sigma^*$ orbitals, as has been proposed for ligands such as triphenylphosphine, but if this occurs, it is with the $\sigma^*$ orbital associated with the S-C bonds, not the S-O bond, consistent with the LUMO shown in the question, which has a large contribution from the py orbital on S (ie the p orbital in the plane of the C atoms). Thus there is no weakening of the S-O $\sigma$ interaction due to $\pi$ acceptance.
As long as the $\sigma$ donation is stronger than the $\pi$ acceptance (which is what we expect), there is a net loss of electron density from S, which further increases the positive charge on the already electron deficient S atom. Increased charge means an increased strength in the ionic S+/O- interaction, so the S-O bond overall is stronger, consistent with the observed increase in the S-O stretching frequency.


How does the splitting in this molecule take place? Consider the following molecule and data:

  $$
\begin{align}
\mathrm{H_a} &= \pu{1.0 ppm} &\quad J_\mathrm{ab} &= \pu{5.0 Hz}\\
\mathrm{H_b} &= \pu{3.0 ppm} &\quad J_\mathrm{bc} &= \pu{8.0 Hz}\\
\mathrm{H_c} &= \pu{6.0 ppm} &\quad J_\mathrm{ac} &= \pu{1.0 Hz}\\
\end{align}
$$

  Draw a splitting diagram for $\mathrm{H_b}.$ $(\pu{1 box} = \pu{1 Hz})$

Tried to draw the splitting diagram and ended up with this diagram:

I am also stuck as to how to draw the spectrum of the whole molecule as I feel $\mathrm{H_a}$ would be triplet of triplet. Am I on the right track and how to draw the sketch of the spectrum?

According to given data, you may consider expected peak patterns are not complicated such as $\mathrm{ABX}$ or $\mathrm{A_2B}$  patterns (more reading look here. This is simply a  $\mathrm{A_2X}$  pattern ($\delta_X = \pu{6 ppm}$ and $\delta_A = \pu{1 ppm}$ without geminal coupling of two $\ce{H_b}$ protons), and hence, peak is a $dt$ for $\ce{H_b}$ as depicted below with relative heights of the peaks:

Similarly, $\ce{H_c}$ is also a $dt$ resonance with $J = 1 \text{ & } \pu{8 Hz}$ and $\ce{H_a}$ is a $tt$ with $J = 1 \text{ & } \pu{5 Hz}$. OP should able to figure out the intensities of relevant peaks.

A valuable reference:

Roy H. Bible, Jr., Ph.D., In Interpretation of NMR Spectra: An Empirical Approach; Springer Science: New York, NY, 1965 (https://doi.org/10.1007/978-1-4684-8288-1) (Note: Most relavent to this question is chapter 4 of the book titled "Higher-Order Spin Patterns and Multiple Resonance").


Hydrate decomposition equation steps When a hydrate decomposes it loses water molecules in steps.
Does each step have a chemical equation?

If so is the following equations for decomposition of cupric sulfate hydrate correct?
$$
\begin{align}
\ce{CuSO4 * 5 H2O &→ CuSO4 * 4 H2O + H2O}\\
\ce{CuSO4 * 4 H2O &→ CuSO4 * 3 H2O + H2O}\\
\ce{CuSO4 * 3 H2O &→ CuSO4 * 2 H2O + H2O}\\
\ce{CuSO4 * 2 H2O &→ CuSO4 * H2O + H2O}\\
\ce{CuSO4 * H2O &→ CuSO4 + H2O}
\end{align}
$$

One can rarely tell a priori what the steps are, and the results obtained experimentally may seem not intuitive at all if one tries to deduce the dehydration steps solely from the crystal structure (e.g. from the arrangement of the aqua ligands and water molecules trapped within the crystal lattice).¹

TG-DTG-DTA techniques, often coupled with powder XRD, are commonly used to determine the exact process behind each step interpreting mass change over time.
Alternatively, you can deduce the possible steps from the initial structure, but this approach is less accurate.

So, for the classical textbook example of thermal decomposition of copper(II) sulfate pentahydrate the following steps of dehydration (1–2) and calcination (3–4) can be distinguished [1]:

$$
\begin{align}
\ce{CuSO4 * 5 H2O &→ CuSO4 * H2O + 4 H2O}\tag{1}\\
\ce{CuSO4 * H2O &→ CuSO4 + H2O}\tag{2}\\
\ce{2 CuSO4 &→ CuO * CuSO4 + SO3}\tag{3}\\
\ce{CuO * CuSO4 &→ 2 CuO + SO3}\tag{4}
\end{align}
$$

You could also assume the loss of four water molecules as a single step since in the crystal structure copper(II) atom has octahedral environment with four $\ce{H2O}$ and two $\ce{SO4^2-}$ ligands.
Writing chemical equation for each act of the water molecule loss is not a sin, but it makes little practical sense as the intermediate crystallohydrates are unstable and rapidly undergo further dehydration.

Notes

¹ From the crystal structure one may assume that during the first step only one water molecule is eliminated from the lattice (Figure 2a) since copper(II) is coordinated octahedrally by four water molecules in equatorial area and has two sulfate ligands in apical positions (Figure 2b).

Figure 2. Crystal structure of $\ce{CuSO4 * 5 H2O}$, data from CIF for Chalcanthite. Color scheme: $\color{#EEEEEE}{\Large\bullet}~\ce{H}$; $\color{#FF0D0D}{\Large\bullet}~\ce{O}$; $\color{#FFFF30}{\Large\bullet}~\ce{S}$; $\color{#C88033}{\Large\bullet}~\ce{Cu}$. a: Unit cell content. b: Coordination environment and nearest neighboring water molecules of the copper(II) center.

However, this is not the case even though lattice water molecules are kept in place via H-bonds and should posses seemingly weaker interaction with the cationic part.
In reality many assumptions suitable for estimation of melting points of molecular crystals, e.g. regarding the number of bonds and their types don't work well for ionic crystals and so the dehydration patterns can only be established exactly experimentally.

References

Gadalla, A. M. Kinetics of Thermal Decomposition of $\ce{CuSO4 · 5H2O}$ to $\ce{CuO}$. International Journal of Chemical Kinetics 1984, 16 (6), 655–668. https://doi.org/10.1002/kin.550160604.


Determining the coupling on a substituted benzene ring I can't seem to find any clear explanation of how benzene ring protons couple with each other when mono and disubstituted with the same constituent.

My professor claims that if a benzene ring is para disubstituted with X, the remaining 4 protons exhibit a doublet interference pattern. This seems strange to me because they are equivalent, and my last two organic chemistry professors said the information isn't useful diagnostically, but never explained why.

Also, he claims that if a benzene ring is monosubstituted the proton in the para position will display a triplet interference pattern. Is this right? Symmetry seems to say that the proton would be a doublet.

Do benzene rings have their own rules for coupling?

Benzene rings follows the same general principles as other spin systems for coupling.

A symmetrically para-substituted benzene, such as 1,4-dichlorobenzene, will have a 1H spectrum which will contain just a singlet. All four proton environments are identical. An XY para-disubstituted benzene will have the aromatic peaks that appear somewhat like two sets of doublets. The appearance is actually incredibly useful diagnostically,as it gives a useful indication of the type of ring substitution you have. They are, however, not true doublets, and consist of second order coupling arising from coupling between magnetically non-equivalent sites. That is, the splitting of the apparent doublets is not a true reflection of the coupling constant.

The para proton on a mono-substituted benzene ring does appear as a triplet. The ortho coupling giving rise to the ~7Hz splitting comes from coupling from two chemically equivalent protons at the meta position. Coupling to 2 equivalent protons will give rise to a 3 line splitting (from your 2nI+1 rule). Strictly speaking, a mono-substituted benzene ring is just a very special subset of a para-disubstituted benzene ring system, but does not typically show the classic splitting expected in a para system, because the 5J coupling is essentially zero.


What is the isomer distribution in monosubstituted fluorobullvalene? Bullvalene (tricyclo[3.3.2.02,8]deca-3,6,9-triene) is a fluxional molecule able to interconvert any two carbon atoms through a series of degenerate Cope rearrangements (for more information, see the Wikipedia article on bullvalene).  At slightly above room temperature bullvalene produces a single line in its 1H-NMR spectrum.  Considering fluorobullvalene, there are 4 possible isomers:

1-fluorotricyclo[3.3.2.02,8]deca-3,6,9-triene, i.e. the fluorine could be attached to any of the 3 equivalent cyclopropyl carbons
3-fluorotricyclo[3.3.2.02,8]deca-3,6,9-triene, i.e. the fluorine could be attached to any of the 3 equivalent olefinic carbons alpha to the cyclopropane ring
4-fluorotricyclo[3.3.2.02,8]deca-3,6,9-triene, i.e. the fluorine could be attached to any of the 3 equivalent olefinic carbons beta to the cyclopropane ring
5-fluorotricyclo[3.3.2.02,8]deca-3,6,9-triene, i.e. the fluorine could be attached to the unique tris-allylic carbon

Again, at room temperature or thereabouts, only a single line is observed in the 1H, 13C and 19F-NMR spectra.  I remember once reading that about 80% of the fluorine is in the tris-allylic position (4), most of the remainder is attached to the two olefinic carbons (2 and 3) and a very small amount is attached at the cyclopropyl position (1). This was often cited as an example to support the argument that fluorine prefers to form bonds to carbon when the involved carbon orbital has high p-content (low s-content).

Does anyone know what the isomer distribution is?  Those 4 numbers and a reference would be very helpful.

Extensive NMR studies of substituted bullvalenes were done in the 1960's and 70's, by Oth et al. Much of their original work was published in German. Thankfully, some of these topics were revisited (in English!) during the 1990's by Luz et al., and these included some rather nice low temperature 19F and 13C NMR studies in both solution and solid state (and make it much quicker and easier for me to read).

Let us assign the following isomers 1–4, as is done in the question:

To answer your question:

Oth et al.[1] studied fluorobullvalene at −25 °C, and found no evidence by 19F of isomer 1. The equilibrium populations of isomers 4:3:2 was 78%:7%:15%. However, they could not distinguish between isomers 3 and 2, so it could be 78:15:7. In any case, predominantly isomer 4. They do state, however, that isomer 1 must exist ["Das C-Isomer muss also durchlaufen werden."], as the Cope rearrangement mechanism for isomer 4 to go to isomer 3 must go through isomer 1.

Luz et al.[2] studied the solution state Cope rearrangement of a number of monosubstituted bullvalenes.
At −30 °C, they identify the ratio of isomers 4:3:2 to be 87%:3.4%:9.6%. Again no evidence of isomer 1 by 19F at this temperature. However, at −55 °C, they observe a peak in the 19F spectrum that they ascribe to isomer 1, with an intensity of 0.2% relative to isomer 4. This peak broadens beyond detection at −40 °C. The ratios of isomers at −55 °C, then, was found to be 1.0:0.018:0.065:0.002 for isomers 4:3:2:1.

And just out of interest:

Of course, the other monsubstituted bullvalenes all have different preferred isomer distributions in solution. In the solid state, all monosubstituted bullvalenes studied by Luz crystallize as well-ordered single isomer crystals. Fluorobullvalene[3] crystallizes exclusively as isomer 4. Cyanobullvalene, bullvalenecarboxylic acid[4], and (ethylthio)bullvalene[5] crystallize exclusively as isomer 3. Bromo- and iodobullvalene[6] crystallize entirely as isomer 2.

Useful references for you:

Oth, J. F. M.; Merényi, R.; Röttele, H.; Schröder, G. Fluor-, chlor- und jodbullvalen. Tetrahedron Lett. 1968, 9 (36), 3941–3946. DOI: 10.1016/S0040-4039(00)72372-0.
Poupko, R.; Zimmermann, H.; Müller, K.; Luz, Z. Dynamic NMR Investigation of the Cope Rearrangement in Solutions of Monosubstituted Bullvalenes. J. Am. Chem. Soc. 1996, 118 (34), 7995–8005. DOI: 10.1021/ja954004t.
Müller, K.; Zimmermann, H.; Krieger, C.; Poupko, R.; Luz, Z. Reaction Pathways in Solid-State Processes. 1. Carbon-13 NMR and X-ray Crystallography of Fluorobullvalene. J. Am. Chem. Soc. 1996, 118 (34), 8006–8014. DOI: 10.1021/ja954005l.
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Predicting sigma bond overlap strengths of s-s, p-p, s-p, sp-sp etc How can we qualitatively predict $\sigma-$bond strengths of overlap between:

$\ce{s-s}$
$\ce{s-p}$
$\ce{p-p}$
$\ce{sp-s}$
$\ce{sp-p}$
$\ce{sp-sp}$

etc.?

My school-book says $\ce{s-s}$ overlap bond strength is greater than $\ce{s-p}$ overlap which is greater than $\ce{p-p}$

Another book Inorganic Chemistry by JD Lee states this:

Are the facts stated in the above books correct or justified? Is there any order of orbital overlap strengths? If so can we predict it qualitatively?

I think the issue might be what orbitals are being referred to. Both books could be correct if they are referring to different $\mathrm{p-p}$ overlap. For example, suppose two atoms are bonding along the z-axis. $\mathrm{p_z}-\mathrm{p_z}$ overlap would be greater than $\mathrm{s-s}$ because the lobes of the p orbitals extend out to meet each other between the two molecules. If however we looked at $\mathrm{p_x}-\mathrm{p_x}$ overlap, the $\mathrm{s-s}$ overlap would be greater because the p lobes don't extend toward each other. Increasing hybridization increases overlap because it combines the directionality of p orbitals (that they extend out away from the atom) and the electronic density of s orbitals (larger space for overlap to occur).


What are the steps of Hartree-Fock Theory? Outline the steps involved in the Hartree–Fock method for the calculation of molecular electronic structure.

I understand the basics of HF, in that it does not account for electron correlation (neither non-dynamical or dynamical) but I do not understand what the question means by steps.

Are the steps of the HF method just those of the self-consistent procedure or is this something difference?

In addition, does Hartree-Fock account for the antisymmetry requirement or not? I have notes that say it does, because it uses a Slater determinant, but I also have notes that say DFT accounts for the antisymmetry neglected by HF.

Yeah, I think it asks you to write the energy as expectation value, vary one orbital and get the Roothan equation.
The Hartree Fock model is the extension of the Hartree model where the wavefunction was schematized by the following guess:
\begin{equation}
\psi=\phi_{a}(1)\phi_{b}(2)...\phi_{n}(n)
\end{equation}
If you work out the equations, you'll see the equations look a lot like the HF ones, but you'll miss the exchange term. This is because the wavefunction is not in the correct fashion to describe a fermionic system not made by infinitely distanced parts.
So you need to plug in the antisymmetric form of the wavefunctions (Slater determinants are a basis in the space of antisymmetric functions). Also, the HF approximation requires you to just approximate the wavefunction with just one determinant, just one piece of the expansion. This is clearly a simplification but it can be correct enough if there is a significative HOMO-LUMO gap and the system is closed shell.
So the HF model gets some correlation (known as fermi hole) meaning that electrons having same spin do avoid each other (you enforced it using antisymmetric wavefunctions) but the electrons don't avoid themselves respecting the Coulomb repulsion (two electrons cannot be in the same place even if they have not the same spin) this is a first piece of correlation you lose, the other being the fact you simply are using one determinant. Unfortunately correlation is described as everything going beyond HF because of Quantum information theory, so you have to be careful when you talk of correlation of HF. As it regards DFT, you'll plug in it functionals, so you may include functionals in it that allow you to recover some more correlation, but remember that DFT functionals are, most of the time, semiempirical.