Small but modest displacements at negligible forces from DFT When looking at a Gaussian log file today, I noticed the following information in the convergence criterion section after an analytical frequency calculation was performed:

         Item               Value     Threshold  Converged?
 Maximum Force            0.000001     0.000450     YES
 RMS     Force            0.000000     0.000300     YES
 Maximum Displacement     0.000190     0.001800     YES
 RMS     Displacement     0.000030     0.001200     YES

This struck me as odd. How could the forces be so low but the maximum displacement still be non-negligible? Does anyone have any hypothesis about why this phenomenon would occur? This makes me wonder if the convergence condition for the maximum displacement is unreasonably low if essentially no acting forces can result in a maximum displacement that's merely 1 order of magnitude lower than the default threshold. I believe the force units are Hartrees/Bohr and the distance units are angstroms.

For clarity I will assume Gaussian performs a generic Newton-Raphson minimization (NR), which should suffice to explain the phenomenon.
In NR, the linear problem
$$
\nabla\nabla^\ast E \Delta + \nabla E = 0
$$
is solved, where $\Delta$ is the displacement. In order to arrive at "large" displacements despite "small" (but non-zero) forces ($-\nabla E$), it suffices for the Hessian ($\nabla\nabla^\ast E$) to have eigenvalues close to 0 itself, because a small number is divided by an even smaller number. This happens when the potential energy surface is very flat.


Is there a way to concentrate the vitamin content of a particular food? I'm looking for a way to extract vitamins out of common food sources of, well, vitamins.

Let's take, as an example, Vitamin A which is commonly found in carrots. A carrot (weighing around 60g) should contain around 500µg of retinoic acid (RA). RA is not soluble in water so in principle if I just dehydrate the carrot (which has around 50g of water) I would be left with around 10g of matter that still contains those 500µg of RA. I think that, with this simple procedure, I would be concentrating the amount of RA in my sample.

But maybe (surely) there are better ways to achieve this, to extract/purify a particular vitamin from food. Any ideas?

Also, some Vitamins are, instead, soluble in water, so we would need to apply another method for those.

And, finally, I would like to accomplish this using simple (household, if possible) techniques. It doesn't matter if I don't reach 100% purity, not even 1%, but if I can extract a 1g sample from my original carrot that has most of the RA still in it that would be huge.

You are correct that retinoids are poorly soluble in water. Vitamins A (retinol), D (calciferol), E (tocopherol) and K (phylloquinone) have high hydrophobicity and therefore should be extracted using organic non-polar solvents.

All these vitamins can be identified using UV/Visible spectroscopy.

Since your focus is on retinoids I shall focus on that (but the principles remain the same for other hydrophobic biomolecules).

Note: carrot has beta-carotene and it is a precursor for retinol and not a retinoid itself.

Crush and homogenise the tissue
With some plant tissues you may need to grind the frozen tissue using liquid nitrogen repeatedly. Tissues containing polyphenols and lignins would need additional treatment but not necessary for carrots.
Precipitate proteins (optional; not required in your case).
Extract using solvents such as hexane or chloroform. Isopropanol has also been used for beta-carotene extraction from carrots (Fikselová et al. 2008)[1]. Gundersen (2006)[2] summarizes the methods of extraction and analysis of retinoids from animal tissue.
To purify, you need to perform chromatography. You can differentiate the retinoids using UV/visible spectroscopy. Beta-carotene is easy to detect; it is bright orange in colour and has an absorption maximum at 450nm.

References:

[1] . Fikselová M., Šilhár S., Mareček J., Frančáková H., Czech J. Food Sci., 2008, 26 (4), 268-274.
[2].  Gundersen T. E., J. Neurobiol., 2006, 66 (7), 631–644.


Effect of increase in volume of container on concentration of chlorine at equilibrium conditions How the concentration of the chlorine would be affected if the volume of the container is increased?
  $$\ce{4 HCl + O2 <=> 2 Cl2 + 2 H2O + $Q$}$$

My initial thought was that the total and partial pressures were decreased on increasing the volume of the container. But I did not account for the fact that the $\ce{HCl}$ did not play a role in this pressure.

So, according to my previous process, the reaction should proceed in the backward direction, concentration of $\ce{Cl2}$ should decrease. On the other hand, if we exclude the $\ce{HCl}$ factor, the reaction should proceed in forward direction increasing the concentration if $\ce{Cl2}$.

Please help me understand what will be the actual effect.

According Le Chatelier's Principle, when a change in concentration or pressure is applied to an equilibrium, the equilibrium changes to counteract the change.

In your equilibrium, the volume of the container is increased, causing a decrease in pressure. To counteract this, the side of the equilibrium with the highest number of moles is favoured. I.o.w. the equilibrium will move somewhat to the left ($\ce{4HCl + O2}$).


Carbon-13 NMR of dimethyl 3,4,5,6-tetraphenylphthalate? I am trying to assign the peaks here to dimethylphenylphatalate but all the benzene rings are throwing me off. All I can get is that the further most signal is the carbonyl carbon (168.71) and the one prior to that (143.27) might be the ester carbon (from the methyl). Can someone help?

Here's a numbered drawing of the title compound to help us keep track of things.  First thing we notice is that there are 13 different carbon atoms in the structure and we see 13 signals in the spectrum, that's a good start.

Let's start with the easy ones.  The upfield peak at 52.28 is in the expected position for C1, the ester methyl carbon.  Down at the other end of the spectrum, the peak at 168.71 is in the expected position for an ester carbonyl carbon (C2).

In the region for aromatic carbons we see 5 low intensity peaks between 143 and 132.  These belong to the (unprotonated) quaternary carbons 3, 4, 5, 6, 10.  Using diethyl phthalate as a model compound, I would guess that the peak at 132.19 is from C3.  The two peaks at 138.73 and 138.60 have the same height and are closely spaced suggesting very similar environments, I might guess them to be C6 and C10, just not sure which is which.  Noting that the peak for the carbon corresponding to our C5 is downfield of the peak corresponding to our C4 in the model compound, I would expect the same trend to hold here, just with each of the peaks shifted downfield due to attachment of a phenyl group.  So I would assign the peak at 143.27 to C5 and the peak at 139.29 to C4.

This leaves the protonated aromatic carbons for us to consider.  The 2 peaks at 125.97 and 126.83 are half the intensity of the other protonated aromatic signals, therefor assign them to C9 and C13 as there are only half as many of these carbons as compared to the other protonated aromatic carbons. It's not clear which peak belongs to C9 and which to C13.  This leaves us with the 4 signals due to C7, C8, C11 and C12.  Using the same resonance structure - delocalization arguments I made in your benzaldehyde cnmr question from yesterday, I would expect C7 and C11 to be downfield from C8 and C12.  Therefor, I would assign C7 and C11 to the peaks at 129.71 and 130.88, not knowing which is which.  C8 and C12 then belong to the signals at 126.88 and 127.41, not knowing which belongs to which.

Obviously there is some guess work in here and if it was an important project I'd do some further experiments and look for some additional model compounds in order to confirm these assignments and remove the discrepancies, but this is a good start.


Any idea why tap water goes stale overnight? We have great tap water here in Hungary. My only problem that it goes stale overnight sometimes in some glasses or sometimes even the 2L water boiler has this problem too. But microbes cannot resist 100°C or 20% vinegar, so I thought it might be rather a chemical compound what causes it. It is really hard to get rid of it. When I use boiling water or vinegar it does not remove it. When I use soap and water several times, it removes it most of the times, but only for a few days. The only working solution was not using the affected glass for a month. It is hard to descibe the taste, but it is something like how new Chinese clothes or paint thinner smell so it is sort of strong plastic taste. The longer I use a glass without washing it every day, the bigger the risk it develops this feature (2 days are enough for certain glasses, while others never develop it), so maybe microbes are producing it after all, but the compound is long lasting even if they die off. Any idea what causes it and more importantly how to get rid of it permanently without having to wash my glasses, boiler, water jug each day?

Exclude reasons one by one
Surface biological contamination: use two vessels, wipe the one of them with H2O2, hydrogen peroxide solution, leave another untreated. Check if time till smell appears is different by a few days. Bacterias can survive 20% vinegar and 100*C for a short time, some of them, if they are in a sleeping stage, it is not quite reliable method i would say.
Tap water biological contamination: use two vessels with water, add a few tablespoons H2O2 with water in one, and just water in other. Check if time till smell appears is different by a few days.
Tap water chemical contamination: buy some bottled water, use two vessels, one with tap water, another with bottled water. Check if it will be different in a few days.
Bulk material of the vessel releasing chemicals: leave water in different material vessels, one is glass, another one is your preferred one. Check if glass vessel has less smell adter a few days. Glass releases the least amount of chemicals into the water.
Dust accumulation: use two vessels, cover one and leave another one open, check if smell will appear faster in one.
Psychological factor: use two identical vessels, ask your friend to give you one or another, randomly, blindfold yourself, do a test 10 times, check if your score is above 8 correct guesses out of 10 tries.
After completing this tests you will narrow down your search significantly to give you some idea about what is going on.


What is parachemistry? What does a parachemist do? I was reading the Washington Post article An atomic town revels in its nuclear past as tunnel collapse raises contamination concerns and came across a quote from someone who was said to have been a "parachemist" at the Hanford nuclear facility in the US state of Washington.

Is parachemistry a recognized subfield of chemistry? If not, what else could this mean?

I'll try to present the best I could find. The titles "para", "meta" and "ortho" were perhaps given by Hermann Kolbe (1818-1884) way back in the nineteenth century. For one thing, Kolbe had the reputation of being a well recognized teacher and a skilled and productive chemist. He is also well known today (partly) as the author of a critique of the foundation of stereo-chemistry, with was established by the Dutch chemist, Dr. J.H. vant Hoff. He had directed bitter remarks not only towards vant Hoff for his hypotheses (which was perhaps the boldest), but at nearly the entire realm of contemporary organic chemistry.
Background

$\ldots$ Once he gave his amanuensis the empirical formulas for three novel compounds for which their discoverer had just assigned structural formulas. The man reported back in a half hour with several more candidate structures, some of which looked more probable to Kolbe. Kolbe concluded that structure theory is a dangerous toy, especially for inexperienced chemists, and that structures are often assigned "in one's sleep," with little or no empirical warrant. $\ldots$

Privately, Frankland "entirely dissent[ed]" from this judgment and Volhard also objected privately; he joined Frankland in expressing regret over the personal character of many of Kolbe's remarks. Kolbe's response was pretty straightforward. As he remarked he must criticize and contest the chemical ideas that he considered false and worthless, he would be happy if his views were contested. He later wrote to Varrentrapp that he would rather be considered sharp-tongued than cowardly.
The crusade

In Kolbe's third and fourth retrospects, those for 1873 and 1874, he went after his favorite example of structuralist excess: Kekulé's benzene theory. He absolved Kekulé himself of much of the blame, for he was convinced that Kekulé regarded the theory merely as an intriguing and useful hypothesis. However, most chemists by this time viewed the hexagon as "infallible dogma," as "the Pope is for Catholics." They were true fanatics, Kolbe wrote, and viewed him (Kolbe) as a rank beginner, of weak understanding. They were right, he thought, for he could not understand arguments built "in the air" or "on loose shifting sand."

The end of their sand castles was not distant. Kolbe continued,

The modern chemist, who knows exactly what a chemical compound looks like in its middle and its end, how the six carbon atoms of benzene are symmetrically linked together in a plane, who then further purports . . . to have a clear conception of the spatial arrangement of the atoms, of their ortho, meta, and para positions, who determines the positions of all of the atoms in the compound, has long since abandoned the solid ground of exact science; the scientist has become a metaphysician.

He then named true physicists as "orthophysicists" as opposed to "paraphysicists" who construct hypotheses on the basis of experimental work of others and "metaphysicists" who believed that they can dispense entirely with empirical evidence.
In his own view, he was an "orthochemist" and Kekulé was a "parachemist. But as the  years went by, he increasingly believed that even the term "parachemist" was too kind; his opponents were "metachemists", to whom "chemical theories are as cheap as blackberries".

References:

Kolbe Versus the “Transcendental Chemists”: the Emergence of Classical Organic Chemistry, A. J. Rocke
The Quiet Revolution Hermann Kolbe and the Science of Organic Chemistry, Alan J. Rocke
Image and Reality: Kekulé, Kopp, and the Scientific Imagination, A.J. Rocke


Is an aqueous solution of potassium bicarbonate acidic or basic in nature? The following is the definition of an acid salt according to my textbook:
The salts formed by the partial replacement of the replaceable hydrogen atoms (H+ ions) of an acid molecule by a metallic or ammonium ion are called acid salts. Acid salts ionise in water to give hydronium ions and therefore, their aqueous solutions show all the properties of an acid.
Going by this definition, potassium bicarbonate should be classified as an acid salt. Hence, an aqueous solution of potassium bicarbonate should be acidic in nature(as hydronium ions would be present in an aqueous solution of potassium bicarbonate).
However, my textbook also says, “The aqueous solutions of salts formed by a weak acid and strong base (as a result of a neutralisation reaction) are basic in nature.”
According to this, potassium bicarbonate solution should be classified as a solution which is basic in nature as potassium bicarbonate solution is formed as a result of a neutralisation reaction between potassium hydroxide, which is a strong base and concentrated carbonic acid, which is a weak acid even in its concentrated solution form.
So is potassium bicarbonate solution acidic or basic in nature?

You could simply calculate the $\mathrm{pH}$ of the solution. $K_\mathrm{a{_1}}$ of $\ce{H2CO3}$ is $\pu{4.30E-7}$ and $K_\mathrm{a{_2}}$ of $\ce{H2CO3}$ is $\pu{4.72E-11}$.The approximate $\mathrm{pH}$ of the solution can be given by:
$$\mathrm{pH}=\frac{(\mathrm{p}K_\mathrm{a{_1}} + \mathrm{p}K_\mathrm{a{_2}})}{2}$$
This yields $\mathrm{pH}= 8.347$, which indicates that the solution is mildly basic.


Mechanism of Acid Hydrolysis I wanted to ask a question about acid catalysed hydrolysis.

I learnt in an introductory course to Physical Organic Chemistry that there are different ways that an ester can be hydrolysed acidically:

  $\ce{A_{AC}1}$
  $\ce{A_{AC}2}$
  $\ce{A_{AL}1}$
  $\ce{A_{AL}2}$

There are different ways that esters can be hydrolysed:
– catalysis by acids (A) or bases (B)
– cleavage of acyl-oxy (AC) or alkyl-oxy (AL) bonds
– the molecularity of the key step (1 or 2).

Using Acids (A) there are 4 different ways to hydrolyse an ester using acid catalysis:

  $\ce{A_{AC}1}$ Cleavage of Acyl-Oxy Bond Unimolecular
  $\ce{A_{AC}2}$ Cleavage of Acyl-Oxy Bond Dimolecular
  $\ce{A_{AL}1}$ Cleavage of Alkyl-Oxy Bond Unimolecular
  $\ce{A_{AL}2}$ Cleavage of Alkyl-Oxy Bond Dimolecular

An explanation can be found here.

But consider this question, with the marking scheme given below:

Our professor did not mention anything regarding how to choose which type of hydrolysis takes place. From this question, he states and goes further onto mention that either $\ce{A_{AC}1}$ or $\ce{A_{AC}2}$ can take place, but no other.

I'm looking at these structures and thinking

Why can't $\ce{A_{AL}1}$ or $\ce{A_{AL}2}$ take place for these two? Why must it be the two given in them marking scheme?

You should have read your explanation material more carefully! It clearly states that $A_{AL}2$ has not even been observed in acid hydrolysis due to water being a poor nucleophile in an $S_N2$ process. As for $A_{AL}1$, that only happens when the esterified radical can leave as a stable carbocation which, in this case methyl is a very very unstable carbocation.


What is an egg emulsion? All good recipes for hollandaise sauce consist of beating egg yolks vigorously and slowly add melted butter over moderately low heat.  Care must be taken to not overheat the sauce or the eggs will scramble and it won't emulsify.  Likewise, the melted butter must not be added too quickly or it will not emulsify. When done properly like a good chef, what is happening here?  Why does the sauce thicken?  What is an egg emulsion and why must the butter be added so slowly rather than faster?

To know what's happening here you should know what "emulsifying" is.

First we need to know that fat is "non-polar" substance. what does that mean?

It means that all fats can't be dissolved in water which is "polar" substance.

So how can we make fat -which is non-polar- dissolve in water or any other polar substances?

We add "emulsifiers" like: egg yolk and many other chemical substances like Lecithin.

These emulsifiers are (polar and non-polar) so they can link fat with water together.

We add butter slowly so that the molecules of emulsifier can link the molecules of butter with molecules of sauce -which is polar-.

If there's no free molecules of emulsifier, every molecule of added butter will not be dissolved in the sauce.


Does inversion of configuration necessitate reversed optical rotation? Say, optically pure 2-bromobutane undergoes $\mathrm{S_N2}$ reaction with an alkali to form 2-butanol. The configuration about chiral carbon will now be inverted, but is it necessary that the sign of optical rotation will be opposite to that of substrate? i.e. if optical rotation of optically pure 2-bromobutane is $+x^\circ$, will optical rotation of the product be $- y^\circ$?

if optical rotation of optically pure 2-bromobutane is some (+X)
  degrees, will optical rotation of the product be some (-Y) degrees?

No, there is no correlation between the configuration at a chiral carbon and the sign of optical rotation.

For example (from the Aldrich catalogue), the following chiral bromo compound, (R)-(+)-2-Bromo-3-methylbutyric acid,

has an optical rotation of [α]22/D +21°, c = 37 (c is the sample concentration in grams/100 ml solution) in benzene.  The inverted hydroxyl analogue, (S)-(+)-2-Hydroxy-3-methylbutyric acid,

also rotates light in the (+) direction.  Its optical rotation is [α]20/D +19°, c = 1 in chloroform.


One half of a nail is exposed to vinegar. Un-exposed half develops a brown film. What is it? I have recently done an experiment for my chemistry EEI (Year 12) where we are testing what factors effect the rate of corrosion. I used vinegar (acetic acid) as one of the acids and Hydrochloric acid as the other.

In the one with the HCl, the results were as expected – the area exposed to the acid will corrode, the unexposed part won't (in the time frame I did).

However, for the vinegar, the results were different. The part exposed to the vinegar didn't show any rust however the part of it that wasn't exposed had a brown coating. Now I understand that the vinegar reacts with the rust, hence the reason we cant see it on the iron nail, but my question is: What is the brown coating on the top of the nail, and what is it caused from? Was there some sort of gas from the vinegar that caused it? If so, what is this gas called and how does it cause corrosion?

The nail is in two solutions. The lower solution is 5% acetic acid, saturated with air (20% $O_2$, 80% $N_2$). At 0.2 atmospheres, $O_2$ solubility in water is 40 mg/L x 0.2 = 8 mg/L, or about 8 parts per million.
The upper solution is a thin film on the nail (perhaps 10 microns thick - or less!) of $H_2O$ with some acetic acid (probably about 3-4%) and 8 ppm $O_2$. This film may have deposited from the lower solution by vapor deposition due to temperature differences, or by capillary action, as the meniscus crept up in a thin layer.
In the lower solution, $Fe^0$ goes into solution as $Fe^{++}$ and electrons go to $O_2$.
4$e^-$ + $O_2$ + 2 $H_2O$ --> 4 $OH^-$
The 8 ppm $O_2$ in the lower solution doesn't go very far, but the meniscus, and any film going higher up, will be continually refreshed with $O_2$, and will develop a higher pH and will precipitate iron hydroxide. Any iron dissolving in the lower solution will stay in solution because of the acidity, and the nail may not even show obvious signs of corrosion damage like pitting. The only obvious sign is the brown film of precipitated iron hydroxide.
This is an example of a differential aeration cell, where the upper solution is an oxygenated cathode, and the lower solution contains the anodic portion of the nail. http://www.corrosion-doctors.org/Corrosion-Factors-Cells/corrosion-cells-aeration.htm
We consider this corrosion in acetic acid to be an electrolytic reaction, and also the reaction of the nail with $HCl$, but the scale of cathodes and anodes in the case of $HCl$ is so small that we can't detect them. And in $HCl$ the $H_2$ is generated so fast that it bubbles off and doesn't need to be depolarized by dissolved oxygen.
Adding a picture suggesting the state of the nail:

https://www.ddcoatings.co.uk/2343/what-is-differential-aeration-corrosion.


Catalyst with zero order kinetics I've read in a book that a catalyst may be included in the rate equation even though it is never used up. It was written as having zero order kinetics however I thought catalyst concentration will always have an effect on the rate, especially since in a lot of biological enzyme-controlled reactions, enzyme concentration is a key element of the rate of a reaction. Is this only the case if the catalyst is above a certain concentration such that adding more will have no effect?

What makes the reaction a zero-order reaction, is it being dependent only on the concentration of one of the elements.

In a system made by a reagent and a catalyst, we expect the reaction to behave in a zero-order fashion when the concentration of the reactant varies in a linear fashion during the course of the reaction.
You could write an equation in the form: $rate = k$ and, implying that only the reactant A varies in concentration, $[A]=[A_0]-kt$.

Now, during a reaction you know that the concentration of the catalyst remains constant.

What you are asking is: what happens if I vary the concentration of the catalyst? Assuming that you are still in zero-order conditions and you run another reaction with a larger amount of catalyst, then you will obtain another equation rate (another value for k).

But that will not make the curve [A] vs t non linear (that is, non zero-order).
More catalyst means a larger rate constant, but doesn't imply a variation in kinetic behaviour of the reaction.

Still, if you were to evaluate the variation of rate in respect to the catalyst's concentration, then you would obtain a reaction rate that is still of order 0 in respect to A, and of order non-zero in respect to the catalyst.

Maybe your book was just pointing out that a set of reactions in which the same amount of catalyst is used, could be studied with a zero-order approach.


How is the magnetic property of an element affected when the element is part of a compound? Data on this site conveys that Neodymium is paramagnetic. Yet the Neodymium magnets out there make extremely strong magnets. I'm guessing these are not pure neodymium magnets but some kind of compound.

Is it possible to predict the change in magnetic property of a compound given the knowledge of it's participating elements?

Magnetism appears in many very different forms that have very different origins.

All elements show some diamagnetism, even though often overshadowed by other, stronger, forms of magnetism, because it is a property of the bound electrons.
The diamagnetic contribution caused by the inner electrons that are not part of a chemical bond will therefore remain unchanged if the compound bonds chemically. The diamagnetic contribution from the electrons participating in the bond will of course change, so even for the most simple form of magnetism, it is not possible to predict the diamagnetic behavior of a compound from that of the elements.

Paramagnetic materials have a permanent magnetic moment from uncompensated magnetic moments, typically electron spins (there are exceptions where a paramagnet is formed from larger units, like magnetic particles).
In a very simplified way, you will find unpaired electrons typically on the outer levels of elements, where they are influenced and participate in chemical bonds. This is even the case for the strongly localized d and f electrons that are characteristic for the strongest paramagnets like $\ce{Gd}$.
So clearly, paramagnetism is influenced by the chemistry.

Both diamagnetism and paramagnetism are extremely weak effects that normally don't play a role in everyday life.

Now for the 'real' forms of magnetism, ferromagnetism (ferri- antiferro and ferri etc.): All of these types of magnetism are collective phenomena, i.e. they are the result of the internal interactions of a large system, through conduction electrons (itinerant), superexchange, double exchange, etc.

Due to the collective nature, it is not possible to predict the behavior of a compound from that of the physical parameters of the elements. (Of course, you can simulate the magnetic behavior of more or less any material, but I think that was not the question)

You could not call $\ce{Fe}$ itself a ferromagnet, the ferromagnetism is a property of a huge number of $\ce{Fe}$ atoms that are acting collectively. This ensemble is very sensitive to changes, often even a small temperature change (e.g. at 20°C for $\ce{Gd}$) will completely change the magnetic properties. A chemical reaction or even simple alloying, will completely change the magnetic properties.

The macroscopic magnetic properties of a material (susceptibilitiy, coercivity, remanence, saturation magnetization, frequency dependence of parameters etc.) do not allow to predict the properties of a compound that contains this material. So you can't predict the coercivity of  $\ce{NdFeB}$ given the coercivities of  $\ce{Nd}$, $\ce{Fe}$ and $\ce B$.

It is possible to simulate the complete electronic structure of a compound with Density Functional Theory. This simulation will tell you the magnetic parameters of the material, including the ordering temperatures in the case of ferromagnetic materials, to a varying degree of accuracy. It is a very computationally intensive approach, to compute the properties of a new material is a research project, requires some ad-hoc assumptions and will need to be experimentally verified as it is not yet 100% reliable.

This computational approach does not make any use of the macroscopic physical properties of the elements in a compound, it directly solves the many body problem of electrons and nuclei.

In summary, it is not possible to predict the change in magnetic property of a compound given the knowledge of the magnetic properties of its participating elements, but you can simulate the electronic structure of a compound, and that will tell you how the compound will behave magnetically.

Note: Neodymium magnets aren't made out of $\ce{Nd}$ metal. They are made from sintered $\ce{Nd2Fe14B}$. $\ce{Nd}$ is only a small fraction of the entire material. This is a very complex material. The exceptionally high coercivity, i.e. the resistance against a change in the magnetization, is the result of high magnetocrystalline anisotropy of $\ce{Nd2Fe14B}$ and the interfaces between the grains that pin the magnetization. This is something you would not think about given just the magnetic parameters of $\ce{Nd}$, $\ce{Fe}$ and $\ce{B}$.


Why does iron oxide form hydrous iron oxide while chromium oxide does not? Both Cr2O3 and Fe2O3 share a similar structure, and both can form some sort of passivating layer, but only one of them hydrates, thus turning into rust and providing a pathway for more water and O2 to further oxidize the iron. I believe the same applies to aluminum and nickel as well. iron seems to be the only one to do this, but I cannot figure out why.

TL;DR: Guignet's green is a green pigment with composition $\ce{Cr2O3.2H2O}$.
Long answer
Chromium oxide is largely used as a green pigment in various industrial and commercial purpose as it is the only green pigment which can withstand both acidic as well as alkaline environment. It is also used as permanent paints for cements, distempers and paints. It can also withstand high temperatures, hence used for coloring glasses, enamels and pottery glazes.
Guignet's green or $\ce{Cr2O3.2H2O}$ is the hydrous form of the oxide which is used for similar purpose as the anhydrous form. This pigment becomes anhydrous when heated at 200 °C, hence not suitable for glass and ceramic industries.
It is prepared by mixing 3 parts of boric acid and 1 parts of potassium dichromate which is heated in reverberatory furnace. It becomes a dull red mass initially but swells up releasing oxygen to give a nice green color. It is then boiled with water whereby chromium borate decomposes to hydrated chromium oxide and boric acid. The precipitate is dried and separated while the mixture is washed with mother liquor to evaporate the boric acid.
$$\ce{2K2Cr2O7 + 4H2B2O4 -> 2Cr2B2O6 + 2K2B2O4 + 3O2 + 4H2O}$$
$$\ce{Cr2B2O6 + 3H2O -> Cr2O3.2H2O + H2B2O4}$$
Reference

S. P. Das Gupta & H. N. Bose (1948) A Note on Preparation of Chromic Oxide Green, Transactions of the Indian Ceramic Society, 7:2, 60-63, DOI: 10.1080/0371750X.1948.10877850


NMR magnetically equivalent protons for a 1,4-disubstituted benzene ring Below I have drawn a molecule with random substituents $\ce{X}$ and $\ce{Y}$:

If the distances from $\ce{H_b}$ and $\ce{H_a}$ to $\ce{H_d}$ are considered, they are different - $\ce{H_b}$ is 5 bonds away and $\ce{H_a}$ is 3 bonds away, so $\ce{H_b}$ sees $\ce{H_d}$ differently compared to $\ce{H_c}$. The same argument can be made vice versa and also between the same pair of protons towards $\ce{H_c}$. This was what the lecturer told me regarding this compound below.

This led me to believe then that all the protons in the molecule are magnetically inequivalent and using up to $\ce{^4J}$ coupling, all protons show a doublet of doublet.

However, when I was in school, I was told that if symmetry was a property of a molecule, as above, $\ce{H_a}$ and $\ce{H_b}$ were equivalent, as are $\ce{H_c}$ and $\ce{H_d}$, and each pair of protons gives a doublet. Researching this matter on Google still held true in my undergraduate studies.

What am I getting wrong here?

I read a quote on another question highlighted here

  For two nuclei to be magnetically equivalent, they need to have equivalent coupling to all other non-chemical shift equivalent nuclei in the molecule.

So as $\ce{H_a}$ and $\ce{H_b}$ are chemically equivalent but $\ce{H_b}$ couples differently to $\ce{H_c}$ than $\ce{H_a}$ does, that therefore means that surely $\ce{H_a}$ and $\ce{H_b}$ are magnetically inequivalent?

when I was in school, I was told that [...] each pair of [equivalent] protons gives a doublet

This is only strictly true for magnetically equivalent protons. If there is magnetic inequivalence, such as in the p-disubstituted benzene ring, then it is no longer true.

It's not as simple as a dd either; when you have magnetically inequivalent protons that see each other, the first-order rules (the so-called $n+1$ rule) doesn't work so well anymore. You can see some real-life examples of how these molecules look like here: https://www.chem.wisc.edu/areas/reich/nmr/05-hmr-15-AABB.htm For the p-disubstituted benzene it sort of looks like a doublet if you zoom out far enough, but if you look closely, there are additional peaks flanking the major peaks.

I say "see each other" because if you think about it, the two protons labelled HA above are also magnetically inequivalent; but since they are so far apart, the two benzene rings are essentially independent of each other, and give rise to the same NMR spectrum.


Is iron pyrite dangerous to handle? I have a cube of pyrite similar to:

(Source: Britannica)

Can pyrite turn into sulfuric acid?

Is this dangerous for me to touch, especially if my hands are sweaty?

Pyrite does not turn into sulfuric acid. You can burn it with oxygen to give iron oxide and $\ce{SO2}$.

$\ce{4 FeS2 + 11 O2 -> 2Fe2O3 + 8 SO2}$

$\ce{SO2}$ with water gives sulfurous acid. To really get sulfuric acid, $\ce{SO2}$ needs to be oxidised to $\ce{SO3}$ (Contact process), and then reacted with water. Actually with more sulfuric acid to give oleum, which decomposes into two equivalents of sulfuric acid upon dilution with water.

The surface of the pyrite will react very slowly in the acidic conditions in your hand. Very, very slowly. There might be a faint smell of $\ce{H2S}$ recognisable if you handle the pyrite for a while in your sweaty hands.

(Also with water and oxygen from the air, pyrite will slowly react, to finally give sulfuric acid, which will speed up further reaction/degradation. But this typically needs decades of very bad ventilation.)


What is the most complex molecule we can manufacture without the help of proteins? (soft queston) What is the most complex molecule we can manufacture without the help of proteins?

This isn't motivated by anything but curiosity about the state of chemical engineering. I realize “complexity” is a loaded term, so I'll defer to the judgment of the respondent: number of molecules, complex arrangement of bonds, or whatever other measure might be appropriate.

I guess in the field of inorganic chemistry polyoxometalates (POMs) are comparable with proteins in terms of structural complexity. Macroanionic POMs such as Keplerates $\ce{\{Mo132\}}$

are also capable of further self-assembly in diluted solutions, producing so-called "blackberries" [1] of a size of around $\pu{0.05 .. 0.1 \mu m}$:

Also, see the post Measure the complexity of a molecule and the corresponding paper [2] of Prof. Lee Cronin, one of the most productive polyoxometalate chemist of current era.

Kistler, M. L.; Bhatt, A.; Liu, G.; Casa, D.; Liu, T. J. Am. Chem. Soc. 2007, 129 (20), 6453–6460. DOI 10.1021/ja0685809.
Marshall, S. M.; Murray, A. R. G.; Cronin, L. arXiv:1705.03460 [q-bio.OT] 2017.


Titration curve of a diprotic acid with a strong base titrant I have wonders regarding the titration curve of the titration of sulphuric acid, or any diprotic acid, with a strong base. I am aware of the fact that a diprotic acid is protolysed in two steps but, how does this affect the titration curve?

The pKa values for sulfuric acid from Wikipedia are −3, and 1.99. So for all practical purposes the first ionization happens completely and would be unobservable with a pH electrode in aqueous solutions. So if you had a 0.1  molar solution of sulfuric acid it would almost look like you were titrating a 0.2N of $\ce{H^+X^-}$ where that acid had a pKa of 1.69 (pH at half NaOH used). This would be a strong acid too and the inflection point would be that for water at pH 7.

Note that the pKa 1.69 is assuming $\ce{H^+X^-}$ and is taken at 1/2 of the $\ce{Na^+OH^-}$ used. Knowing that $\ce{H2SO4}$ has two protons, the second pKa should be taken at 150% on the plot which would give the correct pKa value of 1.99.

If you had an acid with two lower pKa values then you'd see two pH transitions in the titrations. Look at the figure below. With two widely separated pKa values maleic acid shows two nice transitions. With two pkA values closer together the two transitions for malonic acid aren't as sharp. For succinic acid the knee at a volume of about 5 ml is a hint that there are two dissociations but you don't see two nice complete transitions like you do for maleic acid.

To go back to the sulfuric acid tritration, we'd really expect to see a knee at a low pH something like the titration curve we got for succinic acid. Such a knee isn't seen because a pH electrode just doesn't work at such high acid levels (pH range from -3 to 0).

PS - Thanks to the commenters for help to vastly improve the answer.


Justifying the tremendous −I effect of trimethylammonium group by molecular orbital theory The $\ce{-[NMe3]+}$ group is ranked almost at the first position in the −I effect series, which is often given at the very beginning of the teaching of general organic chemistry.

I know this can be justified by hyperconjugation, but I am interested in a rigorous justification by molecular orbital theory.

Powerful electron withdrawing substituents, like $\ce{-NO2}$ can be interpreted as having a large effect on the $\pu{\pi}$-system of an aromatic ring. Resonance structures can be drawn that put a positive charge at the ortho and para positions and put the electron out in the nitro group, on an oxygen.

This must be adjusted for $\ce{-N(CH3)3+}$, because it has no $\pu{\pi}$ electrons, and we can't draw structures with a double bond between the nitrogen and the ring carbon. It works thru a very strong $\ce{-I}$ effect. However, if we use the concept of hyperconjugation, and donate a $\pu{\pi}$ electron from the ring to the nitrogen and then a methyl group (making it $\ce{-CH3}$), unbounded to any other atom, we can imagine the oxygens in the picture replaced with methyls and will get the same distribution of charge in the ring: a positive charge at the ortho and para positions. This will result in meta substitution if any reaction does occur.

For $\ce{-N(CH3)3}$, it's like hyperconjugation in reverse. Interestingly, while $\ce{-N(CH3)3}$ makes it very difficult to substitute, and always meta, $\ce{-NH3+}$ is only a weak meta director.

Reference

Advanced Organic Chemistry, L.F. Fieser, 1961, p628


Understanding energy level splitting of an octahedral complex from spin-orbit coupling, 1st and 2nd order zeeman effects While studying notes from a couple years ago, I encountered a diagram that I don't have further information on but would like to understand.
While I believe I managed to find sources to guide me through the details of the calculation, there are a few conceptual questions I had, which are summarized at the bottom.
The diagram is for a $d^1$ octahedral complex under a magnetic field:

a) ligand field splitting
b) Spin-Orbit coupling
c) First-order Zeeman splitting
d) second-order Zeeman splitting
a)
The ligand-field splitting is straightforward to me.
b)
The spin-orbit coupling I believe I understand.
The $E_g$ level does not have an electron, and therefore does not undergo any spin-orbit coupling.
For the $t_{2g}$ energy level splitting, I followed the procedure covered on pages 9 and 10 here (the formula is also given on the wiki). The formula for spin-orbit coupling is:
$$\Delta E=\frac{B}{2}[j(j+1)-l(l+1)-s(s+1)]$$
For a $^2D$ term, $l=2$ and $s=\frac{1}{2}$. Then $\Delta E$ would be:
$$j=\frac{5}{2}\therefore \Delta E=\frac{3}{2}B$$.
$$j=\frac{3}{2}\therefore \Delta E=-B$$
I assume in the diagram, $\zeta=B$.

Why does the diagram obtain different coefficients ($-\frac{1}{2}\zeta$ and $\zeta$)? Did I do this wrong?

c)
I think I might have understood this one, using the following link. I'll be referencing equations from this link for these two sections. I'm substituting the Bohr Magneton, $\mu _B$ with simply $\mu$, as it renders better this way, and replacing $B_{ext}$ with $B$ for the same reason.
In deriving equation 8.15 from the link, I am confused what is meant by the line

let the hermitian, unperturbed hamiltonian $H_0^\dagger =H_0$ "act to the left" on $\langle n^0|$

I have never used an operator to "act to the left." I've always applied it toward the right.
Assuming it somehow works out to $E_n^0$, I understand how theorem 8.1 is obtained:
$$E_n^1=\langle n^0|H_I|n^0\rangle$$
They apply the above theorem in equation 8.25 with $H_I=H_z$, where
$$H_z=\frac{\mu}{\hbar}(L_z+2S_z)B$$
to obtain:
$$\Delta E =\mu (m_l+2m_s)B$$
In my diagram above, $m_l=2$ and $m_s=\frac{1}{2}$, so
$$\Delta E = \mu (2\pm 1)B=3\mu  B-\mu B$$
$$= 2\mu B$$
That seems to match the diagram (where $B=H$).

Why does the $E_g$ level split into two double degenerate levels, and the ground state $\Gamma _8$ (4) not split at all?

d)
For second-order splitting, the linked PDF provided this formula (theorem 8.3):
$$E_n^{(2)}=\sum_{m\ne n}\frac{\left| \langle m^0|H_I|n^0\rangle \right| ^2}{E_m^0-E_n^0}$$
I was unsure how this relation was true:
$$(\langle n^0|H_I|m^0\rangle )(\langle m^0|H_I|n^0\rangle )=\left| \langle m^0|H_I|n^0\rangle \right| ^2 $$
but otherwise followed the derivation. Accordingly, the numerator indicates the 2nd order zeeman correction is simply the square of the first order correction; however, the denominator I did not follow, namely:
$$E_m^0-E_n^0=3\zeta$$

Where does this $3\zeta$ come from?
Similar to the first-order splitting, why do the energy levels experience different types of splitting?

summary of questions

In section b), why are my splitting coefficients different from the diagram?

In section c) of the diagram, why is the highest-lying energy level in the diagram split into two double degenerate levels, the middle-lying energy level split into two singly degenerate levels, and the lowest-lying energy level not split at all? I assume the answer to this question will also explain why the second-order splittings in section d) are also inconsistent with each other.

In section d) of the diagram, how does the $3\zeta$ term in the denominator for the energy splitting come about?

What is responsible for certain energy levels being doubly degenerate after following splitting from the Zeeman Effect?

Where does the term symbol splitting according to the projection of total angular momentum along the axis of the magnetic field, ie. $j_z=j,j-1,...-j$ fit in here? Aren't these also separated in a magnetic field?

To answer part b it is necessary to consider the symmetry properties in an octahedral field. The spin orbit Hamiltonian is $H_i=\zeta\sum_i\bar l_i\cdot \bar s_i$ where $\zeta$ is a constant characteristic of the radial part of the d orbitals. The orbital angular momentum, which may be written in the form (for the $x$ component) as $\displaystyle l_x =-i\hbar \left(y\frac{\partial }{\partial z}-z\frac{\partial}{\partial y}\right)$ (and similarly for $l_y$ and $l_x$ by rotating subscripts) transforms as does the $T_{1g}$ irreducible representation in an $O_h$ field.

If a term has an orbital symmetry $X$ it will only be susceptible to spin orbit coupling if the symmetry product $X\cdot T_{1g}\cdot X$ contains the totally symmetric representation $A_{1g}$. In other notation $\langle i|H |j\rangle \ne 0$ or $\int \psi_i H \psi_f d\tau \ne 0$). Thus the $E_g$ levels are unaffected in point group $O_h$ because $E_g \cdot T_{1g}\cdot E_g$ does not contain $A_{1g}$ but the $T_{2g}$ interact ($T_{2g} \cdot T_{1g}\cdot T_{2g}$ does  contain $A_{1g}$).

Notes: It is assumed that the potential energy effect of the ligands is only small compared to that of the central field of the ion, thus the spin-orbit coupling is a small perturbation on the energies.

The full derivation is complex and involves using double symmetry groups.

The free ions have SO couplings $^2\mathrm D_{3/2} =-3\zeta_d/2$ and $^2\mathrm D_{5/2} =\zeta_d$. (The formula you use if for the free ion).  In the octahedral field the calculation is rather complicated but the result is that $T_{2g}$ levels are split by $3\zeta/2$ in the ratio you show.

In the remaining parts of your question similar symmetry arguments should indicate what levels are split, but only a detailed calculation will produce the energies that they are split by.


Mass Spectra: 1-pentanol vs 3-pentanol In mass spectrometry, 1-pentanol can be observed to have no peak at $m/z = 59$, whilst 3-pentanol can be observed to have a peak at $m/z = 59$.

Clearly the 3-pentanol peak at $m/z = 59$ is due to the cleavage creating $\ce{CH3CH2CHOH | CH2CH3}$

Why is this the case, as isn't 1-pentanol able to form the fragment $\ce{CH2CH2CH2OH}$ by cleaving at the third carbon?

i.e. fragmenting into $\ce{CH3CH2}$ and $\ce{CH2CH2CH2OH}$ (this should have $m/z = 59$?).

In mass spectrometry, organic compounds can be ionized by different methods. The compounds ionized by electron impact ionization display particular fragmentation patterns. When subjected to mass spectroscopy, cleavage of the $\ce{C-C}$ bond next to the oxygen or nitrogen can usually occur, when the compound is ionized by electron impact ionization.

Let’s look at what happens to 3-pentanol:
$$\ce{CH3CH2CH(OH)CH2CH3 ->[e- beam] CH3CH2CH(O^.+H)CH2CH3 }$$

$$\ce{CH3CH2CH(O^.+H)CH2CH3 -> CH3CH2CH=O^+H + ^.CH2CH3}$$

Now, base peak (most probably), would be $\ce{[CH3CH2CH=OH]+}$, which should be at $ 59 \ m/z$. Note that in general, the molecular ion of an alcohol (here it is $ 88 \ m/z$) is either trace or non-existent, probably loosing $\ce{H2O^.}$.

Now, we'll look at what happens to 1-pentanol:
$$\ce{CH3CH2CH2CH2CH2-OH ->[e- beam] CH3CH2CH2CH2CH2-O^.+H }$$

$$\ce{CH3CH2CH2CH2CH2-O^.+H -> CH2=O^+H + ^.CH2CH2CH2CH3}$$

Now, base peak (again, most probably), would be $\ce{[CH2=OH]+}$, which should be at $ 31 \ m/z$ ($88-31=57$ mass loss). Note that here again, the molecular ion is either trace or non-existent, probably loosing $\ce{H2O^.}$, and possibly $[M-18]$ ion should be significant here, since there is equal chance of cleaving $\ce{C_1-O}$ bond (giving peak at $ 70 \ m/z$) or $\ce{C_2-C_3}$ bond (giving peak at $ 31 \ m/z$):

However, according to the mass spectrum given, $\ce{C_5}$-carbon chain has undergone McLafferty type rearrangement (which is common in MS of aldehydes and ketones) to give the base peak at $ 42 \ m/z$ by loosing a water and ethylene simultaneously from parent ion (see the insert in above diagram).


What do the half-reactions need to be for balancing a redox reaction in an acidic medium? Balance the following reaction:
  $$\ce{CuS + SO4^{2-} -> CuO + SO2}$$

Below, the number mentioned in the parenthesis is the oxidation number of sulfur.
I have found out that the oxidation half-reactions and reduction half-reactions are as follows:

\begin{align}
\ce{CuS^{(-2)}     &-> ^{(4)}SO2}&& \text{Oxidation half reaction}\\
\ce{^{(6)}SO4^{2-} &-> ^{(4)}SO2}&& \text{Reduction half reaction}
\end{align}

Problem:

In this, what about $\ce{CuO}$ in the product side? I have also learned that we need to balance the spectator ions before charges if the spectator ion is other than oxygen and hydrogen.

Here, there is no $\ce{Cu}$ in reduction half reaction. So, I think there is something is wrong in my reduction half-reaction.

Below, the number mentioned in the parenthesis is the oxidation number of sulfur.

Writing down half reactions:
\begin{align}
\ce{CuS^{(-2)}     &-> ^{(4)}SO2}&& \text{Oxidation}\\
\ce{^{(6)}SO4^{2-} &-> ^{(4)}SO2}&& \text{Reduction}
\end{align}
Adding electrons in both reactions:
\begin{align}
\ce{CuS^{(-2)}           &-> ^{(4)}SO2 + 6e-}&& \text{Oxidation}\\
\ce{^{(6)}SO4^{2-} + 2e- &-> ^{(4)}SO2}&& \text{Reduction}
\end{align}
Balancing $\ce{Cu}$:
\begin{align}
\ce{CuS^{(-2)}&-> ^{(4)}SO2 + 6e- + Cu^{+2}}&& \text{Oxidation}
\end{align}
Balancing charges of both half reactions:
\begin{align}
\ce{CuS^{(-2)}&-> ^{(4)}SO2 + 6e- +CuO + 6H+}&& \text{Oxidation}\\
\ce{^{(6)}SO4^{2-} + 2e- +4H+ &-> ^{(4)}SO2}&& \text{Reduction}
\end{align}
Balancing oxygen and hydrogen:
\begin{align}
\ce{CuS^{(-2)} + 3H2O&-> ^{(4)}SO2 + 6e- +CuO + 6H+}&& \text{Oxidation}\\
\ce{^{(6)}SO4^{2-} + 2e- +4H+ &-> ^{(4)}SO2 + 2H2O}&& \text{Reduction}
\end{align}
Adding 3 copies of reduction half reaction with oxidation half reaction, gives finally balanced redox reaction:
$$\ce{CuS + 3SO4^{2-}  + 6H+ -> 4SO2 +CuO + 3H2O}$$

(Thanks to Silvio Levy for helping me to solve this problem.)


can you explain the structure of gemtuzumab ozogamicin? where are the linker boundaries? could someone help me to decipher the structure of this molecule? I am trying to understand the structure of gemtuzumab ozogamicin: The anti-CD33 antibody is acylated with the bifunctional AcBut linker, which allows for hydrazone formation with the calicheamicin derivative. This derivative, N-acetyl-gcalicheamicin dimethylhydrazide (DMH), has a disulfide stabilized by both the two methyl groups and the calicheamicin bicyclic core, and an acetylated amino sugar. References: Joel C. Barrish, Percy H. Carter, Peter T. W. Cheng, Robert Zahler Accou... I would like to know the boundaries of AcBut linker. I got confused because in this picture it is depicted like that:  I thought that DMH was the part after the benzyl ring in the linker. Thank you in advance!

The trick is probably the disulphide bridge which will be cleaved by reduction, once the construct has been internalized into the target cell, liberating the drug. So I'd consider everything proximal from it towards the antibody as linker structure.


Importance of zinc blende and wurzite crystal structures for semiconductors? I am currently studying the textbook Physics of Photonics Devices, Second Edition, by Shun Lien Chuang. In a section discussing the basic concepts of semiconductor band and bonding diagrams, the author gives the following description:

The basic idea is that for a semiconductor, such as $\ce{GaAs}$ or $\ce{InP}$, many interesting optical properties occur near the band edges. For example, Table 1.1 shows part of the periodic table with many of the elements that are important for semiconductors, including group IV, III-V, and II-VI compounds. For a III-V compound semiconductor such as $\ce{GaAs}$, the gallium ($\ce{Ga}$) and arsenic ($\ce{As}$) atoms form a zinc-blende structure, which consists of two interpenetrating face-centered-cubic lattices, one made of gallium atoms and the other made of arsenic atoms (Fig. 1.1).

This seems to be the relevant Wikipedia article for zinc-blende (Zincblende) structure.
I was interested in understanding the relevance of this structure to semiconductors, so I did further research. I then came across this website, which says the following:

Here are the two most important crystal structures for semiconductors.
They are often referred to by the historical names "Zinc blende" from the German "Zinkblende" $= \ce{\alpha-ZnS}$, a rather ubiquitous mineral. The name "Sphalerite" also comes form the German: "Sphalerit", which, as was the custom of the time, stems from the Greek "sphaleros" meaning treacherous or malicious because it is easy to confuse it with other minerals.
Wurtzite was and is the name of the $\ce{\beta-ZnS}$ modification - the hexagonal high-temperature variant. The named after the French chemist C. A. Wurtz (* 1817, † 1884), which gives us an idea of how old those names are.
Zinc blende or sphalerite or Diamond structure

Wurtzite or hexagonal structure
Red lines are not showing bonds.

However, this website does not actually explain why these are the two most important crystal structures for semiconductors. I was wondering if someone would please take the time to explain this.
EDIT:
Thinking about Justanotherchemist's comment and Oscar Lanzi's answer, I wonder if all group $\ce{III-V}$ semiconductors have these $\ce{ZnS}$ structures? Because, if so, then that could very well be the answer to my question here, since, as I understand it, group $\ce{III-V}$ semiconductors are the most important semiconductors for practical purposes? As Oscar Lanzi said in his answer, not all semiconductors have a $\ce{ZnS}$ structure, so the answer cannot be that they are the most important because all semiconductors have a $\ce{ZnS}$ structure.

As suggested I can try to summarize the basic idea in an answer although I wouldn't say it solves the bounty. I'm also unsure why you even set the bounty for it.

For this I'll quote the book that I found a few days ago. As I own the missing pages now as well the answer seems a bit more elaborate then before. As the book is in German I will translate the text.

  The formation of the band-structure of typical element semi-conductors like diamond (C), Si and Ge was already shown in chapter 7.3. By mixing s- and p-wavefunctions we'll get a tetrahedral bonding orbital ($sp^3$), that will lead to a splitting into a bonding and an anti-bonding orbital in the range of the equilibrium-bonding distance. The bonding orbitals form the valence band and the anti-bonding orbitals form the conducting band. The distribution of the four s- and p-electrons into bonding and anti-bonding orbitals leads to a completely filled valence band while the conducting band remains empty.
  The result is therefore an insulator like diamond or, if the energy gap between the valence and conducting band is smaller, to semi-conductors like Si or Ge.

I'll skip the next pages that deal with the formation of the band-structure and the symmetry aspects as I have no experience in these topics and continue with the $III-V$-semi-conductors.

  As the formation of $sp^3$ hybrid orbitals in the chemical bonding of Si and Ge seem to be the a substancial part of the semi-conducting properties it seems reasonable that other materials with a tetrahedral crystal structure, i.e. $sp^3$ hybridisation, should show semi-conducting properties as well.

At this point I will switch to basic solid state chemistry as the book doesn't elaborate this part well enough. As solid state chemist I often have examples like these with my students. A while ago we were discussing the silver-mobility in high-temperature $AgI$, which shows quite some conductitiy through defects and the low temperature modification. At this point I would typically ask for reasons why $AgI$ might be problematic. Typical answer that I expect are:

HSAB-concept  with soft $Ag^+$ and soft $I^-$
A covalent three-dimensional tetrahedral network with ZnS-structure

For the second part they either check these facts before we talk about it or they can use the things they have learned in lectures. One of them being the Grimm-Sommerfeld rule. Basically, compounds, that form between two elements to the left and to the right of another element will likely adopt the structure of said element. So for a group $IV$ element like the silicon structure, it can be $III-V$ or $II-VI$ or $I-VII$. It's one of these rules that are taught in the very first semester but people forget about it although it's super useful for compounds you don't know.

Therefore my idea would be here that if, as the book mentions, $sp^3$ hybridisations and covalent, tetrahedral structures as the main source for the semi-conducting properties it's only natural that $III-V$ compounds show a similar effect. Those bands are basically molecular orbitals. If you start from a di-atomic molecule and increase the amount of bonding partners you will get more and more orbitals that move closer and closer together until you have bands. If let's say Ge has a certain band-gap and you change to GaAs, or course the single contributions to the HOMO and LUMO and later to the bands will be different but not that far away from Ge, so we expect a comparable band-gap.

Source: H. Ibach, H. Lüth, Festkörperphysik: Einführung in die Grundlagen, 7th edition, Springer, 2008.


Converting AlCl3 to Al2O3 Since oxygen is more electronegative than chlorine, if I introduce $\ce{O2}$ gas to $\ce{AlCl3}$ and apply some heat, will the chlorine be replaced by oxygen, resulting in $\ce{Al2O3}$ or $\ce{Al(ClO2)3}$, then evolving the remaining $\ce{Cl2}$ into the air? I suspect not, but why?

Is there a standard way of converting $\ce{AlCl3}$ to $\ce{Al2O3}$?

Yes, you can make aluminium oxide from aluminium chloride pretty much in your way but then again as @andselisk said, it is a pretty inefficient reaction and wastes a lot of fuel and energy and also produces many byproducts. Hence, the reaction is the other way round if considered on an industrial scale. The main preparation method is already described by @andselisk in the comment section. Only to verify its authenticity, here is a patent claim:

  A process for preparing alumina which comprises heating aluminum
  chloride hexahydrate at a temperature of from about 225 to 600 F. For
  a period of time such as to liberate only a portion of its chlorine
  content, commingling a basic reagent with the resultant
  chlorine-containing residue to form aluminum hydroxide, washing and
  filtering the latter and then calcining to produce the oxide.


Why are activities of solids and liquids taken to be unity? First, I should mention that the question was already asked here: Why is active mass of a pure solid or liquid always taken as unity?
However, the treatment there is qualitative and I do not really understand the reasoning for the disclusion of solids and liquids from the reaction qoutient that is presented there. I have attempted to derive this fact mathematically and failed, which why I ask this question again.

Let us look at an example reaction which involves solids and gases. The treatment for other reactions or reactions with liquids shouldn't differ significantly. A very simple reaction for this matter will be the combustion of graphite

$$\ce{C(s) + O2(g) -> CO2(g)}$$

Reaction quotients are derived from the equation for change in Gibbs free energy at nonstandard pressures, so let's find that. Suppose the reaction is carried out in nonstandard pressure, and $\Delta G^o_r$ is known. Let's (I) compress the system isothermally to standard pressure, (II) carry out the reaction and then (III) return it isothermally to the original pressure.

I. $\Delta G$ for the oxygen here is the well known $RT\ln (1/(P/P_o))$. For the carbon: In any isothermal process $dG=VdP$, let's assume for our ideal solid that the volume is constant (I have also tried this in a model where the volume changes with pressure with bulk moduli, the result does not differ significantly), so $\Delta G=V(P-P_o)$.

II. Here simply $\Delta G=\Delta G^o_r$.

III. Here it's just gas compressing so again $\Delta G=RT\ln (P/P_o)$

Adding the contributions from all three stages we have
$$
\Delta G=RT\ln (1/(P/P_o))+V(P-P_o)+\Delta G^o_r+RT\ln (P/P_o)=\Delta G^o_r+V(P-P_o)
$$

This result is different from what a model of an activity of unity for the carbon would predict. What went wrong? And why are activities of solids and liquids taken to be unity? Any help will be appreciated.

It is not that case that the activity of solids and liquids is always taken as unity.

Instead, activity is unity for a pure solid or pure liquid at standard pressure.

This is a matter of definition of "activity" and "standard state".

According to the IUPAC definitions:

activity (relative activity), a

  Defined by the equation
  $a=e^\frac{\mu-\mu ^0}{RT}$
  where $R$ is the gas constant, $T$ the thermodynamic temperature, $\mu$ the chemical potential
  and $\mu ^0$ the standard chemical potential the definition of which depends on the choice of the standard state.

and standard state

  State of a system chosen as standard for reference by convention. Three standard states are recognized: For a gas phase it is the (hypothetical) state of the pure substance in the gaseous phase at the standard pressure p = p °, assuming ideal behaviour. For a pure phase, or a mixture, or a solvent in the liquid or solid state it is the state of the pure substance in the liquid or solid phase at the standard pressure p = p ° ...

So activity is 1 almost by definition when a substance is in its standard state, because $\mu - \mu ^0 = 0$  and $e^0 = 1$


Chemical equilibrium and stoichiometry In reactions that go until the end the stoichiometric number indicates the exact amounts of reactants and products you will get. However, this is not the case in equilibrium reactions and you will get a different amounts (concentrations) than what exhibits the stoichiometric numbers.
$$\ce{Na + Cl -> NaCl}$$
You will get in the end 1 mole of NaCl from 1 mole of Na and one mole of Cl.
$$\ce{2SO3<=>2SO2 + O2}$$
In equilibrium you will not get 1 molar of $\ce{O2}$ and 2 molar of $\ce{SO2}$ from 2 molar of $\ce{SO3}$.
Then what do the stoichiometric numbers indicate in equilibrium?

Let us take the example of your equation:
$$\ce{2SO3<=>2SO2 + O2}$$

The stoichiometric coefficients represent two facts:

If 2 mole $\ce{SO3}$ reacts, it will give 2 mole $\ce{SO2}$ and 1 mole $\ce{O2}$ but one needs to remember here that all of the 2 moles will not react. But the number of moles that do dissociate, give products in the above ratio i.e. if initially have 2 moles of $\ce{SO3}$ and degree of dissociation is 50%, then we will have 1 mole of $\ce{SO3}$ reacting to give 1 mole of $\ce{SO2}$ and half mole of $\ce{O2}$
The stoichiometric coefficients also provide the powers to which the concentrations of each compound will be raised to in the expression for equilibrium constant $$K_\mathrm c = \frac{[\ce{SO2}]^2[\ce{O2}]}{[\ce{SO3}]^2}$$


Emergency shower for nitric acid spills Disclaimer: I'm asking this because I am curious.  Not because I'm actually handling nitric acid.

In another answer I was reading, the poster mentioned that if you spill nitric acid, you shouldn't pour water on anything but a small spill, because it will generate heat (maybe enough to flash-boil the water?), but if you spill nitric acid on yourself, you should have an emergency shower ready.

Except, wouldn't the emergency shower be using water and cause the exact same problem?  Is this just a case of deciding that possible burns from a hot liquid, while under a cold shower are much better than definite acid burns?

If you spill nitric acid onto the table, you yourself are unharmed and you can use a cool head to decide what to do next. If the spill is small, pour water on it to both dilute it and dilute the heat — remember water has a high specific heat capacity. If the spill is larger, I would probably try adding sodium hydrogencarbonate to react away the acid before wiping it up. Adding larger amounts of water will work, too, but you need to get them from somewhere and then need to get rid of the resulting puddle.

If the spill is on you you want to get the acid off your clothes and body immediately. The quickest and most effective solution is to pour water over your head and lots of it. This is what the emergency showers are for. Lots, here, has two functions: it not only quickly dilutes the acid to a less harmful concentration but also (because it is washing down) serves well to actually remove the heat — remember again that water has a high specific heat capacity. Emergency showers have a high throughput for exactly this reason.


Why do green bell peppers become bitter when cooked? Why do green bell peppers taste sweet raw, but become bitter when cooked (steamed)?
What chemical process is responsible for this? (I don't believe the temperatures are high enough for the Maillard reaction)

Please explain the downvote(s). I wondered if this might be off-topic, but since food-chemistry exists, I believe that it's not.

Thousands of different chemicals, previously spatial separated, are mixed together after thermal death of cells, and kept at high temperature to react well. Final taste will depend on original system, but will develop by way unpredictable theoretically from basic principles.
The answer for a particular biological source requires detailed knowledge about particulars of green bell peppers biochemistry, which is rare.
Generally, there are basically few, possibly combined, options:

Bitter tasting compounds - possibly, but not exclusively alkaloids - are already there, but encapsulated in celular structures and released during cooking.
They are actively created by enzymes released after cell deaths ( like $\ce{\mathrm{alliin} ->[\mathrm{alliinase}] \mathrm{allicin}}$ in garlic)
Thermal decomposition of some substances creates such bitter compounds.

Further reading:
bitter-bell-peppers provides rather the opposite opinion:

Why cooked peppers are less bitter
Have you noticed that raw bell peppers taste significantly more bitter when eaten raw (for crudités or in a salad)? That's because most bitter compounds in food break down with heat. The bitterness doesn't go away completely, but it is lessened.

How to Take the Bitterness Out of Bell Peppers:

Select ripe bell peppers. Green bell peppers are harvested before they fully ripen, which causes bitterness. Choose red bell peppers; they are fully mature and slightly sweeter than orange, yellow or green varieties.
Remove the seeds from the cores. The bell pepper seeds are mostly in the cores, and they are naturally bitter. Slice the top of the bell peppers and place them upright. Use a knife to cut out the cores in a circular motion. Cut open the bell peppers and scrape off the remaining seeds.
Peel the skins. Put your bell peppers in a preheated oven until the skins blister. Let them cool for 10 minutes in a paper bag -- the steam will help loosen the skins. Run cold water over the bell peppers and peel off the skins.
Cook the bell peppers. Add soy sauce, salt or other spices to give the bell peppers flavor. The cooking process will eliminate some of the bitterness, and the seasoning will help mask the taste.

chemistry of bell peppers:


Why is hexafluoridonickelate(IV) diamagnetic? As fluoride is a weak-field ligand, it should not pair up the electrons. So why is $\ce{[NiF6]^2-}$ diamagnetic?

This is a common mistake to make when analysing transition metal compounds. The undergraduate or the non-coordination chemist looks at a complex, identifies a ligand, puts it into their box of weak-field or strong-field ligands, deduces an electronic structure and is then completely perplexed if something does not work the way the simple picture would assume.

It is then helpful to remember what the molecular orbital scheme of am octahedral complex looks like; refer to scheme 1. Note that scheme 1 ignores π contributions of the ligands but fret not for I shall briefly explain where (and how) they come in.

Scheme 1: Octahedral $\ce{[ML6]}$ complex with no π interactions. Image copied from this answer and originally taken from Professor Klüfers’ internet scriptum to his coordination chemistry course.

On the left-hand side of the scheme you can see the metal’s orbitals, from top to bottom they are (in nickel’s case) $\mathrm{4p\ (t_{1u}), 4s\ (a_{1g})}$ and $\mathrm{3d\ (e_g + t_{2g})}$. On the right-hand side, we have the six ligand orbitals which transform as $\mathrm{a_{1g} + e_g + t_{1u}}$. Only orbitals with identical irreducible representation interact constructively (or destructively) leading to the corresponding molecular orbitals shown in the middle.

What is commonly termed field split, $10~\mathrm{Dq}$ or $\Delta_\mathrm{o}$ is the difference between $\mathrm{t_{2g}}$ and $\mathrm{e_g^*}$ — the former being entirely metal-centred if π interactions are not considered, the latter being formed by both the metal and the ligand in a bonding-antibonding interaction.

Adding the ligands’ π orbitals, realise that they would transform as $\mathrm{t_{1u} + t_{1g} + t_{2u} + t_{2g}}$. Most of these interactions are either strictly non-bonding or contribute only weakly to bonding since the corresponding symmetry $\mathrm{t_{1u}}$ is already being used to make σ interactions. Only the $\mathrm{t_{2g}}$ p orbitals introduce a new interaction with the central metal by turning what are nonbonding orbitals in scheme 1 into a bonding/antibonding pair of orbitals. Since the interaction is π type, it will be a weak splitting, but it contributes to the decrease of $10~\mathrm{Dq}$ by raising the bottom layer.

I have talked a lot but so far I did not mention the point why hexafluoridonickelate(IV) is a low-spin complex. However, the basis has been laid. It is important to visualise that the field split is — because it derives from the strength (or weakness) of bonding/antibonding interactions in the molecular orbital scheme — dependent both on the metal’s and the ligands’ properties. This can be visualised in the very simple (and thus not satisfying; when approaching a higher level) equation $(1)$:

$$10~\mathrm{Dq} = g_\mathrm{M} \times f_\mathrm{L}\tag{1}$$

In this formula, $g_\mathrm{M}$ represents the contribution of the central metal while $f_\mathrm{L}$ that of the ligands. Obviously, if we modify the metal in a certain way, we can significantly influence the value of $10~\mathrm{Dq}$, i.e. the field split, i.e. whether the complex prefers a high or a low-spin state. Enough circling the issue, let’s get to the point.

The preference for high or low spin can be greatly influenced by the metal’s oxidation state. Remember that a positive charge attracts electrons, so that their respective orbitals will (all!) be lowered in energy by oxidation. Thus, the energies of the metal’s d orbitals and those of the ligands’ σ-symmetric orbitals will move closer together creating a larger stabilisation/destabilisation. At the same time, the π symmetric ligand orbitals will be more removed energetically and thus the weak π interaction will be further diminished. These effects in combination lead to a greater field split for high central metal oxidation states. In hexafluoridonickelate(IV), the oxidation state is $\mathrm{+IV}$ — pretty high for nickel which is probably most commonly found as nickel(II) in coordination complexes. Thus, it is not surprising that this higher oxidation state leads to a low-spin complex. Low-spin $\mathrm{d^6}$ is, of course, diamagnetic.

For further exemplification, I quote two values from a table of $g_\mathrm{M}$ values Professor Klüfers included in the linked internet scriptum (in German):

$$\begin{array}{cc}\hline
\ce{M^n+} & g_\mathrm{M}/ 1000~\mathrm{cm^{-1}}\\ \hline
\ce{Co^2+} & 9 \\
\ce{Co^3+} & 18.2 \\ \hline\end{array}$$

As can be seen from these values, increasing cobalt’s oxidation state from $\mathrm{+II}$ to $\mathrm{+III}$ doubles the metal’s contribution to $10~\mathrm{Dq}$. And that was only an increase of one.


Enantiomers or Diastereomers? The key tells me these are enantiomers.

I however don't think they are; their chiral carbons' configurations aren't inverted - i.e. only one chiral carbon's R/S configuration changes! This suggests diastereomers to me! Plus the Sawhorse projections I draw don't seem to be mirror images.

They are diastereomers.  Both of the "front" chiral carbons (the one with the bromine attached) are "S".  Rotate the one on the right 120 degrees CCW and it is identical to the one on the left.  But the "back" carbon (the one with the hydroxyl attached) is "S" on the left figure and "R" on the right figure - diastereomers.


Why does pH affect fermentation? I know that the lower the pH the faster fermentation occurs. Why does this happen?

pH affects the shape of proteins. In the case of fermentation a collection of enzymes is responsible for the metabolic processes that occur. - an enzyme is a protein which performs a metabolic process. For example sucrase is an enzyme which breaks sucrose down into fructose and glucose. They are kind of like organic catalysts.

Proteins are made of amino acids strung together in a long line, based off DNA. The amino acids bond up into a long polymer. Then cross linking occurs between the R groups on the amino acids. This makes the unique shape of the protein, because the cross-linking turns the protein into a 3-d shape.

Amino acids may be acidic amino acids, or basic amino acids, due to the R group on the amino acids. If the pH is increased, this affects the shape of proteins, by disrupting the bonds in the protein. In the case of fermentation, you say the rate increases when it get's more acidic - when the pH is lower. This is because the organisms - the yeast - producing the enzymes to ferment glucose, have adapted to acidic conditions. This means that through natural selection the enzymes today have been selected because they work best in acidic conditions, which bend the protein into the correct shape to allow fermentation to occur.

Read: http://www.icr.org/article/glycolysis-alcoholic-fermentation/

In this article it says that 10 enzymes turn glucose in pyruvic acid:http://en.wikipedia.org/wiki/File:Pyruvic-acid-2D-skeletal.png

And then two enzymes turn this into ethanol and CO2. The last two enzymes function best in acidic conditions, presumably because they have evolved to work best in the acidic conditions made by the pyruvic cid. However all fermentation stops after the pH drops below about 4.2. The optimum pH is about 4.8 - 5.0. After this we can assume that the pH bends the proteins out of shape too much, and the protein is said to have been denatured.

So to conclude... The rate increases because enzymes are all suited to specific conditions, dictated by natural selection - the conditions they have evolved around - and in the case of fermentation the last 2 enzymes - turning pyruvic acid to ethanol and CO2 - work best in acidic conditions, as they have evolved to those conditions. But increase the pH too much... and the protein denatures, and fermentation stops completely.


Why is fullerene 60 an insulator while graphite is a conductor? Graphite is good conductor of electricity because one carbon atom is bonded only three carbon atoms, which enables the presence of free electrons. In fullerene 60 the carbon atoms are also bonded to three atoms so why is fullerene concidered a bad conductor of electricity?

To reiterate Ivan's comment fullerene is a bad conductor because that's what the measured properties produce as a result. The mechanism that makes it a bad conductor is that it has shorter range continuity than graphite. In graphite the carbon is made of sheets that can be as long as the sample. These sheets have fairly low conductivity, but when the electrons must jump between sheets to continue its path, this is when the conductivity value for the bulk material lowers. Extrapolate this to fullerene. If the fullerene molecule itself had high conductivity, it would still require the electrons to jump between molecules every few nanometers. This can produce a rather large resistivity for the bulk material much more so than observed in graphite.


What is the reasoning behind the resonating structure of halo-benzene? Can anyone explain the resonating structure of halo-benzene? How does the transfer of an electron from the halogen atom to the α-carbon and from there to ß-carbon and so on take place?

I think the misunderstanding here (it's a common one) is over the interpretation of resonance structures.

What you have to understand about resonance structures is that they do not actually exist in the molecule. The separate resonance structures are never formed on their own but they can be used to represent a delocalised bonding situation, more accurately described using molecular orbitals. So electrons do not actually move from the halogen to the $\alpha$ and $\beta$ carbons but what the resonance structures represent is that, only considering mesomeric effects, there is a greater electron density on the ortho and para carbons and than on the meta carbons, the carbon attatched to the halogen and the halogen itself.

However, this is not the full picture because halogens; especially fluorine and chlorine, have a very strong inductive effect and so there is actually considerably more electron density on the halogen than on any other atom.

Electrostatic potential map for chlorobenzene:


How much steric hindrance does a phenyl group offer? I've come across a question in my revision in which I am asked to compare the steric hindrance provided by a number of substituents. Those relating to phenyl are ranked relatively low, despite it being what I thought was a rather large substituent. Is this due to the planarity of the ring, or is it just much smaller than I originally thought; does a tertiary alkyl substituent provide more steric hindrance because it is larger, or because it is spread in all directions?

despite being what I thought was a rather large substituent

I agree, generally speaking the phenyl group is considered a rather large substituent.

One metric commonly used to compare "steric size" is the cyclohexane A-Value.  In the cyclohexane ring system, bulky substituents prefer the equatorial position rather than the axial position. This is due to steric (this can also be described in electronic terms since it is due to the repulsion between interpenetrating electron clouds) destabilization. For example, in methylcyclohexane, the methyl group can exist in an axial or equatorial position. In the axial position, it is destabilized by the repulsive steric (electronic) interactions it has with the axial hydrogens on carbons 3 and 5.

The larger the A-Value, the larger the substituent and the greater its preference for the equatorial location ($\mathrm{\Delta G = -RTlnK_{eq}}$ where $\mathrm{K_{eq}}$ represents the the axial/equatorial ratio at equilibrium).

Here is a link to a Wikipedia article on A-Values.  Look at the table of A-Values in this article.  A methyl group has an A-Value of 1.7 kcal/mol. a phenyl group has an A-value of 3 kcal/mol.  The only substituent in the table with an A-Value larger than phenyl is the t-butyl substituent with an A-Value > 4 kcal/mol. The phenyl group is a relatively large substituent.


When is diffusion steady-state? Fick’s first law of diffusion can be used when $\mathrm{d}C/\mathrm{d}t=0$. If the concentration gradient is time-variant, then we must use Fick’s second law. However, when is diffusion steady-state and when is it nonsteady-state? How do I know when the concentration gradient will change with time and when it won’t?

There is no set answer to your problem, it depends on the geometry of the diffusional process. I give two examples.

(a) Suppose there is a volatile solute in a solvent placed in a beaker and the solute evaporates from the surface, then there is a concentration gradient of solute away from the surface as the solute is lost into the air. Similarly, if there is an electrode at which ions are being deposited as a metal, or a precipitate is being formed, then a concentration gradient is set up, but as material is removed no steady state can be set up as there is a finite amount of substance present.

Solving Fick’s 2nd law produces an equation describing the concentration vs. time which is $c_{x,t}=c_0erf(x/(2\sqrt{Dt}) $ where erf is the error function, D the diffusion coefficient, t time and $C_0$ the initial concentration. (The initial conditions are $t=0, x\ge 0, c=c_0$;  $ t>0,x=0, c=0$;$ x \rightarrow \infty, t=0, c=c_0$ ).

The depletion distance, that is the distance from the sink that the concentration is just becoming less than than $c_0$ is given by $\delta = \sqrt { \pi Dt}$ at time t.

(b) In a closed system, for example, a cell with half of it containing a solution and the other half solvent, equilibrium will become established once diffusion into all the cell is allowed. Similarly an insulated bar half of which is initially hot will effectively come to an equilibrium at the average temperature.

Consider a tube filled with solvent and then closed and some solute injected at some place, diffusion will ensure that equilibrium will eventually be reached. Because the initial distribution of concentration  must level out to a constant value, the solution to Fick’s equation must have both a time dependent and a constant part.

At long times the temperature or concentration will become uniform and if $f(x)$ is the amount initially added at position x and $0 < x < L$, the long time value is the average value;  $\frac{1}{L}\int _0^Lf(x)dx$ . The initial condition is given by the shape of the concentration profile $f(x)$.

The concentration of added solute is $c(x, t)$  at position x and time t. Because no material (or heat)  leaves the concentration gradients at the ends of the tube are zero at all times, or
$$\frac{dc(0,t)}{dx} = \frac{dC(L,t)}{dx} =0, ~~~t>0$$
Starting with the diffusion equation
$$\frac{\partial c}{dt} = \frac{\partial ^2c}{dx^2} $$
and separating variables, gives
$$\frac{\partial c_t}{\partial t} = -Dk^2c_t$$
 and
$$\frac{\partial ^2 c_x}{\partial x^2} =-c_xk^2 $$
where k is a separation parameter to be determined.
The solution is $c=c_tc_x$.

The time dependent equation integrates to  $c_t=e^{-Dk^2t}$ but the spatial part needs special attention to incorporate the boundary conditions.

The separation parameter k can have values of $0$, or it can be positive or positive and imaginary. Only when k is positive are the solutions non-trivial and are $c_x=a\sin (kx)+b\cos (kx)$.

The gradient boundary condition ensures that the solution is the cosine term because the derivative, $kb\sin (kx)$, is zero both at $x = 0$ and at L provided that $k=n\pi/L$, where n is an integer.

Next we use k as if it were an integer and only substitute its actual value at the end. The solution is
$$c_k=e^{-k^2Dt}cos(kx)  ~~~~~~~~  k=0,1,2…$$
and the final solution is a linear combination of the k terms including $k = 0$ and this will give a term that is independent of time. The initial condition ensures that
$$ f(x)=\Sigma _{n=0}^{\infty}b_k\cos (kx)$$
giving
$$ b_k=\frac{2}{L}\int _0^Lf(x)\cos (kx)dx$$
where $k=n\pi /L$. The complete solution is
$$ c = \frac{b_0}{2}+\Sigma _{n=1}^\infty b_ke^{-(n\pi /L)^2Dt}\cos (n\pi x/L )  $$
The first term $b_0/2$ is divided by 2 to make it equal to the average concentration at long times, $\frac{1}{L}\int _0^Lf(x)dx$ because by its defining equation $b_0$ is otherwise 2 times too large.
The results below show how the initial profile eventually reaches a constant, time-independent value, the total amount of material, or heat being conserved. Uniform concentrations are found when $ t > \approx  5/ (n\pi /L)^2D $, n is found to be in the range $10$ to $20$.

figure: One dimensional diffusion in a closed tube. The summation was taken to $n=10 $ terms (left) and $20$ (right).  (Left) The solute is injected in the centre and has a gaussian profile ($\exp (-5000(x/L-0.5)^2)$ at $t=0$  Right. Case where the solution is half the cell. The initial profile is a sigmoid curve $(1+\exp (-100(x/L-0.5))^{-1}$. The diffusion coefficient $D = 1$ and $L=1$. The profile at different times is shown.


Rusting in every other bar I have noticed that, when a structure with bars (such as a gate or a grate) develops rust, this sometimes happens differently across the bars. In particular, it seems that the bars are alternatively clean and covered in rust.
The photos below (taken by me) illustrate this effect (I have highlighted the rusted bars with red arrows). In both cases, from inspection, all bars appeared to be made from the same material and covered in the same kind of protective paint (the rusted parts were exposed to the elements).
My theory is that perhaps the rusting process activates some sort of "circuit" between bars, such that one bar protects the next one from rusting.
Is this possible?

Rebar for concrete is frequently stored in heaps, and rusts. It actually develops a stronger bond if it is not too rusty, because the rebar surface is rougher and better able to adhere to the hardened cement. It's interesting to see a pile of old rebar with a fresh delivery on top - silvery, not a trace of rust - on top of a pile of rebar that looks almost disgustingly rusted. And then you think, yeah, but the stuff on the bottom will have a better bond.
The key word in the observation about the fences is "sometimes".
I'm going to guess that the maker of the fence in the top picture also had a pile (or two) of iron bars, and some were rusty from being stored long or outdoors, or were a teensy bit different from some other pile of iron bars. This fence maker, knowing how his fence weathers, has to make it as uniform as possible, so he takes one bar from one pile, then a bar from the other pile... He prepares them identically, primes them identically, paints them identically... But then nature takes over and the most corrodable bar develops a spot, which spreads until it is all corroded-looking, before the more corrosion-resistant bar has even gotten started!
And then you come along and take a picture just before the other bars begin to corrode! This experiment is in the initial stages, not yet at equilibrium. Check back on the fence every year or two - I believe things will even out in time.


How does hyperconjugation affect bond lengths in alkenes? How does the number of hyperconjugates affect the $\ce{C=C}$ bond length in the compounds propene and 3,3-dimethylbut-1-ene?

In general, is bond length directly or inversely proportional to the number of hyperconjugates?

The higher the number of alpha hydrogens present in a molecule, the more resonance there will be which will decrease the bond order which was earlier 2, thus decreasing its strength and increasing its length.


Why do the lone pairs of electrons first occupy axial positions before occupying equatorial positions for species with octahedral geometry? Why do lone pairs not reside in equatorial positions in species with octahedral geometry?

The main reason is repulsion with the electron pairs. Compare $\ce{SF6}$ &  $\ce{SF4^2-}$ as an example. To get from $\ce{SF6}$ to $\ce{SF4^2-}$, we need to replace two fluorines with lone pairs. The first replacement is arbitrary; in an octahedral geometry, all the positions are equivalent. Once we have replaced that first fluorine however, the bond positions are no longer equivalent. We will have four equatorial bonds (90 degree separation from the lone pair) and one axial bond (180 degree separation from the lone pair). Lone pairs are very repulsive, especially with respect to other lone pairs, so we want the next lone pair to be as separated as possible from the first. The axial bond is further from the lone pair then any of the equatorial bonds, so it makes sense for the lone pair to go there.


Is perovskite symmetric under exchange of A and B atoms? A perovskite is any crystal with the general formula $\ce{ABX3}$ where if the $\ce{B}$ atoms form a cube, an $\ce{A}$ atom sits in its center and the $\ce{X}$ atoms sit between neighboring $\ce{B}$ atoms.

Is the perovskite structure symmetric under exchange of $\ce{A}$ and $\ce{B}$ atoms? That is, if I replace every $\ce{A}$ atom with a $\ce{B}$ atom and vice versa, will I end up with the same exact structure as before?

No.  In the perovskite structure, the "A" atoms are larger than the "B" atoms.  The B atoms are 6-fold coordinated, while the "A" atoms are in 12-fold cuboctohedral coordination.


Is it possible for a group to substitute itself? When I first learned about substitution reactions, I started wondering something that I never ended up figuring out:

Suppose you have a strong nucleophile which is also a good leaving group, such as I$^{-}$. Can such a group substitute itself using the SN1 mechanism? If so, would that mean if such a group were bonded to a chiral carbon, an enantiomerically pure solution could spontaneously racemize?

Yes - a nucleophile can replace itself in a substitution reaction. Under the SN1 mechanism, once the carbocation is formed, it becomes prone to attack from any negatively charged species, including the one that left it in the first place. Likewise, in an SN2 reaction, there is the possibility of a nucleophile replacing itself. Given enough time, I would expect alkyl halides to become racemic in solutions that contain the halide anion; however, substitution reactions remain stereo-selective. This is because when you are preforming a substitution reaction, you normally use a better leaving group than the nucleophile that you are adding. Because of this, the nucleophile in solution will preferentially replace the initial leaving group over its resulting product. This is because the rate laws for both SN1 and SN2 reactions depend on the concentration of the alkyl group - implying that an alkyl group with a better leaving group will react faster than one with a weaker leaving group. In SN1, the rate determining step is the leaving group dissociating from the carbocation, which happens quicker with a better leaving group. In an SN2 reaction, the rate determining step involves both the nucleophile attacking and the leaving group leaving. This - once again - means that the reaction will proceed quicker with a better leaving group. Given enough time, your product should racemize; however, your substitution reaction should be done before this can have a major effect on your yield.


Which reaction occurs first when 5-hexen-1-ol reacts with sulfuric acid? I am trying to solve this question:

Find the major product of the following reaction:

Now we could either do electrophilic addition on the alkene or do dehydration of alcohol. These two different reaction pathways yield different products. So how do we decide which reaction occurs first?
I thought that electrophilic addition could be favored because of the generation of a carbocation that could then undergo $\mathrm{S_Ni}$ reaction with the $\ce{-OH}$ generating a cyclic structure. Now if the cyclic structure were more stable than the open chain product formed by dehydration of alcohol then this reaction could be favored. But I don't know how this could be proven.

In general, intramolecular reactions occur faster than their intermolecular counterparts. Most commonly this is rationalized by an entropic argument: in the case of the intermolecular reaction, two molecules (substrate + water) combine to form one (diol). This is entropically disfavored relative to the intramolecular case, in which the molecule cyclizes.

However, the biggest factor at play in determining which will occur in this case is the size of the resulting ring. If the product contains a 5, 6, or 7-membered ring, these rings are considered be unstrained (bond angles ~ 109.5 degrees). If the rings are smaller than 5 atoms or larger than 8, the rule of thumb is that the thermodynamic (enthalpic) cost of the ring strain associated with cyclization overcomes the entropic driving force to cyclize, and the intermolecular reaction is preferred.


How can the 18 electron rule be used to explain that Fe(η5-C5H5)2 doesn't react with hydrogen but Ni(η5-C5H5)2 does? Fe(η5-C5H5)2 does not react with hydrogen, but its nickel analogue, Ni(η5-C5H5)2 is readily hydrogenated to give Ni(η5-C5H5)(η3-C5H7). Use the 18 electron rule to explain this behaviour.

My answer so far...

Fe(η5-C5H5)2 18 electron rule = valence electrons from Fe is 8, electrons from C5H5 is 5 therefore 8 + 5 + 5 = 18 so it is stable.

Ni(η5-C5H5)2 18 electron rule = valence electrons from Ni is 10, electrons from C5H5 is 5 therefore 10 + 5 + 5 = 20 so Ni is not in most stable form

Ni(η5-C5H5)(η3-C5H7) 18 electron rule = valence electrons from  Ni is 10, electrons from C5H5 is 5.....

how many electrons from (η3-C5H7) and how does this all explain why the iron complex doesn't react with hydrogen but the nickel one does? It is to do with the fact it has two too many electrons (20e) so it bonds with two protons from hydrogen? Also what does this symbol η mean?

Hydrogenation of NiCp2 takes place on the ligand, not the metal, to reduce donor electron count, giving a low-energy closed-shell 18 electron count.

http://en.wikipedia.org/wiki/Nickelocene
http://en.wikipedia.org/wiki/Ferrocene
http://en.wikipedia.org/wiki/Hapticity
http://www.chem.mun.ca/homes/cmkhome/Organometallics_The%20Basics.pdf
   p. 9, 10.  "η" is "hapticity," the number of ligand atoms contributing pi-electron orbitals to the metal center.

$\ce{Cp^-}$ is aromatic (draw the Frost diagram) when coordinated $\ce{η^5}$ on-face, or  $\ce{η^1}$ non-aromatic as a simple anion coordinated on-vertex.  Saturating one localized double bond leaves cyclic delocalized allyl anion coordinated $\ce{η^3}$.

In the case of $\ce{TiCp4}$, two Cp anions are face pentahapto, two are vertex monohapto.  If we take a proton or C-13 NMR, will we see discrete protons or carbons for the $\ce{η^1}$-Cp, or will allowed 1,5-shifts scramble all five protons?  All five carbons?  Will it be temperature dependent if it happens?  If it happens, will H and C have different kinetics?


What are some good references on conductometry as an analytical technique? Non of the standard undergraduate textbooks in analytical chemistry (such as Skoog, Harris) discusses conductometry as an analytical technique. I wonder whether this technique is obsolete. Or is it taught under some other area such as physical chemistry ?

Can someone please point me to some good references on this matter ? I am particularly interested in text books.

Thanks.

I am certainly no expert in the field (in fact, I know nothing about it) but I imagine this 225+ page text will prove to be useful for you:
Oscillometry and Conductometry: International Series of Monographs in Analytical Chemistry, Volume 21 by E. Pungor is available via Google Books. You might be able to find a used copy of this book somewhere (it was published in 1965).


Why does mixing two solutions decrease the concentration of the solutes? The irreversible liquid reaction

  $$\ce{A + B -> C + D}$$

  follows an elementary rate law and is carried out
  isothermally in a plug flow reactor (PFR). The volume of the PFR is 500 litres. The concentration of $\ce{A}$ is $\pu{2 mol L-1}$ and concentration of $\ce{B}$ is $\pu{2 mol L-1}$ before mixing. The volumetric flow rate of each stream is $\pu{5 L min-1}.$

  The streams are mixed immediately before entering. The reaction rate constant at $\pu{20 °C}$ is $\pu{0.05 L mol-1 min-1}$
  and $E = \pu{20 kcal mol-1}.$ (Note: gas constant $R = \pu{1.987 cal mol-1 K-1})$

  Find conversion at $\pu{40 °C}.$

This is what I had:

From the solution given the $c_{\ce{A},0}$ used is $\pu{1 mol L-1}$ which is after mixing. Need some clarification on this. Why is it not $\pu{2 mol L-1}$ before mixing?

Comment turned to answer:

Because it has been mixed.

Each substance is now in doubled volume, compared to volume before mixing.

If you had whiskey and if you mixed it with the same volume of soda, it wouldn't have the same strength as before, would it ?


Explaining temperature change in an exothermic system. In a neutralization reaction, is heat given out to the system, thereby raising its temperature? Suppose we put a thermometer in the acid-base mixture and there is an increase in the thermometer reading. Is this increase simply because part of the reaction's heat given out is supplied to this thermometer, or because the water solvent is actually warmer? In other words, if we measure the temperature of the system before and after the reaction, will there have been an overall decrease in the temperature, or increase (and please explain why)?

In thermodynamics, we like to divide any situation into two parts, the system and the surroundings. In general the system is the part of the universe that is being studied while the surroundings is everything else that interacts with the system. Between the boundary of the system and the surroundings, energy and heat can be transferred (in an open system).

So in an neutralisation reaction, the system is the actual acid and base while the surroundings is the solvent (the water). Since it is an exothermic reaction, this means that energy is released from the system to the surroundings. Since the water gains more kinetic energy from the system as the reaction occurs, its temperature will increase. Also this means that the system loses energy and therefore become colder. So to answer your question, the solvent gets hotter while the system gets colder.


Need to chemically heat water I have a holding tank on rv, non potable water that I need to heat up to drain . I live in northern Alberta, so the average temperatures are -20. I don't want to wreck anything or gas myself out. I am looking for over the counter chemicals.

I'm assuming you only need to liquefy ice to drain a septic/grey water tank. Calcium chloride (sold as damp rid) will melt ice down to $-56^\circ C$.  Just add about 10 Pounds to the tank wait a few hours (perhaps even drive a little to slosh it) and drain.


C-H bond energy in substituted methanes What happens to the $\ce{C-H}$ bond strength in methane as the degree of halogenation is increased (e.g. from $\ce{CH4}$ to $\ce{CH3X}$ to $\ce{CH2X2}$...)

Also, what is the effect of varying the halogens from fluorine to iodine?

According to David R. Lide, ed., CRC Handbook of Chemistry and
Physics, 90th Edition (CD-ROM Version 2010), CRC Press/Taylor and Francis, Boca Raton, FL, the C-H bond dissociation energies for increasing chlorination of methane are as follows:

$D\left(\ce{H-CH3}\right) = (439.3 \pm 0.4) \; \text{kJ/mol}$

$D\left(\ce{H-CH2Cl}\right) = (419.0 \pm 2.3) \; \text{kJ/mol}$

$D\left(\ce{H-CHCl2}\right) = (400.6 \pm 2.0) \; \text{kJ/mol}$

$D\left(\ce{H-CCl3}\right) = (392.5 \pm 2.5) \; \text{kJ/mol}$

The effect of different halogens in H-CX3 is shown below:

$D\left(\ce{H-CF3}\right) = (445.2 \pm 2.9) \; \text{kJ/mol}$

$D\left(\ce{H-CCl3}\right) = (392.5 \pm 2.5) \; \text{kJ/mol}$

$D\left(\ce{H-CBr3}\right) = (399.2 \pm 8.4) \; \text{kJ/mol}$

$D\left(\ce{H-CI3}\right) = (423 \pm 29) \; \text{kJ/mol}$