text:
How can I make a better whiteboard cleaning solution?
I just started work as a lab assistant in a chem lab. The whiteboard the instructor uses for classes gets very dirty very quickly. The woman who started at the same time as I did spent over 3 hours scrubbing it with a solution of a powdered soap-like product in a milk-carton style container to get it anything like clean.
I'd rather not watch her do that again (or, worse, do it myself). Can I make a better cleaning agent without using too much of an expensive chemical?
So far  I've seen large quantities of Ethanol, Acetone, Alcohol, and milliQ water. What else we may have, you folks would probably know better than I.

I found methanol-water mixture to be effective, not critical but about 50-50.

Andselisk's recommendations are great, but you'll save yourself a lot of trouble if you just throw all the erasers out. They get crud on them quickly and after a while, they put more back on the board than they wipe off.
A microfiber cloth is much more effective, and works for a shockingly long time. I would bet on a filthy cloth over a brand new eraser, and unlike erasers, they can be thrown in the wash. You can get them specifically for whiteboards or just stop by an auto parts store.

Scrubbing the surface with an abrasive (powder) only makes it worse -- micro-scratches will accumulate pigments and dirt and overall promote further surface degradation. I would try the solvents (test on a small area first) in the following order: isopropanol, ethanol, acetone. Usually even cleaning wipes for glasses or computer screens do just fine.
I also have a suspicion that this particular whiteboard might need special erasable markers. So, I'd get rid of standard permanent markers in the vicinity, and probably get in touch with a person who makes lab purchases.

"I also have a suspicion that this particular whiteboard might need special erasable markers."  +1, but all this time I was under the impression that whiteboards are compatible with all (non-permanent) markers 8D

@paracetamol Well, yeah, they should be, but it vastly depends on the surface and how it's been treated. There is still some adhesion happening with plastic surfaces, and I doubt we are talking about the glass whiteboard here;)


text:
What is the correct way to clean spectrophotometry cuvettes?
The lab I started working in has a spectrophotometer with cuvettes. The cuvettes need to be inspected and cleaned regularly. The senior people in the lab are in disagreement as to the correct procedure for cleaning them. Some recommend soap and a bristle brush with running water, others just suggest leaving them to soak in an ethanol solution.
The question is twofold;
a) Is there any risk that the brush will scratch the glass?
b) Is there any advantage to the soap method? If not, I am wasting an hour of my boss' time and money.

I would think the thing to do is to consult the manufacturer of the cuvettes...

Not sure what kind of cuvettes you are using, nor what types of solutions you are using. I'd agree with andselisk. No abrasives also meaning no brush which can hold grit. If you're using the square UV/Vis cuvettes with a thin quartz window then I'd certainly stay away from a brush. My thought if the cuvettes are really cruddy, then I'd soak them in a acidic chromate cleaning solution.

I always managed to wash out the remaining solution with the corresponding solvent a couple of times, subsequently rinsing with acetone/isopropyl alcohol and leave it dry on air for a few minutes. The method of spectrophotometry requires that you are using homogeneous solution anyway, so there shouldn't be any precipitation/deposition, otherwise you cannot interpret the data with Beer-Lambert law.
Also, don't soak cells in KOH-bath or similar alkaline media. And I'm always strongly against abrasives, especially when it comes to cleaning optics. Even quartz cuvettes can be occasionally damaged with the brush pick. Disposable PS-cells can be easily scratched. Plus, there is a little risk of hydraulic pressure knocking out the cuvette's bottom (like it happens with test tubes when cleaned too vigorously). Leave mechanical cleaning for the professors of mechanics *.
*From a story published in The Oxford Book of Scientific Anecdotes:

Victor Moritz Goldschmidt, a distinguished geochemist, was reputed to have carried a capsule of potassium cyanide when he was planning his escape from Nazi Germany. When a friend in the engineering department of the university expressed interest, Goldschmidt was supposed to have replied that cyanide was for professors of chemistry; his friend, a professor of mechanics, would have to carry a rope.

Clean sample cells, ATR surfaces etc are imperative in spectrometry and a real problem. My ultimate solution was to buy my exclusive sets of cells and use only one of the set for sample and one for reference and not run an analysis until single beam and double beam matched archive spectra. NO ONE else ever touched these. Today with disposable cells, pipet tips, etc. contamination is a bigger problem than appreciated. It depends on the material but simple rinsing might not be sufficient. I used a "clean" detergent[pure sodium n-lauryl sulfate works good] with a fine brush  and 18meg water after an appropriate organic solvent rinse and it usually worked but blanks should be run before each analysis.

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text:
Finding $c$ with Dulong-Petit law on non-molecular substance
The Wikipedia article states the Dulong-Petit law as follows:
$$c=\frac{3R}M$$
$c$ is specific heat capacity $[\mathrm{\frac J{kg\; K}}]$ and $R$ the gas constant. $M$ is the molar mass in [g/mole], and "grams per mole" makes us think of a number of moles of a substance, as were it a molecular material.
In crystal structures there is no such thing as a "mole" of the substance. An example could be $\ce{CoSb3}$. $\ce{CoSb3}$ is not a molecule, but the simplest formula unit of the endlessly repeating crystal. So which $M$ to plug into the Dulong-Petit formula is not obvious here.
A researcher gave me the following version of the Dulong-Petit formula for this case (which I have used successfully):
$$c=\frac {3R}{M}n$$

The $M$ still means molar mass [g/mol], treating the formula unit as were it a molecule.
The $n$ is the number of atoms in the formula unit. So here, $n=4$. Simple.

I do not see how this version is equivalent with the former version. How does "treating the formula unit as was it a molecule" balance out with "multiplying with the number of atoms"? How is
$$\underbrace{\frac{3R}M}_{molecular}\Leftrightarrow \underbrace{\frac{3R}Mn}_{crystalline}$$
equivalent?

It may be useful to start with the molar heat capacity $c'$ (per mole, instead of per kilogram), which is somewhat more fundamental. Here $c' = 3R$.
This is true even for your crystalline solid, though we have to decide what "moles" refers to in this case. From a physical perspective, the heat capacity is a result of vibrational motion in three dimensions, and each atom will contribute to this motion. Our "moles" should therefore be "moles of atoms"---for the purposes of this calculation, we consider the atom to be the fundamental unit.
The molar mass should then be $\text{mass of solid/moles of atoms}$, which can be interpreted as a weighted molar mass. You can see that your formula $M/n$ calculates exactly this: the mass per mole of a formula unit, divided by the number of atoms in a formula unit.
Hence $$c = \frac{3R}{M/n} = \frac{3R}{M}n.$$

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text:
Why only structures corresponding to valleys on PES are taken into account?
I have had to ask this question again because it was marked as a duplicate but it wasn't. Here it goes..
Suppose we want to estimate $G_f^\circ$ of a molecule. If it has rotational conformers, right way to calculate $G$ should include a Boltzmann average of them.
We will do it as:
$$\large \langle G_f^\circ \rangle = \frac{\sum_{i}G_f(i)e^{-G_f(i)/k_BT}}{\sum_{i}e^{^{-G_f(i)/k_BT}}}$$
where the sum goes over the conformers.
Question:
Why $\langle G_f^\circ \rangle$ is estimated as the sum of $G_f(i)$ of only most stable conformers? (corresponding to minima on PES)
Example
Suppose the following rotational conformers:

Consider the $y-axis$ to be gibbs free energy, we would estimate $\langle G_f^\circ \rangle$ as the sum of structures corresponding to minima (ie 60º, 180º), but there are non-minimum structures around 180º with less energy than the minima in 60º. Why aren't them considered?
EDIT
Someone could argue that a conformer in 150º won't exist because there will be a driving force to 180º, but if we take the relative energies $\frac{G_f(150º)}{G_f(60º)}$ we get that 150º must have almost the same population as 60º. There might be some criteria I am missing or maybe a conceptual error.
Any edit will be welcome.

It could be that the unstable conformers are so low a fraction of the species as to be negligible.

No, the last part of the post tries to explain that point.@a-cyclohexane-molecule

Are you only considering global minima, then? In your question statement, you mention local minima, and I'm not sure to which you refer in the last part of your post.

No, I mean, why estimate mean gibbs free energy as the sum of gibbs free energy of only most stable conformers?@a-cyclohexane-molecule. Thanks for the comments..

I think this is a good question because I would like to see a rigorous explanation of this. However, my intuition from the previous iteration of this question remains the same that the issue in how we define a state. I think the argument at the end of your post is invalid because it implicitly assumes that $150^{\circ}$ is an occupiable state of the system. The only way it would be at that configuration is a brief fluctuation of around the bottom of the $180^{\circ}$ well or en route to the $60^{\circ}$ well.

@Tyberius it is an occupiable state of the system, why not? It is a position of nucleus..

@HernanMiraola by analogy, think of a two state system: you are in an airplane flying at a constant height or you are on a level stretch on the ground. You can pass between the two states and briefly be at a point that doesn't coincide with those two states. But you will rapidly collapse to one or the other. If you jump out of the plane, there is zero probability that you hang in the air somewhere in between the ground and the plane. The same is true if you jump from the ground. The space in between the states is not occupiable in this sense.

When you are using the Gibbs energy, then the vibrational modes within the minimum are already considered, and the nuclei arrangement at angles slightly off the minimum correspond to vibrational distortions.

The vast majority of thermodynamics we consider is at equilibrium, we is the issue here.
At equilibrium we will only find our S states at the bottom of the valleys (or minima) populated.  This this because they have now force acting on them to drive them to another configuration.
An important relationship to remember with potential energy surfaces is:
$$
F = -\frac{dE}{dx}
$$
Force is the negative derivative of potential energy with respect to some coordinate.
This means that the driving force is given by the slope of the PES and is pushing down the PES.
Then a molecule with a $\phi = 150^\circ$ experiences a rotational force pushing it towards the minimum at $\phi = 180^\circ$.
Answer:
The result is that any molecule not at a minimum will be quickly driven towards the nearest minimum, so only the minima will be occupied and need to be summed over.
Side Note - Maxima
Maxima (the peaks) also experience zero force.  However this is only true at the exact peak itself.  Any point infinitesimally far away from the maxima experiences a force pushing it down the hill into a minima.
Thermal fluctuations and collisions with other molecules will continually slightly nudge the molecules around on the potential energy surface.  At maxima this will push any molecules there off the peaks and down into minima.  In minima, the force will bring the molecules back into the minima.

Nice answer. And what about the relation $G_f(150)/G_f(60)$? Why is that reasoning wrong? I mean, we don't consider them because are only a few at equilibria?

That relationship is only valid for comparing the population of different minima, hence not for $G_f(150^\circ)$ as $150^\circ$ is not a minima.  The population of any state that is not a minima will be zero (or at most , briefly, vanishingly small before relaxing back to the minima).

I think there must be a deeper explanation, I mean, I am not sure that because it is not a minima we can't take that fraction of the previous comment. But it is helpful..thanks..

I'm not sure I believe this statement: "At equilibrium we will only find our S states at the bottom of the valleys (or minima) populated." At equilibrium these states should be populated according to the Boltzmann distribution, and certainly higher-energy states have some, if little, population.

@user213305 What about cis/trans isomers when a double is present? Should treat them as rotational conformers?

You are right that this is an approximation. It is, however, a much better approximation than you might think. The assumption is that each of the wells are shaped approximately the same. Furthermore, because the population of a state decreases exponentially with energy, it is really the approximation that each of the wells is shaped the same near the minimum.
It is well known that the smaller the displacements from the minimum, the more harmonic the motion will be. So, a measure of how much error is introduced by this approximation is the difference in harmonic frequency of the torsion in each of these wells. And, in the case that the frequencies in each of these wells is the same, the equation you have written is exact within the harmonic approximation. If the true (experimental) frequencies are identical, then the equation you have written is exact.
To be extra clear, for identically shaped wells, all deviations from the minimum energy structure will cancel out because when you do the Boltzmann averaging in each well, the population above any minimum will follow a Boltzmann distribution. Thus, the contributions from these deviations will be very minimal, and depend on there being different shapes of the well.
Probably the simplest way to include the shapes of the different wells in a calculation like this would be to construct the PES using some kind of ab initio method and then do a dynamics simulation on this surface to get the free energies this way.

The Boltzmann distribution doesn't only take into account the most stable conformers. It can take the energy of any conformer (relative to the energy of the most stable conformer).
I assume that you're calculating the Gibbs energy using $\Delta G = - RT \text{ ln}K$. In this equation, $K$ represents the equilibrium constant between two local minima, i.e. between anti and gauche. So, when you calculate the Gibbs energy in this way, you're actually calculating the Gibbs energy for the rotation between two conformers (you can sort of imagine this as a reaction).

Oh, I am sorry I edit about G, I will add it again.

The standard Gibbs free energy is "the change of Gibbs free energy that accompanies the formation of 1 mole of a substance in its standard state from its constituent elements in their standard states". You only need to consider the elementary components and the minimum energy structure.


text:
How to verify if a rate law fits the experimental data?

A student after doing an experiment on chemical kinetics, decided that the rate law of the reaction is $u = k[\ce{IO3-}]$ which is correct, but do the data below match with this? Or does he have to repeat the experiments?
   $$\ce{IO3- + 8I- + 6H+ → 3I3- + 3H2O}$$
In order to decide the rate law he took into consideration the data below:

experiment: $\pu{1 mL}~\ce{IO3-}$; time: $\pu{11 min}$
experiment: $\pu{0.75 mL}~\ce{IO3-}$; time: $\pu{8 min}$
experiment: $\pu{0.5 mL}~\ce{IO3-}$; time : $\pu{5.7 min}$

Looking at the rate law, the order is supposed to be one.
I know that in order to conclude if the data matches the rate law I have to find out if the concentration is linear combination of time. But I'm confused, because I don't know anything about the concentration and thinking about it I guess I should at least talk about relative concentration? But still I don't know how.

I have no clear idea what you mean by $x~\pu{ml}$ of $\ce{IO3-}$. The equation indicates that the reaction is a system of elementary reaction steps. It is very rare to have 3 body collisions. 4 or more can be safely ignored in almost all chemical contexts. Your equation has 15 reactants and 6 products, hence is not a elementary reaction. It is actually a very well known reaction, especially in introductory chemical kinetics classes, the Dushman reaction. [...]

[...] (cont. alphonse) Without any other information, I'd assume that the volumes (ml) were amounts used to titrate a given (constant) amount of the reaction, thus they are measures of concentration.


text:
Deduce the possible structure given spectroscopic data for the aromatic compound
For $\ce{C11H13NO}$, I have the following $\ce{^1H}$ NMR data:
shift (ppm)

3.00 (6H, s)
6.55 (1H, dd, $J=15.5, \pu{9 Hz}$)
6.70 (2H, d, $J=\pu{8Hz}$)
7.35 (1H, d, $J=\pu{15.5Hz}$)
7.45 (2H, d, $J=\pu{8Hz}$)
9.60 (1H, d, $J=\pu{9Hz}$)

I also know there are two absorptions from IR at $1600$ and $\pu{1660cm^{-1}}$.
From the IR it seems reasonable to suggest an aromatic amide ($\pu{1660cm^{-1}}$). The shift of $\pu{9.60ppm}$ is huge and I think this could really only come from a proton connected to the carbon of a carbonyl. The 6H peak could be two equivalent $\ce{CH3}$ groups, and as it is singlet, I thought they could hang off the nitrogen of the amide, then it wouldn't split. The two sets of 2H atoms I thought could be from the aromatic ring. The problem is piecing all of it together!
The closest structure I have is this:

but I know this is wrong as it lacks a $\pu{9.60ppm}$ shift. How can I arrive at the correct structure?

How many aromatic protons are there?

Well, those with a 15.5 Hz coupling are probably on an alkene trans to one another. And in the 6–8 ppm region there are only 4 other protons. So... . The next step is then to determine the substitution pattern on the ring. Is it ortho, meta, or para - which substitution pattern is consistent with two doublets?

ok so there is 4! and it would be para substituted

Exactly. Now, for the 9.60 ppm proton, you've already correctly said that that's an aldehyde. So, that NMe2 group in your proposed structure should be somewhere else.

ok so 1660cm-1 could correspond to a conjugated carbonyl i.e. an aldehyde off the aromatic ring. I cannot enter an image here but I have a para-substituted benzaldehyde, the substitution being a tertiary ethenamine (-CHCHN(Me)2) where the protons are trans to each other!

That's a pretty good suggestion and as far as the spectroscopic data goes, is plausible. I'd wager, though, that the alkene is between the ring and the carbonyl, not between the ring and the amine. The combination of enamine + aldehyde in your proposed compound probably isn't very stable. You can verify the answer by going onto SDBS and searching with your molecular formula.

am I right in thinking this can be argued because of the coupling constant J=9Hz? One of the trans H atoms has 2 coupling constants and if the alkene part was between the aromatic and amine, 15.5Hz would still be there but it would not be possible to couple at 9Hz with anything?

Excellent point, I missed that. The 9 Hz is clearly a coupling to the aldehydic proton.

Once you figure out it, you might consider posting your solution and your logic as an answer. This may help another person who is struggling with a similar problem.

It always helps to start with the ‘blooming obvious’: the signal at $\pu{9.60ppm}$. This is obviously an aldehyde. It also obviously couples ($^3J=\pu{9Hz}$) with the signal at $\pu{6.55 ppm}$. This proton is either in the range of a dihalogenated carbon, an electron-rich aromatic system or an alkene. We can rule out all possibilities except for the alkene: it can’t be aromatic because there would be no $^3J$ coupling to a non-aromatic proton (there has to be at least one quarternary carbon in-between).
Moving on, we have a very high coupling constant of $J = \pu{15.5 Hz}$. While it would, in theory, be possible that this arises from a geminal $^2J$ coupling of two diastereotopic protons (although those are typically on the order of $\pu{12Hz}$, so slightly less), the more reasonable explanation is an (E)-configured alkene. If it were a $\ce{CH2}$ group, we not only would get into trouble explaining the extremely high chemical shift but also why the aldehyde only couples to one of those protons. Thankfully, an alkene is what we already guessed from the chemical shift. So we now have three protons out of the way and a $\ce{?-CH=CH-CH=O}$ substructure.
Now we might choose to take double bond equivalents into play. The sum formula $\ce{C11H13NO}$ evaluates to 6 double bond equivalents. We have used two for the alkene and the aldehyde. The four remaning double bond equivalents together with the region the chemical shifts are in scream substituted benzene. We may now take a look at the actual integrals and couplings of the protons we deem can still be aromatic (all except for the 6 singlet protons at $\pu{3.00ppm}$). We have two sets of two and each are an apparant doublet. Therefore, we must have a disubstituted with a para-substitution pattern.
The two fragments we have so far, a $\ce{C6H4}$ and a $\ce{C3H3O}$ one, add up to $\ce{C9H7O}$, meaning that $\ce{C2H6N}$ is still missing. We should also note that all these six hydrogens are chemically equivalent. You can play around with this, but the only way to have six chemically equivalent hydrogens and a nitrogen in a group that connects only once is in a dimethylamino group. Thus, the final structure is:

Finally, allow me a few words on why your structure is incorrect apart from the missing proton corresponding to $\pu{9.60ppm}$:

The phenyl protons would show up as a set of three (so far so good). However, they would be a doublet and two triplets (one of the triplets corresponding to the single proton). Your spectrum does not contain any apparant triplets. Also, monosubstituted phenyl groups tend to have much more similar shifts to the point where they often only give a $\ce{5H}$ multiplet.
The six protons on the $\ce{NMe}$ groups must be equivalent. Since they are obviously on different carbons, there must be some rotation that maps one set onto the other. In my structure, you can easily rotate the $\ce{Ph-N}$ bond to transform one group onto the other. In your structure, rotation around the $\ce{C(O)-N}$ bond is restricted due to the partial double bond character.  Therefore, the two methyl groups would show up as two separate signals of three hydrogens each. Look up the spectrum of DMF (dimethylformamide) to confirm this.

Final note: I know nothing about IR and I do not consider it a helpful tool in structure determination. Thus, I cannot comment on the IR frequences and what they can, would or cannot correspond to.


text:
Spontaneous combustion of sulfur in a steel barrel
Got some free sulfur.  Went to the site, and they put a pallet with 3 barrels on it.  One of the barrels was badly corroded, but looked like I could get it home.  I had plastic barrels at home that I could transfer the material to.
About 10 km from home I noticed I was leaving a trail of smoke.  Got in back with a shovel, and found that I had spots of burning sulfur. I smothered as best I could, got home, and spent the next two hours unloading and putting out blobs of burning sulfur.  Did a number on my truck box lining in a few spots.
The lids continued to smoke -- it was a very slow motion fire, with some form of corrosion that would burn across the surface at around an inch an hour.
The smoke smelled strong of sulfur dioxide.
The bulk of the sulfur was unchanged by all this, and I was able to get it into plastic pails and barrels with no problem.
all the barrels had been stored outside, and in the non-corroded barrel had standing water on top of the sulfur.

What happened in storage?  Since sulfur always seems to have an SO2 odour, I assume that either some bacteria makes a living processing S to SO2, or there is a very slow reaction at normal temperatures.
What corrosion product would be flammable?
Why would it start to burn when transported, instead of in the former owner's parking lot?

Sulfur combustion on barrel:

A different lid:

Static electricity?  Maybe a spark happened when you began to move the barrels around?

My guess is tribology. Pieces of the sulfur were banging together as you went down the road which created a spark that started the fire.

Back in the days of sailing ships with wooden hulls, spontaneous ignition was a huge issue.

Why would you need sulfur? I would take a lot of things i get for free, but sulfur is not among them.

Sulfur is used in horticulture to lower soil pH.  There is a bacteria that makes it's living turning S into SO3, and so creates dilute sulfuric acid.  Normally applied at a rate of 1-2 pounds per square meter to drop soil pH by one pH unit.  Used it at heavier applications you can drop soil pH down to about 4.  Now the only thing that will grow is moss.  Easy way to control weeds at edges of things.

Iron sulfide FeS is pyrophoric: flammable, and while it can start just on exposure to air, the presence of UV light will help.  As is usual, one of the key components is the surface to volume ratio.  A massive lump is unlikely to catch fire.  A crusty scale of FeS, such as forms in tanks used to store sour oil (H2S content) is more likely to combust when air is let into the tank.
Deactivation of Pyrophoric Iron Sulfides
From the descriptions in the abstract above, I suspect that I had one of the other iron sulfides.  Whatever I had did not burn fast.

There is an iron sulfide that will burn in air. Presumably this formed in the sealed barrel then moving tore open areas thinned by corrosion and air entered the barrel and ignited the iron sulfide that then ignited the sulfur , maybe.


text:
How is molecular orbital theory used in drug research?
I was reading up on MO theory and came across a particular section:

Application: Computational Chemistry In Drug Design
While the descriptions of bonding described in this chapter involve many
  theoretical concepts, they also have many practical, real-world applications. For example, drug design is an important field that uses our understanding of chemical bonding to develop pharmaceuticals. This interdisciplinary area of study uses biology (understanding diseases and how they operate) to identify specific targets, such as a binding site that is involved in a disease pathway. By modeling the structures of the binding site and potential drugs, computational chemists can predict which structures can fit together and how effectively they will bind.

Whilst I can see from the idea of in-phase and antiphase combinations of wave functions that you could potentially 'match' two atomic orbitals together but I am confused as to how this happens over a large molecule and the binding site of the target?

because you can estimate thermodynamic properties of substances, for example enthalpy, gibbs free energy and so on, which helps to understand the behaviour of substances in some environment..

It would be interesting to hear from an expert what computational drug design has achieved. As I understand it there have been no new classes of antibiotic discovered for two decades and with natural selection making bacteria resistant there is a pressing need, otherwise, before long, all antibiotics will become far less effective if not ineffective.

Common drug design approach does not involve much QC. There are two general approaches - QSAR and docking
QSAR
The idea is that a family of chemicals (training set) with some having desired biological activity, is used to train a pattern recognition algorithm. The trained algorithm is then used on a library of candidates and once recognised as active are then synthesised and tested in a lab.
Docking
The idea is to construct a molecule strongly interacting with a biological important target (a receptor or an enzyme). Since intermolecular interaction are weak, their calculation from first principles is very computationally expensive and is not practical in most cases. Since biologically relevant molecules are usually fairly trivial, even if complex molecules (For example, there is usually no need to consider interactions involving continuous conjugated systems), the intermolecular interactions are easily described using approach of molecular mechanics. So, in this approach a set of molecules is generated and then tested for interaction with a protein target. The tight spot is to find a way two molecules can bind, typically involving search over extremely large conformer space. Billions of potential candidates for ligand-protein complex must be tested, so the computationally cheapest solution (molecular mechanics) is used.
========
As you can see, neither approach normally uses orbital theory or any complex QC, and even if such complex QC is involved, it is clearly subservient. Instead machine learning algorithms and ways to efficiently search conformation space for low-energy conformers are the focus of research.

Well, that's all quite OK, but it shows rather how it's not used ;) Isn't QM-MM sometimes used, for example?

@Mithoron And it often, if not usually, isn't used. It may be a part of generation of descriptors for QSAR and in complicated cases it probably can be used in docking to calcualte interaction energies, but I'm not aware of anything beyond that.

QM is also used for forcefield parameterisation in some cases.

Given a crystal structure of a protein, some people are starting to do QM/MM modelling of the binding pocket. It's necessary to assume that the protein does not change conformation during binding, to keep the computation time reasonable. Unfortunately, this assumption is often false.

Welcome to Chemistry.SE! Take the [tour] to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. For more information in general have a look at the [help]. At the moment this reads more like a comment than an actual answer - could you elaborate a little more. With a bit more rep, you will be able to post comments on any question/answer.


text:
Simultaneous solubility of CaF2 and SrF2
Solubility of two electrolytes having common ion when they are dissolved in solution is called simultaneous solubility.
How do I calculate the simultaneous solubility of the above mentioned salts given that:

$\mathrm {K_{sp}(\ce CaF_2)= 3.9 \times 10^{-11}}$ and $\mathrm {K_{sp}(\ce SrF_2)= 2.9 \times 10^{-9}}$

My attempt:
Let the solubility of $\ce {CaF_2}$ be $x$ and that of $\ce {SrF_2}$ be $y$.
Two equations are obtained:

$3.9\times10^{-11} = x(2(x+y))^2$
$2.9 \times 10^{-9} = y(2(x+y))^2$

Dividing the two we get, $x=0.013y$
Substituting this value of $x$ in equation $1$, I got $y= 5.2\times 10^{-3}$ which should be the solubility of $\ce {SrF_2}$. However, answer given is $\pu{9\times 10^{-4}M}$.
How do I solve this problem then?

The concentration of fluoride ion is not what you think it is. You forgot that little 2 in CaF2.

@IvanNeretin Isn't concentration of fluoride $(x+y)^2$?

Absolutely not. That wouldn't even match the dimension.

Better to solve by letting $\ce{[Ca^{2+}]} = x$ and $\ce{[Sr^{2+}]} = y$

@Abcd Now that's better. Of course this is the square of concentration, but you are right, it is the square that you need.

@MaxW That's precisely what was done.

I'll also point out that using significant figures can make this a lot easier. Think of the ration of Ca to Sr. // You equations are right, but the statement "Let the solubility of CaF2 be x and that of SrF2 be y." is wrong...

I dunno. Guess you're supposed to write an answer yourself. Then again, there is not much value in such an answer, as you were doing almost everything right since the beginning.

Calculate the simultaneous solubility of $\ce{CaF_2}$ and $\ce{SrF_2}$ given that:

$\mathrm {K_{sp}(\ce CaF_2)= 3.9 \times 10^{-11}}$ and $\mathrm {K_{sp}(\ce SrF_2)= 2.9 \times 10^{-9}}$

Let $\ce{[Ca^{2+}]} = x$ and $\ce{[Sr^{2+}]} = y$
So the dissolving $\ce{CaF_2}$ contributes $2x$ of $\ce{F^-}$ and the dissolving $\ce{SrF_2}$ contributes $2y$ of $\ce{F^-}$. So the total $\ce{[F^-]} = 2x + 2y = 2(x+y)$
Substituting into the two $\text{K}_{sp}$ equations leads to "messy" cubic equations which could be solved numerically since there are two equations with two unknowns.
$3.9\times10^{-11} = x(2(x+y))^2 = 4x(x+y)^2$
$2.9 \times 10^{-9} = y(2(x+y))^2 = 4y(x+y)^2$
But going back to the original $\text{K}_{sp}$ equations we have:
$\ce{[Ca^{2+}][F^-]^2 = 3.9 \times 10^{-11}}$
$\ce{[Sr^{2+}][F^-]^2 = 2.9 \times 10^{-9}}$
and by dividing the two we get $\ce{[Ca^{2+}] = 0.0134 [Sr^{2+}]}$ or $x = 0.0134y$. Substituting this into the second equation we get:
$2.9 \times 10^{-9} = 4y(x+y)^2 = 4y(1.0134y)^2 = 4.108y^3$ or $y=8.9\times10^{-4}$
So:
$\ce{[Sr^{2+}]} = 8.9\times10^{-4}$
$\ce{[Ca^{2+}]} = (0.0134)(8.9\times10^{-4}) = 1.2\times10^{-5} $
$\ce{[F^-]} = 2(8.9\times10^{-4} + 1.2\times10^{-5}) = 1.8\times10^{-3}$
Note that this can also be simplified using significant figures a different way. Since the two $\mathrm {K_{sp}}$ values only have two significant figures and  $\ce{[Ca^{2+}] = 0.0134 [Sr^{2+}]}$, we can assume that for all practical purposes that all the $\ce{F^-}$ comes from the dissolution of the $\ce{SrF_2}$. So:
$\ce{4[Sr^{2+}]^3} = 2.9\times10^{-9}$
Thus
$\ce{[Sr^{2+}]} = 9.0 \times10^{-4}$
$\ce{[F^-]} = \sqrt{\frac{2.9\times10^{-9}}{9.0 \times10^{-4}}} = 1.8\times10^{-3}$
$\ce{[Ca^{2+}]} = \dfrac{3.9\times10^{-11}}{(1.8\times10^{-3})^2}=1.2\times10^{-5}$

Note that the statement "Let the solubility of $\ce{CaF2}$ be $x$ and that of $\ce{SrF2}$ be $y$" by the OP  is wrong because it is not $\ce{CaF2}$ that is $x$ and $\ce{SrF2}$ that is $y$ but rather $\ce{[Ca^{2+}]}$ that is $x$ and $\ce{[Sr^{2+}]}$ that is $y$. This leads to the correct value of $\ce{[F^-]}$ which the OP initially had wrong.

Yes, I understood why my statement, "let the solubility..." was wrong. Thanks for pointing that out!


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Is sulfuric acid still hazardous after it dries?
One of our housemates decided to pour battery electrolyte (which is 30-40% sulfuric acid) into a motorcycle battery on our back porch next to our laundry washer and dryer.  He did so extremely sloppily, and we don't know exactly what he might have spilled the acid on.
Is the acid safe to touch after it dries, or is it still a safety hazard?
If it isn't safe when dry, what can we do to neutralize the acid?  I know baking soda can neutralize acid, but I'm not really sure how to easily coat an entire porch in baking soda.
My boyfriend suffers from severe, debilitating OCD with a chemical contamination obsession, so this is an absolute nightmare for him.  Thank you in advance for any help you may have!

Roughly speaking, sulfuric acid never dries (much like oil). It just penetrates deeper down, or possibly gets neutralized with something. Baking soda will do.

Pour water over suspicious places, mop it up and wash out the cloth you used. Use gloves if you feel safer, but it's not really necessary. Or rather have your sloppy housemate do it.

Please do use gloves.

Sulphuric acid won’t dry — meaning evaporate; what water does when one typically thinks of drying — at any noticeable rate. It will, however, given time penetrate deeper into the material if the material is porous or susceptible to degradation.
Depending on how much sulphuric acid was actually spilt, I suggest either mopping up the area with lots of water (if not much was spilt) or first semi-generously applying dry baking soda to the area. Once the baking soda has stopped generating bubbles immediately on application, you can use normal amounts of water for wiping up.
While I would love to suggest you get the sloppy housemate to do it, I think you might be disinclined because they again will do it sloppily.

Definitely hazardous, and also for reasons totally missed by the existing answer and all the comments. Sulfuric acid used in batteries is always contaminated with small but non-negligible amount of dissolved lead, as spongy lead metal and lead dioxide are used as electrodes in those batteries. Sulfuric acid is corrosive but non-toxic; neutralizing it with baking soda removes the property of being corrosive and thus makes it safe. Lead and its compounds, however, is toxic; there is no way to simply neutralize it in a simple way analogous to sprinkling baking soda. Be careful and do such things in remote place away from anyone's belongings to avoid contamination next time.

Based on the description, he was loading electrolyte into a battery not taking it out, so the spilled bit would never have been in the battery and exposed to lead.


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what will be the rate of effusion of oxygen gas at a pressure of 1760 torr and temperature of 227 °C?
The rate of effusion of hydrogen gas at 880 torr and 227°C is 8 torr/min .
what will be the rate of effusion of oxygen gas at a pressure of 1760 torr and temperature of 227 °C?

You cannot find the rate of effusion of a gas with only that little information.
With that information you can only tell the ratio of the rates of effusion at two different conditions.
You should have included what you thought about the question and your thought process. This is not a homework solving website but a site to ask conceptual queries.


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Why do experiments not verify the law of mass action?
I read the answer to Rate Law vs Law of Mass Action and it's written in the answer:

In the rate law, the rate equation consists of actual factors that changes with change in experimental conditions. These may or may not
    be equal to stoichiometric factors. These are generally responsible
    for determining the order of reaction.

Both the laws are correct because both have a wide area of
  application. Rate law is although quite experimental in nature and is
  generally considered to be more appropriate.

Why do reactions not obey the law of mass action? And if that's the case, then how does it even have any applications? Shouldn't all problems be solved using the rate law only?

There is no special "rate law". A rate law is just an empiric mathematical description of a process.

The rate of mass action is true: the rate of reactions is proportional to the activity of its reactants.
On the other hand, from a macroscopic point of view, it is also true that reaction rates may not correspond to apparent stoichiometric proportions.
To understand this disparity, we need to refer to the concept of mechanisms.
The rate of mass action applies to elementary reactions, that is, to reactions that happen the way we write them at the microscopic level.
If the molecular, physical reaction consists on two molecules colliding and reacting, and we write it as
$$\ce{A + B -> C}$$
The rate of that reaction will, per the law of mass action, be
$$r = k · a_A · a_B$$
However, most macroscopic chemical processes aren't that simple, and in fact are the product of multiple steps; for instance, in the previous example, perhaps A undergoes a transformation into an intermediate which then goes on to collide with B to form C:
$$\ce{A -> D}$$
$$\ce{D + B -> C}$$
Macroscopically, we still have $\ce{A + B -> C}$, but the kinetics of the reaction will be very different.
For instance, imagine that the second step ($\ce{D + B -> C}$) is much faster than the first, so that any amount of D created reacts with B almost immediately to produce C. Then the first reaction will act as a bottleneck, and the overall reaction rate will be roughly the rate of the first step:
$$r = k_1 · a_A$$
If we do not know the underlying mechanism, this will be an apparent contradiction between the law of mass action (from which we could naively conclude that $r = k · a_A · a_B$) and the actual rate law ($r = k_1 · a_A$).
Much of the work on chemical kinetics consists in proposing mechanisms, testing these proposals (for instance, looking for experimental evidence that D exists as an intermediate) and deriving rate laws that are consistent with the law of mass action and experimental observation (which applies to elementary steps).


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What makes a bond non-polarisable?
Many chemical bonds are known to be polarisable. Similarly, many ions are also known to be polarisable. Examples include the carbon-iodine bond and the carbonate ion. Recently, I have read that the carbon-fluorine bond is, in fact, non-polarisable. Reasons for this include the high electronegativity of fluorine and the small sizes of the carbon and fluorine atoms.

Do other non-polarisable bonds exist?
What factors would contribute to a chemical bond being non-polarisable?

Polarizability is a property of compounds. Bonds contribute to it relatively little.


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Solubility experiment of different organic compounds
Chemistry lab note says that:1) Aromatic acids like benzoic acid and salicylic acid are NOT soluble in water. But they are soluble in 5% NaOH,  and 5% NaHCO3.  Due to having a large hydrcarbon part (which is hydrophobic), it is understandable that aromatic acids won't dissolve in water. But why would they dissolve in base solvents.2) aromatic carbonyl compounds like benzaldehyde and acetophenone are NOT soluble in H2O, 5%NaOH, 5% NaHCO3, 5% HCl. But they are soluble in highly concentrated H2SO4. Got no idea about the reason behind it. Can somebody clarify?

Carboxylic acids easily react with bases forming salts, which are ionic compounds and are usually readily dissolve in polar solvents such as water, which obviously makes 95% of your base solution.
Carbonyl compounds are easily dissolved in strong sulfuric acid as 1) a reaction may take place, introducing -SO3H group to position 3 of aromatic ring of respective carbonyl compound; and 2) a strong acid can form hydrogen bonds with carbonyl oxygen. In both cases affinity of carbonyl compounds towards sulfuric acid is increased.


text:
Why are electrons added in the ionic counting method?
Why do we add one electron to the ligands that have an odd electron number?
For example the cyclopentadienyl ligand has five electrons that can be donated to the metal but when we count the electrons by the ionic method we add one, so it become total of 6 electrons donated to metal.

In the ionic method of counting, you are not counting neutral compounds. This is reasonable on a molecular basis: the ligand is, in fact, not cyclopentadiene or the cyclopentadienyl radical (with four or five electrons to be donated) but the cyclopentadienyl anion $\ce{C5H5-}$. This species contains six electrons in its π system, meaning that six are donated.

As you've mentioned cyclopentadienyl anion has six $\pi$ electrons, so during complex formation does it donate only one electron or two electrons?

@NatashaJ please reread the last line of my answer.

so basically it has six $\pi$ electrons all of which are donated. I was confused as my professor said that cyclopentadienyl ligand donates only one electron(which is possible only in radical form).


text:
Why is the maximum work done by a voltaic cell negative
The maximum work done by a voltaic cell is given by -nFE. Why the negative sign?  From what I know, a battery does positive work on the load attached to it

It is negative because it's the work done by the system on the surroundings. Here our system is taken as the cell.
Maximum work is also taken as the change in Gibbs free energy $(ΔG)$.

But the system (the cell) is doing positive work on the surrounding (the load)

Work done by the system should be positive, work done by the surroundings on the system should be negative. Does -nFE represent the work done on the battery?

@xasthor Are you sure work done by system is taken positive ;) Are you aware of different conventions ? https://chemistry.stackexchange.com/questions/66088/why-do-chemistry-and-physics-have-different-sign-convention-in-thermodynamics


text:
Moles and chemical equations
Sulfur reacts with oxygen according to the following equation:
2S + 3O2 -----> 2SO3
If 4 moles of sulphur reacts with 9.5 moles of oxygen, how many moles of oxygen would remain after the reaction?

This a very logical question based on limiting reagent concept.
As you can see $2$ moles of sulfur require $3$ moles of $O_2$.
By unitary method $4$ moles of sulfur will require $6$ moles of $O_2$.
Hence $3.5$ moles of $O_2 $ will be left out.
By the way you shouldn't post such "homework" questions without showing any effort.

Posting homework is fine, but we expect to see some kind of effort before we provide help.


text:
To what extent are the radius ratio rules valid for predicting the crystal structure of an ionic compound?
The ionic radii of $\ce{Ba^2+}$ and $\ce{O^2-}$ in barium oxide are $\pu{135pm}$ and $\pu{140pm}$, respectively. The ratio of the radius of $\ce{Ba^2+}$ to $\ce{O^2-}$ is approximately $0.964$.
According to the radius ratio rules the solid barium oxide should crystallise in a body-centered cubic type structure (like $\ce{CsCl}$).

But evidently the prototype structure is $\ce{NaCl}$ type.
What is the reason for this anomaly from general trend?
EDIT: I found that every alkaline earth metal oxides (Of the type $\ce{M_{alk.e}O}$) possesses rock salt (halite) structure but only $\ce{MgO}$ obeys the radius ratio rule (it has a radius ratio of about $0.64$ which is well between $0.414$ and $0.732$).
Certainly as we are going down the group the cationic radius is going to increase and so is the radius ratio.

The mentioned trend is but a crude generalization.

Please elaborate....

What's to elaborate? Tall guys play basketball, short guys play football; that's an empirical correlation. It is mostly true, but exceptions both ways are plenty. On the other hand, $(a+b)^2=a^2+2ab+b^2$; that's a law. Can it be broken? Not a chance. Well, your trend is not a law.

Sir, Any reasons for this "exception"?

The reason is basically that the crystal structure was measured and found to be the NaCl structure. Now good luck finding a reason.

But I thought exceptions do have reasons.

@Serotonin Everything has a reason; it's just that you'll have to wait till someone figures out what the reason is... ;)


text:
Pure substances have only one kind of atom
I read in a book recently that no substance is ever totally pure, even if we take great care to separate it from other substances.
And also that pure substances contain only one kind of matter.
But I actually have some doubts related to both of the notes.
I would rather number...

Pure substances contain only one kind of matter? True for elements but what about compounds? Are compounds not pure substances? They are homogenous at composition, right?
No substance is totally pure? Reactive elements are found in combined state, but what about less reactive or maybe radioactive?
As well, can't we obtain any substance with 100% purity?

I did surf the questions but did not get any reasonable explanation. Also, please do correct me where I am wrong.

Colleagues, please don't close this question, I think it can be salvaged.

I don't know why should the question be closed? Anyways, thanks for the appeal

I think the main issue of the question is more semantic than chemical. Pure has different meanings depending on the context in which it is being used.

@IvanNeretin It could be but all this was asked already.

@Tyberius "Pure" is more or less well-defined. But "matter", it seems to me, is not.

I would rather edit the matter, trying to make the context clear.

Related:https://chemistry.stackexchange.com/questions/61001/what-does-it-mean-for-meth-to-be-pure?rq=1

https://chemistry.stackexchange.com/questions/65180/obtaining-100-pure-metal https://chemistry.stackexchange.com/questions/2870/what-is-a-pure-substance

@Mithoron That's what I was searching for. A satisfactory one. Thanks.

Related :https://chemistry.stackexchange.com/questions/2870/what-is-a-pure-substance?noredirect=1&lq=1

Most of the issues can be resolved with precise definitions.
Let's start with purity. Purity is relative. In many cases (eg table salt) 99% is pure enough. What this means is that for all practical purposes we don't care about the remaining impurities. Chemists tend to describe something as "pure" when there is less of any contaminant than matters for what they are doing with the chemical. Sometimes purity of very high levels matters. Semiconductors behaviour depends on very precise doping with very small amounts of other substances. The wrong contaminant even at levels of 1 atom per million is a problem. So the industry have worked out good ways of making extremely pure silicon, for example. These techniques have been used to make some of the purest substances ever (the silicon sphere now being used to redefine the Kg is pure to parts per billion atoms.) Strictly speaking it isn't pure in the sense of having absolutely 100% of all its atoms as silicon, but that is as pure as it needs to be for its purpose. So, strictly, nothing is ever absolutely pure, but it doesn't matter.
The word "substance" is a bit vague. In chemistry it could be a compound an not just an element. Purity for a compound means that all the molecules in the substance are the same (to the level of purity that matters). Having more than one type of atom doesn't make it impure as long as all the molecules are the same.
Then there are some homogeneous mixtures that are hard to separate by any simple means, some metal alloys for example. We wouldn't normally describe these as "impure" but chemists might use special terminology to describe them (alloy rather than compound for example).
And as for "Pure substances contain only one kind of matter" it depends what you mean by "matter". Chemists would prefer the term "molecule" to "matter". Matter can be interpreted different ways (does it mean atoms or molecules: it's unclear). "Pure substances contain only one kind of "molecule" would be an unambiguous statement chemists would agree with.

Nice answer. Well explained. As per the statement in the book, "matter" meant atom, but yes 'molecule' definitely seems unambiguous. That's why I needed to clear the confusions. And also the example of silicon sphere(new reference standard for kg) was so explaining. So, thanks for the answer.

The Wikipedia article might be a good reference to start.

A pure substance cannot be separated by physical means. This means in principle, so if in practice, you cannot separate physically, but you theoretically could, that's still not a pure substance. A compound would then be pure because you cannot separate out its components by physical means.
That's true, but "pure" here does not refer to purity as you using the word. It refers to purity of essence in that the substance is not constituted from multiple parts by simple mixing.
We can't, but again, that's not what pure here is supposed to imply. Oxygen is a pure substance whether it's mixed or not because it is an element. A mixture of oxygen and something else is not pure because it is a physical mixture.

The first point made some of my points quite clear, driving out a new definition actually in a positive sense. Homogenous solutions, can theoretically, if not actually be separated by physical means. So, including that was reasonable. Thanks.

To really explain the OP's problem you need a number of very precisely worded definitions. You used substance, pure substance, components, multiple parts, and physical mixture for instance.

@MaxW OP seemed pretty clear on most terms, except for "pure."


text:
Mechanism of Halogenation of Alkene

In the above reaction, we use a non-polar compound (such as CCl$_4$) as a solvent. Then, why does the Br$_2$ polarise on attack by the alkene, even though a non-polar solvent does not encourage polarisation of a compound. Is the mutual polarisation enough for the reaction to occur.
Further, in different books the mechanism is either explained by the above mechanism or by the Molecular Orbital Theory (the alkene's filled $\pi$ orbital (HOMO) will interact with bromine's empty $\sigma$* to give the product).
Which one of these is the more popular and/or the officially accepted explanation.

Solvent doesn't have much to do here and all else is more or less complementary.

@Mithoron Then why does polarisation of bromine occur to such an extent

sigh To what "extent" (BTW I don't like untangling reasonings which it looks like)? It's not exactly easy to tell how electron density changes during reaction. It's not like alkene makes something  a Lewis acid - it's Lewis acid because it reacts like Lewis acid.

Just answered your question. At this time what I do is collecting molecular orbital textbooks for organic chemistry. Will you share what textbooks exactly did you use?

Alkene double bond is an area of high electron density and hence high (partial) negative charge. Bromine molecule, approaching double bond, shifts its electrons of covalent bond towards farther Br atom, it becomes polarized and an induced dipole.
That's the simplest explanation in terms of electron movement I would use.

To answer your first question: The out-of-plane areas of an alkene — those areas, where one typically draws the π bond — are areas of high electron density. If a polarisable compound approaches these, the approaching side will be positively polarised.
To answer your second question: You are constructing a dichotomy which does not exist. Both explanations are the same except that one explains it with molecular orbitals and the other with electron-pair arrow-pushing.


text:
What's the difference between free energy and potential energy?
I found different answers for this question. Was wondering to find a way to explain it to my students how to illustrate the two concepts in practical terms.

I'd rather wonder what they have in common.

@Mithoron Both are measured in Joules. Other than that, hardly anything.


text:
Can a single molecule break chemical equilibrium?
If a system is in chemical equilibrium is it possible to put the system out of its current state by introducing novel substances not present in it whose behavior isn't analogous to the one of the substances already in the system and whose interactions can start new reactions unseen by the system till the moment the new substance is added? Can even a miniscule concentration of that substance (so miniscule it could be as small as even 1 molecule) start dramatic changes capable of affecting the entire system?
To be clearer I would like to show an example. Imagine there is a system of substances all of the reactions among which have already reached equilibrium rates and the concentrations of their reagents and products are stable. Then, we add extra reagent not present in the system until that moment capable of interaction with its components to form novel molecule which on its turn is able to further interact with the components of the system to form other novel molecule and it can further interact with components of the system and so on until a molecule so large is formed that it splinters into 2 molecules. However, each of these molecules is further able to interact with components of the system to produce a "new generation" of large molecules which when a certain limit is reached splinter into 2 whose properties, however, are identical to the 2 molecules already described, so in the end novel substances emerge able to dramatically shift the equilibrium in which the system was before their addition. Then, is this example possible proof of the ability of even a single molecule to start a chain of reactions able to shift chemical equilibriums and is it a proof there can be systems appearing to be in the state of equilibrium but actually able to undergo great changes if the appropriate molecules are added to them?
Basically, the purpose of my question is to ask (if it's even possible) to have a system which has reached a state of chemical equilibrium stable in time but which can be radically altered by a reagent not characteristic for it by even a minute concentrations of that reagent. A perfect example of my point is a sterile growth medium. By all means it's in state of chemical equilibrium. All possible reactions able to occur in it are already occurring with equal speeds of the forward and reverse reactions with concentrations of their reagents and products already in equilibrium. As such there is nothing new that can occur in the system if we rely only on the natural processes able to develop in it. However, if we add even 1 bacterium able to grow in the medium many new reactions will start to occur as a result of its metabolism and it can dramatically change the entire system bringing it out of the state it was able to occupy for extremely long periods of time without it. To my mind the growth medium is in a state of chemical equilibrium and there is no way to suddenly develop a bacterium on its own, therefore, our perception of chemical equilibrium entails us with a certain degree of predictability we prescribe to systems actually able to undergo drastic changes if the "right kind" of reagents are added to them even in extremely minute concentrations, even for extremely large systems. Is my interpretation of chemical equilibrium correct? Am I misunderstanding something here?
Furthermore, if it's correct does it means there are possible systems whose path of evolution can be entirely unpredictable but they could appear stable in time? What if the difference between their long-term stability and the myriads of unpredictable states they can occupy is a single molecule of complexity unable to be produced by the system itself? Does it mean, then, that chemical equilibrium alone isn't a measure of the possible states of the system but only an approximation to what is currently possible under certain physical conditions for a mixture of reagents? Then, does it means there is a "limit of opportunity" for any system in equilibrium defining what is possible for the system currently, but not all that is possible for it under the physical conditions it's in? Finally, does it means certain substances can change this "limit of opportunity" by generating novel compounds and starting novel reactions taking the system out of its current state without the need for addition of new reagents for the system? Is that possible for any known chemical (I'm referring to chemistry alone here and not mentioning any equilibrium changes induced by biological agents) systems and what are these examples?
I'm looking for both a system and a reagent capable of displaying such behavior to explain a quest of mine. I'm thinking about the possibility of the existence of "cascade reactions" able to dramatically change systems already in stable, e.g. equilibrium, state but the compound causing the initial reaction in the cascade shouldn't be native to the system and it shouldn't be able to produce it on its own. I'm thinking about the existence of limits to the reactions possible under certain conditions induced not by the conditions themselves but by the levels of complexity the products of the reactions currently occurring in the system can reach. I mean if the reagents are relatively simple substances combining to produce relatively simple products then the system can reach a level of complexity where all reactions are in equilibrium under the given conditions but not all possible reactions that these reagents could participate in have occurred. Then, if we add even a single molecule of novel substance with a higher degree of complexity than the ones in the system, could it trigger this "cascade" of reactions leading it to a higher degree of complexity than it used to have? What if in this new state of more complex chemistry we add 1 more molecule with even a higher degree of complexity? If we repeat the process over and over again how far can we get? Could we react a state where the entire system itself can begin selfreplicating its complexity and even evolving further complexity on its own without further outside intervention? If this is possible then the purpose of my question is to understand in what systems at what state what kind of molecule can trigger such transition? If it isn't possible I want to know why and at what level of complexity such a transition stops? Basically, I my motivation is to understand where is the border when biology and chemistry meet and is concept of a break in chemical equilibrium triggered by even a miniscule amount of the "right" kind of complexity crucial to finding it?
P.S. I hope now my question is clear enough to attract good answers and be unblocked.

Discussion moved to chat.  Please try to keep back and forth to a minimum, even when it's a productive exchange like this one was.

Nobody cares about a single molecule, so let's say "a very small quantity" instead. Well, then the answer will be along the following lines:

Small influences produce small changes. In particular, that small addition will bring your system to a new equilibrium (another state, if you'd like), but it would be infinitesimally close to your initial one.
Alternatively, if one tiny droplet will make the system run out of control and finally end up somewhere pretty far from the initial state, this means it was never in equilibrium in the first place.
In this last case, in particular, the added droplet might not have been necessary. Sooner or later, a huge enough fluctuation would do the same.

As an example of a system in the "seemingly, but not really stable" state, think of those effervescent tablets. They just sit on the shelf and do nothing, but add a very tiny amount of water and see what happens.

To sum it up, out notion of stability is pretty solid. There very much is the state of stability for any system; it is the only one, and it can't be dramatically changed by addition of one molecule of anything. In some systems this state may be hard to reach, but that's another story.
So it goes.

What about a system where all reactions are in equilibriums between the forward and reverse reaction and then if we add just one molecule of some substance this can start a chain reaction of generation of novel substances which spin the system into a totally different state. Then was the initial system not in equilibrium? But then it had all of its reactions at equilibrium concentrations of reagents and products. If that was the case, then, do I missunderstand something about the concept of chemical equilibrium at the first place?

The other way to look at this is significant figures. The mass of a proton is known to about 13 parts per billion. I doubt that any equilibrium constant is known to that precision.

@Yordan If your system had all reactions, then it would have the said chain reaction too, and the chain reaction would develop and carry the system away from that state. If it didn't have the chain reaction (despite its being theoretically possible), then the system was not in equilibrium.

@IvanNeretin, but how can it had this reaction if it didn't had the compound which started it? Do you infer all compounds that can be synthesized under certain conditions are synthesized under the said conditions irrespective of the complexity of interactions required to synthesize them? What if the synthesis of the compound that could change completely the system has such a low probability that it would take a system many times larger than the visible universe to wait for trillions of trillions of years to produce even a 1% of chance to create the change inducing molecule?

For all we can see the system is in equilibrium for all that time, but do you say that our perception is deceiving and system is actually not in equilibrium although all we can measure about it says it is? Are you saying then, that systems which look stable throughout time from our current perception of chemistry aren''t but are just like clockwork devises waiting for their chance to burst into a totally unpredictable states which we couldn't even imagine and we actually never see they turn into our real world? Is that the correct interpretation of the concept of chemical equilibrium,then?

I don't like the "totally unpredictable states" part, for it negates the existence of chemical knowledge as such. You know, more than a few people have put more than a little effort into those predictions. Other than that, you are mostly right. A system may sit out of equilibrium for ages and never tell you a thing. Diamonds, for example, are like that. Then again: fluctuations. If a reaction can be triggered by a single molecule, then it stands a decent chance of being triggered by a random fluctuation, without any extra molecule.

Can you than tell me what kind of fluctuation can turn an ocean of  glucose into a single fully formed Escherichia coli bacterium? Here is a perfect example of a system which is stable for all I can tell about chemical equilibrium but yet a single bacterium can dramatically change this system. I also am quite certain the state of the ocean of glucose forming an E. coli bacterium is totally unpredictable from what we know about the properties of glucose solutions but if you have any other suggestions I am willing to hear them. Please, correct me when I'm wrong

because if I'm not wrong in anything it seems like to me that virtually any chemical having some Gibs free energy can turn into a myriad of things? Does that mean all these chemicals just look stable but they aren't because to me this is something not very well fitting our observations?Am I correct or am I missing something out here?

A single bacterium is much, much greater than a single molecule. Also, an ocean of glucose solution is not stable even without biological influence. Left in contact with air, it will show visible signs of oxidation in a matter of days, maybe weeks. Without air, the decay will take longer, but still nowhere near as long as human life. Have you ever seen old specimens of organic reagents in museums? All except alkanes have turned to black tar. Crystalline glucose might outlast us, but then again, it is no good for E.coli either.

I have seen black tar digged up from specimens of old organic materials and I know it can reach a pretty stable for but then again there are some bacteria which can feed on it, so I guess, the system even in that equilibrium state has enough energy to support life. Plus, what is that black tar anyways? Can you specify its composition given only knowledge of its initial state or is the system capable of turning into a myriad of states? Is the black tar even in equilibrium or is it in some state which can't be exactly predicted so we only have a general notion what is going on it but we ca

can't predict it? Even if we are taking about organic materials if they are sterile (I agree that bacteria can dramatically change everything) we can't know exactly what our sample will turn into-so what is the state of predictability of our research then? Even swamp mud left in anaerobic conditions for millions of years (I guess it pretty stable then) still is a mesh of different reagents although there are no reactions occurring in it and we use it extensively as coal for heat generation.  Only graphite is stable but can we get there without the millions of year of Earth's heat?

Black tar is not at equilibrium, I think. As for predictability, it is somewhat limited. Like you said, there might be a myriad of possibilities.


text:
Is Effective Nuclear Charge the same as Core Electrons?
I'm wondering, what is the difference between the two?
Using the equation for Eff: $\ce{Zeff = Z - S}$
Z being the number of protons in the nucleus and S being inner-shell electrons.
For example:
When calculating Lithium, I first do the electron configuration,
$\ce{1s^2}{2s^1}$
The valence electron configuration is then 1, from looking in the periodic table - it is in the first column. Then to find the core electron configuration, I subtract the valence electron number, 1, from the number of total electrons, 3. I then wind up with 2 as my core electron configuration. Am I just doing the same equation twice if I were to calculate Zeff?
$\ce{Zeff = 3 - 2 = 1}$
They're the same thing, are they not?

S is not equal to the number of inner-shell electrons. You should use the Slater's rules. https://en.wikipedia.org/wiki/Slater%27s_rules


text:
Half-Filled Shells and Stability explanation
I am reading a book about Advanced Chemistry, and it is discussing the subject of half-filled orbitals.
The book notes that Chromium has an electron structure of $1s^2 2s^2 2p^63s^23p^63d^54s^1$

There is also some added stability when a sub-shell is half-full as this minimises the mutual repulsion between pairs of electrons in full orbitals.

Now, I'm confused by this explanation.
Is it correct to conclude that if a sub-shell is half filled, the repulsion between electrons in the half-filled subshell and other subshells with fully-filled orbitals is reduced (e.g. between the $3d$ and $3p$ subshell in Chromium)?

This refers to repulsions between electrons in the half-filled subshell. In this case, the 3d electrons repel each other less.

@orthocresol so what's the reference to "repulsion between pairs of electrons in full orbitals"?

I don't actually know, so I'm confused too by the exact wording. The topic has been explained elsewhere: https://chemistry.stackexchange.com/questions/58625/what-is-the-physical-basis-for-hunds-first-rule You may find it relevant.

I actually just had this in class, and it confused me as well. This site helped me to understand it a bit better:
http://www.chemguide.co.uk/atoms/properties/3d4sproblem.html
The author, a retired chemistry professor, states:

"Many chemistry textbooks and teachers try to explain this by saying that the half-filled orbitals minimise repulsions, but that is a flawed, incomplete argument. You aren't taking into account the size of the energy gap between the lower energy 3d orbitals and the higher energy 4s orbital."


text:
chemical possibility in oil - 8th graders have questions for a project
I have a team of kids (8th graders) that are doing the First Lego League competition this year. As part of that they have to come up with a possible solution to a problem that relates to the human water cycle.  They need to contact experts and share their thoughts and get feedback of their solution.  We would greatly appreciate any feedback.  I've listed what the problem is they are trying to solve and their suggested solution below, and then some questions.
Problem:  Oil spills in water (caused by various things - accidents, hurricanes, etc)
Their proposed solution:  Create an environmentally safe chemical that can be put in the oil, that if there is a leak, as soon as the oil comes in contact with water, the chemical causes the oil to solidify thus sealing the leak or at least slowing it down.  The solidified oil would be buoyant so that it could be easily skimmed out of the water.  The solidified oil could also be reprocessed back into usable oil.
Questions the kids have:
    Does such a chemical currently exist that can do this?
    If so, why is it not being used - any downside to using it?
    If not, is it possible to invent such a chemical?  Now or in the future?
    What kind of components could be used to do create such a chemical?
We really appreciate any response you can give.
Thank you so much!

While idea seems clever, I'm afraid it can't be done in any simple way and therefore nothing like this exists and probably won't be any-time soon.

Thank you for your response!  They understand that it most likely can't be done easily and any-time soon.  what would have to exist to allow for something like this?  are there any chemicals that might be considered as part of this solution?  besides money and time, what would equipment might be used?  this is for a project so any detail given is appreciated.

Hmm, if there was significant fraction of long chain polyunsaturated methylesters and some "miraculous" O2 activated polymerisation catalyst, then maybe fuel could turn into "jelly" on contact with air... There's inherent vice in this idea, though. Fuel has to contact with air at least when used, so after transport this would have to be undone. Also no contact with air would make transport problematic.

A chemical that can be added to crude oil so that the oil solidifies upon contact with water to either seal leaks or to make the oil recoverable sounds like a very good idea. Unfortunately, such a chemical is exceedingly unlikely to exist.
When processing crude oil, you typically want the oil itself and not any additives. That means, the chemical in question would have to be added in concentrations as low as possible. Therefore, it’s reaction with water must be very strong. But this in turn means that the traces of water present everywhere (yes, even inside your sample of crude oil) would almost be enough to cause the reaction. Therefore, outside of water-free atmospheres (used in laboratories under inert gas to perform water and air-sensitive reactions) this chemical would hardly be usable.
A second problem is getting the chemical reasonably dissolved. Crude oil is a very viscous mixture unless heated. To use the chemical effectively, it would need to be added almost immediately and distributed well in the sample of crude oil — which would mean immediately heating the crude oil to such temperatures that efficient mixing is possible (likely under pressure to not lose the more valuable but also more volatile petrol components) right after getting it out of the ground. Especially on oil rigs, I see major difficulties with this process; and it is basically on oil rigs that you would want this chemical the most.
Finally, there is the ‘how would it work’ problem. I have extreme difficulty imagining a reaction with water that would lead to solidification. The only idea I can come up with is a compound that would be liquid at standard conditions but adds water to become a solid which crystallises out of the crude oil phase. However, this solid would likely not do you the favour of crystallising where you need it and you would need so much of it that my first argument becomes invalidated. Therefore, I conclude that such a compound cannot exist.
If such a compound exists, I’m sure we would have heard of it or it would be in use (or have been in use if there are significant and severe environmental issues with it that are worse than the issues with crude oil itself.

The skimming equipment was designed and used many years ago; except it did not involve any chemical additions. Oil floats , A sponge-like continuous belt runs through the water preferentially absorbing oil phase . It then goes through rollers that squeeze out mostly oil by adjusting pressure. Then the belt passes second rollers that squeeze out water.  And the belt passes into the oil slick again. I don't know if it has been a commercial success. I will bet a lot of money the EPA would outlaw the chemical you want as they have for those chemicals used to break-up oil .

While in general a good answer, I unfortunately have to downvote for the final sentence.


text:
Site of first protonation of imidazole and aminoimidazole
I wish to ask where imidazole and aminoimidazole will be protonated on first protonation.
Imidazole:

Aminoimidazole:

I suppose that the nitrogen (without hydrogen) of the ring is the most probable to be protonated, but I am not sure about it. Do you know where it will be protonated first?
Any link to a study where the issue is solved will be good (I need the information for my thesis).

You can predict pKa values with the online webserver Chemicalize https://chemicalize.com/

So you suggest to protonate the different nitrogens and see the result with that tool?

No, just draw the molecule as you did above and Chemicalize will tell you the most probable  de/protonated isomer as a function of pH

@JanJensen I don't know why it doesn't produces a value for -NH2 in 2-aminoimidazole, SMILES is c1cnc([nH]1)N .  Maybe it just can't be protonated because of the resonance.

Yes, Chemicalize doesn't consider all protonation states.  Only the most probable. The $\ce{NH2}$ group won't be protonated for the reason you stated

You already asked about it https://chemistry.stackexchange.com/questions/80504/protonation-of-compounds-with-two-or-three-neighboring-nitrogens

Possible duplicate of Protonation of Guanidine

For imidazole, the lone pair of pyrrole-type nitrogen (the one with hydrogen) is involved in aromaticity of the whole compound, as it contributes two π-electrons in addition to four of those from carbon-carbon or carbon-nitrogen bonds, which makes it 6, which is a classical sign of Huckel aromaticity. In fact, this NH-nitrogen tends more to act as an acidic center than as a basic one, as it loses its hydrogen when treated by strong bases, like $\ce{NaNH2}$.
The pyridine-type nitrogen has its lone pair out of ring conjugation, thus it can act as a acceptor of hydrogen ion, placing its two electrons into a vacant orbital of a $\ce{H+}$.
For reference, you can see any heterocyclic chemistry textbook. I prefer the one authored by Joule, Mills. Or this.

When it comes to 2-aminoimidazole, this publication  (which is unarguably very old) says all three nitrogens are involved in protonation, because the structure is a derivative of guanidine, a strong organic base. Which seems only legit to me if it said that the contribution of pyrrole-type nitrogen is negligible, as aromaticity is ruined in that case. I could also predict that exo-aminogroup gets protonated more likely, as its basicity is somewhat higher: its lone pair is on a sp3-orbital, rather than the lone pair of endo-nitrogen (the one without H), where it is on a sp2-orbital, thus, the contribution of s-orbital is higher and it is closer to the nucleus of nitrogen atom, decreasing its availability to act as a covalent bond acceptor.

(Yes, I know that hybridization theory is false and is not really loved here, but why not use it, when it is a powerful teaching instrument and helps to explain stuff)

Oh no, in the case of organic compounds, hybridisation theory is actually rather well received because it does a good job on explaining this without having to solve the Hamiltonian.

However, the exo-amino group’s lone pair will very likely not be in an $\mathrm{sp^3}$ hybrid orbital until very high $\mathrm{pH}$ values because it takes part in resonance. (Compare with the amino group of aniline.) Also, that third structure you are drawing there in the series of equilibria deserves a downvote.

I do absolutely agree about not-hundred-percent availability of exo-aminogroup lone pair, but still the comparative basicity of those two nitrogens is still a good question, which need more references to be found. But the second scheme, as you can read from my answer, is not my creation, but an illustration for the paper I mentioned.

If the paper you are referring to is the one you linked (B. T. Storey, W. W. Sullivan, C. L. Moyer, J. Am. Chem. Soc. 1964, 29, 3118–3120. DOI: 10.1021/jo01033a537.) I can only suggest you actually read it since it supports none of your claims.

@Jan I think you’re being somewhat unfair. OP already mentioned that the contribution of the pyrrole-type nitrogen is negligible as aromaticity is ruined, which is precisely the third structure you are complaining about (the meaning is that this structure contributes negligibly to the equilibrium mixture). It could perhaps be explained or drawn slightly clearer, but it is not wrong. BTW DMAP is a useful analogy to draw. Where does it get protonated?

Hückel aromaticity in a heterocycle; no.


text:
Reaction between CH3CH2Li and Ethyne
In the reaction CH3-CH2-Li + ethyne
Do the lithium and a hydrogen from ethyne just switch places? And how would i draw the curved arrow notation for this? I'm assuming from the C-Li bond in one to the C-H bond in the other and vice versa? Will upload a pic of my attempt when I get home.

It would be quite difficult to make exact mechanism here, alkyllithiums are oligomers and would react via intermediate pi-complexes.

Terminal alkynes are weak acids, but they are still stronger ones than alkenes and alkanes. This actually means they would readily react with conjugate bases of weaker acids, giving them the proton. So yes, since ethyne is a stronger acid than ethane (and ethyl anion is a stronger base than ethynyl anion - the list for bases is reverse due to obvious reasons), they will exchange H and Li.
Below you can see a mechanism I would use to describe the problem in your question. C-Li bond has significant contribution of both covalent and ionic components, so I wrote it as both in equilibrium:

Terminal alkynes are weak acids, but they are still stronger ones than alkenes and alkanes. || A lot stronger than alkenes and alkanes, I'd say. You can create enough deprotonated alkyne with KOH in non-aqueous media for it to work as intermidiate in many reactions. No so luck with alkanes and alkynes.

Unless the "alkene" is cyclopentadiene, maybe?

Calling cp an alkene is like calling DNA an alcohol, because come on, it has OH group in deoxyribose fragment.


text:
Any program for conformational searching for sugars?
I want to estimate mean Gibbs free energy of formation of glucose and for that purpose I think conformers of ring flip should be taken into account. Do you know if some free program can find those structures automatically? (for example Balloon).
The following image shows the conformers I need to get. I am trying to avoid the possibility of drawing it.

I use RDKit for conformation searching.  More info here https://proteinsandwavefunctions.blogspot.dk/2016/08/conformer-search-with-rdkit.html

I'm sorry (new to chemistry here, but your question is interesting to me and I would like some clarification.) Are you asking for software that will generate conformers for you? (Just thought it was a cool question and wanted clarity.)

@MelanieShebel yes, some free programs that can do that are: balloon and RDkit, you can download them in the web. Also avogadro can do that, although I never try it :)..

Use RDKit
It sounds like you're only considering free or open source conformer generation tools. In that case, I highly recommend using RDKit, since the distance-geometry (and in turn ETKDG) methods will intrinsically generate all possible ring geometries. That includes not just the two chair forms you indicate, but all possible ring-puckering forms.
There are other free tools which do this, but RDKit has been repeatedly shown to be the best or among the best free conformer generation tools. (Based on current benchmarks.)

"High-Quality Dataset of Protein-Bound Ligand Conformations and Its Application to Benchmarking Conformer Ensemble Generators." J Chem Inf Model 2017 57, 529-539.
"Freely Available Conformer Generation Methods: How Good Are They?" J Chem Inf Model 2012, 52 (5), 1146–1158.

I'd advocate the newer ETKDG method with RDKit, since it seems to produce somewhat better geometries.

Hi, thanks for the reply. As far as I know RDkit doesn't move B-A-O-H diedrals(doesn't see hydrogens), and also doesn't find conformers for ciclohexane. Have you tried that?

With most conformer programs, the hydrogens are added at the end. As for cyclohexane, yes RDKit can generate multiple different conformers.

Is it possible to add hydrogens to oxygen and nitrogen before the conf search for those programs? Otherwise, I think some relevant structures will be missed, mainly in sugars.

I am not aware of a program that does that, unless you want to run a PMF yourself, but, on the side, here is a systematic study of the energetics of monosaccharide sugar puckering: Heather B. Mayes, Linda J. Broadbelt, and Gregg T. Beckham, J. Am. Chem. Soc. 2014, 136 (3), 1008–1022. DOI: 10.1021/ja410264d. You may find here the info you are looking for.

Welcome to Chemistry.SE! Take the [tour] to get familiar with this site. We prefer to use human-readable references and non-pdf links, I have edited your post accordingly.

great!Thanks Elisa. Would you clarify PMF?

PMF stands for potential of mean force. Here is a review you may find interesting, http://www.sciencedirect.com/science/article/pii/001046559500053I


text:
When writing out synthetic reaction schemes, why are some reactions combined in one arrow?
An example of what I mean is here in the synthesis of compound 13. The reaction of the first arrow describes a single reaction, whereas the second arrow to form the final compound seems to have five stages combined into one. I understand that the compound is carried on as is without characterization when the parts are combined into a single step.
If one were designing a new synthesis, what factors would lead one to write (and characterize) each step of the synthesis versus combining many individual reactions in one apparent step?
P. S. This is not an organic chemistry question per se, but meta-ochem (sorry if my tags are limited in description).

You seem to be asking about formal rules in the situation where no formal rules exist, nor are they needed. When you design new syntheses, this is the least of your problems.

Even a rough set of guidelines, if anyone has ideas, as opposed to "formal rules" would be appreciated. I'm just trying to make sense of this seemingly haphazard notation.

I want to start off by saying that your example is a particularly bad one. Usually, the arrows in a scheme should at least state a list of reagents and steps. Instead, I would love to discuss this using scheme 7 of Zhang et al.’s Beilstein J. Org. Chem. article.[1] (Beilstein J. Org. Chem. is a open access journal, the full text should be accessible from the DOI link provided below.)
The main reason for combining reactions like this is brevity. For example, every practicing chemist will immediately understand that the homoallyl Grignard under 1. will attack one of the carbonyl functions, that triethylsilane will reduce an oxygen and that TBAF will remove the silyl protecting group on the hydroxy function. None of these steps really warrant any discussion and most people would be able to draw out the missing structures in a few minutes. Likewise, I am not entirely sure why they did not include the preceeding protection into that same reaction scheme because there too there is noting interesting going on.
In more advanced synthetic schemes, the steps that are shortened and written together on a single arrow are often typical name reactions or very easily performed ones:

protections and deprotections
oxidations of alcohols to aldehydes with standard reagents
reductions with DIBAL, LAH or $\ce{NaBH4}$
ozonolyses
double bond isomerisations
simple Wittig/HWE reactions
etc.

They are typically chosen in such a way that the path is obvious and that the structure on the end of the arrow can be drawn up quickly by going through the steps. They will practically never include key steps or steps whose stereochemical outcome is uncommon. Cross-couplings of more elaborate structures or lesser known name reactions are also less common.
However, there is no rule as to what can be combined and what cannot. If anything, the reviewers might mark a scheme as too elaborate (i.e. can be shortened) or too brief (needs elaboration).
Reference:
[1]: J. Zhang, H.-K. Zhang, P.-Q. Huang, Beilstein J. Org. Chem. 2013, 9, 2358–2366. DOI: 10.3762/bjoc.9.271.


text:
Why are the t2 orbitals above the e orbitals in a tetrahedral complex using the crystal field theory?
In case of octahedral coordination entity formation, the $\mathrm{e_g}$ is above in energy level diagram than the $\mathrm{t_{2g}}$ due to more electrostatic force acting when the orbitals are along the axes in case of $\mathrm d_{z^2}$ etc. But in case of tetrahedral, why is the energy diagram inverted and becomes somewhat like this? I can't think of any possible reason. Can anyone please help?

Take into account that $\mathrm{e_g}$ and $\mathrm{t_{2g}}$ (or $\mathrm{e}$ and $\mathrm{t_2}$) are symmetry terms, and they make sense in the context of a certain symmetry. In other words, they do not describe something that is implicit to the orbitals, but, rather, how are orbitals affected by an environment with a certain symmetry.
In crystal field theory, we look only at the electrostatic effect of a certain distribution of charges around the nucleus on the orbitals; the discussion is primarily qualitative and geometric in nature.
In an environment with spheric symmetry (such as an isolated atom), all 5 d-orbitals are degenerate; since electrostatics are the same in any radial direction, and the set of 5 d-orbitals are only distinguished by their angular part (how are they distributed in the solid angle around the nucleus), they are equivalent. [Incidentally, this would be true not only for an isolated atom but also for an atom surrounded by a spherically symmetric charge distribution, such as a uniform charge distribution a certain distance away from the nucleus; however, I can't think of a real system that would have exact spherical symmetry.]
As soon as you introduce point charges around the nucleus, however, symmetry will break down. All 5 d-orbitals will no longer interact equally with the point charge; some orbitals will be closer to the point charge, some will be further away; and, most importantly from the point of view of symmetry, perhaps several orbitals will interact equally with the point charges.
This may be easier to visualize with the three $\mathrm p_x$, $\mathrm p_y$, $\mathrm p_z$ orbitals: if there are two point charges along the $z$ axis, it is clear that the $\mathrm p_z$ will interact more strongly with them than the $\mathrm p_x$ and $\mathrm p_y$ orbitals, but, crucially, $\mathrm p_x$ and $\mathrm p_y$ will interact exactly the same way with the charges; their interactions with the point charges are symmetry-equivalent.
In the extreme case where the distribution of point charges around the atom has no symmetry, all five d-orbitals would in general have different energies, because their interactions with the charges would be different; but if our distribution of charges is symmetric (as happens in both the octahedral and the tetrahedral case), we need to look at how the orbitals behave under the symmetry.
If you do that (and Jan's excellent answer explains what happens with the Cartesian axes under both symmetries), you will see that in the octahedral case two orbitals behave the same way (and are destabilised to the same extent), while another set of three orbitals behave equally (and are stabilised to the same extent). Looking further at their symmetry properties, these sets of two and three orbitals behave in the ways we describe with $\mathrm{e_g}$ and $\mathrm{t_{2g}}$, respectively. In the tetrahedral case, per contrast, we find that two orbitals are equally stabilised while three orbitals are equally destabilised; the symmetry label that fits their behaviour is $\mathrm e$ and $\mathrm t_2$. [In both cases, $\mathrm e$/$\mathrm{e_g}$ is a symmetry label that corresponds to two elements, and $\mathrm t_2$/$\mathrm{t_{2g}}$ corresponds to three.] These "blocs" of orbitals are degenerate (i.e. these two or three orbitals all have the same energy because they interact similarly with the point charges in the environment), while the energy of these blocs is not the same - that's what we call "splitting".
Note that the temptation to think that there are two "$\mathrm e$" d-orbitals and three "$\mathrm t_2$" d-orbitals in an atom is to be avoided: there is nothing intrinsic in the orbitals that produces this particular splitting into two blocs; rather, it's a product of the symmetry of the environment. The same 5 orbitals, if put in a square planar environment, do not form blocs of 2 and 3 orbitals, but 2, 1, 1 and 1; in a trigonal bipyramidal environment, they split into 2, 2, and 1; and so on.
So it just happens that two orbitals are stabilised in a tetrahedral environment (and thus are labelled $\mathrm e$) and three are destabilised (thus labelled $\mathrm t_2$), while it's the opposite in an octahedral environment.
[As a thought exercise: in crystal field theory, we assume that we are dealing with a central cation surrounded by anions, because that's the problem for which CFT was developed. But, what would happen if the point charges around the atoms were positive instead of negative? You'll find that you end up with stabilised $\mathrm{e_g}$ (octahedral) and $\mathrm t_2$ (tetrahedral) and destabilised $\mathrm{t_{2g}}$ (octahedral) and $\mathrm e$ (tetrahedral), exactly the opposite splitting but the same symmetry behaviour.]
There are rigorous mathematical ways to deduce the symmetry labels of d-orbitals (or any orbital, for that matter) under an environment of a certain symmetry, as provided by point group theory, which involve symmetry tables and so on. Most likely, your textbook will include a discussion of the derivation for at least the octahedral case.

I assume you know that in a tetrahedral environment the three d orbitals correspond to $\mathrm{t_2}$, not just $\mathrm t$ (which is not an irreducible)?

Sorry, yes, it was a typo; they indeed correspond to $\mathrm{t_2}$

It is important to start off the answer that what you are talking about is only the very basic crystal field theory. In this, the ligands are assumed to be point-shaped negative charges approaching the central metal from specific directions that correspond to the complex’ symmetry. These negative charges then lead to the destabilisation of orbitals that are pointing towards these charges by electrostatic repulsion.
For an octahedral complex, the ligands approach along the coordinate axes in both positive and negative directions. Therefore, all orbitals that have contributions along these axes will be destabilised. This includes the $\mathrm d_{x^2-y^2}$ orbital ($x$ and $y$ axes) and the $\mathrm d_{z^2}$ orbital (the $z$ axis).
In a tetrahedral complex, the ligands approach from four of the eight corners of a cube whose sides are parallel with the coordinate axes. Without loss of generalisation, we can define one ligand to arrive from $(+x,+y,+z)$, one from $(-x,-y,+z)$, one from $(+x,-y,-z)$ and the final one from $(-x,+y,-z)$. As you may already notice, all of these approach from between coordinate axes. This is almost precisely where the $\mathrm{t_2}$ orbitals ($\mathrm d_{xy}, \mathrm d_{xz}$ and $\mathrm d_{yz}$) have their greatest extent. Therefore, it is these orbitals that are destabilised rather than the ones pointing along the axes (which here corresponds to between the ligands).
As the approach of tetrahedral ligands still somewhat ‘misses’ the orbitals in question, the destabilisation is less than in the octahedral case.
For any more in-depth discussion, typically one has to expand the crystal field model and discuss in terms of molecular orbitals.


text:
How does the use of inert gas in fluorination of saturated aliphatic hydrocarbons make it less violent?
While studying alkanes, I found this written in my book:

Fluoridation can be achieved without violence when alkane is treated with fluorine diluted with an inert gas, or by the action of inorganic fluorides like $\ce{AsF3}$, $\ce{SbF3}$, $\ce{AgF}$, $\ce{HgF2}$, $\ce{Hg2F2}$, etc., or by use of bromo or iodo derivatives.
The picture is given below for clarification:

Inert gases are chemically inactive, except xenon and radon, which are radioactive and may form some compounds.
How do the non-radioactive inert gases help in reducing the violence of fluorination? If I put a certain volume of fluorine and an alkane in a jar for reaction, then even if inert gases are present, they won't react. What is their role?

Let's say the gas mixture has something like 10:1 molecules ratio of inert gas : fluorine. The alkanes in the gas would have 10 times more chance of colliding with an inert gas molecule as compared to a fluorine molecule. This reduces the rate at which alkanes and fluorine react together, making the reaction a lot safer (fluorination is highly exothermic to the point where it can be explosive).
The inert gases are basically a necessity, without which it's impossible to do this reaction safely and responsibly.

The keyword here is dilution ;)

Damn @Martin you answered that question with one word!


text:
Does heat loosen "corrosive bonds" in iron compounds... permanently?
One [obvious] way to remove a seized nut is to heat it [thus enlarge it]... more than the bolt. But that's not the only way to use heat in this application. Apparently you can also let the stuff cool down and the nut will still be significantly easier to remove than before applying theat. Why does this happen? The description of this latter video says something about heat "separating the corrosive bonds". Is there real science behind this claim?

Yes there is some real science behind this, but it is not called "chemistry".

I'm not sure that "corrosive bond" is utilized in a chemical sense here, it's more like a physical problem of increased friction between the nut and the bolt due to metal oxide deposition.

If by corrosive bond they mean a physical bond not a chemical bond then yes, there is a very strong force of dry friction between the two parts.
Bolts and nuts are made so that the two fit together tightly so that they do not wiggle loose but not so tight that they are impossible to fasten. The problem is is that when iron corrodes the oxide has more volume than the iron. In fact when iron oxidizes, its size increases 70% (See: Pillings-Bedworth ration). This expansion puts a lot of force between the nut and the bolt. Since dry friction increases linearly with force, friction between the two increases also making the bolt harder to remove.
When you expand the nut and/or shrink the bolt some of this force is relieved and friction decreases making the nut easier to turn.

That's obvious if only one part is heated. What I don't get is why it is easier to remove even after both parts cool down.

Part of the volume increase in the oxide product is due to water of hydration. When you heat the joint enough, you may dehydrate the iron oxide and thereby cause a little shrinkage or cracking in the joint filler. Torquing the nut to "failure" could be easier.