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How to separate two liquids with both having the same properties?
I was curious if it was possible to separate such liquids e.g. two vegetable oils that happened to have different ingredients, so it is a different liquid, but the same density etc.

Is this one mixture your inquiring about, that has two different vegetable oils (such as soybean and grapeseed) and different solid products mixed all together? If so, then the answer is yes, the oils can be separated regardless of the solid ingredients in them. However, if both oils are actually the same, i.e. both being grapeseed, then there is no way to "separate" them as they are not different liquids, no matter what solids (or other items) are mixed into them. The oils themselves are then equivalent (density, flash points, smoke points, cloud points, etc...).

Strictly having the same properties no, at least not as I know. But they are many great technics for separating mixtures of liquid having very similar properties, i.e., isomers like ortho, meta and para-xylene. But having very different properties is not always a guarantee of separation, if you have an azeotrope, for example, you can't separate the components.

If they have the same properties, they are the same for all practical purposes. Then why bother separating?

@Len_sprague Sorry for being a little vague. They are different oils just as soybean and grapeseed. How could they be separated then?

If they have the same physical and chemical properties then you cannot separate then. If their boiling points are so close they will not distill apart, if they have the same solubilities in all solvents then extraction is not going to work. If they co-elute on different chromatography media then you're pretty stuck with the mixture.

Thank you for your additional information, @Peter. Karl's answer therefore is comprehensive to your question. Hope it helps you!

You could theoretically separate two fatty acids with the same set of physical properties by chemically reacting one to change its physical properties and then reversing the chemical process, for example a polyunsaturated fatty acid and a saturated fatty acid could have the same melting and boiling point, density, solubility etc, but the polyunsaturated fatty acid would be easily oxidized, and the oxidized product would then have different properties and could be removed from the saturated fatty acid.

Also, I believe this to be true.  When you mix two liquids with very similar mepting points, the melting point of the mixture is supressed, so for example, mixing two liquids with MP of 290 might yield a mixed melting point of 280.  The solid mixture resulting from the freezing process has a greater entropy than the two substances would be if frozen separately.  As a result, the mixed solid will TEND to be more stable if like particles are next to like particles.  If the mixture were melted and frozen repeatedly, there would be a progressive trend toward the substances being analyzed.

@Karl where did I say such a thing?

@Karl then we can remove our comments?

Typical vegetable oils are not just one chemical compound, but a mixture (after removing everything that is not) of glyceride triesters. The chance that there is some overlap between your two oils is big, so the answer to your immediate questions likely is no, completely impossible.
Even if there is no overlap in the chemical composition, you would probably still have to separate the oil mixture into all its components, and then know how to mix them together again to get the original two oils.
The latter is possible, of course, with a reasonable analytical effort, the former is close to impossible to perform on a macroscopic sample. A GC(-MS) can easily separate (and identify) the compounds, but that's a few billion molecules or so. Doing liquid column chromatography would be very tricky, with any reasonable apparative effort would give you a few milligrams of each compound, and perhaps there are still a few you can't separate.
Distillation is impossible (boiling points are high and too similar), fractionation by crystallisation would only work well for some of the triglycerides, etc. All in all it's a bad mess. Don't mix two oils and try to separate them afterwards. ;-)

I agree with major point which is that such oils are a mixture. So separating all the individual parts and then trying to put the oil back together is a futile effort.

By saying that you have different materials ("different ingredients"),
an analytical technique which shows a difference could be scaled up to effect a separation. Density was rejected as not giving a difference; chromatography was not recommended (but might work).
The problem with two (different but very similar) vegetable oils is that they are not simply two different materials, but rather two different mixtures of very many of the same or similar compounds. If you could separate the compounds in the mixture (e.g., by chromatography), then you would have to figure out how to apportion the separated individual compounds to form mixtures like the original oils.
In the 1970s, Canadian farmers grew strains of rapeseed with significantly lower amounts of (toxic) erucic acid through crossbreeding. The new plant was called canola, suggesting “Canadian oil, low (erucic) acid.” Canola oil and olive oil are vastly different in source and manufacture, although their composition is similar (from Wikipedia: Saturated fatty acids: 7.4%/13.8%; Oleic acid: 61.8%/71.3%; Linoleic acid: 18.6%/9.8%; smoke point: 238 C/193 C.).
Now if you mixed canola oil and olive oil, you could determine the proportion of each by analysis of all the constituents, but as far as separating the individual compounds and reconstituting the original olive and canola oils… Nah, I don’t think so.
On the other hand, two practically identical materials might be separable if they were only single entities. For example, H2O and D2O are regularly separated even though they are very similar.


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Determination of Urea and Hydrogen Peroxide in Cow Milk by ATR-FTIR(Attenuated total reflection-Fourier Transform Infrared Spectroscopy)
I am doing a research on cow milk adulteration. I am checking Urea and Hydrogen Peroxide content in milk by ATR-FTIR(Attenuated total reflection-Fourier Transform Infrared Spectroscopy). I am using $\ce{1M}$ Urea solution and $\ce{1M}$ Hydrogen peroxide solution. I am adding this to cow milk and testing through FTIR. But i am not getting good IR peaks of $\ce{H2O2}$.
Should i dilute the solutions? As we know, UV spectrophotometer gives better peaks at dilution. But in case of IR what should i do to get better peaks? what concentration of $\ce{H2O2}$ should i use?

In contrast to a typical sample provided by a synthetic lab your sample consists of multiple components; milk -- already consisting of many chemicals for its own, plus urea added, plus hydrogen peroxide added (as aqueous solution).  Using ATR-FT-IR is indeed a good choice for characterizing such samples, both for qualitative, as well as quantitative analysis (even to the point to provide kinetic data) as non-distructive, and potentially on-line method for continous process control.
This being said, you may improve the situation in the way how the ATR data is recorded.  Most often ATR-spectrometers are designed in a single-reflection geometry, wich is fine for not too complex sample compositions.  Potentially better though here is to use a design with a crystal allowing multiple reflections.  (Do not forget to keep the surface of the crystal clean, as (bio)films and absorbed material in general may affect the analysis, esp. with ATR flow cells.)

(source)
As a second hint, after the accumulation of your spectra, you may facilitate recognizing the barycentre of the absorption peaks by looking at their first derivative.  This technique sometimes seen in the field of UV-Vis spectroscopy (ref) is equally helpful if absorption bands are broad, or -- for example because a multi-component system is scrutinized -- partially overlap (e.g., IR-imaging).
At third, data analysis.  Beer-Lambert's Law, likely known from UV-Vis spectroscopy, equally is applicable in IR spectroscopy and you may choose picking an absorption band belonging only to your component of interest.  Because of the complexity of your sample, however, I would suggest that you use Partial Least Squares regression analysis (PLS) as a multi-variate analysis instead.  Very much simplified, instead of performing a regression analysis at one wavelength of interest, you do so for all data points, across the complete spectrum recorded, which equally allows to account for multiple components, too.  Do not worry, there are programs around to assist in this process, either provided with the IR spectrometer (e.g., TQ analyst by ThermoScientific), or independented solutions (e.g., Unscrambler).  Often, PLS allows to lower the detection limit compared to the approach based on a single absorption band quite a lot.
An example of determining hydrogen peroxide in water with PLS by ATR-FT-IR may be found here, of urea in milk here, or milk adultered with Hydrogen peroxide (0.15 g/L) here:

(original source of the article.)

source: Stack Exchange

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Why do inert gas have zero electron affinity?
As we know, we have to apply large amount of energy to cause an inert atom lose electron, so in that case, shouldn't its electron affinity be negative?

Ionization energy has nothing to do with electron affinity.

The inert gases do have a negative electron affinity meaning that the atom has to be in an excited state as well as having an extra electron.

Clarification is likely required for ionization energy versus electronic affinity. Although the trends of both across the periodic table have some similarity, they are not directly related to one another.
Ionization energy is the energy required to remove an electron (or more) from the element (gaining overall positive charge), while electron affinity is the energy difference (required or released) if an additional electron we're to bind to the element (resulting in an overall negative charge). Inherently then, ionization energy quantifies the difficulty of removing an electron already present, whereas electron affinity is the desire of the element to pick up more electrons; a noble element, for instance, is both "tight-fisted" (holding vehemently onto its electrons already present) and uninterested in further pairing/adding of electrons (is fully satisfied electronically), resulting in large ionization energies and small (e.g. zero or even positive) electron affinities.
Check out this page for more on ionization energies and this page for electron affinities, as they do note noble gases in their discussions.

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Acid hydrolysis of ethyl acetate
For my chemistry IA, I am planning to determine the reaction rate and activation energy of the acid catalyzed hydrolysis of ethyl acetate. Where can I find literature values for these, so i can compare my results and evaluate my erros?

hydrolysis of ethyl acetate

Check out this 1953 article from ACS on this exact reaction: https://pubs.acs.org/doi/abs/10.1021/ja01103a521
Does a variety of systems and calculates rate constants for comparison. I'd then recommend using similar terms they use to locate further resources on particular molecule hydrolysis, and even Google Scholar's (or Web of Science, Scopus, JSTOR, etc...) "cited by" option to work your way forward in time to papers that deal with this type of work. Using search terms from any articles you come across in databases (ACS, ScienceDirect, Scifinder, etc...) should get you going with plenty of benchmarking material as well.


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How does the human body deal with free radicals it creates itself?
I know that free radicals are created all over your body all the time as a byproduct of its chemical processes, (or maybe I do not and that is false). However I am confused about the distinction between free radicals formed from UV photons that will affect DNA base pairings, and the ones that the human body creates on a consistent basis.
How does the body deal with the free radicals it created by itself?
And how are those free radicals different from the ones that cause oxidative damage (e.g., from UV rays)?

Antioxidants like glutathione work by reacting with these free radicals before they interact with more delicate things.

Reactive oxygen species, primarily superoxide, may result from one-electron reduction of molecular oxygen. Given that aerobic metabolism involves loooooong chain of electron transfers, it isn't all that rare.
Human body uses specialized enzymes to defang reactive oxygen species. The most well known are peroxidase and superoxide dismutase (see wiki for details).
Aside from this, free radical and other oxygen and nitrogen active species are, apparently, involved in some forms of immune response. In this case they are not dealt with, but instead are expected to do maximum possible local damage.

Some enzyme-catalyzed reactions produce free radicals as short-lived intermediates, but these are likely to be contained within the enzyme active site until they can react further to form product. Example: ribonucleotide reductase - which catalyzes the formation of deoxyribonucleotides and enables the production of DNA.
I can't think of any case where a biochemical process "intentionally" forms a free radical and then allows it to escape and cause damage elsewhere in the body. But biology is wonderfully varied and I would not want to say "never".

Nitric oxide might be a good example of an endogenous radical.

Good point. It is created enzymatically and released systemically as a hormone. However, that particular radical is less reactive than many, and may not cause much, if any, tissue damage.

google://free+radicals+immune+response


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Is the maximum oxidation state of an element equal to it's valency?
I know maximum oxidation state has a sign and valeny doesn't.... But, if we remove the sign then, would it be equal to the valency of an element? I earlier assumed that valency was equal to the oxidation state (without the + or - sign) but later found cases where this wasn't the case.... so I've come up with something new that the maximum oxidation state is equal to it's valency (without the + or -). I'm just making sure if this is right, please do correct me if this is wrong! Thanks in advance.......

Valence refers to the absolute number of electrons lost or gained. Oxidation state refers to number of electrons gained (-) or lost (+). For example, if an element had oxidation states of -2, -1 and +1, the maximum oxidation state is +1.
Since valence is ambiguous, oxidation number is preferred now.
BTW, we can blame Ben Franklin, who arbitrarily assigned electrons a negative charge, for the oddity that gaining an electron reduces the oxidation state -- it wan't until more advanced technology enabled Thomson and Rutherford to show that electrons were the more mobile particles.

Because of the ambiguity of the term valence, nowadays other notations are used in practice. Beside the system of oxidation numbers as used in Stock nomenclature for coordination compounds, and the lambda notation, as used in the IUPAC nomenclature of inorganic chemistry, oxidation state is a more clear indication of the electronic state of atoms in a molecule.
The oxidation state of an atom in a molecule gives the number of valence electrons it has gained or lost. In contrast to the valency number, the oxidation state can be positive (for an electropositive atom) or negative (for an electronegative atom).
Elements in a high oxidation state can have a valence higher than four. For example, in perchlorates, chlorine has seven valence bonds and ruthenium, in the +8 oxidation state in ruthenium tetroxide, has eight valence bonds.


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Gaussian 09 Geometry Optimization: Identical energies printed in output
This question may be slightly off-topic for this site, but since Gaussian is the "calculator" most often used by computational chemists, perhaps it is relevant still.
I have performed a transition state optimization, trying to tightly converge to the TS. I have used the following keywords in my Gaussian 09 rev D.01
# P opt=(calcall,tight,ts,noeigentest)
# freq=noraman
# pbepbe/def2svp
# scf=(MaxCycle=1000,tight)
# int=ultrafine
# EmpiricalDispersion=GD3
# geom=(ModRedundant,Checkpoint)
# NoSymm

The final SCF eneriges in the output are as follows ($ grep "SCF Done" output.out):
 SCF Done:  E(RPBE-PBE) =  -3985.31026080     A.U. after   16 cycles
 SCF Done:  E(RPBE-PBE) =  -3985.31003619     A.U. after   13 cycles
 SCF Done:  E(RPBE-PBE) =  -3985.31009833     A.U. after   11 cycles
 SCF Done:  E(RPBE-PBE) =  -3985.31007900     A.U. after   11 cycles
 SCF Done:  E(RPBE-PBE) =  -3985.31002766     A.U. after   11 cycles
 SCF Done:  E(RPBE-PBE) =  -3985.30992366     A.U. after   12 cycles
 SCF Done:  E(RPBE-PBE) =  -3985.30988992     A.U. after   11 cycles
 SCF Done:  E(RPBE-PBE) =  -3985.30987787     A.U. after   11 cycles
 SCF Done:  E(RPBE-PBE) =  -3985.30987912     A.U. after   10 cycles
 SCF Done:  E(RPBE-PBE) =  -3985.30987915     A.U. after    5 cycles
 SCF Done:  E(RPBE-PBE) =  -3985.30987916     A.U. after    4 cycles
 SCF Done:  E(RPBE-PBE) =  -3985.30987916     A.U. after    4 cycles
 SCF Done:  E(RPBE-PBE) =  -3985.30987916     A.U. after    4 cycles
 SCF Done:  E(RPBE-PBE) =  -3985.30987917     A.U. after    4 cycles
 SCF Done:  E(RPBE-PBE) =  -3985.30987917     A.U. after    4 cycles
 SCF Done:  E(RPBE-PBE) =  -3985.30987917     A.U. after    4 cycles
 SCF Done:  E(RPBE-PBE) =  -3985.30987918     A.U. after    4 cycles
 SCF Done:  E(RPBE-PBE) =  -3985.30987918     A.U. after    4 cycles
 SCF Done:  E(RPBE-PBE) =  -3985.30987918     A.U. after    4 cycles
 SCF Done:  E(RPBE-PBE) =  -3985.30987918     A.U. after    4 cycles
 SCF Done:  E(RPBE-PBE) =  -3985.30987918     A.U. after    4 cycles
 SCF Done:  E(RPBE-PBE) =  -3985.30987918     A.U. after    4 cycles
 SCF Done:  E(RPBE-PBE) =  -3985.30987918     A.U. after    4 cycles
 SCF Done:  E(RPBE-PBE) =  -3985.30987918     A.U. after    4 cycles
 SCF Done:  E(RPBE-PBE) =  -3985.30987918     A.U. after    4 cycles
 SCF Done:  E(RPBE-PBE) =  -3985.30987918     A.U. after    4 cycles
 SCF Done:  E(RPBE-PBE) =  -3985.30987918     A.U. after    3 cycles
 SCF Done:  E(RPBE-PBE) =  -3985.30987918     A.U. after    4 cycles
 SCF Done:  E(RPBE-PBE) =  -3985.30987918     A.U. after    2 cycles
 SCF Done:  E(RPBE-PBE) =  -3985.30987918     A.U. after    2 cycles

Notice that many of the converged SCF energies are identical. My question is of course why has this happened, and is this a valid result?
Am I converging so tightly that the energy change is simply in the decimals that Gaussian has left out? The predicted energy changes are as follows ($ grep "Predicted change in Energy" output.out):
 Predicted change in Energy= 3.362413D-05
 Predicted change in Energy=-6.527689D-05
 Predicted change in Energy= 1.690958D-05
 Predicted change in Energy= 5.120292D-05
 Predicted change in Energy= 1.023774D-04
 Predicted change in Energy= 3.449328D-05
 Predicted change in Energy= 1.361217D-05
 Predicted change in Energy=-1.264148D-06
 Predicted change in Energy=-9.312661D-09
 Predicted change in Energy=-7.213572D-09
 Predicted change in Energy=-5.808562D-09
 Predicted change in Energy=-4.708143D-09
 Predicted change in Energy=-3.818469D-09
 Predicted change in Energy=-3.094391D-09
 Predicted change in Energy=-2.512754D-09
 Predicted change in Energy=-1.889480D-09
 Predicted change in Energy=-1.661426D-09
 Predicted change in Energy=-1.361766D-09
 Predicted change in Energy=-1.110023D-09
 Predicted change in Energy=-8.605687D-10
 Predicted change in Energy=-7.281627D-10
 Predicted change in Energy=-5.759861D-10
 Predicted change in Energy=-4.835488D-10
 Predicted change in Energy=-3.940408D-10
 Predicted change in Energy=-3.130737D-10
 Predicted change in Energy=-2.602865D-10
 Predicted change in Energy=-2.368678D-10
 Predicted change in Energy=-1.560729D-10
 Predicted change in Energy=-1.408281D-10
 Predicted change in Energy=-1.448954D-10

Seeing as these values quickly drop to 1e-9 and 1e-10, perhaps I am right in assuming that the identical energies printed in the output are identical only due to rounding the energy to 8 decimals?

Your SCF energies are only identical to eight decimal places!

That was my suspicion.


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Significant figures in percent error calculation?
Consider a percent error calculation where the measured value has 1 significant figure (XmL), and the actual/true value has 3 significant figures (0.XXXmL). Typically I would think that the percent error would end up being 1 significant figure, but what about the percent conversion factor, 100? This is an exact number, so I assume we don't this into account when thinking of significant figures...
Is it really correct to use 1 significant figure in this case? Or is there a special rule for percent error calculations because it's a percent? I remember my lab instructor last year saying that you simply go to the one's place (no decimals) for percent error, but I'm not sure if that was actually correct/proper or just her specific instructions for us.

You have a number of concepts conflated.

Consider a percent error calculation where the measured value has 1 significant figure (XmL), and the actual/true value has 3 significant figures (0.XXXmL).

If you only have one significant figure then you only have one significant figure. The "true" volume for any quantity of liquid would have a lot more than 3 significant figures. 1 ml of water has $3.45\times10^{22}$ molecules of water.

Typically I would think that the percent error would end up being 1 significant figure, but what about the percent conversion factor, 100? This is an exact number, so I assume we don't this into account when thinking of significant figures... Is it really correct to use 1 significant figure in this case? Or is there a special rule for percent error calculations because it's a percent?

There is no special rule for % conversion. So if you only have one significant figure then you'd have say $7\%$ or $7\times10^1\%$. You can't get $74\%$ conversion out of one significant figure. The whole point behind significant figures is too keep the experimenter from making up extra digits.

I remember my lab instructor last year saying that you simply go to the one's place (no decimals) for percent error, but I'm not sure if that was actually correct/proper or just her specific instructions for us.

That was just some instruction for that lab. If I figure that $\ce{CaCO3}$ weighs 100.1674 grams but the accepted value is 100.0869, then my % error is 0.0802%.


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Wave-particle duality in electron
Electrons show wave-particle duality
Particle (atom) containing a wave (electron) on Quora
Atoms contain electrons. We consider atoms as particles but electrons which are inside the atom as waves (50% chances). When we talk about orbits and orbitals, outer electrons, inner electrons, I sub-consciously assume that we are talking
about the particle nature of electron. But this model is true only half of the time. The possibility of the electron being a wave at any time is also 50% . Particle nature and Wave nature are two different things because scientists came up with wave-particle duality. So we cannot talk about outer electrons, inner electrons, degenerate orbitals in the case of wave nature of electrons.Even if we can say inner part of the wave(close to nucleus) and outer part of the wave(away from the nucleus) ,what we are going to do in that case to solve energy, wavelength, position of the electron (in wave form) in an atom (a particle)?

There are no chances, no possibilities, and no half of the time. Each electron is (kind of) a particle and a wave at the same time.

so how can we measure the energy of a wave by using equations to calculate energy for a particle ? why we need to separate them in first place when we can get the thing going just by confining the energy at one place( in particle form) rather than distributed energy in wave form ?

We can't and don't. You may think of Bohr atom as a lucky coincidence having no practical consequences.

why we don't calculate the energy in wave form of an electron when we have invested our time in finding wave form of the electron ? If results come same by using particle nature equations, then whats different ?

Particle model ceases to make sense when we move past the H atom.

So wave model does if there is one ? if not, how are we gonna collect info about elements other than Hydrogen ?

You might have heard of Schrodinger equation.

Is it for both particle and wave nature of electron ?

We don't actually use this distinction.

Back up. Particles and waves are two abstractions. Whether an electron behaves like classical physics says a particle should, or a waves should depends on the experiment. A zebra is not a white horse 50% of the time and 50% a black horse in the other time, it is a zebra - something different.

neither  completely white nor completely black at one time ?

Interesting question and this is a complicated topic. Actually it is not true that the wave and particle models are each valid 50 percent of the time. I think it is better to say that the wave model is always valid, but under some conditions the wave gets compressed down to a small area, so it looks like a particle. I hope that makes sense, please comment if it doesn't?
I am not a quantum chemist, but I think the normal conditions for electrons in atoms and molecules are not ones that make them act like particle. So therefore we should always think of them as waves. Luckily we can still talk about the "inner" and "outer" electrons while they are waves. It is not "inner" and "outer" exact position, it is an area occupied by the wave, the "inner" one is nearer to the nucleus, the "outer" one if further away. For atoms these are called atomic orbitals - if you look them up on Wikipedia it might be very interesting.
If you want to solve the energy, wavelength and position, it's quite complicated. I would recommend studying quantum mechanics. It will take a long time to get there, although it is very interesting. There might be a quicker but less detailed course just for chemists to do these calculation too. I hope that's helpful.

Solving for the position and wavelength of an electron is not possible since the wavelength is akin directly related to momentum, and solving for both would violate the uncertainty principle. At best, you can have a spread in both.


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Why are combustion reactions spontaneous?
I learned that spontaneous reactions are reactions whose activation energies can already be overcome with the temperature they're already in. Since the temperature they're in already provides enough energy to overcome the activation energy, additional input of energy is not necessary and so it is a spontaneous reaction. So if this is true, why are some reactions like combustion reactions considered spontaneous if the temperature alone wasn't even enough to get over their activation energy? Is this definition of spontaneous wrong?

The notion of spontaneity is commonly associated with thermodynamics and hence has nothing to do with activation energies.

@IvanNeretin Doesn't activation energy determine how fast (spontaneous) a reaction will happen or an equilibrium will establish?

It does. The word "spontaneous" is ambiguous, you see.

Mhm, one of the things to retroactively delete from textbooks, along with  hybridisation, salts, antiaromaticity... On the other hand what "concepts" would stay... ;)

Clarification on the definition of spontaneity must be considered first, as it's meaning is often misinterpreted (and through colloquialisms is further confused). In order to do so, we must separate reaction concepts into thermodynamics (where free energy and spontaneity are discussed) and kinetics (where activation energy and rate are discussed).
Thermodynamics: looking only at the reactants' and products' Gibbs free energies, and the difference between them ( &Delta G ), a reaction is defined as (a) spontaneous if the difference is negative, (b) at equilibrium if zero, and (c) non-spontaneous (a sign that it is spontaneous in reverse) if positive. Thus, thermodynamics is interested only in which form is more stable energetically: the reactants or products. No details about the path (see kinetics below) is included. Check out Khan Academy's "Gibbs free energy and spontaneity" page for some great figures and further detail.
Kinetics: interested in the pathway between reactants and products, and how much energy change is required to obtain an arrangement that leads to one or the other. In other words, this is where the activation energy becomes important, as it is the energy barrier that must be overcome before a reactant could even think about becoming more stable. This barrier impacts the rate at which reactants and products can be formed. A "spontaneous" reaction in kinetics would be a "fast" reaction, one with a nearly zero or negative activation barrier (no representative "hump" in those internal reaction coordinate diagrams).
In summary, thermodynamics is interested in stability of reactants or products, and is not directly discussing the path by which it would arrive at one or the other. Hence, a reaction can be "spontaneous" and still not occur without further prompting because it is kinetically too difficult (read "high activation energy") to get there. However, a "spontaneous" kinetic reaction implies the path is "downhill" (so to speak) and thus is also a "spontaneous" thermodynamic reaction.
Following that understanding, you mention temperature, which is where I believe the confusion builds as it is mixed with the "spark" in combustion. Temperature does effect the Gibbs free energy by the following relationship:
ΔG = ΔH−TΔS
Thus, reactants can be "pushed" with temperature to become products by adding the necessary energy to a non-spontaneous or equilibrium system to breach the activation energy "hump," and be "spontaneous" at that temperature. In other words, they can indeed be triggered just by increasing the temperature. Examples of this are found in spontaneous combustion events as temperatures are slowly raised to a point at which the energy barrier is breached, and reactants are becoming products while releasing heat (because these spontaneous reactions are exothermic, recall negative &Delta G). Heat release then causes surrounding reactants to overcome the same barrier energetically, and result in what we know as fire/burning events.
What about a spark, then? A spark contains the energy needed to begin the combustion process (e.g. from gasoline and oxygen to carbon dioxide and water) but in a sudden shock that triggers this cascade of heat release much more rapidly. A CDC study illustrated the impact of this latent heat given off by the initial sparked reaction by observing a stoppering of combustion by removing the heat with conduction (i.e. blowing out a candle).
So, all in all, "spontaneity" as a definition perhaps isn't wrong, but is used in various circumstances to describe different processes; here, either stability between states or rate at which the states will change.
Be sure to check out the other related stack exchange page mentioned by Mithoron in the comments, with a great answer regarding spontaneity by Brian and Melanie Shebel. Covers more detail about the thermo/kinetic descriptions, whereas here we cover a more broad scope to address your questions about "spontaneity" in general.

authors: Αντώνιος Κελεσίδης


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Redox and acid-base reactions in planetary atmospheres
I'm a chemistry layperson looking for chemistry expertise regarding atmosphere composition.
Earth's atmosphere is oxidizing, but it was somewhat reducing in the distant past before photosynthetic life developed. Venus has an acidic atmosphere primarily composed of carbon dioxide which includes clouds of sulfuric acid. Some planetary bodies have reducing atmospheres containing hydrogen sulfide or carbon monoxide.
How do redox and acid-base reactions interact in the context of planetary atmosphere compositions? Specifically, what combinations of reducing/oxidizing and acidic/basic atmospheres are possible? Are there other planetary atmosphere types that should be considered when categorizing them?
Quick terminology question: Does it even make sense to refer to an atmosphere as acidic or basic?

We most certainly can call atmospheres acidic or basic.  Put some quicklime, or even magnesia, in our air and you see that it undergoes an acid-base reaction with acidic materials (water vapor, carbon dioxide) in our atmosphere.  Over time, our weakly acidic atmosphere gradually dissolves in water and then reacts with alkali and alkaline earth metal silicates too, turning rock into clay.
Our atmosphere became acidic as it picked up oxygen, which is indeed an "acid former" through the volatile (thus atmospheric) compounds it forms with several nonmetals.  In contrast, the gas (or at least, clouds therein) on the giant planets contain some ammonia and would appear basic if we brought typical Earth chemicals there.


text:
Visualizing atoms in XYZ file with colors representing partial charges
I have a file containing the atomic coordinates (in XYZ format) of a structure, and I also have a list of the partial atomic charges for each atom. I'd like to create an image of the structure where the colors of the atoms correspond to the computed charges. What are some good ways for doing this?

(In reply to a deleted comment). I'd be happy if it helps, but I see no benefit in doing so. The XYZ coordinates are a simple $n\times3$ matrix, and the charges are stored as a $n\times1$ vector. The filetype is irrelevant (I can freely convert to any format of your choosing -- CIF, *CAR, XYZ, JMol, etc.).

Alongside the answers below, I found another stackexchange post that is related rather closely, and has a few well discussed options for obtaining your visualization: https://chemistry.stackexchange.com/questions/76592/easiest-way-software-to-visualize-charge-density-from-an-xyz-file-with-point

If you've never heard of IsoDistort you can give it a try. http://stokes.byu.edu/iso/isodistort.php

It's actually very easy to do this with the free visualization software OVITO (www.ovito.org), where you can color-code atoms according to any particle property you import with your structure (like in your case the partial charges).
Simply import your .xyz file with the partial charges appended as an additional column. During file import you can specify which column contains which information, e.g. positions or charge.
Then use the Color Coding modifier and choose "Charge" as "Input Property".
Here's the corresponding manual entry:
http://www.ovito.org/manual/particles.modifiers.color_coding.html
and an example of how that would look like:

Hope that helps!

As far as I know, no software can do this right away but with some work, this should be possible.
Use CYLview to create a povray file, change the atoms to the desired color and render it with Povray. The problem here is that the coordinates of the xyz file and the povray are not the same. But as far as I can tell the numbering stays the same. So if the 3rd atom in your xyz is this:
N       -1.227679000      0.558365000      0.000000000

the 3rd sphere in the povray file would look like that:
sphere {<-1.9762436, 0.9491377, 0.0000000> 0.2812500
   pigment{color rgbt <0.4,0.4,1.0,0.00>}
   finish{F_normal}

All you need to do is change the color here. It should also be pretty easy to automate this color change with some lines of code.
So you can go from this

to for example this

by simply changing the color value shown above to
pigment{color rgbt <0.2,0.8,1.0,0.00>}

That's an interesting example compound you picked. Personally, I'd rather not go anywhere from it, or from azides in general, except maybe straight away to a safe distance. Preferably taking very careful steps while doing so. But then, I'm not actually a real chemist. :)

@IlmariKaronen that was the first one I found and is straight out of our last project. It's a computational model compound. For experimental work we used the 2-azidoethanol and 2-fluoroethylazide. The former one is no problem to handle, the other one is quite tricky.

This is easy to do in Avogadro v1.x. For any display type, you can set the color scheme, e.g. through the settings dialog:

Then in the settings window, you pick a color scheme:

Voila, you have a molecule colored by partial charges:

If the file doesn’t come in with partial charges, you can set them via View ⇒ Properties ⇒ Atom Properties:

Have you tried VMD link? It's free, up-to-date, will work with .xyz format files, and would allow you to set charge (or other property) as a color-sensitive characteristic. Here's an example using the bonding coordination (number of nearest neighbors) in a crystal (although only the surface is loaded to conserve memory)

I know you asked for an xyz file, but it is no problem to convert it into a mol2 file. And from there on, you can do this using Chimera.
I take the example molecule from the linked question in the comment section to your question.
The corresponding mol2 file would look something like this:
@<TRIPOS>MOLECULE
bla2.xyz
 78 84 0 0 0
SMALL
GASTEIGER

@<TRIPOS>ATOM
      1 N          -0.2152   -3.6116    1.0137 N.2     1  UNL1        -0.980426
      2 N          -0.9072   -3.6634   -1.2690 N.3     1  UNL1        -1.159281
      3 H          -1.0201   -4.5161   -1.2392 H       1  UNL1         0.451727
      4 H           0.5924   -3.5332    3.4084 H       1  UNL1         0.466050
      5 H           1.1383   -4.0685   -6.6623 H       1  UNL1         0.121348
      6 ZN          1.0488    0.3616    1.4618 Zn      1  UNL1         1.085568
      7 C           0.8262    0.8063    4.1826 C.2     1  UNL1         0.885122
      8 C           0.4171    1.4943    5.4556 C.ar    1  UNL1        -0.158259
      9 C          -0.1860    2.7336    5.4577 C.ar    1  UNL1        -0.038223
     10 H          -0.3201    3.1852    4.6583 H       1  UNL1         0.111225
     11 N           3.0298    0.2152    1.0137 N.ar    1  UNL1        -0.989702
     12 N           2.9780    0.9072   -1.2690 N.pl3   1  UNL1        -1.323792
     13 H           2.1252    1.0201   -1.2392 H       1  UNL1         0.552920
     14 H           3.1082   -0.5924    3.4084 H       1  UNL1         0.599787
     15 O           0.6110    1.4120    3.0769 O.3     1  UNL1        -0.805465
     16 O           1.3256   -0.3161    4.2420 O.2     1  UNL1        -0.786727
     17 C          -0.8262   -0.8063   -9.1420 C.2     1  UNL1         0.932072
     18 C          -0.4171   -1.4943   -7.8689 C.ar    1  UNL1        -0.116686
     19 C           0.1860   -2.7336   -7.8668 C.ar    1  UNL1        -0.092391
     20 H           0.3201   -3.1852   -8.6663 H       1  UNL1         0.116237
     21 O          -0.6110   -1.4120  -10.2476 O.co2   1  UNL1        -0.908737
     22 O          -1.3256    0.3161   -9.0825 O.co2   1  UNL1        -0.838830
     23 C          -0.4875   -3.0059   -0.1834 C.3     1  UNL1         1.029873
     24 C           0.2178   -2.5742    1.7916 C.cat   1  UNL1         1.084450
     25 N          -0.2869   -1.7042   -0.1394 N.3     1  UNL1        -0.556339
     26 N           0.1740   -1.4213    1.1643 N.pl3   1  UNL1        -0.669532
     27 H          -1.0626   -3.2357   -1.9987 H       1  UNL1         0.479136
     28 N           0.5884   -2.7456    3.0646 N.pl3   1  UNL1        -1.219932
     29 H           0.8222   -2.0668    3.5390 H       1  UNL1         0.535669
     30 ZN         -1.0488   -0.3616   -1.4618 Zn      1  UNL1         1.115312
     31 C          -0.8262   -0.8063   -4.1826 C.2     1  UNL1         0.930235
     32 C          -0.4171   -1.4943   -5.4556 C.ar    1  UNL1        -0.165126
     33 C           0.1860   -2.7336   -5.4577 C.ar    1  UNL1        -0.068129
     34 H           0.3201   -3.1852   -4.6583 H       1  UNL1         0.125910
     35 C          -3.6355   -0.4875    0.1834 C.ar    1  UNL1         1.112801
     36 C          -4.0672    0.2178   -1.7916 C.ar    1  UNL1         1.089920
     37 N          -4.9372   -0.2869    0.1394 N.ar    1  UNL1        -0.669012
     38 N          -5.2201    0.1740   -1.1643 N.ar    1  UNL1        -0.683650
     39 N          -3.0298   -0.2152   -1.0137 N.ar    1  UNL1        -1.010564
     40 N          -2.9780   -0.9072    1.2690 N.pl3   1  UNL1        -1.379275
     41 H          -3.4057   -1.0626    1.9987 H       1  UNL1         0.465316
     42 H          -2.1252   -1.0201    1.2392 H       1  UNL1         0.591458
     43 N          -3.8958    0.5884   -3.0646 N.pl3   1  UNL1        -1.359821
     44 H          -4.5746    0.8222   -3.5390 H       1  UNL1         0.462374
     45 H          -3.1082    0.5924   -3.4084 H       1  UNL1         0.595021
     46 O          -0.6110   -1.4120   -3.0769 O.3     1  UNL1        -0.846291
     47 O          -1.3256    0.3161   -4.2420 O.2     1  UNL1        -0.789904
     48 C           0.4875    3.0059    0.1834 C.3     1  UNL1         1.054807
     49 C          -0.2178    2.5742   -1.7916 C.cat   1  UNL1         1.079690
     50 N           0.2869    1.7042    0.1394 N.3     1  UNL1        -0.596520
     51 N          -0.1740    1.4213   -1.1643 N.pl3   1  UNL1        -0.651619
     52 N           0.2152    3.6116   -1.0137 N.2     1  UNL1        -0.982105
     53 N           0.9072    3.6634    1.2690 N.3     1  UNL1        -1.153898
     54 H           1.0626    3.2357    1.9987 H       1  UNL1         0.472248
     55 H           1.0201    4.5161    1.2392 H       1  UNL1         0.448303
     56 N          -0.5884    2.7456   -3.0646 N.pl3   1  UNL1        -1.192789
     57 H          -0.8222    2.0668   -3.5390 H       1  UNL1         0.511494
     58 H          -0.5924    3.5332   -3.4084 H       1  UNL1         0.458606
     59 C           0.5990   -3.3114   -6.6623 C.ar    1  UNL1        -0.253677
     60 C          -0.7385   -0.8939   -6.6623 C.ar    1  UNL1        -0.060650
     61 H          -1.1769   -0.0744   -6.6623 H       1  UNL1         0.135692
     62 C           0.8262    0.8063    9.1420 C.2     1  UNL1         0.921571
     63 C           0.4171    1.4943    7.8689 C.ar    1  UNL1        -0.136263
     64 C          -0.1860    2.7336    7.8668 C.ar    1  UNL1        -0.033469
     65 H          -0.3201    3.1852    8.6663 H       1  UNL1         0.092701
     66 O           0.6110    1.4120   10.2477 O.co2   1  UNL1        -0.900753
     67 O           1.3256   -0.3161    9.0825 O.co2   1  UNL1        -0.836472
     68 C          -0.5990    3.3114    6.6623 C.ar    1  UNL1        -0.299314
     69 H          -1.1383    4.0685    6.6623 H       1  UNL1         0.134260
     70 C           0.7385    0.8939    6.6623 C.ar    1  UNL1        -0.048643
     71 H           1.1769    0.0744    6.6623 H       1  UNL1         0.130420
     72 C           3.6355    0.4875   -0.1834 C.ar    1  UNL1         1.071205
     73 C           4.0672   -0.2178    1.7916 C.ar    1  UNL1         1.085316
     74 N           4.9372    0.2869   -0.1394 N.ar    1  UNL1        -0.660949
     75 N           5.2201   -0.1740    1.1643 N.ar    1  UNL1        -0.672967
     76 H           3.4057    1.0626   -1.9987 H       1  UNL1         0.459743
     77 N           3.8958   -0.5884    3.0646 N.pl3   1  UNL1        -1.356568
     78 H           4.5746   -0.8222    3.5390 H       1  UNL1         0.457166
@<TRIPOS>BOND
     1    21    17   ar
     2    17    22   ar
     3    17    18    1
     4    20    19    1
     5    18    19   ar
     6    18    60   ar
     7    19    59   ar
     8     5    59    1
     9    61    60    1
    10    60    32   ar
    11    59    33   ar
    12    33    32   ar
    13    33    34    1
    14    32    31    1
    15    47    31    2
    16    31    46    1
    17    57    56    1
    18    44    43    1
    19    45    43    1
    20    58    56    1
    21    56    49    1
    22    43    36    1
    23    76    12    1
    24    27     2    1
    25    49    51    1
    26    49    52    2
    27    36    38   ar
    28    36    39   ar
    29     2     3    1
    30     2    23    1
    31    12    13    1
    32    12    72    1
    33    38    37   ar
    34    51    50    1
    35    39    35   ar
    36    52    48    1
    37    72    74   ar
    38    72    11   ar
    39    23    25    1
    40    23     1    1
    41    74    75   ar
    42    25    26    1
    43    50    48    1
    44    37    35   ar
    45    35    40    1
    46    48    53    1
    47    11    73   ar
    48     1    24    2
    49    26    24    1
    50    75    73   ar
    51    42    40    1
    52    55    53    1
    53    53    54    1
    54    40    41    1
    55    24    28    1
    56    73    77    1
    57    28     4    1
    58    28    29    1
    59    77    14    1
    60    77    78    1
    61    15     7    1
    62     7    16    2
    63     7     8    1
    64    10     9    1
    65     8     9   ar
    66     8    70   ar
    67     9    68   ar
    68    68    69    1
    69    68    64   ar
    70    70    71    1
    71    70    63   ar
    72    64    63   ar
    73    64    65    1
    74    63    62    1
    75    67    62   ar
    76    62    66   ar

We open it with chimera and then choose Tools > Depiction > Render by Attribute, choose the Attribute to be the charge and click on Apply. This results in a structure that is colored by its partial charges.

Another alternative is using Jmol. Once the file is loaded, in the Jmol terminal run the command
color partialcharge

And the xyz file for convenience
5
CH3F
C  -1.94      0.873     0          0.079
H   -0.870    0.873     0          0.058
H  -2.297     0.259     0.800      0.058
H  -2.2974    0.486     -0.932     0.058
F  -2.2974    1.8732     0.12112   -0.253


text:
Why Rutherford was shocked by results of gold foil experiment?
The way Rutherford's classic gold foil experiment has been presented (including by Rutherford himself) doesn't make sense to me.
As many of you know, Rutherford famously described himself as being utterly shocked at the results brought to him by Geiger in 1909:

It was quite the most incredible event that has ever happened to me in my life. It was almost as incredible as if you fired a 15-inch shell at a piece of tissue paper and it came back and hit you.
Rutherford, Ernest; Ratcliffe, John A. (1938). Forty Years of Physics. In Needham, Joseph; Pagel, Walter. Background to Modern Science. Cambridge University Press.

It makes no sense to me that he should have been so surprised.  Rutherford wrote the following the year before, in 1908:

We can conclude with certainty from these experiments that the α particle after losing its charge is a helium atom. Other evidence indicates that the charge is twice the unit charge carried by the hydrogen atom set free in the electrolysis of water.
Ernest Rutherford & Thomas Royds, Philosophical Magazine 17, 281-286 (1909) (Note: Publication date was 1909, but authorship date was 1908.

Thus, at the time of the gold-foil experiment, Rutherford believed (correctly) that his "15-inch shell(s)" (the α particles) were helium atoms that had lost negative charge. So essentially Rutherford, using particles that he knew to be highly concentrated atomic-scale matter derived from atoms (α particles), was shocked to find that atoms contained highly-concentrated atomic scale matter (?!).
I'm not saying he should have known specifically about the nuclear model of the atom prior to the experiment. I'm saying he shouldn't have been utterly shocked to find that atoms contain something that can strongly deflect α particles, given that he knew α particles themselves were simply atoms whose negative charge had been removed. (Plus he knew he was taking particles derived from very light atoms, and shooting them at very heavy atoms.)
To use Rutherford's own metaphor, it's as if he were saying:

I went to an armory and obtained some artillery shells. I then fired these at another armory, and was shocked to find, as a result, that armories contain artillery shells.

I.e., if Rutherford already knew this about helium ions, why wouldn't he expect other atoms would also have their positive charges concentrated as well? Was his thinking that neutral atoms were accurately described by a plum pudding model, yet once you removed the electrons the positive charge would collapse into a small space? If anything, given that he didn't know about nuclear binding, it would seem he should have expected the opposite—that removing the negative charges would cause the positive charges to expand and become even more diffuse.
Can someone familiar with the history of these experiments resolve this?

He knew alpha radiation is helium, he didn't know the particles were single helium atoms. However the elastic backscattering only makes sense if the charge in both target and projectile is concentrated in very small spaces.

In retrospect, most discoveries seem obvious. But that moment you realize that your viewpoint and mental model have to change is quite noticeable.

@Karl You write: "He [Rutherford] knew alpha radiation is helium, he didn't know the particles were single helium atoms."  However, Rutherford wrote the following in 1908, a few years prior to the gold foil experiment (published 1909):  "We can conclude with certainty from these experiments that the α particle after losing its charge is a helium atom. Other evidence indicates that the charge is twice the unit charge carried by the hydrogen atom set free in the electrolysis of water." [Ernest Rutherford & Thomas Royds, Philosophical Magazine 17, 281-286 (1909).]

Knowing that alpha particles are helium atoms does not mean very much. Individual atoms as individual particles had been known for quite some time. The question was what, exactly, were these particles? Knowing that an alpha is a helium atom does not imply that the nucleus is really tiny and strongly holds two protons and two neutrons (which weren't discovered until 1932) together. Knowing that an alpha rebounds intact from gold tells you something new - that there is a single, strongly held ball of mass for each atom (helium and gold).

I meant to say compact helium nuclei. Or at least I should have ...

@JonCuster You write: "Knowing that an alpha is a helium atom does not imply that the nucleus is really tiny and ...."  True, but note that Rutherford knew (or, if you prefer, believed)  (a) that alpha particles were helium atoms that had lost negative charge; and (b)  that alpha particles were super-dense masses that would blow through "plum-pudding atoms".  This means he knew that if you remove negative charge from a helium atom, you are left with a super-dense mass.   So why be utterly shocked to find that atoms contain such super-dense masses?

I don't think it is correct to say that Rutherford believed that the alpha was what we now think of as an atom with a single, effectively point-like nucleus. If the alpha had a smeared-out mass distribution, and other atoms (like gold) had a smeared-out mass distribution, then one would not expect one to bounce off of another in the way that was observed.

OK, suppose Rutherford thought of alpha particles as having roughly the size and mass of He atoms.  Even then, his reaction doesn't make sense to me. This was pre-QM, so atoms were viewed as solid matter.  Given this, why would Rutherford be shocked when a small proportion of very high-speed solid particles, about 1/3 the cross-section of the solid particles with which they were colliding, and with only 1/50 of their mass, bounced back?  Pre-QM, scientists used macroscopic intuition to predict microscopic behavior.  And they certainly knew bullets could ricochet after striking a solid surface.


text:
Calculate the concentration in milimoles per litre (mM) of a solution that is 20% H2SO4 by mass and that has a density of 1.198 g/L

Calculate the concentration in milimoles per litre ($\pu{mM}$) of a solution that is $\pu{20\%} \ce{H2SO4}$ by mass and that has a density of $\pu{1.198 g/L}$.

For the answer, please provide two decimal places.
I calculated an answer but I'm just checking to see if I am correct...
I got $\pu{61.7 mM}$
First, I calculated the molar mass of $\ce{H2SO4}$ which is $\pu{98.086g/mol}$ and then I multiplied it by $\pu{20\%}$ which then got me $\pu{19.6172g/mol}$.
Second, I divided density by mass to cancel out the grams to get $\pu{mol/L}$ which then came out to $\pu{0.06106M}$ and then I divided it by $1000$ to get $\pu{mM} = \pu{61.07mM}$
If someone could check if this is correct that would be great, thanks!

the solution has a density of 1.198 g/L.

This has to be wrong. I assume that the density is 1.198 g/ml.
Thus a liter has a mass of
$$1000\text{ ml/L} \times 1.198\text{ g/ml}= 1198\text{ g/L}$$
20% of the mass is $\ce{H2SO4}$, thus the mass of $\ce{H2SO4}$ in 1.000 L of solutiom is:
$$0.20 \times 1198\text{ g/L} = 239.6 \text{ g/L}$$
$\ce{H2SO4}$ has a molecular mass of $98.079\text{ g/mol}$ so the number of moles is
$$\dfrac{239.6 \text{ g/L}}{98.079\text{ g/mol}} = 2.443\text{ moles/L}$$
Now converting to millimoles
$$2.443\text{ moles/L}\times 1000\text{ mM/mole} = 2443\text{ mM/L} \ce{->[rounding]} 2.4\times 10^3\text{ mM/L} $$

Always best to write these out with unit conversions:

$$\frac {1\  \pu{gram}\ \ce{[H2SO4]}}{\pu{gram}\; [solution]} \mathrm{x}
\frac {1.198\; \pu{gram}\; [solution]} {1\; \pu L\ [solution]} \mathrm{x}
\frac{1\; \pu{mol}\; [\ce{H2SO4}]}{98.086\ \pu{gram}\ [\ce{H2SO4}]} \mathrm{x}
\frac {1000\ \pu{mmol}}{1\pu{mol}}$$

I encourage you to write this one out to check your answer! It looks like you have flipped a fraction somewhere.

Welcome to Chemistry.SE! Good effort, but please note the edits and how much more simple and clean the post is.  Please note that formulas can be better expressed with $\ce{}$ for chemical formulas/equations, $\mathrm{}$ for math term/equations, and $\pu$ for units. More information is available in this meta post Also, take a minute to look over the help center and tour page to better understand our guidelines and question policies.


text:
Rhombohedral second-order Jahn–Teller effect in Au(II) complex?
I was recently listening to a talk about the following work[1] when the author mentioned that they explained the following. In this $\ce{Au(II)}$-porphyrin complex you observe two sets of $\ce{Au-N}$ bond lenghts. Meaning that there is some sort of distortion in the square plane. They explained it with a second order Jahn–Teller effect that would lead to a rhombohedral distortion.
So I started to think about that situation a bit more. Second-order JT effects are of course favored in heavy metals where the relativistic effects cause the $\mathrm{6s}$ orbital and the $\mathrm{5d}$ orbitals to come closer together in energy. But I always assumed that second order Jahn–Teller required two paired electrons. Nevermind, I'm still thinking about how this would look like. I assume, as it is a regular $\ce{d^9}$-system in a square plane geometry it will start from the tetrahedron with a strong first-order JT distortion to raise the $\mathrm{B_{1g}}$ orbital high enough that it can interact with the $\mathrm{6s}$ orbital.
But when I look for square plane distortions what comes up is a paper on rectangular and rhombohedral distortions on molecules. They say the new point group was $D_\mathrm{2h}$ but I cannnot find any $\mathrm{d}_{x^2-y^2}$ in this character table. And furthermore, at least to what I see what will happen in these non transition-metal complexes is a splitting of the two-fold degenerate low-lying orbitals.
I'm not really good with group theory or orbital interactions but would mixing the $\mathrm{d}_{x^2-y^2}$ orbital with the higher s-orbital according to second order Jahn–Teller cause a rhombohedral distortion?

Preiß, S.; Melomedov, J.; Wünsche von Leupoldt, A.; Heinze, K. Gold(III) tetraarylporphyrin amino acid derivatives: ligand or metal centred redox chemistry?. Chem. Sci. 2016, 7 (1), 596–610. DOI: 10.1039/C5SC03429A.

Isn't it more likely that it is Oh going to a D2h? d9 config in tetrahedral complexes should not have any gain since you put either eg electron in each its degenerate orbital for zero gain. I am seriously rusty looking at character tables and sorting stuff like this out, so as a comment - are you certain you can rule out Oh -> D2h ?

In $D_{2h}$,  $x^2-y^2$ should be totally symmetric and so have $A_g$ symmetry label. From a symmetry perspective interacting /mixing with this will not change anything, neither will an S orbital as it is similarly totally symmetric. symmetry.


text:
IUPAC names of imines
How to write IUPAC name for this imine compound?

What do you think it might be?

I don't have an idea about IUPAC nomenclature of imenes.

Try this  http://www.chemspider.com/Chemical-Structure.10400.html?rid=a84a6219-9a90-4a39-a066-8c1d134b4aa3

Related: https://chemistry.stackexchange.com/q/100154/544

You have correctly identified the compound as an imine, which is characterized by the double bond between a carbon atom and a nitrogen atom.
According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), imines are named using the suffix ‘imine’ (in absence of characteristic groups having seniority over imines).

P-62.3.1.1 All imines are named substitutively using the suffix ‘imine’; the presence of several ‘imine’ characteristic groups is denoted by the numerical multiplying prefixes ‘di’, ‘tri’, etc. When there is a choice for numbering, the methodology described in P-62.2.4.1 for amines is recommended to generate preferred IUPAC names

The cross reference to P-62.2.4.1.3 concerns the choice between two or more parent structures in complex polyamines, which is not relevant in this case.
Thus, the parent structure of the given compound is simply methanimine. The corresponding name for the N-substituted imine is N-phenylmethanimine. Further substitution at the carbon atom yields the complete name N,1-diphenylmethanimine.


text:
Identifying Meso compounds

(source: iupac.org)
I can't seem to understand why these compounds are meso compounds..... I don't see any plane of symmetry here. I have trouble identifying meso compounds so can you give me heads up on how to identify them

Let's start from the obvious. Look at the third compound. Apply a plane which is vertical and perpendicular to the screen.

But the chlorine atoms are projecting out of the plane from different sides while the hydrogen are just going down the plane identically... Wouldn't the chlorine mess the mirror images

Of course the chlorine atoms stick out from different sides. The same can be said about the whole left vs right halves of the molecule. That's how mirror planes work, isn't it?

Yea..........Okay then  then how about the second structure. I am pretty sure it doesn't have a plane of symmetry

OK, look at the second structure (I mean the imaginary 3D model, not a picture on the screen). Gently grab it by the horns with both your hands. Now move the left side a little closer to you and the right side farther from you, until the central H eclipses the central OH and you recognize the same vertical plane.

How can central H eclipse OH.. One is into the plane and other is out of the plane (coming towards me). They are at 109.5° if I am correct

The same way Moon can eclipse the Sun, though they are pretty far away from each other and from us.

See: https://chemistry.stackexchange.com/questions/48995/is-2r-3s-butane-2-3-diol-chiral/87415#87415

Why don't use the definition just above the image?

meso-compound
A term for the achiral member(s) of a set of diastereoisomers which also includes one or more chiral members.

Achiral, i.e. the molecule is superimposable with its mirror image. So, if finding a plane of symmetry, or other symmetry elements in the molecule is difficult for you, a simpler method (and according to the definition) might be drawing a mirror image, i.e. all chiral elements inverted

or unchanged but drawn "right-to-left", i.e. flipped

or flipped vertically

.. (just use any of the mirroring methods) and deciding whether it's the same molecule (stereoisomer), by rotating it, in 3D, in your head. (Some internal rotations around single bonds might be needed as well.)
In the first case, the mirrored structure can be superimposed with the original one by in-plane 180° rotation around an axis perpendicular to the paper plane.
In the second case it can be done by 180° rotation around the in-plane horizontal axis.
In the third case it can be done by 180° rotation around the in-plane vertical axis.

How did you draw the second structure....... Did you just flip it. And for checking me so conpound, do So have to just make mirror images and check if they are Distereomers

@Abhinav yes, flipping in fact a mirroring (Reflection(mathematics) see 0th section bottom). Make mirror image and check if such expected enantiomer is in fact  the same compound (the definition is somewhat confusing).

@Abhinav .. but flipping as an image! Some chemical drawig software flips the structure with preservation of the stereochemistry, i.e. it also inverts the configuration, so the configuration is intentionally preserved.

Meso compounds need not only have a plane of symmetry. They can also have a point of symmetry.
The upper left compound has a point of symmetry at the centre of the central C-C bond.
The upper right compound has a plane of symmetry passing through the central H & OH groups.
The lower compound has a plane of symmetry passing through the common side of the two benzene rings.
The ideal way to identify the line of symmetry is to analyse the compound in 3-D. You may use models for that.

That's not quite enough. The $S_{4}$ point group does not have a plane or point of symmetry, but it has an improper axis of rotation.

I don’t understand what compound are you talking about. Please reframe.

This isn't a compound, this is a whole class of compounds, in fact anything that's in a $S_{n}$ point group. example


text:
Wetting agent for H2O2 solution
I need to increase the wetting of an $\ce{H2O2}$ solution on ceramic surfaces. Does anyone know of a common chemical I can use for this?
The properties it would ideally have are:

Increases wetting of $\ce{H2O2}$ - water mixture on ceramics (particularly glazed tiles and tile grout).
Does not react or reacts very slowly with $\ce{H2O2}$ (and water).
Non-toxic / low toxicity. Food safe would be ideal.
Reaction products are non-toxic / low toxicity.
Reaction products are all either soluble in water or evaporate under atmospheric conditions.
Commonly available and reasonably cheap.
Non-foaming or anti-foaming.

Thank you so much to anyone who has a suggestion.

Listerine/Total Care/Stain Remover/Anticavity Mouthwash (4 lines to name the product!) contains H2O2 and sodium lauryl sulfate (and 21.6% ethanol) in a purple plastic bottle. Both the ethanol and the sodium lauryl sulfate would lower the surface tension of the solution.
You might try this mouthwash on the ceramic as is - it is supposed to clean teeth...

Thank you this observation is absolutely perfect. In fact the idea to look at tooth whiteners is absolutely genius the requirements are almost the same as the ones for my application. I will certainly add ethanol. And maybe SLS too although it is foaming I believe that might not be a problem. In terms of using mouthwash directly I plan to use H2o2 at 12% concentration which is probably (!) a bit higher, although I could boost it with a top-up.

You might consider chlorhexidine gluconate. It is a cationic surfactant used as a skin cleanser and disinfectant prior to surgery, and is an ingredient in certain mouthwash products, so it must have low toxicity. Drug stores sell scented and unscented solutions for hand washing -- choose the unscented.

Thank you very much for your suggestion. Just to confirm, since I saw on the wikipedia article that chlorhexidine is prone to precipitation with anions, do you know it won't react like this with the H2O2 peracid group (or in any other way)?
The safety aspect is not quite as cast-iron as I would like, but on the other had in my application the disinfectant effect could be seen as an advantage.


text:
Difference between Henry constant and equilibrium constant
According to wikipedia, Henry constants define the ratio between the aqueous and gas fraction of a chemical species at equilibrium.
Lets consider Henry constants H in atm.L.mol-1 such that $[C]_{g} = H \cdot [C]_{aq}$.
Based on this definition, H is actually the same thing as the equilibrium constant K of the dissolution equation of the chemical species of interest. Indeed, using Gibbs energies found on this site, K provides a good approximation for H in the case of CO2 for example ($\Delta G^0$ of the reaction is -8.4 kJ.mol-1, and $K = e^{-\frac{\Delta G^0}{RT}}$ gives 29.62 atm.mol-1, very close to the value found in Sanders 2015).
However K computed this way is not always close to H (for example in the case of $H_2$ where it is 1211 atm.mol-1 while H is reported as 1300 atm.mol-1). What is then different between using Henry constants and equilibrium constants to compute the saturation concentration of a chemical species?

Henry's law constants can be defined in a variety of ways (see the Henry's law wiki page for a very complete treatment of the subject, with examples.) For example, gas phase concentration can be expressed as moles per liter or as partial pressure; liquid phase concentration as molarity, molality, mole fraction. That may account for the discrepancy.

I would further caution you to write out equilibrium equations fully and to make sure the units are correctly expressed. That way you can avoid inadvertently comparing apples to oranges. Finally, be sure to consider the underlying chemistry: note that hydrogen gas physically dissolves in water but does not react, whereas carbon dioxide both physically dissolves and chemically (reversibly) reacts with water to form carbonic acid.

The K constant is unitless as it is a ratio. If we consider the aqueous phase is in mol.L-1, then the concentration of gas predicted from K will also be in mol.L-1. I guess it is possible to convert such gas concentration to units like Pa or atm using the ideal gas formula (P = nRT/V). To convert mol.L-1 of gas to atm, one then has to use R = 8.21e-2 atm.L.mol-1.K-1, such that [gas] = [aq] * K * n * R * T / V. With n and V being 1, [gas] in atm should then be [aq] * K * 24.47, which still does not correspond to what is found using Henry's law


text:
How to sloppy freeze a dihedral angle within a range in a Gaussian modredundant optimization?
I would like to perform an optimization in Gaussian 09 of my molecule with two dihedral angles being frozen. I would like that the values of these dihedral angles are not fixed, but can relax within a range of ±5°. In Gaussian 03 it seems to be possible using Opt=ModRedundant keyword and the following syntax:
[Type] N1 [N2 [N3 [N4]]] [[+=]value] [A | F] [[min] max]]

I have tried using the same syntax in Gaussian 09, but it doesn't work. Is it possible to perform such calculation in Gaussian 09 or 16 without using the scan approach?
The calculation fails with the following error:
The following ModRedundant input section has been read:
Invalid extra data found.

Do you know it is possible in G03 or do you just think so?

TL;DR: It is impossible (=not documented if it were) with Gaussian 09 Rev. E.01 and older, it might be possible with Gaussian 16, but in all cases not via opt(modredundant).
I am unaware of any sloppy freezing possibility in pre-Gaussian 16, and for the latter one has to dig much deeper in the manual and start constructing it; but at least it works in theory.
Gaussian 03
Unfortunately, you have misinterpreted the commands you found in the manual. You can find the full description for the opt keyword in via The Internet Archive, but I'll share the important parts here for convenience.
The construct
[Type] N1 [N2 [N3 [N4]]] [[+=]value] [A | F] [[min] max]]

indeed exists for a opt(modredundant) input section, but it behaves not as a sloppy freeze, but as a range exclusion criterion, which you can find a little lower in the manual:

[...]
  An asterisk (*) in the place of an atom number indicates a wildcard. Min and max then define a range (or maximum value if min is not given) for coordinate specifications containing wildcards. The action specified by the action code is taken only if the value of the coordinate is in the range.
  [...]

For example, the input
D * * * * F 175.0

freezes all defined dihedral angles that are smaller than 175.0°.
The input
D * 4  5 * F 160.0 175.0

freezes all defined dihedral angles around the bond between atom 4 and 5 in the interval 160.0 - 175.0°.
The input
D 3 4  5 6 F 160.0 175.0

is a syntax error because it does not contain a wildcard.
Gaussian 09
The option modredundant to the opt keyword still exists, but has a slightly modified input. Specifically the [[+=]value]] part has been removed. Again you can look up the full page on The Internet Archive.
The examples above should, however, still behave the same.

[...]
  Lines in a ModRedundant input section use the following syntax:
[Type] N1 [N2 [N3 [N4]]] [A | F] [[min] max]]
  [...]
  An asterisk (*) in the place of an atom number indicates a wildcard. Min and max only apply to coordinate specifications containing wildcards. The action then specified by the action code is taken only if the value of the coordinate is in the min–max range (or below maximum value if min is not given).
  [...]

Therefore you also still get the error "Invalid extra data found".
Gaussian 16
In this version, while opt(modredundant) is still maintained, a new feature is rolled out. As far as I can see, there are no changes to Gaussian 09 regarding the modredundant input section. You can find it live on the Gaussian homepage or a copy on The Internet Archive.
According to this, the above examples should still produce the same results.
The new feature is Generalized Internal Coordinates or GIC. See a description on the Gaussian homepage or on The Internet Archive. You can see the GIC in action in my answer on Constrain bond angle In Gaussian molecular structure optimization and on Gaussian: Relaxed scan with modredundant optimization and dummy atoms.
In theory you should be able to freeze a GIC if it meets certain criteria for example from the manual:

The following example removes an angle coordinate generated by default if ≥179.9°, substituting a linear bend:
A(1,2,3) Remove Min=179.9        !  Remove angle coordinate if too large.
L(1,2,3,0,-1) Add IfNot A(1,2,3) !  Add linear bends only if
L(1,2,3,0,-2) Add IfNot A(1,2,3) !+ the angle coordinate not active.

I have prepared a small example, but I am really not sure if it worked as intended. For cis-butadien the dihedral distortions should be small, so there isn't much visible. There are no syntax errors, but I am not really sure whether the section actually did what it should. Try it yourself and investigate further. (! recognises comments)
#p PM6                ! Semiempirical method for fast testing
opt                   ! Optimise
geom(AddGIC)          ! Use GIC, can also be specified with opt(AddGIC)
scf(xqc,MaxCycle=250) ! Switch to quadratic conv. if necessary, more cycles
symmetry(none)        ! Do not use symmetry

distorted cis-Butadiene (semi-constrained)
optimisation run
[Free title card can have multiple lines]

0 1
 C     0.000000     0.000000     0.000000
 H     0.000000     0.000000     1.089000
 H     1.026720     0.000000    -0.362996
 H    -0.513358    -0.889166    -0.363001
 C    -0.754249     1.306394    -0.533333
 H    -0.738680     2.222526     0.055206
 C    -1.531373     1.266770    -1.931370
 H    -2.044734     2.155933    -2.294370
 C    -1.615181    -0.098364    -2.761662
 H    -2.540288    -0.117779    -3.335874
 H    -0.765588    -0.163525    -3.439797
 H    -1.596709    -0.942933    -2.074433

sloppyfix=D(1,5,7,9)               ! Define an alias for the dihedral
sloppyfix freeze min=5.0 max=355.0 ! freeze in intervall

! Empty line above is important.

That the setup somewhat works can be seen with the following example of dihydrogen. First the atoms are separated 400 pm (=4.00 Å), and the value should be frozen at 250 pm (=2.50 Å), the PM6 optimised value is 76 pm (=0.76 Å).
#p PM6 opt(AddGIC,MaxStep=5) scf(xqc,MaxCycle=250) symmetry(none)

Free title card

0 1
 H    -2.000000     0.000000     0.000000
 H     2.000000     0.000000     0.000000

theHHbond=R(1,2)
theHHbond freeze max=2.5

! Empty line above is important.

Note that it will be frozen after the condition triggers, so it will not be able to the value that triggers it. Here are the values and messages you'll find in the optimisation run:

                           ----------------------------
                           !    Initial Parameters    !
                           ! (Angstroms and Degrees)  !
 --------------------------                            --------------------------
 ! Name      Definition          Value          Derivative Info.                !
 --------------------------------------------------------------------------------
 ! theHHbond R(1,2)              4.0            estimate D2E/DX2                !
 --------------------------------------------------------------------------------

[...]

 Variable       Old X    -DE/DX   Delta X   Delta X   Delta X     New X
                                 (Linear)    (Quad)   (Total)
 theHHbon     4.97283  -0.03796   0.00000  -1.00000  -1.00000   3.97283
         Item               Value     Threshold  Converged?
 Maximum Force            0.037962     0.000450     NO
 RMS     Force            0.037962     0.000300     NO
 Maximum Displacement     0.500000     0.001800     NO
 RMS     Displacement     0.707107     0.001200     NO
 Predicted change in Energy=-4.653046D-02
 Lowest energy point so far.  Saving SCF results.
 Internal coordinate with a value of  3.9728 a.u. has been conditionally frozen
 because it is in Min/Max range.
 The coordinate's label:
 theHHbond
 GradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGrad

[...]

 Variable       Old X    -DE/DX   Delta X   Delta X   Delta X     New X
                                 (Linear)    (Quad)   (Total)
 theHHbon     3.97283  -0.07253   0.00000   0.00000   0.00000   3.97283
         Item               Value     Threshold  Converged?
 Maximum Force            0.000000     0.000450     YES
 RMS     Force            0.000000     0.000300     YES
 Maximum Displacement     0.000000     0.001800     YES
 RMS     Displacement     0.000000     0.001200     YES
 Predicted change in Energy=-0.000000D+00
 Optimization completed.
    -- Stationary point found.
                           ----------------------------
                           !   Optimized Parameters   !
                           ! (Angstroms and Degrees)  !
 --------------------------                            --------------------------
 ! Name      Definition          Value          Derivative Info.                !
 --------------------------------------------------------------------------------
 ! theHHbond R(1,2)              2.1023         -DE/DX =   -0.0725              !
 --------------------------------------------------------------------------------
 Lowest energy point so far.  Saving SCF results.
 Largest change from initial coordinates is atom    1       0.949 Angstoms.
 GradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGrad

Note that in this example there is only one coordinate, ergo the optimisation finishes when the freeze condition is triggered. In this case there is quite a significant deviation from the condition max=2.5 and the actual value when it is frozen 2.1023. I would expect it to behave better with larger systems.
I would assume that the variable remains frozen for the remainder of the optimisation, with no more changes possible.
Standard disclaimer: Geometry optimisations with frozen parameters are critical for production purposes. Always check if you have a well converged molecular structure (stationary point) with no imaginary modes.


text:
Fischer projections

Can somebody tell me how why they changed/choose the groups for vertical and horizontal lines. As far as I know the vertical lines represent group going away from us and horizontal depicts coming towards us but I can't figure out how to choose which groups lie vertically and which groups horizontally... I am hopelessly stuck in Fischer projections.
EDIT:  What I am having difficulty is how to choose where o arrange the groups in the horizontal line.
For eg:

Why is the H bond on the C1 is arranged to the right while the OH is arranged to the left while the opposite is in C2 in the fischer projections It would make more sense for OH to be on the right as it is coming out of the plane and would appear at right when seen from above. Why does dash  line is at right while wedge line to the left. How to interpret?

This depiction tries to show how 3D tetrahedral structure is projected onto the paper plane, so the structure above is not meant to be a formula but a 3D model. Try to imagine you look at the structure from above, you get X−(C)−Z pointing towards you horizontal, and W−(C)−Y pointing down, vertical.

I am confused about the positions of Z and W.... How to know which will be in vertical line?Whever I try I always switch Z and W so it gives me wrong answers. Can you provide an answer on how to view this

.. They did not change anything. For each crossing of the Fischer projection, the molecule had to be oriented (or your point of view changed) such that you see the cross, and the horizontal groups point towards you, vertical outside. On your 1st image, it's already oriented, and you look from above. On the 2nd image it's also oriented, but you look from the right side.  If one can't see that, he/she probably has to train a 3D spatial  imagination.

If I see from right then I see cho-c-c-ch2oh as a line, the oh bond towards me and the hydrogen bond away for the C1, that would be opposite to the given fischer. Can you tell me what I did wrong here

Not enough, imagine you are moving closer and closer to the paper at the right side and looking to left at the molecule, until you cross the paper, then you should see OH (from C1)  pointing left and towards you, H (from C1) pointing right and towards you.

Okay. If I just see from right I can imagine them to be pointimg towards me left and right from C1 reapectively. So I guess then hydrogen would be taken at right and alcohol at left. One last question how  to know from where to look. For the first case we saw from top and here from right

The whole point of the projection is that you project your 3-D structure onto a lower dimension representation, but it doesn't say how you should orient for the projection. In other words, any orientation here is fine. Just like when you look at a hand in any orientation, it's still the same hand. That is, there is no absolutely correct perspective to look at it.
For the example provided, you could cycle any three groups, and the resulting project is valid. For example, W -> X -> Y -> W. This corresponds to a rotation along the axis to the remaining atom, so the stereochemistry is unchanged. Since you're still depicting the same structure, the resulting projection looks different but represents the same thing.
EDIT:
Note that when you have a complex structure, the rest of the structure can and does impose a preference on which groups should go where. For example, for carbohydrates, you should orient each carbon center such that the carbon backbone is vertical.

I'll quibble a bit. Quoting from Wikipedia "The use of Fischer projections in non-carbohydrates is discouraged..." The whole point of Fisher projections for carbohydrates is that the projection does specify how the projection should be oriented. That was the whole point. To draw all the different carbohydrates in a uniform way so that the 2D structures could be easily compared.

@MaxW That's hardly quibbling since you are 100% correct. I'll edit my answer. Thanks for the suggestion. Wish I could donate some of rep for the assist there...

The easiest way to analyze fischer projections is this:

Draw the projection on a page.
Consider the vertical groups to be inside the page and the horizontal groups to be outside the page.

This technique works for me flawlessly. I hope it does for you too.

How do I decide which groups are inside the plane or outside from the wedge dash. Can you elaborate more on this... I think I am confused at that particular point

Any group shown by a wedge bond is always outside the plane. Any group shown by dash bond is always inside the plane. Any group shown by a line is always on that plane. It’s that simple.


text:
Are there still "prairie dog refineries"?
In the old days (meaning the 1940s) there still many small "prairie dog" oil refineries, that were little more than a bunch of hooch stills soldered together, but from what I understand these are all gone now and there are only a handful of oil refineries in the United States.
Are there still small refineries in the United States or have they been regulated out of existence?

They are totally noncompetitive economically, since they can't do all the cracking and rearranging of chain lengths that is now standard. Just separating out the different chain lengths is not enough nowadays.

Probably not in the US, but who knows what has survived in other parts of the world? I'll agree with Jon and doubt it, however.

There are still small operations but they don't have complex  capabilities. They may separate nat gas and LPG and sell the remainder ( may be called a topping plant) . They may separate naptha, diesel, then sell the remainder.  Even among regular "refineries" there are large variations in capabilities.   Some do not have reformers, some do not have hydrocrackers , etc.  So the answer partly depends on what you call a refinery.

The question probably is whether (how much) you can reduce transport costs by such a "preprocessing" refinery.

Any collection of 'hooch stills' that were soldered (?) together would probably have been uneconomical almost as soon as they were built, because welding together a series of serviceable fractional distillation towers is not that technologically advanced and much more efficient.  However, there are certainly a number of relatively small refineries.
For instance, about the largest refinery in the U.S. (for comparison) is Exxon's in Baytown, TX that has a capacity of 560,000 barrels per day.
A couple of the smallest ones are the 5000 bbl/day refinery in Wyoming (Silver Eagle, in Evanston) and the 2000 bbl/day Foreland Refinery in Ely, NV.  The latter (if run all day) is equivalent to around 1.5 barrels a minute.
Also (to respond to a comment) there are still some refineries that don't do catalytic cracking (probably just straight distillation of the petroleum into fractions, which is viable for small refiners because they can save on expensive (and often proprietary) cracking units and are small enough that they can sell all the fractions that they produce.  The recently closed Flint Hill refinery in Alaska used another approach, reinjecting unused fractions back into the TransAlaska pipeline.)
For a list of refineries, the EIA is always a good place to look (pdf warning).
A wiki on refineries is "A Barrel Full" which has detailed information on individual refineries.


text:
Why can't this molecule avoid being antiaromatic by becoming nonplanar?

The peripheral system is clearly antiaromatic if planar, but we know that antiaromatic compounds are more unstable than non aromatic compounds.
So, I think that the molecule will lose its planarity and become non aromatic, but this molecule is antiaromatic according to my professor.
Isn't my reasoning correct?
Also, what would be the IUPAC name of this molecule?

"planar" is relative. this molecule is likely planar enough to give enough pi orbital overlap for antiaromaticity

Yes, this is what my professor further said while explaining. It's not just about the planarity but about the pi conjugation which will be reinforced when the middle carbon lifts out of the plane bringing the pi orbitals closer.

According to chemdoodle the name is 13-Methyltricyclo[6.4.1.04,13]trideca-1(12),2,4,6,8,10-hexaene.

A little information: check out structure 6 on this Wikipedia page and read the description below it. Also, if you have access, see the paper it references.

The preferred IUPAC name for this compound is 2a¹-methyl-2a¹-benzo[]azulene (the former numbering 9b-methyl-9b-benzo[]azulene is no longer recommended). If you have any detailed questions about this name, you might want to consider asking a separate question (see also our guidelines on how to ask nomenclature questions).

The central atom is approximated to be tetrahedral and thus approximately $\mathrm{sp^3}$ hybridised. The three atoms of the outer ring it is bonded to are approximately $\mathrm{sp^2}$ hybridised as they are all part of double bonds.
If you build this, you will notice that while these three atoms are necessarily in a plane they have different orientations. It is simply not possible to make a planar ring without some significant distortions of the geometry. I thus challenge your professor’s assumption.
I would expect this system to have strongly localised double bonds, i.e. significant bond length alteration, in such a way that the double bond are twisted with respect to each other to minimise strain.


text:
What is the Major product in free radical halogenation of Alkanes. (Between 3°,2°,1°)
In Free Radical Halogenation in presence of sunlight, a no. Of products are possible.
Due to higher stability of a 3°(Tertiary) radical, 3° products should be favored. But another factor that comes is that the formation of 3 ° radicals also leads to higher disproportion products. (Radical Disproportion- https://en.m.wikipedia.org/wiki/Radical_disproportionation)
Now, if i look at the intermediate, a higher no. Of Tertiary Radicals should form. But a lot of them will undergo radical disproportion as it is fastest in tertiary radical.
So is there any clear order of what product should be major or does it vary from case to case?

What do you mean by "disproportion products"?

Where did you find this nomenclature with the degree sign for primary, secondary, tertiary?

@Tan Yong Boon https://en.m.wikipedia.org/wiki/Radical_disproportionation

@Karl I frequently saw this nomenclature in Organc Chemistry by Solomans and Fryhle.

It depends on what halogen you use.
Chlorine has a reactivity ratio for $1^o$:$2^o$:$3^o$= 1 : 3.5 : 5.
To calculate the relative yield at that position, multiply the number of Hydrogens at that position to the reactivity at that position. This will give you a comparison of the yield at particular positions.
But in the case of bromine, the reactivity order is 1 : 250 : 6300. Therefore, without any further thinking, the bromine radical will always attack at the tertiary position if available.
In general, reactivity towards tertiary carbon increases down the group as in the case of halogens.

This does not solve my problem. I do know the reactivity order. But i am confused in the major product as much of the Tertiary radical will undergo disproportion.

It all depends on the halogen that is being used. The more reactive ones like chlorine don't really show stereo-selectivity whereas the less reactive ones like bromine do show the expected selectivity: 3°, 2°, 1°


text:
Help understanding the pressure's effect on equilibrium and Le Chatelier's principle
So let's say we have a closed system in the following reaction:
$\ce{N2O4 -> 2 NO2}$
The only way to increase the system's pressure without adding any mole of gas and keeping temperature constant is reducing the system's volume. Once we increase the pressure by reducing the total volume, formation of $\ce{N2O4}$ is observed.
In many text books, the explanation for this is that 2 moles of $\ce{NO2}$ occupy a bigger volume than 1 mole of N2O4, thus the equilibrium is restablished when more $\ce{N2O4}$ is formed. What doesn't make sense for me is that both the substances are contained in the same volume, so how 1 mole of $\ce{N2O4}$ would occupy less volume than 2 moles of $\ce{NO2}$? It only makes sense to think about partial pressures, not partial volumes, right?
I'm trying to use entropy and energy microstates to solve this problem, because it seems more fundamental than this mantra in the general chemistry textbooks. From quantum mechanics, I know (can't explain, just know; need further studying on this subject) that translational energy states become closer when a gas expands in a bigger volume, so I'm assuming that in the reaction above, with a bigger pressure (and thus a smaller volume) these energy states will be more spaced between each other. With energy states more far apart, why are there more possible energy microstates when $\ce{N2O4}$ is formed?
I can't get past this! Why does this implies in the formation of the substances with less gaseous coeficients? How are the energy states of reactants and products and their changes on pressure related with that?
Edit: to make myself clearer, I'll use this example from this textbook I'm reading:

Chem1 Virtual Textbook, Stephen Lower, link
In the endothermic reaction H–H (g) ⟶ 2 H (g), you only get hydrogen atoms at very high temperatures, because you'll have a higher distribution of populated states on the 2 H side, thus you favor the products at higher temperatures for endothermic reactions. I can completely understand the effects of themperature in Le Chatelier's principle, and I'm trying to understand an analogous thought for changes in pressure in the microscopic point of view.

Are you familiar with the concept of the Equilibrium Constant for a reaction?

Yes. I can see from the algebric point of view why it'll increase the concentration or partial pressure on N2O4. But the equilibrium constant comes from gibbs energy which comes from entropy, which is what I can't understand from the microscopic point of view.

It's an interesting questions, what happens on the microscopic scale. One point is that at elevated pressure (and identical temperature), the same species spends more time in close contact. Same speed, more collisions per time. I'm not sure if or when that is already enough for a quantitative explanation.

Which part about the entropy from the microscopic point of view can't you understand (1.  entropy change of a pure ideal gas with changes in pressure, 2.  entropy of mixing of ideal gases, or 3.  entropy change of molecules reacting to produce new substance)?

I think it's a mix of 1. and 3. as you change the pressure causing the equilibrium to be restablished by a chemical reaction.

Well the first step is understanding the derivation of the equilibrium constant, which means understanding the derivation of the Gibbs free energy for a mixture of ideal gases.

What the textbook means when it is saying one occupies more volume than the other is that the reactants (in this case) will occupy less volume than the product under standard conditions (~ atmospheric pressure). Therefore you have to use more force (pressure) to confine the products to the given volume than you need to do so for the reactants.
Therefore there is less enforcement on the reactants than there is on the products. Therefore it is energetically more favorable to be in the reactants-form.
In order to understand this a little better you can have a look at the ideal gas law:
$$
p = \frac{nRT}{V}
$$
As you can clearly see the pressure is linearly dependent on the amount of particles in the system. Given that during the reaction (in the way you stated it in your question) this amount increases (doubles if the reaction did run through completely). Thus the pressure increases as well. However a higher pressure is connected to a higher energy state of the particles in the "box" as they repulse one another.
Why don't the reactants stay as they are then as this is the state that produce the least pressure? Because by forming the products, energy is released ($\Delta_rG < 0$). Therefore the reaction can take place until the increase in pressure results in a "raise of energy" that is equal to the energy released during the reaction. Any further reaction would be energetically unfavorable.
And this point is the one that can be influenced by the volume. Let's say that the reaction can take place until the pressure reaches $1.5p_0$ and $p_0 = \frac{n_0RT}{V_0}$ (and for simplicity let's assume a isothermal reaction). This means that 50% of the the reactants can react until that limit is reached. If we increase the volume, this point is reached only after having a higher particle count than with a lower volume which is why the equilibrium shifts to the side with more gaseous particles if the volume is increased and vice versa.

That makes sense if you, like all of us, have inhaled Le Chatelier as the basis for explaining all chemical experience. I think OP isn't doubting the fact, but is asking "how do the molecules know"? ;-)

@Karl I tried adding an explanation. Hopefully it is correct (and I didn't miss something) and will help to clarify.

You misunderstood. I believe OP is looking for a molecular explanation. Molecules haven't heard of Gibbs. ;-)

Your explanation was great, but it's exactly what @Karl is saying. Also, I can understand ΔrG< 0 since the formation of N2O4 is exothermic, and even a decrease in the system's entropy would cause an increase in the universe's entropy at a certain temperature. But what if it were an endothermic process?

@Dante Rodella The whole Gibbs-thing is merely a "visualisation" for what's going on with the energy. This does apply to moles of particles just as well as to individual particles: They'll always try to minimise their energy (which can only be overcome of something else gets an even lower energy state instead). This is the way we experienced our universe to work and I'm afraid I haven't found any explanation as to why that is. As for the endothermic reaction: It doesn't matter whether the reaction is endothermic or exothermic. It only matters that it is exergonic.

A molecule's rotational states do not become closer as the pressure is increased, neither do the vibrational levels, they are a property of the molecules themselves. The population in these levels is determined by the temperature which in your case is constant.   More collisions per second as the pressure is increased means more possibilities of reaction. This means that there are more chances that at a given total energy (comprised of translational, rotational & vibrational energy) the molecules will have enough energy to cross the energy barrier (transition state) between reactants and products and vice versa, i.e the number of barrier crossings both ways is increased.
Note that this works both ways, a collision between two $\ce{NO2}$ can form $\ce{N2O4}$ as well as a collision destroying it and forming $2\ce{NO2}$, i.e. a collision between $\ce{N2O4}$ and $\ce{NO2}$ (or between two $\ce{N2O4}$ ) can sometimes convert $\ce{N2O4}$ into $2\ce{NO2}$.
The species are in equilibrium $\ce{N2O4 <=> 2NO2}$ and the equilibrium constant has a fixed value at constant temperature and is unaffected by changes in total pressure, because, as noted above, the population of energy levels depends only on temperature.
[Notes: The equilibrium constant can be expressed as the ratio of partial pressures. At equilibrium $\Delta G^\mathrm{o}=-RT\ln(K_p)$ where for a reaction $\ce{xA + yB<=>rC + sD}$ the equilibrium constant is $\displaystyle K_p= \frac{P_C^rP_D^s}{P_A^xP_B^y}$. The change in $K_p$ with temperature is given by the Van't Hoff isochore, $\displaystyle \frac{d\ln(K_p)}{dT}=\frac{\Delta H^\mathrm{o}}{RT^2} $ ]

In terms of reaction rates, it makes sense for me but only because the formation of N2O4 is exothermic. The difference in energy between the transition state and N2O4 is higher than the energy difference between the transition state and NO2, so both reaction rates will be higher as the pressure increases, but the formation of N2O4 will be favored only because the formation of N2O4 has a higher reaction rate constant. What would happen in an endothermic reaction? Does this make sense?

Also I read this in this textbook that I love: "The larger the volume in which the gas is confined, the more closely-spaced are these [energy] states, resulting in a huge increase in the number of microstates into which the available thermal energy can reside" link. And you're saying "A molecule's rotational states do not become closer as the pressure is increased, neither do the vibrational levels, they are a property of the molecules themselves." So is the textbook wrong or those things just don't relate?

Yes; the larger the volume the larger the number of translational energy states, think of this as the number of quantum levels as in a 'particle in a box'; the bigger the box the more closely spaced are the energy levels (density of states increased) and so at a given temperature more levels are populated. The rotational and vibrational states are unaffected by this.

Oh yes, you're right. I corrected exchanging "rotational" by "translational" in my question.

The question is about increasing the pressure by decreasing the volume, and that does change partial pressures, and changes the equillibrium, unless the reaction is pressure-invariant!

@Karl, yes the partial pressures do change and are related to one another by $K_p$ in this case as $K_p=P_{NO2}^2/P_{N2O4}$ . The degree of dissociation of $\ce{N2O4}$ $\alpha$ also changes as total pressure changes, for this example $K_p=4\alpha^2P_{tot}/(1-\alpha^2)$. (If you express the equilibrium constant in terms of concentration then this does change as pressure changes).


text:
Do Sodium Alkynides react with esters in the same way that Grignard reagents do?
We know that Sodium alkynides react with aldehydes and ketones like Grignard reagents. But do they behave the same way with esters?

I no longer have Scifinder access but I can find very few references to the alkynation of esters and the few possible refer to lithium alkynides.

Okay,  I guess they dont.

I don't see why they would not, but they just don't appear to have been used.

I'm not sure though. Grignard reagents are way stronger right?

I don't know about way stronger. I would not use sodium acetylide in ethyl acetate.

I don't understand. Why is that?

I think it would react