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Why does a strong base completely neutralize a weak acid?
For the longest time, weak acid-strong base titration curves completely baffled me.  I understood that the equivalence point of a strong acid-strong base titration is reached when equal numbers of moles of base titrant are added as there are acid analyte, but I didn't realize why the same is true for a weak acid-strong base titration; I incorrectly assumed that a strong base will not completely neutralize a weak acid when a stoichiometrically equal amount of base is added to acid - I thought that since a weak acid only partially deprotonates in solution, it is also only partially neutralized by strong base.  Now that I realize that is an inherently incorrect assumption, my question is why does a strong base fully neutralize a weak acid if only a small percentage of the acid is deprotonated in solution?  Is it an effect of Le Chatelier's principle?

Very short version: a weak acid does a very good job of protecting its acidic protons against removal by water molecules. But hydroxide ions are vastly better at stealing those protons. Look around this site: there are numerous relevant detailed answers or discussions.

A little more. Write the dissociation equilibrium for acetic acid and note that its dissociation constant is much less than one. Next write the auto-ionization equilibrium of water, with hydrogen and hydroxide ions on the left and water on the right. So the k for that is the reciprocal of water’s auto-ionization constant. Finally, add the two equilibria, cancel common species and then you see that the equilibrium constant is acetic acid’s k divided by water’s autoionization constant. Thus it is 100 trillion times larger and much greater than zero.

Equivalence point and neutral point are not the same thing. I think you know the difference but are mixing up the terms.

See the neutralization equilibrium of acetic acid and hydroxide ion, as per my second comment: https://chemistry.stackexchange.com/a/32781/79678 . But note that the k value is $10^{-4.75}$/$10^{-14}$ = $10^{9.25}$, not what the answer mistakenly gave. Still, a value above a billion, so the weak acid almost entirely gets converted to conjugate base.

The simple (high-school chemistry level) answer is that you are correct and it is indeed Le Chatelier's Principle. Consider ethanoic acid, $\ce{CH3COOH}$, which dissociates according to the equation $\ce{CH3COOH + H2O-> CH3COOH + H3O+}$. (If you haven't previously encountered hydronium ions, $\ce{H3O+}$, it's essentially just a somewhat more accurate way to represent $\ce{H+}$ ions in solution.)
As a strong base like $\ce{NaOH}$ is added, it will react with the $\ce{H3O+}$ ions in solution, so equilibrium for the dissociation of the weak acid will move to the right, forming more $\ce{H3O+}$ ions, which react with the strong base. This continues until the weak acid has entirely dissociated.
(Ed V's comment sets out a more quantitative explanation; depending on what level you're currently studying chemistry at, it may be more appropriate.)


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Reaction between copper (II)chloride and aluminium foil
When I reacted copper chloride and aluminum foil instead of getting the brown color I got green color. Can anyone share their procedure where the experiment actually succeeded?

This is the first one that popped up (using Google): https://melscience.com/US-en/articles/battle-metals-experiment/ . Make sure your copper chloride solution is not real dilute. Alternatively, use copper sulfate and sodium chloride. You will know it works when the reaction really gets going: it gets hot and very obvious!

You asked "Can anyone share their procedure where the experiment actually succeeded?" I did exactly that in my answer. Please consider accepting the answer.

This is an easy experiment to do at home, so I did it about an hour ago. The first figure shows the starting chemicals and "equipment":

I did not have any copper chloride, so I used copper sulfate pentahydrate, $\ce{ CuSO4.5H2O}$, and $\ce{NaCl}$. (The sodium and sulfate ions are just spectator ions here.)
I added the two chemicals to the $\pu{25 mL}$ of water in the little beaker, stirred until they were fully dissolved, then poured the solution onto the aluminum foil. The second figure shows what happens about every 2 minutes:

The third figure shows what happens around the 10 minute mark: the solution has eaten through the aluminum and the brown copper is evident:

Finally, Fig. 4 shows the result after the solution has leaked away:

So, the redox reaction works as expected and as shown in various videos on the web. Note that copper sulfate pentahydrate crystals are blue, as are aqueous solution of it. The copper chloro complex ion, $\ce{CuCl4^2-}$, is green.
EDIT: Per the comment from @AJKOER, I ran the experiment again, with powdered iron (Fisher, 100 mesh, electrolytic) and copper sulfate solution. The next figure shows the starting materials. I put a strong magnet under the little beaker to prevent iron, which was deliberately in excess, from going over in the decantation that would follow.

Pouring the copper sulfate solution into the beaker resulted in a vigorous reaction and quite a bit of heat. I stirred the reaction mixture and let it go to completion. The magnet still stuck strongly to the bottom of the beaker, indicating that there was still substantial powdered iron remaining.
The mixture was decanted to another beaker, with some mostly useless filtering (no good filter paper at hand), and allowed to settle for about an hour. The copper particles produced in the reaction are very small and settle out very slowly. Some of the supernatant solution was transferred to a sample cell, illuminated from the left via an LED flashlight and photographed. This is shown in the next figure:

Despite the light scattering, the green color of the ferrous sulfate solution is evident. The photo will be updated after more particulate has settled out.

Upgrade, nice experiment! However, employing CuSO4 implies if there is any Fe presence in your heavy-duty foil, per my analysis below, some green FeSO4 formation. This may also be present per your picture.

Good point, but the blue and green colors from the copper species make it hard to know, without a spectrophotometer at home, what the green ferrous ion contribution happens to be. I have a bottle of powdered iron, so I will do the before and after experiment, using an excess of iron powder to reduce the copper ion to negligible levels. I will add the pics here ASAP. Too bad the question is closed: I see value in actually doing simple experiments.

Good answer and fun experiment. I'm sorry I contributed to the closing but at that time, it was granted because no details on the question.

@MathewMahindaratne No problem at all! You are one of the best folks here, and that is saying something, given how really good the high rep members are! By thge way, I am still trying to get that cell cleaned! ;-)

My guess is that the heavy duty aluminum foil you employed is likely alloyed with iron. To quote a source:

Reynolds Wrap® Aluminum Foil is 98.5% aluminum. The balance is made up of other elements: primarily iron and silicon.

As such, your reaction likely proceeds as follows:
$\ce{2 Al + 3 CuCl2 (aq) -> 2 AlCl3 (aq) + 3 Cu (s)}$
$\ce{Fe + CuCl2 (aq) -> FeCl2 (aq) + 2 Cu (s)}$
Per Wikipedia on Aluminium chloride, to quote:

Aluminium chloride (AlCl3), also known as aluminium trichloride, is the main compound of aluminium and chlorine. It is white, but samples are often contaminated with iron(III) chloride, giving it a yellow color.

However, in the current case, the contamination is ferrous chloride which per Wikipedia:

FeCl2 crystallizes from water as the greenish tetrahydrate, which is the form that is most commonly encountered in commerce and the laboratory. There is also a dihydrate. The compound is highly soluble in water, giving pale green solutions.

Unclear on your targeted product, but you can repeat with a pure Al foil to remove the coloration.

source: Stack Exchange

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

sort: votes


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At what temperature does a polyester release toxic gases?
I've bought a "Shrinky Dinks" printable sheets and now I'm feeling unsure about using it since I've read that heating polyester materials may be very harmful for health, and the final purpose of this material is to melt it at about 325°F.
Searching on the web I've noticed that the information isn't that straightforward about the product. Even on the official website, where they do mention about having specific certifications, I cannot understand them fully since I'm not used to all the technical language they use.
That's why my question is about the temperature I can safely heat this material at.

Your title contains the question I want to know the answer to – about polyester – but your body is about “Shrinky Dinks”. According to Wikipedia, those are made from polystyrene, which is different from polyester.

https://www.ncbi.nlm.nih.gov/pmc/articles/PMC5793520/

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


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What is the IUPAC name of [Cr(H2O)5Cl]Cl2.H2O
What is the IUPAC name of $\ce{[Cr(H2O)5Cl]Cl2.H2O}$?
The net says that the correct name is Pentaaquachloridochromium(III) chloride.
But do we not have to account for the water molecule present outside the coordination sphere?

Yes, you should account for the water molecule outside the coordination sphere. In most cases, you do not need to do so, but if you are looking for specificity and/or want to describe the physical or chemical properties of this particular hydrate, it is advised to also mention the number of molecules of water of crystalliation.
The correct IUPAC name, in that case, would be Pentaaquachloridochromium(III) chloride monohydrate.
For additional hydrate naming rules and prefixes, check out Wikipedia

The formula [Cr(H$_2$O)$_5$Cl]Cl2.H$_2$O describes the crystal. It contains the cation [Cr(H$_2$O)$_5$Cl]$^{2+}$, twice the anion Cl$^-$ and one water molecule in it's lattice.
The IUPAC name you gave Pentaaquachloridochromium(III) chloride only describes the cation, so it does not take the anion and the water molecule.


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Why does raising the surrounding pressure raise the vapor pressure of a liquid?
What’s the qualitative explanation behind the fact that an isothermal increase in surrounding pressure raises the vapor pressure of a liquid? The quantitative expression relating surrounding pressure and fugacity is well-known, but I’d like a qualitative one.

The answer lies in that the chemical potential of the liquid increases with pressure.
Quantitatively, the chemical potential changes with a change in pressure from $p_1$ to $p_2$ as follows:
$$\Delta\mu(\textrm{l})=\int_{p_1}^{p_2}V_mdp$$
A qualitative explanation is possible by linking the change in pressure to the concept of chemical potential. The chemical potential difference is the amount of non-pV work (the "chemical" work) which the system can perform. In essence, increasing the pressure means more of the liquid has the potential to transfer into the gas state (the chemical work in question).   You might visualize the transferred molecules as being "squeezed out" of the liquid by the increasing pressure applied by their neighbours.


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Can a refractometer adequately measure sugar content in fruit pulp?
I am trying to measure the sugar content in mango pulp (used for food processing).
Using a hydrometer to measure specific gravity is not adequate because the pulp thickness prevent the hydrometer from adjusting its level as expected. As a work around I am diluting the pulp to a 1:1 ratio of water and mesure the gravity which I then multiply by 2, though strongly doubting this is an accurate methodology.
I am considering measuring the same with a refractometer, reading a proposed methodology suggesting to squeeze the pulp through a fine filter (cheese cloth) and using the extracted liquid to measure sugar content.
So I have few concerns:

How much solid is too much so as to skew the measurement with a refractometer?
Assuming the extracted liquid is not polluted by solids, how to ascertain this liquid is representative of the sugar content of the pulp (and that solids do not trap more sugar that would be omitted in the measurement)?

Just a comment: indeed your method of dilution alter the results. You should look at the sugar content of the diluted solution then recalculate back. Also in this case you run into the problem that some sugar is within the pulp anyway. I guess that even big industry evaluate a kind of "average“ or the desired value is set to the accessible part, ie the sugar in the aqueous phase. At least for processing purposes it should be enough. For a value to be put in a dietary info box it might be different and analysis should be required. I guess you can find this in a book for food analysis if you won

... won't get an answer here.


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Why is this grignard synthesis incorrect?

This reaction is mentioned as incorrect in my exercise book. Why is this so? My guess was that the Magnesium, instead of forming a grignard reagent with the compound and bromine, would instead form a complex with the N atom. However, wouldn't the complex formed be quite unstable due to the size of the ligand and the steric hindrance associated with it?
Is this reaction incorrect? If so, what reaction would take place here?

It is not about complexes at all. Grignard doesn't sit well with acidic protons like N-H.

@IvanNeretin I knew that it doesn't work with O-H but comparatively, just how acidic is the cyclic N-H compared to an O-H bond?

Much weaker than O-H, but still way too strong to do Grignard anywhere near it.

Another thing is that your compound wouldn't make particularly bulky ligand.

https://chemistry.stackexchange.com/questions/96881/reaction-of-grignard-reagent-with-primary-amines

https://chemistry.stackexchange.com/questions/98551/which-alkyl-halides-can-not-form-grignard-reagent https://chemistry.stackexchange.com/questions/49377/do-grignard-reagents-deprotonate-amines

Product would not form as intended, not proceeding in the path of synthesis usually seen with such a sequence of reagents.
This is due to the (quite) acidic hydrogen atom attached to the heteroatom nitrogen in the ring, which would be readily abstracted by the dense electron cloud on the initial Grignard complex that would form.
The species drawn are highly unstable and would have a really short half-life.

The nitrogen could be protonated subsequently. The desired product simply will not form because a reaction much faster than it will consume the formed reagent instantly.
The Grignard reagent fails to remain in a synthetically useful form.

Your reaction scheme isn't perfect, but, oh well... BTW there's direct reduction of amine hydrogen may be more important reaction then formation and destruction of Grignard reagent.

Well, it is possible that there are lapses in my knowledge (freshman year chemist here), but I'm not sure if piperidine can be reduced any further.. Even a Google search didn't yield anything useful. Could you elaborate?

It's just like reaction of alcohol with sodium - you get alcoholate and hydrogen.

@Mithoron Ahh.. makes sense. But is the H that acidic (N is additionally in a ring, not a simple amine)?

@WilliamR.Ebenezer For what it's worth,the H of a aliphatic alcohol isn't that acidic either(considering the high basicity of the RO- anion). So qualitatively speaking,I think I would give the H in the reagent here a pass as well for this reaction

@Yusuf Well I meant acidic enough for a metal to reduce H to H2 gas. OH is more acidic  than NH, as N- looks like a stronger base than O- (electronegatively speaking). But again, I guess talking these properties is not helping, as the reaction is in heterogeneous phase and in ethereal medium.

@WilliamR.Ebenezer Yeah,I guess both of our answers should be edited with Mithoron's comment to show that this highly simplified picture isn't the complete truth

I would just like to throw in some more perspective here, although your main question has been answered.
First, let's see a postulated mechanism of formation of Grignard reagent in the presence of Mg and diethyl ether.

source
The tentative problems that would may arise by the presence of the N-H bond would be:
1)The R radical made in the first step might begin a homolysis reaction with  the H of the N-H, and thereby forming a kind of amino radical radical. This may lead to further erroneous coupling in the synthesis.
2) Like the answer and comments above state, the major reason for the failure of this synthesis is the reaction of the highly basic Grignard reagent with the acidic H after formation.
3) Non-polar solvents like diethyl ether stabilize the Grignard reagent formed like so:

The N-H bond will have significant polarity due to difference in electronegativities of N and H, so a lot of ether will be expended in stabilizing this bond, so again, it might hinder the formation of the Grignard reagent
Note: This answer is more like food-for-thought on this issue as far as point 1 and 3 are considered, the clear-cut answer is in point 2.
P.S. As pointed out in the commnts by Mithoron, the reaction happens in a heterogenous state, so the mechanism cited above and all these arguments are highly simplified ways of looking at a picture perhaps not completely understood. This reaction probably bears more similarity with that of sodium and an alcohol

I found the first point quite satisfying. Good answer!

@WilliamR.Ebenezer And I didn't. There's no atomic Mg in solution and no radical flying around to react.

@Mithoron I based this answer on the observations given in the quoted article,and the possible side reactions that can occur. From what I saw,I didn't really talk about atomic Mg,but the R• generated does seem like an independent species

@Mithoron Can you elaborate on the actual intermediates then?

Reaction is heterogeneous - reagent adsorbs on Mg surface. Br, or in this case probably mostly N, coordinates with surface atoms  and reduction occurs in bound state. Also @WilliamR.Ebenezer

@Mithoron I see your point..

@Mithoron Ok,so the article I cited is somewhat wrong or lacking in info I guess?

It's simplified and schematic, that's unsurprising, because mechanisms of heterogeneous reactions are poorly known. I think nobody knows what exactly happens on the surface there. I personally doubt it's SET.


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Clarification of textbook concepts relating to "perfect", "ideal", and "real" gases
My textbook, Atkins' Physical Chemistry, 11th edition, by Atkins, de Paula, and Keeler, says the following:

The resulting expression
$$pV = nRT \tag{1A.4}$$
is the perfect gas law (or perfect gas equation of state). It is the approximate equation of state of any gas, and becomes increasingly exact as the pressure of the gas approaches zero. A gas that obeys eqn 1A.4 exactly under all conditions is called a perfect gas (or ideal gas). A real gas, an actual gas, behaves more like a perfect gas the lower the pressure, and is described exactly by eqn 1A.4 in the limit of $p \to 0$. The gas constant $R$ can be determined by evaluating $R = pV/nT$ for a gas in the limit of zero pressure (to guarantee that it is behaving perfectly).
A note on good practice Despite 'ideal gas' being the more common term, 'perfect gas' is preferable. As explained in Topic 5B, in an 'ideal mixture' of A and B, the AA, BB, and AB interactions are all the same but not necessarily zero. In a perfect gas, not only are the interactions all the same, they are also zero.

My questions here relate to this part:

A note on good practice Despite 'ideal gas' being the more common term, 'perfect gas' is preferable. As explained in Topic 5B, in an 'ideal mixture' of A and B, the AA, BB, and AB interactions are all the same but not necessarily zero. In a perfect gas, not only are the interactions all the same, they are also zero.

It seems that the authors here are implying that a "perfect gas" and an "ideal gas" are synonymous, but my research indicates that they are actually different concepts. Are these two concepts the same, or are they different? How should one think about (distinguish) these two concepts?
What do the authors mean when they say that "in an 'ideal mixture' of A and B, the AA, BB, and AB interactions are all the same but not necessarily zero"? And what is meant by the interactions being "zero" here?
What do the authors mean when they say that "in a perfect gas, not only are the interactions all the same, they are also zero"? And what is meant by the interactions being "zero" here?

I would greatly appreciate it if people would please take the time to clarify these points.

The authors attempt to explain that in many sources the term "ideal gas" is used in place of "perfect gas" to indicate a gas following the ideal gas law and which has the property that the molecules do not interact. They equate the absence of an intermolecular interaction with the statement that the "interactions are zero". In other words, there is no intermolecular potential between the idealized gas molecules in a "perfect gas". These idealized molecules lack a volume, so that even steric (excluded volume) interactions expected for hard spheres are absent. They are "point particles".
The authors further explain that it is preferable to use the word "perfect" rather than "ideal" because the word "ideal" is used in other contexts where the condition of no interaction is not observed, ie during ideal mixing. Ideal mixing occurs when different molecules have the same volume and intermolecular potential in the pure and in the mixed states.

Opinion-based aside
BTW I've always been partial to the term "ideal gas". IMHO the term "perfect" seems to imply that the molecules excel in some way, whereas "ideal" simply means that it is the product of human minds, an abstraction, and not necessarily a real thing.
IUPAC does not to have any qualms about continued use of the term "ideal gas", so I don't see reason to stop using this term.

A search for the term "perfect gas" and "IUPAC" leads straight back to IUPACs definition of "ideal gas".

Thanks for the answer, Buck. I'm guessing you're referring to 1A.4 as the "ideal gas law"? The authors actually explicitly label this as the "perfect gas law".

Yes. If you search for the "perfect gas law" online you will obtain a lot of hits for "ideal gas law" (including for the wikipedia) and a few for "perfect gas law". Another opinion-based aside: Atkins et al are heavyweights in the pchem education world but are on what I would consider a quixotic quest in attempting to alter terminology preferences. There are plenty of confusing topics in pchem and I don't think this one merits so much special attention, although it is good to bring attention to the fact that "ideal" can mean different things in slightly different contexts.

Interesting. Thanks for the clarification.

I'm old, but in 2 semesters of intense Chemical Engineering thermodynamics (back in the 1970s), I never once heard of a perfect gas - only ideal gases.  Ideal gas is part of the standard assumptions you make when solving most problems.  You have to get pretty from standard conditions before pV = nRT breaks down.


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What is enthalpy? Why do some sources define it as "heat content" of a system?
So after researching thoroughly on multiple different sources about enthalpy (I'm not talking about the equation but the general concept), many credible sources seem to all contradict themselves.
For instance in a textbook published by Pearson, they define enthalpy as "the total heat content of a system". And other sources say that enthalpy is just the "total energy of a system", which would make it equal to internal energy?
I already know that heat is only energy in transit so cannot exist independently, so saying "heat content of a system" does not make sense since no system can "contain" heat.
Please clarify! Thank you in advance.

Enthalpy is just U + PV.  In practice, it is often a convenient combined physical property of a material to work with.  In my experience, trying to assign fundamental physical significance to enthalpy causes nothing but confusion to the student.

http://goldbook.iupac.org/terms/view/E02141

The answer on a different forum is "Arrgh!": https://qr.ae/T3Ex4r

Enthalpy is a state function, meaning it is a property of a system, and not a process (like heat or work). It is defined as
$$H = U + PV,$$
where $H$ is the enthalpy, $U$ is the internal energy of the system, $P$ is the ambient pressure (assumed to be constant), and $V$ is volume.
Why is it defined in this way? Consider a gas in a closed box. The question is: how much energy does it take to create this system? The gas has internal energy $U$, which is the sum of the microscopic kinetic and potential energies of its constituent particles. But that's not the only energy it would have taken to create the system. We would have also had to do pressure-volume work to create the volume of the gas. Assuming the pressure $P$ is constant, and that we started from an empty volume ($V=0$), the work $W$ is equal to $PV$. So the minimum amount of energy necessary to create this box would be $U + W = U + PV$, which is what we call enthalpy
Why is it sometimes called the "heat content" of a system? Well, as with many quantities in thermo, we are often more interested in how it changes than its exact value. Consider the change in enthalpy
$$\Delta H = \Delta U + \Delta (PV) = \Delta U + P \Delta V$$
(since pressure is constant). The first law of thermodynamics states that, for a closed system, $\Delta U = W + Q$, where $Q$ is heat. The system does work $W = -P\Delta V$, which is negative since the system is doing work against its surroundings. Plugging these equations togethers, we get
$$\Delta H = W + Q + P \Delta V = -P \Delta V + Q + P \Delta V = Q.$$
So the change in enthalpy for a system is simply the heat added to the system (when all of our assumptions hold).


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An ideal gas is undergoing an isothermal expansion against a constant external pressure. Which of the following statements is true?

An ideal gas is undergoing an isothermal expansion against a constant external pressure. Which of the following statements is true?
1) This process is reversible.
2) This process is irreversible.
3) This process is exothermic.
4) The system pressure is always in equilibrium with the external pressure.
5) The information given is insufficient to describe this process.

(source)
I don't get how the answer can be 2, I thought it would be 5. Since it's expansion, $\Delta S$ of the system must be > 0. Now since it's an expansion, work must be < 0. If work is less than 0, q > 0 due to the process being isothermal. So that means $\Delta S$ of the surroundings < 0 since it is -q/T. That means $\Delta S$ of the universe could be > 0, < 0 (non spontaneous) and = 0 depending on the magnitude of $\Delta S$ of the system and surroundings. As a result, I choose 5 since we don't know the magnitudes of both $\Delta S$.
Why would it be 2?

If the gas is ideal, for an isothermal expansion
$$p=\frac{nRT}{V}=\frac{\textrm{constant}}{V}$$
As the volume of the system changes, so must the pressure.
Now a condition of reversibility is mechanical balance between the system and surroundings, that is, $p=p_{\textrm{ext}}$. However, if a process is performed at constant external pressure and temperature, then as shown above the condition of mechanical balance cannot be satisfied, since $p$ will change during the expansion whereas $p_{\textrm{ext}}$ is constant. Therefore the process must be irreversible.

If the temperature and pressure are constant, then the change in entropy of the system is $$\Delta S_{syst}=nR\ln{(V_2/V_1)}$$For constant temperature, the change in internal energy is zero, so the heat transfer from the surroundings is $$Q=P_{ext}(V_2-V_1)=\frac{nRT}{V_2}(V_2-V_1)=nRT\left(1-\frac{V_1}{V_2}\right)$$It then follows that the change in entropy of the surroundings is $$\Delta S_{surr}=-\frac{Q}{T}=-nR\left(1-\frac{V_1}{V_2}\right)$$So the total entropy change of the system and surroundings in this process is:
$$\Delta S=-nR\left[\ln{(V_1/V_2)}+\left(1-\frac{V_1}{V_2}\right)\right]$$
This entropy change is positive for all non-unity values of the volume ratio.  The process is therefore irreversible.


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Why can't Carbon Dioxide be liquid at 1 atm?
Wikipedia says that Carbon Dioxide cannot be liquid under atmospheric pressure.
I thought the phase of an element was related to both pressure and temperature.
If that is so, then should you not be able to set the temperature conditions such that you could have a cup of liquid Carbon Dioxide?

The other answer already shows that this is a solid fact. Carbon dioxide is not (actually cannot stay) liquid at 100 kPa, period.
Why that is the case is much less obvious. I cannot give you a clear, rigorous, quantitative explanation.
But the general reasoning about this and similar observations is that once the crystal structure has broken up under increasing temperature, there is very little that can keep the molecules together.
The linear CO2 molecules are unpolar from a distance, but each bond is quite polar. They have to align carefully to have enough attraction, in the solid. Now if the crystal structure breaks up, all the outer oxygen atoms see are other oxygen atoms, with a negative partial charge. So the condensed phase just blows up.
Only with a significant outside pressure, the molecules in the liquid can jump from one attractive position to another without boiling away.
Naphtalene is a similar case: The molecule is rigid and has a high aspect ratio. It needs a lot of energy to be able to move in the condensed phase at all, but once it is able to turn a bit, it looses 90% of its contact area (for vdW bonding) with its neighbours, and takes off.
Liquid water is kept together by hydrogens bonds, which can break and reform extremely fast. Nitrogen molecules don´t care how their neighbours are aligned, neither does methane or ethane. Alkanes are flexible additionally. They all can move in a liquid phase and still keep up enough interaction.

This is the phase diagram from a better source:

The pressure axis is the vertical one at the left. One bar (which equals 0.986923 atmospheres) is near the bottom, as shown on the diagram. So carbon dioxide is a gas at all temperatures above −78.5 °C when the pressure is one bar. Solid carbon dioxide (dry ice) sublimes at one bar (about one atmosphere), i.e., goes directly from solid to gas.
If you want liquid carbon dioxide, you need a temperature, pressure pair in the liquid region, e.g., −40.0 °C and 20 bar.

OK, so the idea is that it goes from gas right to solid?  It skips the liquid phase at 1 atm?

Below -78.5 degrees Celcius, at one atmosphere, carbon dioxide is solid. Above that temperature, still at one atmosphere, it is gas. There is no liquid phase at one atmosphere. One bar equals 0.986923 atmospheres.

@Vaccano All stable enough substances have a phase diagram for 3 phases. If a line of a given pressure or temperature does not cross the area of the phase of interest, that phase is not at given p/T thermodynamically stable. It can occur temporarily at sudden condition changes, until it transforms itself to other phase.


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Allylic Rearrangement in formation of Grignard Reagent
Is an allylic rearrangement possible in formation of Grignard reagent? Going by the following single electron transfer mechanism, I think it's possible. But, I haven't found any references.

The discussion in the comments here might help:https://chemistry.stackexchange.com/questions/128087/why-is-this-grignard-synthesis-incorrect

Going through the comments, I feel the above SET mechanism might be incomplete. But, I would like to know if any product spread has been observed. References would be helpful.

I do not have references to the rearrangement of Grignard reagent going through single electron transfer mechanism during its formation. However, it is understand that Grignard reaction of an allylic magnesium halide with an electrophile can give a new bond at either carbon of the allylic carbanion, so two isomeric products are possible (Ref.1 & 2). Once formed, allylic magnesium halide is going to be in fast equilibrium with its other isomer.
For example, when generated from either (E)- or (Z)-bromide, (E/Z)-2-buten-1-ylmagnesium bromide (1; $\mathrm{R'} = \ce{CH3}$) would be in fast equilibrium with its other isomer, (2R/2S)-3-buten-2-ylmagnesium bromide (2; $\mathrm{R'} = \ce{CH3}$), favoring 1 in this case (Refs. 1-3). The authors of Ref.3 claims that regardless of starting material (whether it is crotyl bromide or $\alpha$-methylallyl bromide), the resulting major Grignard reagent is 1. When this 1/2 mixture is reacted with a ketone, two isomeric alcohols, 3 and 4, would formed via intermediates I and II:

Increasing the steric bulk of $\mathrm{R'}$- and $\mathrm{R}$- destabilizes I compared to II, favoring 4 over 3 (Ref.1 & 2). For example, when $\mathrm{R'} = \ce{-C(CH3)3}$ (tert-Bu-), the product ratio of 4/3 is $8$ if ketone used is $\ce{Et2C=O}$ (diethylketone) while it is $\gt 100$ if ketone used is $\ce{((CH3)2CH)2C=O}$ (diiso-propylketone; Ref.1). This second instance can be compared to the ratio obtained for the reaction when $\mathrm{R'} = \ce{CH3}$, which is equal to $0.5$.
The allylic Rearrangements have also discussed in earlier paper by Roberts and Young (Ref.4). The suggested mechanism for this rearrangement is discussed in details in Ref.5.
References:

Marc Chérest, Hugh Felkin, Claude Frajerman, “The reaction between ketones and t-butylallyl magnesium bromide. A sensitive measure of steric hindrance in the neighbourhood of the carbonyl group,” Tetrahedron Letters 1971, 12(5), 379-382 (https://doi.org/10.1016/S0040-4039(01)96446-9).
Michael B. Smith, In Organic Synthesis; 3rd Edition; Academic Press: New York, NY, 2011 (Hardcover ISBN: 9781890661403).
J. Eric Nordlander, William G. Young, John D. Roberts, “Nuclear Magnetic Resonance Spectroscopy. The Structure of Butenylmagnesium Bromide,” J. Am. Chem. Soc. 1961, 83(2), 494-495 (https://doi.org/10.1021/ja01463a062).
John D. Roberts, William G. Young, “Allylic Rearrangements. XVI. The Addition of the Butenyl Grignard Reagent to Some Simple Carbonyl Compounds,” J. Am. Chem. Soc. 1945, 67(1), 148-150 (https://doi.org/10.1021/ja01217a051).
Hugh Felkin, Claude Frajerman, “Competitive rate studies for some reactions involving the allyl, butenyl, $\alpha,\gamma$-dimethylallyl, and propyl Grignard reagents. the mechanism of the reactions of allylic Grignard reagents with electrophilic substrates,” Tetrahedron Letters 1970, 11(13), 1045-1048 (https://doi.org/10.1016/S0040-4039(01)97903-1).


text:
Strength of weak acids and bases: Is methane better base than ammonia?
$\mathrm{p}K_\mathrm{a}$ of methane is $\sim 50$ and that of ammonia is $\sim 37,$ thus ammonia is a better acid than methane, which implies methane is a better base than ammonia.
But I don't see how: methane doesn't have any lone pair of electron, neither it has any $\ce{OH-}.$ How is it possible?

What you have stated implies that H3C- is a stronger base than HN2-

@Ivan can you explain "Acidic and basic strength of a compound are unrelated". I thought a strong acid is a weak base and vice versa, do you have some explanation article, data, examples, anything to elaborate.

better is an adjective that is very subjective and therefore inappropriate here

@user251573 No. The conjugate base of a strong acid is not just weak but very weak. https://chemistry.stackexchange.com/questions/115061/why-do-weak-acids-like-ethanol-not-act-as-a-strong-base-in-water/115075#115075

It is generally also true that strong acids are weak bases. HCl loves to give its proton, and hates to take up a second one. But that is not the crucial point of that elementary undergrad lecture about acids and bases which you seem to remember partially. ;)

@Karl The conjugate base of a strong acid is a spectator ion, i.e. it has no conjugate base.

@KarstenTheis For all practical purposes, your "spectator ion" and a very weak base are the same thing. Saying that a strong acid has no conjugate base is however just nonsense.

There are two contradictory conventions in widespread use as to whether the conjugate of a strong acid is referred to as a "weak base" or not.  In Convention I, the conjugate of a strong acid is indeed a weak base.  In Convention II, the conjugate of a strong acid is considered too weak to qualify as a base, so if a source says "the conjugate of a weak base is a weak acid", they are using Convention II.

You got it wrong. Acidic and basic strength of a compound are unrelated, about as much so as the man's name and his weight. Within the definition you and I are currently using (Brønsted–Lowry), acidity is about having a proton on a polar bond ready to ionize, while basicity is about having electron pairs ready for accepting a proton. A compound can have one and not the other, or vice versa, or both, or neither. Why should the two be related? They shouldn't and aren't. Well, there is one obvious exception: a strong acid can't be a strong base at the same time, otherwise it would turn upon itself. Except that, everything is possible.
Indeed, methane is a very weak acid, about 13 orders of magnitude weaker than ammonia (as attested by the difference in their pKa). In fact, it is one of the weakest acids among all compounds. This is not unexpected, given its almost non-polar bonds and the lack of stabilization in the anion. Also, methane is a very weak base. Forcing it to accept a proton was no small achievement in itself (see Methanium). This is not unexpected either, given its lack of lone pairs or negative charge.
Now you must be thinking that I am pulling your leg, because every kid knows that strong acid means weak base and vice versa. It is written in every textbook, said by every teacher, and that with great confidence, too. Can they all be wrong? No, they are right. How so?
Here's the trick: strong acid does indeed mean weak base, but not in the same compound. Look again at that equilibrium:
$$\ce{HA <=> H+ + A-}$$
See that? It is $\ce{HA}$ that acts as an acid here. And now look at the reverse reaction: it is $\ce{A-}$ acting as a base. Now if the equilibrium is shifted to the right, then $\ce{HA}$ is a strong acid, which inevitably means that its conjugate base $\ce{A-}$ is a weak base.
So it goes.

"strong acid means weak base" should be "stronger weak acid means weaker conjugate base", which is a mouthful. It is easier to say (once these are introduced) larger Ka means smaller Kb of the conjugate base.

Also, one must be careful about the vice-versas. A strong acid has a weak conjugate base by simple mathematics and a strong base has a weak conjugate acid also by mathematics. But the conjugate acid of a weak base can be either weak or strong, as is the case for the conjugate base of a weak acid.

When A is considered a stronger acid than B, it does not follow that B is a stronger base than A.  Rather, the appropriate base comparison is between the conjugate bases formed from acids A and B.  Properly, the conjugate base of B is a stronger base than the conjugate base of A.
In terms of the compounds given in the question, the conjugate base of ammonia is amide ion ($\ce{NH3<=>H^+ + \color{blue}{NH2^-}}$) and the conjugate base of methane is methide ion ($\ce{CH4<=>H^+ + \color{blue}{CH3^-}}$).  Therefore, if ammonia is a stronger acid than methane then methide ion is a stronger base than amide ion, regardless of how basic the initial compounds methane and ammonia might be.  Which is true, as anyone knows who has tried to combine a Grignard reagent with a compound having hydrogen bonded to nitrogen.


text:
What is the oxidation state of Boron in Ammonia Borane?
So I was searching about the difference between redox reactions and Lewis Acid-Base reactions and came across classical example of latter,
$$\ce{NH3 + BH3  \rightarrow NH3BH3}$$
Now in this example what is the oxidation state of borane? Is it $+4$, if yes then can this example be considered a type of redox reaction also, since boron went from $+3$ oxidation state in the products to $+4$ oxidation state?

Oxidation states are human convention and not a fact of nature. With that in mind, I think it is +3, so no redox.

As per this source, the idea and definition of oxidation state is based on the following principle:

The oxidation number of an atom in a molecule is based
  on a formalism that forces a covalent compound to possess
  complete ionic character and may be defined as the charge
  that an atom would have if all bonds to it are broken such
  that the ligands retain a closed-shell configuration ;
  an exception, however, refers to homonuclear bonds, in which
  case the bond is broken homolytically and a single electron
  is transferred to each atom.
The oxidation number may
  thus simply be expressed as
  Oxidation number = charge on compound - charge on ligands

So,evidently, for the presented molecule $\ce{NH3BH3}$, we see that $\ce{NH3}$ is a ligand for the $\ce{BH3}$ moiety. So,you can heterolytically cleave the dative bond between $\ce{N}$ and $\ce{B}$ towards $\ce{N}$(as per the electronegativity trends),leaving no charge on the $\ce{B}$ as of now,as after this action,it's valence shell contains 3 electrons.Now,for finding oxidation state on $\ce{B}$,an interesting thing crops up due to the presence of the three $\ce{B-H}$ bonds.
The article further enumerates:

In many cases, the charges assigned to simple monoatomic
  ligands do not vary from compound to compound, as illus-
  trated by $\ce{F-}$, $\ce{Cl-}$, and $\ce{O^2-}$. However, a notable exception is
  provided by hydrogen for which both $\ce{H+}$ and $\ce{H-}$ have per-
  missible closed-shell configurations ($\ce{1s^0}$ and $\ce{1s^2}$
  , respectively).
  In this case, the charge assigned to hydrogen is determined by
  the relative electronegativity of the atom to which it is attached.

So again, due to sligthly higher electronegativity value of $\ce{H}$ than $\ce{B}$,the $\ce{H}$ atom becomes the ligand for the $\ce{B-H}$ bond. Hence,all the $\ce{B-H}$ heterolytically cleave towards $\ce{H}$,each cleavage leading to a +1 charge on $\ce{B}$ and -1 on $\ce{H}$. In totality,$\ce{B}$ ends up with +3 oxidation state, as it has lost all three electrons from it's valence shell.
Note: I highly recommend everybody to read that paper cited above. It's really insightful into the difference between valence,oxidation number and coordination number,which are often used interchangeably

Don't count bonds.  Count electrons.  Here all the bonds to boron are polarized away from that atom as boron is less electronegative than both hydrogen and nitrogen.  Since the boron also has no valence-shell lone pairs we count zero valence electrons dominated by the boron, versus the neutral atom having three.  That drop from three valence electron to zero means an oxidation state of $+3$.
To reach $+4$ the boron would have to engage another electron in bonding to a more electronegative element, but that electron would have to come from the $1s$ core and it ain't happening.


text:
Are there plastic materials that turn into ceramics?
Are there other materials that like clay, are plastic (as in they can be shaped by hand) but set to form a hard ceramic, especially ones harder than the fired clay?

Basically all ceramic materials match that description.

The set of compounds you are looking for are known as Thermosetting Polymers. A few examples of these are:

Bakelite: This is used for making switches and electrical enclosures
Melamine: This is commonly used for making food grade and microwaveable utensils and is an alternative to ceramic (at least in the food packaging industry)

You might also want to take a look at Thermoplastic Polymers, as they behave a bit like wet clay (they become soft on heating and harden on cooling and this can be done as many number of times as required)

Hmm,but organic polymers can hardly compete with ceramic hardness.

"edible" means "can be eaten and digested, without harming your health"


text:
Different force constant in ground vs excited state in spectroscopy
The force constant of iodine in the first excited state is $k=41.78$ N/m, and in the ground state it is $170$ N/m.
I know that the formula for calculating $k$ is:
$$ k= (\frac{ΔE_{vib} *2π}{h})^2 µ$$
What causes this difference?
Is it that the energy difference for the ground state is greater then that of the excited state and so the force constant is also greater?

Likely due to the description of a harmonic oscillator not being as accurate as a Morse (or similar) potential, but you could be looking at something attributable to the Franck-Condon effect if the vibrational transition is accompanied by an electronic one. Where are your force constant data from?

@ToddMinehardt The excited state force constant was determined from experimental data from a lab we carried out ourselves, and the ground state force constant was given to use for comparison.

A bond in the excited state is typically weaker (and longer) than in the ground state, especially in diatomic molecules.

@TAR86 If it is weaker, does that mean that the vibrational energy difference is smaller?

Rearranging the equation you provided should show this.

@TAR86 I calculated the vibrational energy difference for the two and the one for the ground state is almost twice as big as the energy difference for the excited state. But then shouldn't the force constant for the ground state also be twice as big?

Yes ! The energy difference for the ground state is greater then that of the excited state and so the force constant is also greater

Notice that the transition energy (or frequency) is related to the square root of the force constant, not linearly.

At the risk of providing a somewhat circular answer, in the harmonic approximation of an interatomic potential, the force constant is a measure of the force of attraction between the atoms (in a classical description they would obey Hooke's Law $|\vec{F}|=-k(x-x_\circ$)). The answer to the question is therefore that the attractive interatomic force in the electronically excited state of $\ce{I2}$ is weaker than in the ground state.
The following figure from Ref. 1 illustrates the potentials in the ground and excited states:

The potential in the excited (upper) state is broader. In the Morse functional approximation, the potential has the form $$V(r)=D_e\left(1-e^{-\beta(r-r_\circ)}\right)^2$$
where the width of the potential is proportional to $\beta$ and $$\beta = \left(\frac{k}{2D_e}\right)^{1/2}$$
A smaller force constant results in a smaller value of $\beta$ and a broader potential well. This fits nicely with the results noted in the OP and the above figure (the equations are also outlined in the accompanying text). The energy accompanying a vibrational transition is proportional to the square root of the force constant and therefore inversely proportional to the width of the potential well. This is analogous to the behavior observed for a QM particle in a box, wherein the spacing between energy levels narrows as the box size increases. Working backward from vibrational transition energies leads to the conclusion that smaller transition energies are associated with greater oscillatory amplitudes (broader potentials) and smaller harmonic force constant estimates.
References

Bayram, S.B. and Freamat, M.V., A Spectral Analysis of Laser Induced Fluorescence of Iodine. arXiv eprint 1507.02600, 2015.


text:
SMILES vs. Graph representation in deep learning
I have been reading papers on machine learning and deep learning methods for learning molecular space and generating molecules. These methods use different representations of the molecules. The most popular ones in the field include SMILES and graphs [e.g. this and this]. I have seen a shift of interest from SMILES representations to Graph molecular representations in the past couple years. I was wondering what are possible benefits of graph representations over SMILES?
I can think of two reasons:

The uniqueness issue with SMILES: Two SMILE strings might correspond to the same molecule.
SMILES are abstract representations while graphs are more natural representations (although, I can't really see this. I mean, I see what it
means, but doesn't SMILES also contain the same information?)

EDIT:
some paragraph that I found online and I need interpretation for:

it is focused on molecules whose bonds fit the 2-electron valence
model
it handles a limited array of stereochemistry types
there is no standard for handling aromaticity
there is no standard way to generate a canonical representation.

are any of these corrected in graphs? From what I know:
For 2: In graphs, it takes more than just defining atoms and bonds to account for stereochemistry. For example, I should also define chirality for each node.
For 3: While in some graph representations, aromatic bonds have their specific type, I found it better to explicitly use the order of the bonds.
For 1 and 4 I don't know if graphs are any better.

It looks like the 2nd paper about graphs makes some arguments about why they are better (SMILES originate from graph representation anyway, aren't robust to small errors).

Thanks, I missed that. But their arguments are not that strong. For example, at the end of the day, you need to use some sort of chem package, like RDKit, and that's when you will eventually turn graphs into smiles and eventually mol objects, plus that IMHO the overhead is really negligible.

Also, I think when the output is categorical, it won't be robust to small perturbations anyways, i.e. small changes in bond and node classes make output graphs invalid. Think of CO2 graph, now shift every class by one (i.e. instead of selecting C from {C, O, N, F, null} select O and instead of selecting double bond from {single, double, triple, null} select triple) it will be ON2 where O and N have triple bonds (and also a single bond between 2 Ns!).

Please add a citation to the quoted paragraph. If you find a way to canonicalise your representations, then you can fix the problem with aromaticity, too. (It it could be conceptualised accurately in the first place.) You might even be able to fix the stereo chemistry. (That shouldn't be too hard, but it needs to be consistent.) I'm pretty pessimistic about going away from the two electron bond. In the end a string or a graph contain too little information, so you just have to choose which one fits your needs best.

I agree that there seems to be a shift towards using graph representations over SMILES strings. I personally think this is a good thing, and I'll try to explain why, but even if there's nothing inherently better about graph representations of molecules, there is a very practical reason people are moving towards graph representations.
So, first of all, as you point out, both representations suffer from an issue in that they are necessarily a reduction from the complete amount of information needed to specify a molecular structure. For instance, neither a graph nor a SMILES string could distinguish between two isomers which differ only in the direction in which a free $\ce{O-H}$ group points. In fact, a bare graph cannot tell the difference between the boat, chair, or planar hexane molecules as they all have identical connectivity. Nonetheless, a graph is a much more flexible representation than a SMILES string because it is quite common to add weights to edges, which could be bond distances, or add parameters to nodes, which could describe angles to other nodes. So, you can include some information which is in the Cartesian coordinates, without having the problems associated with deriving your molecular representation from Cartesian coordinates.
This hints at the very practical reason why people are using graph representations. Basically, graphs are used in machine learning models in many fields other than chemistry. So, it's very nice to not have to reinvent the wheel, especially when people have so much success with things like transfer learning, where you just take a pre-trained model and re-train it for your own purposes.
Also, very often the first step in training a neural network is making some transformation to your data. For instance, graph convolutional neural networks have been successful in many tasks, so why not just use a convolutional filter on your graph representation of a molecule? You could do this with a SMILES string, but you would probably first just transform the string into something resembling a graph.

As to your specific points about chirality and aromaticity, etc. All of this information can be attached to a graph via parameters belonging to each node, although I would personally avoid giving information that isn't strictly necessary. That is, there is nothing special about a bond which is in an aromatic ring. You need to provide enough data that the model can learn about this on its own. If you tell it that bonds in aromatic rings are special enough to get another parameter, this is probably going to bias the model in some unforeseen way. Chirality is easily handled by a simple parameter attached to each node.
Ultimately, though, the representation depends a lot on the problem you are trying to solve. For instance, if you are trying to learn a representation of the potential energy surface, then graphs can work quite well. What is more common perhaps, is to use so-called atom-centered symmetry functions. In this case, the actual features are abstract vectors which are guaranteed to have the relevant symmetries and smoothness needed in the potential energy surface.
If you're doing something more like a classification problem, then using a representation like a SMILES string might be perfectly suitable.

TL;DR
Graphs are a more flexible representation which are commonly used in fields outside of chemistry. Hence, being able to draw on the knowledge of other disciplines is a huge plus, especially when you're collaborating with computer scientists who know a lot about machine learning and nothing about chemistry.

You mentioned that "neither a graph nor a SMILES string could distinguish between two isomers . . .". Is this also true for the conformers of a single molecule?

@Blade Yes. The example I gave is a conformer. Both can distinguish between isomers where the actual connectivity is different. For instance, the boat and chair isomers of cyclohexane have the same graphical representation.

It depends on how you code your molecular graphs
The idea of a 'connection table' or valence model for molecules, and thus molecular graphs is embedded in chemical thinking.
Let's take your four points:

It's possible to design connection tables that supports a variety of interactions. For example, zero order bonds can encode coordination bonds, delocalized metal-ligand interactions, etc.
You can write a graph that stores a variety of stereochemistry, although admittedly axial chirality, etc. require some work to do so (i.e., it's a property of the molecule itself and not any particular atom or bond). Some formats even support concepts like '55% R, 45% S' stereo centers.
You would need to define an aromaticity model, although many exist and can be adopted (e.g., 'we use the SMILES aromaticity definition')
For both graphs and SMILES, there are many published canonicalization algorithms (e.g., we use the InChI canonical atom order).

In short, people have worried about cheminformatics issues for a long time:

Dietz 1995
Gasteiger 1997

Both papers indicate expansions of the standard molecular graph concepts, e.g. [Gasteiger]:

This representation overcomes the limitations of connection tables
  designed to only represent chemical structures with bonds localized
  between two atoms. The representation introduced is based on the
  separation of the σ- and π-electrons of bonds and the delocalization
  of electrons also across more than two atoms. It also allows the
  description of chemical compounds containing multicenter or
  coordinative bonds.

Alex Clark's zero order bond model linked above gets at many of these issues but in a way that's backwards compatible with the standard SD file format.
It's a long answer, but if you code a good graph representation, you can probably encode a lot of chemistry.


text:
Why ethyl ethanoate and not acetic acid?
In saponification reaction ethyl ethanoate is made to react with NaOH to form sodium acetate as the soap and ethanol as a by-product.
My question is : Why don't we use ethanoic acid instead of ethyl ethanoate to form sodium acetate which is the soap and water as another product?

You could do so. However, saponification is generally applied on fats (basically long-chain esters) which are found in abundance. Finding an acid such as ethanoic acid in bulk in nature is quite hard.

@AniruddhaDeb Please don’t use comments to answer a question.

I doubt that sodium acetate will be an effective soap. A methyl group is not a very effective "greasy" tail.

Sodium acetate is NOT the soap, as Abhirup Adhikary says. It is the result of a saponification rection. But it is not a soap.

I just gave example using acetic acid. Moreover, https://www.quora.com/Is-CH3COONa-a-soap

Saponification refers to a specific class of reaction during which an ester is transformed into its constituent alcohol and an alkaline metal salt of its acid.
While sodium acetate can be produced from acetic (ethanoic) acid and sodium hydroxide, this reaction is known as an acid-base or neutralisation reaction. It follows a very different mechanism to a saponification.
The name saponification, literally soap making is historical; this reaction was a major way of making soaps from various biological fats and oils. The sodium salts of long-chain fatty acids (around $\ce{C18}$) are the main ‘soapy’ ingredient of soap. The corresponding acids were not easy to acquire in large amounts but extracting fatss and oils from biomass was a routine task; hence this specific process was used to generate the salts.
Later, the term was applied to ester hydrolysis using alkaline metal hydroxides in general because these follow the same mechanism and only differ in the specific constitution of reactants and products.

Your answer is quite theoritical but Mr Aniruddha Deb's answer makes it quite clear to me why is it done with ester. But I appreciate your answer

Except Aniruddha Deb didn’t actually post an answer.


text:
Confirmation regarding nucleophilic addition of Grignard reagent

$\ce{A ->[(i)EtMgCl][(ii)H2O] Et3C-OH}$
A can be:
a) $\ce{COCl2}$
b) $\ce{EtCOOEt}$
c) $\ce{Et2CO}$
d) $\ce{All}$

The answer should be (d) according to me, but is marked as (a) in the answer key.
Option (c) obviously gives the required product, as does (b). I derived this scheme for (a)-
$$\ce{2COCl2 + 2EtMgCl -> Et2CO + 2MgCl2}$$
$$\ce{Et2CO + EtMgCl -> Et3C-OMgCl ->[H2O] Et3C-OH + Mg(OH)Cl}$$
Could one please confirm if the above reaction scheme is correct? I personally could not find any flaw in it.

Seems like a goof up by the book: https://www.organic-chemistry.org/namedreactions/grignard-reaction.shtm

I agree, certainly b and c are viable answers

I dont see how this is related to the 'synthesis' of Grignard reagent. Well if we were impractically too critical about the fact that the reagents aren't provided in excess, then also the answer should have been (c) and not (a). So I agree too, unless someone comes to enlighten us

The answer is all of them, thus (d).  In the case of $\ce{COCl2}$ (a), the Grignard reagent displaces both chloride ions in successive nucleophilic substitutions, forming diethyl ketone; then the reaction continues as it would for a typical ketone.
The displacement if a chloride ion in the case of an acyl chloride is described here.  Reaction with $\ce{COCl2}$ would simply involve two stages of this process.


text:
Why does Xenon Hexafluoride exist as [XeF5+][F-] and not [XeF5+][XeF2-] in the solid state
I already referred to this similar question but it yielded no explanation for the reason. Cations and anions of similar sizes stabilize each other through lattice energy effects in the solid state. This explains the stability of the alkali metal carbonates and sulphates as one moves down the group. Applying a similar logic here, the $\ce{[XeF2-]}$ anion is larger than the $\ce{F-}$ anion, yet $\ce{XeF6}$ exists with the $\ce{F-}$ anion.
this paper seems to contain exactly what I need but I am unable to access it. Could someone kindly shed light on the structure of $\ce{XeF6}$ in the solid state?

XeF2− is not a thing, and besides, would give a wrong overall composition.

As has already been pointed out, $\ce{[XeF5+][XeF2-]}$ is a rather problematic formulation; but $\ce{[XeF5+][XeF7-]}$ is more sensible and would be a valid question.

First and most obviously, if something in solid state forms ions by shuffling atoms around, the end result must yield the same ratio of atoms. Take for example $\ce{PF5}$: this exists as $\ce{[PF4+][PF6-]}$. If you add the atoms together, you get $\ce{P2F10}$ which is $2\times\ce{PF5}$. The same is true for the proposed structure $\ce{[XeF5+][F-]}$ which adds up to $\ce{XeF6}$ as it should. Your proposed version would give $\ce{Xe2F7}\ne\ce{XeF6}$.
Second, the proposed anion does not actually work out. $\ce{XeF2}$ exists itself as a neutral compound obeying the octet rule and featuring an all-paired Lewis structure (and I believe also all-paired MO scheme). If you were to create an anion, you would need an additional electron making it $\ce{XeF2^.-}$. This species surely exists but not for long as it will react and degrade to something else.
Without having to go into detail why a different solution is more viable, it is elementary to show how impossible an option your proposed solution is.

You showed conclusively(atleast for me) that the OP's proposition won't work out. However, is there any way to make an educated guess a priori for the formulation of $\ce{[XeF5+][F-]}$ ,given $\ce{[XeF6]}$ ?

The formulation $[XeF_5]^+[F]^-$ is at best a simplification. In reality the various solid state structures consist of a number of oligomers with bridging fluorine see e.g. the abstract for https://www.sciencedirect.com/science/article/abs/pii/S0022113906001266 - halides with bridging halogen are very common, with $Al_2Cl_6$ possibly being the best known example

@IanBush Ohh, I see. But still, is there anyway to guess before hand how many  bridging fluorines to expect(like the one $\ce{[F]-}$ acting as bridge in the above formulation)?

@YusufHasan I don’t know if there’s an easy way to guess. In the $\ce{PF5}$ case, a fluorine atom is essentially transferred from one phosphorus to another. In the xenon case, fluorine atoms bridge between xenon centres. If anything, I would want to assume that octahedral and tetrahedral structures around central atoms might be preferred due to the higher symmetry but take that with a grain of salt (pun intended).

@Jan Just a theory: in $\ce{PF5}$ , the axial bonds are a bit longer than the equatorial ones in the solid state. So, would it be probable to think that one of the axial bonds might break off from one of the monomers, and get transferred to another $\ce{PF5}$, thus allowing $\ce{[PF4]+}$ to adopt a stucture with less bp-bp repulsions?


text:
AChE Aging time of organophosphorus compounds containing hydroxyl groups
Organophosphorus compounds are known to inhibit the enzyme acetylcholinesterase (AChE). This occurs when the OPC phosphorylates the serine-203 residue of the enzyme. If the enzyme is not reactivated in time, a second reaction occurs, whereby one of the alkyl groups on the OP molecule is hydrolyzed off, giving the alcohol and the phosphorylated serine residue. The hydrolysis reaction also protonates the nearby Histidine-447 at the active site of AChE, forming a salt bridge between the OP and the enzyme. This is called "Aging", and makes the inhibited enzyme resistant to activation even by oximes. Depending on the OPC in question, the aging reaction can take between a few minutes (In the case of Soman) to several hours (In most cases). Here is a figure explaining the reaction:

Sources:

https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=7&cad=rja&uact=8&ved=2ahUKEwi0r7T4g8jnAhXCK80KHfFCBIEQFjAGegQICRAB&url=https%3A%2F%2Fwww.mdpi.com%2F1420-3049%2F22%2F9%2F1464%2Fpdf&usg=AOvVaw0WTpEABL2j3q-kXjlvKdQM
https://www.ncbi.nlm.nih.gov/pubmed/8250487

The part which I have circled in red depicts the aging reaction. Firstly, is the information which I have found thus far correct?
Secondly, If an organophosphorus AChE inhibitor which already contained a hydroxyl group, such as EA-2192

or (Presumably) fluoro(methyl)phosphinic acid

were to inhibit AChE, how long would the "Aging" reaction take? My guess is that aging would occur even faster (Seconds to minutes) due to the OPC itself being capable of acting as a proton donor, and thus leading to the formation of the salt bridge without any additional hydrolysis needing to take place beforehand. However, I have not been able to find any information on this particular matter, and so that is just pure speculation. Has any research been conducted into this? Is my speculative guess likely to be correct?

Acetylcholinesterase is important enzyme and its mechanism of function is well known. In active site, it has two pockets: anionic pocket (to attract positively charged Quaternary nitrogen of choline) and esteric pocket. The esteric pocket consists of a serine side chain (a $\ce{-CH2-OH}$ function) as depicted in following cartoon, which pacilitate the deseterification of acetylcholine to choline:

The neurotransmitter, acetylcholine bound to this pockets and $\ce{-CH2-OH}$ function of serine attracts carbonyl group of acetyl group as depicted. As a result, choline group will be released and acetylated serine group would then release acetyl group as acetic acid in a moment by hydrolysis to reactivate the enzyme.
However, if you introduce a competitive inhibitor such as neostigmine (which is a reversible inhibitor), it would bind to some of enzymes and decelerate the enzyme for designed time (e.g., use of this phenomena as a medical treatment). Once the inhibitor was released in an extended time period, enzyme will be reactivated.
Yet, there is a group inhibitors called irreversible inhibitors (and commonly known as suicide inhibitors because they kill the enzyme while get killing themselves), of which do not leave the active site once bound, unless an antidote is introduced.  One group of such inhibitors are called nerve reagents (a.k.a. nerve gases). All most all of nerve reagents are organic phosphorus reagents, which consist of a good leaving group such as fluoride (e.g., Soman and Sarin). The suicide-type inhibition of a nerve reagent is depicted in following diagram (The $\ce{-OR_L}$ group here represents a good leaving group):

For nerve gases such as Soman and Sarin, the leaving group is fluorine atom, which leaves as a fluoride ion ($\ce{F-}$). The strong attraction of phosphorous in nerve reagent to Ser$\ce{-OH}$ group causes $\ce{P=O}$ to break. Reformation of $\ce{P=O}$ bond would have made unacceptable hexavalent phosphorus. To avoid that scenario, one group on attached nerve reagent has to leave, and the best one available to leave is fluorine atom as $\ce{F-}$ (e.g., in Soman and Sarin). That leaves you with irreversibly inhibited acetylcholinesterase, which makes all chaos in central nerve system, until you receive an antidote (you should be given the antidote before infected enzyme goes to the next step, aging). The mechanism of the action of antidote (e.g., 2-PAM) is well documented and been illustrated in following diagram:

2-PAM has high affinity for anionic site (which is free now; see diagram) of the enzyme because it consists of part of nicotine moiety (searh for nicotinic receptor description). Once bound to free anionic site, oxime$\ce{-OH}$ has high attraction for phosphorous atom in infected enzyme site by the proximity. This attraction causes $\ce{P=O}$ to break (see pink arrow). Reformation of $\ce{P=O}$ (see blue arrows) would have again made hexavalent phosphorus. Again, to avoid that scenario, one group has to leave, and the best one to leave is ser$\ce{-OH}$, leaving the free enzyme (note that, all bond making and bond breaking in the active site here are in equilibrium, but once Ser$\ce{-O-P}$ bond is broken, the elimination of 2-PAM bound-phosphorus moiety is faster compared to regeneration of same Ser$\ce{-O-P}$ bond).
The aging depends on how easy to hydrolyze the $\ce{O-R2}$ bond by a water molecule. At least, this bond in Soman, $\ce{R2OH}$ of which is Pinacolyl alcohol (3,3-Dimethyl-2-butanol), can be easily hydrolyze by water in few minutes. This mechanism was studied and following mechanism was given (Ref, 1):

In Sarin, the $\ce{O-R2}$ group is the $\ce{O-CH(CH3)2}$, which would not be hydrolyzed as easy as that in Soman. Thus, for sure, aging of Sarin infected enzyme would take more that minutes, perhaps hours.
The given mechanism is supported by the study of different group on another nerve reagent Tabun, the initial leaving group of which is cyano group (as $\ce{CN-}$ compared to $\ce{F-}$ in Soman) to effect acetylcholinesterase (Ref.2). Aging of the effected enzyme is caused by leaving of ethyl group from $\ce{P-OCH2CH3}$ function ($\ce{-OR2}$) in similar manner to Soman-effected acetylcholinesterase:

References:

Gulseher Sarah Sirin, Yanzi Zhou, Lee Lior-Hoffmann, Shenglong Wang, Yingkai Zhang, "Aging Mechanism of Soman Inhibited Acetylcholinesterase," J Phys Chem B. 2012, 116(40), 12199–12207 (https://doi.org/10.1021/jp307790v).
Eugénie Carletti, Jacques-Philippe Colletier, Florine Dupeux, Marie Trovaslet, Patrick Masson, Florian Nachon, "Structural Evidence That Human Acetylcholinesterase Inhibited by Tabun Ages through O-Dealkylation," J. Med. Chem. 2010, 53(10), 4002-4008 (https://doi.org/10.1021/jm901853b).

Interesting, so it seems that the His440 residue is actually already protonated before aging occurs, is that correct? Can it thus be assumed that OPCs in which the R2 group is a hydrogen atom won't cause aging at all?

No. If it is $\ce{H}$, it's already aged. Yet, $\ce{-OR2}$ group could be either a $\ce{-H}$ group or $\ce{-CH3}$ group.

I added another diagram to assit you understand the making of salt bridge.

Matthew's answer corrected some mistakes which I have made in asking my question. His answer also suggests that OP compounds that contain hydroxyl groups would indeed cause instant aging of inhibited cholinesterase.
I have found confirmation that that is indeed the case. There is a naturally-occurring organophosphate that contains a hydroxyl group, formerly known as Anatoxin-A(s), but now known as guanitoxin.

A study was done regarding the efficacy of oxime reactivators against AChE inhibited by guanitoxin. It was found that acetylcholinesterase inhibited by Guanitoxin is resistant to reactivation by oximes and that Guanitoxin does, indeed, cause instant aging of the enzyme.
Reiter and others have also studied the binding of EA-2192 to AChE and have found that it likewise causes instantaneous aging of the enzyme. This would make sense, since the adduct between AChE and an OPC containing a hydroxyl group bonded to the phosphorus atom is identical to an aged adduct. It would follow that other OPCs containing an OH group bonded to the phosphorus atom would also instantly age the enzyme.
Reference:
Hyde, E. G., & Carmichael, W. W. (1991). Anatoxin-A(S), a naturally occurring organophosphate, is an irreversible active site-directed inhibitor of acetylcholinesterase (EC 3.1.1.7). Journal of Biochemical Toxicology, 6(3), 195–201. doi:10.1002/jbt.2570060305
Reiter, G., Mikler, J., Hill, I., Weatherby, K., Thiermann, H., & Worek, F. (2011). Simultaneous quantification of VX and its toxic metabolite in blood and plasma samples and its application for in vivo and in vitro toxicological studies. Journal of Chromatography B, 879(26), 2704–2713. doi:10.1016/j.jchromb.2011.07.031


text:
Compressibility factor (Z) of a real gas
I was wondering whether the compressibility factor of a real gas (given by $Z = V_{\mathrm{real}}/V_{\mathrm{ideal}}$) is supposed to be measured while keeping pressure constant?
I was attempting to answer the question below and got stumped at part (iii) because I had trouble finding out which was the dominating force being that when I tried calculating the compression factor I got a value of 1 (because $V_{\mathrm{real}}$ and $V_{\mathrm{ideal}}$ are the same, but the pressures are significantly different).

3.0 mol $\ce{CO2}$ behaving as a van der Waals gas, when it is confined under the following conditions: at 750 K in $\pu{150 cm^3}$. For $\ce{CO2}$ van der Waals coefficients, a: $\pu{3.610 atm L^{2} mol^{-2}}$, b: $\pu{4.29 \times 10^{-2} Lmol^{-1}}$. The universal gas constant R is $\pu{8.206 * 10^{-2} L atm K^{-1}mol^{-1}}$
i) Calculate the gas pressure (in atm), [5]
ii) What is the molar volume of the gas (in L/mol), [1]
iii) What is the pressure when it is ideal gas and use it to state the dominating force in the above gas?

Part (iii) is asking which of the van der Waals factors dominates.

It's compressibility factor not compression factor...

Please show us what you did so far,, in detail.

You don't need to know the compressibility factor to solve this problem.  But, in any event, its definition is $$z=\frac{PV}{nRT}$$

From the given data, the molar volume $V_\mathrm m$ is $$V_\mathrm m=\frac{150\ \mathrm{cm^3}}{3\ \mathrm{mol}}=50\ \mathrm{cm^3/mol}=0.05\ \mathrm{L/mol}$$
In terms of molar volume, the van der Waals equation reads:
$$\left(p+\frac{a}{V_\mathrm m^2}\right)(V_\mathrm m-b)=RT$$
So, solving for the pressure gives:
$$p=\frac{RT}{V_\mathrm m-b}-\frac{a}{V_\mathrm m^2}$$
So, to get the pressure you don't need to know the compressibility factor.  Which of these two terms dominates numerically in determining the pressure?

$$\left(p+a\frac{n^2}{V^2}\right)\left(V-nb\right)=nRT$$
$$pV-pnb+a\frac{n^2}V-a\frac{n^3b}{V^2}=nRT$$
cancelling $n$ on both sides
write $pV$ as $znRT$ where $z$ is compressiblity factor
$$(z-1)RT=pb-a\frac nV+ab\left(\frac nV\right)^2$$
it is behaving like ideal gas hence $z=1$ and right hand side is $0$
$$pb-a\frac nV+ab\left(\frac nV\right)^2=0$$
substitute the values and get the result

Welcome on the ChemSE! We have quite good latex support here, I suggested an improvement to your post. After accepting the suggestion, you could end the work. And, please formulate in clear, round sentences.

See https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference for an easy  guide to formatting equations.


text:
How do STO-based computer programs implement the STO integral formulae?
In most cases, STO integrals are evaluated assuming that the orbitals are directed towards each other, using spherical harmonics; however, in GTO-based software Cartesian GTOs are used, with the orbitals being aligned along the x, y or z axis.
Is there any way to evaluate STO integrals using Cartesian STOs, or convert between Cartesian and spherical STOs? For example, if I had a pz-type STO at (0,0,0) and another one at (1,1,1), is there any way to decompose the resultant expression in terms of spherical STO integrals?

I would wager that the relevant recursion formulae are listed in L. Zülicke: Quantenchemie, unfortunately, it is hard to come by.

Have you considered to commit to the Materials Modeling Stack Exchange: https://area51.stackexchange.com/proposals/122958/materials-modeling?referrer=MGY4NjYyM2ZlY2NkNDI0MWRmYjhjMGI1N2RmM2U3ZDk1N2VjOWQ1MGU1MmVhNDZkZjI5MzNiNzY1MmE2ZjRhOYrAzULyFn6juGAbZ7wko5LvlWXZ0iQfr620nQ70gVNl0 ? There's a higher concentration of computational researchers, including basis set expert Kirk Peterson. Several example questions were about basis sets. We will soon be in the private beta phase, where only a select group of users can participate. These users help to determine the style of questions for the future.

Are you able to be specific about which particular computer program for which you're interested in the STO implementation? I know at least 2 programs that use Slaters and the implementation is very different in each of them.


text:
atomic structure ,
one twelveth mass of C-12 isotope is taken to be 1 amu
earlier scientists took the mass of protium isotope of hydrogen as a reference to measure the mass of other atom
And it is also given that when they did so with hydrogen as reference they got fractional atomic mass for many atoms
but what is the difference between the earlier one and C-12
both are same and = 1 amu only so why did they get fraction with hydrogen and integer with carbon
pleae explain

"Was", not "is".

The title and the body are two distinct entities, and each one is supposed to be meaningful in the absence of the other.

“one twelveth mass of C-12 isotope is taken to be 1 amu” – No. The old atomic mass unit (amu) referred to the relative atomic mass of oxygen, which was taken as 16. In chemistry, this unit became obsolete in 1961. It was replaced by the unified atomic mass unit (unit symbol: u), which is equal to 1/12 times the mass of a free carbon-12 atom.

Related: https://chemistry.stackexchange.com/q/77295/7951

All these choices are arbitrary. Nobody can do justice by summarizing 200 years of atomic weights (now atomic masses) history. Atomic reference scales have changed, earlier hydrogen was taken as 1 (exact) and sometimes oxygen was taken as 16 (exact). Other numbers have been chosen as well. When a mass spectrometer was invented, it started to change the story. It allowed physicists to measure "masses" from pure electricity and magnetism principles without relying on chemical reactions. They started their own scale. So chemists had one set of atomic masses, and physicists had another set of atomic masses. All of them were close but slightly different. This issue affected trade, international business and of course chemical industries. In order to overcome this issue, C-12 was suggested as compromise between chemist's and physicist's scale.
Must read for interested readers: The carbon-12 scale of atomic masses by
Abbas Labbauf

earlier scientists took the mass of protium isotope of hydrogen as a reference to measure the mass of other atom And it is also given that when they did so with hydrogen as reference they got fractional atomic mass for many atoms but what is the difference between the earlier one and C-12 both are same and = 1 amu only so why did they get fraction with hydrogen and integer with carbon

Prout's (incorrect) hypothesis was that every element's atomic mass must be an integer multiple of hydrogen's atomic mass. Who stated that if we choose C-12 as the reference, then the atomic masses of other elements become integers?


text:
Diazotization order
I know that diazotization of amines proceeds via the generation of the electrophile $\ce{NO+}$ and thereafter, the $\ce{NH2}$ group acts as a nucleophile and shares it's charge density with $\ce{NO+}$. After a series of rearrangements the diazonium salts are obtained. However, if one had to compare the ease/ rate of diazotization, what should we look at?

The nucleophilic nature of the $\ce{NH2}$ group and it's ability to share it's electron density?
Or something else?

As an example if we had the following compounds (see following diagram):

So how would we compare the Diazotization and what would be the order?
I compared the nucleophilicity and got the order as:
1-aminocyclohexanoic acid < 2-aminoacetophenone < m-acetoxyaniline < aniline.

$\ce{NO+}$ is not the nucleophile here.

It is not just about the nucleophilicity. You need to consider the stability of the diazo compound  produced. You don't see too many alkyl diazo compounds., they're not stable.

Why do you say that NO+ is nucleophile ? It is not

My sincere apologies, I meant electrophile. Editing it right away.

@Waylander can you please answer this question and explain how do we proceed to check the stability of the diazonium salts obtained and also do we not look at the nucleophilic nature of the NH2 group altogether? I understand that there is nothing like ignoring all together in chemistry, but is it like the stability of the product is the dominating factor and there's no need to look into the mechanism?

You are correct in your reasoning in respect of the aromatic amines, that the rate of diazotisation is driven by the nucleophilicity and the same factors that influence that stabilise the +ve charge of the diazonium species formed.
With regard to the cyclohexyl amino acid the diazonium species will not be stable. The reaction will be very slow and the primary product will be the alpha lactone.

Nitrosation of the amino acids RR′C(NH2)(CH2)nCOOH is very slow and kinetic studies are reduced to initial rate measurements. The final products are α-, β-, or γ-lactones for n = 0, 1 and 2, by internal attack by the carboxylate ion at the central carbon atom, which may be concurrent with, or follow, loss of nitrogen, Eq. (99), [168].
  Source:  https://www.sciencedirect.com/topics/chemistry/primary-aliphatic-amine

if you are happy with answer, please accept it


text:
Syntactic vs Semantic validity of molecular graphs
While reading this paper on molecular graph generators using deep learning algorithms, I encountered the terms syntactically and semantically valid molecular graphs:

$\dots$ to ensure the generation of syntactically and semantically valid
  graphs.

The output molecular graph should at least satisfy the following constraints:

Valency: Number of incident edges to an atom shall not exceed the valency of the atom on that node
Connectivity: The output graph should be a single connected graph.

I was wondering what semantic and syntactic validity means in this context? Which type are each of these constraints?

Some concepts from computer science:

Syntax is the structure or form of expressions, statements, and
  program units but Semantics is the meaning of those expressions,
  statements, and program units.

But I can't make an exact connection to molecular graph's case. Also,

In computer science, the syntax of a computer language is the set of
  rules that defines the combinations of symbols that are considered to
  be a correctly structured document or fragment in that language

Then, valency and connectivity, unlike the linked paper in the answer below, should be syntactic constraints.

It would be good to add human readable references.

It is not sarcastic. Are we supposed to know what is syntax-directed VAE or QM9 ?

These are not relevant to the question in hand.

Generative machine learning methods in chemistry are typically tasked with generating new molecular structures. Think of it as 'dreaming up molecules.'
We see these kinds of ML models with predictive text programs.. if I'm typing, my phone attempts to guess the word or complete my sentence. (NB this doesn't work so well with molecule names or scientific terms.)
There are a wide variety of ways to represent a molecule for such a generative ML method. One of the common ways is via SMILES in which the molecular graph is encoded as a text string:
c1ccccc1  benzene
CC(=O)O   acetic acid

In this case, syntax can be a problem. SMILES uses parenthesis to represent branching, and they must match. Numbers are used to represent ring closures, which must match as well.
Consequently, many generative ML method generate syntactically incorrect molecules (ignoring questions of chemical valence, etc.)
c1ccccc2 error
CC(=OO)) error

There are solutions. Noel O'Boyle and Andrew Dalke suggested DeepSMILES which encodes a new syntax that's resistant to such issues.
As you mention in your answer errors in connectivity and/or valence represent semantics - that is what makes a chemically viable molecule (i.e., carbon does not take 5 bonds, isolated oxygen atoms aren't stable, etc.)

The only thing that I can think of is when an empty node is connected to edges. In other words, out of node classes, we select null, and out of bond classes, we select a non-null bond type for the edges incident to that atom. This means bonds that are not connected to any atoms. But in the same aper that I mentioned in my answer, this is again grouped with semantic validity.

So, for SMILES, do these statements look correct?: 1. Syntactic validity is the set of rules that define a correct combination of symbols, regardless of its chemical validity 2. Semantic validity is a syntactically valid representation that also follows the chemical constraints. 3. Syntactic validity is representation validity, while Semantic validity is chemical validity of a syntactically valid representation.

For the second part of question, according to this paper, both valency and connectivity constraints are categorized as semantic validity.

IS this intended as an answer or as a comment to your question? If the latter please delete this answer and add the text as en edit to your question.

I think this is an answer. but you can have your answer as well

That's ok- I don't have an answer, but I thought yours so short that I was confused. I suggest you elaborate?

Sure, I'll elaborate as I find answers online


text:
Why is the first energy of ionization of oxygen lesser than that of nitrogen?
The following question arises from a question I found in my book.

Experimentally it has been determined that the value of the first
  energy of ionization of oxygen is lesser than the first energy of
  ionization of nitrogen. Select the statement which explains this:

The higher electronegativity of oxygen
The lesser electron affinity of nitrogen
The lesser radius of oxygen
The higher stability of nitrogen
The higher electron affinity of nitrogen

According to my book the answer is the fourth option, but I couldn't find a way to justify this. What sort of explanation can do this?.
My guess is that nitrogen forms a triple bond with itself which is higher than the oxygen and this makes the molecule more stable but I don't know if this can be translated into a higher energy of ionization nitrogen rather than the oxygen which does have a sigma and pi bonds. Am I right?
Can someone please illustrate this or bring some sort of diagram with some values to justify this?. It would help a lot if an answer could help me why should I discard the second option?.
Doesn't nitrogen has smaller electron affinity than the oxygen? Wouldn't it mean that having a smaller electron affinity cause it to be easier to strip out an electron from its outer shell rather than the opposite as the question indicates?

But wait it's ionization of atomic element by default. If it was about molecules, it would have to be clearly stated!

@Mithoron Actually it does in part but it doesn't explain more details than stability due half filled orbitals. Could stability be explained using orbital molecular theory or $\pi$ bonding as one of the answerers suggested?.

It's for atoms, so just a matter of  half filled orbitals.

Please cite the textbook from which this exercise is from.

Regardless of whether it’s atoms or molecules (I assume it’s atoms), the fourth option in the question is a meaningless phrase. It... just doesn’t say anything. What kind of “stability” is this? How can the “stability” of lone atoms even be compared? What does it even have to do with ionisation energies?

Two electrons of the oxygen molecule are located on a $\pi $* level. They are relatively easy to ionize. It is not the case in $N_2$ : all electrons are located in binding orbitals.

That was also my thought, but that's barely an idea for an answer.

@Maurice In my comment I guessed that it had to do with the electrons in the $\pi^{*}$ bonds. In your answer you mentioned that this is not the case. Is it because the overlap occurs side to side and this makes the electrons easy to remove?. But doesn't this also occurs in nitrogen?.

If you tried to actually write answers that backup your claims, then you might had realised that what you write is incorrect.


text:
Does it exist a transgression to Hund's rule of maximum multiplicity principle in this given electron box notation?
The following question arises from a set of statements from which I'm asked to state whether they are true or false.
This is as follows:

Indicate which of the following statements are true or false

On $\underset{np}{\underline{\uparrow}\,\underline{\uparrow}\,\underline{\downarrow}}$ there is a transgression to Hund's rule of maximum multiplicity
The given box notation $\underset{np}{\underline{\uparrow\,\uparrow}}$ is a transgression to Pauli's exclusion principle.
The following configuration $1s^22s^22p^13s^1$ is a transgression to Pauli's exclusion principle.
The electron configuration of Silver $(Z=47)$ is $[Kr]5s^{1}4d^{10}$.
Two chemical species with the same number of electrons not necessarily have the same configuration.

Given these statements I found it difficult to make a proper interpretation of what was intended with the fifth and the first option.
The second option is true. As two electrons occupying the same orbital cannot have the same quantum numbers.
The third option is not true due that's not a transgression to Pauli's exclussion principle. But rather to Aufbau's build up principle. Because it states that the electrons will occupy the orbital with the lowest energy possible.
The fourth option is correct, that's the electron configuration of Silver.
But the problem arises from the first option as I don't know if it does make a transgression to Hund's principle of maximum multiplicity?. As I understand there isn't any reason to justify that the electron will always have to be filled with a $m_s=-\frac{1}{2}$. It only states that the most stable configuration will be the one where electrons fill the most orbitals as possible.
Then the fifth option is a bit hard to understand, as I don't get very clear what's the meaning of chemical species. I'm assuming since this question is related with atomic structure chapter, it is referring to atoms. In the given context ions more specifically, and understood in such way it does say that two isoelectronic atoms will not have the same configuration. But is that possible?. How can it be justified?.

If $S$ is total spin the multiplicity is $2S+1$ so in (1) it is $2(1/2)+1$ but it could be $2(3/2)+1$ if the spins were all parallel. In (4) you could change configuration in the C atom of Q3 to be ..$2s^1...3s^2$ with same number of electrons.

5th option let me think of ions and molecules. It shouldn't be there if the course has covered just atoms. You should ask the author :((

@porphyrin I don't think what it was intended was to change the configuration but rather give a judgement solely on what was written, in that sense I assumed that with no allowed modification of the configuration, it is a transgression to Aufbau's rule but not Pauli's rule. Regarding the total spin multiplicy where did you get $\frac{3}{2}$ for the value of $S$?.

3/2 three electrons each with spin ms =+1/2


text:
How does one extract the work characterized by the Helmholtz energy?
From Wikipedia:

Helmholtz free energy is a thermodynamic potential that measures the
  useful work obtainable from a closed thermodynamic system at a
  constant temperature and volume (isothermal, isochoric)

I struggle to understand the Helmholtz more than the Gibbs because of the isochoric nature of its definition. Given that the work extractable by a process on a gas is p * dV, how can one characterize any "useful work obtainable from from a closed thermodynamic system at constant ... volume"?

$p_{ext} dV$ refers to pressure-volume work only.  But many other types of work are also possible, e.g., electrical work.

Not all work is pV work (also called expansion work and a type of mechanical work). non-pV work is possible, e.g. chemical, surface, electrical, magnetic, gravitational (admittedly restricting the system to a gas requires a little creativity in envisioning how such non-pV work might come about).
The first law of thermodynamics can be written generally as $$\begin{align}dU&=dq+dw \\ &=dq+dw_{pV}+dw_{non-pV}\\ &=dq_{rev}-pdV+dw_{non-pV,rev}\end{align}$$
In the last equation the work is performed reversibly, there is mechanical balance between the system and surroundings ($p=p_{ext})$.
The transform to generate the enthalpy function H only alters the contribution  of the expansion work to the internal energy U:
$$\begin{align}dH&=dU+d(pV) \\ &=dq_{rev} + Vdp+dw_{non-pV}\end{align}$$
or for an isobaric ($dp=0$) closed process
$$dH = dq_{rev} + dw_{non-pV}$$
Therefore the non-pV contribution to the total work is retained in the enthalpy change for an isobaric process (not isochoric as I had previously bizarrely written).
Similarly, we can evaluate the differential change in the Helmholtz function A as
$$\begin{align}dA&=dU-d(TS) \\ &=  -pdV+dw_{non-pV} -SdT\end{align}$$
It follows that for an isothermal (constant T, $dT=0$) and isochoric (constant V, $dV=0$) process,
$$dA=  dw_{non-pV} $$
Therefore the non-pV contribution to the total work is retained in the Helmholtz free energy change for an isochoric and isothermal process. In fact, under these conditions the Helmholtz function is equal to this work.


text:
Looking for linear non polar compound which is used heavily due to its boiling or melting temperature or its density
So my son was given an exercise to find some linear non polar compound that is used heavily due to convenient boiling or melting temperatures or its density.
Having no training in chemistry I have no idea how to help him. I tried to look for a kind of search engine for chemicals just to see what kinds of compounds fit the bill, but I cannot find anything.
Any help is appreciated.

Can you arrange your reply as an answer? I understand you are talking about CO2.

OK. I googled and CO2's boiling temperature is just under -78 Celsius. So, does it have to do anything with the process you described?

As I understand, I need to read https://en.wikipedia.org/wiki/Supercritical_fluid_extraction to understand the essence of your answer.

I would not understand much depth anyway. I just want to understand what makes CO2 such a popular SF.

Why not some look up data on hydrocarbon oils, there is a huge choice there.

Warning. We, or the exercise, must be precise. CO2 has no electrical dipole moment but don't qualify as apolar. It is indeed a polar molecule / solvent.

@porphyrin - I really know little about the subject. But I thank you for the feedback, I hope it will be useful to my son.

A possible answer is n-octane,

(image credit)
for what it

consists of linear organic molecules (vide infra)
has a boiling point of about $125\,\pu{^\circ{}C}$ / $257\,\pu{^\circ{}F}$
is a hydrocarbon you either find in, or generate from natural oil

A bit on the n in n-octane:  This is a shorthand for normal, because the carbon atoms are consecutively bound to each other in an open chain.  Which is not the same as if they would form branched molecules, then denoted iso-octane.  A prominent iso-octane (iso as in isomer, i.e. same number of atoms per molecule, but different spatial arrangment) is 2,2,4-trimethylpentane:

(image credit)
Blend in mixtures of the two, they serve as in octane rating of fuels.

Forgive my ignorance, but n-octane does not look like a linear molecule.

I agree with you that even if it was called «linear», n-octane is not as straight as a pen, or like acetylene (https://en.wikipedia.org/wiki/Acetylene).  This zig-zag shape is a result how the carbon atoms bind together, and like thread, these may be twisted, too (e.g. in polymers, https://en.wikipedia.org/wiki/Polymer#Microstructure).  This however contrasts to branching (like the iso-octane), or even forming rings as, e.g. in cyclohexane (https://en.wikipedia.org/wiki/Cyclohexane).

I think I understand what you are saying, but it seems to me the ask is for something like acetylene (geometrically) after all.