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Remove carbon dioxide from solution
What solution could I add to a solution of water with dissolved carbon dioxide in order to get the carbon to precipitate out? I believe the ions are H+ and HCO3-
Thanks, Jerid

There is no solution that is a solution, but, for example, magnesium metal will burn in carbon dioxide gas and produce carbon particles and magnesium oxide. Totally impractical energetically as a way to get the carbon back from carbon dioxide.

do you want to precipitate out the carbonate ions or do you actually want to obtain elemental carbon?

Thank you for the replies. I am not trying to obtain the elemental carbon. I have a process where I have a high concentration of CO2 in a gas sample. The sample also contains 95% Nitrogen and a small amount of oxygen. I am able to extract the CO2 by dissolving the gas through water. This leaves the original gas with a much lower CO2 concentration which is the goal. I am trying to figure a way to get the CO2 out of the water without it returning to a gas state which I am doing now with electrolysis but it is not the ideal setup for this application.

I think you must explain your needs in more detail for a reasonable answer. Is this a batch process in a lab, or is it supposed to be a continuous process on an industrial scale? What sort of flow rate of the gas? What is the starting concentration of CO2 and what would be an acceptable final concentration of CO2 in the gas? Is this supposed to be economical somehow? Is humidity change of gas a problem? Do you really need to ppt the CO2 somehow, or just trap the CO2 in the solution?

You might also consider scrubbing the CO2 straight from the gas. There are many commercial CO2 scrubbers that work by flowing the gas over solid NaOH. That seems more efficient than going into solution and out again.

Adding aqueous Ca(OCl)2 to CO2/H2O will create a precipitate of CaCO3 and Hypochlorous acid (HOCl).
$\ce{Ca(OCl)2 + CO2 + H2O -> CaCO3 (s) + 2 HOCl}$
Per a source, 'A STUDY OF CALCIUM HYPOCHLORITE AS A DISINFECTANT OF WATER', to quote:

The dissolved calcium hypochlorite reacts with the free carbonic acid and half bound carbonic acid in the water and there is formed calcium carbonate, and at the same time free oxychlorid, which is known technically as hypochlorous acid.

Adding Ca(OCl)2 may no be the best idea, because this compound is not pure. It contains also a lot of CaCl2.

The question is 'what solution' and not what pure chemical compound. Here the presence of some dissolved CaCl2, as well, is not relevant.

I would rather add chalk water, which is a solution of $Ca(OH)_2$. At least it will not create a new substance dissolved in water, because the reaction will be : $$Ca(OH)_2 + CO_2 ->  CaCO_3 + H_2O$$ and $CaCO_3$ is an insoluble substance.

Of course it depends on how much $\ce{CO2}$ must be ppt'ed per liter of solution. $\ce{Ca(OH)2}$ isn't greatly soluble.


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Are there other carbon free gasses that can be used for energy transport, apart from hydrogen?
Various gasses are currently used to transport and store energy for heating and electricity generation.
Most of them contain carbon:
Methane, propane, butane, carbon monoxide and natural gas.
It could be of advantage to use gasses that do not contain carbon.
One example for that is hydrogen.
Are there other gases that theoretically could be used, if they could easily be created?
I assume most can not practically used because they are not trivial to generate, but I like to ignore that, and assume they can be created.
Are there other gases that could be used, in principle, as energy carrier?

Ammonia is a possibility, but not particularly practical. Various hydrazines, such as those used as rocket fuels. Also not so practical for various reasons.

I should say that hydrazine itself meets your criteria, but it is quite toxic and unpleasant. The rocket fuel derivatives, such as UDMH (unsymmetrical dimethyl hydrazine) have carbon. But maybe there are amino-hydrazines or hydroxy-hydrazines. You may also enjoy reading John D. Clark’s wonderful book titled “Ignition!”. Overall, ammonia is the best. One last thing: phosphine would be an horrific choice: far too poisonous.

I forgot all about hydrogen sulfide! It is poisonous, smells like rotten eggs (because it is the main smell of actual rotten eggs), and it burns in air. Hydrogen selenide (supposedly smells like rotten onions) and hydrogen telluride (supposedly smells like rotten horseradish) are no-go. Interesting question, by the way.

Interesting question. These are my candidates, roughly in order from most practical to least practical. Some of these are poisonous, smell very bad, or are just plain undesirable. Environmental issues are not considered herein, but certainly would be in a real world potential application.  For details on the chemical substances, see their wiki entries.

As per my first comment and one of the other answers, ammonia, $\ce{NH3}$, is a possibility. It burns in air, is easily liquified for transfort, and is produced in immense quanties already. It has many industrial uses and is not particularly exotic. But is is poisonous and must be handled with respect.
Nitrous oxide, $\ce{N2O}$, decomposes exothermically: this can be initiated with a catalyst and then thermal decomposition takes over. It can be used as a monopropellant and as an oxidizing agent. As 'laughing gas', it has substance abuse issues and does not support respiration. But it is produced in large quantities, is easily liquified for transport and it can be used as the oxidizer in the compustion of, e.g., ammonia.
Hydrazine, $\ce{N2H4}$, has used as a fuel in some rocket engine applications. Hydrazine derivatives such as unsymmetrical dimethyl hydrazine (UDMH) give better performance, but contain carbon. However, maybe amino-hydrazine or hydroxy-hydrazine, assuming they exist, would be viable possibilities. Hydrazine is very poisonous and nasty stuff to deal with. Derivatives probably likewise.
Hydrogen sulfide, $\ce{H2S}$, burns in air, producing water and sulfur dioxide, $\ce{SO2}$. The latter is produced in large quantities when sulfur is burned in the course of producing sulfuric acid, $\ce{H2SO4}$. So the $\ce{SO2}$ product of combustion of $\ce{H2S}$ would have potential commercial value. The downsides are that $\ce{H2S}$ is poisonous and is the main rotten egg smell in actual rotten eggs.

I will add to the list if I think of more.

Hydrazine is a liquid, but it was too interesting to omit.

There are many more, so many the q. is too broad.

@Mithoron No argument from me! But the OP just wanted a few hypotheticals and I only consider ammonia and nitrous oxide to be even notionally practical. Basically just because they are already made in large quaantities.

@Mithoron I see your point - but it could be an interesting answer in the pattern of: There are the obvious candidates a, b and c. There are more candidates in these classes of substances: d based substances, e based substances (and others).

@VolkerSiegel Well, these are mostly hydrides, but lots of them, and well, in most cases it would be unfeasible to use them, or just dumb, or borderline crazy ;)

Yes, ammonia has been cited.
Per a source, 'Ammonia as a Hydrogen Source for Fuel Cells: A Review', to quote:

Like hydrogen, ammonia is carbon free and can be produced from any energy resource. However there are also some significant advantages in terms of storage and transport.

Boranes like $B_2H_6$ or $B_6H_{10}$ are also a possibility. Probably too expensive !

It's interesting that among boranes you picked two of the worst possible candidates ($\ce{B2H6}$ instantly hydrolyses and is prone to self-ignition, and $\ce{B6H10}$ decomposes at the room temperature) and didn't mention pentaborane $\ce{B5H9}$, which was actually proposed and briefly used for this very purpose, the energy storage and transport. Plus, the answer might benefit from citing a patent or a literature source as to what the outcome in terms of energy is and what the hazards are.

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Why aren't all bonds in cyclobutadiene not equal?
Cyclobutadiene is like benzene in that it has alternating single and double bonds in a ring. However, it's bonds are not the same length, the double bonds being shorter than single bonds. The molecule is rectangular, not square.
How would this be explained?
I have one plausible explanation that I tried to give...that is as follows
Cyclobutadiene is anti aromatic with 4n = 4 pi electrons, moreover it has huge strain in the molecule. To avoid such strain the molecule goes out of plane to compensate, but thus loses its ability to resonate.
Thus resonance structures like ones in benzene cannot be drawn for cyclobutadiene.
Is this correct? Please correct me if I am wrong..and some more data about this would be appreciated

@MathewMahindaratne no, I specifically want to know why we cannot draw resonance structures like we do in benzene, for cyclobutadiene

I think this is a question from Solomons and Fryhle; Its answer key states a weird reason - the position of atoms differ in the two resonating structures which can't be the case in resonance phenomenon. And thus no resonance structures are possible. Even I had this doubt and noticed the question linked by Mathew. But it didn't help here.

@GuruVishnu yes it is a question from Solomon's and fryhle. Where did you get the answer key from?

@GuruVishnu The adapted version

Ok. You must login to see the key here - https://www.wileyindia.com/customer/account/login/; Although it contains some errors (like the one I mentioned in my first comment) it's quite useful.

@GuruVishnu What would be the username and password? Do I just create an account?

Comments are not for extended discussions. Let's talk in this chat room - https://chat.stackexchange.com/rooms/74807/jee-chemistry-club

Strictly speaking cyclobutadiene is not antiaromatic but non aromatic, ie the same energy 4A + 4B is attained by applying the Hückel LCAO, as for two isolated double bonds, 2 times 2A + 2B.
Given that cyclobutadiene can only be isolated as non delocalized system when in matrixes and a very low T points to strain as a reason for instability of both forms, the delocalized one remaining hypothetical.
Edit for the readers not familiar with the terms: A system obeying all the requisites (cyclic, pi orbitals on each atoms) is antiaromatic when the hypothetical delocalization destabilizes it as compared to the localized system. If no energy difference arises, the system is called non aromatic.
Beware there are plenty of pages discussing non aromaticity in a trivial way, e.g. aliphatic or linear compounds. Cyclobutadiene is non aromatic within the Hückel picture. Also it seems worth to point out again that antiaromatic refers anyway to an hypothetical structure, not to an actual delocalization. A system which cannot avoid it by distorting would be unstable tout-court. Cyclobutadiene, albeit being non aromatic, is on the same line. Indeed you can find antithetical assertions about it, such as it being "the archetype of antiaromatics“ along with “it reacts as diradical“ as if the two things would be straight forward correlated. Normally a system escape antiaromaticity. Being cyclobutadiene non aromatic results in a distorted rectangular molecule, but only when it cannot escape this anyway unstable situation. See also Resonance structure of cyclobutadiene?. Although the situation of cyclobutadiene is again erroneously described as an antiaromatic case, it gives the answer. Also worth reading is the answer by Georg.

Downvoter should comment though the first version was not particularly clear.

Why when I prepare sketches I am not able to upload the photo? This one was about the Frost diagram and just of 70 kb...

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Enantioselectivity of Noyori Asymmetric Hydrogenation
I'm unable to understand the enantioselectivity of Noyori's hydrogenation reaction, and would appreciate any help with the same:
Here's a mechanism I found on Wikipedia:

but, to my disappointment, the highlighted statement, i.e. The (R)-BINAP-Ru catalyze the synthesis the (R)-Product, and the (S)-BINAP Ru catalyze the synthesis the (S)-product with high ee. contradicts the attached paper about Noyori's hydrogenation.

The last two entries of the table mention formation of S products from R-BINAP.
Also, I feel it's not really possible to determine the absolute configuration beforehand, i.e. without knowledge of surrounding groups.
It'd be great if someone could describe the mechanism based enantioselectivity of this reaction in detail, so that correct predictions about the stereochemistry of the product can be made, given the catalyst (R or S BINAP) and the substrate.
Thanks in advance!

Quoting from Reinhard Brückner's book "Organic Mechanisms: Reactions, Stereochemistry and Synthesis":  "Sadly, there is no vivid explanation why the enantioselective addition to the C=O double bond of the subtrate occurs only if it is flanked by one conjugated substituent (Aryl, C=C)." (translated by me). I should note that in the book's example, Aluminium was used instead of Ruthenium.

I've learned to be extra wary of Wikipedia on highly specific matters like this. For example, my adviser made it a point to include his lecture that the entry on kinetic isotope effects was written by someone who didn't understand kinetic isotope effects.

I think this is easy to answer than you think. Let's look at the stereochemistry of compounds in the question, 1g and 1h (I also put the common product, 1a, for convenience):

The stereospecific reduction is on carbon-3 ($\ce{C}$3) in all cases. Accordingly (see diagram above), if you apply Cahn–Ingold–Prelog (CIP) sequence rules for each, you'd find (3R)- for 1a (methyl group is priority 3 based on $\ce{CHH}$ vs $\ce{HHH}$), (3S)- for 1g (isopropyl group is priority 2 based on $\ce{CHH}$ vs $\ce{CCH}$), and  (3S)- for 1h (phenyl group is priority 2 based on $\ce{CHH}$ vs $\ce{CCC}$). All these assignments are $\ce{C}$2 versus $\ce{C}$4 since priory 1 for all three groups on $\ce{C}$3 is $\ce{OH}$.
Thus, we can conclude that the change of stereochemistry is mainly due to the priority rules (not due to the change in stereochemistry on transition states or intermediates). This is also supported by the example given in the same reference (Ref.1) in the box in depicted diagram. Even when bulkiness at $\ce{C}$2 was increased by adding a methyl group, the reaction (with R-BINAP) has undergone same manner to give (R)-configuration to $\ce{C}$3 center.
References:

Ryoji Noyori, Takeshi Ohkuma, Masato Kitamura, Hidemasa Takaya, Noboru Sayo, Hidenori Kumobayashi, Susumu Akutagawa, "Asymmetric hydrogenation of $\beta$-keto carboxylic esters. A practical, purely chemical access to $\beta$-hydroxy esters in high enantiomeric purity," J. Am. Chem. Soc. 1987, 109(19), 5856-5858 (https://doi.org/10.1021/ja00253a051).


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Am I understanding buffering capacity against strong acid/base correctly?
I just want to make sure I'm understanding buffering capacity correctly.
I've been working on this problem:

Suppose you have an acetic acid buffer, $\mathrm{p}K_\mathrm{a} = 4.74$, at the following $\mathrm{pH}$:
1) $\mathrm{pH} = 4.00$
2) $\mathrm{pH} = 4.35$
3) $\mathrm{pH} = 4.70$
4) $\mathrm{pH} = 5.00$
5) $\mathrm{pH} = 5.40$
6) $\mathrm{pH} = 5.60$
Which buffer solution will have the highest buffering capacity against $\ce{HCl}$? Against $\ce{NaOH}$ ? Which is the best optimal buffer?

From what I understand, optimal buffers are where $\mathrm{pH} = \mathrm{p}K_\mathrm{a}$, since that is when the ratio between conjugate base and acid is equal to $1$. This means that there is just enough acid/base relative to each other to minimize $\mathrm{pH}$ changes whether a strong acid or base is added.
As for buffering against $\ce{HCl}$, we would want our $\mathrm{pH}$ to be as high as possible, right?
We want as much conjugate base as possible in our buffer to react with the $\ce{HCl}$, while minimizing the changes to the acid, which means we want $\frac{[\ce{A-}]}{[\ce{HA}]}$ to be as high as possible. Maximizing $\frac{[\ce{A-}]}{[\ce{HA}]}$ means $\log \left(\frac{[\ce{A-}]}{[\ce{HA}]}\right) \rightarrow \infty$, which means $\mathrm{pH} \rightarrow 14$, since $\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log \left(\frac{[\ce{A-}]}{[\ce{HA}]}\right)$.
The explanation for $\ce{NaOH}$ would be the exact opposite. You want as much acid compared to conjugate base as possible so that the acid reacts with $\ce{NaOH}$ while minimizing changes to the conjugate base, which means $\frac{[\ce{A-}]}{[\ce{HA}]} \rightarrow 0$, $\log \left(\frac{[\ce{A-}]}{[\ce{HA}]}\right) \rightarrow -\infty$, and $\mathrm{pH} \rightarrow 0$.
Is my understanding correct? The thing that's tripping me up is that I was told the exact amount of conjugate base and acid is more important than the ratio (e.g., $0.5 / 0.9$ is better than $0.05 / 0.09$), but since there's no mention of exact concentrations in the problem, I'm guessing we just assume that we have sufficient amounts of $\ce{A-}$ and $\ce{HA}$.
Is this generalization correct: if my buffer $\mathrm{pH} <<< \mathrm{p}K_\mathrm{a}$, then does that mean it is optimal against strong bases (and vice versa)?
Thank you for your help.

User porphyrin is correct that buffer capacity is defined as the number of moles of strong acid or strong base needed to change the pH of 1 liter of solution by ±1 unit.

The buffer capacity is a dimensionless number.
The buffer capacity is a somewhat fuzzy number in that the number of moles of base to cause + 1 pH change may not be the same as the number of moles of acid to cause a -1 pH change.
Any dilution of the starting solution is typically ignored.

Typically a buffer is supposed to protect against the solution becoming more acidic or more basic. Assuming a reasonable concentration of the buffer, at a pH of 4.74 the concentration of acetate anion will equal the concentration of acetic acid. Hence that pH, which is equal to the pKa, will have the optimal buffer capacity against acid or base. (Note that this is a bit fuzzy since there is no universal definition of "optimal buffer capacity.")
However the problem also asks about only acid or only base being added. So:

At a pH of 5.60 there is more of the acetate anion to react with HCl. (starting pH = 5.60, final pH = 4.60) However this buffer would poorly buffer the addition of a strong base since there is very little acetic acid.
At a pH of 4.00 there is more acetic acid to react with NaOH. (starting pH = 4.00, final pH = 5.00) However this buffer would poorly buffer the addition of a strong acid since there is little acetate which isn't protonated to acetic acid.

Further to the answer by Poutnik the buffer capacity $\beta$ of a weak acid - conjugate base buffer is defined as the number of moles of strong acid or base $C_B$ needed to change the $p$H by $\pm1$ unit, where
$$\displaystyle \beta=\frac{d[B]}{d\text{pH}}$$
and the equilibrium concentration base present is  $\displaystyle \mathrm{[B]}=\frac{k_w}{\mathrm{[H^+]}} - \mathrm{[H^+]}+\frac{C_BK_a}{\mathrm{[H^+]}+K_a} $.
(See answer How to set up equation for buffer reaction?)  where $K_w$ is the water ionization equilibrium constant $K_w =\mathrm{ [H^+][OH^-]} = 10^{-14}$, $K_a$ is the acid dissociation constant, and $C_B$ the total concentration of buffer.
After performing the calculation $\displaystyle \beta=2.303\left[\frac{K_w}{[\text{H}^+]}+[\text{H}^+]+\frac{C_BK_a[\text{H}^+]}{ ([\text{H}^+]+K_a)^2 }  \right]$
This can be analysed to find its maximum, but can be simplified first by ignoring the first two terms, because $K_w$ is tiny as is $[H^+]$ compared to the last term close to $K_a$. Plotting this last term with $C_B=0.02$ produces the following curves

where the max buffering is seen to be very close to the $pK_A$. You can see that buffering will be ok only when the $p$H changes by $\approx \pm 1$ and when the $pK_A$ is in the range from about $pK_A $ from $\approx 4 \to 9$. The red curve is the last term in the equation, the blue curve the full equation.
To answer your first question the $p$H should be close to the $pK_A$ and as the total concentration of base ($C_B$) increases the maximum just increases but does not shift.
To find the maximum differentiating the last term only with $[H^+]\equiv x$ produces, after a little algebra, $\displaystyle \frac{d\beta}{dx}=C_BK_a\frac{K_a-x}{(K_a+x)^3}=0$ where the maximum is found when $[H^+]=K_a$

The purpose of buffers is to keep $\mathrm{pH}$, with the differential buffering capacity $\frac { \mathrm{d[B]}}{ \mathrm{d(pH)}}$
If you are interested in the integral buffer capacity across $\mathrm{pH}$ range, than optimal is the buffer with the maximum capacity in the middle of the range. But the useful range for a single pair buffers is usually just  about 2-2.5.
The rest of the answer depends on if more important is the initial $\mathrm{pH}$, the differential or integral capacity, or just the amount of strong acid/base needed for the solution to become strongly acidic/alkalic.

As for buffering against , we would want our  to be as high as possible, right?

No, we would want  $\mathrm{pH}$ to have the desired value, otherwise we do not speak about $\mathrm{pH}$ buffers.

Is this generalization correct: if my buffer $\mathrm{pH} <<< \mathrm{p}K_\mathrm{a}$, then does that mean it is optimal against strong bases (and vice versa)?

It would not be a buffer, but just a weak acid. If $\mathrm{pH}$ is high enough, it would have higher neutralization capacity than a strong acid, as higher molar concentration is needed for given $\mathrm{pH}$. But its initial buffer capacity would be very low.


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Is there a way to create natural Red food coloring from Turmeric in the house kitchen?
As a non chemist I have read in the Wikipedia article Curcumin that:

Curcumin becomes bright red when it interacts electrostatically with
  phospholipid film.

This text appeared without source under this image in the article:
https://en.wikipedia.org/wiki/Curcumin#/media/File:Curcumin_thin_lipid_film.jpg
As an amateur of cooking Indian dishes (which often have bright red "Tandoori" color):
Is there a way to create natural Red food coloring from Turmeric in the house kitchen?
That is to ask; what edible supermarket products contain phospholipid films and how should I use at least one together with Turmeric (fresh-grinned or dry-powdery) to get a bright red solution?

Update
I should note that I didn't intend to ask this question as a "how to make Tandoori colored seasoning paste" question;
About six years ago, I was told by the owner of a large Indian restaurant in Israel, what is the secret (or at least one of the secrets) and the following seems to be that secret from what I understood then and did and replicated several times:
1) Take about 500 gram of dried Kashmiri peppers; clean them from pedicel, placenta and seeds.
2) Put the dried Kashmiri peppers in a metal pan and pour boiled water upon them.
3) Let the dried Kashmiri pepper boil and cool ("water-marinate") about 3-6 hours but don't cook them.
Mere boiling, without cooking, enhances pigmentation but doesn't destroy some pigments as in cooking.
4) Remove the water and let the peppers dry.
5) Grind the processed peppers into a paste with a bit of water and cooking oil and add a tablespoon of that paste as a basis for any Tandoori paste you are making (along with salt, lemon, garlic, ginger and possibly other spices).

Turmeric turns distinctly red at a pH above ~7.8 (we had this experiment in school). Considering that you're cooking tandoori, you would probably use salt and lemon in the seasoning, which would cause the medium to be acidic and revert turmeric into its natural yellow colour.

@AniruddhaDeb in great plea ; please publish this in an answer such as "adding on what Dr. Pippik said"; as you probably know, comments might get deleted with time from various reasons, and yet this is very important. If you choose to do so --- please say if potassium would be a seasoning alternative to salt to prevent yellowization.

Alright will do! Im in the middle of exams so I'm a bit squeezed on time but I'll try to write an answer tomorrow

@AniruddhaDeb if you still have the will (and hopefully if you have spare time), about yesterday the question was reopened and you are now available to publish an answer; kindly,

Yep I'm working on an answer right now. Some practical stuff is involved (and I want to take pictures) So I'll post in around a day from now.

@JohnDoea You just wanted reddish color, what about beet juice? It will impart reddish pink color to foods

@M.Farooq I thought about it a few years back but didn't find any way to implement it to make Tandoori color - perhaps the powder of some species would be helpful.

@JohnDoea, The tandoori color on chicken is mostly a food dye in all ethnic restaurants because most Westerner would not like a very spicy chicken. The reddish color in real tandoori chicken comes from adding dried red chillies to the $masala$ (all species). and sometimes tomatoes. When you fry red chillies in oil or tomatoes, the oil develops a reddish color. It is beyond my tolerance as well. Another source of nice organe color is saffron  and as you know genuine saffron is extremely expensive. https://tasty.co/recipe/homemade-chicken-tikka-masala

@M.Farooq yes, and there is also the Ratan Jog / Ratin Jot (Alkanna tinctoria root) common in Vindaloo; but indeed, some real, non-lousy Indian restaurants use a paste based on the kashmiri boil-marinate paste I described in the update; but it is indeed rare to find in comparison to artificial dye.

@JohnDoea, Interesting I wasn't aware of ratan jog, but I would say it is extremely rare in typical South Asian household foods. Apparently there are side effects.  https://punampaul.com/kashmiri-rogan-josh/

Curcumin(the active ingredient in Turmeric) can turn red in two ways: Either by acting as an acid-base indicator or by forming a complex with a phospholipid film, as you stated.
Colour change due to Acid-Base action
The curcumin molecule is a tribasic acid in basic solution, where it exists as its enol tautomer. The loss of one or more protons from the phenolic hydroxyl groups convert the benzene rings into their quinonoid form, which have a larger absorption/emission wavelength than the benzenoid form, giving a red colour.

This indeed works experimentally: a paper highlighting this can be found here. A simple home experiment would also suffice

The left one is a suspension of turmeric in water; the right one is the same suspension, but with a few drops of soap added to it, which make the medium alkaline and cause the turmeric to turn red. Unfortunately, around my kitchen, I could not find an edible compound basic enough to cause turmeric to change colour. Lime is a possible compound that I could test with but I don't have any foodgrade lime powder lying around. Another issue with using this colouring is that any acidic medium will revert the colour back to yellow: this means that sprinkling lemon as garnishing on your dish would change the colour from an appetizing red to a dull yellow.
Colour change due to Phospholipid complex
A lot of papers have been released on this topic, such as this one and this one. The second paper mentions the following synthesis method to prepare curcumin liposomes (curcumin surrounded by a spherical membrane of phospholipid)

The MFGM phospholipids/soybean lecithin was dispersed in chloroform. Curcumin was dissolved in anhydrous ethanol and thoroughly mixed with the MFGM phospholipids/soybean lecithin dispersion in a 250-mL round-bottomed flask. The mixture solution was evaporated on a rotary evaporator at 35°C, until a thin film was formed on the flask. The film was washed with PBS and completely hydrated for 2 h. After the hydration of thin lipid film with PBS, the lipid dispersion was sonicated for 4 min, and finally the liposomes encapsulated with curcumin were obtained.

For cooking food, this above synthesis method seems unreasonable and overkill. I urge you to experiment with a few methods of forming the red curcumin-phospholipid complex yourself, with soy lecithin/eggs. I would have tried some methods out but I don't have these ingredients at home. see edit.
The advantage of having curcumin in a liposome is that it is much more stable: unlike the acid-base action, curcumin in a liposome remains stable across a wider range of pH's, as shown in this graph. This method does show promise, but as I mentioned, you would have to give this a try yourself.

Other practical solutions
Seasoned Advice would probably give you would get a better answer in this field, but I'll drop in my two cents. Generally, chilli powder is used to provide the red colour to Tandoori. More specifically, Kashmiri chilli powder is specifically used for tandoor preparations, such as tandoori chicken and chicken tikka. This powder is applied as a dressing (in combination with oil, rice flour etc after cooking the food) and during marination (chicken is marinated with curd, red chilli powder and turmeric).
Happy cooking!
References:

Kuntal Maiti, Kakali Mukherjee, Arunava Gantait, Bishnu Pada Saha, Pulok K. Mukherjee; Curcumin–phospholipid complex: Preparation, therapeutic evaluation and pharmacokinetic study in rats, International Journal of Pharmaceutics, Volume 330, Issues 1–2,
Sam-ang Supharoek, Kraingkrai Ponhong, Watsaka Siriangkhawut, Kate Grudpan;
Employing natural reagents from turmeric and lime for acetic acid determination in vinegar sample, Journal of Food and Drug Analysis, Volume 26, Issue 2,
Hong-Hao Jin, Qun Lu, Jian-Guo Jiang; Curcumin liposomes prepared with milk fat globule membrane phospholipids and soybean lecithin, Journal of Dairy Science, Volume 99, Issue 3,
Research gate: Why does curcumin have a red colour in the alkaline condition
Wikipedia: Phospholipids, Curcumin

EDIT: I tried to make a phospholipid complex with turmeric and egg yolks in the kitchen. Around a teaspoon of turmeric was mixed with an equal amount of egg yolk and heated in a water bath: An orangish colour was observed, which was in contrast to the bright yellow of the egg yolk. (I scraped the top layer away which had a larger amount of turmeric to reveal the bottom layer, which had more egg yolk)

A possible conclusion that can be drawn from this experiment is that the complex is indeed forming, but the concentration of phospholipids/curcumin is too small to impart the magnitude of red colour required for food colouring. This pretty much conclusively proves that no easily made derivative of turmeric can be used as red food colouring.

Many foods have high phospholipid content. Eggs and soybean have comparatively high amounts, for example. Phopholipids are essential components of all cell membranes, and are in virtually all living things. They function as emulsifiers, one side of the molecule adhering to water (hydrophilic), and the other to fats (termed lipophilic or hydrophobic).
You can buy food-grade lecitithin, derived from various sources, e.g. sunflower seeds, soy beans, eggs, fish, etc., for about US$7.00/100 g.
So I suggest you to try mixing the turmeric with eggs or soybeans or plain lecithin and to let us know if the color changed to an appealing red! Or, as others stated, use some alkali, such as baking soda, though that certainly would change the taste.

Hello Dr, thank you ; just one question - about using baking soda and turmeric, must it be in an oily solution right? Is there a best oil or group of oils for such solutions (refined/cold-pressed)?

As stated by Aniruddha Deb, the pH is the cause of the color change. See https://en.wikipedia.org/wiki/PH_indicator . Read, and learn.

I assume Dr. Pippik won't be mad if I give the green check mark and 15 rep points to Aniruddha Deb who worked very hard to construct his answer which included direct testing.

Some common materials you might have at home are baking soda (sodium bicarbonate) or washing soda / soda ash (sodium carbonate). Adding them to hot water + turmeric makes a solution that stays bright red for quite a while (as others have said, it will revert over time, but in my experience with dying fabric it's a matter of several hours, so it won't affect food you use in the space of a meal). Avoid using lemon or anything acidic because that will send it back to yellow (even if there's baking soda in it). If you add a small spoonful of baking soda, the flavor won't really be influenced (especially if you're using other spices with a strong taste).

Do you use dying fabric when cooking? Maybe you want to make this part a little more clear.


text:
Do metals form covalent bonds besides ionic and cordinate bond?
My chemistry textbook says that metals form ionic or cordinate bonds whereas non metals form covalent bonds. But in another textbook I read that Lithium, Beryllium, Aluminium, Chromium, Manganese etc, which are metals, form covalent compounds like LiCl, BeH2, AlCl3, H2CrO4, H2MnO4 respectively
Now my question is whether metals form covalent bonds or not? And if yes then how to predict which metallic compound contains ionic and which contains covalent bonds?

There are lots of covalent bonds formed by metals, for example in haemoglobin, chlorophylls, vitamin B12, hydrogenases and plastocyanins and numerous transition metal complexes.

https://chemistry.stackexchange.com/questions/16261/metal-compounds-that-bond-covalently

There's no difference between "coordinate bond" and any other more or less covalent bond.

https://chemistry.stackexchange.com/questions/49076/do-there-exist-metals-that-are-composed-of-molecules?rq=1 https://chemistry.stackexchange.com/questions/104408/gaseous-metals?rq=1

The first thing that should be said is that there's no difference between a coordinate bond (dative bond) and an ordinary covalent bond. Yes, the electrons "come from different places"; but the molecule doesn't actually know this, nor does it care. Once a covalent bond is formed, it is a covalent bond, regardless of where the electrons "come from".
Some of the examples you have given can equally be described as coordinate bonds or covalent bonds. For example, $\ce{MnO4-}$ could be viewed as four $\ce{O^2-}$ ligands forming coordinate bonds to a central $\ce{Mn^7+}$. Or, you could also just view them as plain old Mn=O covalent bonds. It doesn't actually matter.
Having established that there is no real difference between coordinate bonds and covalent bonds, the only real question is how can you tell a covalent bond apart from an ionic bond. The answer is mostly to do with polarisability. If you imagine an ionic bond between a strongly polarising cation ($\ce{Al^3+}$) and a polarisable anion ($\ce{Cl-}$), the cation will pull over some of the anion's electron density: this is exactly what is meant by "polarising". This changes the nature of bonding from an ionic model (where the bonding electrons reside entirely on one atom) to a covalent model (where the bonding electrons are somewhere in between the two atoms). You'll notice that all the cations you've listed: $\ce{Li+}$, $\ce{Be^2+}$, $\ce{Al^3+}$, (formally) $\ce{Cr^6+}$, and $\ce{Mn^7+}$, are very highly polarising. The first three have a very small size, and the last two have exceptionally large positive charges.
That said, covalent bonds between two transition metals are reasonably common, too: a simple example is $\ce{Mn2(CO)10}$ (Wikipedia), which is essentially $\ce{(OC)5Mn-Mn(CO)5}$ with a single bond between the two.
The bond order between two metals can go up to slightly ridiculous numbers. Cr–Cr and Re–Re quadruple bonds are well-known in $\ce{Cr2(OAc)4(H2O)2}$ and $\ce{[Re2Cl8]2-}$ respectively. Even quintuple and sextuple bonds between metals have been described (although some of these accounts are not without controversy).
How can you predict this, though? It's not trivial at all, and in truth you probably don't need to care about these exotic examples until you have enough experience to understand the bonding.

so you mean we cant predict the type of bonding present in metallic compounds by using any general rule but prediction comes with the experience and hence we get remembered about the type of bonding present in a metallic compound ?

Mercury is worth mentioning. And I would like to plug my question: What would follow in the series sigma, pi and delta bonds?

$\ce{Na2}$ and $\ce{K2}$ do exist in the sodium and potassium vapors. This can be proved by analyzing the highly diluted flames produced by hot vapors of these metals in contact with diluted $\ce{Cl2}$ vapor, as proved by Polanyi and co-workers. See the following references:
M. Polanyi, Atomic Reactions, Williams and Norgate, London (1932); M.G. Evans, M. Polanyi, Transactions Faraday Society 35, 178 (1935); C. E. H. Bawn, Ann. Repts. Chem. Soc. 39, 36 (1942).

Your textbook is right.  Lots of metals form covalent bonds.  In the case of lithium chloride such bonding is one explanation for the solubility of this compound in organic solvents (see this answer).
One additional example you might want to know about is Grignard reagents, a class of highly basic compounds in which carbon is covalently bonded to magnesium.  These are widely used in organic synthesis processes.

These have relatively low percent of covalent component of bonding  (are most "basic" though). There's plenty of symmetric metal-metal bonds.


text:
Why is iso-octane different from other iso-s?
While learning nomenclature of organic compounds I found the common  names to be a bit difficult to learn than the IUPAC ones. One of these common name stuff is the iso-,neo-, sec-, etc. So I searched for the ways to differentiate between these on the YouTube and found this video (note that the anchor of the video uses hindi for communication so it might not be intelligible to most people here). To summarize what the anchor said:

Prefix-iso: is used when all carbon atoms except one form a continuous chain.
$$\mathcal {OR}$$
They have a structure like this:

Now in the same video the anchor says that iso-octane is an exception to this rule, as it has the following shape:

Before this came in the video I formed the view that these are just the ways of naming compounds with similar structure but now I think that that isn't the case. So:

Why iso-octane has shape different than the other iso-s?

The anchor says that the structure of iso-octane was determined experimentally. This implies that there is some way to identify these compounds without knowing their structures beforehand so:

How do we know that whether some given compound is iso-, neo-, etc?

What might be the most general guidance for using iso-, neo-, sec- and tert- prefixes?

Do not try to learn chemistry from youtube videos of questionable background.

"secondary" and "tertiary" (+"primary", "quarternary") are clearly defined (by IUPAC) structural descriptors. "iso" and "neo" are not.

@Karl thanks for the suggestion. Actually this was the first time I watched a video from that channel and thought that it might be a legitimate one given that there are 100k+ subscribers. Though you are right and I do try to avoid such things but for now I wasn't able to find any other resources online so I had to reside for that video.

This would be one that is reasonably reputable: https://en.wikipedia.org/wiki/Alkane#Nomenclature

Looks like a slang term that got commonly recognized.  It's definitely unpleasant to the eye, much like $`` \mathrm{C} 0{}_{2} ".$

The usage of non-standard prefixes such as "iso" has never been defined with hard, unambiguous rules such as those in the IUPAC recommendations. Instead, their usage has often more been "by convention". In Organic Chemistry (1st ed.) by Clayden et al., the authors write (p 315):

Strictly speaking, if the standard meaning of ‘iso’ is followed, the name isooctane should be reserved for the isomer 2-methylheptane. However, 2,2,4-trimethylpentane is by far the most important isomer of octane and so, historically, it has been assigned this name.

(This phrase is also found at Wikipedia, where it is copied wholesale without any indication that it's a direct quote. Please don't do this in your work. Even if you provide a citation, you should either paraphrase the information, or clearly indicate that it's a quote, e.g. by enclosing it in quotation marks.)

If I remember right from ~25 years ago quoting my chemistry teacher "Iso-" [equal] indicates a mixture of carb-molecules where the longest chain is the name-giver .. as in iso-octane being an 8-carbon chain. .. Other possible molecules use side-chains of 1 or 2 carbons length.


text:
Why is freezing point depression greater than boiling point elevation for equimolar substances?
I know that the freezing point depression and boiling point elevation are given by the formulas $\Delta\mathrm{T}_{b}=ib\mathrm{K}_{eb}$ and $\Delta\mathrm{T}_{f}=ib\mathrm{K}_{F}.$ It seems that when this formula is applied for equimolar concentrations, the magnitude of freezing point depression is greater than boiling point elevation. Is there a qualitative reason for this? If so, what is it?

To find an answer, let us take a look at how the cryoscopic constant (for decrease in melting point) and the ebullioscopic constant (for increase of boiling point) are defined:

Please note that the Kelvin-scale is used for temperature here. The boiling point is always above the melting point (we keep pressure constant to keep the example simple).
For most substances, the enthalpy of vaporization is greater than the enthalpy of fusion. It follows that the denominator for K_eb is greater, resulting in a smaller K_eb.
This means that the effect of greater enthalpy of vaporization compared with enthalpy of fusion outweighs the effect of boiling temperature being greater than the melting temperature (even though it's the square of absolute temperature).


text:
Why is t-butyl chloride favored over t-butanol at 0°C?
To make t-butyl chloride from t-butanol, you need to bubble HCl gas into a 0°C ether solution of t-butanol.
To make t-butanol from t-butyl chloride, you can dissolve t-butyl chloride into water (not sure the temperature).
Why are the products favored in these two conditions? What is the significance of using a 0°C solution in the first case?

T-butanol tends to undergo an E1 reaction in acidic medium. The 0°C helps to prevent that reaction and favour the SN1 product.

This is basically an esterification equilibrium:
$\ce{t-Butanol + HCl = t-butyl chloride + H2O}$
The term esterification is not limited to esters of carboxylic acids.
You are shifting the chemical equilibrium according to Le Chatelier's principle.
When talking about chemical thermodynamics and chemical equilibrium, the following equation should come to your mind:
$\Delta G = \Delta H - T\Delta S$
This is often referred to as the Gibbs-Helmholtz-equation.
Lower temperatures will cause the -TΔS term and favor the formation of t-butyl chloride + H2O over the formation of t-butanol + HCl. This is because I suggest that the entropy is greater for the t-butanol + HCl side. By using an excess of HCl gas, you further shift the equilibrium.
In the second case, you hydrolyse t-butyl chloride (the ester) by applying an excess of H2O. If the entropy is greater for the t-butanol + HCl side as I suggest, an increased temperature will favor the hydrolysis.
I think for the E1 of tert-butanol, harsher conditions like conc. H2SO4 would be necessary.


text:
Understanding reason behind existence of chemical equilibrium
Many chemistry reactions are reversible in nature.
In such reactions initially reactants react to form products but after some time products also react to give reactants back again. And an equilibrium is established where forward and backward reactions occur with equal rates.
If products are more stable, why will they react again to give reactants?

If the Pacific ocean is deeper (which it is), why wouldn't all water from Atlantic pour into it?

" but after some time ..." No. Directly, always.

This is one version of the central question of thermodynamics. The solution was to introduce the concept of entropy.

Does this answer your question? Why does equilibrium exist?

It is best to think of reversible reactions statistically
A good understanding about equilibrium can be had by thinking about the detailed molecular level processes involved at the molecule level.
Take, for example, the simple equilibrium between liquid water and water vapour in a closed vessel. At room temperature, the equilibrium clearly favours liquid water. But there is always some water vapour present (that's what we call humidity in the wider atmosphere). Why is there any vapour if the equilibrium favours liquid? The reason is that the water molecules don't all have the same amount of energy. some are travelling fast and some some are slow. A small percentage of the fast ones can escape the liquid and become vapour (ie gas-phase water molecules not liquid phase water molecules). In the gas above the liquid, the same is true. Not all the gas molecules have the same speed. Some are travelling slowly and, if they hit the liquid surface, stay in the liquid. Both processes are dynamic and are happening fairly fast. So molecules are constantly exchanging between the vapour and the liquid.
Equilibrium is not reached when everything stops: at room temperature there is a lot of kinetic energy shared out among the molecules of the system. Equilibrium happens when the rates of evaporation and liquefaction are the same. So the number of molecules per second going into the vapour from the liquid equals the number going the other way.
The same is true for other, more complex, reversible reactions. Equilibrium is not stasis but an average state resulting from dynamic reactions in both directions.
This point of view makes it clear why, even when one product is far more stable, the reverse reaction can still happen.

It is no use comparing the stability of the reactants and products.
Maybe you would better understand with an experimental example. Suppose you prepare $\pu{10 mL}$ of an alcoholic solution of cobalt chloride hydrate $\ce{CoCl2.6H2O}$. This solution is blue due to the formation of $\ce{CoCl4^2-}$ according to the following equation:
$$ \ce{2 CoCl2.6H_2O -> [Co(H2O)6]^2+ + [CoCl4]^2-}$$
The ion $\ce{[Co(H2O)6]^2+}$ is pale pinkish, and the ion $\ce{CoCl4^2-}$ is dark blue. So the color of the whole solution is blue. Now you add two or three drops of water. You observe that the solution turns pinkish, because of the following equation, which I will show that it is an equilibrium:
$$\ce{[CoCl4]^2- + 6H2O <=> [Co(H2O)6]^2+ + 4 Cl^-}$$
This equation is written in the exothermic sense.
Now if you heat this pink solution to $\pu{60 ^\circ C}$ or $\pu{70 ^\circ C}$, the solution turns blue. $\ce{[Co(H2O)6]^2+}$ is transformed into $\ce{CoCl4^2-}$. The reaction goes to the left hand side. This is a consequence of the Le Chatelier principle: Heating a system favors the endothermic side, which is also the side where the entropy is maximum.
Cooling the hot blue solution will change its color to pink again, because $\ce{[Co(H2O)6]^2+}$ is favored since cooling favors the exothermic transformation. This can be repeated by successive heating and cooling.

It is no use comparing the stability of the reactants and products. And in the course of your answer you are comparing the stabilities of those reactions...


text:
Chain isomers of hexane
Why does hexane just have 5 chain isomers when I can think of more, like 2,2,3-trimethylbutane?
That's just one of the top of my head, I could think of more.

Trimethylbutane has 7 carbons, not 6.

The list

n-hexane
2-methylpentane
3-methylpentane
2,2-dimethylbutane
2,3-dimethylbutane

is definitely exhaustive.


text:
CuS+HCl is it a reduction or oxidation reaction
How do we identify whether a reaction is oxidation or reduction and please let me know whether the equation given above is oxidation or reduction.

We don't. There is no such thing as oxidation reaction. Ditto for reduction. Every redox reaction is a reduction and an oxidation at once. Then again, there are quite a few reactions which are not redox at all, and yours is one of those.

Ivan is right. Your reaction is probably : $CuS + 2 HCl -> CuCl_2 + H_2S$, which is surprising, because usually $CuS$ does not react with $HCl$. Usually it goes the other way. In qualitative analysis, $Cu^{2+}$ ions react with $H_2S$ to produce a dark precipitate of $CuS$

@Maurice If I remember correctly, CuS precipitates from dilute HCl; in conc. or demiconc. HCl, i think no sulfide is stable.


text:
Understanding the gaff2.dat proper dihedral parameters
I have a project where I'm trying to program my own force field, and I decided to use the AMBER gaff2.dat force field parameters.
However, I am really struggling parsing the gaff2.dat file when it is about the dihedral angle parameters.
I am now trying to replicate the rotation around -O-C-C-O, for which I know the profile :
$$ V_{std} = 0.25 [1+ \cos 3\omega] + 0.15[1+\cos 2\omega]$$
I read that for gaff you have to take every parameters into account, so for $\ce{-O-CH_2-CH_2-O-}$ I have to use the parameters for O-C-C-O, H-C-C-O and H-C-C-H. In the gaff2.dat file I get :
os-c3-c3-os   1    0.000         0.000          -3          p28,suger5ring,suger6ring,coccoc GA AUE=1.1750 RMSE=1.6708 TorType=2
os-c3-c3-os   1    0.000       180.000          -2
os-c3-c3-os   1    0.170       180.000           1

hc-c3-c3-os   1    0.000         0.0            -3.         Junmei et al, 1999 TorType=1
hc-c3-c3-os   1    0.250         0.0             1.         Junmei et al, 1999 TorType=1

hc-c3-c3-hc   1    0.120         0.000           3          m1 SS AUE=0.2420 RMSE=0.2944 TorType=2

Which I guess would mean that :
$$ V_{gaff} = 0.170[1+\cos (\omega -\pi)] + 0.25[1+ \cos \omega] + 0.12[1+\cos 3\omega]$$
However, this gives this profile :

Which is not what I expected to get.
I have spent the past week reading a lot about the AMBER force field parameters and I ran out of ideas. So if anyone know how these dihedral angle are parametrized I could really use some explanation.

In case you aren't aware you might also try the matter modeling stack exchange.


text:
Only valence shell electrons participate in bonding then how is Cr in H2CrO4 forming 6 covalent bonds although it has only 1 valence electron?
The structure of $\ce{H2CrO4}$ is as follows:

Shell configuration of chromium is 2,8,13,1 i.e. it has only 1 valence electron and only valence electron participate in bonding but Chromium in the above structure is forming 6 covalent bonds i.e. it is sharing 6 electrons. How's that possible ? How can it share 6 electrons when it only has 1 valence electron.
The same is the case for $\ce{Mn}$ in $\ce{H2MnO4}$ where it has only 2 valence electrons but is forming 6 covalent bonds and for $\ce{Mo}$ in $\ce{H2MoO4}$ where it has only 1 valence electron but forming 6 covalent bonds.

The definition of valence electrons is a lot more blurry than it seems.

And covalent bonding capacity is based on valence orbitals (which can also get blurry), not valence electrons.  Boron forms compounds where four valence orbitals are involved in covalent bonds even though it brings in only three valence electrons.

The electronic configuration of $\ce{Cr}$ is: $[\ce{Ar}]4s^1 3d^5$. Because of very little energy difference between 4s- orbitals and 3d- orbitals, it can lose 6 of it's valence electrons and thus form 6 covalent bonds.


text:
Can density functional approximations including dispersion correction perform worse than without it?
I am trying correlate some experimental properties of a molecule to its calculated equilibrium geometry. To do so, I calculated the geometry and energy of some conformers.
Now, I obtain mixed results when performing the same energetic calculation with dispersion-corrected (D3) and not corrected functional (B3LYP): when no correction is applied, a conformer has lower energy, while with D3 correction another conformer has lower energy.
The idea of using D3 looked reasonable, as the molecule has two aromatic and two aliphatic wings, which from the D3-uncorrected geometry seemed to promote to some degree an aliphatic-aromatic stacking interaction.
The point is: my intuition drove me to the use of the dispersion correction, and indeed the D3-corrected result would agree wonderfully with spectroscopic results. But is there a case in which a D3-uncorrected result would be more accurate than the corrected one? Or, alternatively, is there a way to understand quantitatively if the uncorrected model does not describe correctly the system?
I saw a lot of reviews talking about the "benefits" of DFT-D3, but could it do worse than DFT alone?
Edit, to answer Marvin's questions:
Yes, the molecule is organic. It all started with a "routine calculation set-up": b3lyp/6-31g*, but it later became evident that more refinement was required, so the current level might be seen as an expansion of the original calculation. Beyond the 6-31+g** level, I don't stress on Pople's basis anymore, and if more refinements will be required, Alrichs basis sets will be employed. Yes, I omitted it, but B-J damping was employed. At the moment I have optimized one structure with D3(BJ), the other one is being optimized. Of both the geometries optimized with B3LYP, the D3BJ-energy was calculated, together with the uncorrected one.
The two molecules arise indeed from a semiempirical conformational search, followed by a fast and inaccurate hf/3-21g* optimization on 100 geometries. A dozen of low energy conformers was optimized with hf/6-31g*, and the two outliers/low lying conformers were subject to further study with the methods discussed. The molecule contains 90 atoms, and the calculated energy difference is less than 0.5 kcal/mol, but as NMR experiments show a net predominance of one geometry, it's likely that the low energy difference is coupled with an high isomerization barrier, which I still haven't calculated: at the moment I am focusing on the equilibrium geometry.
As soon as the "new" geometry calculations will complete, I will probably do more calculations with a functional better suited for dispersion corrections, thank you very much!
Note: an NMR calculation is planned, but as good results require time and a strong basis set, it will probably be the last thing I will do, starting with better geometries.

Very, very unlikely. Especially since version 3 it should very generally yield improvements only. Now the real issue may well be the functional you are using. While B3LYP is popular, its many failures have thoroughly been reported. You have not been very specific about the differences either, that may very well be within the general error of the method. Another thing to consider is basis set. But again: DFT-D3 will very likely be the smallest source of error.

Thank you! I used B3LYP because originally I didn't consider implementing it with D3; as to the basis set, at the moment the basis set is 6-31G(d,p) for the geometries, and energy calculations were run on some geometries at the slightly higher 6-31+g(d,p). The molecule is neutral, non ionic and with no radicals, so the basis set looked to me like a good compromise in terms of performance and time, but I might consider improving it in the next days

I'm not a big fan of Pople basis sets, but the set-up makes sense. I guess your molekel is organic, so it is surprising to see b3lyp fail. Can't make reasonable statements without knowing more specific things. General recommendation: use many functionals for initial benchmarking. If you need something fast, go with M06L (or later, no dispersion needed), BP86-D3BJ, PBE-D3BJ, TPSS-D3BJ. Hybrids: M06, PBE0-D3BJ, B3LYP-D3BJ, etc... You'll probably find an outlier within those... Are you implementing the DFA yourself?

More questions: Did you use Becke-Johnson damping? You have optimised the structure with/without D3, then calculated the energy on the opt. struc.? What is the energy range we're talking about? Another recommendation: Try xtb and crest (https://github.com/grimme-lab/xtb/releases) for automatic conformer search; it's semi-empirical, so it's incredibly fast, produces very good structures.

tldr; I'll check the detailed results, but the overall answer is that dispersion correction is absolutely worth it
We just finished a benchmark over ~700 molecules with up to 10 conformers, so ~6700 single points at the DLPNO-CCSD(T) / def2-TZVP level. I will post a link to the preprint when it's available soon.
In the meantime, here's a quick summary for the effects of dispersion corrections:

Comparing un-corrected PBE, B3LYP, and ωB97X single-point energies to DLPNO-CCSD(T) illustrate a significant effect. The uncorrected median R2 values drop by ~0.12, and the median Spearman correlations drop by ~0.08. For example, the median R2 of B3LYP / TZ drops from 0.920 to 0.706 without the D3BJ dispersion correction.

The metric here is the correlation between a method (i.e. B3LYP) with the DLPNO-CCSD(T) energies.
I'll look to see if there's ever a case where the dispersion correction does worse. But there are many, many papers illustrating that conformer energetics are dominated by non-bonded interactions .. which are improved with dispersion-corrected density functional methods.
Your methodology seems solid, although if you want good energy evaluations of different conformers, I'd consider either RI-MP2 or DLPNO-CCSD(T) single-points. Our timings in the paper show that RI-MP2 can be about as fast as hybrid DFT methods, but more accurate at ranking conformers.

Any consideration for F12 computations at your CC level of theory?

AFAIK, there’s not a DLPNO-CCSD(T)-F12.. the ORCA manual says these will be available “in the foreseeable future.”

When they are available, see how F12 affects any of your structures (if any) that have 2nd row main group elements, particularly P and S.  I've found F12 to DRAMATICALLY improve relative energies with these types of compounds, particularly when 2nd row elements are bonded to elements in the 1st row.


text:
How can there be concentration and pressure terms simultaneously in the Nernst equation?
In the Nernst equation we are taught that the reaction quotient can have pressure terms and concentration terms simultaneously. According to my knowledge this does not happen anywhere else, so why can it happen here?

Look up “Hypochlorite” in wiki: the first equilibrium expression shows it happens. There are many, many more as well. The acidified hypochlorite equilibrium, by the way, is why some careless people gas themselves when they try to clean their toilets.

This topic often causes confusion.
The most correct form of the Nernst equation uses thermodynamic activity of the substances:
$$E = E^{\circ} - \frac{RT}{zF} \ln\frac{a_\mathrm{red}}{a_\mathrm{ox}}$$
For diluted solutions, we can replace the activity of a substance with its concentration in the solution. But how do we express the activity of a gas?
This is basically the question of how to formulate the equilibrium constant of the redox reaction. There are actually multiple solutions to this. The problem is that all of them are correct.
Solution 1: Express activity of the gas as a function of partial pressure.
\begin{align}
pV &= nRT \\
\Rightarrow p &= \frac{n}{V}RT = cRT\\
a(\mathrm{gas}) &\approx c(\mathrm{gas}) = \frac{p(\mathrm{gas})}{RT}
\end{align}
Solution 2: Just use the partial pressure itself:
$$a(\mathrm{gas}) = p(\mathrm{gas})$$
Solution 3: Use the partial pressure divided by a reference pressure:
$$a(\mathrm{gas}) = \frac{p(\mathrm{gas})}{p(\mathrm{reference})}$$

I have updated your post with chemistry markup. If you want to know more, please have a look here and here. Especially \rm will introduce unwanted side effects. Solution 1 & 2 do not conserve the unit of $a$.

You should explain how to get rid of the units, because the logarithm of a number has only a sense if this number has no units. The log of a pressure has only a sense when it is divided by a reference pressure. Which sort of reference do you recommend ?

I don't agree with solution 1. What about the concentration of the dissolved gas? If the reaction is between a gas and a compound in solutions, then it's the gas dissolved in solution that matters not it's concentration in air.


text:
What does (sol) in a chemical equation mean?
I'm reading Shrivers Inorganic Chemistry book and some of the equations have a (sol) state on them. I'm sure the book has already defined this but I cannot find where.
Examples:
\begin{align}
  \ce{[NH3OH]Cl (sol) + NaOBu &-> NH2OH (sol) + NaCl (s) + BuOH (l)}\\
  \ce{SbF5 (l) + HF (l) &-> H2F+ (sol) + SbF6- (sol)}
\end{align}

It is a non-standard form of writing solution, IUPAC suggets (sln) https://chemistry.meta.stackexchange.com/questions/4165/what-is-the-standard-way-to-denote-physical-states-in-a-chemical-reaction

The state (sol) or (solv.) stands for "solvated". Since you are dissolving $\ce{HF}$ in liquid $\ce{SbF5}$, there is no water in that system (which would cause hydrolysis). So it is not correct to use (aq), since it is a nonaqueous system.
Here are some other examples:

Autoprotolysis of liquid ammonia:
$$\ce{2NH3 (l) <=> NH4+ (sol) + NH2- (sol)}$$
Autoprotolysis of nitrosyl chloride:
$$\ce{NOCl (l) <=>  NO+ (sol) + Cl- (sol)}$$
Dissolving sodium metal in liquid ammonia creates solvated electrons:
$$\ce{Na (s) <=>  Na+ (sol) + e- (sol)}$$

I have updated your post with chemistry markup. If you want to know more, please have a look here and here. You may also have a look at my meta answer to 'What is the standard way to denote physical states in a chemical reaction?' and resources linked within. The IUPAC actually suggests (sln).


text:
Given that they contain an OH group, why do carboxylic acids not act as base?
As far I know carboxylic acids have a OH group but during dissociation only hydrogen gets separated. The CO moiety is polar, demonstrating that oxygen is much more electronegative than carbon. Why does the OH group not get separated from carbon and behave like a base?

The same way sulfuric acid, nitric acid, etc do not act as bases despite containing hydroxyl groups, maybe?

The carbon is already polarised from the carbonyl oxygen. It does not want to give away even more electron density, and certainly you cannot get it to take part in a charge separation.

Well it can act as a base depending upon what kind of other substances are there in your mixture. As a matter of fact any substance can act as a base or an acid depending on the substances that are present. Also there are more ways of defining a base (here I think it is a good idea to use the Lowry-Bronsted definition in which a base is anything that can capture H+ and the OH of a carboxylic acid can indeed capture a H+ that may come from a strong acid like H2SO4.And indeed as mentioned before it leaves as water not as HO- the reason why it doesn't act as a normal base (for example NaOH where you have HO- in the solution when you dissolve it in water) is because that one is an ionic compound Na+ and HO- while in a carboxylic acid the bond between C-O is a covalent bond even if it is polarized so it needs some other conditions for its departure to be reasonable (energetically speaking) thus a HO- can not just leave the compound.

Well. Carboxylic acids are acids in water. In other solutions, they may act as a base. For example, in the esterification process, they are mixed with an alcohol in the presence of sulfuric acid (or any other dehydrating agent). It has been possible to show, by using unusual isotopes of Oxygen, that the oxygen bound to H in the -COOH group is eliminated, as if it was a base. The point is that the Oxygen atom used for the -C-O-C- bridge, typical of esters, is coming from the alcohol and not from the carboxylic acid.

The OH likely leaves as its protonated form as water, not hydroxide.


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Hessian Matrix and Physical directions: Potential Energy Surface
How do the eigenvectors of the Hessian Matrix for potential energy surface correspond directly to the physical directions of translation, rotation and vibration?
I see that they might correspond to some physical directions, but it seems fishy that they directly correspond to these very neat directions.

Hi @Mahathi, despite there being a good answer already, I wanted to say that this was a very good question. I would like to invite you to participate as one of the first group of users in the Materials Modeling Stack Exchange: https://area51.stackexchange.com/proposals/122958/materials-modeling?referrer=MGY4NjYyM2ZlY2NkNDI0MWRmYjhjMGI1N2RmM2U3ZDk1N2VjOWQ1MGU1MmVhNDZkZjI5MzNiNzY1MmE2ZjRhOYrAzULyFn6juGAbZ7wko5LvlWXZ0iQfr620nQ70gVNl0. 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.

I think you have the causality backwards here. It's not as if we have the vibrations to start with and they happen to be matched by the eigenvectors of the Hessian. We choose to look at these particular vibrations precisely because they are the eigenvectors of the Hessian.
Consider the example of water. We know that it has it has $9$ degrees of freedom, because each of its $3$ atoms are described by $3$ Cartesian coordinates. However, if we are interested in the energy of a configuration, we know that some of these degrees of freedom don't matter. For example, if we shift the x coordinate of each atom by the same amount, we haven't changed the molecule at all, only its location. This leads us to seek a set of coordinates that describe the potential energy surface without considering parts that don't matter. For water, we can remove $6$ as shifting along or rotating about any axis doesn't change the energy of the molecule.
This leaves us with $3$ remaining degrees of freedom, internal motions of the molecule that do affect its potential energy. What do these look like? I could come up with all sorts of valid variations on this. Off the bat, I might think that a good representation would be individual stretches of each $\ce{O-H}$. While there isn't anything strictly wrong with this representation, it does have the issue that motions along these coordinates are coupled, meaning changes along a given mode affect some or all of the others.
What can I do to avoid to avoid this coupling? Well, within the harmonic approximation, all the coupling between modes is described by off-diagonal elements of the Hessian matrix. So, if I can find a representation where the Hessian is diagonal, I will have a nice set of uncoupled internal motions. This is exactly what we get from the eigenvectors. Now, instead of two coupled $\ce{O-H}$ bond stretches, the normal modes give two decoupled anti/symm vibrations of both $\ce{O-H}$ bond.
The normal modes aren't any more physically correct than my previous example, but they are easier to analyze because they are decoupled. In this way, the "neatness" of the normal modes is really a fundamental mathematical property rather than a fundamental physical one.


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How is silica gel made?
How is silica gel prepared from sand (after extracting it). How many ways can it be used? Is there a way to make it without an oven?

I rather doubt it can be done without resorting to elevated temperatures.  basically you prepare sodium silicate by reacting sand with sodium carbonate or sodium hydroxide (this is what needs forcing conditions), dissolving it in water and adding HCl.  https://www.researchgate.net/publication/251717918_Synthesis_and_characterization_of_silica_gel_from_siliceous_sands_of_southern_Tunisia

Sand may be dissolved in concentrated and hot solutions of NaOH. The dissolution is slow, and pressure may help this action. At the end of the reaction, a viscous solution of so-called sodium silicate is obtained. The formula of the solute is not well defined, and can be described as $(Na_2SiO_3)_n$. It is made of a long polymeric chain similar to the formula $HO-Si-O-(Si-O)_n-Si-O-Si-OH$, with supplementary negatively charged Oxygen atoms above and under each Silicon atoms. The whole chain is negatively charged. And of course, there are Sodium ions $Na^+$ in the vicinity of each negatively charged Oxygen atoms. To make the description more complete, one should mention that from time to time there are side branches of Si-O-Si-O chains, which get fixed on the Si atoms of the main chain. Furthermore there are no $Si=O$ bonds in this formula.
If you add some $HCl$ solution, you produce the precipitation of an insoluble gel whose formula has the same skeleton as the previous solution $HO-Si-O-(Si-O)_n-Si-O-Si-OH$. But the Oxygen atoms above and under the main chain are now holding a H atom, and the whole chain is neutral.
This gel can be filtrated and washed with large amounts of water. This removes the ions $Na^+$ and $Cl^-$. The result is a gel containing more water than silica. Usually this gel is then heated and the water is evaporated. An amorphous white substance is obtained with a formula between $H_2SiO_3$ and $SiO_2$. This is the common "silicagel" used in the labs for dehydrating purposes, because it has a tendency to adsorb water from the atmosphere.
The fresh gel can also be dried in a vacuum, instead of by heating. In this case, the silica gel is transformed into an extremely light substance, lighter than snow, which has a tendency to float in the air. It is a dangerous substance if you breeze it, because it will stay in the nose and the lung, without being able to get out. It may induce cancers, like asbestos.

@Maurice: How do you form proper pellets from the gel?

No idea how to make pellets from the gel.


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Can you approximate the change in entropy of a reaction?
I found a problem in which I am given a hypothetical reaction in which from 2 liquids (reactants) I get 4 moles of gas and 1 mole of a solid substance (products):
$$\ce{A (l) + B (l)  -> 4C (g)  + D (s)}$$
I am asked about the change in entropy (an approximation of it), so I am curious if there is some kind of approximation for the change in entropy of a reaction based on the change in the amount of substance of gaseous compounds (since I know that the highest influence on entropy is that of gases).

The molar standard entropy of formation ($S_{f}^{\ominus}$) for many common substances is documented in tables. You can approximate the overall change in entropy of the reaction ($\Delta S^{\ominus}_{reaction}$) with with Hess's law:
$\Delta S^{\ominus}_{reaction} = \sum S^{\ominus}_{f,molar,products} - \sum S^{\ominus}_{f,molar,reactants}$
You notice that there is no $\Delta$ in front of $S^{\ominus}_{f,molar}$, since entropy is an absolute physical quantity. Every system has an absolute entropy.
Don't forget to multiply every $S^{\ominus}_{f,molar}$ with the number of moles that you have. This is only an approximation because the molar entropy of a substance depends on Temperature as well, and the reaction likely doesn't take place at standard temperature.
As a sidenote please be aware that the molar entropy of formation for elements is not zero, although the molar enthalpy of formation is!

In such a case, you may simply count the number of gaseous molecules on each side of the arrow. In your reaction there are no gas molecules in the reactants, and 4 in the products. The number of moles of gaseous substances increases. So the qualitative entropy increases.

You're oversimplifying even this simplified task.

No , I don't require a qualitative approximation ( I mean it is obvious that the entropy will increase ) , my question is numerically if there is a way of giving some sort of approximation (like for example for the enthalpy of vaporization you can use Trouton's constant for any liquid in which there are no hydrogen bonds or other similar interactions).


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What is the relationship between concentration and voltage in a galvanic cell? How do you calculate the theoretical?
I am a year 12 Chemistry student and I was wondering if you could help me with some questions I have about a Galvanic cell experiment?
I am doing an experimental report in Chemistry. The experiment I decided to do was to change the concentration of the copper sulphate electrolyte in a Daniell cell and measure the voltage produced. The other electrolyte; Zinc nitrate was not changed and kept constant at 0.1M. The copper sulphate concentration increased from 0.25M, 0.5M, 0.75M to 1M. The experiment was performed when the temperature was 24 degrees. The anode was Zinc and the cathode was Copper.
In order to obtain theoretical values - should I use the Nernst equation? Would you be able to tell me how to use the Nernst equation for my experiment?
Also, I understand that voltage increases with concentration, but compared to standard conditions should the voltage produced be higher or lower than the voltage produced at standard conditions and 1M electrolyte?

Look at the Nernst equation. You have all the variables. Watch this video. Write a balanced equation, and then follow the instructions https://www.youtube.com/watch?v=5Z3SlHgrz0o

Your questions are answered here: https://chemistry.stackexchange.com/a/118744/79678 . If you agree, consider upvoting that answer.

There has been a publication in Journal of Chemical Education along time ago, saying that zinc should not be used in a zinc nitrate solution, because it slowly reduces the nitrate to nitrite, so that the concentration of zinc is not constant, according to the equation $$\ce{Zn + NO3^- + H2O -> Zn^{2+}.+ NO2^- + 2 OH^-}$$ Better replace zinc nitrate by zinc sulfate !

The potential of the Copper electrode is given by the Nernst equation :  $$E(Cu^{2+}/Cu) =   + 0.34 V + 0.0295 log[Cu^{2+}]$$ If the concentration $[Cu^{2+}]$ is $0.25 M$ and then $0.50 M$, the potential of these half-cells become respectively $0.32$ V and $0.33 V$.
If you couple this half-cell with a Zinc electrode with $[Zn^{2+}]$ = $0.1 M$, the same sort of calculation gives you a potential $E(Zn^{2+}/Zn)$ =$ -0.76 + 0.0295(-1)$ = $-0.79 V$. If you combine both half-cells, you obtain $1.11 V$ and $1.12 V$ respectively.
This being said, I would like to advise you not to use Zinc nitrate in the zinc half-cell. Please replace Zinc nitrate by Zinc sulfate. The reason is that the ion nitrate attacks the Zinc metal, producing nitrite ions. So the potential of the half-cell made with Zinc metal in a Zinc nitrate solution is not stable.

The reduction by zinc to nitrite is not correct. It only occurs at very low or very high pH.

As a side comment, the concentration of the cupric may be impacted by the equilibrium reaction: Cu + Cu(II) = 2 Cu(l) . This reaction is claimed to be accelerated at lower pH and the nature of the copper salt to undergo hydrolysis may be impactful.

Thank you so much! That makes so much more sense. My chemistry teacher did not even inform me about changing the Zinc electrolyte. Thanks!


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Frequency region and "uncoupled oscillators" in vibrational spectroscopy
I am currently studying the textbook Infrared and Raman Spectroscopy by Larkin. In the first chapter of the textbook, the author presents the following table and accompanying explanation:

The vibrational spectrum may be divided into typical regions shown in Fig. 1.1. These regions can be roughly divided as follows:

$\ce{X-H}$ stretch (str) highest frequencies (3700-2500 cm$^{-1}$)
$\ce{X#Y}$ stretch and cumulated double bonds $\ce{X=Y=Z}$ asymmetric stretch (2500-2000cm$^{-1}$)
$\ce{X=Y}$ stretch (2000-1500cm$^{-1}$)
$\ce{X-H}$ deformation (def) (1500-1000cm$^{-1}$)
$\ce{X-Y}$ stretch (1300-600 cm$^{-1}$)

The above represents vibrations as simple, uncoupled oscillators (with the exception of the cumulated double bonds). The actual vibrations of molecules are often complex and as we will see later, typically involved coupled vibrations.

I have two questions:

What do the $\ce{X}$, $\ce{Y}$, and $\ce{H}$ values mean in the above explanation?
Can someone please elaborate, with some detail (including mathematics), on this concept of "uncoupled oscillators", and its relevance in the above context? To be specific: What specifically is an "uncoupled oscillator", how is it described mathematically, and what is its relevance in the above context?

I appreciate it if someone would please take the time to clarify these points. Thank you.

Despite there being a good answer already, I wanted to say that this was a very good question. I would like to invite you to participate as one of the first group of users in the Materials Modeling & Spectroscopy Stack Exchange https://area51.stackexchange.com/proposals/122958/materials-modeling?referrer=MGY4NjYyM2ZlY2NkNDI0MWRmYjhjMGI1N2RmM2U3ZDk1N2VjOWQ1MGU1MmVhNDZkZjI5MzNiNzY1MmE2ZjRhOYrAzULyFn6juGAbZ7wko5LvlWXZ0iQfr620nQ70gVNl0. 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.

$\ce{H}$ is just hydrogen and $\ce{X}$, $\ce{Y}$, and $\ce{Z}$ are just other elements, in this case likely carbon, nitrogen, or oxygen. They give an example in the table of $\ce{-X=Y}$ where $\ce{-C=N}$ matches this template of two (generally) different elements double bonded together.
They use coupled here in a slightly different way than most literature. I would may switch to delocalized/localized rather than coupled/uncoupled. The modes they mention are mostly localized to a single bond (e.g. an $\ce{-X-H}$ stretch) whereas in general vibrational modes are delocalized/coupled over the whole molecule (e.g. an aromatic $\ce{CH}$ stretch where all the $\ce{C-H}$ bonds of an aromatic ring move in unison).

Thanks for the answer. Your answer for 2. doesn't quite contain the detail that I was looking for. To be specific: What specifically is an "uncoupled oscillator", how is it described mathematically, and what is its relevance in the explanation?

In the way they are using it, I don't think it has a mathematical description. In other places you will see coupled/uncoupled used to describe whether the motion of one mode affects another. Here, I think they are just saying that while sometimes modes correspond to a simple single bond stretches, more often they are distributed motions across the whole molecule and can't be described so simply.

Hmm, I see. And what is your definition of "mode" here? Is it this https://en.wikipedia.org/wiki/Normal_mode ?

And what does it mean to have a bond with no connecting element as in $\ce{-X-H}$?

Yes, I mean normal vibrational modes of the molecule. I wasn't sure what terminology you had encountered before. The unlabelled bond is one way of representing that say $\ce{-X-H}$ is attached to something else. In organic chemistry, you will more commonly see $\ce{R-X-H}$ where R could represent a variety of different molecular fragments.


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How to determine the pH of citrate buffer system
I used this table (http://microscopy.berkeley.edu/Resources/instruction/buffers.html) to make a citrate buffer system. The Pka value is 6.4, and I multiplied the solution by ten to make a larger batch. I used 137ml of citric acid and 363ml sodium citrate in distilled water to make a buffer system with pH of 5.6. Now I'm trying to work backwards to prove this was correct, and using the equation pH = pKa + log(A-/HA) I am getting 6.4 + log 0.0137/0.0363 = 5.97, when it should be 5.6. Anyone know what I'm doing wrong? Thanks!

Remember it is not A-/HA, but A^3-/HA^2-. So for pH=pKa3=6.4, the molar ratio citrate/citric_acid is 5:1. H3A + 5 A^3- -> 3 HA^2- + 3 A^3-


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Is strontium more metallic than sodium? If yes then why?
I got an MCQ in my examination, "Which one of the following is more metallic?" and the options were Sr, Na, Be or Aluminium. I know that sodium is more metallic than Be or Al but my mind stuck on strontium.
-Strontium is the member of the fifth period and has greater number of shielding electrons than sodium (due to greater number of shells). This suggests that strontium should be more likeable to lose electrons.
-But sodium has half filled valence "s" orbital while strontium has fully filled "s" orbital. This suggests that sodium is more likeable to lose its only valence electron.
-I've also looked for their first ionization energy values and they are almost comparable. 495.8 for Na and 549.5 for Sr. Why the I.E of sodium is a little bit less? (And should it be more metallic due to this?)
I've looked everywhere else on the internet but haven't found the answer. My question is, which of them is more metallic and why.
And also are there any members of group IIA that are more metallic than any of the elements of group IA.

What does it even mean to be more metallic?

I always thought sodium was more into progressive rock than metal, but what do I know?

Since the ‘metallic’ nature of a solid is tied to the solid state physics of the crystal bonding, using atomic chemistry is likely misguided.

I wonder if there exists a commonly known definition of "metallicity" or "metalliness" in chemistry such that we could measure it and compare the said elements in order to answer your question.

It would be nice to know what the test administrator says about it.


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Difference between "physical quantities" and "units"?
My textbook, Atkins' Physical Chemistry, 11th edition, by Atkins, de Paula, and Keeler, says the following:

Although it is good practice to use only SI units, there will be occasions where accepted practice is so deeply rooted that physical quantities are expressed using other, non-SI units. By international convention, all physical quantities are represented by oblique (sloping) letters (for instance, $m$ for mass); units are given in roman (upright) letters (for instance m for metre).

I'm confused about what the difference here is between "physical quantities" and "units"?
I would appreciate it if people would please take the time to clarify this.

I think that passage explains it pretty well, but to give another example energy is a physical quantity that can be expressed using different units like Joules and calories. Physical quantities describe some property of an object, units are used to express a particular amount of that property.

@Tyberius So how does the different uses of $m$ and m in the textbook excerpt relate to this?

That's basically just telling you how to read expressions they write in the book. If you see $F=ma$, you know that m is referring to mass. If you see $1600$ m, it's talking about that many meters, not that number times mass.

@Tyberius Oh, so that's what they were referring to. It wasn't even clear to me until you used the $F = ma$ example. I think this could have been expressed more clearly; they managed to make such a simple point into something much vaguer than reasonable. Anyway, thank you for the clarification.

Mass is a physical quantity. Its unit may be gram or kilogram.
Volume is a physical quantity. Its unit may be the liter or the cubic meter

Yes, it wasn't clear to me what they meant until Tyberius clarified it with his example in the comments.


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Discussion about Warren-Cowely Parameter
The Warren-Cowely Parameter (WCP) for short range solution order is defined as:
$$\mathrm{WCP}_{ij} = 1 - \frac{Z_{ij}}{x_j \cdot Z_i}$$
where $Z_{ij}$ is the partial Coordination Number (CN) of  j atoms around i, $Z_{i}$ is the total CN around i, and $ x_{j}$ is the nominal composition of j.
Consider a two component solution of $\ce{A}$ and $\ce{B}$.  If $\mathrm{WCP}>0$, like neighbors are preferred, If $\mathrm{WCP}<0$, unlike neighbors are preferred.
If both of $\mathrm{WCP}_{AA}$ and $\mathrm{WCP}_{BB}$ are positive, which one is preferred?

Could you clarify what $Z_{ij}$ and $x_j$ are in your equations? Does your definition match the one in this paper: https://deepblue.lib.umich.edu/bitstream/handle/2027.42/29021/0000051.pdf;sequence=1

Dear @Tyberius I have updated my question, and added the definitions

@Tyberius do you not have any idea about this question? I checked the parameter with the definition in the paper you introduced, they are identical

I do not know for sure. It could be that $WCP_{AB} is what actually matters.$WCP_{AA}$ and $WCP_{BB}$ both positive seems to me like it implies that both components prefer like neighbors. I had hoped clarifying your question and introducing some terminology for people who aren't familiar (like myself) would encourage an answer.


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What order would these compounds elute in a 3:1 hexanes:ethyl acetate TLC run?
I would figure the order would be (in order of smallest Rf to largest) W, Z, Y, X. Is the carboxylic acid more polar than the acetic anhydride?

Making predictions about elution order is always guesswork but based on my experience I would expect acetic acid to streak off the baseline, The benzyl alcohol to be of low Rf but above the acetic acid, the benzyl acetate Y to run high and I really don't know about the acetic anhydride, I suspect it will hydrolyse on the plate to give you more acetic acid so you will not see a separate spot. How are you planning on visualising these?

I'm given a "plate" with 4 distinct spots and asked to propose an order of elution to the above compounds.

If the acetic anhydride does not hydrolyse then it will run ahead of X. In real life the acetic acid will not give a distinct spot more a smear but it will be the lowest - so my answer is Y,W,X,Z but it is a somewhat unreal example.


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Curtin-Hammett principle to evaluate alkene insertion
Given an unsymmetrical alkene in a simple non-stereoselective catalytic hydrogenation reaction. If one would want to investigate the nuances in the catalytic cycle of a dihydride complex leading to the alkane, there are two important transition-states that have to be considered.

Insertion of the alkene into one metal-hydride bond
Reductive elimination of the alkyl and remaining hydride to give the alkane.

We assume the boundary condition that the transition-state for the second step (TS-2) is always higher in energy than the transition-state for the first step (TS-1).
Moreover, I want to look at the elementary steps of the insertion, e.g. a 1,2-insertion (1-position of alkene binds to the metal and the hydride is transferred to the 2-position of said alkene) or 2,1-insertion (2-position of alkene binds to the metal and the hydride is transferred to the 1-position of the alkene).
If I now have a catalyst where the transition state for the first step (TS-1) lies lower in energy for a 2,1-insertion compared to a 1,2-insertion, but the transition-state for the second step (TS-2) lies higher in energy for the 2,1-insertion product than for the 1,2-insertion product, which pathway is the right one? E.g. does the mechanism proceed through the 2,1-insertion since TS-1 favors this pathway, or does the mechanism proceed through a 1,2-insertion since TS-2 is lower for this pathway.

Mixtures excluded?

Mixtures of what exactly? Lets assume the alkene is 1-propene which is hydrogenated by a M2 complex. The question is based on the relative energies described above, which TS will decide the reaction pathway. So is it the first TS because it is responsible for the actual insertion or the second TS since it is the highest TS in the overall reaction.

Have a look at the energy span model (https://doi.org/10.1021/ar1000956), it is very useful at conceptualising catalytic cycles, and will help you find the most probable path.


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Morse interatomic potential
I am working with Morse interatomic potential, I searched through the Internet but I could not find any Morse parameters for Zirconium element. Is there any table, handbook or something like these which list these parameters? or Is it possible to calculate these parameters for hcp metal elements?

Since the Morse potential was constructed to model diatomic bonding, it seems a bit of a stretch to easily generalize it to a crystalline potential, be it hcp, fcc, etc. So, it seems unlikely that one will find a readily available table.

In principle, one can perform quantum chemical calculations on the zirconium crystal structure and distort it to obtain Morse potential parameters. However, this is a lot of work even for an expert and may even provide the answer OP is looking for, presuming they do not care for the parameters but want to find some property using it.


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Why is it experimentally difficult to accurately find the critical volume of a real gas?
I was reading up on the derivation of the values of various critical state functions( $\ce{Vc}$, $\ce{pc}$ and $\ce{Tc}$) from the Van der Waals equation (Physical Chemistry by G.W. Castellan). There, he was about to express a and b in terms of these critical state functions, when I came across this line:

Since experimentally it is hard to determine $\ce{Vc}$ accurately, it would be better if a and b
  could be obtained from $\ce{pc}$ and $\ce{Tc}$ only.

What is exactly the difficulty in finding $\ce{Vc}$ accurately?
I found this source stating that:

Of the three constants, measurement of critical temperature and critical pressure is very easy. The measurement of critical volume is very difficult because it varies considerably for slightest variation of temperature and pressure near the critical region; hence for the measurement of critical volume, the substance must be kept in perfectly critical temperature and pressure.

My two cents
If $\ce{Vc}$ is so short-lived and unstable, then any variation in the same would almost instantly affect the vapour pressure of the gas phase above it, and thereby $\ce{pc}$ would also not be a very reliably measured quantity. Then maybe we would end up with two variables(a and b) and one equation($\ce{Tc}$).
Can somebody explain the procedure of measurement of critical state function here, and why only $\ce{Vc}$ causes problems,and the other two remain appreciably accurate?

Around the critical point, a variation in volume has very little effect on pressure. Related: https://chemistry.stackexchange.com/questions/97671/why-is-dp-dv-zero-for-a-real-gas-in-critical-conditions/97673#97673

@IvanNeretin Yes

Well, here we go: the first and second derivatives are both zero, hence...

@IvanNeretin Ok, I think I got the idea, that since the situation is a saturated vapour pressure, volume change doesn't change the limiting value of pressure,I feel. But then, why is it saturated vapours at critical point?

Also, why is critical volume so "unstable"? Surely we can measure it quite well when we reach the critical point i guess..

It is not about the saturated vapor. It is about the critical point, which (in a way) is even worse. The volume is "unstable" precisely for the reason that the pressure depends on it very little. As for reaching the critical point, well, see, there is no road sign that says "CRITICAL POINT" in huge letters made with reflective paint. How do you know whether you are there?

@IvanNeretin. Ahh,I see. The Pc and Tc remain kinda constant over a longer range of states than Vc,hence they are more accessible to measurement than Vc. Is that right?

Yes, that's about right.


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Calculating the solubility of cuprous iodide
I am clearly missing something obvious here. I get the $\mathrm{K_{sp}}$ from the CRC for $\ce{CuI}$ as $1.27\cdot10^{-12}$.
So, $\ce{CuI-> Cu+  +  I-}$
So the formula ought to be
$$\mathrm{K_{sp}} = \ce{[Cu+][I-]} =[s][s]= s^2$$
Doing the square root and all gives me a solubility of 2.1 mg/L. Hooray, except that everywhere I look, if the solubility is listed at all, it lists 4.2, exactly double my answer.
I have been wracking my brain on this one. It has to be a scalar but I don’t see how you would introduce it without throwing the entire calculation off by turning the square into a cube or something similar. What am I missing?
Edit: updated to reflect the full Ksp value in the CRC which I was using for my calculations (I had accidentally truncated it to 1 rather than 1.27 when posting this). I am using the online CRC from hbcponline.com, so this is current, not an old edition.

"Kinetics" is a wrong word to use here.

FYI, CuI is a photocatalyst. This could result in the production of solvated electrons and an electron-hole (turning the OH- in water into the hydroxyl radical). Measure in low light conditions.

Your math seems to be wrong. From the $\mathrm{K_{sp}}$, $s = 1\cdot10^{-6}$ molar, and given FW of 190, I get 0.19 mg/liter. Wikipedia lists same $\mathrm{K_{sp}}$, and 0.42 mg/liter (Wikipedia has g/100ml).

As far as the difference, I don't know. You'd have to research where the data came from, and then conditions of the experiments. The concentrations are so low that any number of experimental problems could have affected the data. For example the temperature, and oxygen/CO2 content of the water. // One last point. Much chemical data has been experimentally determined. All in all there isn't any absolute check on consistency. As inconsistent data is found, some researcher will do a literature search, then experiments to correct the data.

In my Handbook of Chemistry and Physics, I find $5·10^{-12}$ for the solubility product of CuI. The solubility is then exactly twice your result.

@Maurice - What edition of CRC? My latest is 92nd edition, 2011-2012, and lists 0.20 mg/L for solubility (page 4-61) and $1.27\cdot10^{-12}$ for Ksp (page 5-196). The CRC lists sources for data which won't be primary sources, but would allow one to track the data back to the original sources. The CRC handbook is probably the best and most consistent source available.

@Broklynite - I should have started off with verifying that yes indeed there is an inconsistency in the data you gave and that the solubility is so low that the problem isn't the activity vs concentration assumption.

Please cite your exact sources, as there seems to be a lot of discrepancy between the value you report with the posted comments and answer.

@MaxW I meant to type 1.27 but only typed 1 when I wrote the question (posting this via cellphone) so please pardon the confusion. I have amended and corrected above to try to address that.

First of all, as MaxW pointed out in the above comment, your math seems to be wrong. From the the $K_\mathrm{sp}$ value given in your question:
$$s = \sqrt{K_\mathrm{sp}}= \sqrt{1.0 \times 10^{-12}}= \pu{1.0 \times 10^{-6} mol/L}= \pu{1.0 \times 10^{-6} mol/L} \times \frac{\pu{190.45 g}}{\pu{1 mol}}\times \frac{\pu{1000 mg}}{\pu{1 g}} \\ = \pu{0.190 mg/L}$$
Other than that, there seems to be a problem with online value of $K_\mathrm{sp}$ of $\ce{CuI}$. Some sites listed it as $1.1 \times 10^{-12}$ while other sites listed as $5.3 \times 10^{-12}$. However, I'd like the value $5.3 \times 10^{-12}$, because at least it's given in one of reliable sources (Ref.1).
If you use this value (instead of $1.1 \times 10^{-12}$), you'd get:
$$s = \sqrt{K_\mathrm{sp}}= \sqrt{5.3 \times 10^{-12}}= \pu{2.30 \times 10^{-6} mol/L}= \pu{2.30 \times 10^{-6} mol/L} \times \frac{\pu{190.45 g}}{\pu{1 mol}}\times \frac{\pu{1000 mg}}{\pu{1 g}} \\ = \pu{0.438 mg/L}$$
This value is in a good agreement with the literature value ($\pu{0.00042 g/L}$ at $\pu{25 ^\circ C}$; Ref.2), which listed the properties of $\ce{CuI}$ as:

Copper (I) iodide is a dense, pure white solid, crystallizing with a zinc-blende structure below $300^\circ$. It is less sensitive to light than either the chloride or bromide, although passage of air over the solid at room temperature in daylight for $\pu{3 hr}$ results in the liberation of a small amount of iodine. It melts at $588^\circ$, boils at $1293^\circ$, and, unlike the other copper halides, is not associated in the vapor state. Being extremely insoluble ($\pu{0.00042 g/L}$ at $25^\circ$), it is not perceptibly decomposed by water. It is insoluble in dilute acids but dissolves in aqueous solutions of ammonia, potassium iodide, potassium cyanide, and sodium thiosulfate. It is decomposed by concentrated sulfuric and nitric acids.

References:

Steven S. Zumdahl, Donald J. DeCoste, In Chemical Principles; 8th Edition; Cengage Learning: Boston, MA, 2017, “Chapter 8: Applications of Aqueous Equilibria,” pp. 241–298 ($K_\mathrm{sp}$ of value $5.3 \times 10^{-12}$ is given for $\ce{CuI}$ in example 8.13, page 288).
George B. Kauffman, Lawrence Y. Fang (Checked by N. Viswanathan, G. Townsend), “Chapter 2: Transition metal Complexes and Compounds – 20. Purification of Copper (I) Iodide,” In Inorganic Syntheses: Volume 22; Smith L. Holt Jr., Ed.; John Wiley & Sons: New York, NY, 1983, pp. 103–107 (https://doi.org/10.1002/9780470132531.ch20).

Quoting data sources is good, but neither of these is the source of the original data, and neither likely references the original source. I think the best that can be said is that the data (solubility vs Ksp)  is inconsistent. Trying to figure of why there is an inconsistency is a research project.

Figured it out, thought I would share the solution to the mystery with you folks:
The 1902 value for Ksp is 5.06 e-12 based on cell measurements and the value for solubility if CuBr.
In the 50’s and 60’s, the solubility product at zero ionic strength, Kso was found to be 1.2e-12.
So, that explains the confusion in K values. My math was not wrong, the reported Ksp values were confused, in part because places never seem to cite their sources.
Speaking of which, I found this by going to the IUPAC Solubility Data Series, volume 65, page 195.


text:
Why are these compounds classified as metamers?

Metamers are compounds having the same molecular formula but different number of carbon atoms,  on either side of functional group.
In the above compounds, the number of carbon atoms around the ester group are 6 on each side.
So how exactly can we call them metamers

I feel that the link doesn't have the answer to my specific query

The answer there seems to suggest that 1. Metamerism is an outdated term 2. Metamer has multiple meanings, the one you are familiar with about the number of carbons and a more general one.

The more general one seems to be constitutional isomerism itself. If I have to precisely state the type of constitutional isomerism, what will it be then?

Kind of positional. Not to be overthink....

This question came in my exam,  shouldn't there be a fixed set of rules to eliminate vagueness