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How does LeChatelier’s principle (which makes qualitative predictions) relate to the quantitative changes observed when an equilibrium is disturbed?
Le Chatelier formulated his principle in 1884; it predicts the direction in which equilibrium is re-established after it has been disturbed by an external cause. How does this qualitative principle relate to the known quantitative changes that occur when an equilibrium is disturbed by a change in temperature, concentration or pressure? Are there any limitations of Le Chatelier's principle as it is stated by modern sources?

Related  https://chemistry.stackexchange.com/a/125159/35806

How does this qualitative principle relate to the known quantitative changes that occur when an equilibrium is disturbed by a change in temperature, concentration or pressure?

Equilibrium constant K and reaction quotient Q
One way to quantitatively describe equilibrium processes is to define an equilibrium constant K and a reaction quotient Q. The equilibrium constant is defined by an equilibrium constant expression which contains all concentrations or partial pressures that are affected by the reaction.
For example, if we look at the following reaction:
$$\ce{A(aq) + B(aq) <=> C(aq) + H2O(l)}$$
The equilibrium constant expression for this reaction is:
$$K = \frac{\ce{[C]_{eq}}}{\ce{[A]_{eq}}\ce{[B]_{eq}}},$$
with $\ce{[X]_{eq}}$ designating the equilibrium activity (or fugacity for gas state) of a species. At a given temperature, the equilibrium constant will have the same value when the reaction is at equilibrium, no matter what the starting conditions.
The reaction quotient is defined in the same way as the equilibrium constant, but using $\ce{[X]_{t}}$ (the current activity) instead of the activity at equilibrium $\ce{[X]_{eq}}$. For the example reaction above, it is:
$$K = \frac{\ce{[C]}_t}{\ce{[A]}_t\ce{[B}_t},$$
Q can have arbitrary values from zero to infinity, depending on what reaction mixtures is found at a given time. Q will change over time if there is a net forward or reverse reaction.
Criterion for equilibrium based on Q vs. K
When Q equals K, the reaction is at equilibrium because the current activities $\ce{[X]_{t}}$ plugged into the equilibrium constant expression match K. Any net reaction in the forward direction would typically increase Q, while any net reaction in the reverse direction would typically decrease Q. This relies on the assumption that the activity of reactants decreases and that of the products (of the species included in the equilibrium constant expression) increases as the reaction goes forward (and vice versa as it goes in reverse).
Change in temperature
When, after the reaction has attained equilibrium, the temperature changes (through an external cause) and consequently the equilibrium constant changes, Q no longer equals K. Concentrations that had been constant over time will change now. In the example, we are changing the temperature at $t = 5$ (arbitrary units) so that the equilibrium constant doubles. The only way to make Q equal K again at this temperature is to have a net forward reaction (this will increase product activities and decrease reactant activities in the reaction quotient expression). An example is shown below:

How temperature affects the equilibrium constant is given by the van't Hoff relation. Stated in qualitative terms, the equilibrium constant of exothermic reactions decreases with an increased temperature, while that of endothermic reaction increases.
Change in activity
If we change the activity of a single species (for example by adding or removing solute) in a reaction that has already attained equilibrium, Q will change. If the temperature is maintained, K will remain constant, and Q will no longer be equal to K. As a result, a net reaction will be observed.
For example, if we double the activity of A in the reaction above, Q will be halved, and the reaction has to move in the net forward direction to attain equilibrium again:

After equilibrium is attained, Q will have the same value as it did before the external manipulation, but the set of activities will be different (necessarily, because there is more matter now).
Limitations and Criticism

Are there any limitations of Le Chatelier's principle as it is stated by modern sources?

The main limitation is that there are multiple non-equivalent ways the principle is stated, so different ways of defining the principle might lead to different counterexamples.
Historically, Le Chatelier wrote (Comptes-rendus de l’Académie des sciences, 99 (1884), p.786-789.) :

Tout système en équilibre chimique stable soumis à l’influence d’une cause extérieure qui tend à faire varier soit sa température, soit sa condensation (pression, concentration, nombre de molécules dans l’unité de volume) dans sa totalité ou seulement dans quelques-unes de ses parties, ne peut éprouver que des modifications intérieures, qui, si elles se produisaient seules, amèneraient un changement de température ou de condensation de signe contraire à celui résultant de la cause extérieure.

This google-translates into:

Any system in stable chemical equilibrium subjected to the influence of an external cause which tends to vary either its temperature or its condensation (pressure, concentration, number of molecules in the unit of volume) in its totality or only in some of its parts, can only experience internal modifications, which, if they occurred alone, would bring about a change in temperature or condensation of a sign contrary to that resulting from the external cause.

Wikipedia states the principle as:

When any system at equilibrium for a long period of time is subjected to change in concentration, temperature, volume, or pressure, (1) the system changes to a new equilibrium and (2) this change partly counteracts the applied change.

Chemistry OpenStax has two versions:

[in text]if an equilibrium system is stressed, the system will experience a shift in response to the stress that re-establishes equilibrium.
[Overview]An equilibrium system subjected to a disturbance will shift in a way that counters the disturbance and re-establishes equilibrium.

ChemPrime by Moore et al states:

if a system is in equilibrium and some factor in the equilibrium conditions is altered, then the system will (if possible) adjust to a new equilibrium state so as to counteract this alteration to some degree.

Limitations of the principle are discussed, for example, in this letter in the Journal of Chemical Education, and in the literature cited within. Another article in the same journal discusses how applying Le Chatelier's principle leads a convenience sample of educators to incorrectly answer questions such as :

The reaction
$$\ce{2NO2(g) <=> N2O4(g)}$$
is at equilibrium in a syringe. If the volume is decreased at constant temperature by moving the plunger, will the concentration of NO2(g) be higher or lower than the original concentration when equilibrium is re-established? Give reasons for your answer.

The misconception is that while Le Chatelier's principle correctly predicts that the reaction will form product after the pressure is increased, this does not completely offset the increase in partial pressure from lowering the volume.
Some authors argue that it makes sense not to teach Le Chatelier's principle to avoid such confusion, and instead treat equilibrium in a quantitative matter, calculating the new equilibrium state rather than predicting it qualitatively.


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How do carbon content and various heat treatment cycles affect the properties of steel at the micro-level?
I watch a YouTube channel by Alec Steele where a lot of metal forging is done. From this I know that varying the carbon content in steel drastically alters the physical properties of the metal, as does different heating cycles (time for cooling and temperature of the forge). I am interested what is happening at the atomic structure during the forging process.
Here are some of my ideas on what is occurring. I think cooling time must affect the grain size of the steel microstructure. I have a vague idea that grain size affects material strength but I don't really know why. I also assume that the crystal structure varies with temperature. At a higher temperature, higher energy polymorphs are probably observed. I know adding carbon to the iron increases the strength of the material but again I don't know why. I assume there is some pi - d bonding occurring which may act akin to semiconductor doping, adding energy states for Fe electrons to populate, increasing metallic bonding?
This does not answer my question:
What chemical characteristics give types of steel their properties?
I am looking for an answer more focused on bonding and energy level population.

To know how steel works you need to know about all this stuff like pearlite, ferrite etc. - it's their structure, and transformations, not some energy states. For example graphite can be dissolved in Fe or not. There can be even stuff like nanotubes found in some steels. It's generally quite complex and seems not exactly as you imagine.

@Mithoron the question is legitimate so either answer it or don't. Posting 'oh you're so thoughtless' isn't what this forum is for.

@Joe, there is a difference between physical and chemical change, and both effects occur in steel making. Chemically, carbides and intermetallic compounds form, and physically, there are differing allotropes, separation of phases, sizes and shapes of crystals, usw.

@Bertram I agree with your core message. I would still like to point out that this is not a forum. Comments (should) aim to improve and clarify a question. Some of the aspects mentioned here are indeed addressed in the other question, albeit not in that clear words. || Joe, this question needs a bit more focus, there are currently a few close votes on it.

source: Stack Exchange

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

sort: votes


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How does sodium borate (Borax) function to promote Fe-Fe bonding?
Borax is used as "flux" during metal forging to help stick two pieces of steel together. I am curious what is happening at the atomic level. How does the addition of borax promote Fe-Fe bond formation?

What is 'flux' usually mean for soldering and brazing? Why do you not consider Borax suitable for that?

"stick two pieces of steel together" isn't equal to  "promote Fe-Fe bond formation". You're making there some assumptions that seem rather too far fetched.

Hint: think about air, and why it could cause issues.

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


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Zinc Ink Availability
does anyone know where I could purchase zinc ink? Sigma Aldrich only sell Zinc Oxide, and I'm struggling to find a method in which I could remove the oxide after I print onto a polymer substrate.
Would be grateful if anyone could point me in the right direction. Thanks.

You can buy "liquid zinc" (my literal translation, "Fluessigzink in German"), for repair purposes. Its about 90% metallic zinc powder in some kind of epoxy binder or similar.

There are many different industrial primers for steel that contain over 90% zinc by weight . Often they are more than one component and some provide zinc powder as a separate addition. Unfortunately they are likely to only be available from industrial suppliers and would not work in a computer printer.


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Why is so different the temperatures of a compressed gas compared to a liquid/gas mixture of the same fluid and they are both at the same pressure?
Why does $\ce{C3H6}$ gas when compressed (in a centrifugal compressor) comes out with a pressure of $\pu{16 bar}$ and $\pu{90 °C},$ and when it is a mixture of $\ce{C3H6}$ liquid/gas stored inside of a vessel reaches also the $\pu{16 bar},$ but at a lower temperature of around $\pu{38 °C}?$
I know that when it is a vessel with stored $\ce{C3H6}$ liquid, some part of the liquid will vaporize until it reaches the equilibrium with the ambient temperature (vapor pressure). I know that if the vessel is exposed to a ambient temperature of $\pu{38 °C}$ and if has enough $\ce{C3H6}$ liquid to vaporize the equilibrium will be reach at a pressure of $\pu{16 bar}.$
But what I cannot understand is why if it is only gas and if I compress the gas to $\pu{16 bar},$ why does it come out with a temperature of $\pu{90 °C}?$ Shouldn't it come out with the same temperature of $\pu{38 °C},$ like the vessel?
If anyone at least could give me the subject/effect/topic so I can study just to get an answer to this question, I would be very grateful.
Note: This isn't a homework question. I work on a LPG storage facility and the other day I started thinking about this, but I cannot understand why this happens.

I don't understand. If you compress any gas to 16 bar, the temperature may have any value. You can heat or cool this compressed gas so as to reach any possible temperature. Why choose 90°C or 38°C ? I don't see your point.

@Maurice my point is that if i can get a pressure of 16 bar with a temperature of 38°C in a liquid/gas mixture inside a storage vessel, why does the gas that goes through the compressor and gets his pressure increase to 16 bar gets a temperature of 90°C after compression?, why is the temperature so different between both, since they get both to 16 bar but at different temperatures?

Generally, compression of any gas transforms mechanical energy to thermal energy and the gas warms up, until heat dissipates.
If you have ever inflated bike tyres, you would notice the air pump gets warm ( MTB ) up to hot ( high pressure road bikes ).
When air drops down by 1500 m ( from 0.85 to 1 bar ) after passing over mountain range, it warms up by adiabatic compression by 15 deg C. Warming up to 90 Deg C to get pressure 16 bar does not seems to be much.
See Adiabatic_process
A gas, with or without its liquid phase, has 3 variables: temperature, pressure and volume. You adjust 2 of them and the gas adapt to it by adjusting the 3rd one according laws of gas behaviour.
When you keep gas/liquid mixture at the given temperature and volume, it adjusts to it by reaching its saturated vapour  pressure for given temperature  by either evaporation either condensation.
When you set a gas to have a particular volume applying a particular pressure ( by pumping it in or out ), it adjusts to it reaching a particular temperature.
For propane gas and adiabatic compression ( without heat exchange with surrounding ):
$$p_1^{1-\gamma} T_1^\gamma=p_2^{1-\gamma} T_2^\gamma $$
where $\gamma=1.13$
If we consider as initial conditions 1 bar and 20 Deg C:
$$293.15^{1.13}=16^{-0.13} \cdot  T_2^{1.13 }$$
$$ T_2^{1.13 }=   293.15^{1.13} \cdot  16^{0.13} $$
$$ t =  T_2 - 273.15 =  {( 293.15^{1.13} \cdot  16^{0.13} )}^{\frac1{1.13}} - 273.15$$
For given data it leads to about 130 deg C.
If you suddenly compress propane to 16 bar, it would get hot. If you compressed it enough and there was some oxygen, it would  self ignite. This effect  is used in diesel engines.
But if the compression of propane is gradual and gas is in contact with surrounding,  heat gets time to partially dissipate.
If the hot compressed propane is kept at constant external pressure 16 bar, it gradually contracts while cooling down.
When it's temperature reaches the temperature, at which saturated propane vapour pressure, it becomes gradually condensation.
Gradual condensation at constant pressure maintains constant temperature, as dissipative heat loses are compensated by releasing the condensation heat.


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Why is the bond angle of sulphur dioxide close to 120°, not 90°?

Figure 1: Structural data of $\ce{SO2}$. Taken from Wikimedia, where a full list of authors is available.
As can be seen from the structural data, the bond angle of $\ce{SO2}$ is almost precisely $120°$, meaning that the σ-orbitals can be described as almost perfect $\mathrm{sp^2}$ orbitals. This means that the sulphur lone pair is also in an $\mathrm{sp^2}$-like orbital while the fourth orbital of sulphur is of p-type to build a π-system not unlike that of the allyl anion. If sulphur were in the second period, this structure would not be surprising and indeed that of ozone is very similar (see figure 2).

Figure 2: structural data of $\ce{O3}$. Taken from Wikimedia, where a full list of authors is available.
However, atoms of the third and higher periods generally have bond angles which are much closer to $90°$ as can be seen in figure 3 with the structure of $\ce{SF2}$. Typically, this is explained by their greater size which results in larger bond lengths and smaller bond angles, allowing for increased s-character of the remaining lone pairs. I myself have argued in this manner in previous answers. This should suggest an $\ce{SO2}$ bond angle much closer to $90°$, maybe around $100°$ degree as we indeed observe for $\ce{SF2}$ (see figure 3).

Figure 3: Structural data of $\ce{SF2}$. Taken from Wikimedia, where a full list of authors is available.
But sulphur dioxide’s bond angle is $119°$ as stated above. Why does sulphur adopt such a large bond angle? Which factors disfavour the approximately $90°$ arrangement?

Ah, Hero Of Time. Nice:D

If you draw the Lewis structures of SF2 and SO2, you notice that in SO2 there is one lone pair on S, while in SF2 there are two lone pairs on S (this is indeed confirmed by NBO analisys on the geometries, which are predicted to be, at the b3lyp/cc-pvqz level, with angles of 118.7 and 98.6 degrees). The geometries which you would expect from VSEPR are those with 120 degrees for SO2 and 90<x<120 for SF2, which indeed agrees with the observation. The thing I would care about, if I had to find something strange, is the reason behind the deviation from the ideal 109 degrees in the case of SF2.

@The_Vinz the ideal for some that are not in the second period is no hybridisation and bond angles of approximately $90^\circ$. Even second period atoms would prefer to get there but they are too small so they have to extend the bonding angle. $\ce{AB4}$-systems are the exception because the tetrahedron is favoured by symmetry.


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Why does protic solvent suggest E1 mechanism rather than E2?
If I understood correctly, having a protic solvent increases chance of E1 mechanism. But I can not rationalize this because carbocation intermediate is a positively charged molecule, and by surrounding that with a lot more positive protons will make the whole system in an unstable state. Hence it should make it less likely for E1 to occur. Is my understanding wrong or did I miss some important fact?

First of all, a protic solvent does not mean a highly acidic solvent; it mainly means a solvent whose molecules contain protons that can reasonably be expected to act as an acid. Take for example methanol: its $\mathrm pK_\mathrm a$ is somewhere around $15$, so calling the hydroxy proton acidic is a stretch by most definitions but there are sufficiently basic compounds that can abstract said proton.
Secondly, the question of how a protic solvent affects the E2 versus E1 ratio has less to do with the intermediate cation and more to do with the leaving group. In an E1 reaction, the leaving group dissociates from the main molecule first taking the bonding pair with it. That typically means this leaving group is now anionic or has a somewhat nucleophilic lone pair on it. The protons of the solvent can ‘block’ that lone pair from reattacking the intermediate cation, preventing the backwards reaction and increasing the odds of proceeding. The resulting cation can then be stabilised by the polar bit of a polar protic solvent: the more negatively polarised part of the solvent molecules will interact favourably with the cation.
In an E2 reaction, the base must interact with the starting material for both the starting material’s proton and the leaving group to be displaced in one step. In this case, having a protic solvent which is capable of protonating the base (reducing its basicity) is harmful meaning the reaction will proceed slower. In E1 reactions, weaker bases can be used that are not able to deprotonate the solvent molecule so this pathway is not much of a problem.

@DDD4C4U I did go into solvation effects stabilising the E1 intermediate.


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Heat of hydrogenation of 1,3-substituted cyclohexanes

From these two molecules in question, (C) should be more unstable as it is in a cis-configuration. The t-butyl and methyl groups being on the same side of the ring would exert a larger steric force on each other as compared to the trans configuration. However, (D) is mentioned as the correct answer.
How is a trans-1,3 disubstituted cyclohexane more unstable than the corresponding cis-isomer? I drew the chair projections of both compounds and the cis one has both groups on the same side, which would cause 1,3-diaxial interaction as mentioned here

Look at that bottom picture and determine whether the methyl group and the protons that are drawn out are cis or trans (better names might by syn or anti)?

If I am not wrong, the groups in the axial projection are cis

If they are so in the axial projection, they are so too in the equatorial one.

So are the groups in the axial position cis? Was my comment right?

Yes, methyl and the two marked protons are cis in both the axial and equatorial pictures.

Well you should draw the substituents in equatorial positions, because they're more stable than in axial positions where there is chance of strong flagpole repulsions with the axial H atoms. in that way in cis(or syn) both of them would be in equatorial positions and are thus more stable than in trans where one is in axial

To my understanding, OP has thought all cis-isomers, substituents are in axial-positions. That is a wrong assignment.

@MathewMahindaratne That’s where I tried to steer OP.

First of all, as Jan pointed out on the comment, better nomenclature for thes compounds is syn- and anti- instead of cis- and trans-, respectively. However, OP is more familiar with cis- and trans- nomenclature I'd continue it with my answer.
Each substituted cyclohexane ring (e.g., di-substituted ones here) has two chair conformers as depicted in OP's bottom diagram of methylcyclohexane (a mono-substituted one). In one conformer, the methyl group is in equatorial position (left hand structure) and the methyl group in the other one is in axial position (right hand structure). Because of two 1,3-interactions cost additional approximately $2 \times \pu{3.8 kJ/mol}$ steric strain energy, the predominant conformer is equatorial one.
Now, if you put tert-butyl group in $\ce{C3}$-equotorial position on equatorial-methylcyclohexane (left hand structure, vide supra), you get cis-1,3-substituted compound (C). On the other hand, if you put tert-butyl group in $\ce{C3}$-equatorial position of axial-methylcyclohexane (right hand structure), you get trans-1,3-substituted compound (D). Because it is too bulky, the tert-butyl group is always in equotorial position in cyclohexane ring (1,3-interaction with one hydrogen would cost additional approximately $\pu{11.4 kJ/mol}$ steric strain energy), as depicted in following diagram (The larger the substituent, the more the equatorial substituted conformer will be favored):

The most predominant cis-isomer (C) has both substituents in equatorial-positions (the left hand side structure of (C)). Other conformer (get by the ring flip; the right hand side structure of (C)) has both substituents in axial-positions. Thus, as a rule of thumb, the conformer with both substituents in equatorial positions is more stable.
On the other hand, the both trans-isomers of (D) have one substituent in an equatorial-position and the other in an axial-position. Yet, the conformer with the tert-butyl group in the equatorial-position (left hand side structure of (D)) is more stable between two, because tert-butyl group is comparatively much larger than methyl group. However, because most predominant cis-isomer (C) has both groups in the equatorial-positions, it is relatively more stable than (D), which has methyl on axial-position, making additional steric strain energy.


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Stability of cuprous iodide in aqueous solution

An aqueous solution of $\ce{Cu+}$ is usually unstable, but $\ce{CuI}$ is stable. Explain. $E^\circ(\ce{Cu^2+/Cu+}) = \pu{0.153 V}$ and $E^\circ(\ce{Cu+/Cu}) = \pu{0.521 V}.$

I found this question while solving a paper and I am confused as to how I should answer. For the first part at least the disproportionation of $\ce{Cu+}$ into $\ce{Cu^2+}$ and $\ce{Cu}$ explains why it is unstable in aqueous solution. (since it is neither oxidised nor reduced by water)
$$\ce{2Cu+ <=> Cu^2+ + Cu}$$
However, for the stability of $\ce{CuI}$ which remains as a precipitate in aqueous solution, I do not understand what reasoning I should provide for its stability in reference to redox reactions and electrode/formal potentials.


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Peptide bond formation involving side chains of charged amino acids
Peptide bonds are formed as such:

Aspartic acid, glutamic acid and lysine all contain either one extra $\ce{-COOH}$ or $\ce{-NH2}$ group in them. Why does that extra group not participate in peptide bond formation? For reference, the structure are given below:

It should be possible for the $\ce{-COOH}$ group attached to $\ce{C}$3 in aspartic acid and $\ce{C}$4 in Glutamic acid to form a peptide bond with $\ce{-NH2}$ of another amino acid. Do these bonds occur in any peptide linkage? If they do, what is the nomenclature associated with them and how do their chemical properties differ?

related: https://chemistry.stackexchange.com/questions/124309/synthesis-of-tripeptide-using-merrifield-peptide-synthesis-procedure-inclusion

Equilibrium in aqueous solution
In aqueous solution (such as our digestive tract), the equilibrium of the reaction as written is on the side of the reactants. Digestive enzymes catalyze the hydrolysis of peptide bonds, showing that they are not thermodynamically stable in water. However, in the absence of catalysts at neutral $\mathrm{pH}$, the peptide bond is very stable against hydrolysis, explaining how it is possible for proteins to fulfill their biological role.
Peptide formation in the cell
In the cell, amino acids form a mixed phosphoric acid anhydride by reacting with ATP, which then reacts with hydroxyl groups of tRNA to form an ester. This ester reacts with the growing peptide chain, catalyzed by ribosomes using mRNA as a template to make a specific polypeptide rather than a random one. All steps are enzyme-catalyzed, ensuring that only the main chain carboxylate reacts while the side chain carboxylate does not.
Peptide solid state synthesis
In a peptide synthesizer, amino acid building blocks start out as esters and amides with suitable protecting groups on the side chains. Only when the entire polypeptide has been synthesized are the side chain protecting groups removed. Below is a schematic of the process:

Source: https://www.biosyn.com/images/userfiles/image/general-scheme-solid-phase.jpg

Why does that extra group not participate in peptide bond formation?

It depends on the context but it does not react because

it is not activated (carboxylic acid in water does not form amide)
it is protected
enzymes are selective for specific functional groups on a molecule

It should be possible for the $\ce{-COOH}$ group attached to $\ce{C}$3 in aspartic acid and $\ce{C}$4 in Glutamic acid to form a peptide bond with $\ce{−NH2}$ of another amino acid. Do these bonds occur in any peptide linkage? If they do, what is the nomenclature associated with them and how do their chemical properties differ?

For side chains with amino groups, there are cases of peptide formation, for example in ubiquitination reactions where the protein ubiquitin is attached to a lysine via its carboxy-terminus. Other side chains are glycosylated, methylated, or attached to lipids. Together, these reactions are described as post-translational modifications. There is a report of glutamate methylation (https://www.ncbi.nlm.nih.gov/pubmed/16707700).
Some cyclic peptide antibiotics form peptide bonds with the glutamate side chain called gamma peptides (instead of alpha peptides). Here is an example:

Source: https://www.ncbi.nlm.nih.gov/pmc/articles/PMC1176672/?page=7

Actually, as far as I am aware the amide bond is the ‘bottom of the thermodynamic well’, i.e. the most stable possible carboxyl derivative. Digestive enzymes (as all enzymes) have a couple of tricks to influence kinetics and make the reaction work.

@Jan Yes, that is why reactions from anhydride to amide (or from amide to carboxylate) work without being coupled to the hydrolysis of ATP or some other source of free energy.

Let me rephrase that: afaik, the equilibrium $\ce{RCO-NHR' + H2O<=>RCOOH + H2NR'}$ favours the reactant side thermodynamically at neutral pH which would be in contradiction to your first paragraph.

@Jan That depends on the concentration of water (just like esters). If you run this in organic solvent, reactants may be favored, certainly when you continuously remove the water. With water as a solvent, the products are favored (enzymes can't change the equilibrium constant, and there is no ATP in the digestive tract for a coupled reaction - we would be unable to digest protein if the products weren't favored).

Regarding the second question, post-translational modification is not the only way. https://en.wikipedia.org/wiki/Isopeptide_bond, and the references therein, answer your questions quite nicely. These are relatively uncommon (compared to the usual peptide bond) because enzymes are capable of exerting control over which carboxylic acid / amine reacts: you will see that the example given, glutathione, requires (different) specialised enzymes to make and to break the isopeptide bond. But it is still sufficiently common that you should not be too surprised to see it.

Thank you very much for the reference. Will read more on these bonds.


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Specific heat for tetraatomic gas molecules
I came across a question which involved non linear tetra atomic gasses.
In it, the value of specific heat capacity at constant volume was given as
$9R,$ added up as
$$\frac 3 2 R + \frac 3 2 R + 6R = 9R$$
Now I know $6R$ is $C_V$ for triatomic gasses, but why do we add the two $\frac 3 2 R$ terms?
Also, how does being linear or non-linear affect the specific heat capacity of a gas?
Explanations with diagrams showing it in 3-D coordinate system would be helpful.

The heat capacity comprises contributions from translational motion (3 terms for each of $x, y, z$) plus (whole body) rotational motion ( 3 axis directions). Each motion accounts for $(1/2)R$ by equipartition theorem making the $(3/2)R + (3/2)R$ above.
Then there is a contribution from the fact that the molecule is vibrating. Non-linear molecules have $3N-6=6$ vibrational modes. Each vibration has $(1/2)R$ contribution from its potential energy and the same from its kinetic energy which in total is the $6R$ in your question.


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What can I use to remove print from hard surfaces?
I have a lot of profile products with the logotype of the company handing them out printed on them. In some situations I don't like to show these logos. However, it is stupid to bin working products for this reason so I would like to remove the print on them.
What solvent can be used for this purpose? What solvent is used in this type of colours?
Currently there are two items I want to remove the printing from:

A clear plastic bottle in BPA free tritan.
A tool in, I think, aluminium (metal anyway) with a surface similar to a Macbook Pro.

Tritan is a co-polyester. It is prone to stress cracking in the presence of a number of common solvents - cyclohexanone and tetrahydrofuran are used as surface bonding agents.
Acetone can induce stress cracking, as can IPA.
I would suggest overprinting if you want to disguise the logo.
As for the aluminium item - take your choice - cyclo, THF, acetone, xylene - lots of options.

Thank you for your answer. Shouldn't the "ink" used on the tritan in itself contain a solvent that could be used to remove it?

It could, but the ink could also be solventless (UV cured) or very fast evaporating. The adhesion of the ink may even depend on the solvent surface etching the tritan. Depositing the ink limits the solvent exposure to the area covered by the ink. Removing the deposited ink will involve adding significantly more solvent over a larger area - and probably repeated washes with the solvent.

I would suggest acetone for both - it's usually very effective for removing colours or inks of all kinds and I used to use it at work (in a lab - make sure you handle and dispose of it safely) to remove permanent marker.


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What is the role of various additives in a titration of vitamin C with N-bromosuccinimide
In a titration to determine the amount of vitamin C in an orange juice sample, the following were mixed with water:

oxalic acid
potassium iodide
starch
acetic acid

Then, the mixture was titrated with a solution of N-bromosuccinimide until the solution turned dark as the starch/iodine/iodide inclusion complex formed.

$$\ce{2I- + NBSucc + H+ -> I2 + Succ + Br-}$$
where NBSucc is N-bromosuccinimide and Succ is succinimide.
Source: https://media.cheggcdn.com/media%2Fd2a%2Fd2a428e0-d1b7-4cf1-bcf9-fb81813660b4%2Fphply3xn7.png
What is the role of the two different acids (acetic acid and oxalic acid) in the titration? How does the outcome of the titration depend on the concentration of potassium iodide?

Related: https://chemistry.stackexchange.com/questions/124856/what-happens-if-i-add-too-much-potassium-iodide-in-titration-of-vitamin-c

Have you seen https://doi.org/10.1021/ac60100a013? I have no access so I don't know if it contains any information, you might though. (I guess adding a reaction equation would be a good idea for some more context to the people coming here because of the general titration technique.)

@Martin-マーチン I added the two reaction equations.

Thanks, Martin, for pointing out the relevant paper with all the answers!
Basic procedure
This titration was first described by Mohamed Zakibarakat, Mohamed Fathy Abd El-Wahab, and Mohamed Mahmoudel-Sadrin in 1954 (doi.org/10.1021/ac60100a013) as an alternative to using 2,6-dichlorophenol-indophenol as titrant. Here is the procedure for a solution of pure acsorbic acid:

Into a 50-m1. conical Erlenmeyer flask, a known volume of the unknown ascorbic acid solution is introduced — i.e., 1 ml. contains 1 mg. of ascorbic acid. Then 5 ml. of 4% potassium iodide solution, 2 ml. of 3% acetic acid, and 2 drops of starch solution as an indicator are added. The aqueous N-bromosuccinimide solution (0.1% weight per volume) is introduced into the microburet and is allowed to run drop by drop, into the ascorbic acid solution, with continuous shaking. The end point is reached when the last drop of the N-bromosuccinimide solution added produces a permanent blue color in the ascorbic acid solution.

For fresh fruit, there is an additional step to prepare the analyte:

Edible Fruits. The new method has been applied to the estimation of ascorbic acid in certain edible Egyptian fruits known to contain ascorbic acid—e.g., oranges, lemons, tomatoes, grapes, watermelons, and water cress. The fruits used in the estimation of ascorbic acid were almost ripe. The sampling and extraction of the material under examination must be carried out with the minimum delay, so that no significant change in ascorbic acid content takes place prior to analysis. Each sample is squeezed or bruised and the juice is directly received in a known volume of 20% trichloroacetic acid, so that the juice is diluted with the stabilizing acid at least twice, to avoid the oxidation of ascorbic acid and also to precipitate interfering substances such as protein and thereby facilitate subsequent clarification of the extract. The acidulated juice is fil-tered and the determination is carried out on a known volume of the clear filtrate as described previously. The N-bromo-succinimide solution used in the titration is 0.01%—i.e., 1 ml. contains 0.1 mg. of N-bromosuccinimide (see Tables XI and XII).

Oxalic acid
Note that they did not mention oxalic acid, but they did use fairly acidic conditions to inhibit redox enzymes and precipitate proteins.
Oxalic acid does not interfere in the determination:

Interfering Substances. The following substances which might interfere have no influence on the titration process: carbohydrates such as glucose, lactose, sucrose, and starch; diketogulonic acid; reductones and reductic acid; urea; uric acid and creatinine; alcohols; formaldehyde; acetone; ethyl acetate and acetoacetic ester; thiamine hydrochloride and ribo-flavine; oxalates, tartrates, and citrates; amino acids such as glycine, alanine, valine, and isoleucine; and ferrous and ferric salts. The only interfering substances that are also oxidized by N-bromosuccinimide before iodine is liberated from potassium iodide in acetic acid medium include sodium sulfite, sulfide, thiosulf ate, and thiourea.

Role of iodide
They nicely describe the two relevant redox reactions used in the procedure:

The fact that ascorbic acid reacts very rapidly with N-bromo-succinimide, whereas many of the interfering substances react more slowly or even do not react at all, provides a reliable titrimetric method for the determination of ascorbic acid. N-Bromosuccinimide is an oxidizing agent and thus can liberate iodine from potassium iodide in aqueous acetic acid medium, but it oxidizes ascorbic acid to dehydroascorbic acid preferentially. Until all the ascorbic acid present in the solution is oxidized, no iodine is liberated from potassium iodide.
The slightest excess of N-bromosuccinimide added, after all the ascorbic acid content has been oxidized, will liberate iodine from potassium iodide, which is easily detected by the blue color developed with a few drops of starch solution added at the beginning of the titration process. The end point is thus easily observed by the appearance of a blue color in the ascorbic acid solution or extract. The presence of other reducing substances does not interfere with the titration with N-bromosuccinimide, since iodine is selec-tively liberated from potassium iodide before reducing substances present, other than ascorbic acid, are oxidized by N-bromosuccinimide, and consequently the end point is easily determined in the presence of starch. The end point is definitely blue in the case of pure solutions and pharmaceuticals, but violet in the case of biologicals and fruits.

So it would seem that as long as there is sufficient iodide to give the color change for the endpoint, the exact amount does not change when the endpoint is observed.


text:
VO2 unexpected conduction
$\ce{VO2}$ is a weird metal which becomes a conductor at temperatures higher than $\pu{67 ^\circ C}$, but below $\pu{67 ^\circ C}$, it is an insulator!
How is this possible anyway? This is against solid state theory and how are we supposed to create band gaps with it?

VO2 is not a metal. it is a metal oxide

Metal and Metal Oxide are not mutually exclusive. VO and VO2 both have metallic phases, for instance.

Metal-insulator transitions are well known in solid state physics. They occur for various reasons. Yes, it is a little weird the first time you encounter one.

It's not really the same vanadium dioxide.  The 67°C temperature corresponds to a phase change.  Compare with the phase change in solid tin.


text:
Has anyone encountered the use of Omega to represent equilibrium?
One of my lecturers last year (a pharmacologist by training) used the symbol $\Omega ^m$ as a shorthand for equilibrium. He implied it was common practice but I've never come across it elsewhere. I was wondering if anyone else has encountered it?

Interesting, what chemistry course was it, a pharmachem? Where exactly was the symbol used: above an arrow like $\ce{A <=>[$Ω^m$] B},$ or in the text and formulas as well?

Yes, it was a Chemical Pharmacology course. It wasn't used with equilibrium arrows, only in text. I've adopted it in all my notes to save me handwriting "equilibrium" every time.

I think the instructor is mistaken. Can you find this notation in any textbook. I haven't seen this notation for equilibrium in my life.

My A level chemistry teacher in 1978 also used this symbol to represent equilibrium but since then I have not come across this.

I would guess $Ω^m$ notation is a personal invention and I couldn't find any sources that would standardize it.
The choice of these symbols for abbreviating "equilibrium" may be justified that the glyph "Libra" ♎ originates from the Greek letter Omega Ω.
Further, from Merriam–Webster Online:

Equilibrium contains a root from the Latin libra, meaning "weight" or "balance". As a constellation, zodiac symbol, and astrological sign, Libra is usually pictured as a set of balance scales, often held by the blindfolded goddess of justice, which symbolizes fairness, equality, and justice. Equilibrium has special meanings in biology, chemistry, physics, and economics, but in all of them it refers to the balance of competing influences.

Uppercase letter "m" is probably because $Ω^m$ is a shortened form (contruction) of a single word that ends in the same letter: equilibrium.
$$\underset{Ω}{\underbrace{\text{equi}\underset{♎}{\underbrace{\text{libr}}}\text{iu}}}\underset{^m}{\underbrace{\text{m}}}$$

Yes, I did think that was the point of the m. I had no idea about Libra - that's really interesting. Thank you!


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Will heavy gases sink in a closed refrigerator?
Let's assume there is no air movement in the refrigerator. Is this enough for sinking of nitrogen? If not in what (low) temperatures we will start to see evident distinction between the composition of air in the upper and lower parts?

No, that's not enough, otherwise there would be industrial processes based on it. You won't have any significant separation before the main components of air start to liquefy.
Meet the barometric formula:
$$\mathfrak p=\mathfrak p_0\cdot exp\left(-{Mgh\over RT}\right)$$
Plug the height of $1\;\rm m$ (which is not unreasonable for a fridge), the oxygen's molar mass of $32\;\rm g/mol$, and the absolute temperature of $253\;\rm K$, because we are in a fridge, after all. You'll get $0.99985$. That's the pressure of oxygen near the top of the fridge, if the pressure near the bottom was $1$.
Now plug the molar mass of nitrogen, the lightest component of air (that's $28\;\rm g/mol$). You'll get $0.99987$.
So does the heavy oxygen sink to the bottom? Er, well... sort of. The gas near the top is enriched with nitrogen by a whopping factor of $1.00002$. Good luck trying to make use of it.
So it goes.


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Ground state elevation for phase transition
I have a question about the illustration from chem.libretexts.org:

the ground states of the liquid and gas are offset from that of the previous state by the heats of fusion and vaporization, respectively.

Why does the change in enthalpy increase the ground state energy?
I would be very grateful if someone can help me with this problem. I have been struggling with it for five days, but Google could not help me.
P.S. My question mark does not work.

In order to make the phase transition, you need to supply your sample molecules with an additional amount of kinetic energy (i.e., the heat of fusion or vaporization) to overcome their intermolecular bonding. This however means that after the transition, your system contains more internal energy than before. That additional energy is encoded in the vertical offset on the energy scale.
Note that if you look at each phase on its own, you could set the their ground states as the zero point on your energy scale. However, if you want to compare the system among different phases, then you need to take that offset into account. Otherwise you'd assume that your system can change from solid to liquid to gas without any effort.

thank you for your answer.
Does this mean the ground state energy itself does not change but because we need an increase in internal energy for having the possibiliy to occupy the new energy levels, we symbolically illustrating this by an elevation of the ground state. I wish you a merry Christmas from austria.

Well, the ground state energy itself does change... that's what the shift means. If you only look at one of the phases though, then all thermodynamics are in reference to its own ground state, so its absolute value doesn't matter much. Only once you take different phases into account. - Thanks, and Merry Christmas to you too!

Well, where could I find a formula or a calculation instruction for translational ground state energy.I would like to get a feeling how much the Eg changes with enthalpy change and on which variables it depends, respectively, and especially why the Eg changes.

The "translational" part was mostly a qualitative explanation as to what's happening inside the sample. In reality it will be more complicated than that, to the point where you probably wouldn't be able to strictly tell apart which energy level "belongs to the liquid" and which one "to the solid". The way I see it, this whole picture is more conceptual than anything else, to give you a broad idea how the phases relate to each other energetically. - Also, the ground state energy change is essentially the heat of fusion/vaporization.

This seems to be a complex topic. Thank you for your helpful answers!


text:
Determining the oxidation state of a bridging platinum complex
I want to ask a question about determining the oxidation state of a Platinum complex.
I was presented today with the following complex and asked to comment on the oxidation state and hence the d-orbital configuration of the central metal ion:

My working was as follows:
The $\ce{Me}$ groups do not contribute to any charge.
The $\ce{SMe2}$ ligands are neutral similar to $\ce{H2O}$ and don't contribute to any charge either.
However, if I take this into account, this would therefore mean that each Platinum central metal ion has the same magnitude of charge, but in opposite magnitudes i.e. $-2$ or $+2$ and this is not valid chemistry.
My other thought was whether the electronegativity of the $\ce{SMe2}$ ligand central S atom compared to the $\ce{Pt}$ meant that the bonding pair of electrons is electronegatively attracted to the S atom, meaning that for each $\ce{Pt}$ atom, the $\ce{Pt}$ atom would lose two of its own electrons, one in each opposite direction giving a +2 oxidation state.
The answer given is +2 for both metal centres but I was unsure using the approaches above whether this was valid.
What other steps am I missing to reach the final oxidation states of the $\ce{Pt}$ centres?

Is it a neutral molecule or an ion ?

Neutral - there is no charge present on the molecule. @Maurice

You can pretty much assume methyl ligands anionic, e.g. there are four $\ce{Me-}$ and two $\ce{Pt^2+},$ and call it a day. This compound is well-known in inorganic synthesis as bis((μ2-dimethylsulfido)-dimethyl-platinum(II)). Also, please note that we prefer not to use MathJax in titles.

Is it just by bad memory, or I remember reading that coordination number of pt is ziesse's salt is not defined. Maybe that's the case here too, not sure could someone help with this

@Zenix be careful there; Pt in Zeise's salt is undoubtedly +2. https://en.wikipedia.org/wiki/Zeise%27s_salt

@orthocresol Thank you for correction, it seems I must go through coordination complexes again..

For organometallic complexes it's often easier to throw the "more electronegative element" idea out entirely and focus on the characteristics of the ligands. This is what @andselisk's comment is referring to. Generally ligands are partitioned into two types, $\ce{L}$ and $\ce{X}$:

$\ce{L}$ refers to neutral ligands: for example $\ce{CO}$, phosphines, $\ce{H2O}$, alkenes (via the pi bond), ... and $\ce{SMe2}$ falls comfortably in this category.
$\ce{X}$ refers to anionic ligands: for example halides, hydroxide, azide, alkyl anions such as $\ce{Me}$, ...

Other ligands can be constructed as sums of these: so for example η4-butadiene would be $\ce{L2}$ (two pi bonds coordinating to the metal), an allyl group would be $\ce{LX}$ (one pi bond plus one anionic centre - ignore resonance), and so on.
The oxidation state of the metal, then, is simply the charge on the metal plus the number of X ligands. In this case each Pt centre is neutral, and has two L and two X ligands. That makes for a total oxidation state of +2.
As another example, the platinum in Zeise's salt has one L ligand (ethylene), three X ligands (chloride), and a negative charge: so its oxidation state is $-1 + 3 = +2$.
Of course, this method will fall apart very quickly when there are non-innocent ligands: $\ce{NO}$ and $\ce{O2}$ are the examples that come to mind, although there are many more. But for relatively simple complexes they suffice.

There is a very similar molecule in the literature this has aryls instead of methyls. (MEYTEP)
M.A.C.Lacabra, A.J.Canty, M.Lutz, J.Patel, A.L.Spek, Huailin Sun, G.van Koten, Inorg.Chim.Acta, 2002, volume 327, page 15. In the Cambridge database this compound is considered as a Pt(II) compound in the paper.
We also have the dimethyl compound with bridging dimethyl sulfide ligands. (XOFNUB)
Datong Song, Suning Wang, J.Organomet.Chem., 2002, volume 648, page 302 which is the compound which the poster was asking about. In the Cambridge database this is considered as Pt(II). The authors regard it as being a Pt(II) compound.
I would vote for it being a Pt(II) compound as the methyl is likely to be an anionic ligand. But there is a big "but".
As it is an organometallic compound we need to be careful about oxidation state. In organometallic chemistry it is often possible to draw the same compound in two different ways with different oxidation states. For example ferrocene can be regarded as an Iron(II) compound with two anionic Cp ligands. As an alternative we can regard the iron as being in the zero oxidation state and the Cp ligands are neutral ligands. Sometimes in organometallic chemistry oxidation state becomes a bit of a joke at times.
Back to our Pt compound, if we consider the methyl ligands to be neutral then the Pt can be in the zero oxidation state the dimethyl sulfide is a ligand which can offer up a lone pair to a metal.

If you go by the "ionic" counting method, we can start by looking at the charge on the ligands:
-The 4 terminal methyl groups are anionic when using the ionic counting method: 4 anions = 4- charge.
-For SMe2, the covalent bond to the methyl groups completes its full shell-no need to assign it a charge as it is happy being bonded with the methyls (doesn't need electrons from the metal).
Total 4- charge divided onto 2 metal centres, means we need 2+ charge at each Pt centre to balance the charge (assuming neutral complex).
Going further, we can assume sulfur is a 2-electron donor with each bond it makes to a metal centre as it still has 2 lone pairs free for donation.
Now, to find geometry let's count the total number of electrons at the metal. 2 sulfurs are 2 electron donors with each bond they make, each metal has 2 bonds from a sulfur = 4 electrons from sulfur. Platinum is group 10 so with a 2+ charge it has d8 configuration - 8 electrons from the metal. As we treated the alkyl groups as being ionic, they are 2 electron donors 2x2=4.
Add them all up 4+8+4 = 16 electron total at the metal.
Although this breaks the 18 electron rule, this is fine for 5d square planar and we know the complex will take square planar geometry at each metal centre from crystal field theory as it is a 5d metal with d8 configuration.

The question is asking about oxidation state, not electron count.

See: "and hence the d-orbital configuration of the central metal ion." I wasn't sure what this meant, but I assume it means geometry/crystal field splitting


text:
Tap water pH adjustment using citric acid
I'm treating tap water to prepare fresh water suitable for the aquarium.
I receive water with $\mathrm{pH}~7.6$ from a dechlorinator and I need to lower ammonia level.
For every unit decrease in $\mathrm{pH}$ the amount of unionized ammonium will decrease 10 times, so I need to lower the $\mathrm{pH}$ to at least $6.5$ to reduce the content of ammonia $(\mathrm{p}K_\mathrm{a}(\ce{NH3}) = 9.24)$ down to
$$\frac{[\ce{NH3}]}{[\ce{NH4+}]} = 10^{\mathrm{pH} - \mathrm{p}K_\mathrm{a}} = 10^{6.5 - 9.24} \approx 0.002\quad\text{or}\quad0.2\,\%.$$
I used citric acid to lower the $\mathrm{pH}$ down to $6,$ but after a couple of hours $\mathrm{pH}$ raised again to $7.$
The $\mathrm{pH}$ was measured using a $\mathrm{pH}$ meter prior to the contact with fish.
What is the reason for this raise? Citric acid overcomes waters buffering  feature then raise the $\ce{H+}$ concentration in water. Why would $\mathrm{pH}$ raise again?

The first part alone is hard to believe. Tap water is not much of a buffer, it's nearly impossible to get it to pH = 6 and not further.

@Ivan Neretin In the Moravian carst area, the majority of water hardness is the bicarbonate hardness. CO2/HCO3- forms a diluted buffer with maximum capacity at pH=pKa1*=6.83, if I remember correctly.

It is possible citric acid ( in any form ) is being consumed by microbial activity, what would increase pH.

Also, note that rainwater in equilibrium with atmospheric $\ce{CO2}$ - assuming aqueous speciation of $10^{-3.5}$ mol / L and given a  Henry's Law constant of $10^{-1.5}$ mol / (L atm) - clocks in with a pH of 5.65. This, btw, famously led researchers at Princeton University towards the (correct) "discovery" of acid rain for the wrong reasons (it's already acidic; the contributions of NOx and SOx were determined later).

If the water contains significant amount of bicarbonates, part of released carbon dioxide may escape and original bicarbonate is finally replace by the dihydrogen or hydrogen citrate.
$$\ce{ HCO3- + H3Citr -> H2Citr- +  H2O +  CO2 ^}$$
$$\ce{2 HCO3- + H3Citr -> HCitr^2- + 3 H2O + 2 CO2 ^}$$
So, the final  $\mathrm{pH}$ may not differ much from the original one, compared to the temporary one.

So if I want to lower it how can I do it? Is there any other ways to lower pH and keep low?

Try adding a citrate buffer Na2HCitr/Na3Citr. pKa3 = 6.39, You can prepare it by citric acid solution neutralisation to pH 6.

Maybe citric acid have had a negligible effect on the $\mathrm{pH}$. Everybody knows that in contact with air, pure water is carbonated by $\ce{CO2}$ from the atmosphere. And the $\mathrm{pH}$ may go down to $5.5$. Later on, the water may loose its $\ce{CO2}$, and the $\mathrm{pH}$ goes back to $7$.

Oxygenated tap water is rich in transition metals including Fe and Mn ions. Citric is a source of H+ and a good chelate and can drive a redox reaction in the presence of oxygen and H+ proceeding as follows:
$$\ce{4 Fe^2+/Mn^2+ + O2 + 2 H+ --> 4 Fe^3+/Mn^3+ + 2 OH-}$$
There is also a likely equilibrium reactions that can be effective in recycling ions (to continue the reaction, as occurs in natural waters) in the presence of citrate acting a chelate for, say, ferric:
$$\ce{Fe^2+ + Mn^3+ <=> Fe^3+ + Mn^2+}$$
$$\ce{Cu+ + Fe^3+ <=> Fe^2+ + Cu^2+}$$
Note: The above are active charge equilibrium reactions, referred to as a metal redox couple, and support the role of transition metals in their lower valence states engaging in notable radical creation formation via Fenton and Fenton-type (non-iron case) reactions. Further, in the presence of lab or sunlight, some further recycling of metal ions is possible.
Reference: See my comments and sources cited here.
The reaction with oxygen is electrochemical in nature with an inception period and proceeds with time.
Note: $\ce{H+}$ is consumed, so the $\mathrm{pH}$ would be expected to rise.
Here is also an electrochemistry reference, in particular, Table 2, where the first listed half-cell reaction corresponds to the above after adding two $\ce{H+}$ to both sides (creating water as a product). However, as this reaction can produce basic salts, as I have discussed previously, I believe it is more informative. Further, I do prefer my rendition, as per my prior link supplied above, it could also display a possible radical chemistry underpinning to the underlying half-cell reaction mechanics.

I doubt chelation would take a couple of hours, in homogeneous aqueous solution it should be much faster. Also, I tried to follow you cited sources, but it appears like you are quoting yourself quoting Wikipedia on copper(I) oxidation, which has a questionable relation to this topic, so it would be nice if you duplicate your sources here, both for clarity and to protect the post from link rot.

There is not enough metal in water. Even traces of metallic ions make the taste of water quite bad. This is not the correct reason. We have to confirm if the author is using citric acid indeed or not!

The cyclic nature of the chemistry (as noted in my metal redox couple comment and possible role of light) can overcome a seemingly small amount of transition metal presence with time. My argument also accounts for a reported pH rise in the presence of air.  An alternate explanation concerning a relatively sudden loss (and not, say a gain, or an established equilibrium situation?) in the water's CO2 content, is in my opinion, not likely, as the pH of tap water is relatively stable.


text:
Hydrogen bonding in B(OH)4 -
How does hydrogen bonding affect hybridisation of species? I mean in $\ce{B(OH)3}$ the hybridization of boron is $\mathrm{sp^2}$ but in aqueous phase, is the hybridization same or it changes?

Good question! Hydrogen bonding in the solid appears to be at least partially responsible for the planar (sp2) structure. ("Crystalline boric acid consists of layers of B(OH)3 molecules held together by hydrogen bonds of length 272 pm." https://en.wikipedia.org/wiki/Boric_acid) If the empty orbital on boron (which is partially filled by donation from the oxygens) were active enough to connect to a fourth oxygen, the hybridization on boron would become sp3.
Boric acid in water might be expected to ionize to a proton and B(OH)4-, which is sp3 hybridized. However, boric acid is a very weak acid, so not much borate ion is created in aqueous solution (pKa:9.24, 12.4, 13.3 for the three OH groups - reference is the same Wikipedia article). Boric acid in water probably has the same kind of structure as it has in the solid; the hydrogen bonding has just shifted from other boric acid OH groups to HOH groups of water - except for a very few molecules that have ionized.
In solid H3BO3 and in aqeous solution, hydrogen bonding is essentially all there is; but if you raise the pH to 11 or 12, you can get hydroxide ions and sp3 hybridization because the extra negative charge on the OH- makes it a better donor into the empty boron p orbital.


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NMR of a bridging Methyl complex
I was provided with the following structure of an inorganic metal complex:

And asked to give information on its NMR spectrum shown below

The highfield signal is caused by a singlet with Platinum ($I = \frac{1}{2}$) due to the terminal $\ce{CH3}$ groups.
As $\ce{^{195}Pt}$ has an abuance of $33.8\%$ this meant that in $66.2\%$ of cases, the singlet will not split, whereas in around $33.8\%$ of cases a doublet would be observed due to the satellites, which would have intensities of $\frac{33.8}{2} = 16.9\%$ hence the "Triplet" was in the ratio of around $16:67:16$.
The low-field signal however for me was easy to deduce but more difficult to calculate ratios for.
The Methyl's could couple to

No $\ce{^{195}Pt}$ resulting in a singlet
One $\ce{^{195}Pt}$ resulting in a doublet
Two $\ce{^{195}Pt}$ resulting in a singlet

The calculations given in the answer are shown:

No $\ce{^{195}Pt}$ resulting in a singlet: intensity $0.66 \times 0.66$
One $\ce{^{195}Pt}$ resulting in a doublet: intensity $2 \times 0.66 \times 0.33$
Two $\ce{^{195}Pt}$ resulting in a singlet: intensity $0.33 \times 0.33$
  This results in a final ratio of 1:8:18:8:1

These calculations confused me.
Why would for a singlet, the intensity be $0.66 \times 0.66$ if both $\ce{CH3}$ give the same signal and hence for the doublet would I need to multiply $0.66 \times 0.33$ by $2$?
These calculations also seemed erroneous.
The calculation for the intensity of the doublet and singlet give the exact same answer and I could not reach the final ratio.
What is the logic behind these calculations?

Seems to me there is an error in the last line of this list:

No $\ce{^{195}Pt}$ resulting in a singlet: intensity $0.66 \times 0.66$
One $\ce{^{195}Pt}$ resulting in a doublet: intensity $2 \times 0.66 \times 0.33$
Two $\ce{^{195}Pt}$ resulting in a singlet: intensity $0.33 \times 0.33$
  This results in a final ratio of 1:8:18:8:1

Simultaneous J-couplings to two $\ce{^{195}Pt}$ should result in a triplet, not a singlet, given that the J-couplings are identical in magnitude.
Therefore the expected relative intensities in the quintet are as follows:
$$\begin{align} &0.34^2\times \frac14 &= 0.029      \tag{1}
\\&0.66\cdot0.34\times1 &=0.224    \tag{2} \\& 0.34^2\times \frac24 + 0.66^2 \times 1 &=0.493   \tag{3} \\& 0.66\times0.34\times1 &=0.224   \tag{4}  \\& 0.34^2\times \frac14 &=0.029\ \tag{5} \end{align}$$
Note there is an implicit assumption that molecules with two $\ce{^{195}Pt}$ result in a 1:2:1 triplet contributing to peaks (1) (3) and (5) above, and that molecules with one $\ce{^{195}Pt}$ account entirely for peaks (2) and (4).
Normalized so that the central peak (3) has an integral of 18 the expected integrals become
$$\begin{align} & 1.0      \\&8.2     \\& 18.0    \\& 8.2     \\&1.0\end{align}$$
Separately I attempted digitizing and integrating the provided image of the spectrum. I obtained the following integrals:
$$\begin{align} & 0.25      \\&7.0     \\& 18.0    \\& 6.2     \\&0.7\end{align}$$
Given the resolution perhaps the disagreement is acceptable.

My answer of course presumes that the nmr spectrum corresponds to the shown compound, and that its existence is even possible. After searching for a while I have not found anything remotely similar, but I cannot claim much expertise in platinum chemistry.


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Regarding units of water potential and chemical potential
If water potential is nothing but the chemical potential of a given solution, then why do their units differ?
If the unit for the latter is taken as 'Energy per molecule' (which makes sense to me mathematically), then why is the former measured is 'Pressure'(Pascal)?

According to German Wikipedia water potential is defined as
$$\psi := \frac{\mu - \mu_0}{\bar{V}},\tag{1}$$
where $\mu_0$ is the standard chemical potential (usually pure water at atmospheric pressure at a specified reference height), and $\bar{V} \approx \pu{18 cm3 mol-1}$ is the molar volume of pure liquid water.
This means that water potential $\psi$ is defined as energy $E$ per volume $V$, which is equivalent to pressure $p$ if you perform dimensional analysis:
$$[\psi] = \frac{[E]}{[V]} = \frac{\mathsf{L}^2·\mathsf{M}·\mathsf{T}^{−2}}{\mathsf{L}^3} = \mathsf{M}·\mathsf{L}^{−1}·\mathsf{T}^{−2} = [p]\tag{2}$$
Megapascal $\pu{MPa}$ is usually used as the unit of measurement; tensiometers are usually graded in hectopascals $\pu{hPa}.$

Chemical potential is a portion of water potential, but not all of it. Factors like gravity and bulk fluid properties also affect the water potential.
Water potential is typically used for macroscopic quantities of water, so it is more natural to consider the amount of water by volume rather than the number of molecules. If you convert the number of molecules in the units of chemical potential to a corresponding volume, you will see that this is equivalent to energy per volume, which is pressure.


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Unsaturation of hydrocarbons and derivatives
What is the definition of unsaturation?
Do we consider it via baeyers test/ bromine test or
do we need to have atleast one degree of unsaturation?
And if that is so then is benzene an unsaturated hydrocarbon?...cause in benzene it just has the degree of unsaturation. And if we go by that defination...even ketones should be unsaturated

There is also degrees (elements) of unsaturation that  quantifies the number of rings and multiple bonds relative relative to a saturated, acyclic alkane. E. g.; benzene (4), cyclohexane (1), acetone (1), ethylene (1), and acetylene (2). http://ursula.chem.yale.edu/~chem220/chem220js/STUDYAIDS/unsaturation/unsat.html

In your question, you are talking about two different things (although they are related). First one is the definition of unsaturation (in organic chemistry). Well, Dictionary.com gives fairly okay definition:

(An organic compound) having a double or triple bond and capable of taking on elements or groups by direct chemical combination without the liberation of other elements or compounds, as ethylene, $\ce{CH2=CH2}$; undersaturated.

However, this definition would mostly apply to non-aromatic compounds. Since theyaromatic compounds are well stabilized by resonance, they would seldomely undergo addition reaction with reactive reagents like $\ce{Br2}$ to give saturated compounds (even though, theoretically, aromatics are unsaturated compounds). Thus, I'd call the definition of unsaturation of organic compound is the one with one or more $\pi$-bonds (e.g., acetone has one $\pi$-bond between $\ce{C-O}$). The number of $\pi$-bonds is the measure of unsaturation (remember, a triple bond is a combination of two $\pi$-bonds).
The second thing in your question is degree of unsaturation. That is the most frequent one used in structure elucidation. The degree of unsaturation of given molecule (or Index of $\ce{H2}$ Deficiency) would only give you the sum of double bonds, triple bonds, and/or rings present within the molecule. For example, if the degree of unsaturation of given compound is 1 (e.g., $\ce{C6H12}$), that means the compound is an alkene (with one double bond; e.g., hexene) or a monocyclic alkane (e.g., cyclohexane).
If you look at the saturated linear alkanes, their molecular forrmula is $\ce{C_nH_{2n+2}}$. Take a $\ce{H2}$ away from the molecule, which gives you $\ce{C_nH_{2n}}$ (like $\ce{C6H12}$, one degree of unsaturation), and so on. Thus, benzene ($\ce{C6H6}$) lacks four $\ce{H2}$ molecules (eight hydrogen atoms; $\ce{C6H14-C6H6}$), and its degree of unsaturation is 4 (3 $\pi$-bonds and a ring).

Unsaturation is a term reserved for double or triple bonds between Carbon atoms, that can add easily $\ce{H2}$ or $\ce{Br2}$ or other reactants. Benzene does not add $\ce{Br2}$ or $\ce{H2}$ easily. Some chemists consider that it is unsaturated. But not all.


text:
Rate constants and their relationship to quantum yields
Question

A dye molecule has a singlet excited state lifetime of $\pu{0.85 ns}.$ The
  quantum yield of triplet formation is $35~\%.$ What is the rate constant for
  intersystem crossing from the singlet to the triplet excited state?

Solution

$$\mathrm{QY} = \tau_\mathrm{isc}/\tau_0 = 0.35 = \tau_\mathrm{isc}/0.85$$
  Therefore
  $$\tau_\mathrm{isc} = 0.35 × 0.85 × 10^{-9} = \pu{0.297E-9} = \pu{0.3E-9}$$
  But $k_\mathrm{isc} = 1/\tau_\mathrm{isc}.$
  So, $k_\mathrm{isc} = \pu{3.3E9 s-1}$

My working
$$\phi_{T} = \frac{k_\mathrm{isc}}{k_0} = \frac{\tau_0}{\tau_\mathrm{isc}}$$
and then substituting the values
$$0.35 = \frac{0.85}{\tau_\mathrm{isc}},$$
hence
$$\tau_\mathrm{isc} = \frac{0.85}{0.35} = \pu{2.43 ns},$$
so
$$k_\mathrm{isc} = \frac{1}{\tau_\mathrm{isc}} = \frac{1}{\pu{2.43E-9}} = \pu{4.11E8 s-1}.$$
But apparently in the solution the quantum yield $\displaystyle\phi_T = \frac{\tau_\mathrm{isc}}{\tau_{0}}$ and then the working follows as above to yield $k_\mathrm{isc} = \pu{3.3E9 s-1}.$
Am I mistaken or is the solution mistaken?

Looks like the solution is mistaken. Not sure what else there is to say about this, really...

I agree. The solution is mistaken, at least in the calculation of the tau for intercrossing.

Your answer is the correct one.


text:
pH of amphoteric species
For a solution of $\ce{HA-}$, I have seen the following approximations for the $\mathrm{pH}:$
$$\mathrm{pH} = \frac{1}{2}(\mathrm{p}K_\mathrm{a1} + \mathrm{p}K_\mathrm{a2})\tag{1}$$
$$\mathrm{pH} = \sqrt{\frac{K_1K_2[\ce{HA-}] + K_1K_\mathrm{w}}{K_1 + [\ce{HA-}]}}\tag{2}$$
How do they differ? Is the second approximation always more accurate than the first? I know that the second is derived from the mass and charge balance equations and the first is the same as the second, with several assumptions.

The second equation should be for $\ce{[H+]}$ not $\pu{pH}$.

The correct expressions are:
$$\mathrm{pH} = \frac{1}{2}(\mathrm{p}K_\mathrm{a1} + \mathrm{p}K_\mathrm{a2})\tag{1}$$
$$\ce{[H+]} = \sqrt{\frac{K_1K_2[\ce{HA-}] + K_1K_\mathrm{w}}{K_1 + [\ce{HA-}]}}\tag{2}$$
Equation 2 is an exact expression (neglecting activities vs. concentrations), but Expression 1 is an approximation.
To dervice the approximate expression start by taking the -log of the second equation:
$$-\log{\ce{[H+]}} = \mathrm{pH} = -\log\left(\sqrt{\frac{K_1K_2[\ce{HA-}] + K_1K_\mathrm{w}}{K_1 + [\ce{HA-}]}}\right)$$
removing $K_1$ from numerator and simplifying...
$$\mathrm{pH} = \frac{1}{2}\left(\mathrm{pK}_1 -\log\left(\frac{K_2[\ce{HA-}] + K_\mathrm{w}}{K_1 + [\ce{HA-}]}\right)\right)$$
Now to reduce $\dfrac{K_2[\ce{HA-}] + K_\mathrm{w}}{K_1 + [\ce{HA-}]}$ to $K_2$ two conditions must be met:

$K_2[\ce{HA-}] \gg K_\mathrm{w}$
$K_1 \ll [\ce{HA-}]$

Derivation of first equation is where Sir ? From where i can see?

The former equation assumes
$$[\ce{H2A}]\simeq [\ce{A^2-}]$$
due reaction
$$\ce{ 2 HA- <=> H2A + A^2-}$$
The is possible with 2 simplifying conditions:

The concentration of oxonium resp. hydroxide ions originated from water dissociation is much lower than concentration of the basic resp. acidic ampholyte form.
$$[\ce{H2A}] \gg \sqrt{K_\mathrm{w}}$$
$$[\ce{A^2-}] \gg \sqrt{K_\mathrm{w}}$$
That assures the effect of ampholyte reaction with water dissociation products
$$\ce{ A^2- + H3O+ -> HA- + H2O}$$
$$\ce{ H2A + OH- -> HA- + H2O}$$
is negligible for the ampholyte  form concentration ratios and you can ignore water dissociation.
This condition is critical for low ampholyte concentration near neutral $\mathrm{pH}$.

The concentration of oxonium resp. hydroxide ions is much lower than concentration of acidic resp. basic ampholyte forms.
$$[\ce{H2A}] \gg [\ce{H+}]$$
$$[\ce{A^2-}] \gg [\ce{OH-}]$$
That assures the effect of production of oxonium resp. hydroxide ions to reach the final $\mathrm{pH}$
$$\ce{ HA- + H2O -> A^2- + H3O+}$$
resp.
$$\ce{ HA- + H2O -> H2A + OH-}$$
does not compete with creation of acidic/basic ampholyte forms.
$$\ce{ 2 HA- + -> A^2- + H2A }$$
This condition is not valid for very low /very high target $\mathrm{pH}$, as respective ions are produced in expense of the deviation from the same ratio of ampholyte form concentration.


text:
Why we can't see 3-d structures of certain compounds on PubChem?
It's common to read statements like one given below on PubChem:

Conformer generation is disallowed since MMFF94s unsupported element.

What's the meaning of the statement, and why are certain compounds unsupported? And what is MMFF94s?
This might be something really basic, but since I have never come across I asked it.

Most classical molecular force fields are parameterized for a set of elements and atom types. The MMFF94 method was designed for standard organic drug-like small molecules, so it has a limited set of elements (H, Li, C, N, O, F, Na, Mg, Si, P, S, Cl, K, Ca, Fe, Cu, Zn, Br, I) categorized into 99 atom types (e.g. mmffprop.par from the Open Babel implementation).
In the cases of the metals, they should be ions, not covalently bonded to any other element.
As in the case of many molecule force fields, there are sets of bond types, angle types, etc. as combinations of the atom types. Molecules that contain elements outside the parameterization, or combinations that aren't in the bond or atom parameters, are rejected.
As an example, the MMFF94 validation set is available:
http://server.ccl.net/cca/data/MMFF94/MMFF94_dative.mol2.shtml
Beyond the limits of MMFF94 and MMFF94s themselves, PubChem3D had several limits indicated in the accompanying manuscript: Bolton et. al. "PubChem3D: a new resource for scientists" J Cheminf. (2011) v. 3, art. 32)

Not too large (with ≤ 50 non-hydrogen atoms).
Not too flexible (with ≤ 15 rotatable bonds).
Consists of only supported elements (H, C, N, O, F, Si, P, S, Cl, Br, and I).
Has only a single covalent unit (i.e., not a salt or a mixture).
Contains only atom types recognized by the MMFF94s force field.
Has fewer than six undefined atom or bond stereo centers.

In the case of molecules with undefined atom or bond stereo (e.g., E/Z) multiple stereoisomers were generated. Personally, I'd use these compounds with extreme care - it's not always obvious what the original PubChem entry represents.
Thus there are other records without 3D versions (e.g, they're large, have multiple covalent units, etc.)

I'll add that the UFF method has rule-based parameters that cover most of the periodic table with various levels of accuracy, which would cover a few of these issues. (I'd personally love to see a newer version of UFF, e.g. with polarizable charges, but that's a research question.)

Since it's been a couple of days, I guess I convert my comment into an answer.
MMFF94s is Merck Molecular Force Field.
From what I understand, the issue is that this FF has certain limitations when it comes to electrostatic interaction, estimating polarizability and intramolecular interactions, which confuses the method trying to predict 3D conformers for ionic compounds, metal complexes or molecules with twisted geometry.
From the Concluding Discussion section of the original publication [1, p. 515]:

Despite encouraging success, certain limitations are evident.
  One of particular importance arises from the fact that MMFF94 uses static atom-centered charges. As such, it neglects both higher order multipoles and electrostatic effects that arise from molecular polarizability.
  Because of these simplifications, MMFF94, like a number of other force fields, employs “enhanced” charge distributions that emulate the effect of polarizability in amplifying electrostatic interactions for favorable contacts in a high-dielectric medium. Unfortunately, these enhanced charge distributions also amplify electrostatically unfavorable iteractions, whereas proper account of polarizability would
  diminish them.
  They also improperly enhance electrostatic interactions in gas-phase or low-dielectric environments. Furthermore, they may not be optimal for describing intramolecular interactions, and may thereby limit the ability of the force field to account for differences in conformational energies.
  Indeed, compounds containing two or more strongly polar functional groups in close
  proximity have proven to be the most problematic in this respect, though good results have been obtained in most cases to date. …  Other significant limitations include: the overly simplistic nature of the bond-charge-increment scheme used to assemble the partial atomic charges …; the lack of conformational dependence of the resultant charges …;
  and the omission of bond-torsion (and certain other) cross terms needed to account for significant geometrical changes that can occur when a torsion angle varies,… an example being the elongation of an amide partial C−N double bond by up to 0.1 Å when conjugation is broken.
  A further significant limitation is that no account is taken of metal-ligand interactions beyond that afforded by a relatively simplistic model that includes only electrostatic and van der Waals nonbonded interactions.

References

Halgren, T. A. Merck Molecular Force Field. I. Basis, Form, Scope, Parameterization, and Performance of MMFF94. Journal of Computational Chemistry 1996, 17 (5–6), 490–519. DOI: 10.1002/(SICI)1096-987X(199604)17:5/6<490::AID-JCC1>3.0.CO;2-P.

Like I wanted to see the structure of Diboron Dioxide, shouldn't the 3-d structure of Diboron Dioxide $\ce{(BO)2}$ accessible?

@Zenix AFAIK, both $\ce{BO}$ and $\ce{B2O2}$ are unstable high-temperature molecular species existing only in gaseous phase somewhere above 1200 °C. It doesn't look like MMFF was designed for conditions like this, but I'm afraid I cannot give you a more precise answer.

@andselisk - your point is good that MMFF94 has significant, known limitations. It was, after all, designed in 1994. It's pretty good, although I'm surprised there hasn't been much effort for a replacement over the last 25 years!


text:
Grouping QM9 molecules
I have recently been reading a paper, in which given dihedral angles of alanine dipeptide molecule $\psi$ and $\phi$, the conformation is catagorized as either $\beta$-1, $\beta$-2, or $\alpha$. I was trying to see if it is possible to extend the same idea to QM9 molecules and cluster them into different groups based on their properties? The properties provided for the QM9 dataset include:

Norm of the dipole moment ($\mu$)
Norm of static polarizability ($\alpha$)
Energy of the electrons in the highest occupied molecular orbital
(HOMO energy), lowest unoccupied molecular orbital (LUMO energy),
and their difference (Gap)
Zero point vibration energy (zpve)
Atomization energy (U)
Electronic spatial extent (R2)

The answer that I'm looking for is for instance something in the lines of: "based on the $\delta$ values, molecules can be divided into to two groups of $\gamma$-1 and $\gamma$-2". In other words does any of these properties inherently divide the dataset into k groups?

This seems a little too broad at this point. For example, you could divide them into molecules without a dipole (or below some threshold) and molecules with a dipole. The same could be done for any of these properties. It would help to know what you hope to accomplish with the groups.

"I was trying to see if it is possible to extend the same idea to QM9 molecules": Why would it not be possible?

@Tyberius Basically, in a number of papers, a machine learning algorithm is used to cluster different conformations of alanine dipeptide molecule and eventually predict it based on the input configuration. Now I was wondering if instead of different configurations of alanine dipeptide molecule, if someone gives different molecules as input to the algorithm, what would be a good indication of prediction capability of this algorithm.

@Tyberius Again, I understand that it is hard to come up with the problem without looking at what we have first, but I just wanted to know that 1) is there a common test used in the literature that I'm unaware of? 2) If not, any [more] suggestions?

@user1271772 Unlike the conformation prediction for alanine dipeptide, from what I see, people didn't try to show a similar type of prediction for their works. I only see an illustration of their latent space and that's pretty much it. Some works color the latent space based on a continuous value, but I was looking for a discrete clustering, similar to my previous question.

@Blade: It doesn't seem you'll get an answer here. Perhaps a Stack Exchange dedicated only to computational physical chemistry or materials science, where there's no questions like "what is an acid?" and the real experts in the field are concentrated and engaged, would be helpful. You might be interested in committing so that you can be part of the private beta of this: https://area51.stackexchange.com/proposals/122958/materials-modeling?referrer=MDhiMGU4YTU1NzNmMjEyZDJhNGRmMGE5MmMzMWE3N2YzN2U2YjIwMWE3Yzk3NWNjMzQ4ZWUxNjdmYjQzZmFkY5vUb8YOVAoVzjAp86f_1qFiBiRoSuyb9efTedyd0hTV0

So you are essentially just looking for something else you can predict and group using the molecular confirmation? The dipole  and/or polarizability should work for this purpose. Alternatively, you could try to generate groups from the properties you have using k-means clustering and then see if you can predict the group just from molecular configuration.

Thanks for the suggestions, I'll try both. Just for future readers, let me correct your sentence : I'm essentially just looking for something else I can predict and group. Previously: INPUT: molecular configuration of a single molecule, PREDICTION: cluster based on conformation type. Now: INPUT: different molecules in QM9 dataset, including atom types and bonds, PREDICTION: this is my question, what people would be interested to see as a prediction, what are common predictions that people address in their works.

@Tyberius Also, I liked the k-means clustering idea very much, the only additional improvement would be if the k was actually meaningful. Meaning that for instant I can divide HOMO energy into "k= any number that I like", but it would have been nicer if HOMO energy could have inherently been divided into k meaningful groups, not a random number that I picked. I guess I'll modify my question now as "if any of these properties inherently divide the dataset into k groups."

IMHO, the problem is two-fold: (1) that the molecules in QM9 are an enumeration of possible molecules up to 9 non-first-row atoms, not a set of known molecules and (2) it's not obvious what the purpose of the clusters would be. In the case of Ramachandran plots, they indicate the shape of the secondary structure. Since QM9 is something of an artificial group, I can't think of an obvious application for clusters from it.

@GeoffHutchison Thanks for the comments. Quick question: What are non-first-row atoms?

Things other than hydrogen (and helium).

Ah, you mean periodic table! My bad


text:
Comparison between solubility of amines and alcohol
Why are alcohols more water soluble than amines for similar molecular weight.
I know this is a duplicate question . The reason i posted it again was that on the original question i posted my answer and requested others to correct me if i am wrong so that i could deeply understand the concept. So I am posting my answer here again please comment on it.

I think it may be due to van der waals force (v.w.f). For both alcohols and amines the v.w.f acting between them and water molecules is keesom force i.e. dipole dipole interaction. In such case v.w.f is directly proportional to molecular mass , but if the molecular mass is also same then v.w.f is directly proportional to surface area of molecule which is inversely proportional to branching. Hence since amines have more branching than alcohol they have less surface area and hence less van der waals force. Due to weaker v.w.f amines are less soluble than alcohol.
Please correct me if I am wrong

If respected people have the time to close my question could they please get some time to solve my query

What is the question ?

Why are alcohols more water doluble than amines for similar molecular weight?


text:
Structure of CCl₂
I read that $\ce{CCl2}$ is non-polar. When I read about the reason on the net it stated that $\ce{CCl2}$ has a linear structure, but according to VSEPR theory it should have bent V-shaped geometry. Please clarify.

Why bother about this compound ? It is not stable. It only exists as an intermediate in some reactions. You will never be able to store it in a flask with label CCl2. So why ?

@Sameer My apologies, I misread the compound formula as dichloromethane, so I deleted my comment as irrelevant. Sorry for the confusion, this is a nice question. Have a look at What is the hybridization state of E (= Sn, C) in ECl₂? and the paper Martin refers to in his answer: https://doi.org/10.1039/b202865d, this should answer your question (yes, CCl₂ is bent).

@Maurice "Science isn't about why; it's about why not".

VSEPR theory is great, but only up to a certain point. You should probably not use it for "exotic" species like carbenes. Even though it happens to give the right result in this case, that's probably more of a coincidence than anything.


text:
Selectivity of diazo-coupling with p-amino phenol

I came across this question and I've been thinking about it since, without being able to find a convincing major product. This question is related to one I found on this site, Diazo Coupling reaction with para-substituted phenol?, but it doesn't totally clear my doubt, and hence I specifically posted this reaction.
My thoughts were, since benzene diazonium is a weak electrophile, it would prefer to attack the most nucleophilic or electron rich site, and $\ce{-NH2}$ being better nucleophile than $\ce{-OH}$, I feel that would be more preferred site of coupling. Also I found out that aniline reacts with diazonium in basic medium to form diazo-aminobenzene which later rearranges to form more stable amino azobenzene; only to support my opinion. Thus my product was:

But the actual answer given was ortho coupling w.r.t. to $\ce{-OH}$, and it supports the product shown in the previously mentioned question, and @orthocresol commented that it occurs due to formation of phenoxide in basic medium:

But still it wasn't very convincing for me, as no one posted a solid answer to the question, neither are their any suitable articles I found on the internet that compare the reactivities of different functional groups towards diazonium.
I would be very obliged if someone could provide a deeper insight to the matter, and clear my misconceptions(if any), and a critical view about the barriers or aids in formation of the products.

I think the right way to approach this question isn't "we don't know the product, let's try to reason it out from first principles"; rather it should be more like "we found that this is the product when we ran the reaction in the lab, how can we rationalise this observation using a model of nucleophilicity that we know?"
So, in that previous comment which you quote, I did not mean to say "phenol will be deprotonated, hence this product will definitely be formed"; rather I meant something more like "if this is indeed the correct product, then a possible reason is because the phenol is deprotonated, making it (or the positions ortho to it) more nucleophilic than the aniline".
That said, I did just look it up, and there are indeed some references from the primary literature which support the claim that the product above is formed.

So, since we have experimental evidence (and insofar as this evidence is sound), you should not doubt what the product is; the only question left is why it is formed. And so I offer you the same explanation: that phenol is deprotonated under the basic conditions, and that phenoxide is more nucleophilic than aniline (at the ortho positions).
As for why phenoxide is more nucleophilic than aniline: I could offer you some wishy-washy explanations, like how a negative charge makes the oxygen atom more electron-donating towards the ring (and that outweighs the natural nucleophilicity ranking of N > O), but I don't think there is a simple one-line answer which is sufficiently rigorous for me to accept. I tried to look for some other evidence, but it is very difficult to compare C-reactivity of phenoxide vs aniline, since most EAS reactions take place in acidic media. There are some studies which conclusively show that phenoxide-O is more nucleophilic than aniline-N, but that isn't what we're looking for either.
So... all in all, it is better to use this experimental observation (that the azo coupling occurs ortho to oxygen) as proof that phenoxide-C is more nucleophilic than aniline-C, rather than to try to do it the other way round, i.e. "phenoxide-C is more nucleophilic than aniline-C, hence it must be true that this reaction will form this product". After all, chemistry is first and foremost an experimental science.
It's a subtle difference, but adopting the correct scientific approach will be better for you as a chemist in the future.

@orthocresol Thank you very much. Yes I do know that chemistry is mostly an experimental science, and that most theories were proposed only to reason the observations. But I was worried because the actual product clashed with my principles, and since I found no literature about the speciality of this reaction, I doubted the answer.
So do you suggest doing the same reaction under acidic conditions(most of which are), it could yield a product with ortho coupling w.r.t. -NH2

@SirArthur7 Are most azo couplings done under acidic conditions...? I'm not aware of that - as far as I know, they are base-catalysed or at least performed in neutral pH. If you acidify too much you will protonate the aniline which kills its nucleophilicity anyway. On top of that, there is another question of whether the diazonium ion reacts ortho to NH2, or with the NH2 group itself to form a triazene (see https://chemistry.stackexchange.com/q/107344/). So I'm afraid I don't quite know how this would turn out, and I don't want to make any suggestions as to what the outcome would be.

Well I just said what I saw on this very site, that aniline undergoes azo coupling under acidic conditions,(https://chemistry.stackexchange.com/questions/29251/what-is-the-role-of-ph-in-azo-coupling-reaction-of-diazonium-with-phenol-and-ani)
And yes I am aware about the possibility of formation of triazene which rearranges to form more stable amino azobenzene(mentioned in the question itself).
I asked the question only to ascertain myself, actually the problem is here at school we don't get apparatus or chemicals to run a reaction, we need to make use of pen paper and basic principles.

@SirArthur7 Ah, ok. I'm not very convinced by Klaus's answer there: if you protonate the aniline, then there is no more lone pair in conjugation with the ring, and the ortho position will no longer be nucleophilic. But I looked it up and you are right that azo couplings do take place at acidic pH's as well. I think it is sensible to guess that under (very mildly) acidic conditions, the preferred product comes about via reaction ortho to NH2. But again, I do not want to definitively claim that to be the case without some sort of evidence.

And talking about acidifying the aniline, I share the same doubt about carrying the reaction under acidic conditions (https://chemistry.stackexchange.com/questions/112468/role-of-ph-in-azo-c-coupling-reaction-continued)...another unanswered question on the same topic. Actually there aren't any solid answers posted on this topic before, for e.g. nobody mentioned about the killing of nucleophilicity of -NH2 via protonation. I wasn't convinced by wishy-washy comments, and I got a feeling that its a topic that most people avoid answering, but couldn't understand why.

"most people avoid answering, but couldn't understand why" - because we don't fully understand it. The truth is that most things don't have a nice, one-line answer which we can rattle off, and if you are taught otherwise, you should probably take it with a healthy pinch of salt.

Let us continue this discussion in chat.


text:
Is boat form of cyclohexane non polar?
In chair form of cyclohexane the C-H dipole moments cancel out each other but in boat form i don't think all dipole moments will be cancelled out. Also if boat form is polar then what will we say overall about cyclohexane is it polar or nonpolar?

You are right in that boat form doesn't enforce zero dipole moment by symmetry. But then again, the dipoles of C-H bonds are very small, so we can consider all hydrocarbons nonpolar, regardless of their conformation.

The boat form is ~6kJ/mol less favourable, making cyclohexane 99% chair at RT. So whatever small polarity the boat form has (after all, its a plain alkane, and still rather symmetric), you need to divide that number by a hundred.
Btw. the twist conformation is not as unfavourable as the boat form, but it is also too symmetric to have any dipole moment.
(https://en.wikipedia.org/wiki/Cyclohexane_conformation says CH is even 99.99% chair)

Being nonpolar does not require having the zero dipole moment.
Enough is being much closer to having a zero dipole moment than to having the a strong dipole moment. It applies also to parts of the molecule.
E.g. carbon dioxide has zero permanent dipole moment, yet liquid carbon dioxide is rather polar solvent due its polar bonds.


text:
Why do transition elements specifically show complementary colours?
When I searched for why the colour absorbed and emitted by coloured transitional elements are different , i found out that the metals absorb the light they need and the remaining is reflected .
But my doubt here is what happens to the absorbed light , i mean will it emit it? If yes, in what form ??
It can't absorb and reflect colours continuosly, right?

Usually a colored stuff absorbs a color, which helps its electrons to go to an excited state. Once excited, the atom or the molecule transfers this energy to the vibrations and rotations of the molecule : the agitation increases, and so the temperature.
The energy may also be reemitted in light after the absorption process. This happens with fluorescent or phosphorescent substances, which are not numerous. In my opinion, no transition element have fluorescent or phosphorescent properties.


text:
Inflection point, slope, and equivalence point in a titration
My textbook, Harris 7th Edition Quantitative Chemical Analysis, says:

In titrations involving 1:1 stoichiometry of reactants, the equivalence point is the steepest point of the titration curve. This is true of acid-base, complexometric, and redox titrations as well.

My question is, does this apply to all acid-base titrations (strong-strong and weak-strong)? My intuition seems to say that it only applies to strong-strong, but I may be incorrect.

Your intuition is not sufficient in this matter. The equivalence point is the steepest point of any titration curve, however weak or strong is the acid and the base.

Within titration precision, it is true, but mathematically, I would not be so sure even for strons/strong case, as even here is not the curve perfectly symmetric. I remember there was similar question recently. I also remember some potentiometric titration chart, where equivalence and inflection points were explicitly marked as not identical.

@Poutnik A discussion of equivalence vs inflection points in titration curves can be found here: https://chemistry.stackexchange.com/questions/124603/equivalence-point-vs-inflection-point-in-the-titration-curve-of-a-weak-acid.

There are two sides to this. We can take a look at the shape of titration curves, but also look into the choice of the end point (where the indicator changes colour).
Let’s first take a look at the titration curve itself. There are several good answers on the internet that explain the shape of a titration curve. See for example The reason behind the steep rise in pH in the acid base titration curve, where several ways of explaining are used. The same reasoning goes for redox titrations (check pages 328-332 of your textbook) or even precipitation titrations (pages 127-131 of the same book). But let’s focus on acid-base titrations for now.
Ok, so we know what titration curves look like and why. But it’s important to know that the end point is chosen to be the steepest part of the titration curve (which happens to be at the equivalence point). That’s because at the steepest point of the curve, the effect of the titration is measured with the greatest precision. Near the equivalence point, a small amount of titrant added will cause a relatively large change in pH. This can be easily observed if a suitable indicator is used. The fact that the equivalence point can thus be measured so precisely makes it especially suitable for analysis.

Then there can be even 3 points: the equivalence point, the end point, the inflection point.


text:
Questions in MO theory
Τhe electrons in the antibonding MO are the unpaired electrons of the molecule? E.g. $\ce{O2}$, by writing down the diagram for the MO we notice that there are only 2 electrons in the top antibonding orbitals thus 2 unpaired electrons, yet by writing down the Lewis structure, the molecule turns out to have 0 unpaired electrons..
So, what do the electrons in the anti bonding orbitals do if they don't represent the unpaired electrons of the molecule?

Refrain from merging multiple questions into one.

For second part, how did you get two $\sigma$ bonds?

The antibonding orbital doesn't necessarily tell you the number of unpaired electrons - it's coincidental that for O$_2$, the HOMO (highest occupied molecular orbital) is a (1$\pi_g)^2 $ orbital - or, if you prefer to write it like you have for N$_2$, ($\pi ^1 _{2px} = \pi ^1 _{2py}$).
$\pi$ orbitals are doubly degenerate, so the lowest energy configuration is for the 2 electrons to occupy different orbitals (one in each of the 2 $\pi$ orbitals). This means they're both unpaired, as shown in this MO diagram of O$_2$ from Wikipedia. For an example of an MO diagram where there are unpaired electrons in an orbital which is not antibonding, try drawing the B$_2$ MO diagram (remember that s-p mixing occurs).

In terms of what they do, antibonding electrons destabilise the bond, "cancelling out" the energetically favourable effect of bonding electrons - try drawing an MO diagram for He$_2$ or Ne$_2$. It should be apparent why they don't form stable molecules.
As far as N$_2$ goes, I think you may have neglected to take into account the effect of the $\sigma ^* _{2s} $ electrons. Remember that antibonding electrons cancel out bonding electrons.

The Lewis model has weaknesses. O2 is a good example of the failure of the Lewis model. It does not give the good structure for the O2 molecule. The Lewis structure of O2 has no unpaired electrons. But the experiment shows that a jet of O2 is deflected in a magnetic field, which can only be explained by the presence of at least one free electron.
The molecule N2 has one sigma bonds (and not two) and two pi bonds. Why do you think that there is two sigma bonds ?

Free electron is not the best term