text:
How do we know a reaction is in equilibrium?
In teaching secondary science equilibrium, how can we actually show there is a backwards reaction occurring during equilibrium rather than no reaction. A lot of equilibrium experiments involve external changes such as temperature or change in pressure. This to me shows that a reaction can occur backwards, i.e reversible but it doesn't show that there is a forward and backwards reaction occurring at the same time at equilibrium.  What's a simple science experiment or everyday life experience that equilibrium can explain where the idea that an reaction just stops taking place doesn't?
Every video just claims that there is a forward AND backwards reaction but doesn't really explain why and is also something I've taken for granted
Thanks

The only way that I can think of to show that there is a dynamic equilibrium would be to use isotopes. Way beyond a secondary science experiment. // To show that reactions are dynamic one could use the iodine clock reaction, but that really doesn't show a dynamic reaction at equilibrium.// Lots of ways to show that a reaction can go backwards and forward, but how to show that happening dynamically is an interesting question.

@MaxW either radioisotopes, either using enrichment by a minor stable isotope. Then the isotope redistribution can be traced by Mass Spectroscopy.

@Poutnik - That was my first thought. However I doubt that any middle schools or high schools in the US has a mass spec. My other thought was Ostwald ripening. Show that a centrifuge can't clear the colloidal size particles, digest for an hour, then show that the centrifuge does clear the particles.

A colour change might be a good way of testing this in the classroom, although not ideal. Perhaps phenolphthalein or methyl orange that have a different colour in acid /base. Adding acid /base should swap the colour back and forth about equilibrium. You would need to find a reaction with a pK v.  close to the indicator colour change pH.

Thanks everyone for the replies, when I couldn't find an answer in the hours of googling, I should've realised it wasn't so simple. It seems to be a conceptual thing, i.e H2 + I2 -> 2HI but also that 2HI can decompose so it makes sense that this should be happening as well.
I will try to investigate the radioisotope method

@MaxW Yes, I agree. Not sure about radioisotopes though, how much better would be available. I may have misread the question, considering just theoretical ways. But how about a precise gravimetry, if e,g. heavy water is inlvolved ?  Like $\ce{HX + D2O <-> DX + HDO}$ Or using other deuterize compounds.

An equilibrium can be demonstrated if the reactants give the same equilibrium mixture as when the products are mixed together.

By definition, the macroscopic observables do not change once equilibrium is reached, and experiments that probe the microscopic domain are usually expensive. I use an analogy to show the microscopic changes:
The model

Take two beakers A and B, and fill one or both with some water (and add some food color for better visibility), and use two straws F and R to continuously transfer from beaker A to beaker B (with straw F) and from beaker B to beaker A (with straw R). The transfer is by having the straw touch the bottom, and then covering the top with your index finger or thumb.
At that point I ask students what they expect will happen if I continue for the rest of the class period. At one point, the level of liquid does not change anymore even though I am still transferring liquid. This is the state of dynamic equilibrium.
To make it more interesting, take straws of unequal diameter (or two vs one straw, like in the looped video).
The analogy explained
The liquid level in the two beakers represent the concentration of reactant A and product B. The diameter of the straws represent the rate constant of the forward reaction (straw F) and the reverse reaction (straw R).
$$\ce{A <=>[k_f][k_r] B}$$
Limitations
There is only one reactant and one product. The reaction will always be first order.
Avoiding misconceptions
If you take a single straw to transfer back and forth, students might think that equilibrium is like a pendulum. If you take straws of equal diameter, they might think equilibrium means that concentrations are equal (rather than correctly, rates are equal).

Every video just claims that there is a forward AND backwards reaction but doesn't really explain why and is also something I've taken for granted

Starting with just A, you learn that the A molecules have the capability of turning into B. Starting with just B, you learn that the B molecules have the capability of turning into A. If equilibrium were static, someone would "have to tell" the molecules to stop reacting. With the dynamic equilibrium model (which has support from expensive molecular-level experiments), molecules keep on doing what they always did - react every now and then. There is no requirement for some kind of switch in behavior, it just is an emergent property of having both forward and reverse reactions going on and influencing concentrations of species.

A reversible equilibrium can be shown in the high school level with cobalt chloride $\ce{CoCl_2·6H_2O}$, which can be written as  $\ce{ [Co(H_2O)_6]Cl_2}$
If a couple of milligrams of usual cobalt chloride $\ce{CoCl_2·6H_2O}$ is dissolved in ethanol in a test tube, a blue solution is obtained. This is due to the reaction :$$\ce{2 [Co(H_2O)_6]Cl_2 -> [Co(H_2O)_6]^{2+} + CoCl_4^{2-}}$$ The two ions produced in this reaction are $\ce{[Co(H_2O)_6]^{2+}}$ which is pale pink, and $\ce{CoCl_4^{2-}}$ which is dark blue. As a result, the mixture looks blue.
Now if you add one or two drops water in this blue solution, the color changes to pale pink, due to the exothermic reaction :$$\ce{CoCl_4^{2-} + 6 H_2O -> [Co(H_2O)_6]^{2+} + 4 Cl^-}$$
Now this equation is an equilibrium that can be easily reversed, according to the equation : $$\ce{CoCl_4^{2-} + 6 H_2O <=> [Co(H_2O)_6]^{2+} + 4 Cl^-}$$
To show that the equation can be reversed, you heat the pink solution. At about $60°$ C or $70°$C, the solution turns blue again : the reaction is reversed, and goes to the endothermic side. By the way, it also favorises the side with the maximum entropy.
This reaction can be reversed by cooling it to room temperature, where it goes back to pink. The same mixture can be reheated and re-cooled, to show that is is reversible.
The same reaction can be obtained with copper chloride $\ce{CuCl_2·2H_2O}$, where the color goes from green to blue.

I think this does not asnwer the question, that is not about reversible equilibrium. It is about to show equilibrium is a dynamic, not static process, where there are ongoing the both opposite reactions, but with the zero net result.

But otherwise nice experiment !

I thought also any simple phase change such as freezing/melting water or solubilization/crystallization of a salt might do. But this is much neater.


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Is Magnesium Oxide alkali
Given that Magnesium Hydroxide is an alkali, can we say Magnesium Oxide is an alkali since dissolving Magnesium Oxide in water can produce Magnesium Hydroxide?

If a question is asked on Chemistry SE site, then, in contrary to sites like Quora, it is expected from the author to elaborate the topic in the question, doing at least basic research oneself, writing what he/she found and understood, and what is the stumblestone. The quick questions without explicitly expressed particular effort are not very welcome, and may be closed.

Tl, dr:  Chemists would generally say "No".  But it's not as cut and dried as one might think.  The alkaline character of magnesia and heavier Group 2 oxides is one of many examples in chemistry where the actual character is intermediate between the extremes a yes/no answer seems to imply.
Usually the term "alkali" is reserved for the hydroxides of Group 1 metals, which (at least among compounds of commonly observed elements) uniquely combine strong basic character with high solubility in water.  Even barium hydroxide, whose solubility in water is limited to less than one molar unless the water is heated, does not get there on solubility.
That being said, there are applications where magnesia (and its congeners with heavier alkaline earth metals, such as lime) functions as an alkali would.  Given the solubility product of magnesium hydroxide being roughly $10^{-11}$ (different sources quote slightly different values), magnesium remains freely soluble up to a pH of about 9 (with lime the corresponding limit for calcium is around pH 11 or 12), which is high enough to displace most transition metal hydroxides from solution.  From Wikipedia:

Magnesium oxide is used extensively in the soil and groundwater remediation, wastewater treatment, drinking water treatment, air emissions treatment, and waste treatment industries for its acid buffering capacity and related effectiveness in stabilizing dissolved heavy metal species.
Many heavy metals species, such as lead and cadmium are most soluble in water at acidic pH (below 6) as well as high pH (above 11). ...
Granular MgO is often blended into metals-contaminated soil or waste material, which is also commonly of a low (acidic) pH, in order to drive the pH into the 8–10 range where most metals are at their lowest solubilities. Metal-hydroxide complexes have a tendency to precipitate out of aqueous solution in the pH range of 8–10.

In effect, magnesia and other alkaline earths act as alkalies in some reactions (displacement of other metal hydroxides) but not in others (failing to form concentrated strongly basic solutions).  They represent an intermediate stage between full-fledged alkalis and the majority of metal oxides and hydroxides which are inert in the absence of added acid or strong base.

So is the premise I took, which Mg(OH)2 is alkali, also right or just intermediate?

Yup, intermediate.  Again, if forced to say yes or no, ignoring the subtleties involved here, chemists would say yes only for alkali metals.

Only for alkali metals? Even Ca(OH)2, Ba(OH)2...?

As I say, even the most soluble alkaline earth hydroxides don't come close in solubility to their Group 1 counterparts.  You're not going to find 1M barium hydroxide solutions sitting on the lsboratory shelf, whereas sodium hydrixide can easily exceed that.

The word alkali is reserved to hydroxides of elements of the first column of the periodic table (Li, Na, K, Rb, Cs, Fr). So oxides are not alkalies. Furthermore Magnesium is an element of the 2nd column. It is an alkaline earth metal. It does not produce an alkali when oxidized to MgO.
Nevertheless, the "Concise Encyclopedia Chemistry" states that "in a broad sense, the alkalies also include the alkaline earth hydroxides and aqueous ammonia, because of their weaker basicity". So Mg(OH)2 may be taken as an alkali, but not MgO.
Concise Encyclopedia Chemistry, translated by Mary Eagleson, ISBN 0-899-457-8, Walter de Gruyter Berlin NRef.: ew York 1994

So, you mean Na2O and K2O are not alkalis but only bases in this regard?

Yes. Oxides are not alkalis.

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How do I import CSVs to OpenChrom?
My end objective is to compare GC outputs of samples in two groups using PCA. I have bene given the data in Excel format but have saved to CSV.
Whilst I can locate CSV files through OpenChrom, they all appear blank when I do so.
Having spend a lof of time trying to find the information, I cannot find the relevant info about how my CSV (or Excel) file should be formatted for OpenChrome.

I have columns for retention time (labelled "RT") and area (labelled "Area").
I have tried a simple file with two columns (RT and Area) for one sample. I have also tried files with multiple samples and columns, both with and without sample no. headers. (see pictures).

Whatever option I try, the charts in OpenChrom appear completely blank. I don't know if it's the data itself, the headings, the process I'm going through, not a clue what's going on to be frank.
Lastly, I would like to apologise that this may be a basic/naiive question for the forum. I've essentially been dumped with the work of running PCAs comparing GC data sets with little to no experience in doing so.
Thank you.

Oh my God! I hope I help you in time. I just spent hours dealing with the same issue, and I finally figured it out. Basically, OpenChrom will only read CSVs formatted in a very specific way, so you will have to change your .csv files to meet these requirements:

The first three columns MUST be have the headers 'RT(milliseconds)', 'RT(minutes) - NOT USED BY IMPORT', and 'RI', exactly as I have written them.
The second column ('RT(minutes) - NOT USED BY IMPORT') does not need any values; as its name suggests, it is not used by the import, but it still needs to be there! You will have to convert your times to milliseconds to populate the first column. The third column ('RI') is for retention indices, if you do not have them you must populate the entire column with zeros (0.0).
Every subsequent column must have an integer header representing the mass of the signal (e.g. '176' for m/z 176).
No trailing comma at the end of the line! E.g. a line of values should look like

1,2,3,4
5,6,7,8

and NOT

1,2,3,4,
5,6,7,8,

In Excel, the only way for my files to save like this is if I save them as CSV (MS-DOS).
I hope this helps!

Thanks, I'll look into this tomorrow. You may have saved me!
So is "retention indicies" my "Area" column?

I do confess in your original post I'm not sure what "Area" means. I am far from being a GC-MS expert (only learned the analysis to help with a class I am TAing) but all of our files are something like retention time vs. net intensity. If your 'Area' represents net intensity measured by the MS, which I think it does, then just put it in the fourth column with some arbitrary integer header.

Okay, I understand now. OpenChrom allows for import of data in either CSD (total area like your data), or MSD perspectives (mass-speciated signals). I have no idea what the .csv requirements for CSD data should be (and frankly don't want to figure it out), but you should still be able to carry out your analysis by formatting your data as I've described above (pseudo-MSD) and opening in the program.

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Rationalising an apparent inconsistency in the determination of reaction spontaneity
It is well-known that a reaction is spontaneous if $\Delta G < 0$. This requirement is derived from thermodynamic considerations, as done here. However, there seems to be two different ways of calculating $\Delta G$. As provided by Orthocresol in the answer here, $\Delta G$ is the rate of change of the Gibbs free energy of the reaction mixture with respect to the extent of reaction. This definition based on the rate of change cannot be disputed as a negative gradient of the graph would mean that as the forward reaction proceeds, Gibbs free energy is being minimised.
However, there appears to be another way in which $\Delta G$ is being calculated. In this approach, we take $\Delta G$ as the absolute change in Gibbs free energy going from reactants to products. One way of calculating this would be using $\Delta G = \Delta G_f [Products] - \Delta G_f [Reactants]$. Another would be $\Delta_rG = \Delta_rH - T \Delta_rS$. In this case, we are not calculating the gradient of the $G$ against extent of reaction curve but instead, we are simply calculating the absolute change in $G$ from all reactants to all products.
Clearly, the two approaches, one based on the gradient of the curve and one based on the absolute change of $G$ between reactants and products, are not the same. However, both approaches can be used to calculate $\Delta G$ in the determination of reaction spontaneity. The second approach clearly being more dubious than the first.
My attempt is rationalise this is as follows:
Both approaches are based on the idea of minimisation of the Gibbs free energy. However, they differ in their analysis of the reaction. The first approach of calculating $\Delta G$ considers an ensemble of molecules going from reactants to products. It looks at small adjustments to the ratio of reactants to products in the reaction mixture as reaction progresses and subsequent changes to entropy of the mixture (due to mixing of reactants and products)
However, in the second approach, we are trying to look at the energetics in the one-step change from a reactant molecule to a product molecule and we are merely concerned with the question: does $G$ decrease when a reactant molecule changes into a product molecule? Clearly, this process does not occur in small steps but occurs at once, unlike the conversion of a mole of reactants to a mole of products. We make use of the $\Delta G_f$ values for the reactants and products to make an approximation for the conversion of one reactant to one product. This 2nd approach is certainly not exact but it is certainly still chemically correct.

I recommend the following article by Juan Quílez if you have access: DOI: 10.1021/ed100477x. The main point is summarized in their figure:

The lower curve shows your first approach, with $\Delta_r G$ as the slope of the curve. The upper curve (i.e. straight line) shows your second approach, where the slope is non-zero and constant. What the second approach leaves out (and it might be appropriate for certain cases, but usually not) is the entropy of mixing (the $R T \ln(Q)$ term in calculating Gibbs energy for given concentrations).
For some more interpretation of the graph and the different definitions, see Orthocresol's excellent answer here: https://chemistry.stackexchange.com/a/41864/72973

Can I say that second approach is thus an approximation while the first is exact?

@TanYongBoon It gives you an exact value of $\Delta G^\circ$. If its value is very high or very low, concentrations don't matter over a large range, i.e. that term is large compared to the $R T \ln(Q)$ term. It is also exactly equal to $\Delta_r G$ for Q = 1.

What I am asking is if the $\ce {\Delta G}$ calculated using the second approach can be used to determine the spontaneity of a reaction... Or is it that this value can only serves as an approximate to determine reaction spontaneity?

@TanYongBoon Yes, you can use the second approach if conditions are such that Q = 1. However, in the absence of products the reaction will always go forward and in the absence of products always in reverse, no matter what the standard free energy of reaction is.


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Why does Chem3D minimize nitrogen dioxide (radical) so badly?
I build from text O-N+O (the SMILES of this is O=[N+][O-]), I delete the hydrogen atom added automatically on N, then I start MM2 Minimization. Both the bond length and the angle differ from what I see in the top right image here: https://en.wikipedia.org/wiki/Nitrogen_dioxide

If you are referring to PerkinElmer's product, I'm afraid we cannot help you since it's a proprietary software and the implementation for the structure optimization algorithms cannot be validated. And even if you contact them, I'd recommend to provide more details as currently it is not clear what exactly the deviation is (at least a screenshot could help). Other than that, this question seems to be off-topic on Chemistry.SE.

@andselisk why exactly is this off-topic?

There are too many reasons, why this happens; your question is not specific enough. The general area is of course on-topic for chemistry.se. However, without knowing the set-up of your calculation, it is impossible to really help you. E. g.: Did you adjust spin and charge to reflect the actual molecule? What is the deviation? Chem3d uses - like stated in the question - the MM2 force field; I think it was designed for neutral hydrocarbons predominantly, no surprise it fails for a very complicated molecule like $\ce{NO2}$.

By the looks of it, Chem3D uses generalised classical force fields to describe the molecular interactions.
Such force fields are optimised to provide a reasonable fit to multiple properties for lots of different molecules. For example, they reproduce reasonable bond lengths not only for $\ce{NO2}$ but also for the various bonds in organic molecules.
They are also constructed to reasonably reproduce solvation properties which depend on the molecule-specific partial charges, which in turn alter bond lengths and angles.
So it is unlikely that you have done anything wrong.
If you require a more accurate model, either apply more accurate force fields (e.g. those fitted specifically to $\ce{}$) or switch to quantum chemistry methods such as DFT.

Thanks for taking the time to explain this!

Yes, indeed, the option explicitly states MM2. This is designed for hydrocarbons and small organic molecules. NO2 is a very complex molecule, it is no surprise that a force field doesn't describe it well. So it is unlikely that you have done anything wrong. There are still plenty of settings that could be wrong though. And I believe this is the case. It makes a huge difference if the program choses a neutral molecule with a doublet, or a singlet cation, or anion. The OP is not specific about that. While I generally agree with your post, this sentence caused me to down-vote.


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Anionic Ring Opening Polymerisation

I figure that t-BuLi has effectively a negative charge on the carbon that is attached to 3 methyl groups. So I'm wondering what is the accurate way to represent this reaction, does the carbon anion attack a silicon atom in the chain like so-

Pretty much this. tBu- is a poor nucleophile but when there is no other pathway...


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Why is sulfuric acid used to seal flask with highly reactive chemicals?
I the video https://youtu.be/saANxD0cqy0?t=200 (time 3:20) NileRed states that he uses sulfuric acid to seal off flask with a mixture of pottasium dichromate and table salt. He says that sulfuric acid is used to seal off flasks with highly reactive chemicals and I could not find any reasons why.
I see sulfuric acid as a very reactive chemical, that could potentionally react with substrate causing some minor problems to purity of the final product. I also wonder why he does not use any other less-reactive seal that would not react with substarte what so ever.
Thank you for your contribution

It´s just a relatively highly viscous liquid with a low vapour pressure, that does not (as opposed to, say, vaseline, which you normally use to lube these joints) react with the content of the flask. Of course it is highly hygroscopic,  I´m not sure if it is advisable to use it.

Can I make a recommendation? Do not watch this NileRed guy to learn chemistry. That grade school dropout seems to have acquired a bit of a god complex because his boss let him work with all this dangerous stuff (before firing him). I´m not saying he doesn´t know what he´s talking, but a guy working with such stuff alone (both physically and regarding feedback from colleagues and supervisors) is a big red flag.

I agree with Karl. I cannot imagine how the guy handles this material and these Chromium vapors which are corrosive and carcinogenic. He must use a rather specially equipped lab to get rid of all that Chromium(VI) which is forbidden in usual labs. Is he still alive ?

I know I should have ..., but ... oh well... This guy is a health hazard to the world.

Sulphuric acid reacts with dichromate and table salt, forming volatile and harmful chromyl chloride.
\begin{multline}
\ce{K2Cr2O7(s) + 4 NaCl(s) + 6 H2SO4 ->}\\
\ce{2 KHSO4(s) + 4 NaHSO4(s) + 2 CrO2Cl2(l,g) + 3 H2O}
\end{multline}
Therefore it serves as a reagent and at the same time as protection against chromyl chloride escape.

Safety considerations
$\ce{CrO2Cl2}$ hydrolyzes to release $\ce{HCl}$ and hexavalent chromium $\ce{Cr^{\mathrm{VI}}}$.
Acute: Exposure to chromyl chloride vapour irritates the respiratory system and severely irritates the eyes, and the liquid burns the skin and eyes. Ingestion would cause severe internal damage.
Chronic: $\ce{CrO2Cl2}$ can produce chromosome abnormality/aberrations and is a human carcinogen via inhalation. Frequent exposure of the skin to chromyl chloride may result in skin ulcer/ulceration.


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How to decrease the percent of ammonia that is converted to the ammonium ion in water?
Ammonia is a weak base that reacts with water according to this equation:
NH3(aq)  +   H2O(l)     ⇌     NH4+(aq)  +    OH −(aq)

Will Fe(NO3)3 decrease the percent of ammonia that is converted to the ammonium ion in water?

@Zenix ecause I'm only considering Fe(NO3)3 or NaOH

I expect $\ce{Fe^3+ + 3 NH3 + 2 H2O -> FeO(OH) + 3 NH4+}$

So it actually increases the percent of ammonia @Poutnik? That doesnt happen with NaOH

Well, if the substance on the left is called ammonia and on the right ammonium, and if the substance on the left is converted to the substance on the right, then the conclusion is ?

(-1) Because wrong answer was accepted.

@EdV I'm not chemistry expert, waiting for a reliable source. That answer was actually getting lots of upvotes. Already fixed it. Feel free to answer of you know.

I reversed my downvote in response. And @MaxW nailed it in his comments: adding hydroxide ions forces the equilibrium to the left.

If ammonia gas is bubbled into pure water, the following reaction takes place:
$$\ce{NH3(aq)  +   H2O(l)   <=>     NH4+(aq)  +    OH−(aq)}\tag{1}$$
How to decrease the percent of ammonia that is converted to the ammonium ion in water?
Le Chatelier's Principle indicates that adding $\ce{OH-}$ would push the equilibrium to the left.
I lifted the following figure from:

HYDROPHOBIC MEMBRANE TECHNOLOGY FOR AMMONIA EXTRACTION FROM WASTEWATERS by Airton Kunz and Saqib Mukhtar Eng. Agríc. vol.36 no.2 Jaboticabal Mar./Apr. 2016
https://doi.org/10.1590/1809-4430-Eng.Agric.v36n2p377-386/2016

As can be seen in the figure, an increase in pH favors formation of $\ce{NH3}$ and a decrease in the pH favors $\ce{NH4+}$.

Since the OP asked only for the percent of $\ce{NH4+}$ to be reduced, there is another way. Dilute the solution.

Will $\ce{Fe(NO3)3}$ decrease the percent of ammonia that is converted to the ammonium ion in water?
Iron (III) forms an relatively insoluble hydroxide as shown by the following equation:
$$\ce{Fe^{3+} + 3OH- <=>[K_{\mathrm{sp}} = 2.79\times10^{−39}] Fe(OH)3(s)}\tag{2}$$
So again according to Le Chatelier's Principle removing $\ce{OH-}$ from reaction (1) favors the reaction moving to the right. Thus the precipitation of $\ce{Fe(OH)3}$ will increase the formation of $\ce{NH4+}$.

In a comment on the question user Poutnik indicated the reaction $\ce{Fe^{3+} + 3NH3 + 2H2O -> FeO(OH) + 3NH4+}$.  I'm sure there is a study somewhere on $\ce{Fe(OH)3 \pu{ vs } FeO(OH)}$ formation, but I didn't find it in 15 minutes of fooling around. I'd expect that strong $\ce{OH-}$ concentrations and higher temperatures (boiling solution) would favor $\ce{FeO(OH)}$.

Short answer lower pH! [EDIT Correction] raise pH.
Chemistry:
$\ce{ NH4+ <=> H+ + NH3 (g) }$
In neutral to basic conditions, this reaction undesirably [EDIT Correction] desirably moves to the right liberating ammonia, especially on warming.
Good news, in acidic conditions the NH4+ ion is hard to chemically attack. To quote a supporting source:

Acidic salts, such as AMMONIUM CHLORIDE, are generally soluble in water. The resulting solutions contain moderate concentrations of hydrogen ions and have pH's of less than 7.0. They react as acids to neutralize bases... They usually do not react as either oxidizing agents or reducing agents but such behavior is not impossible.

So, paths to lower pH, like adding CO2 or salts like Ferric nitrate that undergo a complex formation with water liberating H+. Also, Aluminum sulfate is commonly used to adjust pH and results in a helpful precipitation of Al(OH)3 to quote a source:

Freshly precipitated aluminium hydroxide forms gels, which are the basis for the application of aluminium salts as flocculants in water purification.

Zenix: Seemingly, as Aluminum sulfate in dilute solutions breakdowns liberating Al(OH)3 and acidic H2SO4 which has also been used to treat ground soil that is too alkaline. The 1st comment relates to the stability of the NH4+ anion (very hard generally to attack, but I have accomplished it in practice with HOCl).

WRONG!!! You have it backwards. Lowering the pH favors production of $\ce{NH4+}$. Raising the pH favors production of $\ce{NH3}$. $$\ce{NH3 +H2O <=>[lower pH][raise pH] NH4+ + OH-}$$

I agree with@MaxW

Agree, but my use of the word 'So',  may suggest otherwise, so slight rewording. Deleted 'So' and added 'undesirably'.

The OP asked for less $\ce{NH4+}$ and more $\ce{NH3}$. Lowering the pH means that you add an acid to the solution which favors $\ce{NH4+}$  production. // The OP should raise the pH. $$\ce{NH4+ <=>[raise pH (add base)][lower pH (add acid)] NH3 + H+}$$ look at reaction this way. $$\ce{NH4+ + OH- <=> NH3 + H2O}$$

Here is a source: 'Ionization of Ammonia in Seawater: Effects of Temperature, pH, and Salinity' which may be of value. To quote from the abstract:

Previous studies have shown that the toxicity of ammonia to freshwater fishes depends on the ambient concentration of the un-ionized fraction (NH3) and have quantified the percent NH3 in solution as a function of temperature and pH. If NH3 is also toxic to marine fishes, the effect of salinity on the ionization of ammonia must also be considered. The percent NH3 over ranges of temperature, pH, and salinity common in seawater-culture situations is presented here.

Further:

In the water, ammonia can be found in its ionized (NH4+) or unionized form (NH3); the sum of the two is the total ammonia. Their proportion in the water depends on pH and temperature (Bower and Bidwell, 1978). Many studies confirm that non-ionized ammonia is chemically more toxic due to its ability to diffuse through cell membranes,...Studies show that at lower salinity levels, higher levels of ammonia nitrogen are measured...The reason for these differences should be at least partially, the solubility of ammonia, lower in water containing salt (Bower and Bidwell, 1978). Consistently, sea water (32-40‰) displays~20% less NH3 than in fresh water with pH 7.5-8.5 and at 25°C (Bower and Bidwell, 1978).

I do not have access, but you may wish to gain access to examine the charts.
So, both increasing temperature and pH, increases free NH3 and seawater decreases free ammonia by roughly 20%. This agrees with a temperature vs. pH table on Page 3 of this available EPA report. To quote a comment relating to salt:

In saline or very hard waters there will be small but measurable decreases in the percent NH3.

Note: NH3 (aq) will react with Fe(NO3)3 consuming ammonia and producing NH4NO3 (aq).


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Temperature dependence of emission lines from gases, for example, neon
Emission lines from Ne are frequently used as wavelength standards in spectroscopy. Reading around Wikipedia, I found about the Paschen law regarding the voltage required to start a discharge.
My question, do emission line intensities have dependence on the voltage applied and the temperature of the electrodes. Just to clarify, I am not thinking about the band positions but the relative intensities.

About voltage dependence, at present I have no idea.
About temperature : Since the electronic states are involved, and this is an electric discharge, the temperature maybe quite high so as to affect the Boltzmann population significantly.

This question is related to plasma physics, but also to electronic spectroscopy.

The intensity of a transition from any given atom is determined by the properties of the atom's electronic states. If you have more atoms to excite and/or excite more of them,  the intensity will obviously be greater, but only up to a point. Emission from one  excited atom can be absorbed in an unexcited one as the transition frequency is the same. (You can see this with Hg vapour.) However, there is also Doppler and Collisional broadening which will counteract self absorption to a certain extend.

Your question is too broad. It is not a trivial simple relationship. The emission characteristics even depend on the geometry of electrodes. I think the most well studied systems are hollow cathode lamps, where the effect of high voltage and consequently high current have been studied.


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Does 1-phenyl-1-keto-but-2-ene show positive iodoform test?

It is told often that only compounds having a methyl ketone and corresponding alcohols show positive iodoform test.

I tried to sketch a mechanism to show that 1-phenyl-1-keto-but-2-ene also gives iodoform test. However I haven't found any source that confirms that 1-phenyl-1-keto-but-2-ene shows iodoform test.
Please check if there is any mistake in the mechanism:

This mechanism is very similar to the classical iodoform reaction mechanism.

C-C double bond group not have larger -Ieffect as compared to C=O group and abstraction of H+ from C H bond is not so easy by OH-. This fact can be used here.

However the carbanion at gama position is stabilized by resonance.

Why space between keto and but?

Enones react with with iodine and catalytic base under aqueous conditions to give $\alpha$-iodoenones (Ref.1). Pyridines are most often used for this, the reference quoted also uses $\ce{K2CO3/DMAP}$. They further show that reaction works for a range of enones including those with the $\ce{RCOCH=CH-CH3}$ grouping and no mention is made of $\gamma$-iodination. A "background" reaction was run with $\ce{K2CO3/I2}$ in aq. THF and no iodination was observed.
Proposed mechanism from the referenced paper:

References:

Marie E. Krafft, John W. Cran, "A Convenient Protocol for the α-Iodination of α,β-Unsaturated Carbonyl Compounds with $\ce{I2}$ in an Aqueous Medium," Synlett 2005, (8), 1263-1266 (DOI: 10.1055/s-2005-868495).

@waylander the document in your link requires paid subscription. Could you please provide the link of some other source which shows the reaction?

I got to it directly through Google Scholar by searching "alpha iodination of enones" The paper is by Krafft and Cran and is the top hit on the Google Scholar

Thanks, I saw the reaction. But is there any reaction reported using OH- and I2

The closest is the K2CO3 in aq THF without added pyridine which gives no reaction, I think that is a significant result with respect to your question. I would expect I2/OH- to similarly give no reaction


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Expectation value of an observable in quantum system
In quantum systems, sometimes if the wave function is not an eigen value of an operator then it is expanded in terms of bases states, then we apply the operator over the bases states and we get different eigen values for an observable with different probabilities corresponding to the projection coefficients of the bases states.

In a book, I read that after making a measurement we get an observation, and the probabilities of other observations shrink down to zero. The observation remains the same for howsoever times we make it after the measurement. Like if a coin is tossed, if we get head after first measurement it remains head, even if we make any number of observations.
  My question is that what is the need of expectation value of an observable (given by formula as summation of the product of probability associated with an observable and the eigen value of observable) as after measurement we get only one eigen value of the observable with 100% certainty. I think knowing only the possible values of the observable is sufficient. What is the significance of expectation value and then does the formula for expectation value given in books is for before the measurement?

You perform measurements because you either (1) need to calibrate your measuring device or (2) do not know (and want to know) the value of a property for the system. There is no point otherwise in performing a measurement. Once you perform a measurement (that results say in an ensemble of expectation values for different possible values of the observable), you might want to compare the results to theoretical predictions. Once you have fully characterized the system, you might use it as a reference (as in case #1 above) to compare to other systems in future experiments.

To whoever gave the -1. It's a brand new user. Rather than giving -1, why not edit the question?

Roughly, the significance is that before measurement the system can be in a superposition of states, the wavefunction of which can be represented by a linear combination of the individual eigenfunctions
$\Psi = \sum_i c_i{\psi_i}$
However the very act of observing the system reduces the wavefunction to a single eigenstate due to interaction with the external world
For most cases, the expectation value is just the quantum mechanical way of calculating properties. Remember, the wavefunction is not observable! Only properties like density, dipole moment, etc are and the expectation value allows us a way to go from wavefunctions to properties via operators.
These other answers may also be helpful
https://physics.stackexchange.com/questions/67756/quantum-collapse
https://philosophy.stackexchange.com/questions/48425/is-collapse-of-the-quantum-state-conceptually-misnamed
https://physics.stackexchange.com/questions/151457/how-does-a-wavefunction-collapse


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Electrolysis of concentrated sodium chloride solution using A.C. power supply
I heard that A.C current is never used for electrolysis as the polarity of the electrodes keep changing and this can result in explosive reactions. However, why can't we use this to our advantage?
When electrolysing brine, chloride ions are discharged at the anode to form chlorine gas and hydrogen ions are reduced at the cathode to form hydrogen gas. When using A.C., the polarity of each electrode would change several times every second (depending on the frequency of current), causing both hydrogen and chlorine to be produced at both electrodes. So in theory, a gas syringe could be set up at each electrode to collect the hydrogen and chlorine, which would then react to form HCl gas which can be collected.
What is wrong with this process? We are able to collect both HCl and NaOH, two very useful compounds. Why is this process not used industrially?

If H2 is deposited in the first alternance, the same H2 will be consumed in the next alternation, and redissolved. Sam thing on the second electrode, with Cl2 that get deposited, then redissolved in the next alternation.  Final result : no chemistry at all. Just a Joule effect, to increase the temperature.

Oh that's interesting, I didn't think of that. So all the electrical energy from the A.C. power supply is converted to thermal energy?

Think about the frequency of AC current. It changes its direction so many times in a $second$. Can anyone manually move the syringe that fast?

Technically, there would be no need to move the syringe at all. There will be a syringe at each electrode and both syringes will collect HCl gas


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Would there be need for periodic DFT if we are able to get all the forcefields of all the elements in the periodic table in ReaxFF?
Since ReaxFF combines the advantages of DFT and Classical MD, would there be any need for periodic DFT calculations if we can have all the forcefields of all the elements of all the possible systems?

Some references for ReaxFF would be useful

I would argue there could also be a use for DFT no matter if we have good force fields or otherwise machine learned values. For one, many of these ad hoc approaches require extensive re-parameterization based on the very same values they are replacing. For two, DFT is a formally exact theory, while force-fields are not. Therefore there is always value in re-affirming the model of your model by checking the exact theory.

Quoting from the article The ReaxFF reactive force-field: development, applications and future directions
"Such empirical methods, including reactive force-field (ReaxFF),1 trade accuracy for lower computational expense, making it possible to reach simulation scales that are orders of magnitude beyond what is tractable for QM."
So yes, if you want accuracy, especially for something which the approximations made in generating a ReaxFF force field are not good, you are going to have to get out the Quantum Mechanics.


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Precise definition of cylindrically symmetry in MO theory?
I was watching an MIT OpenCourseWare lecture on MO theory which mentioned that $\sigma$ orbitals are "cylindrically symmetrical." I was a bit confused, because I was under the assumption that the cross section of a $\sigma$ orbital is an ellipse, such that a $\sigma$ orbital is an ellipse that has undergone a full revolution around an axis. This is somewhat of a topology/nomenclature question, but seeing as the context is Chemistry, I though I'd post here first. Could anyone clear up the definition of cylindrical symmetry?

You're right. A $\sigma$ bond would be more like an ellipsoid than a spherocylinder (capsule) and even less like a cylinder. All three of these would have a rotational axis.

They just mean that, like a cylinder, there is an axis of infinite rotational symmetry $C_\infty$


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Molecule or molecular ion that spontaneously breaks up in two different ways
I teach physics, and I'm looking for simple, easy examples from chemistry to use in order to illustrate basic ideas about quantum mechanics for my students. A conceptually simple example is the $\text{He}^-$ ion, which has a metastable state with a mean lifetime of 0.36 ms. This is for the totally isolated ion, i.e., we're not talking about a gas or plasma in which there can be collisions or reactions. (How would chemists refer to this? "In vacuo?") This is easy to relate to in chemical terms (a noble gas needs an electron like a fish needs a bicycle) and provides a simple example of quantum mechanical concepts like exponential decay.
What would be more fun for my purposes would be an example of a metastable molecule or molecular ion that would spontaneously break up in two different ways. I would imagine that for the ions, one could measure the half-life and branching ratios by storing them in a circular accelerator. So we could have something like the decay channel
$$\text{XYZ}^+ \rightarrow \text{XY}^+ + \text{Z}$$
competing with, say,
$$\text{XYZ}^+ \rightarrow \text{X}^+ + \text{YZ}.$$
We can then discuss things like the probabilities (branching ratios) and the fact that if the ion starts out in a well-defined state, presumably its ground state, then we can only predict these probabilities, not the actual outcome.
Part of what's making it hard for me to find examples is that I don't know the terminology to google on. Is this "autodetachment," or is that only used for the loss of an electron? "Autodissociation?"
Note: I really am looking for the process described above, not something else like breakup induced by a collision, nuclear decay, electromagnetic decay of an electronic excitation, electron emission, or chemical reactions that require two reactants to come in contact. Those other examples might be fine for some educational purposes, but they're not what this question is about.

The fragmentation patterns seen in mass spectrometry might be partly what you are looking for: https://www.chemguide.co.uk/analysis/masspec/fragment.html. Not sure how much the rates of decay have been studied, but the patterns of relative abundance are the core of the technique.

Phosphorescence would obey an exponential decay curve.

Any number of radioactive isotopes decay in multiple ways.

x-ray excitation of atoms can decay by fluorescence or Auger electron emission.

many chemical reactions can branch to form multiple products. Copper and nitric acid is one such example.

Almost any aromatic molecule, benzene, naphthalene, amongst many many, others,  show fluorescence from the excited singlet state, this state also decays (intersystem crossing)  to a triplet state at lower energy and then this triplet decays by  phosphorescence, which, like intersystem crossing, is a forbidden transition partially allowed by spin orbit coupling.

This happens all of the time in mass spectrometry following electron-impact ionization. Following ionization, each molecular ion decomposes (or fragments, to use mass spectrometry lingo) into other radicals and ions, usually following multiple pathways. As the pressure inside a mass analyzer is usually near vacuum, each fragmentation event is independent. No collisions or reactions between particles occur. The mass analyzer collects all of the fragments and generates signal that computer software converts to a mass spectrum - a frequency bar graph of the masses of species generated. For some very simple molecules, only one fragmentation pathway may predominate, but for most organic compounds, there are multiple fragmentation pathways.
For example, let's consider the fragmentation of but-3-en-2-one radical cation with the formula of $\ce{C4H6O+}$ and the structure shown below. The mass/charge ratio (m/z) is 70 for this compounds

There are at least two different fragmentation pathways.
The first fragmentation pathway is a loss of a vinyl group ($-\ce{C2H3}$, -27 m/z):
$$\ce{C4H6O+ -> C2H3O+ + C2H3 }$$

The second pathway is a loss of a methyl group ($-\ce{CH3}$, -15 m/z):
$$\ce{C4H6O+ -> C3H3O+ + CH3}$$

Each of these fragmentation pathways is independent. Here is a copy of the mass spectrum from the NIST Chemical Webbook, which is a frequency bar graph of all possible fragments. You can see in addition to the two pathways mentioned above, there are many other possible fragments.

We can estimate the relative rates of these processes from the frequencies of the peaks in the mass spectrum. The first pathway occurs with a relative frequency of 83.19 (based on the height of the m/z = 43 peak) , and the second pathway occurs with a relative frequency of 1.000 (from the height of the m/z= 55 peak). We can use these as proxy for the relative rates of decomposition. We can also see that the parent ion survives this process with a relative frequency of 0.57779 (from the m/z = 70 peak), which indicates that the parent ion has at least some stability.

I was hoping to see a mass spec answer. It seems to fit with what the original post is looking for, with multiple fragmentation pathways and fairly well characterized ratios between those paths based on stability of given types of fragments.

This is nice, thanks. It would be cool to have an example in which there was a known state in the parent ion, and a measurable lifetime.

There are many compounds that will break down into different products when involved in collisions with inert molecules. For example, $\ce{F3-}$ can separate into $\ce{F2 + F-}$ or into $\ce{F. + F2-}$. But atoms or molecules that will breakdown without being involved in a collision are rarer (simply because their instability means they are harder to observe).
One example, though, is the radioactive decay of the bismuth-212 isotope. It decays ~64% of the time by $\beta-$ decay, producing polonium-212: $\ce{^{212}_{83}Bi ->  ^{212}_{84}Po + e- + \nu_e}$. The other ~36% of the time it decays by $\alpha$ decay to thallium-208: $\ce{^{212}_{83}Bi -> ^{208}_{81}Tl + ^4_2He}$.
More info can be found here: http://nucleardata.nuclear.lu.se/toi/nuclide.asp?iZA=830212

Thanks, but this doesn't answer the question. The question is about spontaneous breakup, and it's a molecular physics question, not a nuclear physics question.

The key operation is to make a metastable particle that has at least two decomposition pathways and a short lifetime. Fragmentation is probably the key word you were looking for. Pyrolytic fragmentation of complicated molecules (1-methoxycyclopropylamines) is one way https://www.sciencedirect.com/science/article/pii/S0040403900741402. Photolytic fragmentation of acetaldehyde, a relatively simple molecule, has been investigated https://aip.scitation.org/doi/10.1063/1.4878668 , and many others have been investigated with lasers. Another well known area is mass spectrometry fragmentation https://en.wikipedia.org/wiki/Fragmentation_(mass_spectrometry).  For example, toluene is bombarded with electrons, loses one to form a

cation, then decomposes to form a benzyl cation or a phenyl cation.
An advantage of using mass spectrometric data is that there is a lot out there, and the idea can also be hypothetical, in that a molecule can be selected on the basis of its simplicity and likely decomposition products to allow theoretical calculations, e.g., for educational purposes, even in the face of insufficient actual data.

This looks interesting, thanks. Is the parent ion in a defined state before it breaks up? Is there a measurable lifetime and measurable branching ratios?

Original state for volatile compounds is just the ground state; branching ratios will be measured as output; lifetimes will be more difficult, not really sure how that's done now. I did mass spec work as a grad student, many moons ago.

This is not really a complete self-answer, but I thought it was close enough that it was worth posting. I would be happy if others could come up with better examples.
The ions $\text{N}_2^{2+}$ and  $\text{CO}^{2+}$ have been studied experimentally and theoretically in Pandey, Bapat, and Shamasundar, J. Chem. Phys. 140, 034319 (2014). Pandey's thesis also has a helpful table on p. 122. Their experimental setup is only sensitive to a breakup in which both fragments are charged. They do a total kinematic reconstruction of the two fragments and infer the kinetic energy release (KER). Their resolution is not quite good enough to resolve contributions from individual vibrational states in their KER spectrum, but you can see structures in them which they explain as arising from those vibrational states. They didn't store the fragments, just let them fly through a spectrometer, so they could only see decays that happened within a few milliseconds. With these two dications, you don't get tunneling on these time-scales when the initial state is the vibrational and electronic ground state. They do observe it from the vibrational ground state coupled to an electronic excitation and vice versa. They don't actually seem to have measured any lifetimes, and they don't give references to anyone else who has. There are no branching ratios to measure, because their molecules were binary, and their detector needed to see two charged fragments.
Eland et al., https://arxiv.org/abs/1908.11441.pdf  (2019), have studied the breakup of $\text{HNCO}^{2+}$ and $\text{HNCO}^{3+}$. There are cases where different breakup channels compete, and there seem to be some lifetimes on the order of microseconds. But the experimental setup basically seems pretty messy, and it doesn't seem like they got a lot of detailed data.
Relevant keywords seem to be "dissociative ionization" and "vibrational autoionization."
These are both pretty recent, and I was kind of surprised that the state of the art was so crude. To me as a physicist, these seem like obvious things for a molecular physicist or quantum chemist to want to study. No doubt my ignorance of the experimental techniques is my reason for not understanding why this kind of experiment is so hard. I wonder if there are any systems that happen to be particularly easy to study and that have been better characterized than these.

It's not clear to me how one knows that the cations are all in the ground state before dissociation. Couldn't the breakup channel be dependent on the initial state produced by the ionization collision (eg angle, energy, etc)? I think that's the challenge of finding a system of the sort you are looking for. One needs to have molecules that are unstable enough that they fall apart, but stable enough that they have time to relax to a common ground state as the starting state before dissociation.

@Andrew: In the Pandey experiment, none are in the ground state. See Pandey's table on p. 122, and the text of my answer starting at "With these two dications..."


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DIY Hand Sanitizer Clumping
Hand sanitizer is impossible to find around me and I have 4 family members who are essential and not around a sink or soap during their work day (delivery drivers), and they are out.  I make a lot of soap, shampoo, lotions, things like that and I thought hand sanitizer would be easy - but I was WRONG.  I have 99% anhydrous isopropyl alcohol and aloe vera gel.  Tonight I weighed everything to make it 70% IPA and 30% aloe vera gel.  It formed these solid squishy jelly clumps and didn't form a gel at all.  What am I doing wrong?  I added the IPA in kind of slow, but was stirring the whole time. I saw all these YT videos of people doing this and their gel/IPA combination makes a perfect gel.  HOW?!  I'm thinking about adding some HEC to help thicken it but am not sure to how it will react.  Ideas?

I think the better method that's not clumpy requires glycerin

Sorry, but given the current pandemic I voted to close all questions about sanitizers as personal medical questions.

If your delivery drivers don´t finger their nose or eyes, the already very low chance of getting corona via smear infection becomse zero. And believe me, if you work outdoors, in a dusty delivery van carrying packages, you have internalised to not rub your eyes before you have washed your hands. DONT PANIC.

Err, no, handwashing or handrubbing is important, especially if you have regular contact with non porous, high-touch surfaces such as door knockers or buzzers. The risk may be low, and there are other relevant mitigation measures, but having handrub to hand is one of the most effective in this scenario, so seeking to make some is a very sensible precaution and not panic.

HEC works ok, but it's not that easy to use. You have to hydrate in pure water for a few hours or more, ideally with stirring, and then incorporate the alcohol a little at a time with high shear mixing. You can use a blender, though it will incorporate a lot of air. If you use too much HEC, the "mother" gel (the HEC and water that will become 30% of your final product) will be very firm and it will be difficult to avoid lumps throughout, and the line is very thin between this and ending up with an unusually thin product. When I'm using this, I err on the thin side and use about 0.8% w/w HEC, so the "mother" is 2.67%.
Some other polymers are much easier to work with if you can get hold of them, notably Sepimax Zen aka. GelMaker PH (Polyacrylate Crosspolymer-6). These can be added directly to a 70% isopropanol or ethanol mixture and need no further ingredients to activate or correct the pH afterwards.
I have not tried aloe vera but I think it would be difficult to get a gel with 70% alcohol. I know lots of YouTubers and bloggers are recommending it. Success may depend on what other thickeners have been used in the aloe vera product. I've seen some videos where the result was the extremely thin liquid with a bit of stickiness that you would expect from the recipe.


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Generating hybrid atomic orbitals for Boron
My professor was showing us how to form hybrid atomic orbitals for Boron by taking a linear combination of rotated 2p orbitals and 2s orbitals (of Boron). This was done in a particular manner such that the geometry of BH3 is correct. We were given BH3:

He proposed 3 new hybrid atomic orbitals:

2Py(rot,1): a 2Py orbital rotated 30 degrees clockwise

-2Py(rot,2): take the negative of 2Py, which is then rotated 30 degrees counterclockwise

Can someone provide a visual (picture) showing the "mixing" of the atomic orbitals to create the three hybrid orbitals (while maintaining the same orientation of BH3 as in my first image)?
EDIT: Feodoran helped me clear my confusion regarding the first hybrid wavefunction. If possible, I would still be interested if someone could display the formation of the hybrid wavefunctions with some visuals.

How do you define 2Py(rot,1) and 2Py(rot,2) ?

@Maurice 2Py(rot,1) is a 2Py orbital rotated 30 degrees clockwise. -2Py(rot,2) is achieved by taking the NEGATIVE of 2Py which is then rotated 30 degrees counterclockwise. This is done to match the angles in BH3. The B-H bonds in 2nd and 3rd quadrant is 30 degrees from the y-axis. I have the specific equations describing each rotation, but didn't post them.

@Oscar Lanzi Yes...ultimately those are linear combinations of unrotated Px and Py as I have shown in the equations. But my problem is not with chi2 or chi3 but with chi1.

$\chi_1$ is oriented along the $x$ axis, on which one of the $\ce{B-H}$ bonds lies. What do you mean by "opposite direction"? Rotation of the $p_x$ orbital by 180° around the $z$ or $y$ axis results in the same orbital, just with a sign change. But the sign is arbitrary anyway. So there is not "wrong" direction, both are fine.

@Feodoran  I thought that since the right lobe of 2Px is positive, shouldn't the 2s also be positive (same phase) for the atomic orbitals to combine? I'm not understanding how the sign is arbitrary.

The 2s orbital can be either one, positiv or negativ. Of course the relative phase of the atomic orbitals is important (same sign for bonding interaction), but the overall phase (plus or minus) of the hybrid orbital does not matter. For the observable (the probability density) the sign is lost when taking the absolute square of the wave function.

@Feodoran If we let the 2s orbital be positive doesn't it then interact with the positive lobe of 2Px, making a hybrid orbital with a large positive lobe pointing to the right? Thus allowing the positive 1s orbital of Hydrogen be able to interact with it, and make a bond pointing in the positive x-axis (like my first image). If I were to make the 2s orbital negative, we would get a hybrid orbital with a large negative lobe pointing to the left. How would a positive 1s orbital (of a hydrogen atom on the right) interact in the second case (while maintaining the proper geometry)?

You are only changing one orbitals sign. This will result in qualitatively different molecular orbitals that may not contribute to the bonding. Relative phase needs to be the same, so either you change all signs or none! $\chi_1=\phi_1-\phi_2$ is the "same" (except for the overall sign) as $-\chi_1=-\phi_i+\phi_2$, but something very different than $\chi_2=\phi_1+\phi_2$.

@Feodoran Ok I think I finally see what you mean. So the chi1 I posted is the "same" as 2s-2Px. The general direction seems to be the same, as long as we negate the entire hybrid.


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What will be the major product of the following dehydrohalogenation?
Question:-

My Attempt:-
I believe the answer should be C, because even if 2-butyne is formed $\ce{NaNH2}$ should cause it to rearrange into a terminal alkyne. Hence the answer should be 1-butyne. But the book says B as the answer.
Please correct me.

You are correct, if the last reaction was heated in extended time. However, $\ce{NH2-}$ is not the best catalyst for zipper reaction. Besides, judging from the question, I believe zipper reaction is not intended for the students in this class.

The rearrangement happens at elevated conditions (eg. heating). You may read the first line here: Does a non-terminal alkyne react with sodamide?


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What is the difference between a dπ-dπ and delta bond
I have always been wondering the difference between a dπ-dπ bond and a delta bond. What are the situations that either bond is formed in overlap of two d orbitals.

Why, it's really simple: dπ-dπ has one nodal plane along the bond, while $\delta$ has two.


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How does polarity of a solvent increase conductance of electrolyte?
My textbook says that electrolyte ionize more in polar solvents and greater is the polarity, larger is the conductance. But if polar solvents (lets take water) causes the formation of shells around electrolyte (say NaCl) then shouldn't effective size increase and hence conductance should decrease? What factor is more dominant? Size/Solvation or Polarity?

Gist - To have a current in solutions you need ions. Non-polar solvents don't dissolve ionic salts very well.

What about the viscosity?

@MaxW I understand that well but my question was that if polar solvents create shells around the ion shouldn't that inturn hinder the conductance of the electrolyte?

I wondered whether it could be about solubility or dissociation. Could NaCl be only partly dissociated in some solvents? Then again, as a wild guess, it seems like a polar solvent might become oriented to some degree with an electric field, extending the effect of that field over a greater distance, vaguely analogously to temporary magnetism.

The important parameter is the dielectric constant (relative permitivitty) $\epsilon$ of the solvent because this controls the interaction energy $E$ between two ions at a fixed distance $r$ apart, i.e. $\displaystyle E\sim \frac{1}{\epsilon r}$. If the dielectric constant is large, for water it is 78 but for benzene 2.3 you can see that the interaction energy for benzene will be far larger than for water. This means that any pair of ions formed in benzene will be attracted to one another from a far greater distance (with the same energy) than the same ions in water would be. The chance that the ions separate is given relative to the thermal energy of the solvent which is just $RT$, so if the interaction energy is far larger than $RT$ the chance of separating is small and the ion does not dissolve and vice versa. The thermal energy of the solvent tries to jostle the ions apart.
Notes. The interaction energy is  $\displaystyle E(r,\epsilon)= \frac{Z_1Z_2}{4\pi\epsilon_0}\frac{e^2}{\epsilon r}$ where $Z$ are the charges, $e$ the charge on the electron and $\epsilon_0$ the permitivitty of free space.  The thermal energy per molecule at temperature $T$ is $k_BT$ where $k_B$ is the Boltzmann constant and is $RT$ per mole. The probability of being at separation $r$ with dielectric constant $\epsilon$ is $\sim\exp(-E(r,\epsilon)/k_BT)$.

I would be really interested to know if there's a connection between polarity and the dielectric constant, and, without that, this doesn't seem to answer the original question. Are you saying that a polar solvent is not more likely to conduct more?

A polar solvent will conduct more because more ions will dissolve in it compared to a non-polar solvent. Think of the solubility of sodium chloride in water vs that in oil.  The dielectric constant is proportional to the polarisability of  the molecule (this measures how easily a dipole is induced in a molecule by an external electric field) and additionally by the square of the permanent dipole moment in a polar molecule.

My textbook says that electrolyte ionize more in polar solvents and greater is the polarity, larger is the conductance.

Porphyrin has already given a very useful explanation. I feel this the textbook sentence is quite problematic if taken literally. First of all, polarity is a useless term, it is an umbrella word which is discouraged by the International Union of Pure and Applied Chemistry. Dielectric constant is the right parameter in this context.
The first requirement of ionization in a solvent is indeed solubility. An ionic compound which is dissolved in a solvent will indeed generate ions whether it water or any other solvent. Salt won't dissolve in benzene so there is no change in the conductivity of the solvent. Let us take an example of HCl gas which is not ionic in nature-it is dipolar (i.e. it has a dipole moment). It is soluble in water (large dielectric) and benzene (low dielectric). When HCl dissolves in water it gives highly conducting solution because it can ionize there and when it dissolves in benzene it remains as HCl molecules. It cannot protonate benzene.
Now what you are thinking about it polar solvents forming a shell around an ion. Right, that is a different concept. Read about transport numbers of various ions https://en.wikipedia.org/wiki/Ion_transport_number.


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Possible side reactions in electrolysis of concentrated NaCl
This is a very basic question about which side reactions can take place if the electrodes are not shielded from each other during electrolysis of concentrated $\ce{NaCl}$.
So, this is the desired net reaction for this process:
$$2\ce{NaCl(l) + 2H2O -> 2NaOH(l) + H2(g) + Cl2(g)}$$
At the anode, oxidation of chloride ions to chlorine gas:
$$\ce{2Cl- -> Cl2 + 2e-}$$
At the cathode, reduction of water to form hydrogen gas and hydroxide ions:
$$\ce{2H2O + 2e- -> H2 + 2OH-}$$
The section 'The need to keep all the products separate' on this page gives $ 2$ by-products that are formed when $\ce{ H2}$ and $\ce{Cl2}$ stay in the system:
$$\ce{H2 + Cl2 -> 2HCl}$$
$$\ce{Cl2 + 2NaOH ->NaOCl + NaCl + H2}$$
I wonder if $ 2 $ additional side reactions can take place, simply with the gases diffusing to the other electrode, resulting in the reverse redox reactions:
At the cathode, reduction of chlorine gas to chloride ions:
$$\ce{Cl2 + 2e- -> 2Cl-}$$
At the anode, oxidation of hydrogen gas to form hydronium ions:
$$\ce{H2 + 2H2O -> 2H3O+ + 2e-}$$

Aren't these reverse (and spontaneous, based on the standard potential of these reactions) redox reactions also likely to happen when the electrodes aren't shielded?
Which one of all these side reactions (or maybe others?) are more likely to occur in this scenario? I think $\ce{ H2 + Cl2 -> 2HCl}$ as both are in gas form and are likely to encounter each other

Thanks a lot.

The mixture of gas H2 + Cl2 is stable in the dark at room temperature. The reaction starts if an energetic enough photon enters the mixture : blue, violet or UV. So the main side reaction is the reaction between NaOH and Cl2 producing NaClO. Now if the temperature is nearly at boiling point, NaClO disproportionates into sodium chlorate NaClO3 and NaCl. That's all. –

@Maurice: thank you for the insight. Can you clarify why the reverse reactions on the opposite electrodes also won't happen?

Which reverse reactions ?

At the cathode, reduction of chlorine gas to chloride ions: Cl2 + 2e- -> 2Cl-
At the anode, oxidation of hydrogen gas to form hydronium ions: H2 + 2H2O -> 2H3O+ +

Because of the potential applied to the system from the outside. If you stop applying the potential, then yes (provided the right electrodes and other setup), the reaction will run in the reverse direction, and so will the current. That's how rechargeable batteries work.

@IvanNeretin: Yes, without applying potential, these will be the only reaction that will happen spontaneously. My reasoning in regard to the electrolysis system is that these reactions are still spontaneous (standard potential of Cl2 and H2 do not change) and if they reach opposite electrodes, they will happily react with the electrodes to produce the initial reagents. Perhaps my reasoning is flawed here. If so, why do these reactions not occur when a potential is applied? Thx!


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Desilylation mechanism with fluoride
I want to understand the mechanism of desilylation of trimethylsilyl under fluoride conditions in methanol, because I can't find any useful info that makes me understand the mechanism that takes place.
I did find this:

But my case is a little bit different because a terminal alkyne is bonded to the silyl group instead of oxygen:

Why would that bond be broken and not any other carbon-hydrogen bond?
Thanks in advance!

The mechanism is the same with the generation of the alkynyl anion as the abstract of this paper  ref 1 makes clear:

Tetrabutylammonium triphenyldifluorosilicate (TBAT) can be employed as a fluoride source to cleave silicon−carbon bonds thus generating in situ carbanions that coupled with a variety of electrophiles, including aldehydes and ketones, in moderate to high yields. Among the examples reported is the first instance of fluoride-induced intermolecular coupling between allyltrimethylsilane and imine derivatives. Also, of particular note is the TBAT-initiated coupling of primary alkyl halides with allyltrimethylsilane. TBAT is an easily handled crystalline solid that has several advantages over tetrabutylammonium fluoride (TBAF) as a fluoride source; it is anhydrous, nonhygroscopic, soluble in most commonly used organic solvents, and less basic than TBAF.

Bear in mind that the $\mathrm{p}K_\mathrm{a}$ of an alkynyl proton is approximately 25 (ref 2) so this is a far more stable anion than any possible alternative. Clearly with $\ce{MeOH}$ as the solvent it will be quenched immediately.


text:
Focusing on HOMO when discussing thermal electrocyclic reactions
When taking into consideration the outcome of a thermal electrocyclic reaction, why do we focus specifically on the HOMO of the conjugated polyene? Why does it seemingly have the highest contribution on the product?

Can you give an example of what you mean? Woodward Hoffman rules for example take into account all pi orbitals. Some great work by Gerald Knizia also shows how electrons flow through all the orbitals; see for example DOI 10.1002/anie.201410637


text:
Synthetic route to hydroxychloroquine
This drug hydroxychloroquine has become a new burning topic of discussion as a chemistry enthusiast:

I know that in an industrial preparation we need to account fpr a lot of factors like temperature, solvent, expenditure etc. I tried this, which did not seem good enough:

It seems that the nucleophilic aromatic substitution is not appropriate and the starting products chosen may not be commercially available. Can someone suggest a better yielding and more economical pathway?

Assemble the amino sidechain first and use that in the SNAr with the di-chloroquinoline

@Waylander would both the pathways make a difference?

Yes, I think so.the SNAr will be a high- yielding reaction. I think your reductive am with the 4-aminoquinoline will be poor.

But I had the doubt that may be both the aryl halides would compete for SnAr

What could be other reactions with reductive amination

The other Cl is not activated and will not react. It is comparable to chlorobenzene. See this answer https://chemistry.stackexchange.com/questions/131764/hydroxychloroquine-hcq-synthesis-which-mechanism/131772#131772

Convergent syntheses are higher yielding than linear syntheses. Follow @Waylander.

This took a bit of thinking about but here is a viable route from commercially available bulk materials.
Start with nitroethane, deprotonate ($\ce{t-BuOK}$ in THF would be my choice, but $\ce{NaOEt/EtOH}$ should do it) and react with the commercially available dioxane protected 3-bromopropanal. Reduce the nitro group - alkyl nitro groups are tricky to reduce, I've found Hg/Al amalgam to be pretty good but others may have different conditions. There's a recent review of reductions$\ce{^{[1]}}$. The product of this will do the SNAr with 4,7-dichloroquinoline, I would try $\ce{K2CO3/MeCN}$ for this, but here are plenty of other possible conditions. This is  a good point to purify, then reveal the masked aldehyde by acid hydrolysis and do a reductive amination with the commercially available N-Ethylethanolamine to give the desired product.

Reference

Manuel Orlandi, Davide Brenna, Reentje Harms, Sonja Jost, and Maurizio Benaglia, Organic Process Research & Development 2018 22 (4), 430-445
DOI: 10.1021/acs.oprd.6b00205

@Waylander: A Chem Abst search did not locate effective methods for alkylation of nitroethane with primary alkyl halides. In other cases, C vs. O alkylation is always an issue. However, condensation of nitroethane with aliphatic aldehydes is efficient. (DOI:10.1021/ol800224k). I would offer addition of 2-(ethylamino)-ethanol to acrolein followed by base-catalyzed condensation of the resultant aldehyde with nitroethane. Then, hydrogenate both the double bond and nitro group.

Thank you for that thought. I do not have any database access so cannot check a lot of reactions. I am not a fan of hydrogenating aliphatic nitro groups as they give low yields in my experience.

Note: Organic chemistry is not my field of expertise. This is just a supporting answer to Waylander's excellent answer which is purely based on online research. OP want some canonical answers and if people are interested in this question, I might as well place a bounty.

One route I found is react 1-chloro-4-pentanone with 2-ethylaminoethanol to make aminoketone which undergoes reductive amination making 4-[ethyl(2-hydroxyethyl)amino]-1-methylbutylamine. Reacting this with 4,7-dichlroquinoline* makes the desired hydroxychloroquine.

*There are 3 methods to make 4,7-dichloroquinoline(see ref. below for more details). One of the method is as follows:
Take 3-chloroaniline and ethoxymethylenmalonic ester to make (3-choroanilino)-methylenemalonic ester, which then undergoes high-temperature heterocyclization to make the ethyl ester of 7-chloro-4-hydroxyquinolin-3-carboxylic acid. Hydrolyzing this with sodium hydroxide gives 7-chloro-4-hydroxyquinolin-3-decarboxylic acid, which when heated at 250–270°C is decarboxylated, forming 7-chloro-4-hydroxyquinoline. Treating this with phosphorus oxychloride gives 4,7-dichloroquinoline.

Reference

Hydroxychloroquine and Chloroquine articles respectively of ScienceDirect.com. Original reference is "Drugs for Treating Protozoan Infections by R.S. Vardanyan, V.J. Hruby, 2006"

thanks for taking intrest on my question. I just had one doubt regarding the Decarboxylation  of compound 37.1.1.6. I know that picolinic acid will be decarboxylated on heating will the same happen with this

@Adithya Yes, it should but then again, I am no expert in organic chemistry. Let other expert users clarify this. HCQ is a burning topic nowadays and I believe it will attract more people :)

5-Chloropentan-2-one is accessible from levulinic acid, 4-oxopentanoic acid. https://chemistry.stackexchange.com/questions/126004/how-is-levulinic-acid-formed-from-sucrose/126022#126022


text:
can i synthesise Bis(tri-tert-butylphosphine)palladium(0) from Sodium tetrachloropalladate and tri(t-butyl)phosphine
can i synthesise Bis(tri-tert-butylphosphine)palladium(0) from Sodium tetrachloropalladate and tri(t-butyl)phosphine if so how do i isolate the product

I think it can be done by a two step process via $\ce{(DBA)3Pd2}$:

Prepare Tris(dibenzylideneacetone)dipalladium–Chloroform from sodium tetrachloropalladate by the method of Fairlamb and Tsuji$\ce{^{[1]}}$
Use that to prepare bis(tri-tert-butylphosphine)palladium(0)$\ce{^{[2]}}$

Note $\ce{(t-Bu)3P}$ is extremely air sensitive.
References

Jiro Tsuji and Ian J. S. Fairlamb et. al., The Elusive Structure of Pd 2(dba)3. Examination by Isotopic Labeling, NMR Spectroscopy, and X-ray Diffraction Analysis: Synthesis and Characterization of Pd 2 (dba-Z) 3 Complexes , Journal of the American Chemical Society, 10.1021/ja403259c, 135, 22, (8388-8399), (2013).DOI: https://onlinelibrary.wiley.com/doi/abs/10.1002/047084289X.rt400.pub2
https://patents.google.com/patent/CN101693725A/en


text:
lithiation between thiophene with n-buthyllithium
I just want to know if an electrophilic aromatic substitution reaction between n-buthyllitium and thiophene takes place, where the lithium ion of n-buthyllithium acts as an electrophile and reacts with thiophene during lithiation.
Is this true?

I'm not sure what you mean. 2-Lithiothiophene is well-behaved and stable in solution.

Let me reformulate. I don't understand the concept of lithiation between a thiophene containing molecule (2,2’-(1,3-Butadiyne-1,4 diyl)dithiophene) and n-buthyllithium. My guess is that a directed ortho metalation reaction takes place, but I don't understand how.

pKa of thiophene at the 2 position is approx 33


text:
Polycondensation of PMS

This is the reaction that I'm interested in. I reckon this is what occurs:

But how does the reaction proceed from there, there is a lone pair on the oxygen in right side of my picture, but how do you get rid of the CH3 group on that oxygen?
Although I do see a problem here that the structure of my final product was due to my aforementioned reaction scheme would still have RO-(PMS)-OR, where R is methyl group, instead of RO-(PMS)-R. I don't know what to do...

The more common reaction of this type eleminates acetic acid instead of methanol. Silicon rubber during curing smells like vinegar. I´ll have to look up the specifics ...

"Cationic polymerization can occur with the use of strong or Lewis protic acids. The mechanism of this reaction is yet unknown." https://www.ukessays.com/essays/chemistry/polysiloxanes-synthesis-structure-2201.php
Thereby, we are freed to speculate!
It is more likely that silicon goes pentacoordinate rather than tricoordinate. If a proton attaches to an oxygen on the siloxane, the silicon goes positive, which will attract an oxygen from: 1) a water, 2) a methanol, or 3) a methyltrimethoxy silane. The second and third options can only reform the original reactants. If water attaches after protonation, forming $\ce{H2O-Si-OHCH3}$, methanol could leave, followed by a proton from the water, to give $\ce{CH3Si(OCH3)2OH}$. This becomes a fourth attacking option. If this attacks the silicon of a protonated siloxane thru the $\ce{Si-OH}$ to give $\ce{CH3O(H+)-Si-OHSiCH3(OCH3)2}$, the pentacoordinated silicon can lose methanol, then lose the proton from the oxygen between the silicons, to form $\ce{CH3(OCH3)2Si-O-SiCH3(OCH3)2}$, and that's the first step, a dimer.
Because there are two more alkoxy groups on the silicons which can leave, more reactions can occur, and likely will produce a three-dimensional gel, or even a solid, if you cook it.


text:
Reductions by Na + EtOH
According to my chemistry teacher, sodium in ethanol can reduce ketones, aldehydes, acid halides, esters, cyanides, isocyanides, nitro compounds and acid anhydrides.
Now surfing over the web, I could only verify that it reduces, aldehydes, ketones and esters. Can someone please confirm whether Na in EtOH reduces other mentioned compounds also?

If it opens an ester, and anhydride should be pushover. I would say of all those you mention esters and ketones are the most inert.

$\ce{Na + ROH}$  reduces:

Aldehydes to primary alcohol
Ketones to secondary alcohol
Esters ($\ce{RCOOR'}$) to mixture of two alcohols ($\ce{RCH2OH + R'OH}$)
Cyanides to primary amine
Isocyanides to secondary amine
Alkyl nitro to primary amine
Aryl nitro to aryl amine, can $\ce{LiAlH4}$ do same?

Couldn't verify for Acid anhydrides and for acid halides, but my instinct says it must reduce acyl halides and anhydrides too. As @Waylander states "anhydrides and acyl halides are likely to be converted to ethyl esters by the EtOH solvent and then reduced."

Anhydrides and acyl halides are likely to be converted to ethyl esters by the EtOH solvent and then reduced.

@Waylander if you find any scholarly article for same, I request you to add that in my answer

Thanks alot @Zenix and , I think you both are right, because both acid halides and anhydrides easily get converted into ethyl esters, which is more stable and less reactive so that part logically follows, cant understand how I couldnt make the connection

@Zenix can you please provide any reference regarding cyanides , isocyanides and nitroalkanes

@Zenix as far as i distinctly remember Aryl nitros get converted into azo compounds when treated with LiAlH4, i can confirm this since i have attempted many questions relating to this reactions from reputed books and JEE Advanced papers, if high concentration of nitrobenzene is kept with LiAlH4 it forms azo benzene

@arnav I can provide the reference, but book is not in English, if you still need the reference tell me. You are right about $\ce{LiAlH4}$, I added it as a note, in case someone wasn't aware.

@Zenix Yes please, a reference would be really very helpful!


text:
Elimination of bromine with sodium iodide and electrocyclization
I've been trying to solve the following reaction scheme for a while now and I am having trouble especially with the products E and F.

For product D I got the following, but I'm unsure about compound E, which should be a reactive intermediate.

While trying to figure out the structure of compound E, I drew the following structure, but I'm not sure how the rearrangements will happen to get rid of the charges and get product E. I do expect that E won't be aromatic but that F will be.

All in all, I am wondering what products E and F will be and how to figure them out.

There are no charges in E. Calculate the degree of unsaturation of E. There is no benzene ring in E.

I know the IHD is 5, but how can I determine where the double/triple bonds will be?

For E you have a degree of unsaturation of 5, one of which is a ring. So how do you introduce 4 double bonds (forget triple bonds) into the carbon framework?  Put the two bromine atoms in their most stable orientation.

F arises as a racemate from a conrotatory electrocyclization of E.

Thanks for the tips, this really helped!

I hope you got "I", which was, for some reason, the only species without data. If you want, I'll post an answer.

I think I got the compound "I", but I am a bit unsure whether my compound "H" is correct. Hence an answer would be very helpful, so I could check.

Working on it now!

Pöytä: Your conclusion that D contains four bromine atoms symmetrically distributed is sound. The treatment of D with NaI serves to substitute one, if not two bromine atoms with iodine to form "J". Iodide acts as a reducing agent in a vinylogous fashion. (Compare this reaction with the reduction of a vicinal dibromide with iodide to form an alkene.) The stereochemistry of E is likely to be E,E which arises from a transition state for elimination that avoids steric interactions involving bromine. I am assuming that the "electrocyclization" is thermal since there is no indication that light was involved. Achiral E will undergo a conrotatory electrocyclization in two directions to give racemic F and, eventually, diiodide G via SN2 displacement. (The electrocyclization may be rationalized as involving the HOMO of a butadiene or octatetraene. Look here for a discussion.) Reductive elimination of iodine via Zn-Cu couple affords benzocyclobutadiene H. (The elimination is akin to the formation of ethylene from 1,2-dibromoethane with magnesium.) The strained, reactive double bond of H undergoes a Diels-Alder reaction with cyclopentadiene to form adduct I. The stereochemistry shown arises from a so-called "endo" transition state.I found this question somewhat amusing in that information was provided for all species save the final product. ;)

Thanks! Seems like I did it correctly after your tip about the degree of saturation.

Just a technical note: H should be in squire parenthesis since it is unstable. Also, it would ne nicer to include a reference for relevant reactions, e.g., https://doi.org/10.1021/ja01583a071.

This would be better reference: https://doi.org/10.1021/cr60266a002.

The 50 year old review covers the early steps of the sequence of reactions and it provides an historical perspective. I don't recall seeing any examples of electrocyclizations. Good find nonetheless.

Typo alert, H is actually benzocyclobutadiene not benzocyclobutene.

@Ron: Oops! Done.


text:
Titrating iodine starch solution with sodium thiosulphate - Colour change
I investigated two mixtures with different solvents, one with water and one with n-heptane. Both contained iodine $\ce{I2}$ as a solute. To both solutions I added a bit of starch.
As I remember this resulted in a colourchange. So the solution turned from yellowish to dark blue (if I remember correctly!).
Now according to wikipedia starch and iodine indeed form a structure which has a dark blue colour. But it only forms in the presence of $\ce{I^-}$.
This leaves me wondering, why do I remeber the solution to be dark blue, eventhough I think there was no $\ce{I^-}$ present? Could it be the solution turned dark blue only after I added some sodium thiosulfate? Because in the next step I did a titration with $\ce{Na2S2O3}$.
In this case I don't see which reaction could have produced the $\ce{I^-}$ though. I thought only $\ce{NaI}$ is produced after adding the sodium thiosulfate.
$$\ce{I_2 + 2Na_2S_2O_3 -> 2NaI + Na_2S_4O_6} \tag{1}$$
So at which point did the solution turn dark blue and where did the $\ce{I^-}$ come from, that was needed for the formation of the starch-iodine-compound? Could it be there is an intermediate step to (1) in which $\ce{I^-}$ is formed and this $\ce{I^-}$ was used to produce the dark blue starch-iodine compound?

I don't think your memory is serving you right. That is why we write everything in the notebook, especially color changes.
I think you are doing distribution experiments where iodine is distributed between aqueous layer and an organic layer. When we add indicator for titration, it is not a solid starch but starch which is boiled in water. So when you added starch $solution$ to heptane which contained iodine, I would not be surprised if the starch solution turned blue.
Remember that iodine is strong oxidizing agent as well. A very small fraction of it can easily convert into iodide. You really really need a trace of the triiodide ion to form a dark blue iodine complex.

Your assumptions are correct. This is my first chemistry lab. And yes I should've wrote everything down more carefully. One question for clarification: You think the Iodine interacted with the sodium thiosulphate, forming some $\ce{I^-}$ which then lead to the reaction $\ce{I^-}+\ce{I_2}+\textrm{starch}\leftrightarrow\textrm{dark blue starch}$? And when adding more and more thiosulphate all of the $I_2$ and consequently all of the dark blue starch reacted to the colourless $\ce{I^-}$?

Right, this is what I think happened in your case.

The molecular iodine $\ce{I_2}$ is poorly soluble in water : maximum $0.0011$ M. If starch is added to this solution, the iodine will react with starch and the solution is dark blue. In the lab, this experiment is rarely done with simple $\ce{I_2}$ solutions, because the solutions to be titrated are usually more concentrated than $0.001$ M. Usually $\ce{I_2}$ is dissolved in $\ce{KI}$ solutions, producing $\ce{KI_3}$ or $\ce{I_3^-}$ ions.$$\ce{KI + I_2 <=> KI_3}$$ The "solubility" of $\ce{I_2}$ as combined in $\ce{KI_3}$ is at least $1000$ times higher than $\ce{I_2}$ in water. So in the presence of $\ce{KI}$ in solution, more $\ce{I_2}$ can stay in solution. And if some starch is added to a $\ce{KI_3}$ solution, it will produce a dark blue-black color, due to the small amount of free $\ce{I_2}$ in the $\ce{KI_3}$ solution.
When titrating either $\ce{I_2}$ or $\ce{KI_3}$ by adding thiosulfate  ions $\ce{S_2O_3^{2-}}$, the free $\ce{I_2 }$ is consumed. $$\ce{I_2 + 2 S_2O_3^{2-}-> S_4O_6^{2-} + 2 I^-}$$
$\ce{I_2}$ is consumed by adding $\ce{S_2O_3^{2-}}$. But as the equilibrium $\ce{KI + I_2 <=> KI_3}$ is rapid, new $\ce{I_2}$molecules are continuously regenerated from $\ce{KI_3}$, so that the starch solution stays dark blue up to the end of the titration. Apparently, the titration proceeds as if the solution of $\ce{KI_3}$ is a solution of $\ce{I_2}$.


text:
Pressure of a system of particles
I'll derive the pressure of a system of particles, which would be used in a MD simulation for example, according to Allen's Computer Simulation of Liquids and show where my issue is.
Consider a Hamiltonian system with $N$ interacting particles and generalized coordinates $q_k$. The virial theorem is:
$$
\langle q_k \frac{\partial H}{\partial q_k} \rangle = kT
$$
This implies that
$$
\frac{1}{3} \langle \sum_i \mathbf{r}_i \cdot \mathbf{f}_i^{tot} \rangle = NkT
$$
where $\mathbf{f}_i^{tot}$ is the total force including interparticle and external forces. So we can split this force into internal and external contributions, $\mathbf{f}_i^{tot} = \mathbf{f}_i^{ext} + \mathbf{f}_i^{int}$. The external part of the force defines the pressure on the system via
$$
\frac{1}{3} \langle \sum_i \mathbf{r}_i \cdot \mathbf{f}_i^{ext} \rangle = -PV
$$

What is the reasoning behind this equation?
Why does the sum of dot products between particle position vectors and external forces equal $-PV$?


text:
Dipole moment difference between ch2f2 and ch3f
In CH3F and CH2F2  the net dipole moment increases from 1.82 to 1.97 while in all other halogen substituted carbon atoms the dipole moment decreases. In the case of CH3Cl and CH3F the chlorine substituted methyl group has greater dipole moment due to longer bond length of C-Cl compared to C-F. Then when a compound of formula CH2X2 is formed than dipole moment decreases except in the case of fluorine where it increases.
 Is there any cause to this phenomenon?

Does this answer your question? Why do C-F bond has less dipole moment than C-Cl bond?

Or this https://chemistry.stackexchange.com/q/65965/50406 ?

Or this https://chemistry.stackexchange.com/q/48066/50406?

Or this https://chemistry.stackexchange.com/q/8977/50406?

@Zenix Heheh :D

@Mithoron ik relatable ;)


text:
Ionization and bond energy of isotope
Is there a database with the direct values, or the ingredients to calculate them, for ionization and bond energies of isotopes (i.e. $^{17}O_2$). If I understand it right, ionization energy can be calculated using the atom radius but I cannot find such information for isotopes and I guess it's still slightly different to the one of the abundand forms. Also, for bond energy I read it can be calculated using enthalpies but can't find such data for isotopes either. Thanks in advance!

What's "O_17"? If you have dioxygen-17 in mind, then it should be $\ce{^{17}O2}.$

Ah, good point, I meant dioxygen indeed, changed it in the question, thanks :)

What are you trying to do with the data? You must realize that the internal energy and kinetic of oxygen molecules varies tremendously compared to the difference between isotopic changes. The difference in binding energy  even between hydrogen and deuterium atoms is tiny. // You should be able to get a very reasonable estimate using the difference in the reduced mass.

I don't know of any such tables. For chemistry the values are so close as to make no difference except for very light elements such as hydrogen.
All in all you should be able to get a good idea of the difference by using the difference in reduced mass, $m_\mathrm{r}$.
$$m_\mathrm{r} = \dfrac{m_\mathrm{electron}m_\mathrm{nucleus}}{m_\mathrm{electron}+m_\mathrm{nucleus}} = m_\mathrm{electron}\cdot   \dfrac{1}{1+ \dfrac{m_\mathrm{electron}}{m_\mathrm{nucleus}}}$$
So for $\ce{^{16}O}$
$$m_\mathrm{r,\ce{^{16}O}} = \dfrac{1}{1+\dfrac{5.489\cdot10^{-4}}{15.995}} = 0.9999657$$
for $\ce{^{17}O}$
$$m_\mathrm{r,\ce{^{17}O}} = \dfrac{1}{1+\dfrac{5.489\cdot10^{-4}}{16.999}} = 0.9999677$$
% Difference between $m_\mathrm{r,\ce{^{17}O}}$ and $m_\mathrm{r,\ce{^{16}O}}$ = +0.00020%


text:
Why Nitroso tautomer is more stable than its oxime counterpart?
As I was looking for the answer of the question I found that it is explained using the Electronegativity difference concept.
That's correct but I am unable to prove the same through the bond energy data.
Can someone prove the same through it.

Why is the oxime tautomer more stable than its corresponding nitroso isomer?

Does this answer your question? Why is the oxime tautomer more stable than its corresponding nitroso isomer?

Counter example: Realistically, alkyl oximes are more stable than their nitroso counterparts. For example, computational studies has been shown that formaldehyde oxime is $\pu{15.8 kcal/mol}$ more stable than nitrosomethane when the aqueous solvation correction of $\pu{3.8 kcal/mol}$ is applied to the G2 energies. Unsolvated formaldehyde oxime is estimated to be $\pu{11.1 kcal/mol}$ more stable than nitrone (Ref.1). In solution chemistry, one of the difficulties association with synthesis of C-nitrosoalkanes is their likeliness to convert to the more stable corresponding oxime tautomer (Ref.2), apart from their high reactivity (Rings the bell?). You can find some $\Delta H_f(s)$, $\Delta H_f(l)$, and $\Delta H_f(g)$ of C-nitrosoalkanes in Ref.3. It is also noteworthy that $\ce{C-N}$ and $\ce{N-O}$ bond lengths are quite typical of oximes (overall values are $\pu{128.1 pm}$ and $\pu{141.6 pm}$, respectively), yet $\ce{N-O}$ in oximes are somewhat shorter than that of corresponding hydroxylamines. This indicates that some charge delocalization in oximes. However, the NMR evidence of syn-anti isomerization in oximes (e.g., syn-isome being the one with $\ce{H}$ and $\ce{OH}$ in the same side in acetoxime $\ce{CH3-CH=N-OH}$) confirms the predominance of oxime tautomer.
In closely relevant cases of ortho-nitrosonaphthols and acenaphthenequinonemonooxime, computational studies have been shown that the predominance of oxime form (Ref.4 and 5, respectively). For example, energy diagrams of 2-nitroso-1-naphthol and 1-nitroso-2-naphthol are illustrated in following diagrams:

References:

Judy A. Long, Nathan J. Harris, Koop Lammertsma, "Formaldehyde Oxime $\ce{<=>}$ Nitrosomethane Tautomerism," J. Org. Chem. 2001, 66(20), 6762-6767 (https://doi.org/10.1021/jo010671v).
Brian G. Gowenlock, George B. Richter-Addo, "Preparations of C-Nitroso Compounds," Chem. Rev. 2004, 104(7), 3315-3340 (https://doi.org/10.1021/cr030450k).
Suzanne W Slayden, Joel F. Liebman, "Chapter 3: The organic thermochemistry of hydroxylamines, oximes, hydroxamic Acids and their derivatives," In The Chemistry of Hydroxylamines, Oximes and Hydroxamic Acids, Part 1; Zvi Rappoport,Joel F. Liebman, Editors (Patai Series: The Chemistry of Functional Groups; Series Editor: Zvi Rappoport); John Wiley & Sons, Ltd.: Chichester, England, 2009, pp.53-83.
Galya Ivanova, Venelin Enchev, "Does tautomeric equilibrium exist in ortho-nitrosonaphthols?," Chemical Physics 2001, 264(3), 235-244 (https://doi.org/10.1016/S0301-0104(01)00245-2).
V. Enchev, G. Ivanova, A. Ugrinov, G. D. Neykov, "Tautomeric and conformational equilibrium of acenaphthenequinonemonooxime," Journal of Molecular Structure 1999, 508(1-3), 149-161 (https://doi.org/10.1016/S0022-2860(99)00008-3).


text:
Order of filling electrons in Jahn-Teller energy diagram
When filling electrons in atomic or molecular orbitals we fill first the lower energy orbitals and then the higher, according to Pauli and Hund's rule. But this seem not to be the case when considering metal complexes. According to Wikipedia
For example why the $d_{xy}$ orbital in the $d^4$ high spin is not fully occupied? Is the pairing energy that changes the order of filling the orbitals?

This was just an illustrative example. The tendency of a complex to favor low-spin or high-spin configurations is dictated by the relative magnitudes of the pairing energy, P, and the splitting energy, $\Delta$, between $e_g$ and $t_{2g}$.*
If $\Delta$ is greater than the energy required to pair two electrons, then electrons pair in $t_{2g}$ before occupying $e_g$. In the example you've shown of compression, this would be the $d_{xy}$ orbital.
*For octahedral complexes. Jahn-Teller can also happen for tetrahedral complexes.

Thanks for the answer. So the fact that a complex is low or high spin affect the way of ordering after the Jahn-Teller distortion? For example considering compression for $d^4$ low spin. Should now the $d_{xy}$ be fully occupied?

@adosar exactly.  In the compression example, $d_{xy}$ would be doubly occupied, defining the $d^4$ low spin configuration.

But we can't predict if elongation or compression will happen right? Also when filling atomic orbitals or molecular we neglect pairing energy because it is negligible compared to $\Delta E$ of the orbitals?

@adosar I only mentioned compression because I could name the exact orbital. In the case of elongation and $\Delta$ larger than the pairing energy, then you'd have an electron go into either $d_{xz}$ or $d_{xy}$ as they are degenerate.

@adosar the pairing energy is exactly what we are comparing to the $\Delta$, so you definitely do not ignore it :)

At first I thought you were asking a very astute question about the stabilizing effects from parallel spin (due to exchange symmetry). This is present in the system so you'd really be comparing $\Delta - V$ to the pairing energy, where V is the parallel spin coupling. However, inorganic chemists typically ignore this effect as it is very small for parallel electrons of very different energy.

Two more questions and I accept the answer. Why we don't care about pairing energy when we look at electronic configurations of ground states? Is it so small? I learned the term pairing energy in inorganic chemistry. I didn't hear it even in physical chemistry. Also why the low spin or high spin determines the ordering after Jahn-Teller distorion if the energy levels have changed and $\Delta$ has also changed? Thanks in advance.

I think the confusion may be in how the pairing energy is defined. It is the energy required to place two electrons in the same orbital, and it is not ignored. What is typically ignored is the stabilizing effect of parallel spins in the high spin case (see my previous comment)

@adosar did this clear things up?

So after Jahn-Teller distortion the difference between $\Delta$ and pairing energy is the same and the filling order is the same as before the distortion?

There are two things at play here. $\Delta$ vs pairing energy is what determines whether the high-spin or the low-spin configuration is lower in energy (and therefore occupied by the complex). Jahn-Teller just occurs in certain configurations to lower the energy of the molecule by distorting its geometry to break symmetry. See the bottom figure you posted as an example of when Jahn-Teller distortions wouldn't lower the energy of the complex.