text:
Why wasn't Lu Jeu Sham awarded the 1998 Nobel Prize with Walter Kohn?
Reviewing part of the history behind computational chemistry for my thesis I could not understand why only Walter Kohn was awarded the 1998 Nobel Prize in Chemistry (with John Pople getting the other half) if Lu Jeu Sham was equally important for laying the foundations of Density Functional Theory.
This post from Chemistry World suggests that Sham was excluded because he is not a "middle aged white man". I don't want to be provocative but was it even given a reason why only Kohn's name was immortalized while the 1965 paper is clearly authored by both men?

Who would give such a reason officially?

I know nothing about the development of DFT, but it is conceivable that the committee considered more than just the co-authorship of the original paper. Perhaps later papers were seen as driving the whole thing forward at least as much or more than the original one.

The question may be better suited for SE site History of science and mathematics ( https://hsm.stackexchange.com )

Thank you. I believe that the two answers provided have a good reasoning and I am satisfied. Thank you so much.

Unfortunately, there is very rarely going to be an official statement as to why someone did not receive the Nobel Prize. A number of factors probably played into the decision. One of the big ones is that by rule, a single Nobel prize can be awarded to at most 3 individuals and at most 2 different works. This means that if third person were added to this prize, they would either be getting equal credit with Kohn for creating DFT or with Pople for development of computational methods.
You could make an argument for Sham or Hohenberg to be included on the DFT side, but with Kohn being an author on both the 1964 Hohenberg paper that introduced DFT and an author on the 1965 paper that made it practical through the KS formulation, its not unreasonable to claim that Kohn deserved more of the credit than either Sham or Hohenberg. Its also worth noting that Sham was Kohn's postdoc and Hohenberg had only just begun working at Bell Labs at the time of these publications; there could be a tendency to ascribe the work to the senior-most researcher.
There are a number of other factors you can consider (e.g. why assume Kohn's credit should be split rather than Pople's), but there will likely never be a definitive answer. The selection process is subjective and opaque. As the linked article suggests, it can be controversial and there are probably dozens of figures throughout history who you could argue deserved a Nobel Prize. But the prize is only awarded once a year to at most 2 works/3 people, so inevitably some works/researchers will be missed.

Some elements of the nomination are known (e.g., laureates of previous years may suggest as regularly do chemistry professors in Norwegian universities), but not all - especially not the discussions of the committees to elect up to three laureates per year and division. Thus, complementary to @Tyberius' answer, first paragraph, I would add paragraph 10 of the statutes of the Nobel Foundation as an obstacle here.  At least at present time.
To quote:

§ 10.
No appeals may be made against the decision of a prize-awarding body with regard to the award of a prize.
Proposals received for the award of a prize, and investigations and opinions concerning the award of a prize, may not be divulged. Should divergent opinions have been expressed in connection with the decision of a prize-awarding body concerning the award of a prize, this may not be included in the record or otherwise divulged.
A prize-awarding body may, however, after due consideration in each individual case, permit access to material which formed the basis for the evaluation and decision concerning a prize, for purposes of research in intellectual history. Such permission may not, however, be granted until at least 50 years have elapsed after the date on which the decision in question was made.

(Any emphasize by italics was added by mine.)

Few parts of the archive actually are accessible to the general public here.


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Direction of spins in the singlet vs. triplet state
Suppose we consider the first excited state of the helium atom. We know that the first excited state of helium can exist as a triplet or singlet. The possible functions related to the spin of the two electrons in the triplet state are
$$\alpha(1)\alpha(2)$$
$$\beta(1)\beta(2)$$
$$\dfrac{1}{\sqrt{2}} [\alpha(1)\beta(2) + \beta(1)\alpha(2) ] $$
while the one for the singlet state is
$$\dfrac{1}{\sqrt{2}} [\alpha(1)\beta(2) - \beta(1)\alpha(2) ]$$
The triplet state predicts that the spins of the two electrons are parallel, but according to this equation
$$\dfrac{1}{\sqrt{2}} [\alpha(1)\beta(2) + \beta(1)\alpha(2) ] $$
there is a 50% probability that electron 1 is in the alpha state and a 50% probability that it is in the beta state: the same goes for electron 2.
So, if this function predicts that the two spins are antiparallel, why is it part of one of the triplet states?

I think your confusion stems from the fact that "up" and "down" refer to the sign of the z component of the spin of individual electrons, whereas "triplet" and "singlet" describe the magnitude of the total spin of the combined system, not just the z components. Thus, one of the three triplet states has zero spin in the z-direction (one electron up and one down), but still has non-zero total spin, ie. the combined spin vector is within the x-y plane. The singlet state has zero overall spin, ie. the two spin vectors completely cancel each other, rather than just canceling in the z direction.

@Andrew That fully answers the question I believe... I hope you invest the extra effort to add it as one? That would be appreciated by many, if you find the time.

@Andrew can you see this image?https://upload.wikimedia.org/wikipedia/commons/4/49/ISC_excited_states.png If one of the triplet excited states has one alpha and one beta electron, shouldn't this be the configuration? 1s $\uparrow$ 2s $\downarrow$

This answer has a diagram that you should  find informative.  https://chemistry.stackexchange.com/questions/50277/how-does-spin-flipping-of-triplet-carbenes-occur/54259#54259

@stian yttervik I didn’t write it as an answer because it seemed like a good answer should have images like in the answer porphyrin linked, and I didn’t have time to make them. I think that linked answer covers it though

source: Stack Exchange

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

sort: votes


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Classification model from biological assays with different compounds
If I want to use an undersampling approach to construct the machine learning model (classification model predicting whether the target compounds are active or inactive, using chemical fingerprint as variables), I am wondering if there are any criteria to determine how many times I should sample the data from the majority group (the minority is 14% and the majority is 86%) and build the ML model? I am working with biological assay data and I am recommended not to use oversampling approaches. To determine which sampling approach we should use, are there any criteria to determine before constructing the ML model? Is it really dependent on the field of study?

My similar question has been moved here, https://stats.stackexchange.com/questions/540770/undersampling-approach-in-different-types-of-study

Another option is Matter Modeling Stack exchange. This question isn’t really about chemistry but modeling so it typically isn’t appropriate.

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


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Weerman degradation on alpha-hydroxy amides
I came across this reaction scheme in the solutions of a test paper:

I understand the formation of isocyanate intermediate as the first step. In the second step, I realized that hydrolysis will be more favourable at the benzyl carbon, hence forming benzaldehyde (given as the major product).
However, I cannot understand how aniline is being formed. There must be another migration/rearrangement occurring, but I cannot figure out how. It might be some reaction I am unaware of.

The proton attached to the benzyl position appears to be acidic. It may be removed by OH-. Something similar to the first hoffmann rearrangement (where Br was the leaving group) may occur with the C=O, with phenyl shifting to N atom.

@AdityaRoychowdhury  more acidic than the hydroxy proton?

@Waylander Not more acidic than hydroxy proton, but, nothing else occured to me for the aniline formation.

I think it may be that there is a printing error and benzenamine is shown as aniline..

Is hydrolysis at benzyl carbon obvious? I see that N=C=O carbon also has a substantial partial positive charge due to adjacent electron withdrawing groups, steric hindrance also not so much?

@TRC: The reaction is formally known as Weerman degradation so that I changed the title accordingly. If you don't like what I did, please feel free to roll back.

@Ashish - Even if hydrolysis takes place at the N=C, the compound formed has hydroxyl and amine group on the same carbon, so it won't be stable enough and will convert to benzaldehyde.

@TRC Yeah the leaving group given in the accepted answer is $\ce{H2NCO2H}$ so maybe that is what is happening, but it's a good leaving group nevertheless and aldehyde will form.

@MathewMahindaratne - I'm always glad to learn new reaction names. Saves a lot of trouble and frustration in case I see it in the exam!

Since OP did not give any reference to the rearrangement in the question, I doubted the formation of aniline is true even if as a minor product. The conversion of $\alpha$-hydroxy amides to corresponding aldehydes is a well-known degradation reaction, commonly known as Weerman degradation:
$$\ce{R-CH(OH)-C=O-NH2 ->[NaOBr][\Delta] RCHO + H2NCO2H}$$
Although this degradation is independently discovered (Ref.1), Some authors have regarded it as an extension of the Hofmann rearrangement (Ref.2) because of the similarity in proposed mechanism:

Accordingly, to my knowledge, there is no place to have an rearrangement other than proposed. However, even if it is most unlikely, there could be a possibility to have 1,3-phenyl shift at the intermediate $\bf{I}$ stage:

The resultant phenyl $\alpha$-ketoamide could undergoes base hydrolysis under the reaction condition to give aniline.
Note: The 1,3-phenyl shift has been known for other reactions (Ref.3).
References:

R. A. Weerman, "Sur une synthèse d'aldéhydes et de l'indol," Recueil des Travaux Chimiques des Pays-Bas et de la Belgique 1910, 29(1-2), 18-21 (DOI: https://doi.org/10.1002/recl.19100290104).
C. L. Arcus and D. B. Greenwood, "398. The Hofmann reaction with α- and β-hydroxy-amides: reactions of the intermediate isocyanates," J. Chem. Soc. 1953, 1937-1940 (DOI: https://doi.org/10.1039/JR9530001937).
Colin Eaborn, Karen L. Jones, Paul D. Lickiss, and Włodzimierz A. Stańczyk, "Proportion of 1,3-phenyl migration in trifluoroethanolysis, methanolysis, and hydrolysis of $\ce{(Me3Si)2C(SiMe2Ph)(SiEt2I)}$," J. Chem. Soc., Perkin Trans. 2 1993, (3), 395-397 (DOI: https://doi.org/10.1039/P29930000395).


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$\pi$ Molecular orbital diagram of compounds like urea, guanidine, acetone etc
In my organic chemistry course, we've only been taught about how the $\pi$ MO diagrams of compounds like butan-1,3 diene, hexan-1,3,5 triene and similar single chain compounds look like. I was wondering how the $\pi$ MO diagrams of heteroatomic compounds like urea, guanidine etc. look like, but unfortunately I was unable to find any relevant stuff in google as well as Organic textbooks like Clayden, Solomons etc.

Pretty much the same as in a physics class we learn statics but not to build a bridge

Sure, lol but @Alchimista , can you suggest any resources for the same ?

well I suggest to follow the curriculum and learn the concepts for "simple" molecules bcz they are useful. Note that you are thought systems with no etheroatoms whose chemistry is dominated by the pi electrons. I have no idea how to address you for self learning but perhaps other users can. Note that one molecule you have listed does not really exist ;)


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What determines the kind of Bravais lattice structure (body-centred cubic, hexagonal, etc) a transition metal shows?
With the exception of Zn, Hg, Cd and Mn transition metals most transition metals have only one kind of lattice structure at room temperature. Another trend I noticed was that Groups 3 and 4 have hexagonal close packing; 5 and 6 have Body-centred cubic cells; Groups 7 and 8 (with the exception of Iron) have hexagonal close packing; and the last 3 groups have cubic close packing.
Even at higher temperatures, Groups 3 and 4 usually show usually show body-centred cubic packing (Lanthanum is an exception).
I know that in ionic compounds, we can find the type of packing by comparing the ratio of radi, is something like this going on here? But if that were the case the radi and thus the structure would drastically change down the group (which is not happening); so it must have something to do with the electronic configuration. Can someone help, thanks.

"we can find the type of packing by comparing the ratio of radioactive" ???

All elements show one structure at STP. What are you asking?

@JonCuster edited the question to make it clearer

There is no simple answer. If you think you have one, you are wrong.

@IvanNeretin, I don't expect a simple answer, but I'm in high school so I would prefer something I can understand

That's what would constitute a simple answer.

@IvanNeretin, I'm fine with a complex one too, but I would prefer someone tell me where I can read about those underlying concepts in detail (like some kind of book or a link). Currently I don't even know about the existence of those concepts, forget about understanding them.

I can't comment as I don't have enough rep but these links may help https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Book%3A_Introduction_to_Inorganic_Chemistry/06%3A_Metals_and_Alloys-_Structure_Bonding_Electronic_and_Magnetic_Properties/6.04%3A_Crystal_Structures_of_Metals https://opentextbc.ca/chemistry/chapter/10-6-lattice-structures-in-crystalline-solids/

https://www.semanticscholar.org/paper/Band-structure-theory-of-the-bcc-to-hcp-Burgers-Feng-Widom/178c70dd1e390c508c6f553583dd5302e06b865a might be of interest

Related: https://mattermodeling.stackexchange.com/q/1578/5. Also these: https://mattermodeling.stackexchange.com/search?q=bravais


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Equilibrium chemical potential if the population is not conserved?
The chemical potential, $\mu$, is equal to the free energy $F$, per photon.
$$F = \mu = u-Ts$$
The energy per photon is $u=h\nu$ and $s$ is the entropy per photon. Say we have a process where an incoming photon ($\nu_{1}$) is absorbed by a dye and then emitted at ($\nu_{2}$). Then at equilibrium,
$$h\nu_{1} - s_{1} =  h\nu_{2} - s_{2}$$
What happens at equilibrium if the dye immediately emits two (or more?) photons, so the total number of photons is no longer conserved? Can I still write that the chemical potentials of all the photons are equal as in the equation above?

An attempt at a solution:
The change in Gibbs free energy at equilibrium is $\Delta G=0$. Since this is a homogeneous system, the Gibbs free energy is given by
$$\Delta G = \sum_{i} \mu_{x}x_{i} - \sum_{k} \mu_{y}y_{k} $$
Where $x_{i}$ is the number of particles of product and $y$ is the number of particles reactant. If one photon just becomes one photon then $i=k$, and so
$$\sum_{i} \mu_{y}y_{i} = \sum_{k} \mu_{x}x_{k}  \implies \mu_{x}=\mu_{y}$$
and so the chemical potentials match. Now, if one photon becomes two, so $2k=i$, then
$$\sum_{k} \mu_{y}y_{k} =  \sum_{2k} \mu_{x}x_{k}  \implies 2 \mu_{x}=\mu_{y}$$. This implies that
$$h\nu_{y} - s_{1} =  2 \times (h\nu_{2} - s_{2})$$
at equilibrium.

I think this is more appropriate for physics stack exchange. That said, I would say that the equation should hold for individual photons, irrespective of origin (but tbh that is a guess).

I thought someone would see an equivalence to a chemical reaction here (something splitting/generating two things). Not many things in physics which suddenly become two....

Good point, your intuition seems pretty sound. Sounds like the blackbody radiation problem.

This question is either confused or simply confusing. The concept of equilibrium implies reactant and products and energy interacting so that everything remains constant and cannot be applied to a one photon scenario unless that is intended to be allegorical.  For a photon to interact with anything there must be an interchange of energy and momentum in such a manner that total energy and momentum are both conserved.  This means that the energy and momentum of the photon are changed so the idea that chemical potential is constant must be incorrect. Also can entropy be considered with a single particle? If so the exact energy and momentum must enter the calculation and the resultant would probably be applied to the maximum of a temperature stabilized system.  An example if an inner electron is ejected from an atom by an X or gamma ray. The resulting cascade of photons as outer electrons fell into successive empty orbitals will give a spectrum and the energy of the emitted light will add to the original energy if recoil etc is added in but the entropy of the emitted photons will be greater than that of the one Xray.
This can be applied to any energy flux steady state system where the driving force is the actual entropy change of energy entering and leaving the system since mass, U, H and G are all constant. At equilibrium Delta G = 0. The scenario above seems to describe a steady state not an equilibrium.

So, I do understand your point about equilibrium. For a true equilibrium to exist, photons must be able to interact. However, as photons do not have any first-order interactions, they cannot be in true equilibrium. However, they are still well described by the thermodynamic properties such as chemical potential and temperature (e.g. bose-einstein distribution, and black body radiation). – lets call it quasi equilibrium.

However, the condition of quasi equilibrium in a photon beam must be equivalent to a statement that all photons have, in addition to temperature, an identical chemical potential. Since the chemical potential is equal to the free energy  per photon, ==−, where for a photon =ℎ and  is the entropy per photon. Your point about defining entropy per photon - we can easily determine the entropy of radiation (en.wikipedia.org/wiki/Photon_gas), and so the entropy per photon is simply the number differential of the number of photons $s = (\delta S/ \delta N)$.

Your comment about the allegorical nature I do not understand. It doesn't matter the process going on. A volume of photons in equilibrium can be changed into another volume of photons in equilibrium, however, this brings about irreversibility, which must introduce entropy generation (as required by the second law of thermodynamics). Process doesn't matter. Maybe this point in your answer does solve the question....the transformed beam cannot be in equilibrium, and so we cannot equate chemical potentials....although photons can be in (quasi) equilibrium which each other...

At equilibrium, the number of each species is constant. So if a molecule falls appart into two fragments, these fragments would recombine at the same rate.
Thermodynamics is for looking at bulk properties, not individual particles. Pressure, temperature, entropy are not defined for single particles using equilibrium thermodynamics.
Chemical reactions involving photons are most often non-equilibrium processes, so while it is nice to use analogies, they might not deliver rigorous results in this case.
This reference might be a starting point for studying how to think about the entropy of photons: https://arxiv.org/ftp/arxiv/papers/1306/1306.1993.pdf

The entropy can be determined per photon relatively easily, and we can quite easily think of the chemical potential of an excited charge state or even a photon, so I'm not sure it's true to say that these are not defined for individual particles.
My question is if a reaction goes from A->2B, if the chemical potentials are not the same, what are they? I thought using photons would be easier for analysis....
Agreed regarding the point about non-equilibrium processes. But let us pretend they are!....


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Extraction of copper from chalcopyrite
During the extraction of copper from chalcopyrite ore, the ore ($\ce{Cu2S.Fe2S3}$) is heated to form $\ce{Cu2S}$ and $\ce{FeS}$. Then during smelting it reacts with $\ce{O2}$ to form $\ce{Cu2O}$.
Then it is written that $\ce{Cu2O}$ reacts with $\ce{FeS}$ to form $\ce{Cu2S}$.
My question is that we started off with $\ce{Cu2S}$ then why did we convert it into $\ce{Cu2O}$ and then again get back to $\ce{Cu2S}$?

Chalcopyrite is not transformed into $\ce{Cu2S + Fe2S}$ as you state, because $\ce{Fe2S}$ does not exist.

It rather looks like copper is a catalyst for oxidation of iron sulphides to oxides.

Also, chalcopyrite is $\ce{CuFeS2}$ (not $\ce{Cu2S.Fe2S3}$).

Chalcopyrite $(\ce{CuFeS2})$ is the most common copper bearing mineral on earth (Approximately 70% of the world’s copper reserves are contained in the mineral chalcopyrite), but also the most stable minerals because of its structural configuration, face-centered tetragonal lattice (Ref.1). At present, there are basically two main methods employed worldwide in order to process chalcopyrite for metal production. The most important one is the conventional - pyrometallurgy method: During this process (pyrometallurgy or smelting at high temperature), the sulfur in the metal sulfide is oxidized with air or oxygen to sulfur dioxide and molten matte is produced. This was explain in the Ref.1 as follows:

In the roaster, the copper concentrate (chalcopyrite) is partially oxidized to produce alkaline $(\ce{Cu2S})$ and sulphur dioxide $(\ce{SO2})$ gas. The chalcopyrite ore is heated strongly with silicon dioxide (silica) and air or oxygen in the furnace or series of furnaces. The copper (II) ion in the chalcopyrite is reduced to copper(I) sulfide which is reduced further to copper metal in the final stages. The iron in the chalcopyrite ends up converted into an iron(II) silicate slag which is removed. Most of the sulfur in the chalcopyrite turns to sulfur dioxide gas. This is used to make sulfuric acid via the contact process (copper extraction and purification). Overall equations for these series of steps are:
$$\ce{2CuFeS2 + 2SiO2 + 4O2 -> Cu2S + 2FeSiO3 + 3SO2} \tag1$$
$$\ce{2CuFeS2 + 3O2 -> 2CuS + 2FeO + 2SO2} \tag2$$
$$\ce{Cu2S + O2 -> 2Cu + SO2} \tag3$$
As of 2005, roasting is no longer common in chalcopyrite concentrate treatment. Thus, direct smelting using the following smelting technologies: flash smelting, Noronda, USA smelting, Mitsubishi or EL Temento furnace are in use till date.

The Ref.1 (page 6) further explains pyrometallurgical practice as follows:

Pyrometallurgical practice typically involves smelting converting, anode casting and electro-refining of the anodes to high purity copper metal. The smelting and refining processes used well established technologies are energy efficient and have high metal recoveries including those of gold and silver (Yin et al. 1995). This begin with a dry concentrate containing less than one percent water, which along with flux is contacted in a furnace by a blast of oxygen or oxygen enriched air (McGraw-Hill Encyclopedia, 1998). Pyrometallurgical extraction involves heating the mineral cake in a blast furnace; oxygen pressure and temperature are carefully controlled. The first stage involve the separation of copper and iron ore (Equation 4), followed by the addition of silica $(\ce{SiO2})$ to the blast furnace to convert iron(II) oxide to a less dense liquid layer of slag, iron(III) silicate which is poured off.
$$\ce{2CuFeS2 + 4O2 -> Cu2S + 2FeO2 + 2SO2} \tag4$$
$$\ce{FeO + SiO2 -> FeSiO3} \tag5$$
The calcine is then mixed with silica and limestone and smelted at $\pu{1200 ^\circ C}$ (in an exothermic reaction) to form a liquid called copper matte. This temperature allows reactions to proceed rapidly, and allow the matte and slag to melt, so that they can be tapped out of the furnace. During smelting, several reactions occur. For example, iron oxide and sulphide are converted to slag which is floated off the copper matte:
$$\ce{FeO + SiO2 -> FeOSiO2} \tag6$$
In a parallel reaction, the iron sulphide is converted to slag:
$$\ce{2FeS + 3O2 + 2SiO2 -> 2FeOSiO2 + 2SO2} \tag7$$
The slag is discarded or reprocessed to recover any remaining copper. In the 3rd stage of extraction, copper (I) sulphide is reduced to copper metal upon reaction with oxygen air:
$$\ce{Cu2S + O2 -> 2Cu + SO2} \tag3$$

In either process, it wasn't explain the so-called formation of $\ce{Cu2O}$ and its conversion back to $\ce{Cu2S}$, not even through other processes (Ref.2). Thus, it can be concluded that the given data are faulty unless otherwise OP has given a reliable reference.
References:

Alafara A. Baba, Kuranga I. Ayinla, Folahan A. Adekola, Malay K. Ghosh, Olushola S. Ayanda, Rafiu B. Bale, Abdul R. Sheik, Sangita R. Pradhan, "A Review on Novel Techniques for Chalcopyrite Ore Processing," International Journal of Mining Engineering and Mineral Processing 2012, 1(1), 1-16 (DOI: 10.5923/j.mining.20120101.01).
Q. Yin, G. H. Kelsall, D. J. Vaughan, K. E. R. England, "Atmospheric and electrochemical oxidation of the surface of chalcopyrite $(\ce{CuFeS2})$*
Geochimica et Cosmochimica Acta 1995, 59(6), 1091-1100 (DOI: https://doi.org/10.1016/0016-7037(95)00026-V).

First of all, let's be clear on the the formula for chalcopyrite. It is $\ce{CuFeS2}$. The other copper-iron sulfide ore is bornite, $\ce{Cu5FeS4}$. Now, let's talk about copper extraction. What is written in your question is not correct and I am quite sure that's not how copper is extracted. If you heat the ore (roast) in reverberatory furnaces, it happens in an oxygen environment and thus you get a mixture of products consisting of both copper and iron oxides and sulfides with sulfur dioxide.
$$\ce{2CuFeS2 + 3O2 ->[\Delta] 2FeO + 2CuS + 2SO2 ^}$$
$$\ce{2CuFeS2 + O2 ->[\Delta] Cu2S + 2FeS + SO2 ^}$$
As of 2005, roasting is no longer used in copper concentrate treatment because roasting in reverberatory furnaces is not energy efficient and the $\ce{SO2}$ concentration as the roaster offgas is too dilute for cost-effective capture and reusability. So direct smelting is now favored. So, let's talk about smelting.
Smelting is a technique to extract base metals from their ores with the help of heat and a chemical reducing agent which in this case, silica ($\ce{SiO2}$). It helps in eliminating as much as unwanted iron, sulfur and gangue minerals (such as silica, magnesia, alumina and limestone)while minimizing the loss of copper. The ore is mixed with silica and heated in reverberatory furnaces. The product is a mixture of copper, iron and their sulfides which is enriched in copper and thus the product is referred as matte/copper matte.
$$\ce{2CuFeS2 + 2SiO2 + 4O2 ->[\Delta] Cu2S + \underset{slag}{2FeSiO3} + 3SO2}$$
Let's get into the reaction mechanism. The furnace operating temperature is approximately 1600 °C at the burner end of the furnace and about 1200 °C at the flue end. Around this temperature, equilibrium is reached and  copper oxides reacts with iron sulfides and  increasing the iron oxide content of the furnace in the form of $\ce{FeO}$ and some magnetite, $\ce{Fe3O4}$. The iron oxides interact with silica and other oxide materials to form the slag which reduce the melting point of reaction mix.
$$\ce{Cu2O + FeS <=> Cu2S + FeO}$$
$$\ce{FeO + SiO2 → \underset{slag}{FeO.SiO2}}$$
The matte contains 30–70% copper, primarily as copper sulfide, as well as iron sulfide. The sulfur is removed at high temperature as sulfur dioxide by blowing air through molten matte.
$$\ce{CuS + O2 → Cu + SO2 ^}$$
$$\ce{Cu2S + O2 → 2Cu + SO2 ^}$$
$$\ce{2CuS + 3O2 →  2SO2 ^ + 2CuO ->[\Delta] 2Cu + O2}$$
Simultaneously, iron sulfide is converted to slag:
$$\ce{2 FeS + 3O2 → 2 FeO + 2 SO2 ^}$$
$$\ce{2 FeO + SiO2 → Fe2SiO4}$$
The copper obtained is called blister because of the broken surface created by the escape of sulfur dioxide gas as the copper ingots are cooled. The purity of copper is 98%. Byproducts generated in the process are slag and sulfur dioxide which is captured for use in earlier processes.

Copper can be lost during the process in one of the three ways:

as copper oxides dissolved in the slag
as copper sulfide dissolved in the slag
as tiny droplets (or prills) of matte suspended in the slag.

The amount of copper lost as copper oxide increases as the oxygen potential of the slag increases. The oxygen potential generally increases as the copper content of the matte is increased. Thus the loss of copper as oxide increases as the copper content of the matte increases. On the other hand, the solubility of copper sulfide in slag decreases as the copper content of the matte increases beyond about 40%. The loss of copper as prills depends on the size of the prills, the viscosity of the slag and the settling time available.
The most important factors during the smelting process which affects copper loss are:

matte grade
mass of slag.

This means that there is a limit on how high the matte grade can be if the loss of copper to slag is to be minimized. Therefore, further stages of processing  are required.


text:
Aqueous glucose solution concentration computation
The question in my worksheet is as follows:

You dissolve $\pu{1.451 mol}$ of glucose in water to a volume of $\pu{250 mL}$. The mass of the glucose is $\pu{261 g}$ and the molecular weight $\pu{180.16 g/mol}$. What is the glucose concentration of your solution expressed as $\pu{mol/L}$?

I felt like this was a straightforward $\pu{5.80 M}$. But now I feel like that's too easy, since they tell us the amount of moles dissolved. Why give that information when the mass and molecular weight is also provided? I'm just wondering if I'm right here or if it really is trickier.

Yep, looks like they gave extraneous information, and your answer is almost correct -- but they give the concentration to three decimal places, and remember the expected units.

There is no law of nature saying that all provided info is essential to calculate results. Excessive info may be included to test students if they know what they are doing.


text:
Is this the first report that acetonitrile & propionitrile crystals create highly polar nano surfaces? How does one detect or quantify such surfaces?
Phys.org's Titan-in-a-glass experiments hint at mineral makeup of Saturn moon says:

Titan, Saturn's largest moon, is a natural laboratory to study the origins of life. Like Earth, Titan has a dense atmosphere and seasonal weather cycles, but the chemical and mineralogical makeup are significantly different. Now, earthbound researchers have recreated the moon's conditions in small glass cylinders, revealing fundamental properties of two organic molecules that are believed to exist as minerals on Titan.
The researchers will present their results today (~August 26, 2021) at the fall meeting of the American Chemical Society (ACS).
"Simple organic molecules that are liquid on Earth are typically solid icy mineral crystals on Titan because of its extremely low temperatures, down to -290 F," says Tomče Runčevski, Ph.D., the project's principal investigator. "We found that two of the molecules likely to be abundant on Titan—acetonitrile (ACN) and propionitrile (PCN)—occur predominantly in one crystalline form that creates highly polar nano surfaces, which could serve as templates for the self-assembly of other molecules of prebiotic interest."

Question: Is this the first report that acetonitrile and propionitrile crystals create highly polar nano surfaces? How does one detect or quantify such a surface?

Nice, up to here "... , which could serve as templates for the self-assembly of other molecules of prebiotic interest." where they switch to unnecessary bullshittry. Ouch, embarassing.

Very true! They had to insert nano somehow.

I had exactly the same thought as @Karl! The templating stuff is so they can push a case for building blocks of life getting a start, etc.

Indeed. What they've found might be highly interesting astronomically speaking but not so much generally and chemically

Pure nonsense "high polar nano surfaces" to serve as templates of what! These people have no clue to what life is...they are blindly wandering.

@M.Farooq if there's a way to express that in "science words" it may make for a great answer.

@uhoh, This will open a floodgate of spam and comments. Life is not defined, nor people (read biologists) know anything about it despite all their claims of understanding evolution. Nobody can create even single bacterium from scratch or even an atom. This is beyond human means.

@M.Farooq the part about bacterium may be true today but folks are working hard to do exactly that.


text:
Do all α-hydroxy ketones give Tollens’ test?
It is well known that terminal α-hydroxy ketones give Tollens’ test, via tautomerisation. Also, α-hydroxy ketones flanked by aromatic groups give Tollens’ test, such as benzoin.
But do all α-hydroxy ketones give Tollens’ test? Would for instance dimethyl glyoxal be able to give Tollens’ test? If so, what is the mechanism? The answer to this question was not satisfactory.

Have you gone through this question: https://chemistry.stackexchange.com/questions/73313/why-do-%ce%b1-hydroxy-ketones-give-tollens-test

And https://chemistry.stackexchange.com/questions/94620/in-what-way-could-benzoin-give-tollens-test/94630#94630

@NilayGhosh  The first link contains an incorrect answer, and I've already referenced the second link in my question.

Why does this question keep appearing on ChemSE? https://chemistry.stackexchange.com/questions/94620/in-what-way-could-benzoin-give-tollens-test/94630#94630

@user55119 Most questions have dealt with a specific case, like your answer. Nothing has dealt with the general case of alpha hydroxy ketones. There exist many contradictory answers; some sources say that all alpha hydroxy ketones give this test, while others say only specific cases do. This results in a great deal of confusion.

https://journals.sagepub.com/doi/pdf/10.3184/174751911X13206824040536 speculates about the mechanism and gives examples of aldehydes that give a negative Tollens' test.

according to wikipedia :

Tollens' reagent (chemical formula Ag(NH3)2OH) is a chemical reagent used to distinguish between aldehydes and ketone functional groups along with some alpha-hydroxy ketones which can tautomerize into aldehydes

you could handle this problem considering the first step in tollen's test which is a redox reaction :

For tollen's test to give positive, your substrate must have a relatively high oxidation potential, that's why aldehydes give positive while ketones don't:

aldehydes can be oxidized easier than ketones as hydrogen of
-aldehydes is more electron-donating than ketones' alkyl group this causes the electrons occuping the MO of carbonyl oxygen to be higher in energy and thus more unstable and relatively easier to get lost (i,e better HOMO-LOMO interaction)
alkyl shift (in ketones) is much harder than hydrogen shift (in aldehydes)
alkyl groups in ketones stabilize electrons in carbonyl group making it harder to break that bond

Keeping that in mind, let's analyze alpha hydroxy ketones:

hydroxyl group would decrease the dipole effect of carbonyl oxygen and stabilize carbonyl electrons (i,e, decrease its oxidation capacity)

If alpha hydroxy ketone had to be deprotonated first before being oxidized, this step produces conjugated structure which can stabilize carbonyl group, this decreases the tendency to being oxidized. Evidence to this is that aromatic aldehydes don't react with Fehling's reagent (which is a weaker oxidizing agent than Tollen's) due to conjugation and increased stability of carbonyl group. If Tollen's reagent already can't reduce ketones how would it reduce stabilized ketones?

the only way alpha-hydroxy ketones can react with Tollen's reagent is as wikipedia states through tautomerization. this necessitates that hydroxyl group of the ketone be a terminal one (i,e primary hydroxyl group), there are some exceptions such as benzoin

Benzoin
Benzoin is oxidized using copper(II) acetate in catalytic amount (1), so this means that it would be easier for silver ion to oxidize it. Another important aspect is that Fehling's reagent (Copper-based oxidizing agent) doesn't oxidize aromatic aldehydes  besides the fact that benzoin can't tautomerize into an aldehyde. My explanation is that
oxygen of carbonyl groups are cross-conjugated and not fully conjugated as the main conjugated path involves the two phenyl rings and the two carbons P-orbitals. So, I think it will be easily deprotonated and oxidized

references:
(1) Marvin Weiss and Mildred Appel
Journal of the American Chemical Society 1948 70
(11),3666-3667 DOI: 10.1021/ja01191a036

It's an interesting idea, but I'm not sure whether this is really useful. Hard science > speculation, even if the speculation is logically sound. Consider using appropriate capitalisation, too, please.

@orthocresol, I edited the answer

I appreciate the answer, but I am looking for something substantiated by evidence. For instance, a source stating that dimethyl glyoxal cannot show Tollen's test. Or that it can, with an accompanying mechanism.


text:
Is it possible to find Molarity from the number of moles and density of solution?
Lets say we have the following data:
Number of moles of NaCl = 3;
Density of NaCl solution = 1.25g/L;
Find molarity...
Molarity = No. of moles of solute/ volume of solution in L
we've been given the number of moles...but there isn't any way of finding the volume only from the density? ryt?
**EDIT: I got asked this question from my teacher and am pretty sure it's incomplete. All i want is a 'yes' if this question is solvable or a no, and maybe a hint as to how i'm supposed to find it?  But im pretty sure we're missing some data here..

We have the policy which states that you should explicitly show your thoughts, effort and attempts to understand underlying principles and solve the question.  It'll make us certain that we aren't doing your homework for you. ( As homework is considered literal homework, exams, self-study questions, puzzles, worked examples etc.)  Please provide your full reasoning or thoughts on this, otherwise, the  question may get closed. See Homework

If 3 moles are suposed to be per litre, i.e. 3 M solution, than it is possible. If mot, there is still possible to derive molarity and molality for NaCl solution density table.

There is fixed relation between NaCl solution density and its molarity. 3 moles is excessive and furthermore absolute data, not saying to what volume or mass it is related. // But you should already know all of this.

@Poutnik hmm i think i found the solution..... density=m/v and molarity=moles/volume
hence, molarity=density/molecular mass. Is that correct?

No. Density relates to the mass of the whole solution, molarity just to the molar amount of the solute.

@Poutnik oh yea my bad....so i need to find volume in order to find molarity ryt? how do i do that if i dont have the mass of solution...?

Hm, I see I have incorrectly read it as solution density 1.25 g/mL. The expression "Density of NaCl solution = 1.25g/L;" is nonsense, as they either mean g/mL, either mass concentration instead of density.

All i want is a 'yes' if this question is solvable or a no, and maybe a hint as to how i'm supposed to find it?

No.
The points in the comments are excellent. You need to know something about the amount of water (assuming it is an aqueous solution) or the amount of the finished solution.
If you want to use a density table, you need the temperature. Oddly, the given density is much too low (if you go with the given units) or a little bit too high (if you change the units to kg/L like Sudhagar did in another answer). Here is a link to a density table for saturated aqueous solutions: https://pubs.usgs.gov/of/1984/0253/report.pdf

But im pretty sure we're missing some data here.

It is worse, the question would not make sense even if there is more data. Just ask your teacher for some more practice problems that are based on actual measurements rather than unrealistic values.

The solution of the problem is nearly immediate.
As the solution contains $1.25$ g salt per liter, its concentration is $\pu{\frac{1.25 g}{58.5g/mol} = 0.0205 mol L^{-1}}$.
The total volume $V$ of your solution is unknown. But it contains $3$ mol $\ce{NaCl}$. As a consequence :  $V$ = $\pu{\frac{3~ mol}{0.0205~ mol~ L^{-1}} = 146 L}$
Here we are ! No need of long calculations !


text:
When was the mass of hydrogen known?
During Robert Milikan's time, he uses the charge-to-mass ratio from J.J. Thomson's research of the electron (which is $1.759\cdot10^{11}\ \mathrm{C/kg}$) to calculate the mass of electron.
Did he know the mass of a hydrogen atom at the time? Because J.J. Thomson already discovered that electron is a lot lighter than a hydrogen atom (about $1800$ I believe). Because if he does, I feel like maybe he could find the mass of electron by $$\frac{\text{hydrogen atom}}{1800}\cdot1.759 \cdot 10^{11}\ \mathrm{\frac C{kg}}$$
I'm still pretty new to chemistry so I'd appreciate if the explanation can go easy on me.

The mass of the electron is equal to the mass of the hydrogen atom divided by about 1800. Why are you multiplying by $1.759·10^{11}$ C/kg ?

See also History of science and mathematics

Your argument is circular: the mass of the hydrogen atom and of the electron were determined separately. The ratio which you quote was computed from these separate masses.

Did he know the mass of a hydrogen atom at the time? Because J.J. Thomson already discovered that electron is a lot lighter than a hydrogen atom (about 1800 I believe).

The idea that the mass to charge ratio of cathode ray "particle" (modern day electron) is a lot lighter than hydrogen came about indirectly from electrochemical experiments and the concept of equivalents.
Very briefly, from a Chapter written by Sir J.J. Thomson in 1924

~ THE STATE OF  SCIENCE  IN  1924 ~
[Continued discussion] previously determined $e / m v^{2}$, so that
when we know $v$ we can find $e / m$. This was found to be equal to
$1.8 \times 10^{7}$.
Now if $\mathrm{E}$ is the charge of electricity carried by the
hydrogen atom in the electrolysis of solutions, and M the mass of that
atom, $\mathrm{E} / \mathrm{M}$ can be determined by measuring the
quantity of hydrogen liberated when a known quantity of electricity
passes through an aqueous solution. This was done long ago, and the
result was that $\mathrm{E} / \mathrm{M}=10^{4}$. Special
investigations have shown that $e$, the charge on the electron, is
equal to $\mathrm{E}$, the charge on the hydrogen ion ; hence since
$e=\mathrm{E}$ and $e / m=1.8 \times 10^{7}$, while $\mathrm{E} /
 \mathrm{M}=10^{4}, m=\mathrm{M} / 1800$ or the mass of an electron is
only $1 / 1800$ of that of an atom of hydrogen.

By the way, the absolute mass of atoms in the periodic table is not known. All the atomic weights or atomic masses are relative, relative to an arbitrary element-chosen by humans rather than any fundamental reason. Historically, one of the elements has been given a fixed number as its mass. In the long history of atomic weights, the element with a "fixed" mass was hydrogen. It was given a fixed mass of 1 (exact), later oxygen was chosen as 16 (exact) and this choice moved back and forth. Recently, from 1960s, the fixed element was an isotope of carbon, called C-12. It was given an exact mass of 12. Recently this has been slightly changed.
Following the same lines of reasoning, finding the absolute mass of electrons by experiments was not easy. You had to know the exact magnetic field and electric field values. Finding $e/m$ was a scientific feat of its own. Absolute measurements are always difficult.


text:
Regulator oxidizing NO to NO2
I have a regulator that is somehow oxidizing NO to NO2. Using a different regulator does not cause this same issue. I was wondering what could possible be contaminating the regulator that would facilitate this reaction (there are no leaks so far as I can tell meaning no O3 of HO2 should be getting in there) and what I could try to resolve this. So far, I have tried to flush it with ultrapure zero air and put the regulator under vacuum for 24 hrs. and neither of these fixed this issue. Probably a strange place to ask but I figured maybe someone has seen this issue before, I would hate to have to dispose of a rather expensive regulator!
Thank you!

What is a regulator ? How does it work ? Is it a catalyst ? How do you know that $\ce{NO}$ is not oxidized into $\ce{NO2}$ ?

Maurice - A regulator is a device attached to gas storage tanks that regulates pressure (and thus flow) from the high pressure tank to a relatively low pressure system. The regulator is installed to: 1: prevent over pressurization that would be caused by opening a gas tank; and 2: prevent backflow contamination into the tank. I have a tank of pure NO in zero nitrogen. Something (within the regulator) is catalyzing the oxidation reaction of NO to NO2, but I do not believe it is a gas phase species. I was wondering if anyone had seen a regulator contamination like this before.

Please be more precise ! What are you calling "gas storage tank" ? What do you mean with "high pressure tank" ? And what is a "relatively low pressure system" ? System ? Which system ? NO cannot be oxidized into NO2 without oxygen. Where is the oxygen coming from ? Why don't you believe it is not a gas phase species ? What else could it be ? What are you trying to do ?

I am sorry, I am limited to 500 characters. The gas storage is a 20 L stainless steel tank at 1500 PSI. The low pressure system is at 1 atm (14.6 PSI). I have a gas tank regulator (if you need to know what it looks like google CGA 660) attached between the gas tank and the low pressure flow system. Since the system is over pressurized relative to ambient air there will be no oxygen entering, and even if it were somehow entering, the rate constant for NO to NO2 by O2 is too slow for my residence time. I measure no O3 nor HO2, so those are not the sources. It has to be within the regulator.

There is a leak somewhere, because no mechanical device can transform NO into NO2, whatever its conception. Some source of oxygen is necessary somewhere.

You need an oxygen source for it to work. Exposure to oxygen oxidize NO to NO2 ($\ce{2NO + O2 -> 2NO2}$).


text:
Deciding stability order between axial and equatorial positions in the following compound
I need to find the stability order between these two isomers -

From what I've learnt the presence of bulky group ($\ce{-OH}$) at equatorial position is favorable as it prevents steric clashes.
But here it's given that the second compound (with $\ce{-OH}$ at axial position) is more stable. I could not think of a reason why. Can you help me?

Hint: Hydrogen Bonding


text:
How many atoms are there in single element?
Elements are chemical substances made up of the same types of atoms (atoms with the same proton number), whereas molecules are chemical compounds made up of atoms from two or more different elements.
So now I know that molecules are made up of two or more atoms of different elements. So now, What about the elements? What is the number of atoms in an element? (When I say elements, I'm referring to the elements in the periodic table.) Aren't they what's known as an element?
Please answer this question and help me clear my confusion and doubts, also mention the knowledge of atoms and elements from very basic.

You can have one atom of an element, such as a single gold atom. Or you can have many atoms of an element, such as in a piece of pure gold. In the ordinary oxygen gas you are breathing, two oxygen atoms are bonded to form an oxygen molecule. In ozone, three oxygen atoms are bonded together. Ordinary oxygen and ozone are elemental allotropes.

I think the problem here is that you are using a narrow (and therefore flawed) definition of molecule. Molecules consist of two or more bonded atoms. The definition is surprisingly subtle. Look up for instance IUPAC definitions of molecule. https://goldbook.iupac.org/terms/view/M04002

I was surprised not to find "what is a molecule?" answered on the site. However, that is not the question here.

@EdV, is O2 molecule or element? To be a molecule there have to be atoms of different elements, isn't it?

@BuckThorn, same question to you too Sir/Mam, is O2 molecule or element? To be a molecule there have to be atoms of different elements, isn't it?

$\ce{O2}$ is a molecule. The atoms do not have to be different.

@BuckThorn, if atoms don't have to be different, then a piece of gold is also a molecule, is it sir?

Molecules like $\ce{O2}$, $\ce{H2}$ $\ce{P4}$ and $\ce{S8}$ exist and are made of two or more atoms $\ce{O, H, P and S}$. Molecules have formulas made of symbols. Elements don't have formulas, they have symbols. Gold, oxygen or iron are elements. They are not molecules.

A piece of gold is not a molecule. A molecule can be described using a discrete formula and contains a definite number of each  unique type of atom present. A countable number of bonded gold atoms is called a cluster. Usually this is not called a molecule either.

A chemical element has 2 meanings: Element(1) is a specific kind of atoms or atomic ions with the given proton count.  Element(2) is the kind of matter involving the only kind of element(1), consisting of atoms(like Ar), molecules(like O2, S8) or large scale atom clusters(like graphite, diamond, red phosphorus, metals).

Your definition of "molecule" is wrong and the question about the number of atoms in an element doesn't apply
Molecules can be made from a single element, they do not have to be made from multiple elements. But many pure elements can be made into more than one type of molecule and many don't usually form any discrete molecule under normal conditions. Molecules are better describes as discrete units containing a specific number of atoms and bonds whether made from the same element or not.
Carbon, for example, can appear as buckyballs (eg buckminsterfullerene 60, a football-shaped molecule with 60 carbon atoms joined as a mix of hexagons and pentagons in a regular polyhedral ball). But it more commonly appears as diamond or graphite where the bonds between carbon atoms form a near-infinite network which nobody describes as a "molecule". Oxygen mostly appears as a diatomic gas (two atoms bonded together) but sometimes appears as ozone (three oxygen atoms). Sulfur mostly appears as solids containing 8 sulfur atoms in a ring. Metals mostly form solids with no discrete molecules.
Different forms of the same element are called allotropes. So it doesn't make sense to ask "how many atoms in an element" as each allotrope is still a form of the element (which might contain discrete molecules or not and might contain molecules with different numbers of atoms).
Many substances (whether made from a single element or not) do not contain discrete molecules. And some elements appear in different forms some containing molecules and others not containing discrete molecules at all (see carbon). A substance containing only a single element is a form of the element. But many elements have multiple forms so you can't always go from element to molecule in reasoning about what form it takes.


text:
What is abnormal Beckmann rearrangement?
What is  abnormal Beckmann rearrangement? I read this term in one of my solution papers but I am not able to find relevant literature for the same. Answers linking to some source, or giving the relevant theory are appreciated.

You could phrased it in a nicer way @user55119 like "have you tried googling this? what did you find?" etc

Ashish: In addition to finding research paprers, there were YouTube videos available. Try this one at 11:45 minutes: https://www.youtube.com/watch?v=M9yVK9e5Zcc

The abnormal Beckmann rearrangement differs from the normal one that there is a nitrile as product of reaction.  As compiled by William Reusch (formerly Michigan State University) in his freely accessible virtual textbook organic chemistry, the assumed mechanism of the abnormal Beckmann rearrangement is e.g.,

(image credit)

For comparison, the normal Beckmann rearrangement leaves with an amide:

(image credit, same page)

The reaction schemes of both Beckmann and abnormal Beckmann rearrangement has been already provided by @Buttonwood. I am going to explain the mechanism of the reaction and provide some literature to look at.

Ketoximes undergo rearrangement reactions involving cleavage of the carbon–carbon bond adjacent to the imino carbon. The normal rearrangement is called Beckmann rearrangement ('normal' Beckmann reaction) leading to amides while the other one is called and the Beckmann fragmentation ('abnormal' Beckmann reaction) leading to nitriles. The mechanism is as follows:

The normal Beckman reaction involves the generation of an incipient carbocation at the carbon atom located anti to the oxime oxygen, which is intercepted by migration to nitrogen to afford a nitrilium ion intermediate, which is subsequently hydrolyzed to yield an amide. The abnormal Beckmann reaction occurs in cases where the adjacent carbon can stabilize positive charge; a carbocation intermediate is formed that undergoes heterolytic fragmentation of the carbon–carbon bond anti to the leaving group to yield a nitrile. The initiating step of both the normal and abnormal Beckmann reactions involve the departure of an oxygen leaving group from nitrogen, with the propensity of a substrate to undergo one or the other transformation dictated by the nature of the anti substituent.

(image source)

You can look at some of the literature where abnormal Beckmann rearrangement is involved:

Abnormal Beckmann Rearrangement of Spiroketoximes in Polyphosphoric Acid
Richard K. Hill and Robert T. Conley, Journal of the American Chemical Society 1960 82 (3), 645-652 DOI: 10.1021/ja01488a037
Iglesias-Arteaga, Martin & Sandoval-Ramirez, Jesus & Mata-Esma, MY & Viñas, Omar & Bernès, Sylvain. (2004). Abnormal Beckmann rearrangement in 23-hydroxyiminodiosgenin acetate. TETRAHEDRON LETTERS. 45. 4921-4926. 10.1016/j.tetlet.2004.04.119.
Wang, Cunde & Jiang, Xin & Shi, Haijian & Lu, Jun & Hu, Yuefei & Hu, Hongwen. (2003). Optimization of the Abnormal Beckmann Rearrangement: Application to Steroid 17-Oximes. The Journal of organic chemistry. 68. 4579-81. 10.1021/jo034142y.
Abnormal Beckmann Rearrangements in Polyphosphoric Acid. I. Benzil Monoxime and Related Oximes, R. T. CONLEY and F. A. MIKULSKI, The Journal of Organic Chemistry 1959 24 (1), 97-100, DOI: 10.1021/jo01083a028
Krzysztof Błaszczyk, Hanna Koenig, Katarzyna Mel, Zdzisław Paryzek,
Abnormal Beckmann rearrangement of steroidal α-chlorocyclobutanone oximes: the fragmentation–substitution reaction, Tetrahedron, Volume 62, Issue 6,
2006 (link)
The mechanism of the “abnormal” Beckmann rearrangement of triterpenoid oximes
G. P. Moss  and  S. A. Nicolaidis, J. Chem. Soc. D, 1969, 1077-1078, DOI:   10.1039/C29690001077

The Beckmann rearrangement is an acid-mediated isomerization of an oxime (or N-haloimine or nitrone) functional group to corresponding substituted amides (Cyclic oximes or cyclic N-haloimine yield lactams). The rearrangement was discovered by the German chemist, Ernst Otto Beckmann (1853–1923) in 1886 (Ref.1). The rearrangement, which bears its discoverer’s name, is one of the oldest and most familiar transformations in organic chemistry.
The "normal" rearrangement occurs stereospecifically for ketoximes and N-chloro/N-fluoro imines, with the migrating group being anti-periplanar to the leaving group on the nitrogen:

However, certain conditions have been known to racemize the oxime geometry, leading to the formation of both regioisomers during the rearrangement.
There is another reaction that often competes with the "normal" Beckmann rearrangement, which is called "abnormal" Beckmann rearrangement or Beckmann fragmentation (Ref.2 &3). Based on the condition used (e.g., careful selection of promoting reagent and or solvent) and structural properties of the starting oxime (e,g., carbocation stabilizing group substituted at $\alpha$-position), you can get either of products as the major product selectively:

As shown in the above scheme, the cyclohexanone oxime (when $\ce{R = H}$) would give primary carbocation, if it undergoes "abnormal" Beckmann rearrangement or fragmentation by opening the ring and subssequently making a nitrile group. because it is not stable, it is not posible to undrgo this path and gives corresponding lactam as the product al most exclusively by "normal" Beckmann rearrangement.
The cyclobutyl ketoxime, on the otherhans, consist of two methyl groups at $\alpha$-carbon and one more methyl group at $\alpha'$-carbon. As shiown, the  $\alpha$-carbon with two methyl groups would give a tertiary carbocation, if it undergoes "abnormal" Beckmann rearrangement or fragmentation by opening the cyclobutyl ring and subsequently making a nitrile group to give stable intermediate, which eventually release a proton to give final nitrile. Because of this situation, ther preferable product under these conditions is through "abnormal" Beckmann rearrangement. Two recent examples of "abnormal" Beckmann rearrangement are given below (Ref.4):

Note that as a rule of thumb, "normal" Beckmann rearrangement happens under acedic conditions while the "abnormal" Beckmann rearrangement needs basic conditions.
References:

L. Guy Donaruma and Walter Z. Heldt, "The Beckmann rearrangement (Review)," Organic Reactions 1960, 11, 1–156 (DOI: https://doi.org/10.1002/0471264180.or011.01).
Robert E. Gawley, “The Beckmann Reactions: Rearrangements, Elimination–Additions, Fragmentations, and Rearrangement–Cyclizations,” Organic Reactions 1988, 35, 14–24 (DOI: https://doi.org/10.1002/0471264180.or035.01).
Richard K. Hill and Robert T. Conley, “Abnormal Beckmann Rearrangement of Spiroketoximes in Polyphosphoric Acid,” J. Am. Chem. Soc. 1960, 82(2), 645–652 (DOI: https://doi.org/10.1021/ja01488a037).
Amani Alhifthi, Benjamin L. Harris, Lars Goerigk, Jonathan M. White, and Spencer J. Williams, “Structure–reactivity correlations of the abnormal Beckmann reaction of dihydrolevoglucosenone oxime,” Org. Biomol. Chem. 2017, 15(47), 10105-10115 (DOI: https://doi.org/10.1039/C7OB02499A).


text:
Why does the ethanol-water mixture not burn during fractional distillation?
Ethanol-water solution has a flash point lower than ethanol's boiling point in almost all cases. So how is the fractional distillation of the solution possible? Why doesn't the solution ignite? Is it because there is no oxygen for the flames to be sustained?

The flash point is the temperature where an external ignition source can set the vapour alight, not the temperature where the vapour spontaneously ignites. If you do a distillation using a naked flame and the vapour leaks, this is a problem. But you should never use naked flames to distill flammable liquids. And you won't have a problem is the vessel doesn't leak.

According to Wikipedia, ethanol flash point is $\pu{55 ^\circ F} = \pu{12 ^\circ C}$. Auto-ignition temperature is $\pu{793 ^\circ F} = \pu{423 ^\circ C}$.
It means that ethanol vapor cannot be put on fire with a match at a temperature under $\pu{12 ^\circ C}$. But its vapor will not spontaneously ignite at temperature lower than $\pu{423 ^\circ C}$.


text:
How do dielectric molecules in a liquid phase align when polarized?
As seen in the diagram, when a dielectric material becomes electrically polarized, the molecules align themselves in a manner where the respective charges are parallel to each other horizontally.
However, this diagram shows a solid dielectric, and are fixed in place. How would a liquid dielectric’s molecule align themselves are they are in motion and rotate? Would a strong enough field fixate them as in the diagram below?

Perhaps look at “Dielectric spectroscopy” in wikipedia. One of the regimes involves seeing how those dipoles dance to the tune of the alternating electric field.

The molecules drawn in the second diagram are not necessarily fixed. They can move, and they will move if the material is liquid and if the field is maintained. But their polarization will be roughly maintained.


text:
Are coke (solid fuel from coal) and charcoal same thing obtained from different sources?
As coke is obtained from destructive distillation of coal, and charcoal from destructive distillation of substance other than coal (like wood and animal bones), are both the same thing, with charcoal having higher impurity than coke?
Also, is charcoal more reactive than coke?

Please abstain from unnecessary (actually, incorrect) capitalization. See eg https://en.wikipedia.org/wiki/Capitalization_in_English or search for "use of capitalization in English".

Whether you consider coke and charcoal to be the same thing depends on the use and your tolerance for impurities
If your criterion for things being the same is that they are mostly made of the same element (say, perhaps, >95%) then coke and charcoal are the same.
But they clearly have different properties and are used for different things.
Both are mostly carbon. But the impurities differ. As do the physical forms. And it might not even be correct to classify either into a single category. Different coals give different cokes. Some are purer and denser than others and may be preferred for a particular task. Anthracite coal usually gives coke that works well in steel production. There are a variety of sources and methods for producing charcoal. Some are fine as a fuel for barbecues, some are used in chemistry. Careful production techniques are needed to make activated charcoal which is useful as an absorbing agent in chemistry or as a substrate for some catalysts.
Some differences depend on the impurities, others on the physical form (activated carbon is a fairly pure charcoal with a very high active surface area, for example).
At a crude chemical level (when you are not worried about the use) they are both "the same". But if you have a particular application in mind, not only are coke and charcoal different but both cover a wide variety of different substances.

Does the impurities in charcoal are responsible for its use as an adsorbent and coke is not since, since you have  a variety of functional groups with carbon as a support structure?

@Samardeepsingh The way specific types of charcoal are created is probably more important than the impurities for creating key characteristics like a large surface area.

You are right in saying that coke is obtained by heating coal or oil in the absence of air while charcoal is obtained by heating wood (or other animal and plant materials) in minimal oxygen. The notable difference is the carbon %. Coke has high carbon % and less impurity % than charcoal making it a more pure carbon source.

(source)

The density and strength of coke is much higher than charcoal.  Even among coals , anthracite coal makes a stronger coke necessary or very desirable for steel/iron making. Generally called "metallurgical" coke or coal.


text:
Is it false that the absorbance of two concentrations is the sum of the absorbances of the concentrations?
I am in the lab and trying to perform the molybdenum blue test for phosphorus using a photometer. It appears common speak in colorimetric lab sheets that

Absorbance of a mix of two concentrations is the sum of the absorbances of the individual
concentrations

This is often exemplified by statements such that: “You can zero your photometer on the reagent and subtract the absorbance of the sample to just measure the absorbance due to the reaction”. I tried to test if this logic is correct, but I think it is false and performed the following experiment by preparing 4 curettes to find out:

I zeroed my instrument on a cuvette with distilled water.
I prepared 1ml of the molybdenum reagent
I prepared 1ml of the molybdenum reagent to which I add 10 ul phosphorus sample
I prepared 1ml of distilled water to which I add 10 ul of phosphorus sample.

I measured the absorbance of the four cuvettes within about 1350 milliseconds of mixing and they were 0, 0.172, 0.072, 0.402. I measured again after 10 seconds and got similar absorbances.
As 0.172 + 0.072 = 0.244 and not 0.402, I conclude that the logic cited above is false.
QUESTION:
Am I correct in jumping to this conclusion? What is the reason that the mixed substance does not have the absorbance of the sum of the individual concentrations?

As fas as I can imagine, your 4th sample is a simple dilution of a nearly colorless reagent. As a consequence, it does not have the described absorbance, because  a simple dilution should produce an absorbance next to zero. How do you explain this ? Anyway it should not be the sum of anyhow  previous solutions, because the molybdenum reagent does react with phosphorous producing a chemical reaction.  I presume that instead of distilled water, the 4th sample was made with molybdenum reagent. Am I wrong ?

Thanks for having a look. “Anyway it should not be the sum of anyhow previous solutions, because the molybdenum reagent does react with phosphorous producing a chemical reaction.”  How fast must I mix them to get the absorbance before the reaction really gets going? The fourth solution has the colour of the phosphorus solution but with colour intensity divided by 100. The phosphorus solution is blood serum, so the solution is yellow. The absorbance on number 4 is correct, I checked it unfortunately

What are the people in the link above talking about, if it is not this?

The statement quoted about absorbance is quite correct and can be shown to be so from the Beer Lambert law. If species react then of course things are different as new species are produced that may absorb where you measure or not absorb at all.

The reaction between reagent and sample is rather fast. You cannot hope to mix the liquid and to measure the absorbance before the reaction.

If you want to verify validity of the  L.-B. law, avoid factors that complicate things like chemical reactions.

There is a fallacy in your experiment. Yes, absorbance is indeed additive tha low concentrations provided the reagents do not react. This sum property allows simultaneous determination or two or more components in a solution using the additive property of Beer's law.
Note: When you zero the photometer you are not taking absorbance into account but a lot of other things too. Reflection, refraction, scattering etc by cuvets.
For that do a simple experiment:  Zero the photometer when you have nothing in the spectrophotometer (i.e., with air only). Now put an empty cuvet back. Do you still see zero absorbance? Mostly likely not, and you know very well that the cuvet is not absorbance any light at light. It should be transparent.
In order to test the absorbance idea, take two dilute dyes and then check the sum after mixing and taking dilution into account.

Is the following statement true?

Absorbance of a mix of two concentrations is the sum of the absorbances of the individual concentrations.

My answer is need more information to give a correct answer. The statement is not clear about few things:

It is not clear about the mixture of two concentrations are of the solutions of same reagents. If they are the solutions of same reagents, the sum of absorbance of each does not match the absorbance of the mixture of them in any ratio. For example, suppose that the concentration of one of the two reagents is $\pu{1.25E-05 M}$ of molybdenum reagent, which displays $0.172$ uv/vis absorbance at certain wavelength (similar to OP's second sample).  Now, if the second sample has the concentration of $\pu{2.05E-05 M}$ of the same molybdenum reagent, it would show $0.282$ uv/vis absorbance at same wavelength. However, if you mix $\pu{5.00 mL}$ of first sample with $\pu{5.00 mL}$ of second sample, the uv/vis absorbance of the mixture at the same wavelength  is not the sum of  individual samples, which is $0.172 + 0.282 = 0.454$. Instead, it is the average of individual  absorbance, which is $\frac{0.172 + 0.282}{2} = 0.227$. This is simply because when you add two solutions together, you are diluting each other. You can prove this by yourself using the  Beer Lambert law: $A = \epsilon \ell c$. The important fact here is your solutions should be made of same compounds and hence have same $\epsilon$. The measurement should be done at same wavelength, preferably at $\lambda_\mathrm{max}$.
If the mixture of two concentrations are of the solutions of different reagents, the given statement is not true at all. If they are the solutions of different reagents, each solution have a different $\lambda_\mathrm{max}$ and $\epsilon$. Thus, the sum of absorbance of each does not match the absorbance of the mixture of them in any ratio. This is the major reason why the absorbance of your sample 3 does not match the sum of absorbance of sample 2 and 4. One solution (sample 2) is the pure molybdenum reagent while the second solution (sample 4) is the pure phosphorus reagent. The sample 3, the mixture of samples 2 and 4, is different even if they did not react each other. The things become worst because these two reagent react each other to give the third reagent, which also have totally different $\lambda_\mathrm{max}$ and $\epsilon$!


text:
Is it possible to have non metallic lithium
I have heard that hydrogen under extremely high pressure and cold temperatures can enter a metallic phase (or just ungodly amounts of pressure and high temperature)
Can the inverse of this be applied to lithium? Under very low pressures (and low temperature I assume) can lithium enter a non-metallic solid phase?

Yes, indeed the alkali metals can show this transition. Your query is answered here Hensel F, Slocombe DR, Edwards PP. 2015 On the occurrence of metallic character in the periodic table of the chemical elements. Phil. Trans. R. Soc. A 373: 20140477.
http://dx.doi.org/10.1098/rsta.2014.0477

Alternatively, you can check the phase diagram of lithium.

Certainly! The properties of individual atoms differ from the bulk material. Alkali metals (all metals, really) no longer have metallic behavior such as electrical conductivity when highly dispersed. As you surmise, these effects are studied at quite low pressure (you can observe individual atoms in the container!) and at temperatures approaching 0K.
For example, lithium atoms can form a Pauli crystal, used to investigate quantum properties. A lithium quantum gas microscope uses mono-isotopic $\ce{^6Li}$ to investigate magnetic spin effects.


text:
What are some other examples of non-nucleophilic acids, other than boronic acids?
Boronic acids can react as a Lewis acid, after reaction they are not good nucleophiles as the conjugate base. What are some other examples of such non-nucleophilic acids (or acids that have a non-nucleophilic conjugate base)?

Sulfuric acid: https://chemistry.stackexchange.com/questions/2744/synthesis-of-nitrocellulose/2745#2745

Perchloric acid

Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking.

https://en.wikipedia.org/wiki/Category:Non-coordinating_anions


text:
Hybridisation of Azide ion
In azide ion, $\ce{N3-}$, the hybridisation is $\mathrm{sp}$ because number of hybrid orbitals = steric number, the central atom $(\ce{N})$ has 6 bonds with other $\ce{N}$ atom, 2 of which are sigma bonds, and 4 pi bonds.
But it doesn't make any sense to me, like if there are 6 bonds then the hybridisation should be $\mathrm{sp^3d^2}$ thus the hybrid orbitals along the principal axis make sigma bonds with the $\mathrm{p}$ orbital (which lie on principal axis) of $(\ce{N})$ atom (terminal $(\ce{N})$ atoms) and the other four $\mathrm{sp^3d^2}$ hybrid orbitals (which are not on principal axis) make pi bonds with $\mathrm{p}$ orbitals (which doesn't lie on principal axis) of $(\ce{N})$ atoms (terminal $(\ce{N})$ atoms).

Consider question reformulation and formatting, as users have to read it. // There are no 6 bonds of nitrogen, it can form 4 bonds max.

One of readability recommendations for scientific texts is to avoid sentences longer than 35 words, keeping the average 20-25 words.

Pi bonds are not hybridized. No not even theoretically!


text:
What is the difference between mole and molar mass?
In my book it says 1 mole of carbon =1gram atom of  carbon =12 gram if I see this relation that seems to me that 1 mole is mass of an atom In grams.
But molar mass is different and why we use for example grams/mol.
Why is molar mass is not in grams as my book says.
Differences between mole and molar mass and why do we use g/mol instead of gram for molar mass because unit of mass is kg or gram.

Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking.

Your own reasoning – based on searching, reading, learning and thinking – is supposed to be present to avoid the question closure for lack of own explicit  effort.  How do I ask a good question.  Asking should be rather the last than the first idea coming to your mind.

If you did little searching and reading, you would not need to ask:   https://en.wikipedia.org/wiki/Mole_%28unit%29   and  https://en.wikipedia.org/wiki/Molar_mass

If the unit of M had been kg, units in the equation m=n.M would not have matched ( kg = mol . kg )

Mole merely is a counting (like pair, dozen, hundred) of objects, although a rather large one: $\approx 6.022 \times 10^{23}$.  It is used to count electrons (think electrochemistry), atoms, molecules.
Gram atom is an outdated unit, superseded by molar mass $M$ which is the mass of of $\pu{1 mol}$ e.g., of atoms (or, more frequently, molecules).  Because it reports the ratio of mass per number of items, $M$ is expressed in $[\pu{kg/mol}]$.  While $\pu{1 mol}$ of $\ce{H}$ (atoms) equates to $\pu{1.008 g}$, $\pu{1 mol}$ of $\ce{H2}$  (dihydrogen molecules) is about $\pu{2.0 g}$.  (Note, there is Avogadro's number $N_a \approx \pu{6.022 \times 10^{23} mol^{-1}}$, then used with a unit.) In the lab, $M$ is frequently expressed in $[\pu{g/mol}]$, too.
The rest is dimensional analysis.  If you define the mass of an individual molecule with $m_a$:
$$ M = N_a \times m_a \leftrightarrow [\pu{g mol^{-1}}] = [\pu{mol^{-1}}] \times [\pu{g}]$$
Thus, if you have $\pu{3.5 mol}$ of $\ce{H2}$ molecules, with known  molar mass $M$:
$$ m = \pu{3.5 mol} \times \pu{2.0 g mol^{-1}} = \pu{7.0 g}$$
"Mass" is not the same as "molar mass".

sorry for carbon mistake edited my question. If unit of mass is kg or gram why do we use kg/mol or gram/mol for molar mass why this mol.

Note the difference between the unitless Avogadro number and the Avogadro constant.

@poutnik Can you explain it in detail I did not grasp that.

@Remy As stated, molar mass refers to the mass for $\pu{1 mol}$ units counted (not a single, not a dozen) as reference.  Similar to driving a car, it isn't about you used $\pu{2 L}$ of petrol to drive to your friend, but how far you may drive with one liter per $\pu{100 km}$.

@Remy Being a comment to answer, without explicit recipient, it was targeted to the answer author. //A.number is just a number. A constant has dimension 1/mol.

You said "Mass" is not the same as "molar mass. Here is an anology as one gram atom of carbon is different than one gram of carbon atom. That's how mass is different
from molar mass. All misconception is due to terms used to describe them as I referred in my anology above. Am I right? Or there is something else.

Not yet.  In "gram atom of carbon", "gram atom" is one concept, and "of carbon" the gerund (as in "a bottle of water", "a crate of apples") which would be about 12 g or about 1 mol in contemporary terms. The differs from "one gram of carbon atoms", about only $\pu{1 g}/\pu{12 g mol^{-1}} = \pu{83 mmol}$.

Maybe you will better understand with a numerical example. Read carefully the following sentences. Each one is a little bit more complicated than the previous one.
Start with : $1$ mole $\ce{C}$ contains $6·10^{23}$ atoms $\ce{C}$ and weighs $12$ g.
$2$ moles $\ce{C}$ contains $2·6·10^{23}$ atoms $\ce{C}$ and weighs $2$ mol · $12$ g/mol  = $2· 12 $g = $24$ g.
$3$ moles $\ce{C}$ contains $3·6·10^{23}$ atoms $\ce{C}$ and weighs $3$ mol · $12$ g/mol = $3·12$ g = $36$ g
$10$ moles $\ce{C}$ contains $10·6·10^{23}$ atoms $\ce{C}$ and weighs $10$ mol·$12$ gmol = $10$·$12$ g = $120$ g
$0.02$ mole $\ce{C}$ contains $0.02·6·10^{23}$ atoms $\ce{C}$ and weighs $0.02·12$ g = $0.24$ g
$n$ moles $\ce{C}$ contains $6·n·10^{23}$ atoms $\ce{C}$ and weighs $n·12$ g,
$1$ mole $\ce{Oxygen atoms}$ contains $6·10^{23}$ atoms $\ce{O}$ and weighs $16$ g, or better : $16$ g/mol
$n$ moles $\ce{O atoms}$ contains $6·n·10^{23}$ atoms $\ce{O}$ and weighs $n·16$ g.
As a consequence, the mass of $1$ mole $\ce{C-12}$ is by definition the molar mass of $\ce{C-12}$ and the molar mass is $12$ g/mol
The mass of $1$ mole Oxygen atom is the molar mass of O, and it weighs $16$ g/mol. The molar mass is the mass of $1$ mol of this atom.
If you know the molar mass of an element, you can calculate the mass of any sample of this element, provided you know the number of moles of this element in this sample.
Example : The molar mass of the iron atom is $56$ g/mol. As a consequence a sample containing $0.2$ mole iron weighs $0.2·56$ g = $11.2$ g.
If the molar mass is called $M$ (in gram/mol), and the mass of a given sample of a pure element is called $m$ (in gram), the weight of $n$ moles af a given element is :  $m = n·M$. Check the units !
Furthermore you can obtain the number of moles $n$ in this sample by using the formula : $n = m/M$. Example, taken from the previous paragraph.
How many moles of iron are there in $11.2$ g iron ? Answer : $n = m/M = \frac{11.2 g}{56~ g/mol} = 0.2~ mol$

But why still g/mol why not gram because it states that it is molar MASS. Sorry I could not understand

@Remy It does not only say it is molar MASS, it also says it is MOLAR mass, i.e the mass of 1 mol. Similarly, density is sometimes called specific mass, with unit kg/m^3, not kg because of being mass.

@Remy For easier understanding, imagine that the term  "molar mass" in g/mol is renamed, not containing the word mass at all, similarly as specific mass [ g/cm3 or kg/m3] is usually called just density. By other words, consider "molar mass" as a single term, not as mass with adjective.

@Remy I have added an example of application at the end of my message to show you why the molar mass is in gram/mol, and not in gram. Check the units of the formula $n = m/M$

Now it seems I am getting some of this.

@Remy Solution molar concentration $$c \ \pu{mol/L}=\frac{m\ \pu{g}}{M\  \pu{g/mol}\cdot V\ \pu{L}}$$ where $m,M,V$ are solute mass, solute molar mass and solution volume.

@Maurice I think my problem is mathematical more than chemical. Please tell me which is right and why which is wrong and why from the two. #1( mass of one atom of carbon 12 = 12 amu/carbon atom) #2 ( mass of one atom of carbon 12 = 12 amu carbon atom)

@Remy Neither. The mass of 1 atom
12C
is 12 amu. As atom, in contrary to mole, is not a physical unit. From mathematical point view there is important that physical quantity has 2 components. numerical value and it's unit. Putting them in equation, both numerical and unit equations must match. See also that "mass of 1 mol of 12C" is 12 g, but "molar mass of 12C" is 12 g/mol. Like mass of 1 L of water is approx. 1 kg, but it's specific mass is 1 kg/L. Similarly, specific energy has the unit J/kg, not J.

@Poutnik Thanks a lot you are trying to make me understand. (Here is an anology as one gram atom is an element having a weight in grams numerically equal to the element's atomic weight. And one gram of an atom is different that is "converting atomic weight from amu to gram). And we take" gram atom"as single term and a different concept and do not take "gram" in the term as adjective. That's how molar mass is single term and different concept and we should not take it as mass with adjective. Am I right?

@Remy I highly recommend you reading wikipedia articles for any term of interest before asking. // I would fully abandon the very obsolete quantity name "gram atom". It is not used in western chemistry for decades, even if it may be still taught in India. //  1 gram atom = 1 mol of the respective atoms. 1 gram of atoms, the same as 1 gram of anything is still 1 g. And yes, mass conbined with certained adjectives forms terms, qualitatively different to mass. // Note mole=unit name like kilogram, mol=unit symbol like kg.

No, you are right. Anytime when you are in doubts about units, take an equations with quantities of interest and plug there the supposed units instead of numbers. Evaluate the equation and if there is not match, something is wrong. For more, see Dimensional analysis // As homework, derive units for conductance, conductivity and molar conductivity

Accidentally deleted my comment which was before @Poutnik,s above comment writing it again for better understanding .[@Poutnik I am justing drawing an anology that how we should not be confused of terms due to words that are used in them like in obsolete "gram atom" we should not confuse our self with gram. Because it is 1gram atom =1mole not the gram of 1atom of some element. That's how mass and molar mass are different. Or am I again wrong?]

@Remy. You are right. The term "gram atom" has been replaced by the word "molar mass" in order to avoid confusions similar to yours. A "gram atom" is not the mass in gram of an atom. It is the mas of 1 mole of an element.


text:
Why is it so difficult to crystallise anything in a glass capillary?
I'm talking about very fine glass capillaries with a diameter under 1 mm. I've found that if I carry out melting point analysis, the material in the capillary doesn't crystallise again. I did wonder whether it was forming a glass, and I experimented using a larger volume of material and cutting/tilting the capillary. Definitely a liquid and definitely supercooled — it would often crystallise as I cut the capillary. If I put it in the fridge or freezer it would also give crystals.
I'm really struggling to understand this phenomenon — I've looked in textbooks and online, but I'm struggling to find anything, probably because capillary is a very common word.
I understand classical nucleation theory (I think), so I tried to relate that, but I'm struggling to see how the shape of a container would influence the clusters forming — surely even a very narrow capillary is huge to an atom! And as touched on in some of the answers, is it just nucleation that's slow to proceed, or crystal growth as well?

In addition to the answers you got, it would be interesting to see if the resistance towards crystallisation is retained in general if you let the capillar in the bath. It is quite possible that removing it when the mp is established results in a relatively fast cooling even in air. The capillar is very small and whatever warm it cools down rapidly. Plus one for investigating a common fact.

It is well known that imperfections on the surface of glass often act as nucleation centres (and sometimes when using larger vessels, chemists actively scratch the surface to promote nucleation). What is less-well known is that the process of creating small capillary tubes tends to create glass with very few surface defects. This is probably a significant factor.

I'm no expert, but I think it likely IS related to both the shape and dimension.  If I understand how the capillary tube method (I referred to 2.2.15. MELTING POINT - OPEN
CAPILLARY METHOD in https://file.wuxuwang.com/yaopinbz/EP8.0_1_00025.pdf , Ph. Eur. 8.0, 20215 (01/2008).  , cited as described in their FAQ: https://faq.edqm.eu/pages/viewpage.action?pageId=1377105, I quote: "The official abbreviation is ‘Ph. Eur.’ and not ‘EP’ (which is registered for the European Parliament). A good example of a reference would be: Ph. Eur. 6.0, 1742 (01/2008).") of determining melting point works in the first place, it is essentially by seeing when it is liquid enough to move up the capillary.  To do the reverse means one will need to be able to overcome the capillary action forces per se.  At least that is my intuitive (and so quite possibly wrong) interpretation, i.e. we have found where something comes apart by slowly adding energy while under a tension of sorts, now we still have that tension, but want to put it back together per se.  I found a paper that goes into more depth:

The temperature at
which the crystal is thermodynamically stable is also a func-
tion of the capillary diameter. As the diameter decreases, the
freezing point of water is lowered, which is called the capil-
lary freezing point depression. The relation between the
freezing point depression and the curvature of the interface is
described by the Kelvin equation [1]. The original Kelvin
equation was a liquid-vapor version that quantifies the devia-
tion in equilibrium vapor pressure above a curved surface
from that which would exist above a plane surface at the
same temperature. It was later extended to solid-liquid
phases and became well known as the ‘‘Gibbs-Thomson’’
relation in solidification that defines the local displacement
of equilibrium temperature on a curved solid-liquid interface
[2,3]. According to this relation, the freezing point depres-
sion due to curvature is

(1) Equation that doesn't copy paste easily unfortunately

where Kappa is the mean curvature of the interface (m^-1 ), v_S is
the specific volume of the solid phase (m 3 /kg), L is the
latent heat of fusion (J/kg), sigma_LS is the interfacial excess
energy per unit area of ice-water interface (J/m^2 ), and T_C
and T_E are the freezing point and the normal equilibrium
temperature of the ice-water interface (i.e., 273.15 K), re-
spectively.

The paper is available here: https://www.researchgate.net/publication/7525813_Measurement_of_freezing_point_depression_of_water_in_glass_capillaries_and_the_associated_ice_front_shape/link/00b7d52bd2dd8da068000000/download
APA Style Citation:
Liu, Z., Muldrew, K., Wan, R. G., & Elliott, J. A. (2003). Measurement of freezing point depression of water in glass capillaries and the associated ice front shape. Physical review. E, Statistical, nonlinear, and soft matter physics, 67(6 Pt 1), 061602. https://doi.org/10.1103/PhysRevE.67.061602
You may also look at the related phenomenon of capillary condensation, e.g. https://www.nature.com/articles/s41586-020-2978-1 , (Yang, Q., Sun, P.Z., Fumagalli, L. et al. Capillary condensation under atomic-scale confinement. Nature 588, 250–253 (2020). https://doi.org/10.1038/s41586-020-2978-1) governed by the same Kelvin equation.  https://www.dora.lib4ri.ch/empa/islandora/object/empa%3A7600/datastream/PDF/Nowak-2008-Thermodynamic_and_kinetic_supercooling_of-%28published_version%29.pdf (APA: Nowak, D., Heuberger, M., Zäch, M., & Christenson, H. K. (2008). Thermodynamic and kinetic supercooling of liquid in a wedge pore. The Journal of chemical physics, 129(15), 154509. https://doi.org/10.1063/1.2996293) is potentially also a useful read though I think the math is different for the wedge pores they examine vs. the more classical capillary you're working with.
I wouldn't discount the probabilistic approach given in other answers, but I think there may be more at play here.  Wikipedia article on the Kelvin equation, mostly as it relates to condensation: https://en.wikipedia.org/wiki/Kelvin_equation In that context they state: "It is unlikely, however, that new phases often arise by this fluctuation mechanism and the resultant spontaneous nucleation. Calculations show that the chance, e−ΔS/k, is usually too small. It is more likely that tiny dust particles act as nuclei in supersaturated vapours or solutions. In the cloud chamber, it is the clusters of ions caused by a passing high-energy particle that acts as nucleation centers. Actually, vapours seem to be much less finicky than solutions about the sort of nuclei required. This is because a liquid will condense on almost any surface, but crystallization requires the presence of crystal faces of the proper kind."

It is preferable to include human readable citations to also prevent any fallout from possible link degradation. See for more details: https://chemistry.meta.stackexchange.com/q/2944/4945

This seems the more to the point answer. Tough the other are also valid as general to crystallisation.

It has not (so much) to do with the shape of your container, rather than the accessible volume and thus, probability of events.  To quote Meldrum and O'Shaughnessy:

"Nucleation in small volumes usually gives rise to a reduction  in the nucleation rate, where this can be attributed to a number factors. i) The creation of small volumes is typically associated with the exclusion of impurities that promote nucleation in bulk solution [...].  Nucleation rates can therefore approach homogeneous rates, where it is of course impossible to eliminate all interfaces.  This effect can be seen in finite volumes in the µL regime and below, provided the number of droplets vastly exceeds the number of impurities present."
(loc. cit., p. 3 of 64; emphasis added).

Of course, «the other» known (and unknown) parameters affecting crystallizations interfere here as they would do for a crystallization in a bulky beaker, too (e.g., concentration, temperature/temperature gradient, etc.).  A recurrent difficulty for the crystallization of organic compounds from melt indeed is the tendency to yield supercooled liquids which spontaneously may solidify/crystallize.  Benzoic acid is a cheap and experimentally easy example (mp $\pu{122 ^\circ{}C}$) for this sometimes showcased in demonstration lectures.  Flexible alkyl chains tend to ease the formation of supercooled liquids and subsequently, glasses instead of crystals, too.
Note that conditions optimal for the creation of crystal seeds often are not optimal for crystal growth.  Energetically speaking, it is an uphill battle for flexible molecules of the melt to orderly assemble in a crystal.  Even if successful, small particles formed are (if normalized for their volume) of higher energy, than larger ones, too and heat of crystallization has to be dissipated which may inhibit crystallization, too.  One experiment in physical chemistry is e.g. to follow the temperature when crystallizing water to ice: at one point, temperature actually rises briefly before crystallization continues.  This reflects that a crystal seed needs to grow beyond a critical radius r before there is an overall gain in free energy:

(image credit, Meldrum et al.)
The caption of this diagram reads:

"Schematic of the free energy ($\Delta{}G$) of a growing crystal nucleus as function of the radius ($r$).  The energy profile is a result of the favorable volume energy ($\Delta{}G_\mathrm{V}$) and the surface energy ($\Delta{}G_\mathrm{A}$).  The maximum value ($\Delta{}G_\mathrm{C}$) is achieved at the critical radius ($r_\mathrm{crit}$)."  (loc. cit., p. 9 of 64)

As the authors show, the critical nucleus size $r_\mathrm{C}$ can be described by
$$r_\mathrm{C} = \frac{2\gamma{}V_\mathrm{m}} {\Delta{}\mu{}} = \frac{2 \gamma{} V_\mathrm{V}} {kT \ln S}$$
dependent on the interfacial free energy ($\gamma{}$), the molecular volume ($V_\mathrm{m}$), the chemical potential of the molecule ($\mu{}$).
Cutting/tilting the capillary just disturbed the (metastable) equilibrium reached and possibly introduced new sites where crystallization may start.  (As if you scratch the inner of a beaker with a glass rod when recrystallizing a product from hot solution.)  This being said, crystallization in capillaries often is a necessity if (e.g., biochem) materials of interest are scarce or/and do not form large crystals:

(image credit, access on ResearchGate)
especially if the crystals may be obtained by counter-diffusion of reagents.
Reference
Meldrum, F. C. and O'Shaughnessy, C. Crystallization in Confinement. Adv. Mater 2020, 32, 201068; doi 10.1002/adma.202001068  (an open access publication).

Thank you, that's helpful. Would you be able to expand a little bit more on the accessible volume/probability point please? I think that's where I'm getting stuck.

I wonder if surface tension may play any role in triggering of the phase change.

@BeginTheBeguine In first place, it boils down to the reduction of nucleation. The answer now includes this aspect.

@Poutnik The now added literature reference (open access) targets crystallization in confinement (well, not the one of 2020/21) and surface interactions e.g., of the crystal and the solution equally are mentioned there.

Thank you, and thank you for the reference I will read that now. One last question (and I might edit the main question to add this), does that mean that if I added a nucleating agent it should then freeze at about the right temperature inside the capillary?

@BeginTheBeguine The addition of nucleating agent might ease the crystallization.  If it dissolves in the compound, however, the melting point of the mixture may differ from the one of the pure compound. Typically, the melting point is lowered (known as freezing-point depression, or cryoscopy). It is more practical to melt e.g., benzoic acid in a sealed Pasteur pipette, get a nice seed in the narrower capillary first, and slowly grow a larger crystal in the wider part by directed crystallization (zone melting-like).  The shape of the crystal isn't a parallelepiped and it requires some tests.

Here is a simple explanation, which is valid for elementary chemistry classes.
When a liquid or a solution is cooled under its melting point, it must solidify. But it sometimes happens that the crystallization is delayed. Everything looks as if no particular molecule (or atom) feels decided to start the first crystal and show how and where the other ones must be alined. Usually the first crystal appears on an impurity (dust), or any irregularity in the flask. And an irregularity may be created by scratching the flask or by a cosmic ray crossing the liquid. Cosmic rays are able to remove an electron in an atomic target, which is immediately recombined. But this creates an irregularity for a while. We are continuously crossed by cosmic rays at a rate of about $1$ ray per cm2 of skin and per second. We are used to it. But if the sample is small enough, there are no cosmic rays crossing the liquid during the first seconds. The liquid stays liquid even though the temperature gets lower than the melting point. The crystallization starts with delay. Some substances (sodium acetate, sodium thiosulfate) display this strange property more easily than others.
But this explains why crystallization starts with delay in a tiny glass capillary. The probability of being touched by a cosmic ray is rather small.

If the explanation by cosmic rays did work, we wouldn't be able to have bottles of supercooled water for such long times as to be able to pour it (and watch its freezing on hitting the dish).

It is easy to get supercooled liquid water down to -$5$°C in small test tubes, if they are dipped in a mixture salt + ice, whose temperature can be $-20$°C. It usually lasts about half a minute. And suddenly when the first ice crystal appears, the temperature suddenly goes up to $0$°C, and stays there as long as some liquid water is visible

In my opinion the cosmic rays factor is overemphasised at least respect to the fact that a capillar cool down fast (due to size shape mass), to the role of surface tension (see ttbek answer) and the reduction of nucleation rate (answer by Button) which of course seems dictated by the all the reason mentioned in this thread, including the cosmic rays one that you propose. It is also interesting, to be clear.

@Alchinmista. You are right. But in my experience of high school chemistry teacher, I have observed that the explanation with cosmic rays is immediately understood by students, faster than the mechanism with nucleation or surface tension.