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How to arrive at the preferred IUPAC name for 1,2-diphenylethane?
I am reading a textbook that mentions the molecule 1,2-diphenylethane, $\ce{C14H14}$. A Google search leads to PubChem, where it is mentioned that the IUPAC name for the molecule is 2-phenylethylbenzene. I'm trying to understand how one names the molecule, using IUPAC nomenclature. However, I'm struggling to find any material that describes how apply IUPAC nomenclature to these "double benzene" (polycyclic aromatic hydrocarbon?) molecules. I would greatly appreciate it if someone could please redirect me to a page that describes this and describe the naming process for this particular molecule.

Naming these compounds can be tricky; I am afraid there is probably no website available for the most recent nomenclature recommendations due to copyright. In any case, the Blue Book is the one with these recommendations, and you might be lucky enough to find it at a library close to you.

@Martin-マーチン Surely there's somewhere (a university webpage, for instance) that provides information on how to name these types of molecules?

The book has about thousand and five hundred pages (https://doi.org/10.1039/9781849733069). There is a website with the earlier and deprecated recommendations from 1993, but they explicitly state that the cannot publish the 2013 rules due to copyright. And the naming can be significantly different from the earlier versions. You can wait, maybe someone with access to the book will answer the question for you.

The essential problem in this case is finding the parent structure. According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), a structure with two or more identical units that are linked by di- or polyvalent groups can be named in two ways:

by ordinary substitutive nomenclature, in which one of the parent structures is chosen as the senior parent structure and the remainder of the structure is expressed by substituent prefixes
by multiplicative nomenclature, in which two or more parent structures are connected by symmetrical or unsymmetrical single or concatenated substituent groups

Both methods yield correct names, which may be used in general nomenclature.
1. Substitutive nomenclature
In many related simple cases, the compound consists of just a ring and a chain so that the parent structure could be the ring or the chain.
The corresponding rule in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows:

P-44.1.2.2 Systems composed of rings and chains (exclusive of linear phanes)
Two methods are recognized to name systems composed of rings and chains
  (exclusive of linear phanes).
(1) Within the same class, a ring or ring system has seniority over a chain. When a ring and a chain contain the same senior element, the ring is chosen as parent. Rings and chains are chosen regardless of their degree of hydrogenation. As a consequence, this approach prefers the choice of a ring over a chain in systems composed of cyclic and acyclic hydrocarbons.
(2) The context may favor the ring or the chain, so that, for example, substituents may be treated alike or an unsaturated acyclic structure may be recognized, or the one chosen has the greater number of skeletal atoms in the ring or in the principal chain of the acyclic structure.
(…) For selection of a preferred IUPAC name, see P-52.2.8.

P-52.2.8 Selection between a ring and a chain as parent hydride
Within the same heteroatom class and for the same number of characteristic groups cited as the principal characteristic group, a ring is always selected as the parent hydride to construct a preferred IUPAC name. In general nomenclature, a ring or a chain can be the parent hydride (see P-44.1.2.2).

These rules apply to compounds that consist of one ring and one chain. The compound that is given in the question, however, consists of a symmetrical chain with two rings. Extending the use of the above-mentioned rules to this case, we could pick one of the rings as the parent structure in accordance with Rule P-52.2.8. This approach would result in the correct name (2-phenylethyl)benzene. Since this method usually yields the preferred IUPAC in many simple cases, it should not be a surprise if some computer-generated names follow this pattern also in this case. Note, however, that the name “2-phenylethylbenzene” suggested in Pubchem is not correct and ambiguous (“2‑phenylethylbenzene” could be interpreted as “(2‑phenylethyl)benzene” or “2‑phenyl‑1‑ethylbenzene”).
In accordance with P-44.1.2.2, the context may favour the chain as parent structure. For the compound that is given in the question, the context may favour the chain in order to highlight the symmetrical shape of the structure, in order to treat the equivalent rings alike, or because a following reaction happens at the chain. This approach would result in the correct name 1,2-diphenylethane.
2. Multiplicative nomenclature
The IUPAC recommendations explain when multiplicative nomenclature is used to generate the preferred IUPAC name (PIN) as follows.

P-45.1 MULTIPLICATION OF IDENTICAL SENIOR PARENT STRUCTURES
P-45.1.1 Multiplicative nomenclature is senior to substitutive nomenclature for generating preferred IUPAC names to express multiple occurrences of identical senior parent structures, other than alkanes, in the name of the parent structure (see P-51.3.1). In most cases multiplicative names are shorter than regular substitutive names. A preferred IUPAC name is generated by multiplicative nomenclature when the following criteria for its use are met (see P-51.2.3).
  (1) the linking bonds (single or multiple) between the central substituent group of the multiplicative group and all subsequent structural units are identical; and
  (2) the multiplicative groups, other than the central multiplicative group, are symmetrically substituted; and
  (3) the locants of all substituent groups on the identical parent structures, including suffix groups, are identical.
When these conditions are not met, substitutive nomenclature generates preferred IUPAC names.

The compound that is given in the question fulfils these requirements; therefore, the preferred IUPAC name is generated by multiplicative nomenclature as follows.

P-15.3.1.3 Multiplicative name formation
  Multiplicative names are formed in accordance with the number of occurrences of identical structural units as defined in P-15.3.1.1 and the relationship of the linking multiplicative substituent group to the identical structural units.
  When a compound contains identical structural units as defined in P-15.3.1.1 linked by a symmetrical simple, compound, complex, or concatenated multiplicative group (a di- or polyvalent substituent group), it is named by stating successively:
(a) the locants for the positions of substitution of the linking multiplicative substituent atom or group to the identical parent structural unit (the locant 1 is omitted when alone in the name of a mononuclear parent hydride);

Here: 1,1'

(b) the name of the linking multiplicative substituent atom or group;

Here: ethane-1,2-diyl

(c) the numerical prefix ‘di’, ‘tri’, etc.; and/or ‘bis-‘, ‘tris-‘, etc., with no elision of the final vowel before the name of the identical parent structural unit;

Here: di

(d) the name of one of the identical structural units including the principal characteristic group and substituents, if any, enclosed in appropriate enclosing marks (see P-16.5).

Here: benzene

The numbering of the identical parent structural unit is retained and, when there is a choice, the locants of the point of substitution by the linking multiplicative substituent groups on the identical parent structure are as low as possible.

Therefore, the preferred IUPAC name is 1,1'-(ethane-1,2-diyl)dibenzene.


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Use of Concentration for pH
When we find the pH of a solution, do we use normality or molarity? My teacher said that if the compound is present by itself, we can use normality but in all other cases we should use molarity. Can you explain why?

Current definition implies that $\mathrm{pH}$ is a function of relative activity.
Originally, the amount concentration of  $\ce{H+}$ in $\pu{mol L-1}$ was proposed, which is also often used these days as an approximation [1].

$\mathrm{pH}$ was originally defined by Sørensen in 1909 … in terms of the concentration of hydrogen ions (in modern nomenclature) as $\mathrm{pH} = −\lg(c_\ce{H}/c^\circ)$ where $c_\ce{H}$ is the hydrogen ion concentration in $\pu{mol dm–3},$ and $c^\circ = \pu{1 mol dm–3}$ is the standard amount concentration. Subsequently …, it has been accepted that it is more satisfactory to define $\mathrm{pH}$ in terms of the relative activity of hydrogen ions in solution
$$\mathrm{pH} = -\lg a_\ce{H} = -\lg (m_\ce{H}γ_\ce{H}/m^\circ)\tag{1}$$
where $a_\ce{H}$ is the relative (molality basis) activity and $γ_\ce{H}$ is the molal activity coefficient of the hydrogen ion $\ce{H+}$ at the molality $m_\ce{H},$ and $m^\circ$ is the standard molality. The quantity $\mathrm{pH}$ is intended to be a measure of the activity of hydrogen ions in solution.
However, since it is defined in terms of a quantity that cannot be measured by a thermodynamically valid method, eq. 1 can be only a notional definition of $\mathrm{pH}.$

So, if the numerical value for normality coincides with the one for molarity, and one can omit the activity in favor of molarity, i.e. when both equivalence factor $f_\mathrm{eq}$ and activity coefficient $γ$ are approx. 1, then yes, normality can also be used to estimate the $\mathrm{pH}.$
I'm not sure how to interpret "present by itself" part of your question, but if it means there is a single solute, then it seems to be incorrect.
The simplest counter example would be a solution of di-, tri- and polybasic acid/base, say $\ce{H2SO4},$ for which the equivalent concentration is double the value of amount concentration.
References

Buck, R. P.; Rondinini, S.; Covington, A. K.; Baucke, F. G. K.; Brett, C. M. A.; Camoes, M. F.; Milton, M. J. T.; Mussini, T.; Naumann, R.; Pratt, K. W.; et al. Measurement of pH. Definition, Standards, and Procedures (IUPAC Recommendations 2002). Pure and Applied Chemistry 2002, 74 (11), 2169–2200. DOI: 10.1351/pac200274112169. (Free Access)

So in every scenario we would be using Normality rather than Molarity? If so then what do we use Molarity for?

@Natewilliam No, you got it backwards, it's amount concentration, or molarity, that is used in the absence of activity.

Isn't it the same? If there's no activity, then normality is equal to molarity ( as Normality = Molarity*activity(or n-factor) ) So isn't it correct to say that we always use normality for any sort of calculation rendering molarity unnecessary?

@Natewilliam No. Normality is $c/f_\mathrm{eq}.$ If $f_\mathrm{eq} = 1$ (monobasic substance), then the numerical values of normality and molarity are the same. Activity has nothing to do with $f_\mathrm{eq}$ (or $n$-factor, s you call it), it's an unrelated quantity.

pH is not defined versus normality or molarity of an acid: Whatever the normality or the molarity, the pH is defined from the activity (or the concentration) of the ions H+. A given value of the normality or of the molarity does not give you the activity (or the concentration) of H+. So it does not allow you to calculate the pH.

The statements "pH is not defined versus normality or molarity" and "pH is defined from the activity (or the concentration)" seem to contradict each other as both normality or molarity are concentrations. You either have to be more specific, or stick with only one nominal definition. However, if you do the latter, this won't answer the question. Besides, I don't see how this adds anything to an existing answer.

I think Maurice’s point is that OP seems to be referring to normality or molarity of an acid, not of protons. For a strong acid, the molar amount of protons is equal to normality of acid (which is all in the form of conj base). For a weaker acid, that is not the case, so pH cannot be calculated using only the acid molarity or normality.

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Difference between neutral and amphoteric oxides
What is actual difference between neutral oxides (e.g. $\ce{NO},$ $\ce{N2O},$ $\ce{CO})$ and amphoteric oxides (e.g. $\ce{Al2O3})?$ I got a hint of it from the comment of Ivan Neretin.

Neutral oxides do not manifest either acidic either alkalic behaviour ( but eventually very weak in extremely alkalic/acidic environment).
Amphoteric oxides manifest, more or less balanced, both behaviours in reasonably strong acidic/alkalic environments.

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Have IH₃ and IH₅ already been synthesized?
I've seen a paper by Sikalov [1] that says that $\ce{IH3}$ and $\ce{IH5}$ should be observable. Has anyone actually tried to synthesize such molecules?
References

Sikalov, A. A. Hypervalent Halogen Hydrides $\ce{HalH_n}$ ($\ce{Hal}$ = $\ce{Cl},$ $\ce{Br},$ $\ce{I};$ $n = 3, 5, 7)$: DFT and Ab Initio Stability Prediction. Theor Chem Acc 2019, 139 (1), 8. DOI: 10.1007/s00214-019-2524-0.

I've done some literature research but couldn't find anything...

Under high pressure, maybe? Also is there a reason to put iodine first not hydrogen as in ordinary hydrogen iodide?

Oscar Lanzi , in this case, some of Hydrogen atoms are negatively charged. That's probably why the author uses such notation.

Not the specific compounds $\ce{IH3}$ and $\ce{IH5}$, but hydrogen-iodine compounds with nonclassical stoichiometriws have been synthesized under high pressure.
Binns et al[1] report two high-pressure, hydrogen-rich hydrogen iodides. The more widely stable one, with empirical formula $\ce{H_{27}I}$ consisting of a hydrogen iodide molecule isolated in a shell of thirteen dihydrogen molecules, is stable from 9 to at least 130 GPa pressure. The second compound, stable up to 12.5 GPa, has two hydrogen iodide molecules per one dihydrogen molecules for an empirical formula $\ce{H2I}$. Both compounds are characterized by synchrotron XRD and Raman spectroscopy, and both have only secondary bonding between the hydrogen iodide component and additional hydrogen rather than hypercoordinated iodine.
Reference

Jack Binns, Philip Dalladay-Simpson, Mengnan Wang, Graeme J. Ackland, Eugene Gregoryanz, and Ross T. Howie (2018)."Formation of H2-rich iodine-hydrogen compounds at high pressure".
Phys. Rev. B 97, 024111. https://doi.org/10.1103/PhysRevB.97.024111


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Preferred Lewis structure for sulphate anion and XeO4
In a section reviewing Lewis structure, the textbook Atkins' Physical Chemistry, 10th Edition, by Atkins, Paula, and Keeler, gives the following illustration of hypervalence and octet expansion:

I have two questions regarding this illustration:

With regards to the structure of the $\ce{SO4^{2-}}$ ion, notice that the lowest energy structure (1b) has the double covalent bonds placed at angles that are closer than had the double bonds been placed at opposing sides of the sulfur atom. I'm confused as to why this is supposedly the lowest energy state, since one would expect the lowest energy state to be when the two double bonds are at farthest angles from each other (which would be when they are on opposing sides of the Sulphur atom in the diagram), since this would be the case when there is the most distance between the repulsive negative charges of the electrons in the two double covalent bonds? Based on my research, it seems that what I am referring to here would resemble a linear molecular geometry (although, I am obviously not saying that the molecule would actually have a linear molecular geometry), where the we can imagine the red atom in the following image as being the sulfur atom, and the two white atoms as being the oxygen atoms that have the double covalent bond:

My next question is with regards to the $\ce{XeO4}$ molecule. It seems to me that the arrangement of all of the free electron pairs on the oxygen atoms would be in a more energetically favourable state (the molecule would be in a lower energy state) if they were farther away (at greater angles) from the double covalent bonds? It seems to me that their current arrangement would make sense in a scenario where there were an additional 2 free electrons on the oxygens, as is the case for two of the oxygen atoms of structure 1b, but since these oxygen atoms only possess 6 electrons instead of 8 (as is the case for two of the oxygen atoms of $\ce{SO4^{2-}}$), then wouldn't it be more energetically favourable to have the electron pairs at greater angles from the double covalent bonds? Based on some research, it seems that the geometry that I am referring to is a trigonal planar molecule geometry, as shown here (although, I am just referring to the Lewis structure of the molecule -- not the 3-dimensional geometry). In the following image from the same Wikipedia article, imagine the red atom as being the oxygen, and one of the white atoms as being the double covalent bond, and the other two white atoms as being the lone pairs of electrons, so that you have a $60^\circ$ angle between them:

I would greatly appreciate it if people could please take the time to clarify this.

https://chemistry.stackexchange.com/questions/19836/how-can-compounds-like-xeo4-form-tetrahedral-structures-despite-having-double-bo https://chemistry.stackexchange.com/questions/39396/are-the-phosphate-and-sulfate-ion-diagrams-with-double-bonds-an-invalid-picture

@Mithoron I don't see how those answer my questions?

And what are they really? You made your post as unclear as possible. As far as I can tell  your reasoning doesn't make much sense. Even book you cite is wrong - actually single bonds reflect reality much better.

@Mithoron If my post is unclear, then it certainly isn't because I intentionally made it to be so. I am a novice when it comes to chemistry, and I'm trying to learn, which is why I am studying this textbook. The textbook itself is one of the most highly regarded, if not the most highly regarded, physical chemistry textbook of the present. If my questions do not make sense, or if there is something incorrect in the textbook, then feel free to explain.

Atkins may be most common book, but it doesn't make it impervious to outdated theories and octet expansion is one of them. Making a clear post doesn't have much to do with being novice in chemistry

@Mithoron What I mean is that maybe it is my lack of understanding that has led to an absurd and/or unclear question. What about my question is unclear? I will attempt to clarify.

@Mithoron I have made some edits in an attempt to clarify my questions. What I have essentially done in thinking about this section of the textbook is I have used my novice-level understanding of chemistry to critically analyze what the authors have presented as the most energetically favourable Lewis structures for the two molecules, and then proposed a Lewis structure that, using my novice-level understanding of chemistry, seems would be more energetically favourable, and then asked why this isn't the case (presuming that the textbook is correct).

You are reasoning as if the ion $\ce{SO4^2-}$ is planar. It is not the case. It has a tetrahedral geometry. And what you see in Atkins' book is a projection of the tetrahedron on a horizontal plane. So the angles between the bonds are not 90°, but larger. It is the same for $\ce{XeO4}$. This reality will solve all your questions simultaneously.

I don't think what you've described answers my questions. I am aware that the physical geometry of these molecules is 3-dimensional and not 2-dimensional (planar). But this is a case in which the textbook describes the molecular geometry of the molecules and their energy states using Lewis structures, which are 2-dimensional diagrams. And given the comparison the authors made between structures 1a and 1b, they are obviously using the diagrams to also convey the (lowest) energy states of the molecules; which leads to my question of why the Lewis structures are not as I wonder they should be.

His point is that in a tetrahedron the oxygens are equidistant from each other, so there is no distinction between the double bonds being on the Os that appear in 2D to be adjacent vs opposite. Likewise, in the XeO4 Lewis structure the position of the dots is not intended to represent the positions of the electrons in a rigorous geometrical way. Independent of your misinterpretation of the Lewis structures, expanded octets on sulfur complexes is an outdated and debunked idea.

@Andrew Hmm, I see. Thank you for taking the time to clarify. So, out of curiosity, is the entire idea of expanded octets outdated or debunked, or just the idea of expanded octets on sulfur complexes?

For larger atoms, d and even f orbitals can be involved in bonding, which means the octet rule is no longer obeyed. Note, however, that the conventions for drawing Lewis structures have not changed, so the standard practice, for example, is to use formal charge rather than octet expansion on nitrogen oxides, but with sulfur and phosphorus oxides, we still draw 6 bonds. As a general statement, Lewis structures are not intended to rigorously represent the electronic or geometric structure of molecules. They are useful simple representations.


text:
What is a word for "atom or molecule"?
What is a word for "atom or molecule"?  As in: "The entry of an atom or molecule across a cell membrane into a cell is dependent on its size and solubility."  This keeps coming up, and it's really irritating (on par with the lack of a good word for "he or she" in clinical writing talking about patients).  English majors come up with frustrating responses like "elemental particle".  I just need a regular word for normal scientific communication (intro sections of papers, science education docs, etc.).

Why not "solute"?

I would understand "the entry of a species across..." as well as "the patient...." or "the subject...".

How about: Particle?

@Alchimista: "Species" might not be the best choice here, given that the OP is dealing with cellular biochemistry and a reader would probably would be thinking of the word "species" in the biological context.

@Sean out of the dark perhaps.. Surely not within context. Like "... ions of calcium. This species moves.....". This is rather a question on how to write rather than in chemistry, anyway. That is why I have just commented.

I think the trouble is the very definition of molecule, which seems to have a clause specifying that it consists of two or more atoms. Dump that clause, allow 1 atom in the definition of a molecule, and your problem is solved.

One option is to write "The entry of a [insert term here][1] across a cell membrane into a cell is dependent on its size and solubility." And then in footnotes, say "[1] In this article, [insert term here] is taken to mean an atom or molecule." You only have to do that once for the article.

Nit: Not sure if you're writing this yourself, but "across" is not really the right preposition with "entry."

The classic catch-all term is molecular entity, often abbreviated to just entity. There's a Wikipedia page for it, which references the IUPAC Gold Book entry. Quoting from the latter:

Any constitutionally or isotopically distinct atom, molecule, ion, ion pair, radical, radical ion, complex, conformer etc., identifiable as a separately distinguishable entity.
Molecular entity is used in this Compendium as a general term for singular entities, irrespective of their nature, while chemical species stands for sets or ensembles of molecular entities. Note that the name of a compound may refer to the respective molecular entity or to the chemical species, e.g., methane, may mean a single molecule of $\ce{CH4}$ (molecular entity) or a molar amount, specified or not (chemical species), participating in a reaction. The degree of precision
  necessary to describe a molecular entity depends on the context. For example, 'hydrogen molecule' is an adequate definition of a certain molecular entity for some purposes, whereas for others it is necessary to distinguish the electronic state and/or vibrational state and/or nuclear spin, etc. of the hydrogen molecule.
Source: PAC, 1994, 66, 1077. (Glossary of terms used in physical organic chemistry (IUPAC Recommendations 1994)) on page 1142.

As you can see, the definition is exceptionally general. It also specifically includes atoms. Usage of the term is widespread, even if it seems a bit exotic when first encountered.
Another option which can work is chemical species (often abbreviated to just species), which is even mentioned in the definition above. Again, there is a Wikipedia page. IUPAC actually has two definitions for chemical species, but the relevant definition is the following:

An ensemble of chemically identical molecular entities that can explore the same set of molecular energy levels on the time scale of the experiment. The term is applied equally to a set of chemically identical atomic or molecular structural units in a solid array.
For example, two conformational isomers may be interconverted sufficiently slowly to be detectable by separate NMR spectra and hence to be considered to be separate chemical species on a time scale governed by the radiofrequency of the spectrometer
  used. On the other hand, in a slow chemical reaction the same mixture of conformers may behave as a single chemical species, i.e. there is virtually complete equilibrium population of the total set of molecular energy levels belonging to the two conformers. Except where the context requires otherwise, the term is taken to refer to a set of molecular entities containing isotopes in their natural abundance. The wording of the definition given in the first paragraph is intended to embrace both cases such as graphite, sodium chloride or a surface oxide, where the basic structural units may not be capable of isolated existence, as well as those cases where they are. In common chemical usage generic and specific chemical names (such as radical or hydroxide ion) or chemical formulae refer either to a chemical species or to a molecular entity.
See also: chemical species (of an element)
Sources:
  PAC, 1994, 66, 1077. (Glossary of terms used in physical organic chemistry (IUPAC Recommendations 1994)) on page 1096.
  PAC, 1996, 68, 2193. (Basic terminology of stereochemistry (IUPAC Recommendations 1996)) on page 2202.

In this case, there is a notion of an ensemble (a collection/multiple) of whatever is being referred to. This is why "chemical specimen" (singular) is never used.
In short, a (chemical) species is composed of a collection of (molecular) entities. Both are umbrella terms covering virtually everything, and therefore can be used in almost all contexts. Perhaps in your case, (molecular) entity is the preferred term, as it emphasizes the differences in identity of what interacts with the cells.

This gives new meaning to the term Crystalline Entity!

I think you're looking for the word: particle.

... a particle (or corpuscule in older texts) is a
  small localized object to which can be ascribed
  several physical or chemical properties such
  as volume, density or mass. They vary greatly in size or
  quantity, from subatomic particles like the electron, to microscopic
  particles like atoms and molecules, to macroscopic
  particles like powders and other granular materials.

Source: https://en.wikipedia.org/wiki/Particle

How often is "particle" used when describing (bio-)organic reactions though? I've seen "chemical species" used most often, and less commonly "entities".

Chemical!

any basic substance that is used in or produced by a reaction involving changes to atoms or molecules

Hydrogen is an atom, it's also a chemical, Water is a molecule, and is also a chemical
There are further subdivisions, but a chemical covers both atomic and molecular substances. If you can find matter that is not composed of chemicals, there is a man offering a $1m price.
The obvious example of non-chemical substances being anti-matter, but in your example that would lead to immediate cell death on a city wide scale

IUPAC has no definition for chemical, though it has one for chemical substance, Interestingly, the term entity is used in the definition! I definitely understand the point though, precision and clarity can be a tradeoff, and chemical substance is certainly clearer.


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Flame test for copper salts
I have read from sources here, here and even in my textbook that Copper (specifically $\ce{CuCl2}$ ) burns with a bluish green flame. All the sources mention that this is due to electron excitation and de-excitation in the orbitals of Cu. However, Why is copper the only transition element to give a coloured flame?
Most other flame tests are given by alkali metals and alkaline earth metals, which is understandable and is a common trend across the group. Copper stands out as an exception. I would like to know if there are any other transition elements that can give flame tests and the colours of the corresponding flames. I would also appreciate if someone could explain the electron excitation, either via Valence bond theory or by a Molecular Orbital treatment of a copper compound.

"Why is copper the only transition element to give a coloured flame?" — but this isn't true, plenty of other transition metals provide characteristic flame color…

There is a famous saying that paper never refused an ink, which means anyone can write anything on the paper and the paper won't say anything back to the writer. Internet has made this situation worse. One should develop a healthy level of skepticism when reading website and forums.

Why is copper the only transition element to give a coloured flame? Most other flame tests are given by alkali metals and alkaline earth metals, which is understandable and is a common trend across the group. Copper stands out as an exception.

This is not true. Boron compounds, when burned in alcohol, burn with a similar green color. Now here is the fundamental point to understand. When we see colors in the flames, usually it is due to very simple molecules of those metallic compounds in the flame. It is a molecular emission. If you can buy a pocket spectroscope, you will just a broad bands of color. So a flame spectrum, specially a low temperature flame, is not equal to an atomic emission spectrum. If I introduce copper into a very hot flame (>2000 Kelvin), you will not see anything, because copper atoms emit light mainly in the UV range.
Therefore all the common flame tests with calcium, barium, strontium show the flame emission spectrum of simple diatomic molecules of Ca/Ba/Sr in the flame. Copper is no exception. You may also wonder why a hydrocarbon flame is blue, there is no transition element there?
Vanadium is yellowish green
Cerium makes the flame yellow
Lead is blue white
Yttrium is beautiful crimson, which often appears on textbook covers on inductively coupled plasma spectroscopy (very high temperature atomic emission). Wikipedia has a big list.
Atomic spectroscopists use Grotrian diagrams to assign the transition lines the emission spectrum. Nobody uses VBT to explain the spectrum. Predicting the correct emission spectrum even of a diatomic molecule in the flame, will perhaps require the knowledge level of a PhD in physics.


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Effective Bohr Radius in Transition Metal Complexes?
I am playing around with a toy model of a transition metal complex where the HOMO are $d$-electron states of predominantly transition metal character. Let's say this is a $d^1$ (or $d^9$) system and the states are split in energy by ligand field effects. The relevant HOMO orbital is some linear combination of $d$-orbitals that correctly respects the symmetry of the transition metal complex.
I want to estimate the transition probability for a quadrupole $d$-to-$d$ transition from HOMO state. In other words, calculate $\langle f\vert r_i r_j \vert i \rangle$, where $\vert i \rangle$ and $\vert f \rangle$ are some $d$-orbitals I care about.
I can roughly estimate this transition amplitude by just using the Hydrogen $d$-orbital wavefunctions with an effective Bohr radius $a_{eff}$. My question is: what effective Bohr radius should I use?
For example, Copper is said to have a 3d-orbital radius of $\sim 0.6$ Angstrom (which is not the same as $a_{eff}$ I believe), but using the Slater estimate of the effective nuclear charge, I would guess $a_{eff} \approx a_0/Z_{eff} = a_0/13.2 \approx0.04$ Angstrom.
What effective Bohr radius should I actually use for very weakly hybridized transition metal orbitals?

If you inspect Table II in Ref. 1, which you will find referenced in the link you provide to the WebElements website, you'll see that AU units are used (also in the linked page), not pm, so the radius of the 3d Cu orbital in Ref. 1 is actually
$$ r_{\textrm{Mann}} = 0.613 \times 0.529 Å = 0.324 Å $$
or 32.4 pm.
As for Slaters approach, it is not enough to simply divide $a_0$ by the effective nuclear charge. You can only do this for an unshielded hydrogenic 1s orbital, otherwise it is generally necessary to account for the principal quantum number (in Slater's method, this may be an effective value not equal to the principal quantum number, see Slater's original article$^2$). The location of the maximum follows from the functional form of a Slater wave function:
$$\psi \propto r^{n^*-1}\exp\left({\frac{Z_{\textrm{eff}}}{n^*a_0}r}\right)$$
In general, if you change the nuclear charge of a hydrogenic atom you can perform the transformation $a_0 \rightarrow a_0/Z_{\textrm{eff}}$ to obtain the new orbitals$^3$.
The effective quantum number is n*=3 for a 3rd shell electron so that
$$r_{\textrm{Slater}} = \frac{n^{*2}}{Z_{\textrm{eff}}}=\frac{9}{13.2}\times0.529 Å = 0.361 Å $$
or 36.1 pm.
The difference in $r_{\textrm{max}}$ obtained by the two approaches is therefore not that dramatically different.
Of course there are two other things that seem to be important here: (1) whether using an accurate value of $r_{\textrm{max}}$ can be expected to provide you with an accurate value of the property you seek and (2) how to obtain an accurate estimate of $r_{\textrm{max}}$for a d-orbital of Cu in the transition metal complex.
References

J.B. Mann, Atomic Structure Calculations II. Hartree-Fock wave functions and radial expectation values: hydrogen to lawrencium, LA-3691, Los Alamos Scientific Laboratory, USA, 1968.
J.C. Slater. Atomic Shielding Constants. Phys. Rev., 36, 57, 1930.
W. Greiner. Quantum Mechanics: An Introduction. Springer 1994.

So if I want to get the actual estimated wavefunction, I would need to choose $a_0^*$ such that $\sqrt{\langle r^2\rangle}$ is equal to the Slater radius of the orbital under question?

@user157879 I'm not sure I understand your question but maybe my updated answer addresses it.

Ok I see, so the effective Bohr radius is what zi thought originally, with the addition  of an effective $n^$. Is it just a coincidence for the 3rd shell that $n^=3$? Do you have a non-paywall link to a table of $n^*$?

@user157879 You can probably get Slaters original article from another source if you search long enough, since its pretty foundational. I would start with the wikipedia and links therein: https://en.wikipedia.org/wiki/Slater%27s_rules


text:
Difference between liquid and solid stearic acid
I had some stearic acid in solid pellet format. It is slippy and in white color. I used this to make some experiments. I recently notice that some manufacturer can supply liquid stearic acid. I checked the books. Stearic acid should be in solid form at room temperature. I was confused.
Would someone help me to understand what is the liquid form stearic acid? What is the difference with the solid form? Thanks.

It might help if you link to the products in question or provide more information about them.

https://mateossl.es/stearic-acid-2/?lang=en. Click on the Stearic acid supply format. I found other Chinese suppliers (no English available) supply liquid format too.

Another one in India https://m.indiamart.com/proddetail/stearic-acid-5004037191.html

The liquid form must be shipped in heated tanks...

One should ask the supplier. Unless a SE user has direct knowledge, it is hard to say what is used to make what would be basically a solution or an emulsion.


text:
QM9 property based clustering
I have used a machine learning algorithm on QM9 dataset to extract some features. To visualize this feature space, I used a PCA algorithm to project it to a 2D space (in simple words, every point in the figure below represents a molecule). Then, I colored the points by property values of these molecules. For instance, in the figure below, I have colored the points by their LUMO energy value.
In my visualizations, I have used 10000 data. I was trying to figure out if each cluster below has any meaningful relation with molecular properties. To do so, I plotted another figure and this time I colored the points by their index in the QM9 dataset.
Roughly speaking, molecules with index i in [4000, 10000] have 8 atoms, with i in [800, 4000] 7 atom, [200, 800] 6 atoms, and the rest have 5 atoms or less.

What is the QM9 dataset sorted according to?
I was wondering if anyone has any suggestions for possible property
that each cluster does represent. Any suggestions for tests that I
can try would be highly appreciated as well!

Edits:
I tried to find the smiles for a group of these molecules. I selected the molecules in the red rectangle below, applied PCA on this group alone, and colored them inside the red circle based on LUMO energy (indices were all over the place for these sub section alone.)

Then I printed out smiles for two groups:
1) Ones to the right of dashed line:
CC(C)(C)CC=O
CC(=O)C(C)(C)C
CC(=O)CC(C)C
CC1(C)CC(=O)C1
C#CC(C)(C)C=OCCC(C)(C)C=O
CC#CCC(C)=O
C#CC#CC(C)=O
C#CCCC(C)=O
CC(C)C#CC=OCC(C)CCC=O
CCC#CC(C)=O
CCCCC(C)=O
CC1(C)CCC1=O
CC1(C)CC1C=OC#CCC(=O)C#C
C#CC(=O)C#CC
CC#CC(C)C=O
C#CCC(C)C=O
C#CC(C)CC=O
CC#CC(=O)CC
C#CCC(=O)CC
CCC(C)CC=O
C#CC(=O)CCC
CCCC(=O)CC
CCCC(C)C=O
C#CC(C)C(C)=O
C#CC(=O)C(C)C
CC(C)C(C)C=OCCC(=O)C(C)C
CCC(C)C(C)=O
CC1(C=O)CCC1
CC1(CC=O)CC1CC1CC1(C)C=O
CC(=O)C1(C)CC1
CC1=CC(=O)CC1
CC1=CCC(=O)C1CC1C=CC(=O)C1
CC1CCC(=O)C1
CC12CC(=O)C1C2
C#CC1CC(=O)C1CC1CC(C=O)C1
CCC1CC(=O)C1
CC(=O)CC1CC1
O=CC12CC(C1)C2O=C1CC2(CC2)C1
O=C1CC=CCC1
O=C1C=CCCC1
O=C1CCCCC1
CC1=CCCC1=O
CC1C=CCC1=O
CC1CC=CC1=O
CC1CCCC1=O
CC1CC(=O)C1C
CC1CC(C)C1=OCC12CC1CC2=O
O=CCC1CCC1
CC1CC1CC=O
CC(=O)C1CCC1
CC(=O)C1CC1CCC1C2CC1C2=O
O=C1CC2CC2C1
O=C1CC2CC1C2
C#CC(C#C)C=O
C#CC(C=O)CCCCC(C=O)CC
C#CC1(C=O)CC1
CCC1(C=O)CC1
C#CCC#CC=O
C#CC#CCC=OC#CCCCC=O
CC#CC#CC=O
CC#CCCC=O
CCC#CCC=O
CCCC#CC=O
CCCCCC=OC#CC1CCC1=O
CC1CCC1C=O
CCC1CCC1=O
C#CC1CC1C=O
CCC1CC1C=OO=CC1=CCCC1
O=CC1CC=CC1
O=CC1C=CCC1
O=CC1CCCC1
C#CC(=O)C1CC1CC(C=O)C1CC1
CCC(=O)C1CC1
O=CC#CC1CC1
O=CCCC1CC1
CC1C(C)C1C=OO=CC12CCC1C2
O=C1CCC12CC2
CC1C(=O)C2CC12
O=C1CC2CCC12O=CC1CC2CC12
O=CC1C2CCC12
CC1C2CC(=O)C12
O=C1C=CC2CC12O=C1CCC2CC12
O=C1C2C=CC1C2
O=C1C2CCC1C2
O=CC1C2CC1C2
O=C1C2CC3C1C23

2) Ones to the left of the dashed line:
Oc1ccccc1
CC#CC(C)(C)O
C#CCC(C)(C)O
CC(C)(C)CCO
CCCC(C)(C)O
CCOC(C)(C)C
COCC(C)(C)C
CC(C)C(C)(C)O
CC(O)C(C)(C)C
CC(C)CC(C)O
CC(C)OC(C)C
CC1(C)CC(O)C1
CC1CC(C)(C)O1
CC1CC(C)(O)C1
CC1(C)CC1(C)O
CC1(C)OC1(C)C
CC12CC(C)(C1)O2
CC12CC(O)(C1)C2
C#CC(C)(O)C#C
C#CC(C)(C)CO
C#CC(C)(O)CC
C#CC(C)(C)OC
CCC(C)(C)CO
CCC(C)(C)OC
CCC(C)(O)CC
CC#CCC(C)O
C#CC#CC(C)O
C#CCOC(C)C
C#CCCC(C)O
CC(C)C#CCO
CCC#CC(C)O
CC(C)CCCO
CCCCC(C)O
CCCOC(C)C
CCOCC(C)C
COCCC(C)C
CC1(C)CCC1O
CC1CCC1(C)O
CC1COC1(C)C
CC1OCC1(C)C
C#CC1OC1(C)C
C#CC1CC1(C)O
CC1(C)CC1CO
CCC1CC1(C)O
CCC1OC1(C)C
COC1CC1(C)C
CC1(O)CC=CC1
CC1(C)C=CCO1
CC1(O)C=CCC1
CC1(C)CCCO1
CC1(C)CCOC1
CC1(O)CCCC1
CC1C(C)C1(C)O
CC1C(O)C1(C)C
C#CC(O)C#CC
C#CCC(O)C#C
CC#CC(C)CO
CC#CC(O)CC
CC#CC(C)OC
C#CCC(C)CO
C#CCC(O)CC
C#CCC(C)OC
C#CC(C)CCO
C#CC(O)CCC
C#CC(C)OCC
C#CC(C)COC
CCC(C)CCO
CCC(C)COC
CCCC(C)CO
CCCC(C)OC
CCCC(O)CC
CCOC(C)CC
C#CC(O)C(C)C
C#CC(C)C(C)O
CC(C)C(C)CO
CCC(C)C(C)O
CCC(O)C(C)C
COC(C)C(C)C
C#CC1(C)CCO1
C#CC1(C)COC1
C#CC1(O)CCC1
CC1(CO)CCC1
CCC1(C)CCO1
CCC1(C)COC1
CCC1(O)CCC1
COC1(C)CCC1
CC#CC1(C)CO1
CC#CC1(O)CC1
C#CCC1(C)CO1
C#CCC1(O)CC1
CC1(CCO)CC1
CCCC1(C)CO1
CCCC1(O)CC1
CCOC1(C)CC1
COCC1(C)CC1
C#CC1(C)CC1O
C#CC1(O)CC1C
C#CC1(C)OC1C
CC1CC1(C)CO
CCC1(C)CC1O
CCC1(C)OC1C
CCC1(O)CC1C
COC1(C)CC1C
CC(C)C1(C)CO1
CC(C)C1(O)CC1
CC(O)C1(C)CC1
CC(C)(C)C1CO1
CC(C)(O)C1CC1
CC1C=CC(O)C1
CC1C=CC(C)O1
CC1=CCC(O)C1
CC1=CC(O)CC1
CC1=CC(C)OC1
CC1CCC(C)O1
CC1CCC(O)C1
CC1COC(C)C1
CC12CC(C1)C2O
CC1C2CC1(C)O2
CC1C2CC1(O)C2
CC12CC(O)C1C2
CC1CC2(C)OC12
CC1CC2(O)CC12
CC1OC2(C)CC12
CC12C=CC(C1)O2
OC12C=CC(C1)C2
CC12CC(CO1)C2
CC12COC(C1)C2
CC12CCC(C1)O2
OC12CCC(C1)C2
CC12COC1(C)C2
CC12CCC1(O)C2
CC12CCC1(C)O2
C#CC1CC(C)O1
C#CC1CC(O)C1
CC1CC(CO)C1
CCC1CC(C)O1
CCC1CC(O)C1
COC1CC(C)C1
CC(C)CC1CO1
CC(C)OC1CC1
CC(O)CC1CC1
C#CC12CC(C1)O2
CCC12CC(C1)O2
COC12CC(C1)C2
OCC12CC(C1)C2
CC1CC2(CC2)O1
CC1CC2(CO2)C1
OC1CC2(CC2)C1
CC1(C)CC2OC21
CC1(C)OC2CC21
CC1(O)CC2CC21
CC1(C)C2CC1O2
CC1(O)C2CC1C2
CC1(C)C2COC21
CC1(O)C2CCC21
CC1=CCCOC1
CC1=CCOCC1
CC1CC=CCO1
OC1CC=CCC1
CC1C=CCOC1
CC1C=CCCO1
OC1C=CCCC1
CC1CCCCO1
CC1CCCOC1
CC1CCOCC1
OC1CCCCC1
CC1=C(C)COC1
CC1C=CCC1O
CC1CC=CC1O
CC1=CCCC1O
CC1=CCOC1C
CC1CCCC1O
CC1CCOC1C
CC1COCC1C
CC1CC(C)C1O
CC1CC(O)C1C
CC1OC(C)C1C
CC12CCC1CO2
CC12CCC1OC2
OC12CCC1CC2
CC12CCC1C2O
CC1C2CCC12O
CC1C2COC12C
CC1C2OCC12C
CC12CC=CC1O2
OC12CC=CC1C2
CC12C=CCC1O2
OC12C=CCC1C2
CC12CC1CCO2
CC12COCC1C2
CC12CCOC1C2
CC12CCCC1O2
OC12CCCC1C2
CC12CC1CC2O
CC1CC2CC12O
CC1CC2OC12C
CC1OC2CC12C
C#CCC1CCO1
C#CCC1COC1
CC#CC1CCO1
CC#CC1COC1
CCCC1CCO1
CCCC1COC1
CCOC1CCC1
COCC1CCC1
OCCC1CCC1
CC#CC1CC1O
CC#CC1OC1C
C#CCC1OC1C
C#CCC1CC1O
CC1CC1CCO
CCCC1CC1O
CCCC1OC1C
CCOC1CC1C
COCC1CC1C
CC(C)C1CCO1
CC(C)C1COC1
CC(O)C1CCC1
CC(C)C1CC1O
CC(C)C1OC1C
CC(O)C1CC1C
C1CC2(C1)CCO2
C1CC2(C1)COC2
CC1CC12CCO2
CC1CC12COC2
CC1OC12CCC2
OC1CC12CCC2
CC1C2CC1C2O
CC1C2OC1C2C
CC12CC3C1OC32
CC12OC3C1CC32
OC12CC3C1CC32
CC12CC3OC1C32
CC12OC3CC1C32
OC12CC3CC1C32
C1=CC2CC(C1)O2
C1OCC2CC1C2
C1CC2CC(C2)O1
C1CC2CC(C1)O2
CC1=CC2OC2C1
CC1CC2CC2O1
CC1CC2OC2C1
OC1CC2CC2C1
CC1=CC2CC1O2
CC1CC2CC1O2
CC1OC2CC1C2
OC1CC2CC1C2
CC12CC3C(O1)C32
OC12CC3C(C1)C32
C#CC(C#C)OC
C#CC(C#C)CO
C#CC(CC)CO
C#CC(CC)OC
CCC(CC)CO
CCC(CC)OC
C#CC1(C#C)CO1
CCC1(CC)CO1
CCC1(OC)CC1
CCC1(CO)CC1
C#CC1(CC)CO1
C#CC1(OC)CC1
C#CC1(CO)CC1
C#CCOCC#C
CC#CC#CCO
C#CCC#CCO
C#CC#CCOC
C#CC#CCCO
CCC#CCCO
CCC#CCOC
CCCC#CCO
CC#CCCCO
CC#CCOCC
CC#CCCOC
C#CCOCCC
C#CCCOCC
C#CCCCOC
C#CCCCCO
CCCCCCO
CCCCCOC
CCCCOCC
CCCOCCC
C#CC1OCC1C
C#CC1COC1C
C#CC1CCC1O
CC1CCC1CO
CCC1CCC1O
CCC1COC1C
CCC1OCC1C
COC1CCC1C
C#CC1OC1C#C
C#CC1OC1CC
C#CC1CC1OC
C#CC1CC1CO
CCC1CC1CO
CCC1CC1OC
CCC1OC1CC
C#CC1=CCOC1
C#CC1C=CCO1
C#CC1CCCO1
C#CC1CCOC1
CCC1=CCOC1
OCC1=CCCC1
COC1CC=CC1
OCC1CC=CC1
CCC1C=CCO1
COC1C=CCC1
OCC1C=CCC1
CCC1CCCO1
CCC1CCOC1
COC1CCCC1
OCC1CCCC1
C#CC(C)C1CO1
C#CC(O)C1CC1
CC(CO)C1CC1
CCC(C)C1CO1
CCC(O)C1CC1
COC(C)C1CC1
C#CC#CC1CO1
C#CCCC1CO1
C#CCOC1CC1
CC#CCC1CO1
CCC#CC1CO1
OCC#CC1CC1
CCCCC1CO1
CCCOC1CC1
CCOCC1CC1
COCCC1CC1
OCCCC1CC1
C#CC1C(C)C1O
CC1C(C)C1CO
CCC1C(C)C1O
COC1C(C)C1C
C#CC12CC1CO2
C#CC12COC1C2
C#CC12CCC1O2
CCC12CC1CO2
CCC12COC1C2
CCC12CCC1O2
COC12CCC1C2
OCC12CCC1C2
CC1CCC12CO2
CC1COC12CC2
CC1OCC12CC2
OC1CCC12CC2
C1=CCC2(C1)CO2
C1=CC2(CC2)OC1
C1=CC2(CC1)CO2
C1COC2(C1)CC2
C1CC2(CC2)CO1
C1CCC2(C1)CO2
CC1(C2CO2)CC1
CC1(C2CC2)CO1
OC1(C2CC2)CC1
C1=CC2C=CC1O2
C1=CC2CCC1O2
C1=CC2CC1CO2
C1CC2CC1CO2
C1CC2CCC1O2
CC1C(O)C2CC12
CC1C(C)C2OC12
CC12CC1C1OC12
CC12OC1C1CC12
OC12CC1C1CC12
C1CC1CC1CO1
C1CC1OC1CC1
C1CC12CC1OC12
C1C2OC3(CC3)C12
C1C2CC3(CO3)C12
C1=CCOCC=C1
C1=CCCOCC1
C1=CCOCCC1
C1CCCOCC1
CC1CC2COC12
CC1CC2OCC12
CC1OC2CCC12
OC1CC2CCC12
C#CC1CC2OC12
C#CC1OC2CC12
CCC1CC2OC12
CCC1OC2CC12
COC1CC2CC21
OCC1CC2CC12
C1=CCC2OC2C1
C1=CC2CC2OC1
C1=CC2OC2CC1
C1COC2CC2C1
C1CC2CC2CO1
C1CCC2OC2C1
C1=C2CCC2OC1
C1=C2COC2CC1
C1=CC2OCC2C1
C1=CC2COC2C1
C1CC2COCC12
C1CC2CCC2O1
C1CC2COC2C1
OC1C2C=CCC12
CC1C2CCOC12
CC1C2COCC12
OC1C2CCCC12
C#CC1C2COC12
CCC1C2COC12
COC1C2CCC21
OCC1C2CCC12
CC1C2CC(O)C12
CC1OC2C(C)C12
CC1CC2C(O)C12
CC1=CCC2OC12
CC1C=CC2OC12
OC1C=CC2CC12
CC1CCC2OC12
CC1COC2CC12
CC1OCC2CC21
OC1CCC2CC12
CC1C2C=CC1O2
OC1C2C=CC1C2
CC1C2COC1C2
CC1C2CCC1O2
OC1C2CCC1C2
CC12C3CC1C2O3
OC12C3CC1C2C3
OC12C3C4C3C1C42
C1C2CC3C1OC23
C1C2OC3CC2C13
C1C2CC3C1C3O2
C#CC1C2CC1O2
CCC1C2CC1O2
COC1C2CC1C2
OCC1C2CC1C2
C1CC(C2CC2)O1
C1CC1C1COC1
C1CC(C2CO2)C1
C1C2OC1C21CC1
C1C2CC1C21CO1
CC1C2C3OC2C13
OC1C2C3CC2C13
C1=CC2C3OC2C13
C1OC2C3CC2C13
C1CC2C3OC2C13
CC1C2OC3C1C23
OC1C2CC3C1C23
C1C2OC1C1CC21
C1C2CC1C1OC21
C1OC2C1C1CC12
C1CC2C1C1OC21
CC1C2C3OC3C12

Is there any distinctive feature between these two groups?

For curiosity: May you provide additional details than «to extract some features» (e.g., using pymatgen?, what properties in addition?)?  Was your algorith allowed to pick additional properties deposit in the QM9 set, or to derive secondary properties prior to clustering? +1 for bringing such a topic to ChemSE. A second +1 would be due to leave the jet/rainbow scale and to use a viridis palette instead.

The implementation is original, it's not a package, it is graph based, so it picks atom types and the weighted adjacency matrix (essentially bond orders), nothing else.

The PCA should tell you what features (or linear combination of) each axis represents. From that you should be able to answer the question yourself, by seeing what the values are for each cluster. Sounds like you are doing some cool work, maybe you should find a collaborator.

Perhaps my comment was a little naive. If the "machine learning algorithm" you are using to generate features is not transparent (for instance a neural network?) then I guess you have to take a further step and try to disentangle what the original features mean. Associating abstract ML features with human-understandable ones by crowd-sourcing on chem SE is kinda fun but maybe ultimately kinda a crapshoot. Maybe try translating those SMILES into structures.

https://chemistry.stackexchange.com/questions/43299/is-there-a-way-to-use-free-software-to-convert-smiles-strings-to-structures

By the way: http://www.cheminfo.org/Chemistry/Cheminformatics/SMILES_to_svg/index.html#

I recommend using openbabel as suggested on a number of other answers. I edited the list of SMILES to make it easier to input into another program.

@Blade: I'm trying to build a Stack Exchange where the community is nearly 100% made of the users that are experts in computational/physical/quantum chemistry type questions, where users would not have to see general chem questions like "what is an acid?". There's machine learning questions already proposed there. We need only a few more questions to reach 10 upvotes. In case you were unaware of it, and wouldn't mind helping us, here it is: https://area51.stackexchange.com/proposals/122958/materials-modeling

It seems that the clustering is based on the functional groups. The first group of SMILES are ketones. The second group is a combination of ethers and alcohols.


text:
Oxidation of ascorbic acid (Vitamin C)
What would you suggest me to use for oxidation of vitamin C to dehydroascorbic acid instead of bromine water?

Iron(II), see https://www.sciencedirect.com/science/article/pii/S1319610310001389?via%3Dihub

Why not iodine? It is a standard and quantitative reaction.

Would you describe me procedure with iodine, M. Farooq? Thank you.

@Karsten Theis: Fe(III)?

@user55119 yes, ferric i e iron(III)


text:
Why is NOₓ conversion low at fuel rich conditions in a three-way catalytic converter?

$$
\begin{align}
&\text{Oxidation:} &\quad \ce{2 CO + O2 &-> 2 CO2}\\
& & \ce{``HC" + O2 &-> CO2 + H2O}\\
& & & \\
&\text{Reduction:} &\quad \ce{2 CO + 2 NO &-> 2 CO2 + N2}\\
& & \ce{``HC" + NO &-> CO2 + H2O + N2}
\end{align}
$$

$\ce{NO_x}$ conversion is low at oxygen rich conditions (lean operation) because no $\ce{CO}$ or $\ce{HC}$ are available for the reduction.
But why is $\ce{NO_x}$ conversion also low at fuel rich conditions, so an optima exists? Rich conditions lead to high $\ce{CO}$ and $\ce{HC}$ emission and that should be perfect condition for $\ce{NO_x}$ reduction.

It may be affected by the reaction mechanism and kinetics. More fuel rich conditions mean lower NO concentration. If NO reduction rate decreases with NO concentration faster than concentration itself, e.g. if the reaction kinetic is of the 2nd order wrt NO, it will lead to lower conversion efficiency.

N2 +  02 -> NO this reaction is endotherm.

why do u say fuel rich conditions mean lower NO concentration. Is it cooler or why?

do you have some reference i need to read it myself

For similar reasons, why there is low CH for oxygen rich mixture. No reference available, just applying basic principles.

hm not sure. HC: there is an other reason. Its because  HC are only partly oxidized, because there is not enough oxygen for total conversion. On the other hand if you don't have good HC conversion. NO conversion should be great, if you look at Reduction reactions of my picture.

fuel rich means i dont have total complete oxidation of my fuel: as a result i get more CO and HC. And those molecules are needed for the reduction of NO. This is the reason why NO conversion drops at high oxygen level (lean conditions). A lot of oxygen meaning that i already have only H20 and CO2 and no CO and HC

For fuel rich mixture there is not enough oxygen to create NO, as there is kinetically competing CH oxidation, so NO content is low. If kinetics of the catalytic reduction of NO is d[NO]/dt=k.[NO]^2, then a twice lower NO concentration has four times slower conversion rate.

kinetics should be something like d[NO]/dt=k.[NO][CO] or am i wrong. Ok sure we can say we get less NO because there not enough O2, but this doesnt explain  the conversion of it.

We have lower NO concentration but CO concentration is higher?

Rich mixture=partial only combustion=more CO

Remember that the catalytic NO reduction need not to be a reaction of the 1st (kinetic) order. If it is higher than 1st order, then efficiency of NO conversion is smaller for lower NO concentrations, which occur for rich mixtures.

maybe i have a better explaination.. it is a Langmuir–Hinshelwood mechanism
A + S ⇌ AS
B + S ⇌ BS
AS + BS → Products
S= Surface
The best reaction rate is when both AS and BS concentrations are 50/50 A = NO and B = CO so if our NO concentration is low the surface concentration is also low and its not optimal. this would also explain the optima?

https://en.wikipedia.org/wiki/Reactions_on_surfaces?wprov=sfla1


text:
What does environmental factor include?
Does E-factor include all of the reagents including catalysts, solvents, and reactants that go into making the product? Further, in the E-factor calculation should all the solvents used in work-up (such as water and ammonia solution) also be included?
$$\text{E-factor} = \frac{\text{total waste}~(\pu{kg})}{\text{product}~(\pu{kg})} = \frac{(\text{total reagents} - \text{product})~(\pu{kg})}{\text{product}~(\pu{kg})}$$

What is recycled ( like destilled solvents , or purified reagents) or reused ( like catalysts ) cannot be taken as waste.


text:
Le Chatelier's principle and van 't Hoff equation
When temperature is increased, then according to Le Chatelier's principle the reaction will move in the direction where energy is absorbed. I can understand this, but there is also another fact that the $K$ value depends on temperature. So, if my temperature changes, then $K$ value should also change (van 't Hoff equation).

So does k value changes ?

Find out answer in here.

Pls can u give a simple explanation, I am unable to understand

@Yash The K remains the same if and only if T remains the same. . added  notes to the answer, together with the link to related Q/A.

van 't Hoff equation

$$\frac{\mathrm d}{\mathrm dT} \ln K_\mathrm{eq} = \frac{\Delta H^⦵}{RT^2}$$

= Le Chatelier's principle in a particular context of  temperature and chemical reaction equilibrium.
What the Le Chatelier's principle says  qualitatively as a general principle, the van 't Hoff equation says quantitatively in context of temperature change for chemical reactions.
Van't Hoff equation is the way, how Le Chatelier's principle applies on equilibrium constants of chemical reactions at different temperature.
By other words, attempts to apply both, van 't Hoff equation and Le Chatelier's principle  effectively mean applying Le Chatelier's principle twice.
Generally, all equilibrium constants are temperature dependent, it is not limited to chemical reactions.
$K$ is an equilibrium constant and does change with temperature, quantitatively according to the van't Hoff equation.
Read the Wikipedia articles about the both following the above links.

Consider the Haber-Bosch ammonia production
$$\ce{N2 + 3 H2 <=> 2 NH3}$$
The equilibrium constant is with application of the equilibrium partial pressures:
$$K=\frac{(p_{\ce{NH3}})^2 }{ p_{\ce{N2}}\cdot  (p_{\ce{H2}})^3   }    $$
The reaction quotient  is with application of the current partial pressures:
$$Q=\frac{(p_{\ce{NH3}})^2 }{ p_{\ce{N2}}\cdot  (p_{\ce{H2}})^3   }    $$

Now, application of Le Chatelier's principle at higher pressure of the mixture at the same temperature:

$K$ is the same, as $T$ is the same.
$Q$ is at the same $T$ and composition lower (Check that $\frac{2^2}{2 \cdot 2^3} < \frac{1^2}{1 \cdot  1^3}$).
Products are more preferred for $Q$ to reach $K$.
The reaction decreases pressure, as products have lower pressure for the same volume, acting against increasing the pressure.
As  T remains the same, K remains the same. Q changes temporarily due the change of the pressure, but then it is approaching the K value back again.

Now, application of Le Chatelier's principle at higher temperature with the same pressure of the mixture:

$K$ is lower as $\Delta H_\mathrm{r} < 0$, according to van 't Hoff equation.
As result, reagent side of nitrogen and hydrogen is more preferred.
The forward reaction is exothermic, while backward reaction is endothermic.
Being more pushed toward the endothermic side, it partially consumes thermal energy provided by increasing temperature.
The Q is initially the same as before T changes. But K changes, as T changes. Q is then approaching  the new K value at the changed temperature.

See the related question with some nice simple charts.


text:
Structure of dihydrogen peroxy-phosphate ion
In structure of Dihydrogen peroxy-phosphate ion $\ce{H2PO5-}$, shouldn't negative charge be on oxygen linked directly to phosphorus, rather than on oxygen linked through peroxy linkage with another oxygen? Since in former it will be in resonance with $\mathrm{p}\pi-\mathrm{d}\pi$ multiple bond between phosphorus and oxygen, while in latter these is no resonance. But my friend argues with me that it should be on latter, since there will be much higher stabilisation by inductive effect and there are only two canonical structures of $\ce{H2PO5-}$.


text:
Energy Levels of Molecular Orbitals formed by combination of Atomic Orbitals
In general when we talk about chemical bonding, we say that it is "nature's way to stabilize the systems" and the energy of the molecule formed by atoms will be lower than that of the individual atoms; hence leading to stabilization.
However, while reading about the Molecular Orbital Theory in my NCERT Chemistry Textbook, I came across a statement -

The total energy of the molecular orbitals formed remains same as that of the original atomic orbitals.

How can we then justify through the Molecular Orbital Theory that the process of chemical bonding leads to stabilization, if no energy is released?

Is true just for the orbitals They probably don't have made clear that the statement has no physical content, or when it happens it does not lead to a stable bound species. Indeed is valid, and under assumptions/approximation, only for molecules with bond order equal to zero. Take it as "the bonding and antibonding orbitals are roughly equally spaced from the energy levels of the atomic orbitals". It is all about the energy of empty orbitals. You can order them. Is the same as to say that eight 100 m has more potential energy than sea level, tough there is no stone at that altitude yet.

@Alchemista thanks for the comment. However could you please elaborate your points further in more detail ?? I could not properly understand your point, especially about the stone.

They are speaking about energy of empty orbitals. A stone lift up would have more potential energy. But unless you actually bring it there, no physical energy is involved. The same for mo. The statement is roughly valid for the orbitals formed. If you equally fill both bonding and antibonding then the molecule does not occur.

In other words molecular orbitals exists independently of the real existence of a molecule, they can be used to predict that the molecule does not exist /, is not stable. This should remove your doubt. No energy release no stable bonding. This is the use to be made, not the contrary.

Thanks for the clarification @Alchemista. I do get your point now.


text:
Why is molality per mass solvent rather than per mass of solution?
Knowing the moles of substance per mass of solution would be useful. Pour some solution and you'd know how much substance you have. Why is molality defined not this way, but as moles of substance per mass solute instead? It seems like this unit was engineered to have a value similar to molarity rather than to be maximally useful.
Why was molality defined this way? Is there an advantage I'm missing? This measurement seems easy to prepare but more cumbersome to use, and I can't see why that would be an advantage.

"Why is molality defined […] as moles of substance per solute instead?" — no, molality is the amount of solute per mass of the solvent. And if I understand the question correctly, your proposition to use the amount of solution would make preparation of such solution unnecessarily complex. Besides, what you describe as "moles of substance per amount of solution" is already known as mole fraction.

@andselisk I edited to clarify—I meant mass in every case of unspecified unit. Mole fraction is not per kg but per moles of solution.

To clarify, you seem to be asking about usefulness similar to that of molarity where the solvent is generally considered an inert carrier that is only used for ease of measuring a dose. If molality was defined per unit of solution it would basically be a temperature insensitive molarity that could also be used with solid solutions. Is this rewording accurate?

Now I wonder if there is a mass based version of molarity, or if the need for that level of temperature compensation, or solid solutions, is rare enough that it is simply handled as a custom procedure/unit on an as needed bases. It is easy enough to say each Kg of solution contained 5 moles of solute.

This all probably dates back before digital balances when volumes were quick and masses were cumbersome.

Molality $b$ is defined its way because being proportional for the given solvent to the solute/solvent molecular ratio.
Molarity or molar/mass fractions or molar_amount/solution_mass are not proportional to it.
The boiling and melting point shifts are proportional via ebullioscopic/kryoscopic solvent constants to molality.
$$\Delta T_\mathrm{b}=K_\mathrm{e} \cdot b$$
$$\Delta T_\mathrm{m}=K_\mathrm{k} \cdot b$$

The number of solvent (s) molecules per 1 solute molecules
$$\frac{n_\mathrm{s}}{n}=\frac{m_\mathrm{s}}{M_\mathrm{s}\cdot n}=\frac 1{ M_\mathrm{s}\cdot b}$$
If we provide as the input the solute mass and solvent volume and density, the above molecule ratio is
$$\frac{n_\mathrm{s}}{n}= \frac{m_\mathrm{s}\cdot M}{M_\mathrm{s}\cdot V \cdot \rho}$$

The molality is defined by moles of solutes per mass of the solvent, There is a practical reason for this choice. This concentration unit is mainly used in cryoscopy. In cryoscopy, you start with a given mass of solvent, and you don't want to know about the volume. Then you dissolve a known amount of solute. The final volume does not matter. You just want to know at which temperature the solution begins to solidify or the solid begins to melt. This temperature allows you to determine the molar mass of the solute. And this is why cryoscopy is useful.
For example, adding n moles of a solute to 1 kg solvant makes the freezing point of water decrease by n · 1.86°C. So if you dissolve m grams of a solute in 1 kg water, and if you observe that the freezing point of water is -3.72°C, you know that there was 2 moles of your solute in the m grams used. This allows you to determine the molar mass of the solute. The total volume is of no interest in this determination.
The same measurements can be made in other solvents. In naphtalène, melting at 80°C, the freezing point constant is not $1.86$ , but $6.8$ K L $mol^{-1}$, if the unit molality is used.

The question as it is currently written is not asking about volume.


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Why doesn't H₄CO₄ exist as a stable compound?
Why doesn't $\ce{H4CO4}$ exist as a relatively stable compound? Its heavier homologue $\ce{H4SiO4}$ seems to exist although it is unstable to some extent but new methods have been able to stabilise orthosilicic acid using choline chloride.

It is called orthocarbonic acid and it does not exist.

https://en.wikipedia.org/wiki/Orthocarbonic_acid

@NilayGhosh I just couldn't explain what I mean so I changed the question

Imho, this is a perfectly fine question and there is no reason for it to be closed.

I also think it shouldn't be close

Perhaps the question should be changed to "why is H4CO4 -> H2CO3 + H2O a favorable reaction but H4SiO4 -> H2SiO3 + H2O is not?"

Wikipedia reports that calculations show orthocarbonic acid spontaneously decomposes to the ordinary (meta)carbonic acid $\ce{H2CO3}$ plus water.  However, orthocarbonate esters $\ce{C(OR)4}$ do exist, and more recently salts with the alkaline earth metals have been synthesized under conditions matching the deep mantle of Earth.
The orthoacid may, however, become stable at high pressure [1, with a nontechnical summary given in 2] and possibly exist within Uranus and Neptune.  It is further posited that $\ce{H2CO3}$, ordinarily only a minor species in aqueous solution, becomes a stable solid in the interiors of some icy moons, possibly impacting interactions thought to be necessary for the development of subterranean life.
A final note: the swastika-like shape of the orthocarbonic acid molecule has led unfortunately to the name "Hitler's acid".  Actually the swastika long predates the Nazis and retains non-Nazi usages to this day.


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Acid-catalysed isomerisation of phenyl epoxide to aldehyde
How does $\ce{H3O+}$ cause the formation of an aldehyde from the following epoxide?

I thought that the epoxide would undergo opening to form a 1,2-diol, as illustrated below. Am I wrong?

You are correct if this epoxide is a alkyl peroxide. However, this is a 2-phenyloxirane (styrene oxide or phenylethylene oxide),  which can stabilize a protonated intermediate by resonance, specifically with 2-hydroxyphenyl group. Thus, it can isomerize to phenylacetaldehyde through 1,2-hydride shift (similar to pinacol-pinacolone rearrangement). The following diagram illustrate how styrene oxide converted to phenylacetaldehyde (Ref.1):

Note that styrene oxide is not stabilized by extra hydroxy group here. That's probably the reason the authors have used slica-supported phosphotungstic heteropoly acid in cyclohexane (no other nucleophile there).
The identical isomerization has been used in the synthesis of 2-phenylnaphthalenes from styrene oxides (Ref.2).
References:

Vinícius V. Costa, Kelly A. da Silva Rocha, Ivan V. Kozhevnikov, Elena V. Gusevskaya, "Isomerization of styrene oxide to phenylacetaldehyde over supported phosphotungstic heteropoly acid," Applied Catalysis A: General 2010, 383(1–2), 217-220 (https://doi.org/10.1016/j.apcata.2010.06.005).
Kishor V. Wagh, Bhalchandra M. Bhanage, "Synthesis of 2-phenylnaphthalenes from styrene oxides using a recyclable Brønsted acidic $\ce{[HNMP]+HSO4−}$ ionic liquid," Green Chemistry 2015, 17(8), 4446-4451 (DOI: 10.1039/C5GC01170A).

In the first line, is it peroxide or epoxide?


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Recommendations for textbooks with correct and up-to-date information
I recently began studying chemistry, only to find out that, apparently, concepts such as “octet expansion” and Lewis structures are outdated and incorrect. And I don’t mean they’re approximations of reality, as is the case for numerous theories taught in physics, but, apparently, actually incorrect and deprecated concepts. However, commonly promoted textbooks still teach these concepts as though they are true, with no reference to them being outdated or incorrect. I find this both disappointing and disturbing, since I want to learn concepts that are actually correct. And when I purchase (expensive, recently published) textbooks, I expect to be taught correct and up-to-date concepts — especially since the people who wrote these textbooks are in an academic position to know better.
At the moment, my goal is to study both physical and organic chemistry. I was wondering if people could please recommend good textbooks in these areas that do not teach incorrect, outdated, and deprecated concepts.

All recommendations are listed in Resources for learning Chemistry. Knowing obsolete concepts is somewhat necessary. I don't know how it's these days, but a decade ago "obsolete" systems like MS-DOS and UNIX were still taught  during the computer science course at the high-school, and must I say it helped a lot in understanding higher-level stuff later on.

@andselisk Hmm, I see. In that case, maybe I should stick with those "commonly recommended" textbooks. Thanks for the advice.

Lewis structures are neither incorrect nor deprecated, and commonly used with the assumption that practitioners of chemistry are aware its conceptual role and limitations. It would be very hard to read eg organic synthetic papers without using them.

I disagree with the idea to forget about Lewis formulae because of their age. If you browse across the table of contents of scientific journals, however, you see Lewis formualae are widely used.  From the first December issue of Organic Letters, for example is taken the following screen photo:

If Lewis formulae were so off and wrong, chemists would use other forms to describe molecular structures in their publications.  As so often, however, it is a question about what you want to describe, and to know what limitations of the tools at disposition inheritated.
By convention (and training), for example, chemists do not read the rings in formula about nitronaphthalene (on the left) as an alternation of isolated single- and double bonds, but conceptionalize this as to depict aromatic systems.  Counting electrons, for example, this is easier this way as with the Thiele ring which accentuates the delocalization of the $\pi$-electrons.*)

(source)
So it is better to recognize where Lewis formulae may be misleading, with some instances listed e.g., here, and some you will encounter in study, e.g. diborane

(source)
It equally is helpful to know that dashes in formulae, depending on convention, may represent different aspects:

(source)
*) In 1899, Thiele's perspective was to emphasize that benzene's molecular structure is between the two mesomeric structures expressed as Kekulé-structure and thus used the (initially dashed) cricle in the structural formula.  Chemists' perspective of delocalization of $\pi$-electrons today obviously is influenced by quantum physics.


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Writing a system of differential equations with the given arbitrary reactions
We have the following elementary reactions:
\begin{align}
&\overset{k_1}{\longrightarrow}A\overset{k_2}{\longrightarrow}{B+C}\\
&C \overset{k_4}{\longrightarrow} 2D \overset{k_5}{\longrightarrow}{C}\\
&B \overset{k_3}{\longrightarrow}\\
&D \overset{k_6}{\longrightarrow}
\end{align}
Let $\alpha(t), \beta(t), \gamma(t), \delta(t)$ denote the concentrations of $A, B, C, D$, respectively.
\begin{align}
\frac{\mathrm d}{\mathrm dt}\alpha(t) &= k_1 - k_2\alpha(t)\\[3pt]
\frac{\mathrm d}{\mathrm dt}\beta(t) &= k_2\alpha(t) - k_3\beta(t)\\[3pt]
\frac{\mathrm d}{\mathrm dt}\bigl(2\delta(t)\bigr) = 2\frac{\mathrm d}{\mathrm dt}\delta(t) &= k_4\gamma(t)-k_6\delta(t)^2 \quad \text{or} \quad \frac{\mathrm d}{\mathrm dt}\delta(t) = \frac{k_4}{2}\gamma(t) - \frac{k_6}{2}\delta(t)^2\\[3pt]
\frac{\mathrm d}{\mathrm dt}\gamma(t) &= k_5\delta(t)^2 - k_4\gamma(t) + k_2\alpha(t)\\
\end{align}
Is my systems of equations correct?

No. There must be an instance of $\delta^2$. Also, where did the $\frac{d}{dt}(\beta + \gamma)$ thing come from?

Where is the law of mass action (I'm surprised you are invoking it for a non-equilibrium system)?

@IvanNeretin: It came from $A \overset{k_2}{\longrightarrow}B + C$?

@KarstenTheis: Oh no. I must have misunderstood what it meant. Sorry, my background is not in chemistry. I will remove it. Thanks!

@TheLastCipher Well, this reaction does not produce B or C; instead, each time it runs, it produces B and C, so it should be represented by two addends, both in $d\beta\over dt$ and $d\gamma\over dt$.

@IvanNeretin: What do you mean addend? I did write $\operatorname{\frac{d}{dt}}\beta(t) + \operatorname{\frac{d}{dt}}\gamma(t)$. Is that different? Thank you for your time answering.

@IvanNeretin: Also, should I have $\operatorname{\frac{d}{dt}}\gamma(t) = 2k_5\delta(t)^2 - k_4$, instead? Would love to know why it is suppose to be $\delta(t)^2$. Thank you very much!

I meant an addend in the RHS. On the left you shouldn't have any addends at all, just $d\beta\over dt$. As for $\delta^2$, well, it is because the law of mass action says so. We are probably going to apply it somewhere outside of its rightful domain, but then again, what choice do we have?

@IvanNeretin: I've simplified the equation with an addend on the left to be the equation to the left of the $\Leftrightarrow$ symbol. Maybe I should just remove the "unsimplified" form?

Maybe. Also, the $k_4$ term (the one without the $\gamma$) seems wrong.

@IvanNeretin: Turns out they aren't the same. I have edited it now. My brain was stuck applying the rules on derivatives, but I now get where went wrong, conceptually. Thanks!

Now I don't like the derivative of $2\delta$. This is not how it works at all. You run the reaction once, you get 2D for one C. The change in D is twice greater than that in C. The derivative of D should be greater, not smaller.

$\displaystyle\frac{\mathrm d}{\mathrm dt}\alpha(t) = k_1 - k_2\alpha(t)$ is correct.
Just to show you the way, here are the next equations:
$$
\begin{align}
\frac{\mathrm d}{\mathrm dt}\beta(t) &= k_2\alpha(t) - k_3\beta(t) \\
\frac{\mathrm d}{\mathrm dt}\gamma(t) &= k_2\alpha(t) - k_4\gamma(t) + k_5\delta(t)^2 \\
\frac{\mathrm d}{\mathrm dt}\delta(t) &= 2k_4\alpha(t) - k_6\delta(t) - k_5\delta(t)^2
\end{align}
$$

we are left with $2\operatorname{\frac{d}{dt}}\delta(t) = k_4\gamma(t) - 2k_6 - k_5\delta(t)^2 \overset{equivalently}{\Longleftrightarrow} \operatorname{\frac{d}{dt}}\delta(t) = \frac{k_4}{2}\gamma(t) - k_6 - \frac{k_5}{2}\delta(t)^2$?

I appended the last equation from your another answer and cleaned up syntax a little. Please note there is a gray edit button on the bottom of the posts, which allows to commit changes to the Q&As, which is a preferred way over adding a new post if the addition is marginal. Also, please avoid breaking math expressions in the middle with dollar signs for no reason: $x + y$ is correct, $x$ + $y$ is wrong. Math operators are usually typed in upright typeface to avoid confusion with variables.

@Maurice: Why the $2k_4\alpha(t)$ term on the rate of change of $\delta(t)$?


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Comparing basic strength order of amines

In the above question I was asked to compare the basic strength order.
So I did it like this:
The Nitrogen in I and II will not participate in resonance so the only effect it could produce is inductive (-I) .
Now the compound IV has less resonating structures than the other compounds so the the Nitrogen atom will have larger electron density.
So IV should be most basic then III > II > I because of (-I) effect.
But my book mentions (C) as the answer.
How? Any help will be appreciated

You don't have problem with assigning compound IV (1, 4, 5, 6- tetrahydropyrimidine) as the most basic compound among given four amines. I agree with you, but not the same reasons. According to Ref.1, the 1,4,5,6-tetrahydropyrimidines (THPs) are extremely basic compounds, as might be expected from their amidine structure (e.g., 1,8-Diazabicyclo[5.4.0]undec-7-ene or DBU). The parent THP, 1,4,5,6-tetrahydropyrimidine (IV) has a $\mathrm{p}K_\mathrm{a}$ of about $13$ at $\pu{20 ^\circ C}$, which is even reacted with water (Ref.1). This highly basic nature of IV can be attributed to its stabilization by resonance between two equivalent structures, V and V', when protonated:

Also, keep in mind that it can also get protonated on 3-nitrogen atom (VI; inside the blue box) as well (Ref.1).
Among the rest of other 3 amines, two are aminopyridines (I and II), and the third one is aniline (III). Since the lone pair of nitrogen in pyridines does not contributed to aromatic system (hence it is readily available for proton) while that of aniline is involved in aromatic system by resonance (not readily available), generally pyridine is more basic than aniline. Even though I and II both have extra amino group, they are expected to behave like pyridines. By this analysis, we can conclude that the least basic amine among given four amines is aniline (III).
Now, let's look at what happens when I (4-aminopyridine) is protonated:

From above illustration, you can clearly see conjugate acid of I is well stabilized by resonance including full contribution of the lone pair of 4-amino-nitrogen. If you visualize the same with II (3-aminopyridine), it is not stabilized by the extra amino grop as depicted in following diagram:

Thus, by these facts, we can conclude that II is less basic than I. Therefore, the order of basicity strength of four given amines is: IV > I > II > III.
References:

D. J. Brown, R. F. Evans, “100. Hydropyrimidines. Part I. 1, 4, 5, 6- Tetrahydropyrimidine and its derivatives,” Journal of the Chemical Society 1962, 527–533 (https://doi.org/10.1039/JR9620000527).

How to decide which is the basic site of a compound?

The lone pair of pyridine $\ce{N}$ is fully available than that on 3- or 4-amino $\ce{N}$.

@MathewMahindaratne I had some confusion regarding (I) and (IV). Both have lone pair of electrons which are localized hence good basic sites. The (+) charge on both is stabilized by nitrogen. They both suffer (-I) due to nitrogen and the benzene ring respectively. So how do you decide between them? Infact (I) has more resonance structures hence its (+) charge should be better stabilized without compromising the electron density since lp are not in the plane

@Ashish: (IV) is not aromatic amine, and everything else are. Thus, most basic amine is (IV) due to that fact alone. (III) has on ring nitrogen (not a pyridine), hence it is the least basic. Your argument applies between (I) and (II), which I distinguished in my answer.

But thats my doubt, why does being an aromatic amine reduces its basicity even though the electrons of the amine are not delocalised in the ring ?

@Ashish: You should have read my answer very carefully. The question about basicity between aromatic vs aliphatic amines is a separate question. I'm sure someone in CSE has answered that question. You can search CSE for that, and if there is none, you should post it as a separate question. Until then I think this discussion is over.


text:
Role of phosphoric acid in redox titrations
For redox titrations of $\ce{Fe^2+}$ versus $\ce{KMnO_4}$ or versus $\ce{K_2Cr_2O_7}$  , role of phosphoric acid is quite unclear to me. In dichromometric titration, $\ce{H_3PO_4}$ complexes $\ce{Fe^3+}$ and thus lowers the  potential of the $\ce{Fe^3+/Fe^2+}$ system, but if it is a permanganometric titration, $\ce{H_3PO_4}$ strictly complexes to mask the yellow colour of $\ce{FeCl_3}$ but plays no part in lowering the potential of the $\ce{Fe^3+/Fe^2+}$ system. (role of ZR solution if $\ce{Cl-}$is present in the system )
My professor strictly mentions this as a key point of difference between roles of phosphoric acid in these two types of titrations and we cannot write the lowering potential point for the latter but that makes no sense to me. I do not understand why $\ce{H_3PO_4}$ in titrations versus $\ce{KMnO_4}$ will not reduce the potential of $\ce{Fe^3+/Fe^2+}$ system.

I agree with you. I don't see why $\ce{H3PO4}$  should react with $\ce{Fe^3+}$ ions after the oxidation of $\ce{Fe^2+}$ by $\ce{MnO4-}$ and should not do it after the oxidation of the same ion by $\ce{Cr2O7^2-}$

I don't know why your professor is unnecessarily creating an example which has no real value! No one would do potentiometric titrations of iron in the presence of phosphate.
(i) Potentiometric titrations have nothing to do with color or need masking.
(ii) One would need to do a preliminary separation of iron via ion exchange resin.


text:
How would I measure the Carbon Dioxide content in Coca Cola over a period of time?
I'm sorry if this is a broad question, but I am trying to plan a simple experiment. I am wondering is there a somewhat simple way of continuously measuring $\ce{CO2}$. I already know the initial carbon dioxide content is around $\pu{2.2g}$ from research, but I don't know how this result is reached.

Monitor pH continuously and stop when you hit the pH that corresponds to atmospheric $P_\ce{CO2}$ equilibrium in solution.

You want to know how the initial CO2 content of 2.2 grams is determined or do you want to plan an experiment to monitor the loss of CO2 from Coca Cola over time?

@ToddMinehardt, there is citric acid and phosphoric acid too. Not sure if the pH will change significantly.

@M.Farooq - True, good point. That Coke might go well with a Twinkie, reported to have a shelf life of 25 days...

With some control measurements (tap water, carbonated water and flat soda) and calibration curves, IR spectroscopy might provide a contactless and direct quantification of CO2 content.

Measure the change in mass over time of the remaining liquid. Though some water will also evaporate, you can control for that by keeping the humidity near 100%.
If you have to be precise, collect the outgassed $\ce{CO2}$ in a liquid-nitrogen cold trap. Check the mass of the condensate, which should equal that lost from the soda. Check purity, to be really finicky.

You can also get flat Coke to compare, to get an approximation of how much mass loss is due to water evaporation (although the evaporation rate of water from flat soda may not be exactly the same as from carbonated).

@Acccumulation, the question asks for a , "way of continuously measuring CO2." The flat soda only gives an end-point... and what determines flat? One hour? One day? One week?

Measure concentration in a fixed volume of air
Easiest way might be to put it in a sealed impermeable bag or box and measure the CO2 concentration over time. Home hobbyists use something like the MG-811 CO2 sensor or a simple non-dispersive optical sensor and connect it to a Raspberry Pi or Arduino using an analog to digital converter chip (ADC). For (a little) more on that see this answer to Why are CO2 sensors so expensive when CO sensors aren't?

...CO2 sensors are either solid electrolyte chemical cells like the MG-811 or newer versions (see this question) or they are Non-dispersive Infrared Light sensors (NDIRs) based on optical absorption of a certain wavelength of infrared light, using a hot filament or other light source to produce IR and a narrow multilayer optical filter to select one of the strong absorption lines of CO2. The more CO2, the more the reduction in the transmission of light in the optical path of the sensor.

You would estimate the total volume of trapped air and multiply by the fraction that is CO2 (at each time point) to get the accumulated amount of CO2 released by the liquid.
Measure amount collected in a CO2 scrubber
Building on @DrMoishe Pippi's proposed technique though I can't guarantee precision, you could try to use something else to absorb CO2 besides a cold trap.
I propose that you consider a carbon dioxide scrubber.
However, lithium hydroxide may not be suitable for home use as it is very dangerous to touch or otherwise contact! So you should do this with supervision.
From PubChem; Lithium hydroxide (Compound)

Lithium hydroxide, solution appears as a clear to water-white liquid which may have a pungent odor. Contact may cause severe irritation to skin, eyes, and mucous membranes. It may be toxic by ingestion, inhalation and skin absorption. It is used to make other chemicals.

Regenerative carbon dioxide removal system
The regenerative carbon dioxide removal system (RCRS) on the space shuttle orbiter used a two-bed system that provided continuous removal of carbon dioxide without expendable products. Regenerable systems allowed a shuttle mission a longer stay in space without having to replenish its sorbent canisters. Older lithium hydroxide (LiOH)-based systems, which are non-regenerable, were replaced by regenerable metal-oxide-based systems. A system based on metal oxide primarily consisted of a metal oxide sorbent canister and a regenerator assembly. It worked by removing carbon dioxide using a sorbent material and then regenerating the sorbent material. The metal-oxide sorbent canister was regenerated by pumping air at approximately 400 °F (204 °C) through it at a standard flow rate of 7.5 cu ft/min (0.0035 m3/s) for 10 hours.

Exactly how to do it will require some input by an actual chemist. That might be best asked as a new question "How to use off-the-shelf LiOH (pellets or powder) as a home-made CO2 scrubber?" for example.
If you put the open container in a larger container (sealed, impermeable plastic bag or box) with something else that absorbed CO2 and held on to it at room temperature, you could weigh it before and after as a lower limit at least to the CO2 lost by the liquid.
If you have ever seen the movie Apollo 13 or know the story of the CO2 scrubber canister (1, 2) you might recollect that it removed the astronaut's exhaled CO2 from the cabin air and held it within the solid LiOH pellets it contained.
Because they needed it to be compact but to scrub a large volume of air, the spacecraft used a fan to circulate air through the scrubber. But in your case it might be possible to use lithium hydroxide or another material like silver oxide (see this answer and EMU METOX Performance Testing).
Here is some data from that report. You can see that the ~14 kg of silver oxide could hold roughly 0.8 kg of CO2.

History, from NASA's Building an Apollo 13 LiOH Canister Adapter:

Pretty much any strong base can be used to scrub $\ce{CO2}$ from the air. Sodium hydroxide, for example, is commonly available in general stores. While it's also caustic (like any strong base) and needs to be handled with care, the precautions needed are merely the general ones for safe handling of laboratory chemicals: basically keep it out of your eyes and mouth, don't inhale the fumes if you heat it, try not to get any on your skin and if you do, rinse it off promptly.

For a direct measurement of $\ce{CO2}$ content, it is possible to use infrared (IR) spectroscopy. Infrared radiation can be absorbed by molecules, causing them to vibrate. The exact frequency of the radiation absorbed is often characteristic of certain bonds or functional groups.
In particular, the fundamental antisymmetric stretching mode of $\ce{CO2}$ in aqueous solution (specifically the carbon-12 isotopologue $\ce{^{12}CO2(aq)}$, not carbonic acid $\ce{H2CO3}$) is excited by photons of wavelength $\mathrm{4268\ nm}$ (more commonly reported as $\mathrm{2342.9\ cm^{-1}}$ wavenumbers for historical reasons). This vibrational mode is useful for the following reasons:

The absorption peak is quite sharp, with a "width" (FWHM) of only $\mathrm{9\ cm^{-1}}$
The molar absorptivity is strong, at $\mathrm{1.5 \times 10^6\ cm^2\ mol^{-1}}$
The absorption intensity varies linearly with concentration
Few other substances have strong absorption in the same region

Altogether, these properties make the peak suitable for analytic measurements. This has been discussed in Vib. Spect., 1992, 4 (1), 105-108. Though the article did not explore the possibility, it's likely the technique could be adapted for use in sodas. Measurements could potentially be performed directly in unopened bottles. Furthermore, similar benefits exist for measuring gaseous $\ce{CO2}$, so an independent measurement of the space above the liquid could be performed with the same instrument, and the results cross-checked for additional accuracy.
This wouldn't be very easy for a hobbyist doing kitchen chemistry (though handheld/compact/portable IR spectrometers exist and aren't tremendously expensive), but such a study would be very well within reach of most chemical laboratories.

You might want to use a $\ce{CO2}$ ion selective electrode, such as PS-3517 from PASCO:

This Carbon Dioxide Ion Selective Electrode (ISE) Probe is a gas-sensing electrode that allows fast, simple, eco-nomical, and accurate measurement of dissolved CO2 in aqueous solutions

The advantage over other methods is that you measure the concentration in situ and therefore do not have to concern yourself with how to capture $\ce{CO2}$ quantitatively (weighting small amounts), how to discriminate $\ce{CO2}$ from other gases (for instance water, which must be removed from the gas or prevented from condensing in the $\ce{CO2}$ trap), on complicated intermittent sampling, and on how the rate of capture might affect your measurement.
It should work with Coca-Cola:

pH Range: Samples and standards must be adjusted to below pH 4.00

+1 This could potentially (pun intended) be the best overall solution for the OP, but what's the minimum equipment necessary to read the probe? Can it be done with a variety of meters or electrometers that could be borrowed (something with very high impedance that reads millivolts), or would this require a purchase specifically of a "PASCO box" of some kind? Would a calibration solution also be needed as well or does this kind of probe yield at least approximate results without using a standard solution?

@uhoh If I understood correctly the probe can be operated connected to a standard pH meter, see eg an instruction manual here: https://d2n0lz049icia2.cloudfront.net/product_document/Carbon-Dioxide-ISE-Probe-Manual-PS-3516.pdf You do need to calibrate it, this requires appropriate calibration solutions which can be purchased separately.

Actually, a CO2 standard solution is included with the probe. The meter would be calibrated with regular pH calibration solutions.

A very different method which should work well given the high initial level of carbonation in Coca-Cola (141 mM by my estimation based on this post) is $\ce{^{13}C}$-NMR. See for instance Reference 1. Spectra of liquid solutions are high in resolution and one dimensional data can be acquired continuously in a span of a few minutes per spectrum.
Note the spectra might be acquired at natural $\ce{^{13}C}$ abundance (about 1%) or spiked (enriched) with $\ce{^{13}C}$, which of course will be (a lot) more expensive. You might on the other hand be able to capture for reuse the released $\ce{^{13}C}$ with one of the methods suggested in one of the other answers.
References

Qi Yang,Mark Bown, Abdelselam Ali, Dave Winkler, Graeme Puxty and Moetaz Attalla. A Carbon-13 NMR Study of Carbon Dioxide Absorption and
Desorption with Aqueous Amine Solutions. Energy Procedia 1 (2009) 955–962


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Is specific rotation defined for mixtures of enantiomers or meso compounds?
I recently had to do an experiment where we had to make a racemic mixture of D and L complexes and then resolve the enantiomers, and measure the optical rotation of the racemic mixture, and the individual salts. After that we had to calculate the specific rotation in case.
However, when reading about optical activity, I came across the definition:

For an optically active substance, defined by $\ce{[α]^θ_λ = α/γl}$, where α is the angle through which plane polarized light is rotated by a solution of mass concentration γ and path length l. Here θ is the Celsius temperature and λ the wavelength of the light at which the measurement is carried out.

Source: CRC handbook of Chemistry and Physics
Also, specific rotation is treated as an intensive property of a compound (i.e. one enantiomer). Given that, does it make any sense to calculate the specific optical rotation for a mixture of enantiomers (whether racemic or not)?
And applying the same logic, can we say that specific rotation has no meaning for a meso-compound ?

It would help if you could please put references when you answer, because my confusion is about the definition itself.

If you knew for certain that you only had a mixture of two enantiomers in a solution, but didn't know the relative concentrations, you could use OR to determine them. The specific rotation for a meso compound would be zero, so it does have meaning, at least in the sense that it tells you the compound is not optically active.

An observed rotation of zero degrees does not necessarily mean a racemate, meso or achiral compound. If zero is obtained, the sample should be diluted and the rotation repeated to assure that the original reading was not a multiple of 360 (0) degrees.

@user55119, Is there any real example of 360 degree rotation due to a mixture? Never heard of it.

@M. Farooq: In part it is a philosophical issue. Observed rotation depends on concentration. I was not referring to specific rotation.

@user55119, My gut feeling that 360 degree is theoretically impossible.

@M.Farooq: Read this: Large and small rotations. https://en.wikipedia.org/wiki/Specific_rotation

My apologies, but I don't trust Wikipedia as a scientist especially when there is no authentic reference in that section. As you know it is open to editing by everyone. A polarimeter cannot measure above 180, if I remember correctly, but molar optical rotation can be large, but that is a calculated number. An interesting question is how to observe a >360 degree rotation experimentally?

Nothing stops you from taking the equation and calculating a specific rotation of a mixture. Indeed, as far as I am aware when establishing a new asymmetric reaction, the e.e. is measured by something like chiral HPLC (which will unambiguously give two peaks that can be integrated and compared) while the specific rotation is measured for the resulting mixture of enantiomers as is. So the paper will say something like $99~\%$ yield ($98~\%\ \text{e.e.}$) but below there will be a standard calculation of $[\alpha]_\mathrm D^\pu{20^\circ C}$.
So obviously you can measure a specific rotation of the racemate or meso compound but whatever value you measure should fall into the error of the polarimeter so that the final value of $[\alpha]_\mathrm D^\pu{20^\circ C}=0$.