text:
What replaced lead-based pigments?
In the 1960s and before, even black and white letters had some lead from the actual metallic letters used by typesetters—some rubbed off. Same thing with books printed before 1970 or so and libraries were concerned about kids' books and started to take the older ones off the shelves.
But what had a really large amount of lead was the funny pages, the pigment itself had lead and other metals. I assume this is no longer the case and if so, what is now used for color in newspapers and magazines?

Presently, the white color of the paper is due to $\ce{CaCO3}$ for ordinary paper, and by $\ce{TiO2}$ for precious documents.

Documents were not a major source of lead in the environment. Paint and some ceramic glazes were far more likely to cause danger to people. I actually doubt the idea that lead from typesetting was ever notable, pigments possibly so but I'd need to see real evidence lead was widely used.

Into the 70's the better white paints were tri-metal; lead, zinc and tin (oxides). Titanium oxide (rutile) replaced the lead and the zinc and tin adjusted. The titanium has excellent hiding power so the whiteness actually improved. In addition in urban locations, the lead yellowed with age because of sulfur pollution. So a win/win for lead removal. Calcium carbonate is primarily a filler to reduce cost. Answer for house paint not printing inks.


text:
Is the cell in an isothermal titration calorimeter open to the atmosphere?
I'm just trying to understand how they can model an isochoric process.

Please include everything in the question body, which is not just an addendum to the title.

It models an isobaric process:

The titration is performed at constant pressure and temperature, meaning that a single ITC experiment offers data on the binding enthalpy, the equilibrium association constant and the stoichiometry, from which the entropy of binding and Gibbs energy can be computed. Hence, a single ITC experiment offers direct access to the key thermodynamic potentials related to the interaction process — Gibbs energy, enthalpy and entropy.

Source: https://www.azom.com/article.aspx?ArticleID=12258
To keep a constant pressure, you need to keep the cell open. You can limit the heat transfer by having a small opening and quickly adjusting the temperature of the reaction mixture to the ambient temperature.
The thermodynmanic parameters extracted from the raw data are Gibbs energy, enthalpy and equilibrium constant, all appropriate for an isobaric process.

Yes, in my experience with a old GE-branded ITC-200 the cell is not sealed. The syringe for titration does cover a lot of the opening, but it is not sealing the cell from the atmosphere

source: Stack Exchange

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text:
Confusion with when is asking about impure mass of the component and pure mass
I've had this problem just to illustrate my confusion better:

An impure mixture of $Fe_2O_3$ that reacts with commercial $HCl$ with a density of $1.19 g/mL$, that contains a 35% of mass of the pure acid.
a) Find the percentage of pure $Fe_2O_3$ if 5 grams of this component react exactly with 10mL of the acid.

One of my main confusion is if the density given will be of the pure or impure mass, but I decided it to be referring to the pure mass, same with the volume, I assumed they were talking about pure volume of the component. Please let me know if this is wrong logic. So with what I said: $$m_{pure}=d_{pure}V_{pure}$$ and then since I assumed that $m_{pure}=0.35m_{total}$ then $$m_{total}=0.35d_{pure}V_{pure}$$ plugging the info above: $$m_{total}=4.165g \text{ of } HCl$$ then here comes another one of my big conceptual questions:
When you operate between different components, are you reacting with the $m_{total}$ or with the $m_{pure}$ mass of, in this case, $HCl$?
I assumed you react with the whole $m_{total}$? Please do let me know if this is wrong logic. Then with what I just assumed: $$\frac{159.6g \ Fe_2O_3}{219g \ HCl}=\frac{x \ g \ Fe_2O_3}{4.165g\ HCl}\implies x=3.035g \text{ of the total impure }Fe_2O_3$$ and hence $\%=\frac{3.03}{5}\cdot 100=60\%$ the fun fact is that I get the correct answer, but the logic I used seems completely flawed and what confused me most, is that I even contradict myself at some times I think, specially at the end since the formula is pure over impure and not otherwise, but still gets me the correct answer. Anyone can shine a light on all my questions, specially the ones I specifically wrote in bold?
Please don't post an alternative solution, I already have tons of those, just want to solve my doubts with the method I just mentioned and all the questions I wrote, thanks.

The density relates to the acid solution, it has no relation to purity/impurity.

I am sorry, your approach is quite incorrect and this is why you are confused. However, at the end you state that you do not want an alternative solution, which would be the rational solution.

"35% of pure acid" is misleading expression, there is no such thing as pure HCl acid. HCl is a gas.

@Poutnik okay that's clearer, what about the second question I wrote in bold?

Dear editors, please refrain from incomplete edits. Especially serial ones, and especially to the posts of new users. This causes more harm than good because it bumps the question to the main page over and over again and the question, still not being properly corrected properly, attracts downvotes and annoys other users. Instead, make a single solid edit that does it all for OP and community.

As Poutnik has been hinting at, density has no relationship with the purity of an HCl solution. There are density vs. concentration tables for acids but you don't need them. From the density, you cannot extract the moles of HCl directly. You will have to go through the molarity route and life will become much easier.
Also, ignore the 35% pure acid statment, probably that is adding more problems. It is a very simple problem and the logic is as follows. You will see where density of the acid is used.
You have 35% w/w HCl solution (the question should specify weight/weight). All that means is in 100 g of acid solution, you have 35 g HCl (dissolved gas, Mol. wt 36.458) or 35 /36.458  = 0.96 moles of HCl.
Now we need molarity of the HCl solution: We know the solution mass, and we know the solution density, 1.19 g/mL.
The volume of 100 g of acid solution = 100 g/1.19 g/mL = 84 mL
The molarity of HCl solution is 0.96 moles/ 0.084 L = 11. 43 M.
Now write a balanced equation between $\ce{Fe2O3}$ and HCl, and see what the products should be.
From the balanced equation determine the moles of HCl consumed  ( Molarity= moles/Volume)!
From the stoichiometric coefficients, determine the moles and hence the mass of $\ce{Fe2O3}$.

−1 for 1. jargonic and physically wrong term "moles" for the amount of substance; 2. improperly modified unit symbol (% w/w); 3. incorrect abbreviation "Mol. wt"; 4. missing units; 5. inconsistent spacing; 6. "textual" formulas; 7. half of the answer is not the answer, rather directions and suggestions that should've been sorted out in the comment prior to posting.

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text:
Is there a difference in electrolsys rate limiting effects between electrodes of different area if one reverses the electrical polarity?
If one has two concentric stainless steel tubes with the small one being 1/2 the circumference and inside the other, and KOH for the electrolyte, then given the different surface areas between the electrodes, will the cell perform the same if the applied voltage is reversed, or will one way conduct better than the other at low overvolt potential and would same be true at a high overvolt potential.
I suspect that one rate limiting effect arises from hydrogen production being twice the volume of oxygen, that this particular effect would more greatly limit high ionic currents due to more restricted access to the electrode surface area. Is this so, and are there other rate limitting effects that may cause a performance difference between the current direction between two different sized electrodes.

Your assumption is reasonable: an electrolytic cell becomes "polarized" from gas buildup on an electrode surface, and the greater amount of hydrogen, as opposed to oxygen, on the smaller electrode would polarize it more quickly.  There also might be secondary effects due to greater current density. [Dissimilar metals, such as Al and Cu, can be used to make an electrolytic rectifier.]


text:
What would happen if I increase the pressure and temperature of a gas present in a container
Assume the gas to be a real gas and the container to be made of diamond or anything stronger than that. Take the amount of gas to be 2mL and the pressure is increased with the help of strongest pistons on earth. And the temperature is beyond the critical temperature of that gas in order to prevent phase change

https://chemistry.stackexchange.com/questions/60984/will-supercritical-nitrogen-stay-supercritical-if-we-raise-temperature-and-press

The gas would become solid, as the gas to solid transition does not have critical temperature, at least for usual orders of temperature. The needed pressure is just a question of the gas temperature. If we can achieve such a pressure is another question.

Is it applicable for all the gases?

All (real) gases. Just curious side info: helium is the only gas that cannot get solid at room pressure at any temperature. It needs several atm for that.


text:
Characteristics of cations with KCN and excess KCN

How many cations from $\ce{Ag+,Pb^2+,Cu^2+,Cd^2+,Fe^2+,Fe^3+}$ form precipitate with $\ce{KCN}$ but also get dissolved in excess $\ce{KCN}$?

I have just been introduced to Qualitative Analysis of a few cations and came across this problem. All of this is rather memory based applications for me because I'm preparing for a test, so I knew a few of the reactions but I'm not sure about them:
(I have a feeling that $\ce{Pb^2+}$ would not satisfy the conditions given in the problem)
$$\ce{Ag+ ->[KCN] \underset{White ppt}{AgCN} ->[excess KCN] K[Ag(CN)2]}$$
$$\ce{Cu^2+->[KCN] \underset{Yellow Solution}{Cu(CN)2} ->[KCN] \underset{White ppt}{CuCN}  ->[excess KCN] K3[Cu(CN)4]}$$
$$\ce{Cd^2+->[KCN] \underset{White ppt}{Cd(CN)2} ->[excess KCN] K2[Cd(CN)4]}$$
$$\ce{Fe^2+->[KCN] \underset{Green ppt}{Fe(CN)2} ->[excess KCN]K4[Fe(CN)6]}$$
I'm not quite sure about $\ce{Pb^2+}$ or $\ce{Fe^3+}$ , I would appreciate if someone could help me out with the equations for these two and if the equations written above are right or not?

Browsing through the Net, it seems that lead cyanide and ferric cyanide do not exist. They are not commercially available. $\ce{Pb(CN)2}$ is mentioned as being hydrolyzed in water in $\ce{HCN + Pb(OH)2}$. Ferric cyanide $\ce{Fe(CN)3}$ is mentioned as being rather ferric ferricyanide with a formula like $\ce{Fe^{III}[Fe^{III}(CN)6]}$.

Converting my comment into an answer.

You are correct with your analysis. Lead(II) is the only cation to not form a complex with excess cyanide. It forms the white $\ce{Pb(CN)2}$ and is insoluble in excess cyanide which helps is distinguishing with mercury(I) and silver(I) which forms the complex.
Copper(II) and cadmium(II) also forms the cyanide complex (with the said path you wrote) and the good thing is that this test can be used to separate them. $\ce{[Cu(CN)4]^3-}$ is so stable that it hardly has any free copper(I) ions and hydrogen sulfide won't precipitate out copper(I) sulfide while $\ce{[Cd(CN)4]^3-}$ is not so stable and it will precipitate out yellow cadmium sulfide on adding hydrogen sulfide.
$$\ce{[Cd(CN)4]^3- + H2S -> \underset{yellow ppt.}{CdS \downarrow} + 2H+ + 4CN-}$$
Both Fe(II) and Fe(III) will form hexacyanoferrate salts on excess cyanide through intermediate iron(II)/iron(III) cyanide.
$$\ce{Fe^2+ + 2CN- -> \underset{yellow-brown ppt.}{Fe(CN)2 \downarrow} ->[4CN-] \underset{pale-yellow soln.}{[Fe(CN)6]^4-}}$$
$$\ce{Fe^3+ + 3CN- -> \underset{reddish-brown ppt.}{Fe(CN)3 \downarrow} ->[3CN-] \underset{yellow soln.}{[Fe(CN)6]^3-}}$$
Free $\ce{Fe^3+}$ might react with ferricyanide to form brown iron(III) ferricyanide (un)commonly known as "Prussian brown" but it is hard for free $\ce{Fe^2+}$ to react with ferrocyanide to form iron(II) ferrocyanide as it quickly oxidize in air to form Prussian blue.

Potassium ferrocyanide and potassium ferricyanide are historically known under the empirical names yellow and red Prussiate of potash, respectively. Czech names are in translation yellow and red "blood salt", referring to source material.


text:
What is the chemical result and resulting product of putting alabaster (calcium sulfate dihydrate) in water?
As a sculptor, I use alabaster. I have tried using flowing water and I know it dissolves slowly. I want to try exhibiting a sculpture immersed in water either in a sealed or non sealed container. Are the resulting gases or solutions hazardous?

Calcium sulfate dihydrate has a solubility in water of 0.26 g/100ml at 25 °C, so it dissolves slowly - you already know this. No gases will be evolved on dissolution so there is nothing to worry about there.
With respect to the resulting solution this is from the Wikipedia page on Calcium Sulfate:

The calcium sulfate hydrates are used as a coagulant in products such as tofu. For the FDA, it is permitted in Cheese and Related Cheese Products; Cereal Flours; Bakery Products; Frozen Desserts; Artificial Sweeteners for Jelly & Preserves; Condiment Vegetables; and Condiment Tomatoes and some candies. It is known in the E number series as E516, and the UN's FAO knows it as a firming agent, a flour treatment agent, a sequestrant, and a leavening agent.

Not much to worry about there either, just dilute it with water for disposal and rinse off any that gets on your skin.


text:
Why water doesn't break anionic part of oxyacids?
Recently, I was studying about oxyacids and my  teacher took example of a general compound $\ce{X-O-H}$. He told that that on dissolving it into water ,water would break apart either $\ce{O-H}$ bond or $\ce{X-O}$ bond depending upon the electronegativity difference and hence they would be called an oxyacid or oxobase.
But I want to know whether water would also separate the remaining part as the left out atoms also have electronegativity difference in $\ce{X-O}$ and $\ce{O-H}$. Is the explanation hidden in negative charge of remaining part? If not why?

That's a huge simplification. Continuing in the same vein, though, it would suffice to say that the difference in electronegativity is not great enough.

Let's take the simplest case, hypochlorous acid : $\ce{H-O-Cl}$. If the bond $\ce{H-O}$ is broken, it produces the ions $\ce{H+ + OCl^-}$ like every acid in water. If the bond $\ce{Cl-O}$ is broken, it produces the ions $\ce{Cl^+ + OH^-}$. What could these ions make later on ? If they react with water, could they make new substances ?

I don't think we're getting what you're imagining here. In first approximation nothing happens to conjugated bases.

Let me put it simply : Will water break XO- in a similar fashion it broke OH bond in XOH

To answer simply: Only if X-O bond visits the same fashion salon as the O-H bond, and even then, the extra electron purse makes the difference.


text:
Does benzene sulphonic acids show steric inhibition of resonance / ortho effect?
I have read about ortho effects in aniline (steric inhibition of protonation) and benzoic acids but have not seen anything about Sulphonic Acids ($\ce{-SO3H}$) do they also show considerable ortho effect?


text:
nBuLi and tBuLi can take part in halogen metal exchange OR deprotonate. Is there any way to predict which it favours?

I have included a synarchive pathway: https://synarchive.com/syn/144 where BuLi favours deprotonation over halogen metal exchange

No direct answer from me, but Clayden's book on organolithiums is an excellent read for these things, and I suspect may contain an answer for you. I don't mean the general org chem textbook, I mean this: https://www.elsevier.com/books/organolithiums-selectivity-for-synthesis/clayden/978-0-08-043262-5

I recommend looking for Prof. Hans Reich (RIP) and his work covering a broad range of organolithium chemistry, with lots of NMR kinetic studies. He condensed much of his life's work in his personal webpage, which has since been repurposed as part of organicchemistrydata.org. There are multiple sections on organolithium chemistry, which you can access on the dropdown menu on the left. My suspicion is that deprotonation is generally faster, and only Li/I and Li/Te exchange can compete with deprotonation.

You need to look at the pKas of the protons that can potentially be removed. If there are any below ~50 then you will get deprotonation.

The reagents should be written as discreet steps: 1)t-BuLi, 2) 4-iodobutene.

According to Bordwell $\mathrm{p}K_\mathrm{a}$ Table, $\mathrm{p}K_\mathrm{a}$ of $\ce{PhCH2COSPh}$ is $16.9$, which is compatible with alcoholic $\ce{OH}$. We all know that Grignard reagents and organolithium reagents are very susceptible to acidic hydrogens because they are strongly basic as well as nucleophilic. Between $\ce{t-BuLi}$ and $\ce{n-BuLi}$, $\ce{t-BuLi}$ is considered to be acting as base than nucleophile compared to $\ce{n-BuLi}$, due to its bulkiness. However, both of them are known for their capability of metal-proton exchange (e.g., see Wikipedia).
On the other hand, lithium-halogen exchange reactions are kinetically controlled. The position of the equilibrium varies with the stabilities of the carbanion intermediates involved $\mathrm{(sp >> sp^2 >> sp^3)}$, rather than whether it is $\ce{n-BuLi}$ or $\ce{t-BuLi}$. Keep in mind that lithium-halogen exchange reactions using alkyl-lithium typically employ two or more equivalents of alkyl-lithium reagent. The first equivalent is used for the exchange and the second equivalent reacts with the alkyl-I produced to form corresponding alkane, alkene, and lithium iodide (e.g., Ref.1). Nevertheless, if acidic hydrogen is available, the reaction goes through metal-proton exchange.
Note that the scheme given in Synthesis of Aplysin, part of which OP has copied is somewhat misleading. When you look at the original reference (Ref.2), it is clear that their intention wasn't halogen-metal exchange. It is clearly to use the lithium reagent as a base:

It indicated that $\ce{t-BuLi}$ is added to the compound $\bf{12}$ at $\pu{-78 ^\circ C}$, waited $\pu{10 min}$ before adding 4-iodobutene, and then let the solution warm up to $\pu{0 ^\circ C}$. After $\pu{5 h}$ at room temperature, the yield of the compound $\bf{13}$ was still only $48\%$ (see the red box). The ref.2 did not mention what was the other $52\%$. It may be a combination of nucleophilic addition and halogen exchange. I'm just speculating.

An example for use of $\ce{n-BuLi}$ for halogen-metal exchange is given in Ref.3.
References:

Helmut Neumann, and Dieter Seebach, "Stereospecific preparation of terminal vinyllithium derivatives by Br/Li-exchange with t-butyllithium," Tetrahedron Letters 1976, 17(52), 4839-4842 (DOI: https://doi.org/10.1016/S0040-4039(00)78926-X).
David C. Harrowven, Matthew C. Lucas, and Peter D Howes, "Total syntheses of aplysin and debromoaplysin using a diastereoselective, sulfur mediated radical cyclisation strategy," Tetrahedron Letters 1999, 40(23), 4443-4444 (DOI: https://doi.org/10.1016/S0040-4039(99)00768-6).
Donald F. Hoeg, Donald I. Lusk, and Alvin L. Crumbliss, "Preparation and Chemistry of α-Chloroalkyllithium Compounds. Their Role as Carbenoid Intermediates," J. Am. Chem. Soc. 1965, 87(18), 4147–4155 (DOI: https://doi.org/10.1021/ja01096a025).


text:
How do we find the SALCs?
I am trying to understand exactly what IRREPs are and, in order to know more about it, I started reading the chem.libretext course on molecular symmetry which up to this point seemed fairly comprehensive and understandable to me. However, I blocked right at the moment the irreducible representations were introduced in the context of this course.
What I understand:

Symmetry operations that can be applied to molecules form point groups, with "point" being there because there is always a point in the molecule that is not modified by the symmetry operations, as opposition to "space" groups in crystals where translations become important
Symmetry operations can be represented as matrices that act on vectors of basis functions, leading to the calculation of linear combinations of the basis functions
A set of matrices that represent symmetry operations is called a "representation"
There is a process of reducing representations into various "pieces" called irreducible representations which involves decomposing the rotations matrices into a set of smaller matrices which when "directly-summed" (with this odd ⨁ symbol) reconstitute the original matrices.

But then, at This page, the author suddenly, out of the blue, introduces that it is possible to somehow replace the basis functions used as reference with what are called "symmetry adapted linear combinations" (SALCs) of basis functions. Suddenly, she specifies the actual SALC, but I don't understand how she gets there, and the SALC that are depicted seem to be completely random, they do not make any sense to my eye.

Why are these specific basis functions SALC for the symmetry group of ammonia? This is completely puzzling to me. Especially $s_{2}'$ and $s_{3}'$ do not make any sense to my intuition.
How do we get these SALCs allowing us to decompose the transformation matrices the most?
Does this have anything to do with "block diagonalization"?
As an annex question, what exactly is an IRREP? Can we have a molecular feeling of what an IRREP is? Can we have a visual depiction of an IRREP? Everytime I try to google this question, I end up with abstract character tables and I get confused by the last column of these character tables with all the x, y, z which seem to be out of place here.
Thanks in advance!

You may find this corresponding question on Math.SE useful.

The answer to this is very deep, and is rooted in mathematics, not physics. I don't think any chemistry textbook seriously explains what an irreducible representation "really" is, precisely because it's almost impossible to teach it to the average chemist. To properly understand this, I'd suggest reading about group theory, and then representation theory if you're interested.
However, let's see if I can try to provide a brief-ish overview. As will become very clear, covering this in full is not feasible in a Stack Exchange answer. Also, you will likely need some linear algebra background to understand this. I simply can't explain everything from the ground up, so apologies to any readers left behind.
We start from the definition of a group. This is a set of objects $G$, equipped with a binary operation $\cdot$, which obeys the four group axioms, namely:

closure (for all $g, h \in G$, $g\cdot h \in G$)
associativity (for all $g, h, k \in G$, $(g \cdot h) \cdot k = g \cdot (h \cdot k)$)
identity (there exists some element $e \in G$ such that for all $g \in G$, $e\cdot g = g \cdot e = g$)
inverse (for all $g \in G$, there exists some element $g^{-1} \in G$ such that $g\cdot g^{-1} = g^{-1} \cdot g = e$)

In molecular symmetry, the objects $g$ are symmetry operations. That is, the point group consists of all symmetry operations which preserve the molecular geometry. These elements form the top row of the character table.*
Often in (basic) group theory we are concerned with the elements of the group $g$, i.e. the symmetry operations. However, in chemistry we are really more interested in the molecules or the orbitals, which are not elements of the group! These are actually objects which the group elements act on.
Mathematically speaking, these objects are vectors which inhabit a vector space $V$. For example, in ammonia, we might have a vector space which consists of the three hydrogen 1s orbitals, as well as the 2s and 2p orbitals on nitrogen. It is left as an exercise to the reader to convince themselves that the group elements (i.e. symmetry operations) convert an element of the vector space (e.g. a hydrogen 1s) to another element of the same vector space (e.g. a rotation converts it to the 1s orbital of a different hydrogen).
Vector spaces are defined by a few more axioms, which can be easily found in any reference. Each group element $g_i \in G$ is therefore a map: it acts on a vector $v \in V$ and produces another vector $w \in V$. Of course, this map is different for every group element. So each group element $g_i$ is a different map $\rho_i: V \to V$. The act of "going from group elements $g_i$ to vector mappings $\rho_i$" is called a group homomorphism, or a representation. $V$ is called the representation space. There is a condition on this, namely that all of the $\rho_i$'s are linear:
$$\begin{align}
\rho_i(v_1 + v_2) &= \rho_i(v_1) + \rho_i(v_2) \\
\rho_i(kv_1) &= k\rho_i(v_1)
\end{align}$$
where $v_1, v_2 \in V$ and $k$ is some complex number.†
A different representation simply means a different set of $\rho_i$'s, which may well act on a different vector space $V'$.
If our vector space has finite dimensions, let's say $n$ dimensions, then each vector $v \in V$ can be associated with a $n \times 1$ (complex-valued) column vector. The exact numbers in the column vector depends on the choice of basis. (Again, I can't explain all of this here; this is basic linear algebra.) Then, the fact that the $\rho_i$'s are linear maps on $V$, means that the $\rho_i$'s are actually $n \times n$ matrices. Again, the matrix elements will depend on the basis chosen for $V$. Often you will read that these matrices are elements of the general linear group $GL(n; \mathbb{C})$. Maths textbooks don't like to choose an explicit basis for $V$, but it's necessary for our purposes here.
Let's take stock of where we are. We've explained how group elements may be represented by $n \times n$ complex matrices (assuming a finite-dimensional representation space). These matrix elements $\rho_{i,pq}$ depend on the basis chosen.
$$g_i \to \rho_i = \begin{pmatrix}
\rho_{i,11} & \rho_{i,12} & \cdots & \rho_{i,1n} \\
\rho_{i,21} & \rho_{i,22} & \cdots & \rho_{i,2n} \\
\vdots      & \vdots      & \ddots & \vdots      \\
\rho_{i,n1} & \rho_{i,n2} & \cdots & \rho_{1,nn} \\
\end{pmatrix}$$
The act of decomposing a representation refers to finding a basis of $V$ which make all $\rho_i$'s have (the same) block diagonal form. Mathematicians don't like to speak about bases of $V$, and so this is often framed in terms of "subspaces" and "quotient spaces"; you may see that in books. But we will stick with our explicit matrices and bases.
Let's talk about a three-dimensional basis for now; so the $\rho_i$'s are $3 \times 3$ matrices. (Essentially, $V = \mathbb{C}^3$.) And let's suppose that we find a basis in which
$$g_i \to \rho_i = \begin{pmatrix}
\rho_{i,11} & 0           & 0           \\
0           & \rho_{i,22} & \rho_{i,23} \\
0           & \rho_{i,32} & \rho_{i,33} \\
\end{pmatrix}$$
and that every matrix $\rho_i$, for all group elements $g_i$, has the same block-diagonal form in this basis. Essentially, what a block-diagonal matrix does is it disentangles different basis vectors from one another. If the matrix $\rho_i$ acts on the vector $v_1 = (a, 0, 0)^T$, then we can guarantee that the resulting vector $\rho_i v_1$ is of the form $(a', 0, 0)^T$ (and furthermore $a' = \rho_{i,11}a$). Likewise, if we take $\rho_i$ and act on some other vector $v_2 = (0, b, c)^T$, then the resulting vector $\rho_i v_2$ is $(0, b', c')^T$ where the first element is still $0$.
This means that we have essentially obtained two different representations from this.

The first representation is defined by the vector space $V_1 = \{(a, 0, 0)^T\}$ for some $a \in \mathbb{C}$. We can get rid of the two zero entries such that $V_1$ is basically a 1-dimensional space.

The $\rho_i$'s in this representation are then just $1 \times 1$ matrices with the single element $\rho_{i,11}$.

The second representation is defined by the vector space $V_2 = \{(0, b, c)^T\}$ for $b, c \in \mathbb{C}$. Again, we can drop the zero entry at the beginning.

$\rho_i$ in this representation is then the bottom-right $2\times 2$ block of the previous matrix.

Note that the old $3 \times 3$ $\rho_i$ matrix can be obtained through a direct sum of the $1 \times 1$ and $2 \times 2$ matrices. This is where the direct sum comes in.

Alternatively, you could leave in all the zeroes that we took out; and instead of the direct sum we would just have the ordinary sum of two matrices.

These two new representations are called subrepresentations. An irreducible representation is then a subrepresentation which cannot be reduced further in this manner. It remains to make a few points:

The SALCs that you see are the basis vectors of $V$ which allow for this decomposition.
The character listed for each irrep in the character table refers to the trace of the matrix $\rho_i$. Note that each group element corresponds to a different matrix with (in general) a different trace, so the character of an irrep under each symmetry operation is (in general) not the same.
If an irrep is 2-dimensional (for example), then the basis in this 2-dimensional vector space can be arbitrarily chosen (the fact is that it can't be further decomposed into block-diagonal form, regardless of basis: so it doesn't matter what the basis is). So, the SALCs for 2- and higher-dimensional irreps are actually arbitrarily chosen. This is true for the $s_2'$ and $s_3'$ irreps in the question: any linear combination of $s_2'$ and $s_3'$ is a valid pair of SALCs.
Changing the basis within an irrep does not change the characters of the irrep, because the trace of a matrix is invariant under a change of basis.

Footnotes
* Note that in a character table, elements that are conjugate to one another are grouped into the same entry. Two elements $g, h \in G$ are said to be conjugate if there exists some element $t \in G$ such that $g = tht^{-1}$.
† We are dealing with complex-valued vector spaces. Most of the simple character tables don't have complex numbers. But counterexamples are easy to find: see e.g. the character table for $C_3$. http://symmetry.jacobs-university.de/cgi-bin/group.cgi?group=203&option=4

Thank you for this answer; it's nice to have a perspective grounded in chemistry but oriented around the math. Could you confirm if the following summary is accurate?

A point group is a group of isometries with at least one fixed point.
All molecules can be classified under a point group.
A representation of a group is the action of the group elements on a specific vector space V (here, of molecules).
A representation can be reduced by partitioning $V$ into disjoint subspaces that are left invariant by the representation.

The restriction of a representation to such a subspace is a subrepresentation.
A subrepresentation that cannot be further reduced is an irreducible representation (irrep).
The original representation can be recovered as a direct sum of irreps.
The set of SALCs is a basis for $V$ in which a representation is maximally block diagonal.
Each block corresponds to an irrep and an invariant subspace of $V$.

Still have no idea what character tables are, but that's a question for a different time. ;)

@a-cyclohexane-molecule (1) Excluding translations, I think. (3) Given a particular molecule, I think the orbitals form a more useful vector space than the molecule itself: because under the group operation the molecule always maps to itself, the molecule should always transform under the totally symmetric irrep. It's a valid vector space but just not particularly useful. (6) You'd probably like the formal definition, which is that an irrep is a representation that has no proper subrepresentations (i.e. no subrepresentations for which the vector subspace $V'$ is a proper subspace of $V$).

The rest agrees with what I understand of the topic (which is not all that much beyond what I've just written here...!) The character tables just collate the traces of the matrices $\rho_i$ for each irreducible representation of the group.


text:
What are "molar equivalent amounts" in a description of an immune cross-reaction assay?
From "Folate Receptor Alpha Autoantibodies in Autism Spectrum Disorders: Diagnosis, Treatment and Prevention" (2021):

Figure 2. (A) Folate receptor concentrations in milk. (B) Immune cross-reactivity of blocking autoantibodies against various FR antigens. The blocking assay was performed by testing a known amount of blocking activity in serum samples from 8 different patients with molar equivalent amounts of FR antigens from milk. Blocking activity was determined as pico moles of 3HPGA blocked and was compared to blocking in human milk antigen.

The whole sentence in italic is cryptic to me, for instance, how can there be "a known amount of blocking activity"? Is activity measured in amounts?
But especially what are "molar equivalent amounts"? Samples of milk containing the same number of moles of the FR receptor as the number of moles of anti-FR antibodies contained in the samples taken from the 8 different patients? Thus it could be measured precisely how these antibodies react with FR receptors belonging not to the human, but to a different specie (a goat, a cow)?

I'm afraid we chemists are not good enough biochemists. We don't know what is a folate receptor, what is an autoantibody, and what is cross reactivity.

See also https://en.wikipedia.org/wiki/Equivalent_%28chemistry%29


text:
Aromaticity in cyclic ketones
I have read that, cycloprop-2-ene-1-one is aromatic and cyclopent-2,4-dien-1-one is antiaromatic.
Basically this is visualized by visualizing the resonance structures of this.

Similarly if I apply this concept in para benzoquinone, I get that it is antiaromatic because of 4 conjugated $\pi$  electrons. But many sites on net say it is not aromatic. Where am I going wrong ?

The concept is called resonance and the contributing structures are called resonance structures; they are not called resonating structures.

link does this clears your doubt?

Para-benzoquinone is antiaromatic, and tries to get out of antiaromaticity by disrupting the conjugation; i.e. lengthening the "single bonds" so that pi-overlap across these bonds is not very feasible. This effectively results in two seperate pi-bonds in the ring; hence the observed "non-aromaticity".
It does have antiaromatic characters though, in the sense that it is a strong oxidant, easily taking away two electrons from other molecules so that it can become the aromatic(6pi) hydroquinone diide anion.

Antiaromaticity is a purely hypothetical concept, such compounds cannot exist. By distorting the structure the symmetry of the molecule is reduced, therefore breaking degeneracy. A small distortion can lead to a significant change there, while overlap will probably be still very similar.

I would argue that parabenzoquinone is easily reduced not hecause it is antiaromatic as such, but because the reduction generates an aromatic compound. So we are generating aromaticity, which very much exists, rather than getting rid of antiaromaticity which may not really be so.


text:
Why is the reaction between calcium oxide and silicon dioxide neutralization?
I get that calcium oxide is a basic oxide and silicon dioxide is acidic. However by definition, neutralization is "when an acid and a base react to form water and a salt and involves the combination of $\ce{H+}$ ions and $\ce{OH-}$ ions to generate water". Yet I don't see any water as a product when calcium silicate is formed.
$$\ce{CaO(s) + SiO2(s)-> CaSiO3(l)}$$
So since water is not a product, how can it be considered neutralization? Or if it isn't neutralization, what sort of reaction is it?

acid base reaction does not necessary involve water formation. I will another example: HCl (gas)+ NH3(gas) what do they form- a solid salt in gas phase.

The statement that "neutralization is the reaction of an acid and a base producing a salt plus water" is a concept developed in the 19th century. The reaction of $\ce{CaO + SiO2}$ is not really a neutralization. It is a synthesis.

It's only an acid base reaction according to Lux–Flood definition

According to wikipedia, neutralisation is just a rxn btw acids and bases/alkali; it doesn't talk about the production of water. CaO is a basic oxide while SiO2 is an acidic oxide, hence it is similar to acids & bases reacting i.e. neutralisation.


text:
What is the chemical composition of Ferrous Ferrocyanide?
To this day, iron-cyanide complexes confuses me. So, I want to settle this once and for all. After scavenging the internet, I gathered the following result:
$$
\ce{Ferr}
\color{blue}{\ce{ous}}~\ce{Ferr}\color{blue}{\ce{i}}\ce{cyanide -> Fe3^{II}[Fe^{III}(CN)6]2}$$
$$
\ce{Ferr}
\color{blue}{\ce{ous}}~\ce{Ferr}\color{blue}{\ce{o}}\ce{cyanide ->} \color{reD}{\ce{???}}$$
$$
\ce{Ferr}
\color{blue}{\ce{ic}}~\ce{Ferr}\color{blue}{\ce{i}}\ce{cyanide (Prussian}~\color{green}{\ce{Green}} \ce{)-> Fe^{III}[Fe^{III}(CN)6]}$$
$$
\ce{Ferr}
\color{blue}{\ce{ic}}~\ce{Ferr}\color{blue}{\ce{i}}\ce{cyanide (Prussian}~\color{brown}{\ce{Brown}} \ce{)-> Fe^{III}Fe^{III}(CN)6.2H2O}$$
$$
\ce{Ferr}
\color{blue}{\ce{ic}}~\ce{Ferr}\color{blue}{\ce{o}}\ce{cyanide (Prussian}~\color{blue}{\ce{Blue}} \ce{)-> Fe4^{III}[Fe^{II}(CN)6]3}$$
There are some mentions of "ferrous ferrocyanide" on the internet but I couldn't find its formula.

Prussian blue (also known as Berlin blue or, in painting, Parisian or Paris blue) is a dark blue pigment produced by oxidation of ferrous ferrocyanide salts (Prussian Blue - Wikipedia)

Comment section of this question "What's the chemical formula of "Everitt's salt"?"

Evidence in the literature on the synthesis of ferrous ferricyanide is critically discussed. Pyrolysis and pressure effects on Prussian
Blue lead to ferrous ferrocyanide together with decomposition
by-products, and not to ferrous ferricyanide. The latter compound
could be a precursor in the formation of Turnbull's Blue or an
excited state of Prussian Blue, but it is not a stable chemical
species.
Reguera, E., Fernández-Bertrán, J. & Balmaseda, J. The existence of ferrous ferricyanide. Transition Metal Chemistry 24,
648–654 (1999). https://doi.org/10.1023/A:1006942415737

Memoirs and Proceedings of the Manchester Literary & Philosophical Society: (Manchester Memoirs.)., Manchester Literary
and Philosophical Society, 1909

Question: What is the chemical formula for "Ferrous Ferrocyanide"?

 To be honest, I don't even know the correct formula for Prussian blue. There are so many formula for PB, some of which has potassium included. Moreover, "Turnbull Blue" has similar kind of formula adding more confusion.

It is not common to find the ferrous ferrocyanide because it oxidizes rapidly. As you can see the remaining salts or pigments are pretty stable in air. This one is white. I quote from a very old Analytical Chemistry book by Treadwell, pg. 155, this should address your query.  It states:

Potassium Ferrocyanide, $\ce{K4[Fe(CN)6]}$, produces in solutions of ferrous salts, with complete exclusion of air, a white precipitate of potassium ferrous ferrocyanide or of ferrous ferrocyanide, depending upon whether one or two molecules of ferrous salt react with one molecule of potassium ferrocyanide:
$$\ce{K4[Fe(CN)6] + FeSO4 <=> K2SO4 + K2Fe[Fe(CN)6]}$$
$$\ce{K4[Fe(CN)6] + 2 FeSO4 <=> 2 K2SO4 + Fe2[Fe(CN)6]}$$
Although both of the above salts are white, a light-blue color is almost always obtained, because the precipitate is immediately oxidized somewhat by the air, forming the ferric salt of hydroferrocyanic acid (Prussian blue):
$$\ce{6 Fe2[Fe(CN)6] + 6 H2O + 3 O2 <=> $\underset{\text{Prussian blue}}{\ce{2 Fe4[Fe(CN)6]3}}$ + 4 Fe(OH)3}$$

@bittonwood short of time for a major format change.


text:
Why does the PE becomes negative in potential energy versus intermolecular distance graph?
In the potential energy versus inter molecular distance graph, we know that atoms/molecules/particles want to be at optimum distance from each other ie $r_0$ and to the left of this position in the graph, we get the repulsive force and to the right would be the attractive force acting between them.
Note: Inter-molecular distance is denoted by ‘r’.
However, I have the following questions regarding this graph:
Why does graph represent negative Potential energy after a certain inter-molecular distance ?
What does negative potential energy mean in this context since the repulsive energy at r=0 was positive?
What happens at the point when P.E. becomes zero for a certain inter-molecular distance?
If the P.E. at that point has already reached zero, why is there a further need to go into negative potential energy?

@Poutnik, I do understand that part but this analogy doesn't seem to fit into what I was wondering. Why does even some negative PE show repulsion as well? Shouldn't it have been that all negative PE=attraction and bonding while all of the positive PE=repulsion?

What's happening where the PE=0?

Well, a marble ball does roll down even along the pit wall. it does not need to be a hill slope. For attraction/repulsion,  important is the PE gradient along the path.  What happens at PE=0 depends on the sign and value of d(PE)/dx.

The zero energy level is by convention the situation where the particles are infinitely far away from one another.

Helpful will be reviewing the very basics about potential and potential energy around classical central electrostatic and gravitational force.

@Maurice, The PE is zero at a short distance too. Why is that?

@M. Farooq. No. At short distance, the PE is small, nearly zero. But it is not zero !

@Maurice, I am talking about the point where PE curve crosses the zero line before becoming too large.

@ M. Farooq. Which point ? Never heard of it.

@Maurice You cannot be serious. You have it directly before your eyes if you look at the picture the OP posted. When atoms get too close, they start to repulse each other, the negative PE starts climbing, crossing the zero point and finally becomes positive.

@Poutnik. OK. Thank you. I was not thinking about this point. I was thinking of some strange phenomena at longer distances, a little bit like the maximum happening when ionic $\ce{NaCl}$ is extended and transformed into neutral atoms. There is a PE maximum at about three or four times the equilibrium distance.

@Maurice Thanks. I did suspect you might meant something else, as you could not miss this triviality.


text:
In the LCAO method for $H_2^+$, why do we only add or subtract orbitals but never alter one of the phases of the orbitals?
TLDR:
Atkins' physical chemistry contains the quote "In general, from N atomic orbitals we can build N molecular orbitals". In the case of $H^+_2$, we use a combination of N=2 orbitals to create the bonding and anti-bonding orbitals as shown in eq 1. But if we allow one of the orbitals to differ by a phase $C_0$ so that $\Psi_{0}(r)=N\left[\psi_{H1s_A}(r)+C_o\psi_{H1s_B}(r)\right]$, then from two atomic orbitals we should be able to create an infinite amount of molecular orbitals. Atkins is correct. I am incorrect. But why?
TLDR Over
Using the LCAO method for the $H_2^+$ ion, we use as our trial functions
$$\Psi_{\pm}(r)=N\left[\psi_{H1s_A}(r)\pm\psi_{H1s_B}(r)\right] \tag{1}$$
where $N$ is a normalization factor, and $\psi_{H1s_i}$ is the 1s orbital centered around the $i$'th nucleus. The wave function $\Psi_{+}$ results in a significant reduction in the systems energy whilst the $\Psi_{-}$ wavefunction significantly increases the systems energy. For this reason, $\Psi_{+}$ is called the bonding orbital whilst $\Psi_{-}$  is called the anti-bonding orbital.
This phenomenon is often described as playing a role in the origin of the band structure of solids. When N independent potential wells are brought near each other, the initially N-fold degenerate ground state "splits" into N energy eigenstates. This is simply an N-potential generalization to the $H^+_2$ case described above. In that case, we simply have $N=2$ and the $N=2$ energy eigenstates that arise as the nuclei approach each other are simply the bonding and anti-bonding states.
My issue is that I can't seem to think of any reason as to why we should restrict ourselves to only the two states  $\Psi_{+}(r)$ and $\Psi_-(r)$ when performing the LCAO method. Surely the state $\Psi_{0}(r)=N\left[\psi_{H1s_A}(r)+C_o\psi_{H1s_B}(r)\right]$ (where $C_o$ is a complex number) should be just as valid as either of the $\Psi_{\pm}$ states? (Note that anti-symmetrization is not a requirement here since we are calculating 1 electron wave functions).
I also feel like depending on the value of $C_o$, the energy of this new state should vary continuously between the energy of the bonding state and the energy of the anti-bonding state? If this is the case, then the coupling that occurs when $N=2$ hydrogen nuclei are brought to within bonding distance of each other should result in an infinite continuum of energy eigenstates. Not merely two eigenstates. Similarly, when  $N\approx 10^{23}$ atoms are brought within bonding distance of each other, we shouldn't get $N$ energy levels but rather an infinite amount of energy levels. This would then ruin the whole notion that energy bands in a solid of N atoms should each contain $N$ energy levels.

Atkins simplifies things a lot (and rightly so, otherwise they would be impossible to digest). Indeed, N atomic orbitals can produce infinitely many combinations, but only N of those are eigenfunctions of the Hamiltonian, and thus deserve to be called molecular orbitals. As luck would have it, they all use real coefficients.

"an infinite continuum of energy eigenstates" - You can calculate the expectation value of the energy for any state $\psi$, via $\langle E \rangle = \langle \psi | \hat{H}
| \psi \rangle$. And indeed if you calculate that you will find that this depends continuously on the number $C_0$. But that doesn't mean that the state $\psi$ is an energy eigenstate for any $C_0$. What you've shown is that you have an infinite continuum of states, not energy eigenstates.

Crossposted from https://physics.stackexchange.com/q/665364/2451

Your problem seems to stem from confusing the number of linear independent basis functions(which is the same number as the size of the atomic orbital basis with which we started) and the number of all possible functions that can be built using this basis. Just as in a two dimensional vector space, where you have a maximum of two independent basis functions, but an infinite number of possible vectors by adding those two basis vectors by varying the coefficients.
A way to obtain eigenfunctions is to express the Hamilton operator in a basis, for example, the atomic orbital basis. If the basis is finite, we obtain a finite matrix representation for the Hamilton operator which can be diagonalized to obtain its eigenvectors. These eigenvectors are the molecular orbitals expressed with respect to the atomic orbital basis, which means that we get for each a molecular orbital a set of coefficients that tells us how to linearly combine the original atomic orbitals to form each molecular orbital.
In the finite case, we obtain as many eigenvectors/molecular orbitals, as basis functions that we originally started with. So using a linearly independent atomic orbital basis of size N, will yield a NxN Hamiltonian matrix, which will upon diagonalization yield N eigenvectors=molecular orbitals. These molecular orbitals are not all functions that can be formed. But they are a specifically useful basis set of functions that span the N dimensional space of functions. You can add and subtract these basis functions and generate different functions that are no longer eigenfunctions of the Hamilton operator but it is trivially to calculate the energy as long as you work in this basis since you know the  action of the Hamilton operator on each molecular orbital.
The complex number field is simply the field for our vector space of functions but it doesn't change any of the linear algebra or how many eigenvectors/molecular orbitals we obtain.
Now, as mentioned in the comments already, you can build an infinite number of functions/vectors with your basis simply by varying the coefficients. But these functions are not eigenfunctions/molecular orbitals. And you are correct that the energy of these functions varies.
Answer to question in comments:
You can indeed use more basis functions. Most quantum chemistry codes do that actually. The basis where we have one function per electron is the minimal basis set. You may get lower energies for your bonding orbitals with a larger basis. However, you will not neccessarily get  more bonding orbitals. Most of the newly obtained molecular orbitals will simply be anti-bonding and energetically way above the "bonding" band. The newly added basis function is, loosely speaking, used to optimize the bonding orbitals but most of its contribution will end up as part of an antibonding molecular orbital as long as your initial smaller basis was not totally inadequate to begin with. They sort of get stacked upon the orbitals that you had already at higher energies. There should only be a little mixing of the new function unless your initial smaller basis was chosen very poorly. Figuratively it should look like this, where the left side uses 2 basis functions and the right side three basis functions,

The lowest molecular orbital should be almost the same in both cases and if we have a system of only two electrons, we would only occupy this one. Likewise with bands in solids. You would fill your band up to the number of available electrons. The new basis functions would mostly lead to an extension of the band at higher energies that won't be occupied. And in this sense the number of bonding orbitals is constant even if you increase the number of basis functions.

Thanks for the great answer!  Clears up most of my confusion. Just before I accept your answer, I think part of my confusion stems from the fact that the LCAO method is an approximation. Thus, the true $H_2^+$ ion bonding orbital  is not exactly of the form $\Psi_{+}(r)=N[\psi_{H1s_A}(r)+\psi_{H1s_B}(r)]$. The true orbital could actually be $\Psi_{+,true}(r)=N[\psi_{H1s_A}(r)+\psi_{H1s_B}(r)+\psi_{correction}(r)]$ where the last psi is a linearly independent wavefunction correcting the approx to obtain the true orbital. Now, however, we should be able to create 3 linearly independent ...

... bonding orbitals instead of 2. It seems that we obtain 2 (or N for generality) molecular orbitals precisely because we started our approximation as an LC of 2 (or N) atomic orbitals. But there isn't any fundamental reason that we should make this initial choice in our approximation (?) . Surely we could always make our approximation better by increasing the amount of terms in our initial linear combination approximation? This would then ruin the whole notion that energy bands in a solid of N atoms should each contain N energy levels

@SalahTheGoat I have added a disscussion to the answer regarding your comments.


text:
Why exactly do we need formal charge?
Formal charge is considered to be the charge present in one atom by considering all the bonds to be 100% covalent.What is the significance of 100% covalent here? Why are we considering the bonds to be covalent and what is formal charged used for ?

https://en.wikipedia.org/wiki/Formal_charge // In contrary to some other Q/A or forum sites, answers on CH SE site are figuratively paid by the user's own effort. When you ask, it is expected you have thoroughly searched and thought about the topic, providing explicit summary of partial answers/ideas/thoughts you have got until then. Effort not shown may be considered as effort not done and such a question may get closed.
How do I ask a good question.

What is the source for your first sentence (which I don't agree with...)?

@Karsen Theis This is the definition available in most reference texts. Kindly mention your views about it

@KarstenTheis 100% covalency = zero electronegativity difference = zero polarity = even bond electron distribution. // I do agree definition the OP used is unusual. Alternative is charge assigned to an atom in a molecule, assuming that electrons in all chemical bonds are shared equally between atoms, regardless of relative electronegativity.

Formal charge is considered to be the charge present in one atom by considering all the bonds to be 100% covalent.

The "charge present in one atom" is not a clear concept. A better way is to say "formal charge is the charge assigned to an atom symbol in a Lewis structure". This acknowledges that the formal charge depends on the choice of Lewis structure. For some structures, you can write multiple plausible Lewis structures, which might show distinct formal charges.

Why exactly do we need formal charge?

We don't. It is useful in deciding which Lewis structures make sense, and to get a rough sense of the charge distribution (and the net charge). There are better but more difficult methods to get at the actual atom-by-atom charge distribution , e.g. https://en.wikipedia.org/wiki/Voronoi_deformation_density. However, the charge is not really distributed atom by atom, but different parts of space have different charge.

@Gowri If this answer helped you, consider accepting it by clicking the checkmark.

Formal charge, like oxidation state, is fundamentally just a bookkeeping device (with a different counting method). This being so, formal charge can be correlated with an unequal sharing electrons between like atoms. In such cases it points to molecular polarity in situations where we would not ordinarily expect it.
The classic example is ozone, $\ce{O3}$ (and it's congener, $\ce{S3}$ which is known in sulfur-rich gases such as Io's volcanic eruptions). If we assume that the bonding is nonpolar, then because the central atom shares three of its electrons and the others share only one or two, we see a separation of formal charges and therefore the bonding is not really nonpolar. The middle oxygen has a formal charge of $+1$, the others $-1/2$ apiece. The electrons in the bonds are likely polarized towards the middle oxygen to compensate for this inequity, but the bent ozone molecule still retains some dipole moment (0.53 D) correlated with the formal charge separation.


text:
Why are sulfoxides usually stable as ylides (S charged positively and O negatively)? What's the BDE for S-O?
I can't find any info on BDE for singularly bonded S to oxygen. Moreover, I'd like to understand, why is it quite usual to find sulfoxides structure written with charge separation?

What BDE are you asking about? Energy for the diatomic molecules won't address the question regarding ylides, and BDEs in polyatomic molecules vastly depend on the neighbors and structure (there is no single value whatsoever). What exactly are you trying to compare here? Finally, I'd suggest to reformulate "I'd like to understand" part into a straightforward question since this makes an impression the question isn't about chemistry and is about cognition.

If you are looking for the latter set of vetted values for various compounds, check out p. 9-73 and onwards of Haynes, W. M.; Lide, D. R.; Bruno, T. J. CRC Handbook of Chemistry and Physics: A Ready-Reference Book of Chemical and Physical Data, 97th ed.; Taylor & Francis Group (CRC Press): Boca Raton, FL, 2016. ISBN 978-1-4987-5429-3.


text:
Stereochemical relationship between the given compounds

Are the given compounds enantiomers or diastereomers? Or are they identical? The compound is trans-1,2-dimethylcyclobutane:

My reasoning is that they are identical. They are not mirror images, and on rotating 180° give the other compound.
Is my answer and reasoning correct?
Note: Ignore the confirmations of cyclobutane.

$$ $$ Yes $$ $$

Rotate 180 deg. about an axis perpendicular to the plane of the paper/screen.

The given compound, 1,2-dimethylcyclobutane has two chiral centers, and hence, has maximum of $2^2 = 4$ stereoisomers. However, as shown in the scheme below, the cis-isomer has a plane of symmetry, and as a consequence, it is a meso-isomer:

However, trans-isomer can exists in two forms: (1R,2R)-1,2-dimethylcyclobutane and  (1S,2S)-1,2-dimethylcyclobutane (see Scheme $\bf{A}$).
The two isomers given in the question are both trans-isomers. Thus, they woul be either identical or enanthiomers as shown in the Scheme $\bf{A}$ (cannot be diastereomers). Although, the OP's visualization is correct and can use to identify simple molecule like this, it is hard to visualize if presented complex molecule. The best way to identify stereoisomers are assign their appropriate (R/S)-configurations as I have done for OP's structures (see Scheme $\bf{B}$).
Accordingly, both structures are (1R,2R)-1,2-dimethylcyclobutane. Thus, the compounds are identical.


text:
What do natural hybrids of peptides and nucleic acids look like?
Until recently I thought that there was a clean distinction between polypeptides and oligonucleotides, but apparently there are recent studies of therapeutic compounds that could be described as "chimeras", "hybrids", or "conjugates" of nucleic acids and peptides. Usually, these are linked by a synthetic spacer group.
This got me wondering if there are hybrids of polypeptides with DNA or RNA that occur naturally. I know about aminoacyl-RNA, but my understanding is that this is only a single amino acid covalently bonded to RNA.
So I want to ask what these hybrids might look like in nature, if they do exist. (For example, would an amino acid residue be directly attached to (deoxy)ribose, or somehow connected with the phosphate?)
My attempts at searching keep turning up peptide nucleic acids, which are interesting in themselves, but are fully synthetic and thus not what I am asking about.
Pointers to the literature would be highly appreciated (especially survey articles).

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

I mention them in the fourth paragraph.


text:
Reaction of 3-chloro-2,2-dimethylbutane with sodium ethanolate in EtOH, which mechanism?
When we let 3-chloro-2,2-dimethylbutane react with sodium ethanolate in EtOH, which reaction do we expect? SN2, E2, E1 or SN1?
I would say it's an SN2 reaction because ethanolate is not that strong of a base, but is that correct? A carbocation would also be stabilized insufficiently, because it's a secondary one, so either E2 or SN2 would be probable, but I don't know if [OEt]- is sufficiently basic to favor the E2 reaction. I would be glad over any tips on this question, thanks in advance.

How different are ethoxide and hydroxide in basicity? What are the pKa's of their conjugate acids?

This is a very hindered system so you are not going to get very much SN2. Neo-pentyl halides are very poor electrophiles and this is yet more hindered. E2 is the most likely event.


text:
Manganese and valence layer
Why does manganese make $4$ bonds in $\ce{H_2MnO_4}$ when it has only two electrons in the valence shell?

There are two things wrong with the premise. First, no one atom really makes any covalent bonds. A covalent bond requires at least two atoms, more if orbitals are delocalized.  Second, manganese has not only two but seven valence electrons, which are plenty for bonding covalently to as many as four oxygen atoms. The additional valence electrons come from the partially filled 3d orbitals that define all transition metals in that row of the Periodic Table.
I may also add that $\ce{H2MnO4}$ does not exist. $\ce{MnO4^{2-}}$ is known only as salts which are in turn stable only in highly alkaline environments.


text:
Bond type in Grignard reagent
I just want to confirm if the $\ce{R-Mg}$ bond is a covalent bond which undergoes heterolytic fission during reactions while the bond between magnesium and halogen atom is ionic. Am I right or wrong?

The R-magnesium bond in a Grignard reagent is polar covalent with carbon being the negative end of the dipole, which explains its nucleophilicity. And the magnesium-halogen bond is largely ionic.

https://chemistry.stackexchange.com/questions/429/why-do-magnesium-and-lithium-form-covalent-organometallic-compounds

Yes, you are correct. The carbon-magnesium bond in a Grignard reagent is polar covalent with carbon being the negative end of the dipole, which explains its nucleophilicity and the magnesium-halogen bond is largely ionic.

(image source)


text:
Why is carbon tetrachloride more stable than methane?
When a mixture of methane and chlorine is exposed to ultraviolet light – typically sunlight – a substitution reaction occurs and the organic product is chloromethane. However, the reaction doesn't stop there, and all the hydrogens in the methane can in turn be replaced by chlorine atoms.
As in the end carbon tetrachloride is formed, it means that $\ce{CCl4}$ is more stable than $\ce{CH4}$.
My doubt is that we know the $\ce{C-H}$ bond is stronger than the $\ce{C-Cl}$ bond, then how can we say that $\ce{CCl4}$ is more stable than $\ce{CH4}$? So why does the chlorine radical replace the hydrogen radical from methane? As this happens it means that $\ce{CCl4}$ is more stable than $\ce{CH4}$, but why?
This are the reactions I am talking about:
\begin{align}
\ce{CH4  +  Cl2 &->  CH3Cl  +  HCl}\\
\ce{CH3Cl  +  Cl2  &->  CH2Cl2  +  HCl }\\
\ce{CH2Cl2  +  Cl2 &->  CHCl3  +  HCl }\\
\ce{CHCl3  +  Cl2  &-> CCl4  +  HCl}
\end{align}
Here methane is taken in excess, why is methane taken in excess?

Not sure - but does that depend on the proportion of Cl:H? Is this with a large excess of Cl2?

Then by the law of mass action, https://en.wikipedia.org/wiki/Law_of_mass_action, the reaction is forced towards CCl4. Consider that UV will knock out H atoms, which get lost in the volume of Cl2.

If what you are saying is correct then any reaction should happen if we take the reactant in excess even if the product is not stable ? Let's suppose we haven't taken chlorine in excess , then the reaction won't happen ?

CH$_4$ and CCl$_4$ are both non-polar solvents. London dispersion forces dominate in them which increase with increase in molecular weight. therefore CCl4 is more stable due to stronger London dispersion forces.

London dispersion forces are intermolecular forces . Ofcourse in  CCl4 London dispersion forces are stronger than CH4 but that doesn't prove the stability of molecule itself . It only means that CH4---CH4 bond is weaker than CCl4--CCl4 bond . But CH bond is stronger than CCl bond and London dispersion forces are very weak even .

This reaction is fueled by the stability of hydrogen chloride!!

DrMoishe Pippik actually this reaction is in excess of methane . And that's the reason why only one chlorine is getting attached to carbon at one time . Chlorine isn't in excess here .

The relative amounts of the various products depend on the proportion of the two reactants used. In the case of methane, a large excess of the hydrocarbon favors formation of methyl chloride as the chief product; whereas, an excess of chlorine favors formation of chloroform and carbon tetrachloride. See here and here.

Yeah that's true that chlorine isn't taken in a small quantity but the point is that methane is in a large excess as compared to chlorine . If this wasn't the case then carbon tetrachloride would have formed in the first place with 2 moles of hydrogen gas getting released.

It is impossible to compare the stability of methane and carbon tetrachloride directly as the do not have a common frame of reference. At the same time you cannot claim that the dichlorine molecule is less stable than the hydrogen chloride molecule.
The only true observation that you can make is that the reaction of methane and chlorine to chloromethane and hydrogen chloride is (under certain conditions) favourable.
$$\ce{CH4 + Cl2 -> H3CCl + HCl}$$
Let's take some average bond strengths from Chemistry Libre Texts.[1]
\begin{array}{lr}
\text{Bond} & \text{Energy}/\pu{kJ mol-1}\\\hline
\ce{H-Cl}   &   427\\
\ce{C-H}    &   413\\
\ce{C-Cl}   &   339\\
\ce{Cl-Cl}  &   239\\
\end{array}
You break one $\ce{C-H}$ and one $\ce{Cl-Cl}$ bond and you form one $\ce{C-Cl}$ and one $\ce{H-Cl}$ bond:
$$\pu{+413 kJ//mol} + \pu{+239 kJ//mol} + \pu{-339 kJ//mol} + \pu{-427 kJ//mol} = \pu{-114 kJ//mol}$$
This reaction releases energy (within this crude approximation).
Within this framework a mixture of the reactants would be (thermodynamically) less stable than a mixture of the products.
This would (in this averaging approximation) be true for every following chlorinations.
Another part of this is the mechanism, which is a radical mechanism.
Without going into much detail and here are some of the steps that will be involved:
\begin{align}
\ce{Cl2 &->[h\nu] 2Cl}\\
\ce{CH4 + Cl &-> CH3 + HCl}\\
\ce{CH3 + Cl2 &-> H3CCl + Cl}\\
&\vdots\\
\ce{H3CCl + Cl &-> H2CCl + HCl}\\
\ce{H2CCl + Cl2 &-> H2CCl2 + Cl}\\
&\vdots\\
\end{align}
It becomes a matter of probabilities and concentrations and you would typically expect a mixture of all species.
Using excess methane will probably give you the highest amount of chloromethane. The following reactions may also happen to some extent, providing that path:
\begin{align}
\ce{H2CCl2 + CH3 &-> H3CCl + H2CCl}\\
\ce{H2CCl + CH4 &-> H3CCl  + CH3}
\end{align}
There are probably studies about this reactions with very accurate numbers, but for the time being and for the purpose of this answer this should be enough of a demonstration.

Bond Energies. (2020, August 22). Retrieved September 15, 2021, from https://chem.libretexts.org/@go/page/1981

Check the sentence that begins "Within this framework...".


text:
Separation of Racemic Mixture
In the racemic mixture below, if it is reacted with Mosher's Acid Chloride, it would form X and Y. Can I ask why my book state that X and Y are not enantiomers and thus can based on differences in their boiling points? I thought X and Y are enantiomers since they are non-superimposable mirror images of each other.

In this case, You now have two chiral centers ( in X and Y). The stereo of the Mosher's Acid Chloride did not change during the reaction while, regarding the other chiral centre: one is R and other is S. So say Mosher's Acid Chloride has a stereo S, then X has SR and Y has SS configuration. These are not enantiomers and instead are diastereomers.
For example, RR and SS or RS and SR are enantiomers but SR and SS or RR and RS are not enantiomers, these are diastereomers.


text:
How come bromine adds meta to phenol in this reaction, which is catalysed by HSbF6

This is a rxn in the synthesis of Tramadol. I found in in IChO prep problem 2001, Q19...
Edit: I have seen in clayden a rxn with HSbF6 whereby the benzene ring gets protonated so I followed that path of thinking and got stuck. Thanks for the answers and enlightenment!

Fluoroantimonic acid - $\ce{HSbF6}$ is made by combining $\ce{HF}$ and $\ce{SbF5}$. In this system, $\ce{HF}$ releases its proton ($\ce{H+}$) concomitant with the binding of $\ce{F-}$ by antimony pentafluoride, which is a Lewis acid. The resulting anion ($\ce{SbF6-}$) is both a weak nucleophile and an extraordinarily weak base.
Two things to note are:

$\ce{HSbF6}$ is a very strong acid.
Its conjugate base is a very poor nucleophile.

The hydroxyl group of phenol attacks $\ce{H+}$ and makes phenylhydronium ion which would behave like a nitro group. Therefore, the substitution would occur preferably on the 3 position.
Since $\ce{HSbF6}$ is such a strong acid, you can safely assume that virtually all of the phenol is protonated all the time.


text:
A gaseous mixture contains hydrogen atoms in the 4th excited state and helium ion; spectral lines emitted when all these return to the ground state

A gaseous mixture contains hydrogen atoms in the 4th excited state, $\ce{He^+}$ ions in 3d excited state, and $\ce{Li^{2+}}$ in 2nd excited state. Find the number of distinct spectral lines emitted when all these ions return to the ground state.

Number of lines emitted $=\displaystyle\frac{n(n-1)}{2}$, where $n$ is the number of the shell the electron is in.
If we simply add the number of lines, without considering repetitions, we get $10+6+3=19$ lines.
However since there are 3 atoms, there will be some cases of repetition. Now we can always find out the repetitive cases by listing down the energy of each transition, but that will be very long, and there is potential for error since there are 19 lines. Is there a better method to solve this question.
Energy of transition $=\displaystyle13.6z^2\left|\frac{1}{n_1^2}-\frac{1}{n_2^2}\right|\ \mathrm{eV}$

Involve algebraic evaluation. if/when for given z1, z2, n11, n12, n21 exists n22, matching equality, for given n constraints.

The best method is to write down each transition, just like you said. Is there a better method? Yes, but it is actually way worse, immensely long and heavily involved in math.

Enumeration is definitely faster, especially if one involves a simple Excel table.

Also, repetitions are theoretically possible even with one atom (not in this case, though).

Can such a mix even exist? an excited H and He+ would ionize at the energies needed to  form Li++ ions.


text:
What is the role of magnesium nitride in ammosamide B synthesis?
The following is a snapshot of the synthesis of ammosamide B from Synarchive:

In the final step $\ce{Mg3N2}$ is used to convert esters to amides or amides to amines. What is the mechanism of this process? How does the $\ce{N3-}$ get converted to $\ce{H2O}?$

Your last phrase is cryptic, as I see no H2O here.

$\ce{Mg3N2}$ is magnesium nitride. It reacts with $\ce{MeOH}$ to generate magnesium methoxide and ammonia according to online report by Buske [1].
The $\ce{NH3}$ thus generated is acting as nucleophile to react with the methyl ester to give the primary amide. It also removes two trifluoracetamide groups (or the $\ce{MeO-}$ does) to reveal the amino groups.
Quite why the authors have chosen to use this over more conventional sources of $\ce{NH3}/\ce{MeOH}$ (which is commercially available) is for them to explain.
Reference

Gary Buske. Chemical Safety: $\ce{Mg3N2}$ Hazard. Chemical & Engineering News 2009, 87 (28).


text:
Where do the chloride ions go during saponification precipitation?
During saponification hydroxide ions from the lye join to the triglyceride to form glycerol, and the sodium bonds with the fatty acids making the soap molecules. However, it is sometimes the case that $NaCl$ is needed at this point to precipitate out the soap. The sodium ions are used in this precipitation. But, what happens to the excess chloride ions?

Adding a concentrated $\ce{NaCl}$ solution to a soap solution produces the precipitation of sodium alkanoates (indeed it is a mixture of alkanoates and alkenoates). This is a consequence of the law of mass action. The ions $\ce{Cl-}$ and $\ce{Na+}$ added in this operation remain in solution. The only products to precipitate are the sodium alkanoates which were present in the solution before adding $\ce{NaCl}$


text:
How to prepare a solution of Ferrous Iron?
I need to prepare a solution of Ferrous Iron (Fe2+). This solution will be used to test a water filtration system by simulating the Ferrous Iron that is sometimes present in well water. The system is supposed to convert the Ferrous Iron to Iron Hydroxide precipitate via a Manganese Oxide catalyst and filter it out.
What is the best way to get the Iron 2+ ions into the solution? I have seen suggestions to add Iron(II) Sulfate or Mohr's Salt. Something that is confusing me is that the Wikipedia articles for these compounds suggest they form a complex when dissolved in water. For Iron(II) Sulfate and Mohrs Salt the aquo complex [Fe(H2O)6]2+ is formed. I am wondering, are Iron(II) ions always in this aquo complex, and perhaps in some discussions of chemistry this fact is ignored for simplification? I question if the Iron 2+ ion will be available to participate in the reactions and get converted to Iron Hydroxide if it is in the aquo complex.
To summarize, what is the best way to make a solution with Fe 2+ ions that will simulate well water? And an explanation of the chemistry of the aquo complex formed by Mohrs salt and Iron(II) Sulfate and how that differs or is the same as a Fe 2+ ion alone in solution would be appreciated if someone is so inclined.
Thank you for sharing your knowledge!

Have you searched the literature? Certainly, there must be previous examples. You are right, it is $\ce{[Fe(H2O)6]^{2+}}$. Consider it synonymous with $\ce{Fe^{2+}}$.

If you got your answer, feel free to self answer it. Self-answers are appreciated and it also helps future readers.


text:
Why is Amphotericin B insoluble in water in the pH range that makes it zwitterionic?
Here is the structure of Amphotericin B, with its carboxylic acid and amine groups circled in red:

According to this source, The carboxylic acid group has a pKa of 5.5 and the amine group has a pKa of 10. It says there and in other sources as well that Amphotericin B is insoluble in water in the pH range of 6-7, but is more soluble in pH < 2 or > 11.
My question is why is it that in a pH range between the pKa values, where both the carboxylic group and the amine group are charged (a zwitterionic state), the molecule is considered insoluble in water? With both groups charged, I would assume solubility should be higher then when only one of them is charged, as is the case in pH < 2 or > 11.
I couldn't find anything about this issue online. I went back to basic organic chemistry textbooks and read about zwitterions, isoelectric points etc., but couldn't find anything useful regarding such a complex molecule as Amphotericin B. I did find this question here on Chemistry.SE regarding water solubility of amino acids in their zwitterionic state, but the answer given there does not suffice for me, as I still do not understand why would zero net charge mean lower solubility, when we have two charged moieties on top of many hydroxyl groups that also contribute plenty to water solubility?

Molecules with no net charge aggregate and crystallize more easily than those that are charged and need a counter ion. Non polar amino acids like valine are also much less soluble as zwitterions than as cations or anions

Do molecules with no net charge necessarily not require counter ions? It seems that for Amphotericin B, the charged groups are not very close to each other, so doesn't each of them require a counter ion of its own?

You are asking too much when expect to dissolve a compound with at least 14 carbon non-polar chain (without a counter ion).  Even, sodium dodecyl sulfate with only 12 carbon and a counter ion has critical micellar concentration, meaning not completely soluble.

But why is it more soluble in extreme pH values when it is in ionic state?

One would need a little ball & stick model but I suspect it can even be an internal salt as for the charged sites can be brought in contact


text:
How big a role does electron repulsion play in regard to the bond length of a covalent bond?
In a covalent bond, there are several forces acting together to determine the bond length.
Electron-electron repulsion increases the bond length.
Proton-proton repulsion increases the bond length.
Proton-electron attraction decreases the bond length.

When I try to understand that a triple bond is shorter than a single bond, I have a hard time trying to comprehend how big a role does electron-electron repulsion play in the bond length. Since a triple bond has three pairs of shared electrons, I would reason that the repulsion is stronger than a single bond and therefore the bond length should be longer. Clearly this is not the case. The proton- electron attraction is also ruled out. As a single bonded c-o has the same number of protons as a triple bonded c-o. Triply bonded c-o however, has a shorter bond length of 113 pm while single bonded c-o is 143 pm.
While I can rote memorize this fact, it will be really appreciated if there is a rational explanation.

Bonding means lowering energy of valence electrons and that effectively means electron attraction.

Try to approach your both hands, with fingers stretched. When your thumbs are touching one another, it is similar to a single bond : the distance between the centers of the hands is large. When the thumbs and the forefingers are touching one another, this looks like a double bond : the distance between the centers of your hands is smaller. When the thumbs, the forefingers and any other third fingers are touching one another, it looks like a triple bond : the distance between the centers of your hands is still smaller.

The logic you follow is correct. The three assertions are somewhat correct as well. The point is that chemical bond does not relies on classical electrostatic. And not even the existence of H does.

@Mithoron that article is a bit advance for me to comprehend at my current level, but it will be kept on my reading list.


text:
What is the hybridization of copper in blue vitriol?
My book says that the structure of blue vitriol is the following:

From the structure, I figured that since there are 4 water molecules coordinated to the cuprate ion, the hybridization should be $sp^3$. But when I googled, I saw that people were saying its hybridization is $sp^3d^2$. Why would that be true?

Can the downvoter explain what mistake I made?

The picture you provided shows half the story.
In the diagram above you can see that copper has a charge of +2 i.e. Cu$^{2+}$ which leads to a 3d$^9$ configuration. Since H$_2$O will here act as a weak field ligand, no pairing of electrons will be done in d-orbitals which will lead to sp$^3$d$^2$ hybridization.

Not a believer in hybridization of transition metals, but the structure is worth a +1.

I think sulphate ion doesn't form any coordinate covalent bond with cupper. But there exists an ionic bond between [Cu(H20)4]+2 and (SO4)-2. Because coordination number of Cu is 4 not 6.

@OscarLanzi How can we explain the average number of water molecules being 5 in blue vitriol? 2 H-bonded water molecules are bonded to 4 sulphate ions but on average 1 sulpahe ion is considered bonded with Cu(II) ion.


text:
How to do work up of reactions (Problem of Emulsion) where NO2 group is reduced using SnCl2.2H2O?
I did a reaction where I have reduced Nitro group to -NH2 group. I have added solid NaHCO3 and shifted pH to 10 as reported in procedure. While extraction emulsion is forming so how to overcome the emulsion issue?
Procedure

I have run this reaction many times on a variety of substrates.
The solution to this is to keep adding base, preferably hydroxide rather than carbonate/bicarbonate (I used 50% aq NaOH), until the tin salts start to go back into solution and you get good phase separation.
And don't forget to extract your aq. phase a couple of times.
note: the reaction works perfectly well using EtOAc as solvent

Or, chemically speaking, the OP has to overcome the formation of amphoteric $\ce{Sn(OH)2}$ to $\ce{[Sn(OH)3]^{-}}$ and $\ce{[Sn(OH)4]^{2-}}$.  Though $\ce{[Sn(OH)3]^{-} + 3 OH^- -> [Sn(OH)6]^{2-} + 2 e^-}$ with a potential of $\pu{- 0.93 V}$ is a remarkable reducing agent, too.


text:
What molecule is pictured on the cover of Dead Sara's album "Ain't It Tragic?"
An album released today by Dead Sara, titled Ain't It Tragic, has a chemical structure on the album cover. Here's a close-up of that structure:

As the structure is partially obscured, it's difficult to identify what molecule might be depicted. However, I was able to identify some features that may be useful:

As an undergraduate chemistry student, I haven't encountered any structures similar enough to this to give me an idea of what I need to be looking for. Here are the things I've been able to establish so far:

One aromatic ring with a hydroxyl group
Two rings fused to the aromatic ring, one of which is an ether ring and the other of which has a tricyclic structure (bridge carbon marked #16)
Another fused ring with two bridge carbons (#8 and #17)
A methyl ether extending from this ring
A cyclopropyl group connected in some way to the main chain
The structure has at least 26 carbons: the 24 numbered carbons and two carbons in the lower right that are part of side chains.

What molecule is this?

I see two possible candidates: buprenorphine and thienorphine.
Both are strong opioid painkillers fitting the “art” theme.
Depending on how one draws the location of what appears to be a methoxy and a methylene group, it could be either one of those drugs.
I leave it to the reader to decide how to unveil the structural part covered by the middle finger (literally) by posting a triptych for comparison:

Fig. 1. A: cropped molecular fragment from the cover art; B: Buprenorphine; C: Thienorphine.

Worth noting that, though buprenorphine is itself an opioid painkiller, one of its major medical uses is in treating opioid addiction caused by more common opiates like heroin.

Thank you for the thorough answer with imagery! I think I'll go with buprenorphine because it has 24 carbons (not counting the methoxy group), which matches the numbers shown on the cover art. It's difficult to see how the cyclopropyl group on thienorphine might be numbered 22, 23, 24 without using nonconsecutive numbers on the thiophene ring.


text:
Lewis acid catalysis with BF3
I am a beginner of Lewis acids/bases reactions and found Lewis acid catalysis. There is an example, which is the opening of epoxide by an alkyne anion activated by $\ce{BF3OEt2}$:

According to Wikipedia, $\ce{BF3OEt2}$ serves as a source of $\ce{BF3}$. What are the factors to select a molecule that coordinates with the Lewis acid $\ce{BF3}$? For example, the same Wikipedia page also describes methanol and even propanol can be used (seems the same principle as the lone pair electron on $\ce{O}$ donates into empty p orbital in $\ce{B}$) Can I say any Lewis bases can coordinate with $\ce{BF3}$ for this Lewis acid catalysis?

Furthermore, according to Wikipedia:

Lewis acid catalysis often involves binding of the catalyst to Lewis
basic heteroatoms and withdrawing electron density.

So, just to confirm my understanding: The electronegativity of boron and fluorine differs greatly. When $\ce{BF3}$ forms an adduct to the oxygen atom in epoxide, the electron density of oxygen is drawn to the 3 fluorine atoms in $\ce{BF3}$, hence increasing the electrophilicity of oxygen atom. Therefore, rate of the attack by the nucleophile (alkyne anion) to open the epoxide is enhanced.

Note, chemical information may be advantageously formatted using on ChemSE with mhchem. Take moment to familiarize with this. You are encouraged to use it in the body of questions, answers, and comments. Because it is something special not all web browsers understand well, do not use it in the title of questions or answers.  2) Within reason, include pictures directly into the question because links to other sites than chemistry.se maye become dysfunctional.

As a general rule for the English language, if not used at the start of a phrase, the names of elements,  are not capitalized.  This equally applies to chemical names, too (e.g. iron chloride).


text:
What antimicrobial can be readily synthesized in a school laboratory for a project on antimicrobial resistance?
I am currently a student studying at a high school hoping to do a project on antimicrobial compound synthesis to use in investigating antimicrobial resistance in E. coli strains. I am hoping to synthesize an antimicrobial to use in in vitro testing.
I have looked at sulfonamides, but the involvement of benzene in some of the procedures (for example, in the synthesis step of 3-amino-5-methylisoxazole of sulfamethoxazole), is banned because of health and safety regulations in high school labs.
Is there an alternative for a compound that could be synthesized with high school lab chemicals under guidance with maybe one or two external orders of specific compounds?

Is benzene just a solvent? You could try simply replacing it with toluene…

Thank you for the suggestion. Unfortunately benzene is suggested in the paper as both the solvent and something used to recrystallize the residue formed from the procedure in the end, so it might not be possible.The paper also said ether could be used as a solvent.

I am pretty sure toluene could be used as solvent instead of benzene.

The point is that is not clear if the project is about chemistry or somewhat biology. I do not see why a project on antimicrobial resistance involves the synthesis of the antimicrobial agent. Not that I am against a multidisciplinary approach, but perhaps you should focus on the antimicrobial resistance in E. Coli and test different antibiotics

Well... that's tougher. Is it possible for you or for someone to test the recrys with toluene before committing to the full synthesis?

I do not think that it is possible. Thank you for the suggestion.
Thank you for your reflections on the aims of my project too, but it was simply a nice idea to test what I've synthesized on the E. Coli K12 strains as a proof of concept that it works. I have also tried to expand my idea to include multiple variants of antimicrobial compounds including testing Sulfamethoxazole/Trimethoprim mixes in a 20:1 ratio to see if it would be more effective compared using them alone. However, that also seems too ambitious due to the limitations of a school laboratory on certain compounds and also cost.

Metronidazole perhaps

Thank you for your reply, would you possible have details concerning synthesis of metronidazole...? I'm quite new to synthesis and the only paper I can find is its synthesis from ethylenediamine.

Metronidazole can be prepared from the commercially available 2-methylimidazole (100g for £24 from Sigma-Aldrich) in 2 steps. Nitration https://patents.google.com/patent/US4209631A/en then reaction with chloroethanol (£30 for 100g from Sigma Aldrich)

Thank you for your help. I will check the patent out and check with relevant teachers to hopefully see if this can be done.

If the nitration is a problem then the 5-nitro-2-methylimidazole is also commercially available, and fairly cheap

@Waylander I think metronidazole would work well. Thank you for the suggestion. On a side note, would you have any data regarding the synthesis of Trimethoprim..? Hoping to synthesize that too hopefully but the molecule probably looks too complicated. I have found multiple patents describing various different methods of synthesis of intermediates, but it is really hard to organize into a single coherent synthesis sequence.

@CrownEther  The syntheses of trimethoprim I have seen involved the use of acrylonitrile which is a fairly nasty allergen and suspect carcinogen. I suspect that may be too much for a high school lab.

Thank you for your reply, I will remove that from my list.

@Waylander would you also have the steps regarding the second step of the reaction for metronidazole regarding chloroethanol? Deciding on asking a professional before purchasing the document that probably details the reaction https://link.springer.com/article/10.1007/BF00764821.

I have never done this chemistry, but a search of the patent literature yields a method whereby 2-Me-5-nitroimidazole is heated with an excess of chloroethanol in the absence of solvent.  See example 1 of https://patents.google.com/patent/US2944061A/en   Chloroethanol is a hazardous material by inhalation or skin contact. Take appropriate precautions.

Thank you for your help. I will be adequately protected.

There are many common antimicrobials in use, and one that might be comparatively easy to sythesize would be a quaternary ammonium compound. For example, this patent claims the "reaction is carried out in water."
That said, the reactants and the resulting antibacterial might have intrinsic environmental and health hazards.
BTW, why not investigate some traditional antimicrobial compounds used for food preservation, such as sodium chloride, calcium chloride, acetic acid, or various sugars? No synthesis is needed. And must you use E. coli, or can you investigate other organisms, such as fungi, that cause food spoilage?

Thank you for your reply. I would prefer using E.coli because it links to a previous project I have done, but that seems like a good idea to consider doing in the future too.
This project would also explore the mechanism of action of the drug on the bacteria, as an example sulfamethoxazol blocks the biosynthetic pathway of folic acid in bacteria., which is quite a specific and interesting topic to learn and write about.
It just would be quite fascinating and be a valuable experience to synthesize it yourself and see it have an action on bacteria based on theory.


text:
Practical use of the partition function in molecular simulations
In the second chapter of Understanding Molecular Simulation, Frenkel and Smit derive an equation for the partition function and the thermal average of the generic observable A, stating that these equations "are the starting point of virtually all classical simulations of many-body systems".
I understand the theoretical significance of the partition function, and I can see how, in theory, by knowing the partition function one would know the energy of the system, and if I'm not mistaken, also all the other thermodynamic quantities one could be interested in.
What leaves me confused is the practical use of the partition function in the computational chemist's work. I have some experience with running simulations with programs such as LAMMPS and GROMACS, and yet have never had a real first-hand encounter with the partition function. Is it used by the programs to compute the thermodynamic quantities they output? If so, would it be possible to provide a general description of how this is done?

The partition function is impractical to compute, so the reason you never encounter it in simulation is because you can't work with it; the fact that the partition function is also the normalization constant for probabilities means that simulations have to work around this inconvenience instead. This is what motivates the use of Monte Carlo methods in simulation: you never have to work with absolute probabilities (hard to compute because of the partition function), but only their ratios (the partition function cancels out).

The actual partition function is unimaginably formidable. For just $N$ point particles in a 3D box, it's already got $3N$ dimensions. If the box is length $L$, GROMACS would probably divide the box into a machine-precision grid and calculate the energy from $\sim L \times 10^{10}$ values in each dimension. The partition function would incorporate all of that information, the energy at each of those $L \times 10^{30N}$ points; even for two particles in a one unit box, that's already $10^{60}$.
When you speak of "the partition function" you're talking about the entire thermodynamic state of the system and all of its possible permutations. If we truly knew that function, we would be omniscient and could calculate anything about the system! Since we can't really calculate at the level of the entire partition function, we use simulations (MC and MD, etc.) to sample regions of interest (usually low-energy areas).
To a computational chemist, the partition function is most practical for its mathematical basis. It's the way to get from statistical mechanics (a bunch of particles moving around in a box) to actual thermodynamic properties (Helmholtz energy, pressure, etc.).
As a practical example, imagine you want to calculate the surface tension for some fluid interacting according to some potential function. Using LAMMPS or GROMACS, you can set up your vapor-liquid interface in the simulation box. Then by simply changing (perturbing) the simulation box shape and measuring the energy change, you can calculate the surface tension. The whole derivation of this method starts with the configuration integral of the partition function and then, via mathematical manipulation, eventually leads to the quantity we want to calculate and variables we can actually measure.
Test-area method - Gloor and Jackson, J. Chem. Phys. 123, 134703 (2005)
Free energy perturbation - Zwanzig, J. Chem. Phys. 22, 1420 (1954)

But to your last point: is it used in practice in a quite direct way? if so, how?

@simulation_engine I have edited and hopefully addressed this now!

thanks for the edit, it did address my question more fully. it would be great if you could provide a couple additional examples that have to do with some other properties besides the surface tension. in any case i'll mark your answer as the accepted answer.

Partition function, by definition can be written,
\begin{align}
Q&=\frac{1}{N!h^{3N}}\int_{-\infty}^{\infty}d\textbf{P}\int d\textbf{R} e^{-\beta U(\textbf{P},\textbf{R})}.
\end{align}
In most cases, the integral over $\textbf{P}$ can be calculated analytically, leaving only $\textbf{R}$ contribution left inside the integral. It is easy to see that direct integration of the Boltzmann factor($e^{-\beta U(\textbf{R})}$), is impractical; you need to sample all possible configurations. Roughly speaking, the partition function in statistical mechanics is like wave function in quantum mechanics.  If you know analytically form of the functions, you know everything about the system, but the function itself does not have clear meanings. Of course, knowing the analytical form of the functions is out of reach except for few simplest examples. That brings us to your question of what practical use the partition function has.
My answer to that would be "free energy".
Free energy not only directs in which direction a system will evolve over time but also relative probabilities of different states of the system when it reached equilibrium. For example, under fixed number of particles, volume, and temperature, system's partition function, called canonical partition function is related to Helmholtz free energy, $A$,
\begin{align}
A&=-\frac{1}{\beta} \ln Q.
\end{align}
I am not going to prove the equation above, but maybe you can look it up somewhere online. What I want to show is how such relation connects free energy differences between states to relative probabilities between the states when the system has reached equilibrium. First of all, the range of integral over $\textbf{R}$ in $Q$ is considered along the volume of the system but can be limited to the states of interest. For instance, consider two states, $a$ and $b$, where all particles are limited in $R_a$ or $R_b$ for each state, where both of them are subspaces of $R$. The free energy of state $a(b)$ can be written,
\begin{align}
\label{eq:twopartitionfuctions}
A_{a(b)}&=-\frac{1}{\beta} \ln Q_{a(b)}\\
\text{where, }Q_{a(b)} &=\frac{1}{N!h^{3N}}\int_{-\infty}^{\infty}d\textbf{P}\int _{a(b)}d\textbf{R} e^{-\beta U(\textbf{P},\textbf{R})}.
\end{align}
The range of integral for position vector $\textbf{R}$, is limited to each state $a$ or $b$. The difference between the two free energies can be written,
\begin{align}
\label{eq:twopartitionfuctions1}
\Delta A&=A_{b}-A_{a}\\
\label{eq:twopartitionfuctions2}
&=-\frac{1}{\beta} \ln \frac{Q_{b}}{Q_{a}}\\
\label{eq:twopartitionfuctions3}
&=-\frac{1}{\beta} \ln \frac{\frac{Q_{b}}{Q}}{\frac{Q_{a}}{Q}}.
\end{align}
The denominator inside the logarithm, $\frac{Q_{b}}{Q}$ can be expanded,
\begin{align}
\label{eq:twopartitionfuctions4}
\frac{Q_{b}}{Q}&=\int _{b}d\textbf{R} \frac{e^{-\beta V(\textbf{R})}}{\int d\textbf{R} e^{-\beta V(\textbf{R})}}\\
\label{eq:twopartitionfuctions5}
&=\int _{b}d\textbf{R} P(\textbf{R} )\\
\label{eq:twopartitionfuctions6}
&=P_b,
\end{align}
where $P(\textbf{R})$ is Boltzmann probability, and $P_b$ is probability of system to be at state $b$ in equilibrium. Using the result, the free energy difference between state $A$ and $B$, $\Delta A$ is related to there relative probabilities in equilibrium,
\begin{align}
\label{eq:twopartitionfuctions7}
\Delta A&=-\frac{1}{\beta} \ln \frac{P_{b}}{P_{a}}.
\end{align}
In summary, partition function is related to free energy and free energy is related to equilibrium probability. Knowing the probability is critical on understanding a physical or chemical phenomenon like in the example of drug binding. When designing a drug, we want the drug to favor the target site most, but not other sites in the target protein and sites in other proteins. The probability of the drug binding to many possible binding sites will tell if the drug is a good candidate or not. That is just one example of why free energy calculations are important topics in the field of computational chemistry.


text:
Is it correct to say that benzene is more stable than cylcohexatriene due to delocalization?
I'm virtually all chemistry books I own say that benzene is more stable than the theoretical cyclohexatriene, by 150 kJ to be specific, due to the delocalization of pi electron density around the ring.
I've come across many explanations for this such as "spreading of charge minimizes repulsion" and "particle-in-a-box from quantum mechanics", but these explanations don't really seem satisfactory. If delocalization or spreading of charge does is in fact stabilizing, why doesn't it occur in every molecule? (I am aware that electrons are actually delocalized in every molecule but I'm referring to the uniform spread of charge that occurs in benzene.)
I feel that the statement that "delocalisation stabilises benzene" is misleading at the very least. The actual electronic distribution of any molecule is, by definition, the one with the lowest possible energy. So any inaccurate lewis structure must have a higher energy than the actual one. So shouldn't the correct statement be "benzene is more thermodynamically stable than cyclohexatriene because the actual benzene has delocalised π electrons while cyclohexatriene does not, so cyclohexatriene is an inaccurate and thus less stable electronic configuration"?
I've asked this question from my teacher but he says he's never heard of this.

Not all molecules have proper conjugated pi bond system.

What you say of the lowest possible energy is correct. But I think the statement in question is less misleading than you feel. This is where it really leads: "Delocalisation stabilises benzene as compared to the theoretical cyclohexatriene, thus making the latter structure impossible, or in other words, inaccurate". How about that?

I think my main issue is the implication that delocalization of charge is somehow a universal stabilizing factor, stemming from statements like "benzene is stabilized due to delocalization" that do not mention that this "stability" is only relative to an inaccurate description

Yes. Common statements often have implicit parts. See Ivan Neretin. But the distilled juice is already in your question. So you are (were) over thinking. And yes delocalisation always take place, just not the one intended, that as per the Poutnik comment, needs structural requisites.

As you suspect, delocalisation isn't always more stable: see cyclobutadiene (4 pi electrons).