text:
Is the 3D spacing arrangement of atoms in chemical bond always unique?
I would like to ask if the 3D spacing arrangement of atoms in chemical bond configurations are always unique, please?
For example, glycine consists of the set of atoms HHNCHHCOOH. Does there exist a different stable 3D arrangement of this collection of atoms, other than:

What causes this arrangement to be unique? If we throw in and react the ingredients in a different order, such as HHHHHNCCOO, will they all still end up as the same unique shape of regular glycine?
Thank you.

It is complicated. If we throw in the same atoms in different order, most likely we'll get a bunch of different molecules.

There are plenty of other possibilities, like $\ce{H2N-CHOH-CHO}$, or $\ce{HO-NH-CH2-CHO}$, or $\ce{H2N-CO-CH2OH}$, or $\ce{HO-NH-CH=CHOH}$ or $\ce{H2N-CO-CH3}$, etc.

The word you are looking for is isomerism: it is the phenomenon that given number of atoms can form different stable molecules. Depending on how they are different, there are different types of isomerism.

https://www.sigmaaldrich.com/GB/en/product/aldrich/480959

https://www.sigmaaldrich.com/GB/en/product/aldrich/159034

From a given number of atoms, you can usually construct multiple molecules (isomers), so we shall just concentrate on the case where the bond connectivity is the same.
The terms you are looking for are conformation and configuration. The latter is most usually related to double bonds and E/Z isomerism. If we take but-2-ene, for example, it can exist in 2 forms, E and Z. Both are stable and have the same atom connectivity. However, they cannot interconvert between each other - the activation energy is too high.
Two conformations of the same molecule, however, can interconvert between each other (at least at room temperature) and they are both stable (i.e., potential energy minima). A classic case of a molecule is butane, whose conformations can be seen in the picture below.

The anti- and gauche- conformations are both 'butane' and both are stable (even though the anti- is more stable), but have a different 3D arrangement of atoms.
As a sidenote, these can be distinguished at a low enough temperature, at which they cannot interconvert.
Another famous molecule with various conformations is cyclohaxane, with the two stable ones being the chair and the much less common twist-boat.

Please include a source and an image for the image. || Related: Conformations of butane

The arrangement doesn't have to be unique.
For instance, using your notation, propanol is HHHHHHHHCCCO, but it exists in two forms:
Propan-1-ol:

and Propan-2-ol:

The difference is where the OH is attached, to an end-C or the middle-C.


text:
Mechanism for the synthesis of benzimidazole
I have thought up a mechanism for the synthesis of benzimidazole from o-phenylenediamine and formic acid. I would like to know if it's correct. I have attached it below.
My rationale for the protonated intermediate is that methyl formate can run without an acidic catalyst (Lewis or Bronsted) and still provide fair yields. Formic acid must be protonating itself for this to work.

Such protonated carboxylic acid doesn't seem likely intermediate in reaction with amine. I'd rather write proton transfer after attack of amine on neutral acid.

Huh. I guess that makes sense, but then what kicks off water?

I don't see which part. So, would formic acid protonate the intermediate and then water get kicked off? Explain yourself a little more please.

Huh, I don't think I get your way of writing mech. Reaction is autocatalysed, you don't have some magical substrate that works out everything. There's surplus of the acid and every possible proton transfer happens all the time. So no, it wasn't OK :(

Okay. So no self-protonation is necessary?

source: Stack Exchange

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

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text:
Difference between SN' and elimination reactions
My professor has recently taught us about SN' reaction example in which 3-bromo-3-methyl-but-1-ene was reacted with with KCN to yield 4-methyl-pent-3-ene-nitrile as the major product.When I asked him why elimination didn't take place,my professor replied that it was based on experimental data.
Can anyone please clarify whether  the reaction is correct?If yes,then why elimination is not taking place.
Also,I was very surprised that I couldn't find any article on SN' reactions online..

Many factors contribute as to whether SN1, E1, SN2, E2 occur. By that way, it is not SN'. It is SN1

@M.L No, I am talking about SN '( prime) reactions, not $SN _1$.

What are SN1' and SN2' reactions?

@andselisk  Thank you for the related link. Can you please explain why in my case and also in the related answer why E2 elimination is not taking place to form the major product?

Honestly, I have not heard of this type of reaction before. But why can't we logically say this is the major product because SN1 leads to allylic carbocation where you can actually insert the CN in the specified position leaving the double bond in a more stable position.

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text:
Difference between the specific heat capacity of potassium chloride solution and sodium chloride solution
I am curious about the difference between the specific heat capacity of a potassium chloride solution (KCl + water) and a sodium chloride solution (NaCl + water) - particularly the specific heat capacity of which one is greater and by how much (a number would be nice!). You may assume that all variables for the two solutions are the same - volume, concentration, etc.
I assume that this would have to be linked to ionic bonding to understand which molecules are affected by heating the solution but I am not aware of the differences in bonding in the two solutions, and more importantly, how to represent it mathematically.
A simple one-liner such as "Solution X has a greater specific heat capacity than Solution Y by a factor of n" would be perfect (some reasoning/explanation would be great as well).

The Handbook of Chemistry and Physics gives the following heat capacities for $\ce{KCl}$ and $\ce{NaCl}$ solutions at $\pu{18°C}$ :

Mole fraction x
0.04
0.02
0.01

$\ce{KCl}$
3.46 $\pu{J g^{-1} K^{-1}}$
3.78 $\pu{J g^{-1} K^{-1}}$
3.966 $\pu{J g^{-1} K^{-1}}$

$\ce{NaCl}$
3.68 $\pu{J g^{-1} K^{-1}}$
3.895 $\pu{J g^{-1} K^{-1}}$
4.025 $\pu{J g^{-1} K^{-1}}$

Would you know how to find the (approximate) value of any mole fraction and at any temperature using this table?


text:
Size of Orbitals, Making Intuitive Sense of Quantum Model, Nomenclature of Subshells in the Quantum Model
Alright, so I am doing the Quantum Mechanical (or what some people call the Wave Mechanical model) of an atom. There's this part where we have different zones of the probablity of finding electrons, the orbitals. The orbitals are a part of our imaginary concept of shells and subshells. (I know that it might not be completely true at the higher academic levels, but that is the vocabulary I know for now.) I have a few different questions about this atomic model.
A. What is the size of an orbital, or a subshell, or a shell in this model? Is it somewhere related to the Heisenberg postion uncertainty? Like, the size of an orbital being equal to how much uncertain we are about the postion of an electron?
B. My teacher keeps saying that all these shapes of orbitals and even their existence is purely mathematical. Is there really no intuitive sense we can make of the reason of their presence as such without involving any differential equations?
C. Why are the names of subshells based on completely random letters (s, p, d and f)? I tried searching for some source on the internet but it simply dwells deeper into their shapes and planes and graphs rather than explaining their names.

Hey. I posted the question and would just want to add a few more things. First thing, I know my understanding of the Quantum Model might be flawed, simply because of how little we are taught about it and how difficult a concept it is. Please give any suggestions you might have about that. Second, please answer my questions based on the academic level of a typical high school student. Thanks a lot.

The format of this Qustions/Answers site is intended rather for filling the gaps, not for laying the foundations. Consider studying available textbook and online materials and rasing follow up questions. // Search valuable sites with the term  like site:libretexts.org or  site:hyperphysics.phy-astr.gsu.edu or site:en.wikipedia.org.  See also Chemistry SE: resources-for-learning-chemistry

Thanks @Poutnik for the suggestion, I will definitely look up these resources. But I am afraid that most books don't answer these questions about size of orbitals and other stuff. I looked up a few books before posting the question myself, but in vain. It would be helpful if you could lead me to some specific text. Thanks, again.

Especially the libretexts site is specifically focused of education at level you can understand. Hyperphysics format is then done by very usable form of a "cheat cards", very easy to digest. See googled topic orbitals on the respective sites Libretexts and Hyperphysics

You may remember the order of the subshells $spdf$ by the following reasoning : $s$ is the first letter of "spherical", because the $s$ orbitals have a spherical symmetry. $p$ is the first letter of  "plane", meaning that the $p$ orbital has a planar symmetry. $d$ is the first letter of "different", because these orbitals have different symmetry surfaces. Other orbitals are defined by alphabetic order after "$d$", avoiding the letter $e$ which is used for exponential. Historically speaking, the letters $s, p, d, f$ were named after other criterias, which are harder to explain.

Addressing the last part of your post

Why are the names of subshells based on completely random letters (s,
p, d and f)? I tried searching for some source on the internet but it
simply dwells deeper into their shapes and planes and graphs rather
than explaining their names.

These letters are not random and as you can see all over the internet (e.g., Wikipedia) they stand for sharp, principal, diffuse and fundamental. After that the story ends without telling what is sharp, principal, etc. What are those adjective for and what do they describe? These letters described the nature of the lines in an atomic spectrum and this system was not universal (Taken from "A Report on Series Spectra by A Fowler).

Basically, long before the concept of orbitals originated, people were fascinated by the spectrum of alkali metals. They have a rich spectrum if you excite them under an electric arc. When you view the arc through a spectroscope (=a device that separates wavelengths), you see a number of lines these letters described those lines as they appeared to the viewer! Pure visual labels.

Now you don't see fundamental there because the fundamental series was discovered in the infrared region. So no visual label, it is just a label. Nothing fundamental about it.
Now, spectroscopists, mathematicians, physicists wanted to find out a mathemtical relation among the wavelengths of those lines. They came up with mathematical series that described the wavelengths as terms of a series. When quantum mechanics was developed this nomenclature was retained since they were able to assign the electronic transitions in orbitals corresponding to those wavelengths. I will have check original papers from 1920s to see who first assigned them.
P.S. Saw an interesting comment below the original post. I want to correct it since some student later on start to spread this reasoning in future on other websites.

You may remember the order of the subshells spdf by the following
reasoning : s is the first letter of "spherical", because the s
orbitals have a spherical symmetry. p is the first letter of "plane",
meaning that the p orbital has a planar symmetry. d is the first
letter of "different", because these orbitals have different symmetry
surfaces. Other orbitals are defined by alphabetic order after "d",
avoiding the letter e which is used for exponential. Historically
speaking, the letters s,p,d,f were named after other criterias, which
are harder to explain.

I am afraid this is completely imaginary and there is no historical work which justifies the choice of the names orbitals using the above reasoning. The letters do not stand for spherical, plane, and different. The letter f was not chosen to avoid "exponential".

Thanks a lot @M. Farooq! This was exactly the kind of answer I was looking forward to my question. I have a few counter questions, and will be really glad if you answer them too. First, isn't the spectrum you mention the same thing we do for hydrogen and what Bohr used for justifying the quantisation of energy? Why don't we have different bands like these there as well? Second, can you please tell what source are you referring to, since that will be really helpful to me for further information. Thanks. (:

Ayush, Welcome, the source is already mentioned, "A Report on Series Spectra" by A Fowler. Go to Internet archive and locate the book. And post a separate question for Bohr.

The orbital sizes and shapes represent the probability of finding an electron in that space with 95% confidence. You are correct that it is related to the uncertainty principle - we can only describe electron positions as probabilities.
s, p, d, f are short terms for the funny names physicists give to things when they first observe them but don't know what they are. In the case of orbitals, these are "sharp," "principal," "diffuse," and "fine" (or "fundamental").
Based on your question and curiosity abut these things, I think you may enjoy reading Richard Rhodes' book The Making of the Atomic Bomb. The first half of the book is a fascinating and accessible history of quantum physics. That's where I learned what s, p, d, and f stand for.

The probabilistic interpretation of QM (see Born rule) has not all that much to do with the Heisenberg uncertainty principle at all. IMO, the HUP is often invoked in places where it shouldn't be, or needn't be. Basically: orbitals have a size because they represent a probability distribution -- that's true. The HUP doesn't state that, though; the HUP only enforces a constraint on that probability distribution (which the orbitals satisfy).

@orthocresol, that is exactly I want to know. If it isn't the uncertainty principle which predicts the size of an orbital, then what does? All I have read about quantum numbers only tells about "how many" electrons are located in a specific orbital and not about where they are located.

@AyushRoy That depends on how you define the "size" of an orbital. Orbitals don't have a "volume"; they actually extend out to infinity (the conventional pictorial form essentially truncates it), so if you wanted to go by where their furthest extent is, then all orbitals are the same (a very lame conclusion). A more sensible metric might be the mean distance from the nucleus, which you can obtain through integration once you know the mathematical forms of the orbitals (e.g., google '1s wavefunction'). I don't want to go into the details here, but none of this has to do with the HUP, really.

Here is the connection between Born, Heisenberg, and the hertofore not mentioned Jordan. It's a bit easier to comprehend than the previous link. https://aapt.scitation.org/doi/10.1119/1.3009634

@orthocresol, I meant orbital to the region of the maximum probability, something like 90-95 percent. Anyways, thanks a lot orthocresol and LynneKLR. I will definitely look up these sources.


text:
Why can't there be a gas-dispersed-in-gas colloid?
Colloids, a special type of heterogeneous mixtures, typically have two components. One is the dispersed phase (DP), which is the "solute-like" part and another one is the dispersion medium (DM), which is the "solvent-like" part. The different interactions of the DP and DM define the properties of a colloid.
My textbook has a few examples based on the different types of DP and DM, but it doesn't have any reference to a colloid where both DP and DM are gaseous. I asked my teacher about it and he told me that it was because gases can flow into each other and hence no heterogeneous mixture, and hence no colloid, of them can ever be possible.
That makes sense, but doesn't it also rules out the existence of colloids with DP and DM both being liquids? Liquids can also flow fairly easily right? Why is the science behind both of them not the same? And, WHAT exactly is the science behind both of them?

https://chemistry.stackexchange.com/questions/43320/why-do-we-consider-all-gasses-to-be-in-one-phase https://chemistry.stackexchange.com/questions/43118/can-gases-be-immiscible

https://chemistry.stackexchange.com/questions/58420/why-does-a-distinction-between-homogenous-and-heterogenous-mixtures-exist

The nature of gases is their molecules are separated while the nature of heterogenous dispersed mixtures is they are not separated. Therefore a heterogenous ( in equilibrium ) mixture of gases is an oxymoron.
Liquids are different case, forming droplets, kept by the surface tension. In contrary to gases, mutual dissolving of liquids depends on intermolecular interactions. such droplets may even have micelar nature, like n-amylalcohol in water.
Once I have tried to replace unavailable i-amylalcohol by available n-amylalcohol for extraction from water and deeply regretted it, seeing the stable emulsion. i-amylalcohol does not have such behavior, as the $\ce{(CH3)2CH^{}-}$ group is an obstacle for building such structures.

If I am getting it correct, what you want to say is that liquids don't exist seperately as molecules, but as droplet like entities and that is the reason why it isn't possible for them to behave like gases while forming solutions and mixtures?

Non miscible liquids form droplets. Miscible ones dissolve each other into true solutions with separated molecules, interacting all mutually by van der Waals forces and eventually forming hydrogen bonds. // Why not to read what gas and liquid mean on molecular level? All this is direct conclusion of that.

Let's take gas 1 as the dispersed phase (which is already at molecular level) which disperses evenly throughout the dispersing medium which is  gas 2 (also at molecular level). Thus, the system is already homogeneous at the molecular level. So, we can consider it "solution". For e.g. The atmosphere is a solution. Take the major two gases namely nitrogen and oxygen and we can consider the atmosphere to be "oxygen dissolved into nitrogen".
For a detailed discussion, see: Name of particles in a suspension


text:
How to find ion concentrations at various points in a solution that will precipitate multiple insoluble salts?

The solubility product constant of lead (II) iodide is $1.2 \times 10^{-5}$ at 273 K.
The solubility product constant for lead (II) carbonate is $1.5 \times 10^{-13}$ at the same temperature.
50 milliliters of a solution at this temperature contains an unknown concentration of carbonate ions and 0.050 M iodine ion concentration.
A secondary solution of 0.10 M lead (II) nitrate is added. After 32 milliliters of the lead (II) nitrate solution is added, the lead (II) iodide begins to precipitate.
Calculate the original concentration of the carbonate ions in addition to the concentration of carbonate ions in the solution when the lead (II) iodide begins to precipitate.

I attempted to find the concentration of carbonate ions when the lead iodide started to precipitate by dividing the lead carbonate's solubility product constant by the concentration of lead ions at this time, since the lead carbonate precipitates first.
So I concluded that it will be at equilibrium at this point ($1.5 \times 10^{-13} / (0.10 \times 0.032 / 0.082))$ and then multiplying this by the current volume (0.082 L) and dividing by the original volume (0.05 L) to get the original concentration of carbonate ions, at $6.3 \times 10^{-12}$ M.
To get the concentration at the time when the lead iodide started dissolving, I took my answer from the first part of the earlier calculation ($1.5 \times 10^{-13} / (0.10 \times 0.032 / 0.082))$, for a final carbonate ion concentration of $3.8 \times 10^{-12}$. However, I am unsure if the process I used was correct.
Any help is appreciated. Thank you.

When writing a text, the writer must  remember others will have to read it. Your text is a challenge for reading and orientation.  // Good practice is starting with symbolic algebraic expressions and keeping it this way until all is ready to plug in literal numbers. It helps focusing on principles and spotting mistakes. It also improves orientation, Q/A reusability and the permanent value.
You may find useful  formatting mathematical/chemical expressions/formulas.

Make yourself sure the process is correct before trying to apply it.

Hint: Define an algebraic variable for the unknown carbonate concentration and the function calculating the PbI2 precipitation point from this unknown concentration. The inverse function will then provide you the result.

Your reasoning is too much complicated. If all the data are calculated in moles, the result is immediate. Let's try !
$32$ mL of $0.1 M \ce{Pb^{2+}}$ contains $0.0032$ mol $\ce{Pb}$. As a consequence, the initial carbonate solution contains the same amount of $\ce{CO3^{2-}}$. As these $0.0032$ mole are dissolved in $50$ mL. It means that the initial concentration $\ce{[CO3^{2-}]}$ is $\ce{0.0032 mol/0.05 L = 0.064 M}$


text:
Comparison between Atomic Spectrum of Hydrogen and Alkali Metals
I posted a question on Chemistry Stack Exchange a while back. It was related to the naming of atomic orbitals. One of the answers to it mentioned a research experiment. The link to the question is given below.
Size of Orbitals, Making Intuitive Sense of Quantum Model, Nomenclature of Subshells in the Quantum Model
As mentioned in the answer, alkali metals show various different bands of spectrum, on the basis of which we named the S, P, D and F orbitals.
But I didn't get why we don't have these seperate spectra in Hydrogen. (I've studied Hydrogen Spectrum in my classes, and it doesn't talk about anything like this. It only talks about Bohr using them to justify the quantisation of energy.) So what makes Hydrogen different?
Also, do elements other than alkali metals show a similar spectrum pattern or is it a single spectrum like Hydrogen? Which ones do? What's the defining criteria for a single spectrum and multiple spectrum?

The fact a hydrogen atom has just a single electron is a deal breaker. Similarly as for the classical gravitational analogy, the general analytical solution of the system state exists only for systems with 2 (point-like or spherically symmetric) objects.

The basis for orbital naming was the hydrogen spectrum not the alkali metal spectrum. And I have no idea what you mean by "we don't have these separate spectra in hydrogen". The biggest difference between the hydrogen spectrum and others is that hydrogen has only one electron making it far simpler (and analytically soluble).

The H and alkali atomic spectra are very similar in structure. The main difference being that the upper levels of the alkalis are different in energy between the s, p, d orbitals because the inner electrons shield the nucleus. For H the levels with the same n are of equal energies. Herzberg, Atomic Spectra and Atomic Structure, Explains this nicely and in excruciating detail

As mentioned in my previous answer to your first query, sharp, principal, diffuse, and fundamental notation for modern orbital "labels" comes from alkali and alkaline earth metal spectra. This is very well established in historical spectroscopy. It did not come from analyzing hydrogen spectrum. Why? The reason is that these visual "labels" for spectrum lines come from the time when the electron was not discovered and it was very easy to study alkali and alkaline earth metals by electric arcs or sparks. It was just pure mathematical analysis (hence the name series spectra...the series is the empirical mathematical series).
Hydrogen atom is the simplest atom with only one electron. How did they find out why hydrogen has only one electron is another story. This was the first atom that was tackled with sound and rigorous physics principles known to man at that time. Those wonderful years were the mid-1920s for physics but perhaps not for mankind. In fact, hydrogen atom spectrum lines also showed "mathematical series" behavior, their names were Lyman, Balmer, Paschen, Pfund series etc, named after European spectroscopists. They did not call them sharp, principal, diffuse, fundamental series.
Today these s,p,d,f labels just have a notational meaning related to orbital angular momentum $l$ values in Schrodinger's equation. If $l$ =0, you label it as $s$, $l$ =1, it is $p$ etc. This is pure algebraic use. Do not associate any other historical meaning to them now.

Okay right, being a multielectron system makes a lot of difference over here. That is exactly why the Bohr model failed and we had to look up to Quantum Menchanics right, because it couldn't explain the spectra of multielectronic species. Thanks a lot, @M. Farooq!

Yes, multielectron atoms are still a challenge.


text:
Why is Pb3O4 called lead(II,IV) oxide?
We know that iron(II,III) oxide has the formula $\ce{Fe3O4}$ which is a combination of iron(II) oxide and iron(III) oxide with formulas $\ce{FeO}$ and $\ce{Fe2O3}$ respectively.
So why isn't this the case in $\ce{Pb3O4}$? Why can't we call it lead(II,III) oxide just like the iron example?
On the other hand, lead(II,IV) oxide should have the formula $\ce{Pb2O3}$ which is the same with lead(III) oxide.
Lead(II)oxide: $\ce{PbO}$ and lead(IV) oxide: $\ce{PbO2}$.

Because lead (III) is not a thing.

@IvanNeretin According to Wikipedia, the oxidation states of lead are $−4, −2, −1, +1, +2, +3$, and $+4$; with $+2$ and $+4$ being dominant.

These are not combinations of 2 oxides, but they are rather single oxides where metal atoms alternate their oxidation numbers ( II,III for Fe, II,IV for Pb ), often not exactly in stoichiometric ratios of small integers. This can be determined from X-ray crystallographic analysis. $\ce{Pb3O4}$ is AFAIK actually $\ce{Pb2[PbO4]}$

The name of both cases is based on what is known, not what is liked. // Chemistry SE site strongly recommends plain text titles for index/search reasons and due possible displaying issues in question lists.

lead(II,IV) oxide should have the formula Pb2O3 It would be true if Pb(II)/Pb(IV) ratio were 1:1, but it is 2:1. There is $\ce{Pb2^{II}Pb^{IV}O4}$ versus  $\ce{Fe^{II}Fe2^{III}O4}$

@Poutnik What would the iron oxide formula be if the Fe(II)/Fe(III) ratio were 1:1?

In the old valency notation (19th century), $\ce{Pb3O4}$ would have been  $\ce{O=Pb-O-Pb-O-Pb=O}$

@Maurice it can be considered as the salt lead(II) orthoplumbate(IV). // From wikipedia: Plumbate
"Lead tetroxide" $\ce{Pb3O4}$ may be thought of as lead(II) ortho-plumbate(IV), $\ce{Pb2^2+[PbO4]^4-}$. "Lead sesquioxide" $\ce{Pb2O3}$ is also known, and has the structure lead(II) meta-plumbate(IV), $\ce{Pb^2+[PbO3]^2-}$

@Poutnik Yes. It can be considered as the salt you mentioned. Bit in the 19th century ions were not yet known.

On Chemistry mathematical and chemical expressions can be formatted using MathJax (and LaTeX Syntax).  If you want to know more, please have a look here and here. We prefer to not use MathJax in the title field, see here for details.

@IvanNeretin technically, lead(III) is known in hexamethyldiplumbane. But yes, the oxide structure better matches with a combination of lead(II) ans lead (IV).

Both crystal structure and periodic characteristics favor the formulation $\ce{Pb^{II}_2Pb^{IV}O4}$.
Below is a picture of the crystal structure from Wikipedia.

Source
This structure has at least two features that would be consistent with a lead(II,IV) identification:

One third of the lead atoms have higher coordination number (six) to the electronegative oxygen and the rest have lower coordination number (two). The more highly coordinated lead atoms would naturally fit with a higher oxidation state, and $\ce{Pb^{II}_2Pb^{IV}O4}$ fits with the ratio of high to low coordinate lead atoms.

The six-coordinate lead atoms noted above have the same coordination as they do in the pure lead (IV) oxide $\ce{PbO2}$. Note the contrast here with iron, where six-coordination is typical for oxidation states +2 and +3.

In addition, $p$-block elements — unlike transition metals — strongly favor paired-electron structures, which means even-$Z$ elements favor even oxidation states and odd-$Z$ elements favor odd ones. Where the oxidation state has the "wrong" parity it's usually accompanied by homonuclear bonding which allows the electrons to be paired without being polarized (like peroxides with their oxygen-oxygen bond or lead(III) in $\ce{(CH3)3Pb–Pb(CH3)3}$). No such homonuclear bonding is evident in $\ce{Pb3O4}$, so the lead atoms ($Z=82$) would be assigned even oxidation states to fit the general trend.

@Poutnik gave us the Wikipedia link which basically contains the answer:

Lead tetroxide ("red lead"), a mixed oxide with formula $\ce{Pb3O4}$, may be thought of as lead(II) ortho-plumbate(IV), $\ce{[Pb^2+]2[PbO4]^4−}$. Lead sesquioxide, $\ce{Pb2O3}$, is also known, and has the structure lead(II) meta-plumbate(IV), $\ce{[Pb^{2+}][PbO3]^2−}$.

Since $\ce{Pb3O4}$ contains $\ce{Pb(II)}$ and $\ce{Pb(IV)}$, it is lead(II,IV) oxide. Magnetite is actually $\ce{[Fe^{2+}][(Fe^{3+})2O4]}$, so it is iron(II,III) oxide.


text:
Under what conditions does manganese dioxide lose its catalytic property?
We know that under standard conditions (room temperature and 760 mmHg air pressure) we have this reaction:
$$\ce{2H2O2(l) ->[MnO2] 2H2O(l) + O2(g)}$$
Here, manganese dioxide is behaving as a catalyst, accelerating the decomposition of hydrogen peroxide into water and oxygen.
If we want manganese dioxide to instead react with hydrogen peroxide what should the conditions be? Also, what would the products be if this reaction occurs?

By catalysing hydrogen peroxide decomposition, it does react with it, but this reaction cyclically regenerate it to the initial state, with manganese reversibly switching oxidation state.
One of possible models is:
$$\begin{align}
\ce{H2O2(aq) + 2 MnO2(s) &-> H2O(l) + Mn2O3(s) + O2(g)}\\
\ce{H2O2(aq) + Mn2O3(s) &-> H2O(l) + 2 MnO2(s)}
\end{align}$$
The point is, hydrogen peroxide has both oxidative and reductive properties, with the latter being oxidized to oxygen.
Variants ( with the $\ce{MnO2}$ reduction part ongoing in the Leclanché cell) :
$$\begin{align}
\ce{H2O2(aq) + 2 MnO2(s) + H2O(l) &-> H2O(l) + 2 MnO(OH)(s) + O2(g)}\\
\ce{H2O2(aq) + MnO2(s) &-> Mn(OH)2(s) + O2(g)}\\
\ce{H2O2(aq) + 2 MnO(OH)(s) &-> 2 H2O(l) + 2 MnO2(s)}\\
\ce{H2O2(aq) + Mn(OH)2(s) &-> 3 H2O(l) +  MnO2(s)}
\end{align}$$
Other model may involve also $\ce{Mn^2+(aq)}$ at low enough $\mathrm{pH}$.
The way of suppress the catalysis may be catching eventual intermediates e.g. by a reagent forming stable complexes with $\ce{Mn^{II}}$ or $\ce{Mn^{III}}$. It would cause majority of $\ce{Mn}$ ending in these complexes after oxidation enough $\ce{H2O2}$ to $\ce{O2}$.
Altenative would be using acidic environment where $\ce{MnO2}$ would end as $\ce{Mn^2+(aq)}$.
Reaction of $\ce{MnO2}$ with citric acid combines both, with acid undergoing oxidation and complexation.

I suppose, it could stop being a catalyst if there were a set of conditions which inhibited one of the two reactions, while allowing the other one to proceed. I don't know if such conditions exist...!

@orthocresol I suppose blocking it, e.g. by forming some stable manganese complex,would block also eventual reactions forming other manganese products.

@Poutnik, Is there a reference or experimental proof that supports those equations?

@M.Farooq Not sure. Possibly. They have been meant as model reactions, following general concept of redox reactions, frequently homogenous, being catalyzed by transition metals with clipping oxidation numbers, like iron, copper or silver. // It could be written more schematically like Mn(III) <-> Mn(IV)

@Poutnik, I highly doubt that there are redox reactions on MnO2. The above reactions imply that the decomposition is not quantitative (i.e., one molecule of H2O2 is wasted in the second equation) but as far I remember it is quantitative.

@M.Farooq I do not see your implication. Check it more carefully. The net reaction is the same as eventual direct decomposition: $\ce{2 H2O2 -> 2 H2O + O2}$ // But I do agree that in context of heterogen catalysis, it would be more complicated than the model.

@Poutnik, You are right. The net result is the quantitative decomposition equation.

Mn2O3 is frequently mentioned for Leclanche galvanic cell. On the other hand, Mn^2+ is not, AFAIK, even if there is mildly acidic pH due NH4Cl.// Unfortunately, I do not have access.

Catalyst is a substance which alters the rate of reaction without being consumed in it. It doesn't mean that catalyst doesn't participate in the reaction. It does react with reactants as Poutnik already stated in his answer. I came up with an another mechanism from [reference][1] and [reference][2]:
$$ \ce{ MnO2  + H2O2 + 2H+ -> Mn^{2+} + 2H2O + O2}$$
$$ \ce{ Mn^{2+} + 2H2O <--> Mn(OH)_2 + 2H+ }$$
$$ \ce{ Mn(OH)_2 + H2O2 -> MnO2 + 2H2O }$$
giving an overall reaction:
$$ \ce { 2H2O2 ->[MnO_2] 2H2O + O2}$$

References:
[1]: D. B. Broughton, and R. L. Wentworth, "Mechanism of Decomposition of Hydrogen Peroxide Solutions with Manganese Dioxide. I," J. Am. Chem. Soc. 1947, 69(4), 741–744 (DOI: https://doi.org/10.1021/ja01196a003).
[2]: D. B. Broughton, R. L. Wentworth, and M. E. Laing,"Mechanism of Decomposition of Hydrogen Peroxide Solutions with Manganese Dioxide. II," J. Am. Chem. Soc. 1947, 69(4), 744–747 (DOI: https://pubs.acs.org/doi/10.1021/ja01196a004).

I think the relavent equations are given in reference 1. Reference 2 is their further studies. If you don't agree feel free to delete my addition.

@M.Farooq: Do you think these references are good enough to backup Poutnik's answer?

@MathewMahindaratne, I am not an expert, but the references are very outdated, not because they are not reliable, but the formation of Mn(OH)2 sounds a little dubious. In those times, chemical kinetics was not very advanced. What proof do the authors provide about the formation of Mn(OH)2 as an intermediate?

@M.Farooq: I don't have access to the papers at home but will look at them on Monday. Yet, it was stated that catalysis does not occur until the solution is saturated with manganese hydroxide. Their reserch is based on radioactive mangenese compound as well. Not sure how they come to the conclusion.

@MathewMahindaratne, I had a quick look. They have several hypotheses, but Poutnik's  is not supported anywhere in those papers. They insist on the role Mn (OH)2...but keep in mind the pH effects. In short a more recent and detailed study is needed. I did not downvote this answer.

The 2nd equation is unbalanced and it is not really Redox equation..There should be rather H2O instead of H2O2. I would say formation of Mn^2+ is questionable.

The main question is not catalytic mechanism, but catalysis suppression.

Manganese dioxide loses its catalytic activity if there is a strong acid in the reaction your mentioned in your post. Consequently, manganese dioxide simply reduces itself to colorless/light $\ce{Mn^{2+}}$ and hydrogen peroxide converts itself to water. I do not exactly recall evolution of oxygen. For the sake electron balance in a redox process, $\ce{O2}$ formation is necessary. However, the dissolution is very fast.
In fact this reaction was widely used indirectly (via $\ce{KMnO4}$ route)  for cleaning glassware long time ago in classical analytical work.

If MnO2 is reduced to Mn(II), and H2O2 reduced to water, what gets oxidised?

@orthocresol, MnO2 almost instantly dissolves in acidic solution in the presence of H2O2. I cannot find an equation at this moment. If you ever had a KMnO4 stain on the skin, a small amount of vinegar and medicinal H2O2 (< 1.5-3%) instantly removes it.


text:
Isolating potassium chloride from sodium chloride and potassium hydroxide
Assume that you have an aqueous solution of sodium chloride and potassium hydroxide. Is it possible to isolate potassium chloride from this?
$$\ce{Na+(aq) + Cl-(aq) + K+(aq) + OH-(aq) -> Na+(aq) + OH-(aq) + KCl(s)}$$
Since all involved salts/salt combinations are highly water-soluble, I'm not sure how one could isolate the potassium chloride. Is there maybe a way to selectively precipitate the sodium via a non-water soluble salt?

Comments are not for extended discussion; this conversation has been moved to chat.

I suggest to buy $\ce{KCl}$, or to prepare it by neutralization of $\ce{KOH}$ and $\ce{HCl}$.
If it happens you have mixed solution of $\ce{NaCl}$ and $\ce{KOH}$, neutralize it first by $\ce{HCl}$ (pH meter or pH indicator paper strips).
From Wikipedia solubility table one can conclude that while $\ce{NaCl}$ solubility almost does not depend on temperature, solubility of $\ce{KCl}$ is almost twice at 100 °C, compared 0 °C.
My suggestion is fractional crystallization:

Partially evaporate the solution until significant mass of salts crystallizes.

Heat the mixture to near boiling and mix well - part of $\ce{KCl}$ would dissolve.

Separate the precipitate enriched by $\ce{NaCl}$.

Cool down solution and let crystalyze salt enriched by $\ce{KCl}$.

With both precipitates and residual solution, you can recursively repeat the procedure.

You can set a system of N batches, where $\ce{NaCl}$ enriched precipitate would go to direction of the batch more enriched by $\ce{NaCl}$ and vice versa.

If it looks complicated to you, buy $\ce{KCl}$ or remember the Pierre and Marie Curie  struggles with the isolation of traces of $\ce{RaCl2}$ from $\ce{BaCl2}$, where their solubilities are much closer.

Another option is using about 3 times better solubility of $\ce{NaCl}$ in methanol, compared to $\ce{KCl}$. But as it is like $\pu{5 g/L}\ \ce{KCl}$ and $\pu{15 g/L}\ \ce{NaCl}$, there would be extensive solvent manipulation with price and safety issues.
As useful info, here is the question/answer about the solubility diagram of mixtures of both salts

The solubility of a salt like $\ce{NaCl}$ or $\ce{KCl}$ is not the same in pure water and in a solution of another salt.  Please look at solubility diagrams $\ce{NaCl - KCl}$ !

@Maurice I am well aware of it. :-) In fact, I have answered some question with such a diagram last year. The solubility in pure water is just a guidance, there is not totally messed up relation of their temperature dependencies.(seen at the diagram as well).

Ion exchange chromatography could be an option, too. A couple of (late) transition metals are separated from each other this way; the value attributed to them justifies the investment.  It very well may be a different story for NaCl, in economic perspective.

NaCl is about three times as soluble as KCl in methanol at room temp. That said, while it might barely be feasible to separate their salt that way, assuming nearly 100% solvent recovery, and near-zero energy cost (perhaps waste geothermal?), the most efficient commercial method to separate various salts is with evaporation ponds.
That said, this is not an answer for your question on separating a few liters of solution, but I've written it as one to add the links. If others think this should be placed as a comment, I'll try to do so.

I was thinking about methanol too, in the deleted comment, but decided against it.


text:
Systematic IUPAC name of buckminsterfullerane
Many of you have heard of buckminsterfullerEne, which is C60, also called the "buckyball".
Well, there's also buckminsterfullerAne, which is C60H60. This is a buckyball that's fully saturated (all carbon-carbon bonds are single bonds) and fully hydrogenated.
I'm aware that the nomenclature for polycyclic hydrocarbons can get pretty messy. For non-aromatic hydrocarbons, I've already determined why the systematic IUPAC names (the ones with numbers inside brackets that indicate the "backbones" and bridges) for tetrahedrane, cubane, and dodecahedrane are what they are:
Tetrahedrane: Tricyclo[1.1.0.02,4]butane
Cubane: Pentacyclo[4.2.0.02,5.03,8.04.7]octane
And from the Hamiltonian path of a regular dodecahedron, I was also able to verify the systematic name for dodecahdrane.
Dodecahedrane: Undecacyclo[9.9.0.02,9.03,7.04,20.05,18.06,16.08,15.010,14.012,19.013,17]icosane
As you can see, with a dodecahedron it starts getting pretty messy, but in the linked image above, if you number the rightmost point as #1, then whichever way you go the point that is connected to #1 via a dashed line would be #11. From then I was able to verify the locations of all the secondary bridges. From my intuition, it seems that you want to number the vertices such that at least one pair of vertices with a bridge has the vertex numbers differing by one-half the total number of vertices.
Now, that got me wondering, what would the same name for a truncated icosahedron be?
I started looking for Hamiltonian paths through a truncated icosahedron, and came across this paper. On Page 2, Figure 2, six possible Hamiltonian paths are shown. I numbered the first two of these and found out that the red figure has two bridges where the two vertices' numbers differ by 30 (10/40 and 25/55, the 32/52 in the circle is incorrect). I've numbered them as follows:
And then after locating the two bridges, I re-numbered the diagram with vertex #1 as one of the vertices in the black rectangle:
And from there, I tried to deduce the systematic IUPAC name for buckminsterfullerane. This is what I came up with:
Triacontakaihenacyclo[29.29.0.02,28.03,26.04,23.05,60.06,22.07,58.08,20.09,57.010,18.011,55.012,17.013,53.014,51.015,48.016,46.019,45.021,44.024,43.025,41.027,39.029,37.030,35.032,59.033,56.034,54.036,52.038,50.040,49.042,47]hexacontane
(man, what a mouthful)
where triacontahena- means 31 and hexaconta- means 60.
My question is whether this name is correct. And if it's not correct, where are the correct locations for the bridges?

What makes you think fully hydrogenated buckyballs exist?

https://www.science.org/doi/10.1126/science.253.5017.330 suggested that you'd need hydrogens inside the shell you not make it very strained. Such compound would have a lot of isomers you don't take into account.

@Mithoron Stereoisomers?


text:
My book's claim about the shielding effect of s,p,d and f electrons
It's a relatively unknown Bangladeshi book. It's called "Chemistry-First Paper (Class XI-XII)" by Professor Haradhan Nag
According to my book, "Electrons of s and p orbitals produce more shielding than electrons of d and f orbitals. As a result, atoms of s and p-block elements experience lower effective nuclear charge. Consequently, their atomic sizes are greater than those of atoms of d and f-block elements".
My book's claim makes no sense to me. It's true that the electrons of s and p orbitals produce more shielding than electrons of d and f orbitals due to penetration, but why would that make atomic sizes of s & p block elements greater than d and f-block elements?
Let me take s-block as an example. In period 4, $\ce{K}(220 \pu{pm})$ is an s-block element and $\ce{Sc}(160\pu {pm})$ is a d-block element. It's true that the atomic size of K is greater than Sc; however, I don't think it's because s-orbitals have a stronger shielding effect than d-orbitals. If anything, we could ironically argue that Sc should have a greater atomic radius than K as s-orbitals have a stronger shielding effect than d-orbitals because the 4s-orbital is inner in Sc. In contrast, the 4s-orbital is situated outermost in K. So, why is my book making the claim that an outermost orbital's shielding power will determine the size of the atom? Outermost orbitals have no contribution to shielding!

Related

Edit:
It turns out that the 4s orbital is not actually inner in Sc. However, my point still stands. Why is my book making claims about shielding based on the outermost orbital (which has no contribution to shielding!)

See also Wikipedia: Slater's rules

Let's have a look at some illustrations:

Size of atoms by atomic number:

The size of Li, C and Ne with their orbitals (blue is completed and yellow is empty):

The shielding effect, illustrated with the density of probability of the electrons' presence (green):

Hydrogen probability densities of 2p and 2s orbitals:

We will need to use quantum physics to answer this question, so it gets further than the classical chemistry problems. I therefore consider electrons as matter waves, their position being non-deterministic, the probability of their presence at such and such a place determined by the Schrödinger equation.
On the illustration n°1, you notice that the atomic radius decreases with the number of electrons until reaching a new orbital, which necessarily increases the range of the electronic cloud (each orbital being more extended than the previous one).
The illustration n°2 shows you that the atomic radius decreases with the stability of the electron (number of completed orbitals).
The third and fourth ones just illustrate the consequences of the shielding effect.
Now, here is the answer to your question: within a set of neutral atoms that have the same number of orbitals, the charge of the nucleus and the electronic cloud increases with the atomic number Z. As the shielding effect of the first sublayers is limited to the maximum number that these orbitals can hold, the charge felt by the farthest electrons will be stronger and stronger, which will attract them more towards the center and decrease the atomic radius. Concerning the shielding effect of the higher orbitals, I have a little trouble understanding what your book means since the s and p orbitals always have a shielding effect (never replaced by the shielding of the other orbitals, but added to it). Generally speaking, a smaller shielding effect gives a higher charge of the nucleus and therefore a smaller radius. Atoms with d- and f-layers have a larger radius because the range of these orbitals is simply larger according to the Schrödinger equation (you can compare the range of 2s and 2p on picture n°4). I hope this helped you!


text:
Is collision theory applicable only for molecules?
$$\ce{Na (s) + Cl_2 (g)->NaCl(s)}$$
Is collision theory applicable for the above reaction, or is collision theory applicable only for molecules? In other words, do chlorine atoms collide with sodium atoms to produce sodium chloride , or does the chlorine atom get near the sodium atom, electron transfer takes place (as it is energetically favorable), and $\ce{Na+}$ and $\ce{Cl-}$ get stuck?
Edit after @BuckThorn's comment:

or does the chlorine atom get near the sodium atom, electron transfer takes place (as it is energetically favorable), and
$\ce{Na+}$ and $\ce{Cl-}$ get stuck?

In this process, the Na atom and the Cl atom don't touch, and electron transfer occurs at a distance (due to attractive/repulsive forces). Only after electron transfer takes place do $\ce{Na+}$ and $\ce{Cl-}$ touch and get stuck. That's why I don't think it is akin to collision.

You may want to explain why the two alternatives you present might or might not agree with collision theory (in particular the second). Collision theory combines a description of the rate at which reactants collide with a model of the fraction of those reactants that have enough kinetic energy to result in a productive collision. The details will change but, provided the assumption that a collision between reactants exceeding the required energy is appropriate, the theory should be able to describe reactions between gases and solids.

Note that $\ce{Na(g) + Cl(g) -> Na+(g) + Cl-(g)}$  is energetically unfavourable, unless ions end bound close enough ($\ce{Na+Cl-(g)}$) by coulombic force. Saying that, not speaking about energy needed for formal $\ce{Na(s)}$ atomization and $\ce{Cl2}$ dissociation.

@Poutnik On a related note: how do we determine if a reaction is energetically favorable or not? Do we add the ionization energy of Na (496kJ/mol) and the electron affinity of Cl (-349kJ/mol), and if the answer is positive(147kJ/mol) then we say that the reaction isn't favourable?

Yes it is like that. We would say such a reaction is endothermic, consuming thermal energy. Strictly speaking, I should use thermodynamically favourable via negative Gibbs free energy change, not enthalpy change. ( $\Delta G = \Delta H - T \Delta S$ ), but at these big $H$ changes the entropy term plays just minor role. When energy is released, it is usually dissipated elsewhere, so it cannot be reused to go back. OTOH, if energy is needed, you have to gain it somewhere.

@Poutnik "unless ions end bound close enough (Na+Cl−(g)) by coulombic force."- sir, I couldn't understand exactly what you mean by this. Do you mean that large amounts of energy are released (energetically favourable) when Na+ ion and Cl- get closer and get stuck together?

@TomHardy Energy outcome of producing ions depends on how far from each other the are  If far enough, it is endothermic. If close enough, it is exothermic. The best if many opposite ions are close enough, forming solid ionic lattice.

Your example is one  of many that have been studied by molecular beam scattering techniques so this is a rather general answer.
Collision theory calculates the rate constant averaged over all geometries and energies at at given temperature. It is the product of cross section, average collision velocity and the Arrhenius factor. Thus it depends on the cross section. The simplest model assumes hard spheres so will work for atom-atom, atom -molecule etc. but less well as the species become more complex. But actually the cross section is very complicated especially as the potential between molecules varies with distance and orientation perhaps as coulomb potential or Lennard Jones or something more complicated. In this case The trajectories of the interaction between species has to be calculated and these averaged to get the collision theory rate constant.
In fact the rate constant is not that important what is important is the potential energy profile between species as this reflects the electronic properties of the molecules and can be calculated from Quantum Theory. The rate constant can then, in principle, be calculated from the potential although experiment is always essential.
Experiments of this sort are studied using atom/molecular beams and numerous reactive scattering reactions have been studied such as H+H2, D+H2, N+O2, F+H2,H+F2, O+CH, O+Cs and many more.
You should look at two excellent books on this topic Levine & Bernstein 'Molecular Reaction Dynamics, and Chemical Reactivity' publ OUP 1987, and Steinfeld, Francisco & Hase, 'Chemical Kinetics and Dynamics', publ Prentice Hall 1999.
The picture below shows how complicated the trajectories become when there is a Lennard jones potential between two atoms as one approaches the other from the left but at different position vertically. You can see that the species get pulled together at a larger separation than hard sphere (the inner circle) but can orbit and escape if there is enough energy. In this picture no reaction occurs but you can appreciate if the species get close enough with enough energy a condition could be made to make them react, for example being within the wide grey ring.

The picture below shows the Cl+H2 $\to$ HCl+H reaction. The species approach with the same energy in each case but collide at different point in the H$_2$ vibration and this leads to different outcomes, i.e no reaction or more of less vibrational excitation in HCL. The blue colour is low energy and the transition state can be seen at the bend. To calculate the rate constant the number of successful transition state crossings vs. total number at a given energy must be calculated. This could then be compared with the collision theory rate calculation and an effective cross section derived.

"The species approach with the same energy in each case but collide at different point in the Cl2"-You probably mean Cl instead of Cl2.

"The picture below shows the Cl+H2 → HCl+H reaction"-how does Cl remain as Cl? Isn't it a diatomic molecule?

"The simplest model assumes hard spheres so will work for atom-atom, atom -molecule etc."-Just to be clear, you agree with me that collision theory should be applicable for the reaction Na(s)+Cl2 (g)⟶NaCl(s), right?

you seem to be considering a solid /gas reaction which is different.

BuckThorn made a comment about this matter. Do you agree with him, sir?

@user119245 I just now noticed, it is a typo it should be H$_2$  vibration not CI$_2 I will edit it shortly$


text:
Spontaneous reaction between chlorine and a hypothetical atom
Suppose, my hypothetical element is $\ce X$, whose first ionization energy is $\pu{200kJ/mol}$. An atom of this element will react with an atom of chlorine (assume that I've already dissociated a chlorine atom from a $\ce{Cl2}$ molecule). The first electron affinity of chlorine is approximately $\pu{349kJ/mol}$.
We take those two atoms inside a non-reactive box in a controlled room. Inside the box is vacuum. I intend to make the following reaction happen:
$$\ce{X(g) + Cl(g)-> XCl(s)}$$
Now, I'm curious as to the course of events.
Hypothesis 1:
One hypothesis is that the reaction will not proceed even if you get the atoms really close together. For the reaction to proceed, you must provide the required ionization energy to  $\ce X$ first. Then the removed electron from X will be acquired by Cl promptly. Energy will be released in this process. The energy released in this process will be greater than the ionization energy that was provided to $\ce X$ previously. In the final step, $\ce {X^+}$ and $\ce {Cl^-}$ will get stuck together due to the electrostatic force between them. In conclusion, the reaction will process in the following steps (in order):
$$\ce{X(g) -> X+ + e-}\tag{1}$$
$$\ce{Cl(g) + e- -> Cl^{-}(g)}\tag{2}$$
$$\ce{X+(g) +Cl^{-}(g) -> XCl(s)}\tag{3}$$
In this hypothesis, the reaction isn't fully spontaneous since heat/energy needs to be added in the beginning for the reaction to proceed.
Hypothesis 2:
Another hypothesis is that once you take the two atoms close together, electron transfer will take place i.e. the reaction will proceed even if you don't add any heat. In other words, in this hypothesis, heat/energy does not need to be added for the reaction to proceed: the reaction is fully spontaneous. Ionization energy will not need to be provided to $\ce X$ as the reaction is energetically favourable. The reaction will take place in the following steps:
$$\ce{X(g) -> X+ + e-}\tag{4}$$
$$\ce{Cl(g) + e- -> Cl^{-}(g)}\tag{5}$$
$$\ce{X+(g) +Cl^{-}(g) -> XCl(s)}\tag{6}$$
The commonality between hypothesis 1 and 2 is that there will be a net release of energy after $(1)$ & $(2)$ and $(4)$ & $(5)$. The net release of energy after $(1)$ & $(2)$ will be equal to the net release of energy after $(4)$ & $(5)$. $(3)$ and $(6)$ will be the same in both hypotheses. However, unlike hypothesis 1, $(4)$ and $(5)$ take place simultaneously while in hypothesis 1, $(1)$ takes place first and $(2)$ takes place second.
In this hypothesis, the reaction is fully spontaneous since heat/energy doesn not need to be added in the beginning for the reaction to proceed.
My question:

Which hypothesis is correct?

In context of this question, one should distinguish thermodynamical and kinetical spontaneity. It can be thermodynamically spontaneous even if there is needed such a high activation energy the process is never observed. Like some isotopes are thermodynamically radioactive with thermodynamically spontaneous decays, but kinetically stable.

Both hypotheses are conditionally correct, as each describes a different possible scenario. I assume hypothesis 2 would still have some activation energy, but much lower than X ionization energy.

Reaction's simply $$\ce{X(g) + Cl(g)-> XCl(g)}$$ That's it! No magical ionisation, no flying electrons, no ions.

@Mithoron "That's it! No magical ionisation, no flying electrons, no ions."- I'm not confident about the ionization and flying electrons part, but I'm positive about the formation of ions: in the formation of NaCl, Na+ and Cl- ions are definitely formed and they get stuck together.

Answer you linked is incorrect. https://chemistry.stackexchange.com/questions/14174/what-is-sodium-chloride-like-in-gas-state https://chemistry.stackexchange.com/questions/2511/will-gaseous-ionic-compounds-be-free-moving-ions

@Mithoron Sir, AmritanshSinghal used the Born-Haber cycle in his answer.

@Mithoron  It is correct I think (pic found in chemlibretexts). I'm having difficulty understanding why you think otherwise. Could you please explain it to me kindly?

Born-Haber cycle has hardly anything to do with mechanisms. Atoms of Cl and Na hit each other and NaCl molecule is created (and surplus energy needs to be released, but that's besides point). Bond it this molecule may have more ionic than covalent component, though.

@Mithoron, Sir, just to understand your position better, do you think this answer is incorrect as it wrongly presents the Born-Haber cycle as the reaction mechanism of $$\ce{Na(s) + Cl_2(g)->NaCl(s)}$$?

Well, it only says "following steps are involved in the formation" not exactly "mechanism", but that's still incorrect.

@Mithoron According to wikipedia, "In chemistry, a reaction mechanism is the step by step sequence of elementary reactions by which overall chemical change occurs." So, I think that AmritanshSinghal is describing the reaction mechanism as reaction mechanism just means a step by step sequence of elementary reactions, and AmritanshSinghal presented a step by step sequence of elementary reactions. Thoughts?

In this hypothetical case, formation of free ions is exothermic, in contrary to real endothermic cases. As consequence, X(g) +Cl(g) -> X+(g) + Cl-(g).

@Poutnik So, ionization energy will not need to be provided?

At contact of both atoms - no, it need not, as it is a different system than 2 separate atoms. Thete will be some activation energy, but lower than X ionization energy. Similarly as a planet could eventually escape a star easier, if some other star would incidentally come close to the first star.

@Poutnik What is the difference between when two atoms are in contact and when they are separate?

I agree with @Poutnik. It depends on the shape of the potential energy surface. If you get the atoms close together, you actually add energy to the system. Imagine you have to go from one valley to another one through a saddle point among two mountains and the two valleys. The ionisation energy moves you to the saddle and then you do not need energy to go downhill. If you put the atoms together, you just move closer to the saddle and the height you have to overcome is lower. This makes really high probability of quantum tunelling to the other side of the saddle and then downhill again.

@TomHardy What is the difference if there is another star near the Sun at the distance of the planet Neptun or not ? There is different electrostatic field for atom case and different classical gravitational field in the star case. // By other words, atoms at contact are not like billiard balls at contact. They are different and behave different.

When an electron moves from one atom to another -- at the contact -- to form a ion pair, it is not a free electron like in the IE context, but it is a bound electron all the time, with the energy lower than a free electron would have.

With respect to the Born-Haber cycle answer, the wonderful thing about state functions like enthalpy is that they are path independent, so we can use any steps we want in a calculation as long as the end points are the same. That is the fundamental logic behind the Born-Haber cycle - the calculation is not intended to represent the actual reaction steps but rather is a path comprised of steps whose energies are available.

Poutnik and Mithoron are disagreeing, so I'm not sure who is correct (point 1). Moreover, Mithoron disagrees with the answer I linked too (point 2). Light shed on both point 1 and point 2 would be much appreciated. (...)

(...) Moreover, I think that hypothesis 1 might be the correct one. This is because my book said, "The ionization energy necessary to remove one or two or three electrons from metal atoms is available in the environment, but to remove 4 electrons from atoms much more ionization energy is required, which is not readily available in the environment, so atoms of many elements fulfill their octet by sharing electrons. These bonds are called covalent bonds." My book seems to imply that ionization energy is non-negotiable.(...)

(...) For ionic bonds to form through electron transfer, one must provide the requisite ionization energy, without which the reaction will not proceed at all! Hypothesis 1 seems to be consistent with this idea, so probably hypothesis 1 is correct. I might be wrong.

@Poutnik If the ionization energy of X would've been greater than the electron affinity of Cl, would the provision of ionization energy then for the reaction to proceed be non-negotiable? [Assume that the two atoms are not in contact and are a measurable distance apart]

Should this be crossposted in Physics SE?

In such a case there are 2 separate operations, with the first ionization independent on eventually following electron capture // Crossposting is frown upon on SE network.

This thought experiment needs more thinking.  Both scenarios are essentially Born-Haber cycles not possible mechanisms. Forming a positive ion always requires energy so that makes equations 4,5,6 not relevant to the second concerted mechanism. In the case of the concerted mechanism the X and Cl atoms must have sufficient KE to overcome electron repulsions as they approach and to continue to distort the atomic orbitals to form a molecular orbital. Ionic bonding requires a lattice energy to complete the cycle and two atoms will not suffice. Finally there must be a mechanism to remove the bond energy from the incipient bond. Radiation is a possibility.
The first proposal is not mechanistically reasonable. It requires external energy sources and the energy loss from the electron affinity is just lost and does not contribute to the overall reaction and the same applies to the final bond formation.   Considering an initiated reaction such as a free radical chlorination where final lattice heat provides energy to maintain the reaction is probable but such a mechanism would not necessarily involve free chloride ions.

There is a flaw in hypothesis 1 - getting the atoms "really close" will provide activation energy.
At sufficiently close distance the combined electron Pauli repulsion and nuclear core electrostatic repulsion will add significant energy.
For example, $2$ protons $7$ angstroms apart have about $200$ kJ mol$^{-1}$ of electrostatic repulsion between them.
Also, $200$ kJ mol$^{-1}$ ionization energy is the absolute maximum activation energy required.  As the two potential wells - the $\ce{X}$ nucleus and the $\ce{Cl}$ nucleus - approach, eventually the wells are close enough that an electron can tunnel from a $\ce{X}$ to a $\ce{X}$ to form $\ce{X+}$ and $\ce{Cl-}$ ions.
Alternatively, other bound electronic states across both atoms (essential a covalent bond) will form at much lower energies than $200$ kJ mol$^{-1}$ above the separated states.
Before that even, at much larger distances very weak dispersion interactions will draw neutral atoms towards one each over.  The long-range potential is attractive and the first minimum is likely close enough to transfer an electron.  The reaction is very likely to have no activation energy.
So diffusion will dominate to create dimers.  Depending on the temperature I would expect it to look like hypothesis 2.
These dimers will essentially be entirely ionic - pulled as close together as possible until the Pauli repulsion of the electrons stabilises the electrostatic attractions.
As the ions are more thermodynamically stable, the balance of the depth of the attractive potential against entropy will determine the size of aggregate species - from free ions (high temperature plasma) to dimers (NaCl gas) to larger oligomers or condensation/sublimation.

If the ionization energy of X would've been greater than the electron affinity of Cl, would the provision of ionization energy then for the reaction to proceed be non-negotiable? [Assume that the two atoms are not in contact and are a measurable distance apart]

If the atoms are very (infinitely) far apart and the ionization of X was greater (as is the case with all real elements), then the thermodynamically stable configuration is 2 neutral atoms.
They have to be close enough for tunneling, or for some activation energy to be added as I mention - then the situation is only stabilised by the electrostatic attraction between the two charged ions.


text:
Shielding vs electron-electron repulsion
Example of shielding:
 (source)
The last electron in the 6s subshell of $\ce {Cs}$ is shielded from the nucleus by the inner electrons.
Example of electron-electron repulsion:
The electron affinity of $\ce F$ is surprisingly lower than that of $\ce{Cl}$. This is because, in $\ce F$, there are 5 electrons in the small-sized 2p subshell, so the electron density is very high. Therefore, the incoming electron is heavily repelled by the 5 electrons in the 2p subshell through electron-electron repulsion. So, the electron affinity of  $\ce F$ is lower.
My comments:
Can the example of electron-electron repulsion of $\ce F$ be called shielding? I previously knew that electrons in the same shell don't contribute to shielding; however, I found a comment on the contrary.
My questions:

Do electrons in the same shell contribute to shielding?
Can the example of electron-electron repulsion in the case of $\ce F$ be called shielding?

As Q1: If you read wikipedia.org : Slater's rules - Rules, you would see they do. Q2: In a way, yes, but the shielding is rather considered in context of affecting the mean radial force, while repulsion involves substantial tangential force.

E.g. an electron in the lobes of the orbital 2p_x  partially tangentially repulses electrons in 2p_y and 2p_z, and also eventually its mate in the same orbital, especially when in the same lobe.


text:
Does heat increase the hydrolysis of urea hydrochloride?
The preparation of ammonium chloride by reacting urea with HCl produce urea hydrochloride the reaction take too much time to get ammonium chloride depending on ratios.

So can heat increase the process?

Thermodynamically or kinetically? For the latter, it speeds up all processes. Some more than others, so it can shift equilibrium both ways.

I have cropped your image to highlight the relevant text but it's better to write the text as pictures are not searchable.

You may look here to understand the reaction.


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What does it mean by pruning and enrichment in the case of Rosenbluth method?
As far as I understand:

Pruning means deleting something.
Enrich means to enhance/increase weight.

Now my question is, in the case of PERM algorithm,

Are we deleting some polymers chains? Or, are we deleting some beads in all polymer chains?
Are we increasing the weights of some polymers? Or, are we increasing the weights of some beads/atoms of all polymers?

When we are modifying the pruned weights, what are we replacing them with?

You posted a duplicate of this in mattermodelling? If so, could you please link each question to the other so that once it is answered on one site, the answer can be accessed from the other site as well.

https://link.springer.com/article/10.1007/s10955-011-0268-x


text:
Can chloride donate an electron pair to silver(I)?
According to the Lewis dot structure of $\ce{Cl-}$, $\ce{Cl-}$ has octet fulfilled and should be able to donate a pair of electrons as it is a Lewis base. The Lewis dot structure of $\ce{Ag+}$ seems weird. As it is a Lewis acid, I should assume that it can accept a pair of electrons.
However, now the problem arises. If $\ce{Cl-}$ donates a pair of electrons to $\ce{Ag+}$, there should be a coordinate covalent bond between them. In reality, $\ce{AgCl}$ is an ionic compound, and there is no such coordinate covalent bond, so I'm a bit confused.
In simpler words, $\ce{Ag+}$ is a lewis acid.$\ce{Cl-}$ is a lewis base. Can $\ce{Cl-}$ donate an electron pair to $\ce{Ag+}$?
Could someone please point me in the right direction?

Cl- ion is capable of donating electron pairs. See the question (...)

(...), and then see the answer.

If $\mathrm{Cl^-}$ donates a pair of electrons to $\mathrm{Ag^+}$, there should be a coordinate covalent bond between them. In reality, $\mathrm{AgCl}$ is an ionic compound, and there is no such coordinate covalent bond, so I'm a bit confused.

Yes, there is no dative bond between them. $\mathrm{AgCl}$ is an ionic compound with partially covalent character. So, the bond between $\mathrm{Ag^+}$ and $\mathrm{Cl^-}$ will be a simple ionic bond.

In simpler words, $\mathrm{Ag^+}$ is a lewis acid. $\mathrm{Cl^-}$ is a lewis base. Can $\mathrm{Cl^-}$ donate an electron pair to $\mathrm{Ag^+}$ ?

Yes, $\mathrm{Ag^+}$ can act as Lewis acid and $\mathrm{Cl^-}$ can act as Lewis base. But actually $\mathrm{Cl^-}$ can't donate an electron pair to $\mathrm{Ag^+}$. This is because both are weak Lewis acids and bases.

If $\mathrm{Ag^+}$ can't accept an electron pair, how can it be a Lewis acid?

It may not accept an electron pair in this case. It doesn't mean that it isn't a Lewis acid. Lewis acid doens't mean that it can accept an electron pair from any type of ligands. For example, $\mathrm{Na^+}$ is also a Lewis acid. Eventhough it is a Lewis acid it frequently forms ionic compounds rather than forming covalent compounds.
I can give an example where $\mathrm{Ag^+}$ can act as a Lewis acid. $\mathrm{AgCl}$ is insoluble in water but it is soluble in ammonia. This is because $\mathrm{AgCl}$ forms complex with $\mathrm{NH_3}$ which accounts for its solubility. During formation of complex $\mathrm{NH_3}$ will donate an electron pair to $\mathrm{Ag^+}$. So here it acts as a Lewis acid. I can give its equation -
$$ \ce{ Ag+ + 2NH3 -> [Ag(NH3)2]+}$$

According to the Wikipedia definition, a Lewis acid is one that can accept an electron pair. If Ag+ can't accept an electron pair, how can it be a Lewis acid?

Sorry, it is not actually a question posted here. It just for your clarification.

Okay I will include answers to this comments in my post.

@Infinite is it always true that weak lewis bases don't form dative bonds with lewis acids?

@Arpan no, it is not true. If cation has high charge density for example Cr(III) ion can act as LA in presence of Cl- ion.

@Mithoron I wonna know where I was incorrect ?

Easier to tell where you were correct, just about last paragraph. Cl does coordinate Ag all the time.

Correct? You said it was thoroughly incorrect, right?

Yeah, Cl will coordinate to Ag all the time. If it doesn't coordinate, then formation of the compound would be senseless.

@Mithoron Is there a coordinate covalent bond between Ag+ and Cl- in AgCl?

Oh, why you even try to answer such a lame question. AgCl is more covalent then ionic. Coordinate bonds are no different then "non-coordinate". Cl− can donate an electron pair to Ag+. And they aren't "weak" Lewis acid and base. There's no single scale of strength of them anyways.

Chlorine $\ce{(Cl)}$ needs to gain one electron in order to complete its octet (forming the $\ce{Cl-}$ ion), while silver $\ce{(Ag)}$ needs to lose one electron (forming the $\ce{Ag+}$ ion). This is why it is an ionic reaction. Unlike in a covalent bond, the electrons aren't shared; they are donated, forming dissociated ions in solution:
$$\ce{Ag^+(aq) + Cl^-(aq) <=>> AgCl (v)}$$
Note that $\ce{AgCl}$ generally forms as a precipitate. The reaction is technically reversible, but only proceeds to a very limited extent in solution.
Indeed, as you noted, $\ce{Ag+}$ is a Lewis acid, and $\ce{Cl-}$ is a Lewis base. Therefore, a Lewis acid-base adduct can form between the two. In fact, this is precisely why $\ce{AgCl}$ precipitates in solution! The $\ce{Ag+}$ and $\ce{Cl-}$ ions hold tightly together, forming a bond that has an extremely high covalent character. The bond between them is very difficult to break apart, which is why $\ce{AgCl}$ forms as a precipitate rather than being solubilized in water.

According to Wikipedia, "A Lewis acid (named for the American physical chemist Gilbert N. Lewis) is a chemical species that contains an empty orbital which is capable of accepting an electron pair from a Lewis base to form a Lewis adduct." If Ag+ can't accept an electron pair, how is it a Lewis acid?

Thank you for responding, sir! However, I'm genuinely confused as to why Ag+ is a Lewis acid if it can't accept an electron pair.


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Can general purpose compressors be used for a specific gas without modification?
Air compressors are commonly used for all sorts of purposes related to compressing air above 1 atm. My question is, could an air compressor be used for a specific gas without modification?
For example, could I use an air compressor to compress Hydrogen, Oxygen, or Nitrogen, abet at a lower pressure?

Specific gas properties have to be considered. E.g. compressed oxygen cannot come into contact with organic substances like oil or grease to avoid explosion.

And hydrogen can leak through fine openings, past compressor seals, and is very easy to ignite over a wide concentration ratio. Using it for H2 compression involves the Hindenburg certainty principal.

Having seen the result of using a non-oxygen-rated fitting in an oxygen line (potentially mislabeled by the supplier!), extreme caution and care is needed. Do not use any equipment outside of its instructions and labeling. Pressurized hydrogen has potential excitement written over it as well.

Yes, without further specifics. Assuming you mean a modest size reciprocal type. As commented, there are special consideration for primarily hydrogen or oxygen, and of course if there are specific corrodents. Large recips typically have cage/disc/flutter valves. I have seen many valves because they are a problem. I remember only one type- hard 410 stainless , so they are not altered for various gases ( refineries, chem plants and pipelines pump a variety of stuff). Get information for anything extreme. One story; Motion compensator for off shore drilling; They use 2000 psi air. They operate 24 hours/day. Some lube oil gets in the piping. Rarely it can cause a detonation of the air piping ( as a detonation ,it ruptures the pipe in several places  and throws shrapnel around).

So you are saying "yes, but really no"?

Tentative "yes", depending on specifics.

Can't help, but notice you should be more careful with punctuation and capitalisation. You should edit posts to fix such errors, otherwise it's like you scribbled it in a hurry.


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Validating Resultant Structures of Computational Quantum Chemistry Optimisations
I have tried finding sources (publications, websites, blogs) that outline the logic of accepting the resultant structure of optimisation. I understand using a Hessian to determine if the structure is a local minimum. I am more interested in understanding what steps I should take to make sure that my functional and basis set are accurately determining the bond lengths and angles. I would need to know for benchmark testing so I can choose a functional and basis set that balances accuracy and cost.
Are there any resources that explain this type of process in-depth?
EDIT:
I found these answers on this forum from LordStryker and Martin that both give pretty thorough explanations for how to validate structures and choose a good functional for a project
Dipole moment of cis-2-butene
DFT Functional Selection Criteria

Do you have questions that go beyond the ones you have already found? If so, please be more specific. Currently it's hard to add to that,


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How are membranes for Polymer electrolyte membrane electrolysis produced?
Polymer electrolyte membrane electrolysis is a process used for creating hydrogen with a thin metal membrane. This membrane is often includes a thin metal sheet of Platinum. My question is, how these ultra-thin membranes are produced?

Please be more clear. Are you wondering how thin platinum sheets are fabricated, specifically? Otherwise this appears to be a rather broad question. The technology undoubtedly has evolved and production methods might vary. References in the Wikipedia article you linked to provide some information on the subject, for instance https://www.sciencedirect.com/science/article/pii/S037877531630012X?via%3Dihub

If it's Pt thin films you are interested in, there is this for instance: https://www.sciencedirect.com/science/article/pii/S0360319920312532


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Monochlorination of (R)-2-chlorohexane: Number of possible chiral dichloro products

How many possible chiral dichloro products can be formed when (R)-2-chlorohexane reacted with one equivelent of $\ce{Cl2}$ in the presence of UV radiation (monochlorination only)?
$$\ce{C6H11Cl + Cl2 ->[$h\nu$] C6H10Cl2 + HCl}$$

I am getting 8 as the answer but given answer is 9:

(2R)-1,2-dichlorohexane
(2R,3R)-2,3-dichlorohexane
(2R,3S)-2,3-dichlorohexane
(2R,4R)-2,4-dichlorohexane
(2R,4S)-2,4-dichlorohexane
(2R,5R)-2,5-dichlorohexane
(2R,5S)-2,5-dichlorohexane
(5R)-1,5-dichlorohexane

Am I missing something? Will the stereochemistry of the already chlorinated C change during the reaction?

Don't put your whole question in the title.

There are nine total dichloro products. Seven are chiral. the eighth does not have a chirality center. The ninth is a meso compound.

Related: https://chemistry.stackexchange.com/questions/93766/how-many-products-are-possible-when-chlorination-of-cyclohexane-takes-place ... https://chemistry.stackexchange.com/questions/153943/possibility-of-further-chlorination-of-hexachlorocyclohexane


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Why don't secondary alcohols react with HBr/HI via SN2?
In polar protic solvents, $\ce{Br-}$ and $\ce{I-}$ are considered to be very good nucleophiles. Now, here the solvent is alcohol which is polar protic. So, why don't secondary alcohols react with $\ce{HBr/HI}$ via $\mathrm{S_N}$2 and go by $\mathrm{S_N}$1 instead?
Notably, primary alcohols do undergo $\mathrm{S_N}$2, and secondary carbons can definitely undergo $\mathrm{S_N}$2 as well, so something must be special about water as a leaving group that makes us say that secondary alcohols primarily undergo $\mathrm{S_N}$1.

The hydroxyl group is not a good leaving group. Thus, the reaction is not simple substitution reaction. It is acid catalyzed and go through a cabocation intermediate based on the alcohol syructure (e.g., a primary alcohol does not go through carbocation intermediate).

@MathewMahindaratne Yes, but both SN1 and SN2 can occur after protonation of alcohol( as we know because SN2 seen in primary and SN1 in tertiary)...now, in secondary alcohol SN2 should occur because Br- and I- are strong nucleophiles?

You said it is in alcohol medium so alcohol is also a nucleophile. Weak but has plenty. So product is a mixture. Nonrtheless, the secondery alcohol reaction is whether SN1 or SN2 depends also on solvent. If it is polar protic SN1 is predominate due to stabilization of carbocation intermediate.  If it is polar aprotic, SN2 predominate.

@MathewMahindaratne But aren't Br- and I- strong nucleophiles in polar protic? So does that mean all nucleophiles(no matter how strong) will show SN1 in polar protic?


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Which quantum numbers does the orbital angular momentum depend on?
I learned from a question that I recently solved that the orbital angular momentum depends on both the azimuthal and magnetic quantum numbers.
I did not expect this because the formula for the orbital angular momentum is $ \sqrt{l(l+1)} \frac {h}{2π} $ and so I expected it to only depend on the azimuthal quantum number.
So, why does orbital angular momentum depend on the magnetic quantum number? A detailed explanation will be appreciated.

What keeps you from searching for what the magnetic quantum number is about? // The site expects that you write explicit compact summary of your prior effort to answer the question, based on your knowledge and on searching for existing related  info or answers. It would prevent others to tell you what you already know or what you could easily find yourself.

@Poutnik I can't find a detailed and understandable explanation online.

Try harder and describe what you have tried.

@Poutnik I tried "why does the orbital angular momentum depend on the magnetic quantum number" on google. That's the most I can do.

You can do much more than you say you can. See Wikipedia page for magnetic quantum number. You have nice picture there illustrating relation of orbital angular momentum and magnetic quantum number.

Note that relation of magnetic quantum number to orbital angular momentum is  the same as relation of spin quantum number to spin angular momentum.

Orbital angular momentum, like any other form of angular momentum, is a vector quantity.* Thus, it has both a magnitude as well as a direction.
The magnitude of the orbital angular momentum is given by the formula you wrote, and is thus only dependent on the quantum number $l$:
$$|L| = \hbar\sqrt{l(l+1)}.$$
However, the orientation of the angular momentum vector is partly described by the quantum number $m$. In particular, the $z$-component of the orbital angular momentum is given by
$$L_z = m\hbar.$$
It's a quirk of quantum mechanics, however, that you can't simultaneously know all three components $(L_x, L_y, L_z)$, so this is the best we can really do. Ultimately, this is because the operators $L_x, L_y, L_z$ don't commute with one another.
So, given that we know the magnitude $|L|$ as well as the $z$-component $L_z$, we can say that the orbital angular momentum lies somewhere along a cone, as the following picture from Atkins' Molecular Quantum Mechanics (4th ed.) shows. Atkins uses the general symbol $I$ to refer to some kind of angular momentum; the fact that it's orbital angular momentum doesn't change anything. Indeed, the Wikipedia page on magnetic quantum number has a very similar picture, as pointed out in the comments.

* Pseudovector, for the picky.

Pseudovector, for the picky. I was always humbled by my naive chemist approach to vectors in Usenet sci.physics and sci.physics.relativity in the past. Pseudovectors, covariant vectors, contravariant vectors, "not at all" vectors.... Oh my, the good old classic of old time volumetry. :-)

@Poutnik To be honest, I don't profess to understand it anywhere near as well as the physicists do ;) They talk about dual spaces and Hodge stars and something like that. To me, it's just called "reflect it in a mirror and it changes"!


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Why doesn't molecularity ever exceed 3?
Recently, I learnt that molecularity of a reaction is limited to 3 and even that is quite rare. I searched online and asked my teacher about this but the explanation given was that it is difficult for more than 3 molecules to collide simultaneously.
However, I find this explanation to be superficial. I have no deep knowledge myself but I can think of a superficial counter-argument to this explanation.
Consider the following diagram:-

Here, we can observe that 8 other molecules are in contact/colliding with the inner molecules. So, going by this superifical model, a molecularity of 8 should be possible.
There, are obviously several flaws to this elementary model but it still proves the need for a better explanation than the trivial one that I found. What is it?

@Poutnik so is the reason not linked to geometry? Is it linked to probability?

Reaction molecularity has nothing to do with how many molecules can be simultaneously in contact with central one, geometry-wise. Typical 3 molecular case $\ce{3 ^4He -> ^{12}C}$ is $\ce{2 ^4He <=> ^{8}Be^{} <=>[+ ^4He]^{12}C^{} ->^{12}C + \gamma}$

Usually, only 1 key can be inserted into the lock at the same time, regardless of how many keys can simultaneously touch it. Reactions are more about properly oriented collisions than touches. even trimolecular one are usually hidden forming of intermediate, reacting with yet another one.

@Poutnik then why is a molecularity of 3 possible considering the key argument?

Some locks may be made for 2 keys. Asking after sufficient prior thinking improves thinking. Some molecules have 2 reaction centers, some  trimolecular reactions are de facto 2 fast subsequent bimolecular reactions with short-time intermediate.

"Key" can be made of two parts.

@Mithoron Would not it make the 2 parts 2 keys ? But I would not dwell too much in the key analogy. (In my experience, some people have problems with analogies not being differently pictured identities.)

Higher molecularity is possible, fourth and fifth order reactions are known. See this answer and its comments.

@NicolauSakerNeto Order versus molecularity. Complex reactions can have high order, consisting of several bimolecular reactions.

To the picture - no molecule is spherically symmetric and no molecule is eager to simultaneously react toward all directions reactants come from. Neither atoms are.

This figure is in a textbook showing unit cell types, namely, this is body-centered cubic. I.e. it is nothing to do with molecules or chemical reactions, you have just inserted a random cat photo as an "argument".

Imagine vigorously shaking a box of tiny marbles. Pairs will collide often enough, but how often will three collide simultaneously? Or four? Add in the fact others have mentioned that chemical reactions require certain orientations of the colliding molecules, and certain speeds/energies, and hopefully you can see how rare a tetramolecular or even termolecular reaction is. A termolecular reaction requires simultaneous collision of three molecules, of the right relative speeds, at the right angles to each other, in the right spatial orientations.

Often, there is no right speed, energy, orientation nor place for the 3rd molecule to react at the same time. OTOH, for big molecules, especially polymers like proteins or polysaccharides, multimolecular reactions are possible, as there is multiple potential reaction centers.

@Poutnik That's a good point, I didn't consider macromolecules. I think perhaps one could regard that sort of thing as multiple reactions occuring in different places on the large molecule, rather than a single reaction involving several molecules.

Yes, I suppose that is the case, too. Truly trimolecular reaction is hard to find. I guess this one counts  H + H + M -> H2 + M ,as H + H -> H2* -> H + H.

Energy-wise, truly bimolecular reactions of atoms or small molecules, producing a single product are rare. As they have to deal with released energy not to break the product back. It cannot be translational energy due momentum conservation. It cannot usually(unless distributed to other bonds) be vibration energy as it would break the formed bonds. It cannot be rotational energy due angular momentum conservation. It may be electron excitation energy in some cases. But the 3rd reagent is an inert reagent, with which the kinetic energy, linear and angular momentum are shared.

Usually reactions having 3rd order behaviour on further examination show that they form an intermediate complex that then reacts to form products. The chance of three species simultaneously coming into contact in less than a nanosecond being vanishingly small.
Experiments by Nobel prize winner Porter (Proc. Roy Soc. v261, p29,1961) on iodine atom recombination in the vapour phase show that while $\displaystyle \frac{d[I]}{dt}=-k[I]^2[M]$ is observed, M being an inert gas or iodine vapour  ($I_2$) itself, the mechanism is
$$\displaystyle I+M\rightleftharpoons IM, \quad I+IM \to I_2+M$$
The observed activation energy is negative which is not possible for a single step reaction.


text:
Why doesn't an acid-base reaction occur in basic amide hydrolysis instead of nucleophilic acyl substitution?
In the reaction of Grignard reagent with a primary and secondary amide, nucleophilic acyl substiution doesn't happen. The reason given is that R- is a strong base as well as strong nucleophile, an acidic H is present in amide and finally, acid-base reactions are much faster than nucleophilic reactions. However, in basic amide hydrolysis, nucleophilic acyl substitution is able to occur despite OH- also being a strong base and nucleophile. Why doesn't the acid-base reaction occur here?

OH- is nowhere near as strong a base as a Grignard. Look up the pKa of a typical amide proton.

@Waylander AFAIK, amides are weak acids as well as weak bases. OH- is a(relatively) strong base and so the amide should react like an acid. Even, if the reaction is slower than that with Grignard, shouldn't it still be much faster than SN2th?

The pKa of a typical amide is around 20, OH- is not stronger enough to deprotonate it


text:
Why are amides weaker acids than water?
Based on a a recent question
I asked on ChemSE, the answer to my question was that amides are weaker acids than water with a typical pKa value of ~20. Although, this answers my question, it got me wondering why amides are weaker acids than water. After deprotonation, the negative charge formed on the amide is delocalized by resonance and on two electronegative atoms (N & O). After deprotonation in water, the negative charge is localized on a single electronegative atom (O). So why are amides weaker acids than water? Is it linked to the fact that, on average , the negative charge is on a more electronegative atom in OH- than it is in the conjugate base of amide? Isn't resonance delocalization a bigger factor?

Consider that even if amides are very weak acids, they are much stronger acids than ammonia.

How is ammonia relevant to amides? An amide is in one formal way just ammonia with 1 H substituted by acyl.  $\ce{H-NH2}$ versus $\ce{H-NH-CO-R}$


text:
Quantity vs quality in determining acidic strength
Consider the following order of acidity based on pKa values:-
ethanoic acid > phthalimide >phenol >water >acetamide
Looking towards similar questions on stack exchange such as this
, the recurring theme is that both quality(the electronegativities of the atoms on which charge is delocalized)as well as quantity of resonance(the number of resonating structures) should be seen when determining the stability of conjugate base and quality is given precedence over quantity. However, this doesn't seem to work here. Let me categorize the conjugate bases of the aforementioned compounds:-

Ethanoic Acid- Delocalisation over 2 oxygen atoms
Phtalimide- Delocalisation over 2 oxygen atoms and 1 nitrogen atom
Phenol-Delocalisation over 1 oxygen atom and 3 carbon atoms
Water-Localised over 1 oxygen atom
Acetamide- Delocalisation over 1 oxygen atom and 1 nitrogen atom

Now, based on the linked question, we get ethanoic acid > phenol > water where we have seen quality of resonance first then quantity. However, this doesn't work for phtalimide which should be placed above carboxylic acid and also acetamide which should be placed above water and possibly even phenol.
So how would you explain this order?

I think you have this whole thing backwards. From the experimental acidity, we give these rule-of-thumb explanations, which fit to varying degrees. You shouldn't expect the other way around to work in general. New chemistry students get too caught up in these rules and overestimate their explanatory power and precision.

@NicolauSakerNeto Agreed, but there must be some explanation given for experimental results, right? Is the explanation in this case too advanced(like based on the schridinger equation) or can it be explained by basic principles?

Indeed. The accurate computation of substance acidities purely from first principles, especially outside of the gas phase, is still a matter of active research.

https://chemistry.stackexchange.com/questions/94870/why-is-ethanoic-acid-more-acidic-than-phthalimide?rq=1

(...) quality is given precedence over quantity. However, this doesn't seem to work - doesn't it? See you're comparing apples and oranges here. N acid is not O acid. If imidic acids were major tautomers and not amides then your logic could been OK.

BTW this is pretty much your earlier q. all over again https://chemistry.stackexchange.com/questions/165706/why-are-amides-weaker-acids-than-water

@Mithoron so is the atom to which H is directly connected to of greater importance than all the atoms upon which the charge resonates?

More precisely it depends which mesomeric structure structure is the major contributor. You were assuming there were equal, while apparently the one with charged nitrogen is more important.

@Mithoron Why is the one with charged nitrogen more important when the charge is more stable on oxygen?

I assume it the matter of stronger C=O then C=N


text:
Confusion regarding order of reaction
The following points are given in my textbook about order and molecularity:-
(i) Order of a reaction is an experimental quantity. It can be zero and even a fraction but molecularity cannot be zero or a non-integer. Its values are limited to 1,2 or 3.
(ii) Order is applicable to elementary as well as complex reactions whereas molecularity is applicable only for elementary reactions.For a complex reaction, molecularity has no meaning.
(iii) Order is given by the slowest step(also known as RDS) and molecularity of the slowest step is same as the order of the overall reaction.
(iv) Molecularity is equal to the sum of stoichiometric coefficients of reactants in the elementary reaction.
However, I have observed an inconsistency in the above  points and would like a clarification.
Let us consider 2 cases:-
Case 1: Complex Reaction
By (iii)-order of overall reaction=molecularity of RDS. Since molecularity is 1,2 or 3, order must also be 1, 2 or 3.
Case 2:Elementary Reaction
By (iii)-order of overall reaction=molecularity of the reaction(since there is only 1 step so that must be the RDS). Since molecularity is 1,2 or 3, order must also be 1,2 or 3.
Now, combining these two cases, order =1,2 or 3. However, (i) mentions that the order can be fractional or zero. How is this possible?

Authors of the rules have not thought through all possible cases. E.g. nonlinear reaction schemes with a reaction feedback, regenerating reactants of some prior reaction steps.

@Poutnik Can you elaborate on that in an answer?

No resources for that now, but see e.g. https://www.colby.edu/chemistry/PChem/notes/ChainMech.pdf

The combustion of a candle is of order zero, in a first approximation. The amount of paraffin burned in a second is practically the same from the beginning to the end of the combustion.

Note that many processes have their pseudo-order lowered, if one component content is about  invariant within the process. E.g. A + B -> C with  [A] >> [B] is reaction of pseudo-order 1, with [A] implicitly involved in reaction rate constant. This applies to the above candle example too.

Kinetics of most reaction have little to do molecularity. BTW there's many issues with this post, see for example: Are you confused by confusing question titles including the word "Confusion"?, not to mention punctuation

@Mithoron punctuation edited. Could you suggest a better title? I'm unable to frame one without it being too long.

Well, earlier one was better. Your post is kinda "what is this kinetics all about", though.

@Mithoron How, I specifically asked what my question was in my post? How can it be improved?

Some reactions have no order. For example, the reaction $\ce{H2 + Br2 -> 2 HBr}$ has a reaction rate given by the formula : $$\ce{\frac{d[HBr]}{dt} = \frac{k[[H2][Br2]^{1/2}}{1 + k'[HBr]/[Br2]}}$$ This reaction has no order ! Ref. Arthur A. Frost, Ralph G. Pearson, Kinetics and Mechanism, John Wiley & Sons. New York, 1961, p. 236.

Here is some clarification.

You are correct that order of reaction is an experimental quantity.
Just like order of reaction is is an experimental quantity, molecularity is purely theoretical quantity and as of today, I have not found anyone who can observe it. And, it only work for elementary (1-step) reaction.
Order of reactions do only measurable for RDS. But just by stating that there exist an RDS, you are also stating that the reaction is not elementary, thus the idea of molecularity simply does not apply for a reaction with more than one step.
Molecularity have to be positive whole number the same reason you cannot have less than one atom, it is simply not make sense.

I hope this clear things up.


text:
What does the word "dynamics" mean in the context of proteins or biomolecules in general?
What does it mean by “dynamics” when we say “dynamics of protein” or “dynamics of biomolecules”? For instance, McCammon's paper has the title "Protein Dynamics" [1]. What does the word “dynamics” mean in this case?
Please, supply me with at least one reference.
Reference

McCammon, J. A. Protein Dynamics. Rep. Prog. Phys. 1984, 47 (1), 1–46. DOI: 10.1088/0034-4885/47/1/001.

It just means movements

And why not read the paper, or at least Wikipedia https://en.wikipedia.org/wiki/Protein_dynamics

@Mithoron, en.wikipedia.org/wiki/Protein_dynamics --- this is a total mess and utterly useless.

From Lottspeich' Bioanalytics, subsection Determination of Protein Dynamics [1, p. 463] (strong emphasis mine):

Determination of Protein Dynamics
Proteins are not rigid, static entities, but rather exist as ensembles of conformations.
The internal motions, which give rise to transitions between different structural states, are collectively referred to as protein dynamics and occur on a wide
range of time scales (ns–s).
Protein dynamics play crucial roles in protein functions particularly for the interaction with binding partners, in enzyme catalysis, and in allosteric regulation.
Luckily, many NMR spectroscopic parameters depend both on the mobility of the whole molecule (translational or rotational motions) and on internal motions (transitions between different conformations or bond rotations), providing insight into a wide variety of dynamic processes ranging from fast ﬂuctuations (lasting picoseconds) to slower conformational changes (lasting microseconds or more).

Protein dynamics studies have been used in biochemistry for studying genomics [2], enzymology [3] and many other areas, including GPCRs activity research which lead to the 2012 Nobel Prize in chemistry.
References

Bioanalytics: Analytical Methods and Concepts in Biochemistry and Molecular Biology, 1st ed.; Lottspeich, F., Engels, J. W., Eds.; Wiley: Weinheim, 2018. ISBN 978-3-527-33919-8.
Misteli, T. Protein Dynamics: Implications for Nuclear Architecture and Gene Expression. Science 2001, 291 (5505), 843–847. DOI: 10.1126/science.291.5505.843.
Klinman, J. P.; Kohen, A. Hydrogen Tunneling Links Protein Dynamics to Enzyme Catalysis. Annu Rev Biochem 2013, 82, 471–496. DOI: 10.1146/annurev-biochem-051710-133623.