text:
How does the same electron orbit both atoms in covalent bond?
During elementary school chemistry, when they teach about the covalent bond, they say that both atoms need the electron but they can't lose or gain them so they share it.
And for instance they show us the Lewis dot structure to clarify how do they share an electron.
If we move further into the question about how electrons actually get shared among two atoms, then we learn about orbitals, overlapping, sigma bond, pi bond etc.
But the actual question remains unsatisfied...
They say that the orbitals of two atoms overlap each other and that the electron will move there, but how?
How can the same electron that is shared in a covalent bond orbit nucleons of both atoms simultaneously? How is it actually shared between atoms? And what will be the path of its trajectory around two atoms?

The electron does not orbit both nuclei, it envelops them.

Even If electrons do "orbit" anything in there (like in Bohmian mechanics), such orbits are literally unknowable, so you're wasting your time. Just learn about orbitals.

@Mithoron Quite an unlucky term, this orbital, isn't it?

@Poutnik Kinda yeah, I guess.

A good analogue to the puzzle of how an electron in a molecule (say in the $\ce{H2+}$ cation) can be "shared" between two nuclei is a circumbinary planet. This gravitational equivalent involves a planet orbiting a binary star system. The planet moves along a trajectory about the center of mass of the two stars. Kinetic energy keeps the various bodies from crashing into each other, while centripetal force due to the mutual gravitational attraction keeps them in orbit. It should be possible to construct an even closer analogue involving electrostatic charges, a set of charged spheres, say, where two are positively charged and the third negatively.
While a classical analogue misses important details regarding the delocalized distribution of an electron in a molecule, something that requires quantum mechanics, it explains how it is possible for one particle to move in the basin of two attractors. One important difference with the gravitational example is that in that case all of the bodies attract, whereas in the molecule the electron functions as a sort of "glue" that keeps the nuclei, which otherwise repel, from flying away from each other.

How can the same electron that is shared in a covalent bond orbit nucleons of both atoms simultaneously?

The above suggests how a body can be in orbit about two others. It is important to add that a covalent bond is a fiction created to simplify the representation of molecular structures in  a form that is understandable to humans (and presumably a form that can be conveyed visually). As a concept it implies that the electron is "shared", yes, in the sense that the electron is trapped in a stable arrangement due to mutual attraction to the two nuclei, and that due to the attraction the electron can be said to be found with a high probability in the region between the nuclei. That high density between the nuclei is what we like to call a covalent bond.

How is it actually shared between atoms? And what will be the path of its trajectory around two atoms?

The electron doesn't really belong to either atom, in the same way that Earth does not belong to the Sun (well, it is trapped by the gravitational pull of the Sun within the solar system). The electron is doing what electrons do, which is "spread itself out" about the atoms due to wavelike properties while remaining bound to the nuclei due to electrostatic attraction. It helps to think of an electron as a fuzzy (probability) cloud rather than a concrete particle.

The circumbinary planet analogy is not really accurate fo two reasons. First, in the planetary case the orbiter would have to be far away compared with the dustance between the parent bodies to make the orbit stable. This works with gravity but does not do so well for chemical bonding. Second, as noted elsewhere we don't really have orbital motion. Enveloping is more accurate.

@OscarLanzi I agree that there is no ideal classical analogue to chemical bonding, but a circumbinary system with a planet in stable orbit about two as opposed to a single star provides a starting point to grasp how attractive forces can lead to this shared orbital distribution. Note the analogy works in the classical sense that F=ma, such that the planet (or electron) moves much more quickly than the more massive body or particle.

There are at least two basic reasons it is a flawed analogue and I attempt to address these later in my answer. Maybe I downplay the importance of the differences but I am trying to convince the OP that a shared orbit is not so mysterious.

An electron doesn't have a trajectory, a path. The electron is not a small planet turning around the nucleus-sun. The electron has no speed, no position. If it had a velocity anywhere, it would progressively loose energy by synchrotron radiation, and it will slowly crash down on the nucleus. The electron has no speed, and this does not mean that its speed is zero. No ! It does not have any speed at all, as it does not have any religion. The electron is only described by a wave function. And this wave function can cover or envelop one atom of course, but it may also cover or envelop two or more atoms in covalent bonds.

Can't help but wonder how you'd explain relativistic effects with your "no speed".  Not to mention particle/wave duality.

I can't visualise/imagine what does it acutally mean by enveloping atoms

@Yug Ahuja. An electron has no dimension, but it looks as is it is about as big as the atom. Its probability of presence goes to zero when the distance to the nucleus tends to infinite. If it had only two dimensions, it would be like a thick three-dimensional tissue becoming infinitely thin at large distances from the nucleus. It is not possible to represent it at three dimensions.

@Mithoron I have objected to this Maurice's view too on a related post. Considering Dirac himself was evaluating electron speed is (at least for light atoms) much slower than c and therefore SR (speed-wise) corrections to a wave equation is negligible. Hard to imagine Dirac is not qualified for statements in quantum mechanics.

@Poutnik. Dirac is right. And Pekka Pykkö also. They are able to calculate, but they are unable to explain the relativistic correction to electrons in atoms having no speed. .

@Mithoron. I am afraid I cannot explain relativistic effects with the model of no speed. But if you read Pekka Pykkö's article about relativistic effect in chemistry, you will soon read that they do not have a simple way of explaining that an electron has no speed, but can have a relativistic contraction when the electron is moving nearly at the velocity of the light. I agree that it is contradictory. Ref. : Pekka Pykkö, Relativity and the Periodic System of Elements, Accounts of Chemical Research, Vol. 12, No. 8, (1979), p. 276 - 281.

No speed? I've taken an electron out for a spin and they get around an atom in no time at all. ;-)

Can't you define a speed as the momentum divided by the mass, at least in absolute value terms, if not vectorially?

@Buck Thorn. The idea of defining the speed as the momentum divided by the mass is a good idea. Thank you for the idea.

I should add: the expectation value of the velocity (not the speed!) of an electron bound in an atom is zero, since it is confined to the atom and has zero average velocity in the reference frame of the atom. You might say the same of the Earth averaged over the reference frame of the solar system (I think). Otherwise it would escape.


text:
Mechanism of ring expansion of epoxide
The below transformation is from

J. Am. Chem. Soc. 1990, 112, 14, 5583–5601

What does the treatment with lithium anion of ethoxyacetylene in the presence of the lewis acid do to epoxides? How does it lead to the compound 56? Wouldn't the below be obtained after oxymercuration?

And what do you think?

Draw out the intermediate that is the product of step 1 then search "oxymercuration"

@Mithoron edited.

@Waylander but how does oxymercuration lead to the cyclisation?

Did you read the experimental section of the paper? After step 1 one has "crude bis-acetylenic alcohol."

Ignoring the OTBS and EtO groups, the product has 9 carbons, the same as the reactants (5 + 2 +2). Your product has only 6 carbons. Your enols tautomerizes to $\ce{-CO2Et}$.

BF3 facilitates the nucleophilic opening of epoxides by binding to the oxygen. Under its influence the lithium anion of ethoxyacetylene adds at the least hindered sites to both the epoxides opening them to secondary alcohols. The t-Butyldimethylsilylether is preserved.
In the second step Hg2+ adds to both ynol ether triple bonds to form a three-centred cationic mercurinium species (see here and here for the general mechanism). The mercurinium species are nucleophilically attacked by the secondary alcohols to give [5] rings. The C-Hg bonds cleave releasing Hg2+. The enol ethers hydrolyse to give the lactone carbonyl and you have your product.

source: Stack Exchange

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

sort: votes


text:
From chemfig to .mol file/SMILES
I usually draw molecules using the markup language known as LaTeX. In particular, I use the chemfig package. I much prefer this method over WYSIWYG apps because it allows me a greater degree of customization.
Yes, I know this package can generate structure from .mol and SMILES files. I'd like to know: is it possible to follow the reverse path, i.e. once the structure has been drawn with chemfig, generate the .mol file and/or the SMILES?

I'm not sure how much chemistry is in the chemfig descriptions, but if a machine can understand it, it probably can convert it. It you are working off an image alone, then some kind of pattern recognition is needed. I'm sure there's going to be a tool for that. There might already be a question on this platform about this.

Working backwards is not always easy (actually, rarely) but my guess is it might not be so hard to look at a chemfig text file and interpret the commands to generate a molecule in mol format. SMILES might be harder but a format that generates bonds based on proximity should be doable. You'll need programming experience, a library of standard bond lengths (or rather ranges), and some geometry commands to rotate atoms in space.

Maybe just the task for ML.

The answer is a definitively yes there are programs to scan papers with the aim to retrieve chemical information from the illustrations.  One of them is OSRA hosted by NIH (project landing page) with a demo page (requires JavaScript enabled); below with a a «glucose»* structure from the ChemFig's manual (French edition provided here, version 1.6e/June 30th, 2023, p. 42) submitted as .png (322 x 223 px, 8-bit/color RGB, non-interlaced)

(screen photo of NIH' OSRA)
to retrieve SMILES and .sdf about glucose (here: the program corrected the OH/HO bonding issue, but there is no stereochemistry retained).  It isn't the only player in the field (e.g. ChemOCR) and there are multiple references in the literature (below just about OSRA and a more recent review).  Think about digitization of printed journals and patents to build literature databases and machine learning as potential fields of application.
Some of the potential pitfalls are conceptual similar to the ones encountered with OCR on text: prior use, there is need for training; and low input quality requires some sort of correction. See for example the adrenaline structure by the same manual, page 43, captured as 330x200 px .png

vs an input of better quality based on Wikipedia's depiction of α-tocopherol (default .png, upper part of the property box):

Better input quality can be any combination of file type (e.g., png vs gif), is the figure a color/half-tone/black-and-white illustration, presence/absence of background, the optical resolution, font type used, etc.

* Personal note: I once was a user of ChemFig but glitches like the bonding of an hydroxyl group to the back bone via H instead of O a sketcher can warn you while constructing the formula was a reason to step away.  A second one is that ChemFig's input format, though documented, is not used widely enough (in my perception vs .cdxml, .sdf, .mol, .mol2, etc.) by other programs (other editors and viewers, databases, programs like RDKit and openbabel, etc) to ease quick and reliable and smooth exchange with colleagues.  (True, if it were used by many, more programs would support the format.)  It is one thing to write a thesis / prepare a presentation with much time to tinker on your own vs. the fast paced interaction with your in colleagues (possibly not so much aware of LaTeX, especially in organic chemistry [I know, it is a stereotype ... no pun intended]) on the bench.  But have a look on chemobabel.
So yes, I'm on the lazy side here to assemble the structure / reaction and apply one of the Journal's style sheets (e.g. ACS, RSC for one or two columns, Wiley, HCA) many editors ship with by default / a house style the group adopted.  Or substitute it by an other as after a few mouse clicks every label and bond is reset consistently.  (It can go south and corrupt alignments, but this is corrected quickly via the editor's GUI.)  The export of the illustration in a vector format (.svg, .[e]ps, .pdf) is supported for long by many editors (see Gunda's review).  The synchronization of labeling numbers below the compounds in the illustration and the LaTeX based compuscript can be addressed for instance with temporary labels in chemscheme, too.
(1) Filippov, I. V.; Nicklaus, M. C. Optical Structure Recognition Software To Recover Chemical Information: OSRA, An Open Source Solution. J. Chem. Inf. Model. 2009, 49, 740–743. doi 10.1021/ci800067r. and author's copy on pubchem.
(2) Park, J.; Rosania, G. R.; Shedden, K. A.; Nguyen, M.; Lyu, N.; Saitou, K. Automated Extraction of Chemical Structure Information from Digital Raster Images. Chem. Cent. J. 2009, 3, 4. doi 10.1186/1752-153X-3-4 and copy on pubchem.
(3) Rajan, K.; Brinkhaus, H. O.; Zielesny, A.; Steinbeck, C. A Review of Optical Chemical Structure Recognition Tools. J. Cheminform. 2020, 12, 60. doi 10.1186/s13321-020-00465-0 (open access)

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


text:
Deriving population of $t_{2g}$ orbitals in transition metal oxides
I have recently studied ligand field theory but there is one thing I do not understand.
Consider an octaedral geometry of ligands around a transition metal with a $t_{2g}$ - $e_g$ splitting of atomic $d$-orbitals due to bonding with the ligands. Suppose that the gap between the two is very large and electrons can only populate $t_{2g}$ orbitals.
My question is: how do I understand how many electrons populate the $t_{2g}$ orbitals knowing the number of electrons in the atomic d orbitals?
Originally I thought they were just the same, but apparently they are not! For example, in this paper https://journals.aps.org/prl/pdf/10.1103/PhysRevLett.107.256401 the authors consider e.g. SrMnO$_3$ and they say that in this case the $t_{2g}$ is half filled with 3 electrons. However Mn has 5 valence electrons.
Why is that?
I am a layman in chemistry, so please be patient if this is a silly question!

There is more to this question than meets the eye. First I answer with the OP's assumption that the metal ions are "low spin" — meaning because all the $t_{2g}$ orbitals are fully occupied before any of the hugher $e_g$ orbitals start to be populated. Then I will assess the accuracy of this assumption. (Hint: it's about as accurate as my typing. I had to make a lot of corrections and probably missed some.) In this discussion I assume octahedral coordination, which covers at least most oxides. I'll let the reader work out the corresponding situation with tetrahedral coordination, if that comes up.
There are three $t_{2g}$ orbitals, and in the low-spin case they would hold the first six $d$ electrons from the metal ion. Thus we render the count of $t_{2g}$ electrons based on the number of valence $d$ electrons in the metal ion:
$d^0$-$d^6$: $t_{2g}$ electrons = total $d$ electrons
$d^7$-$d^{10}$: $t_{2g}$ electrons = $6$
For instance, with $\ce{Fe^{2+}}$ we have the valence electron configuration $3d^6$, so all six of yhose $d$ electrons go into the $t_{2g}$ orbitals, filling them completely and leaving none for the $e_g$ orbitals. Thereby all electrons are paired, and there is zero spin — as low-spin as you can get.
Now how does that compare with the real world? If iron(II) were really low-spin in its oxide $\ce{FeO}$, then the oxide would be diamagnetic with no unpaired electrons, or at most only weakly paramagenetic (actual "$\ce{FeO}$" is not quite stoichiometric and contains a littke iron(III)).
Not so: Wikipedia renders a magnetic moment of $+7200×10^{−6}$ cm$^3$/mol, making it powerfully paramagnetic. Such a result is consistent instead with the "high-spin" model, where both $t_{2g}$ and $e_g$ get the first electron before any orbital gets another (thus maximizing unpaired electrons and total spin). So, at least for $3d$-valence metals, we don't get any more than the first three electrons into the $t_{2g}$ orbitals until the $e_g$ ones get a share:
$d^0$-$d^3$: $t_{2g}$ electrons = total $d$ electrons
$d^4$-$d^5$: $t_{2g}$ electrons = $3$ as these two electrons go into the $e_g$ orbitals
$d^6$-$d^8$: $t_{2g}$ electrons = total $d$ electrons minus the two that went into $e_g$
$d^9$-$d^{10}$: $t_{2g}$ electrons = $6$
We should expect this in $3d$ oxides because the pi-donor character of the oxide ligands tends to lower the $t_{2g}–e_g$ splitting.
Now we calculate that $\ce{Fe(II)}$ with six total $d$ electrons in $\ce{FeO}$ has only four electrons in the three $t_{2g}$ orbitals and two in the two $e_g$ orbitals — giving four unpaired electrons that match well with the observed paramagnetism of the oxide.
Thus still isn't the whole story. Ionic $3d$ transition metal oxides are indeed high-spin undrr the ambient pressure conditions in our everyday experience, but they may turn into low-spin under the gigapascal pressures deep inside Earth, with significant implications for our planetary interior and thus our planet's geological activity. See Sherman[1] for more details.
Reference

Sherman, D.M. (1988). "High-Spin to Low-Spin Transition of Iron(II) Oxides at High Pressures: Possible Effects on the Physics and Chemistry of the Lower Mantle". In: Ghose, S., Coey, J.M.D., Salje, E. (eds) Structural and Magnetic Phase Transitions in Minerals. Advances in Physical Geochemistry, vol 7. Springer, New York, NY. https://doi.org/10.1007/978-1-4612-3862-1_6

Thanks for your very clear answer :) With this logic I expect SrVO$3$ to be $t{2g}^3$, but apparently they say it's $t_{2g}^1$, how is that possible?

Vanadium works out to an oxidation state of $+4$. That leaves only one $d$ electron with vanadium. So you can have only $t_{2g}^1$.

This is where I get confused: 1. the population of $t_{2g}$ orbitals also depends on the oxidation number of the metal, is that correct? 2. You say "that leaves only 1 $d$ electron with vanadium": that's because you are also counting the two electrons in the $4s^2$ shell, correct?

Vanadium starts out with only five valence electroms. $+4$ means only one is left, period.

Ok, I think I got it now. Thanks again


text:
Why do vinylic carbocations have an empty sp2 orbital instead of an empty p orbital?
Assume a tetrahedral carbon with a leaving group reacts via the SN1 mechanism. This carbon is initially sp3 hybridized, and after the departure of the leaving group, the resulting tertiary carbocation is sp2 hybridized with an empty p orbital. My understanding of why this carbocation has an empty p orbital instead of an empty sp3 orbital is that the carbocation's 3 valence electrons "rehybridize" into 3 sp2 orbitals, leaving an empty p orbital behind. I believe this occurs because sp2 orbitals are more stable than sp3 orbitals since orbitals with more s character are lower in energy. However, I just learned that the solvolysis of vinyl halides yields a carbocation with an empty sp2 orbital (https://chemistry.stackexchange.com/a/14483). Based on my understanding of tertiary carbocations, this vinylic carbocation would "rehybridize" into 2 sp orbitals, nothing would happen to its filled p orbital, and it would leave behind another empty p orbital. Therefore, I've concluded that my understanding of "rehybridization" must be wrong. Can someone help me understand what's actually going on in terms of orbitals during the formation of carbocations?
Clarification: If I wasn't completely clear, my main question is why vinylic carbocations have an empty sp2 orbital while tertiary carbocations have an empty p orbital. I think my problem here is that I have a misunderstanding of "rehybridization" during the formation of a carbocation that I explain above.

https://chemistry.stackexchange.com/questions/78672/are-carbocations-necessarily-sp2-hybridized-and-trigonal-planar

Does this answer your question? Do vinyl cations adopt a classical or non-classical structure?

@Mithoron Thanks for these. However, they don't necessarily answer my primary question of why vinyl carbocations have an empty sp2 orbital while tertiary carbocations have an empty p orbital.

Answers that ron wasn't right this time. Your thinking was better, but you still didn't foresee it could be non-classical instead of linear.


text:
Radius of Ag+ and K+
According to the database of ionic radius
Database of ionic radius
The ionic radius of silver(I) cation is totally smaller than potassium(I), regardless of their coordination number.
It seems quite contradictory to me as potassium(I) possesses argon electron configuration which is in third shell. While silver(I) valent electron belong to the fourth shell
What is probably the reason for this issue?

You have forgotten to consider the electron cloud contraction along each period.

I wonder if the contraction is so powerful to acquire such decrease in radius while considering 4s and 4p electron shielding effect.

it is very powerful. Search for periodic table with displayed atomic or ionic radii. See figure 4.5.2 at https://pressbooks.online.ucf.edu/chemistryfundamentals/chapter/libreaclone-ionic-radii-and-isoelectronic-series/


text:
Reduction of benzylic alcohol and halide using H2 and Pd
I am studying drug synthesis, and I found some examples that uses $\ce{H_2}$ and Pd/C or Pd/$\ce{BaSO_4}$ as catalytic hydrogenation reagents for reduction of alcohol or halide in benzylic position. But I have never heard this situation in organic chemistry textbook, and also I have tried search about it but I couldn't found generalized explanation. Is it unusual case of catalytic hydrogenation? Below are the examples.

In the synthesis of Zolpidem, chloro group is reduced by $\ce{H_2}$ and Pd/$\ce{BaSO_4}$.

In the synthesis of Methamphetamine, hydroxyl group is reduced by $\ce{H_2}$ and Pd/C or Pd/$\ce{BaSO_4}$. Also, bromo group is reduced by $\ce{H_2}$ and Pd/$\ce{BaSO_4}$.

Hydrogenation is a standard method of removing benzyl ether protecting groups so there is ample precedent of cleaving O-Bn bonds whether ethers or alcohols. The stability of both the benzyl cation and radical facilitate this.

"Unusual" is a function of experience.

This recent article explains Zolpidem synthesis, which does not use $\ce{H2/Pd/BaSO4}$ after $\ce{PBr}$ reaction with allyl $\ce{OH}$ group as described in your methamphetamine synthesis route. There might be a mistake somewhere.


text:
Do PH3F2 and XeS2 exist (in the light of hybridisation and Valence Bond Theory)? [Doubt in JEE question]
Do $\ce{PH3F2}$ and $\ce{XeS2}$ exist, in accordance with Valence Bond Theory?
According to Drago's Rule, if the central atom of a compound is an element of the third period, and the attached atom has an electronegativity lesser than 2.5, then hybridisation does not occur, and bonding occurs solely in atomic orbitals.
In the case of a compound like $\ce{PH3F2}$, the fluorine atoms occupy the axial positions, while the hydrogen atoms occupy the equatorial positions. However, the electronegativity of fluorine is 4, hence the EN. difference with phosphorous is larger, making the axial bonds very strong, meanwhile the electronegativity of hydrogen is 2.1 (below 2.5), hence the equatorial bonds are very weak, and hybridisation shouldn't occur ie. $\ce{PH3F2}$ shouldn't exist, according to Valence Bond Theory.
Searching for answers online just led me to this Quora post, contradicting the answer given, which I found inconclusive.
And I couldn't find any explanation for $\ce{XeS2}$ not existing, at all.
This doubt stemmed from the following JEE (competitive exam) question:
Identify the number of molecules which do not exist.
$\ce{SBr6}$, $\ce{PH5}$, $\ce{XeF^-_3}$, $\ce{PH4F}$, $\ce{PH3F2}$, $\ce{OF4}$, $\ce{XeF^-_5}$, $\ce{XeO4}$, $\ce{XeS2}$
The answer given was: (7)
$\ce{SBr6}$, $\ce{PH5}$, $\ce{XeF^-_3}$, $\ce{PH4F}$, $\ce{PH3F2}$, $\ce{OF4}$, $\ce{XeS2}$
I understood the rationale behind all, except for $\ce{PH3F2}$ and $\ce{XeS2}$.
Edit: If anyone could answer this question with respect to VBT, VSEPR and hybridisation I'd gladly give you my upvote :)

A one minute Google search turns up https://pubs.acs.org/doi/pdf/10.1021/ic00322a013

I suggest you read one of the many answers here that explain why hybridisation should not be used here, or arguably anywhere. One example https://chemistry.stackexchange.com/questions/18427/why-does-f-replace-the-axial-bond-in-pcl5

For future reference: for the body of questions, answers, and comments, chemistry.se offers to use mhchem as a comfortable method to add chemical equations.

@Poutnik Considering that Chemistry is infamous for exceptions, I've gotten used to expecting the unexpected. Why don't you leave an answer explaining the rationale of your "chemist guts" (even if it doesn't incorporate theory)?

https://chemistry.stackexchange.com/questions/64613/what-is-dragos-rule-does-it-really-exist

I don't see why $\ce{PH4^+ F-}$ wouldn't exist. Iodide does https://en.wikipedia.org/wiki/Phosphonium_iodide

There's no way you get Xe compound without electron withdrawing groups. Compare electronegativity of Xe and S for starters.

@Mithoron This is perplexing. I did some research online, and found this: [https://www.quora.com/How-does-PH3F2-exist-whereas-PH4F-does-not]
I know that Quora is known for being unreliable, but the majority of the answers have the same point (PH3F2 exists, but PH4F doesn't), and are given by graduates from some of the top Indian universities, so I trust them.

@Mithoron  The OP probably means pentavalent PH4F like PF5, not the phosphonium fluoride.

@Poutnik So I highlighted its structure. With some "PH3F2", even being ionic wouldn't help because of this remaining P-F bond.

@Mithoron I do not object.

@Poutnik Not sure $\ce{PH4^+F^-}$ can make it. Fluoride might accept a proton displacing the extremely weak base phosphine. Substituted phosphonium fluorides are known.

The notion stated in the "Edit-statement" is simply not true and quite rude actually. Hybridisation is a necessary construct of Valence Bond Theory, which is an actual theory and not that handwavy stuff that is written in some textbooks. As such, hybridisation is not a concept that is frowned upon, at least when you try to really understand it. Sure, there are other ways of looking at things and you might not need the concept. Anyway, there is a time to use the concept of hybridisation and then there is a time where it simply is wrong. This has nothing to do with preference.

@Martin-マーチン Thanks for your support. People on this site don't often side with newbies like me +1.

I don't understand your comment. I'm saying that the last paragraph of your question is in my opinion wrong and rude. I'm also not supporting your question. All of the referenced molecules should not be explained with hybridisation. And what you call Valence bond theory is unfortunately just a simplified and broken down view of the real theory. I'm sorry to burst your bubble, but the jee stuff is notoriously bad at teaching the right tools to understand the world.

@Martin-マーチン For now, I just want to get into a good college and make my parents proud ;), I can delay a little on understanding the world. But very well, I'm a man of honour, deleted the edit.

In theory, all of the mentioned compounds in the answer don't exist expect for $\ce{PH3F2}$. Difluorophosphorane has been characterized through spectroscopical method long time ago. Here is an excerpt from a paper1:

$\ce{PH3F2}$, first reported in 1971 by Seel and Velleman, is now accessible in pure form and has been unambiguously characterized by high resolution infrared spectroscopy. At present it marks the borderline between the well-established fluorophosphoranes $\ce{H_{n}PF_{5−n}, n = 2−5}$, and the unknown species $\ce{PH4F}$ and $\ce{PH5}$.

I am going to also shamelessly plug the paper2 mentioned by @IanBush in the comments.
Regarding $\ce{XeS2}$, it doesn't exist. However, oxosulfides of form $\ce{XeO_{n}S_{m}}$ are known3. Also, excimer $\ce{XeS}$* is known4.
References

Gas-phase structure of difluorophosphorane determined by high resolution infrared spectroscopy, H. Beckers, H. Bürger, A. Rahner, Vol.54, 1991, DOI:  10.1016/S0022-1139(00)83877-4
Preparation and spectroscopic characterization of difluorophosphorane, PH3F2. Phosphorus-31 NMR spectrum of protonated diphosphine, P2H5+, Rolf Minkwitz and Andreas Liedtke, Inorganic Chemistry 1989 28 (23), 4238-4242
DOI: 10.1021/ic00322a013
Xenon oxides, sulfides, and oxysulfides. A theoretical ab initio investigation, Aristotle Papakondylis, Vol. 1015, 2013, DOI: 10.1016/j.comptc.2013.03.030
Xenon oxide and xenon sulfide emission systems at 234 and 227 nm, J. Xu, D.W. Setser, J.K. Ku, Vol. 132, 1986, DOI: 10.1016/0009-2614(86)80640-6

Nice to have an Indian, who understands what a fellow JEE aspirant is going through, answer my question without relentlessly bashing me. +1 and marked as answer.

As if to refute the hybridization theory once and for all, $\ce{PH3F2}$ (difluorophosphorane) exists! Andrews and Withnall[1] produced it and trapped it in an argon matrix by cocondensing phosphine and elemental fluorine into this matrix. The authors also mention that their result is the first experimental evidence for $\ce{PH2F}$, which "should be" more stable and easier to access according to the hybridization-based argument.
Reference

Lester Andrews and Robert Withnall (1989). "Cocondensation reaction between phosphine and fluorine: matrix infrared spectra of difluorophosphorane, difluorophosphine and fluorophosphine". Inorg. Chem. 28, 3, 494–499.
https://doi.org/10.1021/ic00302a023

Meh, that only proves they are actual intermediates in reaction between PH3 and F2. Many really surprising things can be trapped like that.

Still existence for me.

Says guy who was doubting phosphonium fluoride ;p

But I did not see PH4F in the reference. Ionic PH4F would decompose and PH3F2 apparently did so partially in that reference.

I'm saying it's a different league completely. In argon matrix you can have like anything, starting with single atoms of any element. You could probably even have a non-ionic PH4F.

https://pubs.acs.org/doi/pdf/10.1021/j100331a015

https://pubs.acs.org/doi/pdf/10.1021/ic00322a013

Well, maybe you were right with PH3 not being basic enough to react with HF.  Apparently HSFO3 isn't enough but HTaF6 is.


text:
Friedel-Crafts alkylation of fluorochloropropane
Consider benzene reacting with 1-chloro-3-fluoropropane, with $\ce{AlCl3}$ as the Lewis acid. My question is, is this reaction feasible? If so, what is the product formed and do cationic rearrangements occur?
I found this on a post:

Now, I understand why the original question (Friedel craft alkylation halo-carbon bond cleavage) states the $\ce{C-F}$ bond breaks. However, I'm unsure if this reasoning is completely accurate as this would imply the cationic nature of the ion-pair formed would be higher. As a result won't that cause more rearrangements? So, why don't rearrangements occur here and is this reasoning accurate?
Edit: Mathew Mahindaratne pointed out that the product will depend on reaction conditions, which I agree with. However, I would appreciate if someone explained the reason why the unrearranged product is major here.

In the aforementioned question, Soumik Das has given acceptable reasoning (at least within our community, nobody has questioned his answer for five years). However, beside the fact that the Lewis acid used in your reaction $(\ce{AlCl3})$ is much stronger than $\ce{BCl3}$ in that question, but you haven't given all reaction conditions used (such as temperature and reaction time) in the reaction. Thus, it is hard to predict the results as you expect. For example:

When 1-chloro-3-bromo-propane is reacted with benzene at $6$-$\pu{12 ^\circ C}$ in the presence of $\ce{AlCl3}$, (3-bromopropyl)benzene forms in 60% yield. But at a higher temperature, diphenylpropane is the product (Ref.1).

Just for your information, the reasoning for the higher reactivity of $\ce{C-F}$ bond over bonds of carbon and other halogens is given in Ref.1 as follows:

The substantially higher reactivity of the $\ce{C-F}$ bond over that of the $\ce{C-Cl, C-Br,}$ and $\ce{C-I}$ bonds in Lewis acid catalyzed reactions of fluorohaloalkanes is at first sight unexpected in view of the carbon-halogen bond energies, which show increasing strength from iodine to fluorine. However, the base strength of halide reagents in Friedel-Crafts systems is a consequence of a number of contributing factors involving both the donor base and acceptor acid.

The most important contributing factors are: (1) the high polarity of the $\ce{C-F}$ bond to be cleaved (electronegativity of the halogens: $\ce{F \gt Cl \gt Br \gt I}$); (2) the high bond energy of the bond to be formed by the interaction with the Lewis halide catalyst ($\ce{B-F}$  bond); and (3) an obviously small steric hindrance. These facts must contribute substantially to the high reactivity of the $\ce{C-F}$ bond. Soumik Das' answer gave some of these.
It is also interesting to note that, if $\ce{BF3}$ in place of $\ce{BCl3}$ is used as the catalyst in those reactions of straight chain fluorohaloalkanes (as in the pointed reaction), almost complete isomerization of the alkyl chain would occur (Ref.1). For example:
$$\ce{Ph-H + FCH2CH2CH2Br ->[BF3][81\%  yield] CH3CH(Ph)CH2Br + HF}$$
Isomerization would occur with other boron halide catalysts (such as $\ce{BCl3}$ and $\ce{BBr3}$) as well, but only 5-15%. That also based on conditions used:
$$\ce{Ph-H + FCH2(CH2)2CH2Br ->[BCl3 or BBr3][61\%  total yield] PhCH2(CH2)2CH2Br + CH3CH(Ph)CH2CH2Br + HF}$$
The rearranged product (4-bromo-2-phenylbutane) in above reaction is 30% of total yield while that of unrearranged product (4-bromo-1-phenylbutane) is 70% (Ref.1).
However, in current reaction, the all $\ce{C-X}$ bonds $(\ce{X }= \ce{F, Cl, Br, I})$ are reactive towards $\ce{AlCl3}$ (depending on the temperature used) so that I'd expect the products as fused cycloalkylarene (e.g., indane) if not rearranged (temperature as given in Ref.1, kinetic product) or 1,2-diphenylpropane if rearranged (at higher temperatures, thermodynamic product).
References:

George A. Olah and Stephen J. Kuhn, "Selective Friedel—Crafts Reactions. I. Boron Halide Catalyzed Haloalkylation of Benzene and Alkylbenzenes with Fluorohaloalkanes," J. Org. Chem. 1964, 29(8), 2317–2320 (DOI: https://doi.org/10.1021/jo01031a051).

I am not aware of the reaction conditions but this is the product I was told is formed by my teacher... could you provide some information as to why that is the case - like why are there no rearrangements occurring in the major product

At room temperature, I'd expect rearranged product with $\ce{AlCl3}$ (at least in higher percentage than non-rearranged product).

Yeah I was expecting that too but allegedly this is what's formed which mildly confused me


text:
Crystallography - unit cells
According to what I've read, in 2 dimensions, a unit cell is a parallelogram whose vertices are all one of the lattice points. In addition we define the conventional cell as the unit cell which demonstrate the symmetry properties of a lattice in the best way. Now my question is: why do we consider a hexagon as a conventional cell while its not a parallelogram?
Thanks for your help.

A paved surface covered by consecutive hexagons can also be considered as paved by consecutive lozenges or diamonds, where each lozenge or diamond is made of two small equilateral triangles.

The primitive cell is also a parallelogram for hexagonal systems: https://chemistry.stackexchange.com/a/162878

In theory, a unit cell can be anything at all in which all pairs of opposite edges and (in three dimensions) are parallel. Among such shapes the parallelogram in two dimensions, or the parallelepiped in three dimensions, is the simplest to drae, to interpret and to do calculations with (e.g. cell volume, number of formula units).


text:
Why Michaelis Menten Assumption of equilibrium is necessary to integrate rate of ES production
I read in my textbook that the assumptions of equilibrium and of steady state used by Michaelis and Menten were simplifying assumptions intended to make the following equation one that can be integrated:
d[ES]/dt = k1[E][S] - k_-1[ES] - k_2[ES] (1)
They go on to assume that k_-1 is much bigger than k_2 so that equilibrium is achieved and can now write:
Ks= k_-1/k_1 = ([E][S])/[ES].
According to the textbook, the first equation can now be integrated. Can someone explain why it could not be integrated prior, why it can be integrated now and why it was necessary to integrate in the first place? The textbook is Fundamentals of Biochemistry third edition by Voet and Pratt Chapter 12 p.368

You can integrate the equations, but only numerically, the nature of the equations is such that an exact algebraic solution is not easy or feasible. It turns out that mathematically many types of differential equations are very, very hard to integrate.

I should add that to integrate you also need $d[E]/dt$ and $d[S]/dt$, by substituting for $k_s$ the term in $[E][S]$ can be changed to one in $[ES]$ alone so $d[ES]/dt$ can now be integrated by standard methods. Whether this is valid in a chemical sense is another matter.

I read in my textbook that the assumptions of equilibrium and of steady state used by Michaelis and Menten were simplifying assumptions intended to make the following equation one that can be integrated: d[ES]/dt = k1[E][S] - k_-1[ES] - k_2[ES]

No, you are assuming a steady state (i.e. d[ES]/dt = 0) so you can calculate the concentration of the enzyme substrate complex. The alternative would solve the system of differential equations ("to integrate them"), but that is more complicated and arguably gives you less insight.

They go on to assume that k_-1 is much bigger than k_2 so that equilibrium is achieved and can now write: Ks= k_-1/k_1 = ([E][S])/[ES].

That is not the classical derivation. I would expect a definition of $K_M$ and an expression of the initial rate dependent on $k_\mathrm{cat}, K_M, E_\mathrm{tot}$ and [S].

(+1) Nice answer! For what it is worth, there is no particular difficulty in integrating the coupled ordinary differential equations with initial conditions and there is no need to simplify by assuming steady state or equilibrium or quasi-equilibrium, etc. There is even an analytic solution in terms of the Lambert W function. Despite all that, the standard simplifications do exactly what you say: they provide insight.

@EdV There is a historical subtlety I was not aware of. While the approximation is named after Michaelis and Menten, they did have an equilibrium approximation, and the steady state came a bit later by Haldane (according to Voet, Voet and Pratt textbook).

Ah, J.B.S. Haldane! Nice to know!


text:
Structure of AgNO2 and AgONO
We had a discussion in class today about the substitution reaction of RX with AgNO2 and KNO2. Pretty unimportant, since those seem just fine. But there were 2 arguments.
The root problem was, that we couldn't figure out what the structure of AgNO2 was.
Online, almost all images depict the structure as an Ag atom bonded with the Oxygen atom of the ONO group for instance...
The question raised was whether this is AgNO2 or whether should this be called AgONO.
In my opinion, the true structure for AgNO2 should be

except there should be Ag instead of R.
The structure in the first image, according to me should be that of AgONO.
Hence when talking about reaction with AgNO2, N would be bonded to Ag, making only oxygen available to attack RX, hence forming RONO as a product.
However, some argue that the first image does indeed depict the structure of AgNO2, hence it is in fact Oxygen that is bonded to Ag, making Nitrogen available for attack and hence forming RNO2 (second img) as a product. In fact, many resources online as say this as well, which as per my knowledge is incorrect.
Which structure for AgNO2 is correct? If it is the first one, what then would be different in AgONO if there is any difference between the two.
As a bonus question, do tell the products for both the reactions, i.e. RX with AgNO2 as well as KNO2.
I guess I do understand that Ag compound, regardless of which one it is would not entirely dissociate as is the case with AgCN group. Hence, whichever atom is bonded to Ag, the R group would be attached to the other atom that is available. that is... if Ag is attached to N, RONO would be formed whereas if Ag is attached to O, RNO2 would be formed. KNO2 confuses me a little too. Since there is a complete dissociation, we have the NO2- group. In this case, would it be the Nitrogen atom that would attach to the R group (since it is less electronegative and would thus donate electrons more easily) or would it be the Oxygen atom that would attach (since it simply has more available electrons to form bonds)? In other words, for KNO2 or KONO, whichever one it is, would the product be RONO or RNO2?

Alkyl nitrites (R-ONO) can be made by the reaction of alcohols with sodium nitrite (NaNO2) and acid (forming nitrous acid HNO2) see http://orgsyn.org/demo.aspx?prep=CV2P0108

The nitrite ion is ambident; alkyl halides react specifically with silver nitrite to form both nitroalkanes and alkyl nitrite esters. https://pubs.acs.org/doi/10.1021/ja01641a029

Does this answer your question? What is the difference between the structure of nitrito-O and nitrito-N?

Based on all the answers and comments, Here's the picture I have in my mind :
As per Saytzeff rule, no bond is completely covalent or completely ionic. In case of RNO2, the bond is predominantly covalent whereas, with K and Ag, the NO2 bond is predominantly ionic. Thus in RNO2, there is a fair possibility of NO2 acting as an ambidentate ligand, both NO2 and ONO structures are possible.
NO2- is a polyatomic ion; its charge distribution has been controversial but largely, at least in case of metals like Ag, Oxygen is what interacts during bonding.

--continued--
KNO2 completely dissociates into K+ and NO2-, making Oxygen available to make bonds with R, reacting with RX to form RONO.
AgNO2 however doesn't dissociate as much due to it being slightly more covalent. Thus, while oxygen is still loosely bonded to Ag+ due to Electrostatic attractions as part of NO2- group, Nitrogen, with its lone pair, is the next best nucleophile to interact with the R group in RX, forming RNO2.

@Maddy this seems about right.

I'm glad! Thank you, everyone, for your help.

The reference[1] mentioned in the comments shows that the reaction of silver nitrite with alkyl halides produces predominantly the nitro compound, $\ce{R–NO2}$ with over $80$% yield. This suggests that regardless of its crystal structure in the solid, silver nitrate in organic-solvent solution has the structure
$\ce{Ag–O–N=O}$
with the nitrite group covalently bonded through oxygen. This structure is also given in this answer by Nilay Ghosh. It renders the lone pair on the less electronegative nitrogen atom as the primary nucleophilic site, producing the nitro compound as the major product.
The bonding to oxygen in the molecular/monomeric structure may be explained in terms of ligand-field theory. The nitrite-ion ligand is a sigma donor primarily through the oxygen atoms, which formally share the negative charge, and a pi acceptor through the nitrogen atom where a relatively low-lying pi antibonding orbital has most of its electron density (similar to the molecular-orbital structure of ozone or sulfur dioxide). In many complexes, such as cobaltinitrite, the pi-acceptor interaction dominates and coordination through the nitrogen is favored; but in Group 11 cations like $\ce{Ag^+}$ the $d$ electrons are effectively sunk to the core and the pi interaction is poor. So the silver coordinates predominantly by sigma bonding through the better sigma donor (oxygen), neutralizing the formal charge and leaving the nitrogen as the next best nucleophile for reaction with the alkyl halide.
Reference

Nathan Kornblum, Bernard Taub, and Herbert E. Ungnade (1954). "The Reaction of Silver Nitrite with Primary Alkyl Halides". J. Am. Chem. Soc. 76, 12, 3209–3211.
https://doi.org/10.1021/ja01641a029

okayy! that was really helpful and detailed. So in a nutshell, you suggest that oxygen and nitrogen are preferable at 2 distinct levels of pi interactions. In case of Ag, oxygen is more involved in the interaction as compared to nitrogen, making nitrogen the next best nucleophile available to react with the alkyl halide. Yeah? How about KNO2? I believe in that case, the covalent character is too weak that even the oxygen is available for the reaction with alkyl halide. Right?

Yes, there is essentiallu no metal-nitrite association in KNO2 so the oxygen is the main nucleophile.

thank you! can you please read my comment on the main post and see if I have understood everything correctly?

I will attempt to answer what I think is an underlying question, "How is charge distributed in the nitrite ion?". If this question is answered, then which is the correct structure for silver nitrite between your suggestions will follow.
According to Long and Marsh the oxygen and nitrogen atoms are both about the same distance away from the silver atoms in the solid crystal structure. They refer to the work of Kay and Frazer on sodium nitrite, which has the same crystal structure, wherein they estimate the charges on the atoms of the nitrite anions based on neutron scattering data. The charge estimates for oxygen and nitrogen were -0.32 and -0.36 charge units, respectively. So my best guess is that real life lies about in the middle between your two structures.
As an example or consequence of this "just about even" charge distribution, I will consider a classic named reaction, the Victor Meyer Reaction, wherein silver nitrite is reacted with an alkyl halide (R-X) to form a nitroalkane. Note that in the nitroalkane, the nitrogen has been alkylated: R-NO2. A competing reaction is nitrite ester formation, where the oxygen of the nitrite ion is alkylated: R-ONO. In practice both of these reactions occur, e.g. Plummer and Drake, which is another indication that the charge is distributed similarly between nitrogen and oxygen.
Long and Marsh, Acta Crystallographica, 15, 448 (1962)
Kay and Frazer, Acta Crystallographica, 14, 56, (1961)
Plummer and Drake, JACS, 76, 2720 (1954)

The crystal structure of silver nitrite has been published at least three times, with different silver nitrogen bond lengths leading to different conclusions:
In the 1981 paper (Structure of silver (I) nitrite, a redetermination, S Ohba and Y Saito, Acta Crystallographica Section B), the abstract reads:

AgNO2, Mr= 153.9, orthorhombic, Imm2, a= 3.528 (1), b= 6.172 (1), c= 5.181 (1) A, V= 112.8 (1)/~ 3, Z= 2, Dx= 4.53 Mg m-3, g (Ag Ka)= 4.26 mm-k The structure of AgNO2 has been redetermined using a four-circle diffractometer with Ag Ka radiation. The final R value is 0.024 for 1128 observed reflections. The Ag-N distance is 2.304 (2) A, suggesting some covalency in the Ag-N bond. A large anisotropy in the atomic thermal vibrations was found.

So they say it is ionic with some covalency. The discussion of the Ag-N bond without the discussion of the Ag-O bond suggests that nitrogen is the closest neighbor of nitrogen.
The introduction talks a bit about the other two papers:

The crystal structure of AgNO2 was first determined by Ketelaar (1936) from rotation and powder photographs and it was reinvestigated by Long & Marsh (1962) on the basis of Weissenberg photo-graphs. Covalent radii of Ag and N atoms are 1.34 and 0.74,~, respectively, the sum being 2.08/~. The Ag-N distance derived by Ketelaar was 2.07 (30)] k, indicating complete covalency, whereas Long & Marsh reported it to be 2.47 (8) A, which suggests a primarily ionic bond. It is probable that the Ag-N bond has appreciable covalency because AgNO2 possesses the character of nitro compounds [...]

Neither showing a covalent bond only, or showing just one anion and one cation (instead of the complete crystal structure) does the compound justice. What the data shows, though, is that the nitrogen atom is pretty close to the silver.


text:
Change in internal energy at constant pressure versus heat released at constant volume
I am trying to solve a question that involves studying the heat released of a combustion process at constant pressure versus at constant volume.
The question is

The  heat of combustion of ethanol into carbon
dioxide and water is $\pu{-327 kcal}$ at constant pressure. The heat evolved (in cal) at constant volume and $\pu{27^\circ C}$ (if all gases behave ideally) is?

The solution to this problem is given using the equation
$$ \Delta H = \Delta U + RT\Delta n (\mathrm{gas})$$
where $\Delta H$ is the heat released at constant pressure, and $\Delta U$ is stated as the heat that is released for this reaction at constant volume.
What I am unable to understand is how we can place the values of enthalpy and internal energy changes for the reaction taking place under different conditions in the same equation. Does this mean that the enthalpy change that would take place under constant volume would be the same as the one at constant pressure? Doesn't placing the value of $\Delta H$ as $\pu{-327 kcal}$ give us the internal energy change for the reaction at constant pressure?

You are first trying to find the internal energy change for the reaction at constant temperature and pressure, given the enthalpy change at the same temperature and pressure.  So we are using $$\Delta H=\Delta U+\Delta (PV)$$  For a reaction of ideal gases at constant temperature and pressure, $$\Delta (PV)=(\Delta n)RT$$So, $$\Delta U=\Delta H-(\Delta nRT)$$for the reaction at constant temperature and pressure.  Of course, this applies only where the ideal gas approximation is valid.
Next, we reduce the volume of the product mixture back down to the original volume.  We know that the internal energy of a gas mixture is a function only of T, and independent of mixture volume. So there is no further change in internal energy in reducing the product mixture volume back down to the initial reactant mixture volume. So the $\Delta U$ of the reaction we calculated for constant temperature and pressure is also the $\Delta U$ of the reaction for constant temperature and volume.

Will $\Delta U$ not change as the reaction involves phase change too? For example the internal energy change of boiling water at 100 deegrees celsius is not 0

Futhermore, if we say that $U $ only depends on temperature for this reaction, then we obtain that $\Delta U$ is zero for both the reactions as it takes place under constant temperature, implying that the heat evolved at constant volume is also zero

We are talking about ideal gases, so there is no phase change involved.  $U$ depends on temperature and chemical composition (not just temperature) for a reacting system.  However, for the final product mixture, the chemical composition is fixed, and its U depends only on its temperature.  But, for the entire process, certainly the chemical composition of the reaction mixture is different from the product mixture.  So going from the reaction mixture to the product mixture at constant volume will involve a $\Delta U$.  For a process where $\Delta U=0$, temperature would have to change.

If I understand correctly, the internal energy change associated with the change in chemical compsition can be represented by the energy changes involved in each step of the reaction mechanism, i.e breaking and forming bonds. Does this energy change not depend on the pressure/volume?

Not for an ideal gas mixture.

What about ethanol which is liquid at the given T

We are assuming that the initial ethanol is at a low enough pressure and mole fraction in the reaction mixture that it is in the gas phase.

If we assume that all products and reactants are ideal gases, then there will be no interactions between the molecules, is the reaction possible under ideal gas laws?

What is the mean free path of air molecules at 1 bar?

It can be calculated by using the relation 1/(sqrt(2) * n * pi * d²),

Let us continue this discussion in chat.


text:
Why is nickel-56 not a stable nuclide?

Why is $\ce{_{28}^{56}Ni}$ not a stable nuclide?

As far as I studied radiochemistry, magical numbers of protons and neutrons correspond to the more stable elements, due to their perfectly spherical form: $2,$ $8,$ $20,$ $28,$ $50,$ $82$ and $126.$ Double magic nuclei are those with both $Z$ and $A - Z$ magical, like $\ce{_2^4He},$ $\ce{_8^{16}O},$ $\ce{_{82}^{208}Pb}.$ All of these end points of decay series and very stable elements.
However, $\ce{_{28}^{58}Ni}$ seems to be favored, alongside $\ce{_{28}^{61}Ni}$, instead of the more obvious choice $\ce{_{28}^{56}Ni}$. What is the reason?

https://en.wikipedia.org/wiki/Isotopes_of_nickel#Notable_isotopes

After calcium-40, elements start to deviate from initially ideal 1:1 nucleon ratio, as more neutrons are needed for the most stable isotopes of the element.  1:1 iron-52 is not stable either.

@Poutnik iron-52 1:1 does not have the double magic property

I am well aware of it. But there is multiple factors determining the isotope stability. In this case ratio 1:1 is already out of the valley of stability , what beats the effect of magical numbers. Tin-100 is double magic number isotope too, but the lightest stable isotopes are tin-112(observationally stable) and tin-114.

You might want to search around in the physics stack exchange. Lots of related stuff, like this: https://physics.stackexchange.com/a/460464/313612. Or just ask over there.

Not all stable nuclei with a "magic" number of protons or neutrons are "doubly magic". Perhaps they would be (actually the same number of protons and neutrons) if only internuclear forces were acting within the nuclei, but in practice electromagnetic forces become increasingly important as we move to heavier nuclei. The competition between electromagnetic and internuclear forces leads to stable isotopes generally not being the most "magical" with respect to internuclear forces alone. For example, with 50 as the magic number:

Zirconium-90, the most abundant isotope of that element, has 50 neutrons but only 40 protons.

Tin forms several relatively abundant (>10%) isotopes with 50 protons but 66, 68 or 70 neutrons.

Given this circumstance, the fact that lead-208 is doubly magic is something of a coincidence. This point is emphasized by the familiar uranium-238 decay chain ending with lead-206 which is also stable (the alpha and beta decay mechanisms commonly seen in radioactive decay of heavy elements favor the mass number remaining congruent modulo 4).
It becomes evident that "doubly magic" has only limited utility in deciding nuclear stability. Scientists who are seeking such nuclei among superheavy isotopes are (or at least should be) cautious about predicting their actual stability against radioactive decay.


text:
What is electrovalency and covalency?
I know that valence is defined in the Gold Book, and valency, electrovalency, and covalency are not defined. However, there are some terms that are confusing me. According to the given definitions, a metal shows positive electrovalency and a nonmetal shows negative electrovalency in the bonding of a metal and a nonmetal. For example, for $\ce{MgO}$, electrovalency of $\ce{Mg}$ is $+2$ and that of $\ce{O}$ is $-2$. I am not comfortable assigning a sign to valence:

Valence: The maximum number of univalent atoms (originally hydrogen or chlorine atoms) that may combine with an atom of the element under consideration, or with a fragment, or for which an atom of this element can be substituted.

According to IUPAC, valence or valency (which is not defined) should always be positive.
Further the textbook says that valency of $\ce{H}$ is always $+1$. What confuses me further is the concept of covalency, where the shared electrons result in a positive valency regardless of the electronegativity of the atoms involved.
References:
Valence, Variable Valence, and Valency

The whole concept of valence, with whatever pre- or suffix, is old and flawed and has probably changed meaning throughout the centuries a couple of times. There is a reason for why there are no official definitions: the terms are ambiguous and no longer useful in scientific communication. See them as prose that helps a paper come together for your reading pleasure. What dies it matter what hydrogen's valence is when it bond to two fluoride ions?

Electrovalency or electrovalence were two synonynous terms used in the first half of the 20th century to describe the charge of a metallic cation. Some chemists used electrovalence, and some electrovalency. Today both terms have disappeared.

@Martin-マーチン you raise a vital issue regarding the concept of valence, I wonder why valence is still defined in the Gold Book. I guess it only works as a rule of thumb.

@Maurice clearly, since it is still part of textbooks and the Gold Book, the terms and related concept has not disappeared and is used regularly by chemists to explain several ideas.

It's used by teachers to waste students' time ;>


text:
Why does the answer given consider octahedral splitting even though there is no specification of the complex
My textbook (NCERT Chemistry Part 1) gives this question as an example:

Why is $\ce{Cr^2+}$ reducing and $\ce{Mn^3+}$ oxidising when both have $d^4$ configuration?

And the solution given along with it says the following:

$\ce{Cr^2+}$ is reducing as its configuration changes from $d^4$ to $d^3$, the latter having a half-filled $t_{2\text{g}}$ level. On the other hand, the change from $\ce{Mn^3+}$ to $\ce{Mn^2+}$ results in the half-filled $d^5$ configuration which has extra stability.

Now my question is: why do they consider octahedral crystal field splitting, even though we don't know anything about the complex formed? If $\ce{Cr}$ underwent tetrahedral or square-planar splitting, it would have a different configuration which may not justify its reducing ability.


text:
Nutrients in translucent bottles/boxes
Nutrients e.g vitamin C, calcium etc are sensitive to light and that is the reason that major companies have juices and milks in boxes and opaque packages.
But there are some companies that the recent years use a transparent package/bottle (plastic or glass).
Does glass or plastic offer any kind of protection to the contents? What's the point of buying e.g. an orange juice in a transparent plastic bottle if the light is diminishing the vitamin C while in storage?

It's not like you put it direct sunlight, I hope? Otherwise I don't see a big deal.

I don't think that calcium is sensitive to light.

@DrMoishePippik: in the grocery stores, super markets, convenience stores the light is always on

@Mithoron:do you think that during transport there is precautions to avoid sunlight? I have seen from time to time freeges in convenience stores that are exposed to light

Ascorbic acid absorbs mainly below 300 nm, and when inside a building light at shorter wavelengths than 300 nm will be absorbed first by window glass and also a little by the packaging. Artificial lighting has little or no uv content, so you would be okay inside. If you keep the juice outside in sunshine it will be much more liable to under go some photochemistry as only the packaging will absorb and its much brighter. I suppose most folk will keep the juice is a cupboard or fridge anyway so it seems to be ok to use transparent packaging.

But I am pretty sure during transportation, there are no extra precautions to avoid sunlight or light. So I don't understand why companies use translucent packaging

So basically if artificial light has little uv content then having the item stored in a lit storage is ok?

Yes, provided that there is no visible light induced change to whatever else is in the bottles.

What about when consuming the content (juice, milk) in a glass outdoors where there's sunlight?


text:
Find cell configuration notation for permanganate in acidic medium
They gave us an unbalanced reaction of permanganate and iodide in acidic medium, and ask to balance the reaction using redox methods and the write the cell configuration notation for the resultant cells. The last part is where I get into trouble.
Unbalanced reaction:
$$\ce{MnO4- (ac) + H+ (ac) + I- (ac) <=> Mn^2+ (ac) + I2 (s)}$$
I balanced the two half reactions and got:
\begin{align}
2 \times&&
  \big(\ce{MnO4- (ac) + 8 H+ (ac) + 5 e-
    &<=> Mn^2+ (ac) + 4 H2O (l)}\big)&&
  \text{(red./cat.)}\\
+ 5 \times&&
  \big(\ce{2 I- (ac) &<=> I2 (s) + 2 e-}\big)&&
  \text{(ox./an.)}\\\hline
&&\ce{2MnO4- (ac)+ 16 H+ (ac) + 10 I- (ac)
  &<=> 2Mn^2+ (ac) + 8 H2O (l) + 5 I2 (s)}
\end{align}
Now, for writing the half cell notation for the cells, we are to use platinum as electrode. The question is, should I put it in both the cathode and anode? Because I heard people saying that since the anode already has a solid substance (solid iodine) it's not necessary to add platinum to the anode. I digress because I think platinum is necessary for the electrons to flow from anode to cathode.
I would write
$$\text{anode} || \text{cathode}$$
$$\ce{I- (ac) | I2 (s), Pt (s) || MnO4- (ac), Mn^2+ (ac) | H2O (l) | Pt (s)}$$
I'm not sure what the correct cell configuration notation is, would anyone clarify this for me please?

In this redox reaction, using platinum electrodes in the cathode and anode is correct, regardless of if there is solid iodine.

Chem+Math Expression formatting reference:
MathJax Basics / Chem+Math expressions/formulas/equations / Upright vs italic / Math SE Mathjax tutorial
// MathJax is preferred not to be used in CH SE Q titles.

I understand it as they ask for half reaction notation, not for cell configuration notation.

@Poutnik i meant the configuration, thats what i want to know

Review the results for search site:chem.libretexts.org electrochemical cell notation

Welcome to Chemistry! States of aggregation should not be subscripted, it is not wrong, but the recommendations (Sec. 2.1.) are different.

@Poutnik thanks, but i overall know how to write the notation, its just that i have this doubt with this particular exercise, some people have told me that is not necessary to add platinum electrode in the anode since the anode already has a solid substance (I2(s)). my question is, should i put platinum there or not?

@Martin-マーチンoh, i didnt know, thanks!

Are platinum conductors parts of the cell or not? Material of conductors in contact with redox systems is important, regardless of the fact, if it is part of the redox system or not.

@Poutnik yes, the question says "write the cell configuration considering that the reactions on anode and cathode occur with platinum electrodes". we are to use platinum as electrodes in the cell, but one of my friends who is also a student said that some professor have said that "if you already have some solid in the cell then its not necessary to add an electrode" which i don't believe because i still think that the electrode in the anode its still necessary to transfer the electrons. i asked the professor but got no answer. maybe my friend understood wrong, or maybe i'm wrong.

Because I heard people saying that since the anode already has a solid substance (solid iodine) it's not necessary to add platinum to the anode. I digress because I think platinum is necessary for the electrons to flow from anode to cathode.

Your reasoning is correct. Iodine will be oxidized in the anode, as iodide will enter the electrolyte and electrons will flow to the cathode. The way to accomplish this, is to have an electronic conductor for the electrons to reach the cathode. If this is absent, it is impossible for the reduction of permanganate to take place, since the charge transfer happens at the solid/electrolyte interface. If you can't ensure one reduction and one oxidation, then the cell cannot operate, because one cannot occur without the other taking place at the same time.
This phase can be $\ce{Pt}$ or any other solid phase. The requirement is that it should be an electronic conductor, and do not interfere with the reactions (it is “inert”). It would be pretty embarrassing to use a solid phase, for which its reduction has more tendency to occur than the one you seek (in this case the reduction of permangante).
I will make some corrections to your cell notation:

Every time you identify a potential difference between two phases, you must use a bar | or a slash /. Writing “$\ce{I2(s),Pt(s)}$” is not correct. Commas are used for separating different species in the same phase, e.g., ions in an electrolyte, like you did with permanganate and manganese ions.
Although it seems that $\ce{l}$ and $\ce{aq}$ are different phases, both are at the same liquid phase, or the same potential. They are all together in the electrolyte at the cathode.
The double bar $||$ represents a potential difference which is negligible to the other potential differences in the cell. I imagine that you have put those in order to separate both electrolyte phases from the cathode and the anode. If you have that information in the problem, then put both bars.

With this in mind, the cell configuration is the following:
$$\ce{Pt|I2|I^-||MnO4^2-,Mn^2+,H^+|Pt}$$

You still need an electrode in the solution hence Pt since it might be nonreactive.

I looked it up in Lange's Compilation of potentials. the potentials for I2[s] and I3- ion both require platinum electrodes. The potentials are the same.

@jimchmst You are totally right, and $\ce{Pt}$ (or another inert electrdoe) is needed. Huge mistake from my part, thanks!


text:
How to perform NBO calculations at each point while scanning the bond length?
For a given bond, I would like to perform an NBO calculation for each structure while scanning the bond lengths.

Can I use Opt=ModRedundant in combination with pop=nbo? Simply specifying both keywords would only perform the NBO calculation for the last structure.

Is there an easy way to retrieve the structures obtained from the scan as separate input files (e.g. .xyz or .gjf)?

I can use Gaussian 16 and Gaussview 5 or Gaussview 6.

This is question probably better suited at [mattermodeling.se]. I'm almost confident it would be on topic there. Please don't repost the question there, instead we can migrate the question. Please leave a custom flag.

I'm not too sure how NBO does behave for non-converged structures, it is probably dependent on what values you look at. Population analysis should still be quite okay. | Instead of using the modred interface, you might want to look at GIC, it is the superior interface! There might be the possibility to access different structures from a checkpoint file, and you could simply run a pop analysis over it in a second job. Any well-behaved molecular viewer (probably not Gausview) should be able to extract the steps of your scan in xyz coordinates. Have a look at Chemcraft or Avogardro.

Look at this: https://chemistry.stackexchange.com/q/86646/4945 Try pop(always,nbo6). Never ever use the built in nbo module of Gaussian. This is incredibly old, has no support and plenty of know issues. This is only ever suitable for reproduction of values and showing that they are wrong. Also, part 2 of the question: https://chemistry.stackexchange.com/q/62530/4945

You may find helpful the book F. Weinhold and C. R. Landis, Valency and bonding : a natural bond orbital donor-acceptor perspective, Cambridge University Press (2005).


text:
Why does the two OH bonds in H2O have different bond disassociation enthalpies? How does the bond cleavage happen here?
I came across this question, and in the text it was written as 'change in the chemical environment'. I wasn't satisfied with this explanation, so I am looking for a more detailed answer on this matter but didnt find much about it/

Review the guide How to ask and Asking FAQs to prevent clarification requests, objections,  downvoting or closure.

Neither books nor magazine articles nor SE questions should try to put near all the story to their titles.
With comparable lengths of the question title and the question body, the former is too long and/or the latter too short.

Two $\ce{OH}$ bonds in any molecule have different dissociation enthalpies, as only the first one is related to the original molecule.
The second one is related to the original molecule with the cleaved first $\ce{OH}$ bond.
If both bonds have the same chemical neighborhood and are far enough from each other, then their dissociation enthalpies converge to each other with the growing distance.


text:
When optimizing AMBER molecular dynamics parameters for protein-ligand interactions, how do I balance accuracy and efficiency?
Are there established practices for fine-tuning AMBER MD parameters in scenarios like mine, involving specific factors like mutant proteins or non-standard ligands? Has anyone successfully tackled challenges in a similar protein-ligand system?
Additionally, I'd appreciate insights on recommended approaches for addressing specific challenges and optimizing the simulation setup. The ultimate goal is to ensure that MD simulations precisely capture the nuances of protein-ligand interactions while maintaining computational efficiency.

It's a bit open-ended, but it should be fine. I'm not sure there are many experts on the topic her, so maybe transfer to [mattermodeling.se] would be better. Please do not repost the question there without deleting it here first. Alternatively, raise a flag and we'll migrate it.

Also see the AMBER Mail Reflector, which is quite active and very likely has addressed your question at some time.


text:
Why would the chlorine atom be assigned a positive partial charge in chloroform and carbon tetrachloride?
Looking at electronegativity tables, chlorine consistently has a higher electronegativity than carbon. However, when I use simple algorithms (molcalc, acc2), I get a positive partial charge on the chlorine atoms in $\ce{CCl4}$ and $\ce{CHCl3}$. This correlates with chloroform having a lower dipole moment than chloromethane (see this question and this duplicate question).
Is this an artifact of the simple algorithms? Or is this a reflection of different electron distribution around the carbon:chlorine bond in the different chloroalkanes?

https://en.wikipedia.org/wiki/Sigma_hole_interactions ?

I mean this positive partial charge may represent this sigma hole.

Molcalc appears to use a polarizable continuum model to model a solvent (assumed to be water) and a PM3 semi-empirical model for the solute embedded in the solvent. A PCM models the solvent as a dielectric, with counter-charges in the solvent mirroring those of the solute. The use of a PCM model would change the charge distribution in the solute. However I don't see how this would lead to the observed partial charges. One way to check this is to perform a calculation on a bigger molecule?

@BuckThorn According to https://doi.org/10.1021/ed400164n in MolCalc "These properties are computed using the GAMESS program at either the RHF/STO-3G (orbitals and orbital energies) or PM3 level of theory (all other properties) " From the question it's unclear what is meant by a "positive partial charge on the chlorine atoms", but my guess would be from a Mulliken analysis on the output of the SCF - in which case given the miserable quality of the basis set I wouldn't trust a word of it. Population analyses are notoriously basis set dependent, especially for poor quality ones.

@IanBush So is there a good source of reliable partial charge values for these molecules? From the difference in dipole moments, we know that polarization of the bonds is different in chloroform vs chloromethane, but it is not clear which bond "to blame" for this difference.

@Karsten Well given the orbitals permeate in principle the whole molecule determining a partial charge requires an arbitrary separation of the electron density into contributions from the different atomic centres. Different methods give different results, and quite what is a good way of doing this is a research area in itself, see e.g. https://www.osti.gov/servlets/purl/1712684 , so one has to ask quite what you mean by "reliable" . It's not really my thing, but from my understanding a Bader analysis on a high level calculation in a big basis set would be at least respected.

@IanBush I wrote my comment based on what I found in the source code. My interpretation is that the default is a call using PM3 theory and PCM solvation. Regarding the partial charge calculation that is handled by Jmol following a solvation computation. I didn't look into how it computes charges. See: https://github.com/jensengroup/molcalc/blob/master/molcalc_lib/gamess_calculations.py Solvation calculations appear to be performed by default.

Calculation issues aside, chlorine may become positively charged through pi back-donation similar to the role of the oxygen in trifluoramine oxide.
In this interaction, formally nonbonding electron pairs on the chlorine atoms overlap antibonding orbitals on adjacent sigma bonds, weakening the sigma bonds but creating pi bonding in return. We may render this as a delocalization of the bonding, as in carbon tetrachloride:
$\ce{Cl–CCl2–Cl <-> [\overset{+}{Cl}=CCl2]Cl^-}$
Because of the polarity of the carbon-chlorine bond, the carbon-chlorine antibonds are a better pi acceptor than carbon-hydrogen ones. So the additional chlorine atoms in more chlorinated molecules strengthen the pi interactions of each one. Since the molecular-orbital model (unlike the valence-bond structures given above) has both carbon and chlorine accepting the charge from this pi back-donation, the net effect is to transfer more electronic charge from chlorine to carbon in chloroform or carbon tetrachloride versus chloromethane.


text:
What is the SMARTS string for a benzene ring (any substitution pattern)
I would like to use RDKit to highlight all benzene rings (regardless of substitution pattern - e.g. phenol, chlorobenene, benzoic acid type should all be highlighted) in any given molecule. I cannot come up with a general SMARTS pattern that simply matches 'benzene ring'. This seems like it should be a lot simpler than I found it to be :<

just "[c]1:[c]:[c]:[c]:[c]:[c]1" should do the job. lowercase c means aromatic carbon, i also include aromatic bonds ":" explicitly though even without it should work quite well for your purpose.
demonstration:
from rdkit import Chem
patt=Chem.MolFromSmarts("[c]1:[c]:[c]:[c]:[c]:[c]1")
chlorobenzene=Chem.MolFromSmiles("c1ccccc1Cl")
phenol=Chem.MolFromSmiles("c1ccccc1O")
benzoicacid=Chem.MolFromSmiles("c1ccccc1C(=O)O")
for m in [chlorobenzene,phenol,benzoicacid]:
    print(m.HasSubstructMatch(patt))

spits out True three times

Thanks so much wikke!


text:
How can enthalpy of an ideal gas be independent of pressure?
I know that when one applies a manipulation of the Ideal Gas Law to the Maxwell Relations the result that enthalpy is independent of pressure tumbles out of it, i.e., $(\mathrm dH/\mathrm dp)=0$.
I don't understand what this actually means though, because how could it be? If $H=U+pV$ where $U$ is internal energy and $pV$ is the pressure-volume work, then the pressure clearly must play a role. Imagine some other planet where their atmosphere is 10 times the pressure of the Earth atmosphere. It takes more energy to do the $pV$ work to create the space for the system to exist. So for an equal volume is there not a greater enthalpy as a result of increased pressure? More heat at constant pressure, i.e. enthalpy, would need to flow into the system to reach that final state under that larger pressure regime. What is meant that enthalpy is in fact independent of pressure?

PV is not work. It is just multiple of P and V.  Ať constant T, PV is constant for ideal gas, in spite of nonzero volume work (and heat compensating this work).

The statement means that enthalpy is independent of pressure, for an ideal gas, if temperature and the amount of matter are both held constant.
For example, if you have 1 mol of idealium on Earth's surface at 273 K, it has a volume of 22.4 L. Take that same 1 mol of gas to your exotic planet, where the temperature is still 273 K but the pressure is 10 atm, and the volume will now be just 2.24 L. Although P and V have both changed, PV is the same. The internal energy U is also unchanged because the temperature is unchanged. Therefore the enthalpy is unchanged.
I think you are imagining a system that has a pressure of 10 atm but takes up the same 22.4 L on the exotic planet. But to achieve that, either the temperature would have to be higher, or you would have to have more moles of idealium, or some combination of those. Either way, one of the assumptions is violated.


text:
How do I calculate the fuel consumption inside a cylinder-piston stoichiometrically?

Consider I have car 4-cylinders with total cylinder capacity is 1,496cc. Per piston, the cylinder is 374cc. The piston diameter is 72.5mm, height=90.6mm, and compression ratio is 11.5:1. Hence we may calculate the annulus (A) = 41.28cm^2, and the swept volume = 321.57cc with its height=7.79cm, clearance volume = 52.4256cc with its height = 1.27cm. The detail just as in the simple picture, just in case you need it (even I am not sure that it is required).
The car, according to manufacturer, is consumes C10H24 gasoline (IUPAC name: ethane;methylcyclopentane). As the air consists of 21% only oxygen, the 21% translated to the 374cc cylinder volume, it is 78.54cc or 78.54mL. Assume that the reaction is occurs at Normal Temperature and Pressure (NTP), then how many cc or mL is C10H24 is required for every combustion?
With the help of chatGPT, I got that the calculation is like below:

The combustion reaction: C10H24 + 16O2 --> 10CO2 + 12H2O.
Number of mole of O2 = Volume of (O2/MolarVolume) = 78.54mL/24.0548L/mole = 0.003265 mole O2, or 3.265 mmole.
To balance the equation, the ratio of C10H24:O2 = 1:16. So, the mole of C10H24 = moles of (O2/16) = 0.003265 mole/16 = 0.000204065 mole, or 0.204065 mmole.
Convert that C10H24's mole to volume at NTP: Volume of C10H24 = mole of C10H24 * MolarVolume =  0.000204065 mole * 24.0548L/mole = 0.0049087L or 4.91cc.
When I put into real condition, seems like that fuel consumption is too little. As the 4.91mL is for single cylinder and the car is 4 cylinder, then the total fuel consume for every single combustion is 4.91mL/cylinder * 4 cylinder = 19.63mL or 19.63cc.
If I run my car at 3000rpm transmission 5 (I Googled, the gear ratio for transmission 5 is 5:0.838 or every single crankshaft will rotate the wheel 5/0.838 or 5.966587 ~= 6 times).
My wheel diameter is 16", or 40.64cm. ==>Hence, the circumference = pi * d = 1.2767m. Means, every single crankshaft rotation the wheel will travel as long as 6 * 1.2767m, or 7.6602 meters.
If I drive that car at constant RPM @3000rpm on a flat road, then the distance it can travel for one minute is 22,980.6 meters or 22.98km.
The fuel that it consumes for that 3000rpm is (rpm/2) * (volume of fuel/combustion)=1500rpm * 19.63mL = 29.45L This is too much.

My question then,

Is that my calculation correct? how many cc or mL is C10H24 is required for every combustion?
What is the correct calculation if mine is not correct?

Comments have been moved to chat; please do not continue the discussion here. Before posting a comment below this one, please review the purposes of comments. Comments that do not request clarification or suggest improvements usually belong as an answer, on [meta], or in [chat]. Comments continuing discussion may be removed.

You have made a number of assumptions, some of which are mostly right, but others are seriously wrong. Or maybe ChatGPT did it on your behalf. (Stay away from ChatGPT if you want to learn -- it's quite harmful.)
Because your assumptions have diverged seriously from what happens in real cars, the fuel usage will not make sense.
The two most important issues are:

You assumed a gear ratio of 1:6, with the engine turning 1 time for every 6 wheel revolutions. In real cars, the wheels turn more slowly than the engine. In my car, the gear ratio, in top gear, is about 3:1, meaning that the engine turns 3 times for every time the wheels turn. You have it the other way around. If your car's wheel turns 6 times per engine revolution, and there are 50 engine revolutions per second, the car wheel spins 300 times per second and the car travels at 383 m/s. That is faster than the speed of sound.
(As a side note, your rims may be 16 inches, but unless you are running on the bare rim, the wheel with the tire is bigger than that.)

You assumed that the pressure in the cylinder was "standard" (atmospheric pressure). In reality, power in an automobile engine is managed with a throttle. Unless the throttle is "wide open," it restricts air flow into the engine and causes the intake stroke to fill the cylinder at a pressure below 1 atm. Varying the pressure is what allows the engine to burn a near-stoichiometric mixture whether you request low power or full power. (This is the traditional method; some modern cars use other methods too, like variable valve timing or cylinder disabling.)
If you assume that the cylinder is filled at atmospheric pressure, that implies a wide-open throttle. A wide-open throttle will cause the car to accelerate. Eventually it will reach its top speed, probably at more than 3000 RPM. The fuel consumption will be much higher than you're used to, unless you always drive with the pedal to the metal.
Because the pressure in the engine is unknown, this calculation is not particularly useful for estimating fuel consumption.

I am more concern about that stoichiometry. To make it easy, just make a SINGLE combustion. So, what is the required gasoline/fuel to make an ideal (not lean or rich fuel) combustion? Regarding that gear ratio I have to check it again. I am more concern about that stoichiometry, so please comment about it.

Well, as Poutnik observed, C10H24 is not a chemical. At best, it's a mixture of two chemicals in a precise ratio. This is important because the volume of a gas depends on the number of actual molecules. A 2:1 ethane/methylcyclopentane mixture has an "empirical formula" of C10H24 but takes up three times the volume as a gas.
If you did have one molecule of the imaginary C10H24 it would take 16 dioxygens to completely burn it, so in that respect your 16:1 ratio is correct.

The C10H24 actually is a fuel in our country, it is RON 92. Just think be really exist, so what is the required C10H24 if I have 78.54mL 16 mole O2? I have compared if I replace that C10H24 with C10H22, then the O2 will be just 15.5 moles, and the volume of the required C10H22 changes slightly. If you have basic in chemistry, I believe you can do it easily.

Gasoline is a variable mixture of dozens of different compounds with variable molar masses and composition, mostly hydrocarbones:

alkanes/paraffins $\ce{C_nH_{2n+2}}$,
alkenes/olefines $\ce{C_nH_{2n}}$, $\ce{C_nH_{2n-2}}$, .... ,
cycloalkanes/naphtenes  $\ce{C_nH_{2n}}$, $\ce{C_nH_{2n-2}}$, .... ,
aromatics $\ce{C_nH_n}$, $\ce{C_nH_{n-2}}$, .... ,

with $\ce{C4-C12}$ straight, branched or cycled carbon chains.
Additionally, gasoline contains typically 0.5 to 4 mass percents of bonded oxygen, partly as additives improving the Octane number, like MTBE ( $\ce{CH3-O-C(CH3)3}$ ) in EU or ethanol ($\ce{CH3CH2OH}$) in US.
Note that the formula $\ce{C10H24}$ does not belong to any existing molecule of hydrocarbons. There is not enough bonding opportunities for hydrogen atoms to attach to the chain of ten carbon atoms. It is at the best just empirical formula and even in that case it is rather off.
Being an empirical formula, it cannot be used to accurately predict fuel vapor volume for given temperature, pressure and liquid fuel mass, volume or molar amount. E.g. vapor of the fictitious $\ce{C10H24}$ would have at the same conditions the half volume of pentane $\ce{C5H12}$ vapor of the same mass, or 1/3 of volume of fictitious "ethane:methylpentane 2:1".
Note that the typical gasoline elemental composition is 84% C, 15% H, 1% O, what is equivalent to the empirical formula $\ce{C112H240O}$. The molar ratio $\ce{H:C}$ is $\approx 2.14$, while for $\ce{C10H24}$ it is $2.4$. Be aware that all such empirical formulas have no relation to any particular molecule nor its respective molar mass nor the vapor density at given temperature and pressure.

If we consider the gasoline empirical model formula as $\ce{C_aH_b}$, the combustion reaction is then
$$\ce{C_aH_b + (a + b/4)O2 -> a CO2 + b/2 H2O}.$$
The oxygen : fuel formal molar ratio is then
$$R_\mathrm{m,\ce{O2}}= a + \frac{b}{4}.$$
The volume(= molar) fraction of oxygen in air is approximately 0.21, the effective formal air : fuel molar ratio
$$R_\text{m, air}= \frac {100}{21}(a + \frac{b}{4}).$$
The consequence of undefined true fuel molar mass is, that for well defines air : fuel mass ratio is undefined volume ratio air : fuel vapor.

The car industry is used rather to evaluate mass ratios, as molar mass of gasoline is not defined. It can be just formally involved for empirical formulas reflecting elemental composition.
As the effective formal molar masses of the fuel and air are approximately

$M_\text{fuel} \approx 12a + b\ \pu{ g mol-1}$
$M_\text{air} \approx \pu{28.8 g mol-1}$

The mass ratio air : fuel is then
$$R_\text{mass,air} = R_\text{m,air} \frac {M_\text{air}} {M_\text{fuel}} =\frac {100}{21}(a + \frac{b}{4}) \frac {28.8}{12a+b}=\\
=\frac {100}{21}(4a + b) \frac {7.2}{12a+b}=\frac {720}{21} \frac {4a + b}{12a+b}$$
For octane $\ce{C8H18}$ it is about 15.0,
for decane $\ce{C10H22}$ too,
for "$\ce{C10H24}$" it is 15.2.
Standardized value for gasoline is taken by automotive industry as $14.7$, with correction for components containing oxygen (mostly ethers or alkohols) etc.), that leads to slightly lower oxygen demand for total combustion.

According to Wikipedia - Air-fuel-ratio - Engine_management_systems

The stoichiometric mixture for a gasoline engine is the ideal ratio of air to fuel that burns all fuel with no excess air. For gasoline fuel, the stoichiometric air–fuel mixture is about 14.7:1 i.e. for every one gram of fuel, 14.7 grams of air are required.

Any mixture greater than 14.7:1 is considered a lean mixture; any less than 14.7:1 is a rich mixture – given perfect (ideal) "test" fuel (gasoline consisting of solely n-heptane and iso-octane). In reality, most fuels consist of a combination of heptane, octane, a handful of other alkanes, plus additives including detergents, and possibly oxygenators such as MTBE (methyl tert-butyl ether) or ethanol/methanol. These compounds all alter the stoichiometric ratio, with most of the additives pushing the ratio downward (oxygenators bring extra oxygen to the combustion event in liquid form that is released at the time of combustions; for MTBE-laden fuel, a stoichiometric ratio can be as low as 14.1:1).

The input pressure of air is controlled by throttling valve or equivalent devices, so the fuel input is decreased accordingly.
The actual air / fuel mass ratio is usually kept in range
12 : 1 or less (rich mixture) to 16 : 1 or more (lean mixture).
There is considered direct fuel injection in further analysis.

The air mass in a cylinder is  $$m_\mathrm{air}=\frac{pVM_\text{air}}{RT}$$
The fuel mass in a cylinder is  $$m_\text{fuel}=\frac{pVM_\text{air}}{RTR_\text{mass,air}}$$
The typical gasoline density $\rho = \pu{0.74 g mL-1}$.

The liquid fuel volume provided to a cylinder for one cycle is
$$V_\text{fuel,l}= \frac{m_\text{fuel}}{\rho}=\frac{pVM_\text{air}}{\rho RTR_\text{mass,air}}=\\
\frac{(\pu{e5 Pa}) (\pu{0.000374 m3})(\pu{0.0288 kg /mol})}{(\pu{740 kg m-3})(\pu{8.314 J K-1 mol-1}) (\pu{298 K}) (15.2)} \approx \pu{3.87E-8 m3}=\pu{3.87E-5 L}=\pu{38.7 \mu L}.$$

With four cylinders and engine crankshaft revolutions $\pu{300 RPM}$, the hourly fuel consumption is (considering 2 revolutions per ignition) :
$$V_\text{fuel,hour}[\pu{L/h}]=(\pu{60 min/h} ) \cdot (\text{3000/2  }\pu{min-1} )  \cdot 4 \cdot (V_\text{fuel,l}[L])= 13.9\ \pu{L h-1}$$
This seems a lot, but consider these factors:

It is at full power. Idling or at limited power, the throttling valves bring to cylinders less air, and proportionally less fuel is provided to maintain about stoichiometric ratio.
It seems like effectively less volume should be considered, regarding the ratio $\frac{\pu{322 mL} }{ \pu{374 mL}}$
There are used variable air : fuel mass ratios, 12 : 1 or less for rich mixtures, 16 : 1 or more for lean mixtures.
The true air temperature before compression is unknown. For given pressure, the air mass in the cylinder is reciprocal to absolute temperature. If e.g. the air in valve has $T=\pu{400 K} (\approx \pu{127 ^\circ C})$, the cylinder would contain 1/4 less air, compared to $T=\pu{300 K} (\approx \pu{27 ^\circ C})$
Some engines (mostly 6+ cylinders, but some 4-cylinder too) deactivate some cylinders when idling or at cruising mode with minimal power.

With the gear ratio $G \gt 1$ toward a slower rotation with a stronger torque and with the wheel diameter $D$, the car passes per hour the distance
$$L[\pu{km/h}] = (\pu{0.001 km/m})(\pu{60 min/h}) \cdot (\frac{\pu{3000 min-1}}{G}) \cdot 3.1416 \cdot (D\ \pu{m})$$
The car fuel consumption $C$, if evaluated in $\pu{L/100 km}$ as usual in EU:
$$C = 100  \frac{V_\text{fuel,hour} }{L}\ \pu{[L/100 km]}$$


text:
Advise on performance vs cost of the Berkland-Eyde Process for direct synthesis of Nitric Acid from atrmospheric air
I am new to chemistry, having gone down an electronics route in my career, so I decided to pick up where my 'O' levels ended (many years ago) and make a better understanding of the topic a sort of hobby.
As it seems it's almost impossible, at least as an individual, to obtain strong acids in this country. I am forced to create my own....
The research I've done so far pointed me to the Berkland-Eyde Process, being a means to directly synthesise Nitric Acid from air and being 'safer', with less also hard to obtain ingredients, than most other methods.
I built a Berkland-Eyde apparatus, and in about 8 hours managed to create a rather paltry yield of 250ml of 6% concentration HNO3.  Which I could concentrate to ~60-70% through distillation, though that would be a very small yield for a rather large effort in time and equipment.
I notice that articles on this use significantly larger currents (I am limited to just 20mA at 10kV without spending a lot on a more powerful PSU)
I came here to search for answers and was surprised to find this question or anything like it does not exist - so here it is!
I could try to obtain an old style Neon sign transformer, but they are rare.  Modern powerful ones are very expensive and I am looking for a cheap solution, i.e. This is just for self education and not worth investing in.
I have found a cheap 600W/50kV/DC (adjustable) PSU, huge compared to my current 200W (probably less) 20mA/10kV/400Hz/AC one (which is not happy about what I am asking of it, as it gets hot and is starting to arc internally).
Other options I have found:

Diesel/Oil heater ignitors (electronic) or (iron)
CO2 laser PSUs, like or
Cheap, crude 'ZVT' fly-back transformer based PSUs
Old school transformer based Neon sign PSUs
Modern lower current electronic Neon sign PSUs (one issue here is that they often are USA mains, so 110V, the mains here is 230V).

Firstly, and the main purpose of this question:

Would using DC rather than AC make a significant difference?

There are some other related questions (perhaps I should write more than one question?):
From a physics and chemistry perspective what attributes optimise the result, e.g. Any combinations of:

Geometry:

Area of air flow vs arc
Arc length
Electrode shape

PSU:

Current
Voltage
Power
Polarity
Frequency
Duty cycle (big regular pulses vs smaller contiguous arc)

Electrode temperature
Flow rate
Pressure
Additional attributes, e.g. Presence of magnetic field

What is the best 'balance' between voltage and current?  Is it the hot electrode or the gas plasma that does the work?   I ask this as the PSU is adjustable so at a lower voltage I can have more current (within limits).

A higher voltage means a longer (or hotter) arc.  Would constraining the air path and forcing it past the electrodes be better than slowing the air down to have it spend longer in the reactor?

My current electrode are Tungsten TIG welding rods, which seem to work and I don't get any noticeable discolouration of the product, but:

I have been advised that the red hot Tungsten reacts with the small quantity of acid present in the reactor due to water vapour being pulled in. This is true and what should be used instead.  I first tried copper but it immediately melted.

Is there any data on the rate of generation I should expect for a given air flow rate and voltage?

UPDATE:
I read up on the current state of law and found that I was given incorrect information.  Now a license for certain substances can be obtained by individuals rather than just corporate entities.
In searching for this I also found a list of other controlled chemicals, to which I linked for completeness.
Substances on that list now require a license to possess (costing £40 and lasting 3 years) this is now to control (rather than prevent purchase as was the original intent from 2014) as a non-corporate entity - so actually an improvement!!

2016 Regulations
License application process
FAQ (PDF)
Additional advice for members of the public

I expect this is the topic of a PHD, any takers?

The process is so inefficient that you'd be paying far more for energy use than it would cost to buy ammonia water (or liquid NH3, if available) and convert it to $\ce{HNO3}$ by the Ostwald Process: https://en.wikipedia.org/wiki/Ostwald_process.

It is surprising to hear that you are able to find Tungsten TIG welding electrode, Neon signs PSU, CO2 laser PSU, and that you are unable to get ordinary nitric acid...

@Maurice: This UK legislation stems from the historic problems in Northern Ireland. I understood that unless a trading company, any strong acid (and for lots of other stuff) you need a government issued license to purchase any quantity and the seller has to keep records of the identity of the buyer -  no company - no buy.  I just read today that recent legislation (Oct'23) makes it illegal to possess certain items - which means I have to destroy my pathetic attempt :-|  but it also opens things up as anybody can get a license if they pass a set of tests....

As an electronics engineer I already have a PSU and I own a TIG welder (three in fact).

As it seem relevant, and perhaps useful to others; I updated my question to include pointers to the relevant UK Government legislation, including the list of controlled substances, that require a License and also another list covered by other legislation that puts restrictions of precursors and ingredients that could be used to make illegal substances. However I think the original question is still relevant as even after a lot of searching, I could not find any concrete information on much of this.

@DrMoishePippik.  With an input power of just 200W I got 6% in <8hrs.  My assumption is that I am a complete novice so I am probably doing it in at best an average way. And more like the worst possible. That is 1.8kWh @ 30pence/kWh (the most I can pay) = £0.54.  At even 100 times that it would still be cheaper, surely?

@JayM, ingteresting -- you did not do as poorly as I would have expected. However, 1. That solution is not pure $\ce{HNO3}$ in water, but contains other nitrogen oxides. 2. Add in the energy cost of distillation. 3. $\ce{NH3}$ is comparatively cheap, and likely available in UK. Currently, about US$0.42/kg (https://www.chemanalyst.com/Pricing-data/ammonia-37), FOB New Orleans, LA).

It does seem the new legislation does not (yet) control NH3, though from what I've tried so far it's not that easy to buy in solution above 33%. And I would expect liquified gas is even harder? I have a means to generate the high temps required. (induction) What you say about purity is interesting as I believed that any oxides break down into smaller ones which then find a H+ ion in the water. What methods are there to test the purity of the product, I.e. other disolved nitrites, etc.  Fairly sure I am not in need of amazing quality, but I guess any would affect my apparent yield?

I've decided to go buy that 600W DC PSU from China in the hope that it helps answer this question....