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Electron coordinates in QM programs
I find it somewhat unusual that one does not need to define electron coordinates in an electronic structure calculation. Taking Hartree Fock for example, the Fock operator includes the one-electron operator that depends on electron coordinate. However, the matrix representation of the Fock operator in the basis of $\{\phi_{\mu}\}$ is
$F_{\mu v}=\int d \mathbf{r}_{1} \phi_{\mu}^{*}(1) f(1) \phi_{v}(1)$
So because the entries in the Fock matrix are all integrals over all space $\int_{r_{1}=0}^{\infty} d \mathbf{r}_{1}$, it doesn't need the electron coordinate. Is that correct?

Electron is not a thing that has coordinates.

Yes, very much so

@IvanNeretin I'm very much aware of the quantum treatment of electron as a wave rather than a particle with definite coordinates. I'm more concerned with the numerical implementation of electronic structure method!

The answer is yes and no. Electronic coordinates are needed when calculating the Fock matrix elements $F_{rs}$, which consist of one-electron integrals and two-electron integrals. Once you get the Fock matrix elements, then the Hatree-Fock-Roothaan equations are solved by matrix algebra through diagonalization without using electron coordinates. Therefore, electron coordinates are used explicitly in the step of Fock matrix elements calculations but not in the step of matrix diagonalization.

Please, if you do not think that it does help, i.e. it answers the question, do not post the answer; you may consider a comment instead. If you think your answer is helpful, don't use tag lines like "hope this helps". The voting system is designed to judge the helpfulness of an answer based on reader's opinions. Avoid unnecessary fluff.

It is my understanding that it is humble and polite way to end the answer. Now it is removed.

I'm sorry to tell you, but it has a negative connotation on the Internet. You might want to look it up on the urban dictionary.

It seems to be fine from English stackexchange at https://ell.stackexchange.com/questions/51076/hope-this-help-or-hope-this-helps

I really do not want to go into discussion on this. On Stack Exchange in general, we do not use salutations, greetings, tag lines, signatures, etc. Some of this will be even filtered, others will be removed by editors. It's not meant to be unfriendly, is just the way this network is designed: it is content oriented. You can and should show appreciation through voting. Here's some more light reading: https://chemistry.meta.stackexchange.com/q/593/4945 (And you should really check the urban dictionary.)

Point is well taken. I am new here and still learning through mistakes. Thanks.

While we're not all new here, we can always still learn a thing or two every day. So don't worry about that. You can also use [chat] to talk to others, even though it's a bit deserted lately. You can still give it a try, and the system is also right now reminding us to avoid discussions in the comments. So better stop here. Have fun and thanks for your contributions.

Think about the integration as solving the Schrödinger (or Fock, or whatever) equation for all coordinates in space at the same time.
The whole point of this is to obtain a wavefunction which yields the energy eigenvalue to your operator no matter what electron coordinates you plug in. In other terms,
$\hat H \Psi ( \mathbf{x} ) = E \Psi ( \mathbf{x} ) $
for any $\mathbf{x}$.
So, once you have your final wavefunction, nothing stops you from "putting your electrons in specific locations", but that's not really all that interesting because we know that this will still yield the same energy eigenvalue - because that's how we constructed the wavefunction.


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How do the orbitals distort when adding more protons?
How do the orbitals distort as you move along the series of chemical elements with greater protons. I cant imagine, the first 1s orbital remains a similar shape at Kr 36 protons as compared to He 2. And is the addition of protons responsible for the orbital energy imbalance occurring at the transition metals at Sc>

"How do the orbitals distort[...]" They get smaller; that's about it.

The single-electron wave functions distort in ways directly related to the number of protons, since the nucleus charge changes. The real trick is that the wave functions of an atom filled with all its electrons don't change that much, but do get a bit squirrely, and the d-block is one such manifestation.

Do you have any reference to academic material to support your what you're saying please

The simplest approximation you can make is to describe each orbital by a one-electron wavefunction like the ones we find in the hydrogen atom. For those wavefunctions, the radial distribution is described by an exponential factor of $exp(-Zr/n)$ (with $n$ the principal quantum number) and a polynomial with mixed powers of $Z$ and $r$. If you increase $Z$, this function becomes "narrower" along the $r$ axis. Thus, increasing the number of protons in the nucleus contracts the orbitals. We can intuitively explain this due to the increased electrostatic attraction between the nucleus and the electrons.
This is ignoring all effects from actually having multiple electrons in your atom, where inner electrons can "shield" outer ones from feeling the full nuclear charge. In those cases, the outer electron shells are affected less by that contraction. In general however, it's difficult to say exactly what a multi-electron wavefunction will do because we don't have a closed analytical expression for them like we do for the one-electron case.

source: Stack Exchange

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Any Open-Source, Free Software That Can Be Used to Computationally Calculate Binding Affinity Between Two Molecules?
There are programs out there which calculate the binding affinity between a ligand and a target protein (Autodock 4, Autodock Vina, Chimera... etc). Is there any analogous program that can be used to calculate the binding affinities between two molecules?
For my purposes, I'd like to computationally determine the binding affinity between crystalline hydroxyapatite and an analogue of a bisphosphonate molecule.

Software recommendation is not really on topic here, but it might be on [mattermodeling.se]. Please check their guidelines before asking there. I'm afraid this question has a high chance of being closed here.

Thank you for the heads up! I will switch forums immediately.

Before you do that, I've already asked in chat there. So we'll might get a hint soon. And if it's in topic we can simply migrate it

I’m voting to close this question because it was cross posted to a different site: https://mattermodeling.stackexchange.com/q/8489/49

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


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Why is the major product of an elimination reaction an alkyne and not the diene?
1,2-Dibromopropane and KOH react together to form propyne. But I wondered if 1,2-propadiene was also being produced since the there is also a proton on the $\ce{CH3}$ group to be used in the elimination reaction.
Does it just come down to Zaitzev’s rule (in an elimination reaction the major product is the more stable alkene with the more highly substituted double bond) that all I see in propyne or is there something I am missing?

The acidic character of the vinylic hydrogen is much higher than that of allylic hydrogen.

The concept here is that after the formation of the alkene (propene, in this case), it is much more easier to abstract the vinylic hydrogen as compared to the allylic hydrogen and thus preferring the formation of alkyne (propyne, in this case) over the formation of a diene (prop-1,2-diene, in this case).
Remember, $\mathrm pK\mathrm a$ of vinylic systems is around $43$ and that of allylic systems is about $50$. This webpage might be helpful to you in future.

@DrunkenTaurus If you think this answered your question, you can upvote and accept this answer.

It turns out that propyne and allene do equilibrate with the equilibrium mix in favor of the triple bond. I looked it up and think that Keq is about 20. Both geminal and vicinal dihalides seem to give predominately the alkyne.  It might be possible that this is a situation of thermodynamic control. Maybe these reactions should be examined a bit more carefully?


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Hydrogen safety during electro-chemical machining
I'm interested in doing some experiments with electro-chemical machining in my shop. The process involves running electricity through an electrolyte (salt water) to break down a metal anode (aluminum/steel), and it releases hydrogen gas. I'm comfortable with electricity safety, but I have very little chemistry experience, and I'm worried about generating hydrogen in my shop. There's obviously an explosion risk, and I want to understand the safety concerns of what I'm doing.
My research so far has found a lot of high school chemistry experiments, which seem to simply ignore the hydrogen and release it into the room. I will be running processes for hours, so I don't think this is an option for me. I've also found descriptions of large scale industrial processes and proper chemistry labs, both of which use pre-made systems generally out of my reach. Chemical safety datasheets have storage and handling info, but I don't see anything I can use for build guidance.
In the absence of a pre-built lab and fume hood, what is the best practice for dealing with medium amounts of hydrogen gas? Is the answer simply "you need a fume hood"? Can I contain and/or compress the gas and dispose of it safely? If ventilation is the only way, is there any advice/guidance for constructing it within safety guidelines? Ideally I'd love a guide for what materials are safe to use, what's a safe way to vent (if that exists), etc.
Edit: it occurred to me, can I just burn the hydrogen immediately as it's generated? It seems like maybe some kind of contain and combust strategy might make things nice and simple

Prudent Practices in the Laboratory (National Research Council) has general parameters for what you request. NASA's 389-page tome Safety Standard for Hydrogen and Hydrogen Systems is what I would defer to, as NASA is charged with keeping humans alive in all kinds off adverse environments. Plus, I think 389 pages probably covers all the bases.

This is exactly what I was looking for, thank you so much @ToddMinehardt!

The primary concern for hydrogen in a room is that it can rise to the celling with limited dilution and reach explosive concentrations.

A fume hood is the most practical way to handle it. We commonly used hydrogen and vented it to atmosphere with laboratory hoods.  My first employer  ( very long ago) did not have a hood and blew out a large brick wall

If you know the current, you can accurately calculate the rate at which gas is produced (e.g. in liter per hour). Knowing that number will give you a much better idea about the possible danger, and sensible countermeasures.

Yes, NASA has good information on Hydrogen Safety.
Here is the key take away, to quote:

Hydrogen has a very broad flammability range—a 4 percent to 74 percent concentration in air and 4 percent to 94 percent in oxygen; therefore, keeping air or oxygen from mixing with hydrogen inside confined spaces is very important.

So, how does one practically stay out of this range, particularly, with small amounts that could tally just over 4%?
The only practical safe answer is a powerful ventilation system. As exactly how one effectively accomplished this is the main question. Perhaps, as hydrogen gas is a very light, one idea suggests the room has a high ceiling which, to a degree, has a fire place like design (noted for safely venting carbon monoxide), complete with a chimney.
Good luck and stay safe.


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Tetrachlorophosphonium
I am aware that PCl3 and PCl5 both react with water. In these reactions, water breaks the chlorine atoms off of the phosphorus atom and hydroxyl groups take their place:
$\ce{PCl3 + 3H2O -> H3PO3 + 3HCl}$
$\ce{PCl5 + 4H2O -> H3PO4 + 5HCl}$
So, judging by these equations, my prediction is that the tetrachlorophosphonium ($\ce{PCl4+}$) ion should readily react with water as well. This is my guess at how the reaction should go:
$\ce{PCl4+ + 4H2O -> H+ + H3PO4 + 4HCl}$
Is this correct? If so, I predict the tetrahalophosphonium ions (with fluorine, bromine, and iodine) would have the exact same reaction but with HF, HBr, or HI formed instead of HCl.
If this is not correct, what would the resulting hydrolysis products be?

The reaction would form not only phosphoric acid but also phosphorus acid. Not sure about other halogens. Fluorine will most probably only partially hydrolyze, whereas in case of iodine reducing property of HI might be too high and so you might observe I2 vapors.

$\ce{PCl4^+}$ reacts like $\ce{PCl5}$, because of the equilibrium $$\ce{PCl5 <=> PCl4^+ + Cl^-}$$ So the reaction with water will produce $\ce{H3PO4 + HCl}$ as stated in the original post, and no phosphorous acid.


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Electronic states of VO vanadium oxide in $C_{2v}$ point group
The electron configuration of $V^{2+}$: ... 3$d^{3}$ with 3 d electrons in 3 d orbitals. In $C_{\infty v}$, $d_{xz}$ and $d_{yz}$ have $\Pi$ IRREP; $d_{z}^{2}$ has $\Sigma^{+}$ IRREP; $d_{x^{2}-y^{2}}$ and $d_{xy}$ have $\Delta$ IRREP. So I think the 3 lowest lying  orbitals are $\Sigma^{+}$ and $\Pi$. The 2 possible spins are doublet and quartet. The 3 spatial occupations are

$\Sigma^{+}$ x $\Pi$  x $\Pi$ = $\Sigma^{+}$ + [$\Sigma^{-}$] + $\Delta$
$\Sigma^{+}$ x  $\Sigma^{+}$ x $\Pi$ = $\Pi$
$\Pi$ x $\Pi$ x $\Pi$ = ($\Sigma^{+}$ + [$\Sigma^{-}$] + $\Delta$) x $\Pi$ = $\Pi$ + $\Pi$ + $\Pi$ + $\Phi$

I only know that  $\Sigma^{+}$ is symmetric and $\Sigma^{-}$ is antisymmetric. But I'm not sure about other IRREPs' symmetry ($\Pi$, $\Delta$, $\Phi$). Also I don't know how the sum of many IRREPs which should also takes into account the various occupation permutations is transformed into $C_{2v}$. And since the $O^{2-}$ is fully occupied, only $V^{2+}$ contributes to the MOs. (Or is the concept of MO invalid in ionic bonding?) Is that correct?
Also a book recommendation for group theory in transition metal complex would be very much appreciated.

If you know you’re looking at C2v symmetry, why do you look up the $C_{\infty v}$ irreps? You can find how the d orbitals transform in the C2v character table.

I think you should try to figure out the electronic states of VO by using the electronic configurations of the molecule instead of $\ce {V^2+}$.

Please check this paper with information on the electronic states of VO molecule: https://aip.scitation.org/doi/pdf/10.1063/1.4926393.

Since 19 months have gone by with no answers or votes, I think it's fair to let you know about Matter Modeling Stack Exchange which also answers group theory questions, but make sure to delete the question from here if you're cross-posting there (or to at least put a note at the top of your question, saying that it's cross-posted).


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Is chemAxon check of validity of a smiles the same as RDKit check?
I am working on an AI-based drug design application, I want to verify the validity of the produced molecules (which are in smiles representation). I found a solution to this using the RDKit library the following way:
from rdkit import Chem

def check_validity(smi):
    m = Chem.MolFromSmiles(smi, sanitize=False)
    if m is None:
        print('invalid smiles')
    else:
        print("valid smiles")
        try:
            Chem.SanitizeMol(m)
        except:
            print('invalid chemistry')
    return v

However, most similar work in the literature use the ChemAxon-checker software (which I think is a closed-source software):
https://docs.chemaxon.com/display/docs/structure-checker-getting-started.md.
This might make the comparison of performances unreliable.
So my question is: Is the ChemAxon-check of validity of smiles the same as my RDKit checker function?
Note: I am not a chemist, I come from a computer science background

Not a proper answer, but you might be interested in this : https://depth-first.com/articles/2021/09/22/beyond-smiles/ Richard Apodaca is trying to better 'formalise' SMILEs here

One potential pitfall working with SMILES is that there are multiple dialects of SMILES. Not obvious for small molecules like benzene, pyridine, etc.; however with growing complexity of molecular structures, it is better a) to stick with one form, or b) to convert all into one form, and then sanitize them. E.g., OpenBabel lets you choose canonical, universal, or inchified; PubChemPy isomeric ones. And Marvin/ChemAxon?

Maybe «reliable» doesn't fit what you intend to express for a meta analysis how well (e.g., in terms of describing a structure without ambiguity [bijective relation between structure and reduced representation/SMILES] and chemical reason [e.g., charges, valences, tautomerism]; rate of processing, ease of scaling) ChemAxon vs. RDKit performs. The discussion board or issue tracker might be resonate earlier (by RDKit's developers/user community) to this interesting question.


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Burning vegetable oil as fuel
I heard just recently that vegetable oil (i.e palm oil) does burn.
I am also aware that burning used motor/engine oil, does emit hazardous gases such as nitro or sulfur oxide, hence prohibited by government.
How about burning used cooking oil?
I am not saying convert them first to methyl ester, but straight put them directly on a burner/fire. What gases do they emit and is it harmful too?

There is a good reason why vegetable oils are usually converted to esters before use as a fuel. The resulting compounds are far more similar to existing diesel and can be used without substantial adjustments to the engines (and fuel pumps) required to compensate for different physical and chemical properties.

By "fuel", do you mean as in for an internal combustion engine, or for cooking, or for light, or what?

The short answer: with technical adjustments, it is possible (cf. e.g., Wikipedia's article).
Note, the levels of sulfur in modern gas/petrol are low; for one, to limit the generation of sulfur oxides during the combustion (e.g., $\ce{SO2}$) contributing to acid rain.  However, desulfurization equally is economically interesting to access sulfur which is used as such for vulcanization, or the controlled combustion to yield $\ce{SO3}$ and eventually sulfuric acid.  One of the commonly known processes of desulfurization is the Claus process.
The potential problem with burning used cooking oil is its chemical composition.  Like gas/petrol for a car, vegetable oil is a blend of chemicals.  Gas/petrol mainly consists of hydrocarbons (compounds only consisting of carbon and hydrogen) and some additives (like methyl tert-butyl ether (MTBE or/and methyl tert-butyl ether (ETBE)):

(image credit to Wikipedia)
Vegetable oils, on the other hand, are esters of glycerol. Because each of the three hydroxyl groups may be esterified differently, the chemical composition of the oil however may vary much more (plus effects by season of harvest/how it is processed lining up for their use e.g., in the kitchen):

(image credit to Wikipedia)
In comparison to molecules in gas/petrol for a car, the molecules in vegetable oils tend to consist of a larger number of atoms per molecule. This and aforementioned greater variability of chemical composition of vegetable oil requires an adjustment of the air intake during the combustion to yield an exhaustive combustion while reducing the generation of nitrogen oxides and soot.  (Some of the later passes the exhaust pipe, some may deposit next and on the burner/oil atomizer nozzle.)  It has been possible for long time (e.g., youtube video about an old omnivore combustion engine) and vegetable oils seem to attract again interest (cf. e.g., Wikipedia's article) to contribute as a form of energy.  Technology aside, it equally is a topic of discussion to balance production of e.g., canola/oilseed for food vs. combustion.


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What is the structure of sodium chloride in the liquid state?
I understand that sodium chloride when solid, is a lattice, so NaCl there is an empirical formula,(so, expresses that the ratio of Na to Cl is 1-1), and not a molecular formula expressing exact numbers of atoms.
I was looking at this question which covers what sodium chloride is like in gas state. What is Sodium Chloride like in gas state?
And it mentions that in gas form, you get an NaCl monomer, and $\ce{Na2Cl2}$ dimer. That's mentioned in the abstract https://pubs.acs.org/doi/10.1021/jp073328o. And those are molecules, not like with the sodium chloride lattice.
But what about for sodium chloride as liquid. What form does sodium chloride take there?

@OscarLanzi so would you say it's not a monomer https://en.wikipedia.org/wiki/Monomer  or dimer? (or would you say that a monomer or a dimer, isn't necessarily a molecule)?

You can certainly have monomeric ion pairs.

A good starting point to check is https://moltensalt.org/whatIsMoltenSalt.html. It is essentially high-temperature ionic liquid, whereas the topical room-temperature ionic liquid (RTIL) is essentially ambient-temperature molten salt. I think a key property to understand the ion associations in molten salt is radial distribution function. There are lots of molecular dynamics simulations, even from decades ago.

@OscarLanzi Well, I would sure call them molecules. Ion pairs are in cases like here: https://chemistry.stackexchange.com/questions/136299/what-happens-to-boiled-bmimpf6

More to the point, liquid NaCl isn't much different then solid, but average coordination number is lower.

https://chemistry.stackexchange.com/questions/136857/when-you-melt-an-ionic-compound-do-you-break-its-electrostatic-force-of-attra

https://chemistry.stackexchange.com/questions/59462/what-are-bonds-in-molten-nacl

For an answer you might want to look at Karsten Theis's answer in https://chemistry.stackexchange.com/questions/136857/when-you-melt-an-ionic-compound-do-you-break-its-electrostatic-force-of-attra for instance.


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Nature of Bisphosphonates Binding onto Hydroxyapatite in the Aqueous Phase
In many studies I've read, bisphosphonates (first image) are said to bind remarkably onto bone hydroxyapatite (second image) crystals. It was reported by J. Roelofs et al., (2008) that this mechanism occurs in a bidentate fashion, where both phosphate groups act as a monodentate ligand to coordinate with the hydroxyapatite. But how specifically does this occur? In other words, what atoms in the bisphosphonate act as the electron donors, and what atoms in the hydroxyapatite act as the electron acceptors? I suspect that the electron donor in the bisphosphonate are the oxygen groups after being deprotonated, though verification would be appreciated!

￼ ￼ ￼

Image 1 depicts the structure of a generalized bisphosphonate
Image 2 depicts the ball-and-stick structure of hydroxyapatite, from Rajitanapanich et al., 2014.
Image 3 depicts an additional representation of hydroxyapatite, from Selvaraj et al., 2019.

One question too much. also 2nd picture is pretty bad.

@Mithoron Apologies. The post has now been updated to only contain the primary question, and additionally, a clearer 2nd and 3rd picture.


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Identification of endo and exo products in this Diels-Alder reaction
This synthesis example is shown in the book Cycloaddition Reactions in Organic Synthesis by  S. Kobayashi and K. A. Jorgensen (link), on p. 15:

8 is a chiral catalyst, which can be ignored, in the context of my question. The product shown is the exo product. However, I am unable to understand or visualise why it is the exo product. It does not seem apparent to me that the transition state leading to this product would have the aldehyde group being further away from the $\pi$ system of electrons. I would greatly appreciate assistance in helping me understand this.

This post on the DA reaction may be of interest to you.

The product you have drawn looks like an exo-product. Conformation 1 leads to an exo-transition state that produces exo 2 as the enantiomer shown. With an achiral catalyst ent-exo 2 would also be formed as a racemic pair. To obtain the racemic endo product, flip the diene 180o in conformation 1 to obtain endo 2. For the enantiomer, flip the unsaturated aldehyde 180o. If you have access to the literature, research papers by Herbert House or William Roush[1] on this type of Diels-Alder reaction. I believe they saw trans-fusions (endo products) for the ring system.

Addendum: (12/7/2022)
It is likely that the formation of the exo-product you cite is a typographical error.
In this example,[2] a chiral ruthenium catalyst forms the endo-product, which has the trans-fusion.

Reference:

Roush, W. R.; Peseckis, S. M. Intramolecular Diels-Alder reactions: the angularly methylated trans-perhydroindan ring system. J. Am. Chem. Soc. 1981, 103, 6696-6704; doi 10.1021/ja00412a027.
Thamapipol, S.; Bernardinelli, G.; Besnard, C.; Kundig, E. P. Chiral Ruthenium Acid Catalyzed Intramolecular Diels-Alder Reactions. Org. Lett. 2010, 12, 24, 5604-5607; doi 10.1021/ol1019103

@Buttonwood: Thanks for the fix. The hour was getting late. ;)

Thanks for pointing out that the product shown is in fact the exo product. I have made the necessary edits to my question.

Though not part of every citation style, I perceive the provision of a more complete set of bibliographic data -- doi to get to the landing page of the publication plus indication of the title of the article -- as helpful for learning. In this regard (my point of view) chemistry.se benefits from the pattern equally seen e.g., on Master Organic Chemistry (example), or papers in J.Chem. Educ. (example).

With the Chemistry Reference Resolver living in the browser, it often only is a single click to eventually get to a place to collect the missing data, too.

For me, the most foolproof way to identify the endo and exo products is to look at the stereochemistry in the product. Consider first a standard intermolecular Diels–Alder reaction:

I labelled the substituents on the diene $\mathrm{R^t}$ and $\mathrm{R^c}$, for trans and cis respectively, to describe their position with respect to the single bond in the middle of the diene.
What we can see from this is that the endo product has the electron-withdrawing group (EWG) placed cis to $\mathrm{R^t}$ (and trans to $\mathrm{R^c}$). (OK, sorry, the naming is a bit confusing; but I'm sure you can make sense of it.) The opposite would be true of the exo product: in that case, the EWG would be cis to $\mathrm{R^c}$ (and trans to $\mathrm{R^t}$).
The intramolecular case is harder to draw a transition state for (user55119 has done it nicely!), but this doesn't change the conclusion drawn above. In your case, the diene only has a single substituent which is of the $\mathrm{R^t}$ type (the substituted double bond is trans configured). This substituent ends up becoming the five-membered ring:

It is quite clear from the final product that the EWG is trans to the $\mathrm{R^t}$ substituent. So, in agreement with user55119's analysis, the product shown is actually exo.

"What we can see from this is that the endo product has the electron-withdrawing group (EWG) placed cis to $\ce {R_t}$ (and trans to $\ce {R_c}$)." Is this just like a statement of fact or is there supposed to be an explanation too as to why it is like that? Is it supposed to be explained by the orientation of the groups in the transition state?

@TanYongBoon That's a literal description of the stereochemistry in the product, which in turn follows from the transition state.

So I can imagine the two $\ce {R_c}$ groups to be like a methylene group in cyclopentadiene and hence, we always see the methylene bridge above the 6-membered ring formed in the DA reactions of cyclopentadiene? I am just trying to find some way to help myself remember this. Is what I suggested regarding cyclopentadiene and the relation to the $\ce {R_c}$ groups correct?

Also, what if $\ce {R_c = R_t = H}$?

@TanYongBoon First comment: In cyclopentadiene, yes, the CH2 group would take the place of R_c. When drawn in this way, then yes, the bridge will always end up "above" the 6-membered ring. However, that's not necessarily true, because if the dienophile approaches from above the diene rather than below as I've drawn, then the bridge will actually end up below the six-membered ring. That, therefore, isn't a reliable way to remember this. What is reliable is the relative stereochemistry between the EWG and the groups. Regardless of where the dienophile approaches from, [...]

[...] the endo product will always have the EWG trans to R_c, i.e., the EWG is opposite the methylene bridge. Now, to be honest, I don't suggest memorising this at all. Why bother? You can draw it out whenever you need to. If you aren't comfortable with drawing out the TS of a D–A reaction and obtaining the stereochemistry from that, I'd strongly recommend getting better at that, before trying to look at more complicated examples. Understanding it is better than memorising odd mnemonics like mine. Lastly, if R_c = R_t, then there are no endo or exo diastereomers, there is only one product.

Is there a rotation of the $\ce {R_c}$ and $\ce {R_t}$ groups during the reaction? Certainly there must be a rotation from being in the plane of butadiene right? Wouldn't the orientation of the groups in the product depend on how the two groups rotate?

@TanYongBoon Yes, they have to rotate, but it's physically impossible to rotate "the other way". The moment the single bond is formed, the stereochemistry is fixed. You just need to "unfold" the six-membered ring, and there aren't two ways to go about it, there isn't any conrotatory/disrotatory thing going on here.

One last question, can I say that for the example reaction you have provided, whether the dienophile approaches from top or bottom, the endo product is the same?

To be absolutely honest, the rotation either way seems physically possible to me. So I am not able to understand why you say that the rotation in one of the directions is physically impossible

@TanYongBoon Whether the dienophile approaches from top or bottom affects which enantiomer you get. It doesn't affect which diastereomer you get, i.e. the relative stereochemistry, and that's what exo/endo are about, they are labels for diastereomers. The endo product will always be the same diastereomer, although in general you will get both enantiomers of it. I'm afraid, I cannot give you a tutorial on Diels–Alder stereochemistry. Please look it up in a textbook. It's an important enough topic that any decent textbook will cover it. I vaguely recall Clayden has a good section on this.

@TanYongBoon Maybe this might help: https://i.stack.imgur.com/NLxaf.jpg (1) Reactants. (2) The DA product. Note that I haven't dont any "rotation" of the groups yet; this is just what happens when you form the necessary bonds. (3) Same DA product but viewed from a different angle. (4) Rotated product / redrawn stereochemistry of the bottom carbon to match conventional chemical structures. At no point in this, is it possible for C and T to change places. In particular, before the reaction happens they can't interchange (double bond) and after the reaction happens the stereochem is fixed.


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Balancing OH- reduction to H2
I have to balance the following redox equation:
$\ce{B + NaOH -> NaBO2 + H2 + Na2O}$
Here I identified B as the species which undrgoes oxidation and wrote the oxidation half-reaction:
$\ce{4OH- + B -> BO2- + 2H2O + 3e-}$
The only element whose oxidation number is reduced is H from +1 in $\ce{OH-}$ to 0 in $\ce{H2}$. So the (unbalanced) reduction half-reaction should be something like:
$\ce{2e- + 2OH- -> H2}$
How does one balance that?

If you write NaOH as OH- why not towrite Na2O as O^2- ?

Does this answer your question? Balancing reaction between copper and nitric acid given the ratio of formed nitrogen oxides

I think it could be: 2$\ce{B}$ + 6$\ce{NaOH}$ $\rightarrow$ 2$\ce{NaBO2}$ + 3$\ce{H2}$ + 2 $\ce{Na2O}$

Per this source: "Boron reacts with fused sodium hydroxide to form sodium borate and hydrogen."  https://chemistry.stackexchange.com/questions/67434/reaction-of-boron-with-alkali

It would be rather Na3B3O6, as metaborate is a cyclical trimer with alternating O and BO-

Your last equation can be balanced by writing : $\ce{2 e- + 2 OH^- -> H2 + 2 O^{2-}}$

As suggested by Poutnik in the comments I considered $\ce{O2-}$ in the reduction half reaction which then became:
$\ce{2OH- + 2e- -> H2 + 2O^2-}$
After multiplying each half-reaction by the number of electrons exchanged by the other one and adding back the spectator ions the balanced equation becomes:
$\ce{14NaOH + 2B -> 2NaBO2 + 3H2 + 6Na2O + 4H2O}$

Subtract from it reaction $\ce{8 NaOH -> 4 Na2O + 4 H2O}$ that does not happen in a condensed phase and you will get $\ce{6 NaOH + 2 B -> 2 NaBO2 + 3 H2 + 2 Na2O}$


text:
Reaction between iron oxides to form magnetite
In the notes of a course it is indicated that the reaction of formation of magnetite from iron oxides can only take place if the partial pressure of oxygen is higher than 1 nanobar.
$\ce{FeO}$ + $\ce{Fe2O3}$ $\rightarrow$ $\ce{Fe3O4}$
I have tried to search in the scientific literature for the origin of this fact, but I have not found anything. I assume it will be a kinetic limitation, since the reaction is thermodynamically favored at any temperature. Am I correct?

..but not in any oxygen partial pressure due $\ce{2 Fe3O4 <=> 6 FeO + O2}$.

@Poutnik And, from this equilibrium that you propose, how could you calculate the minimum pressure for the magnetite formation reaction to take place?

You cannot without more data.

What data would I need? I would be interested in learning how to perform such calculations. @Poutnik

The is not so simple. You might want to look at the equilibrium phase diagram.. Moreover, $\ce{FeO}$ is thermodynamically unstable below 575 °C and disproportionate: $\ce{4FeO ->[\Delta] Fe + Fe3O4}$


text:
what's the importance of "s" character in hybrid orbitals?
I am a high school student and I am learning about hybridization, my teacher told me that it is due to the "s" character in the hybrid orbital that they can arrange themselves so that the molecule faces minimum repulsion {lone pair-lone pair, lone pair-bond pair and bond pair-bond pair} but I am a little confused as "s" orbital is uniformly distributed throughout the space in a spherical form so then even if the hybridization is say "sp2" involving mixing of "pz","px" and "s" orbital then why those three orbitals can arrange in x-z plane only? I know they have "pz" and "px" character but still due to s character can't it do so? or is it a rule that hybrid orbitals can show its s character by only arranging in planes of p or d orbitals involved? as I am learning it at my level it seems that its only qualitative talk but does it have any mathematical explanation also  proving that it can't arrange in x-y or y-z plane? or I have to just memorize it?

Sorry, but your second half of your question doesn't really make much sense. Better start at the beginning. The s-orbital has no influence on where the hybrid orbitals point. It's the p-orbitals which point in certain directions, and the hybrid orbitals inherit this "directionality" from whichever p-orbitals they comprise.

The coordinate system is arbitrary, you are assigning the axis. So if you assign x and y along the p orbitals you are using for the hybridisation, it follows that all orbitals involved are within that plane. Whenever you have a planar arrangement, you will be able to assign the coordinate system in a way that the plane is spun by two axis. Also important: hybridisation models involving d-orbitals are most likely based on outdated model systems and can be explained with simpler approaches.

basically if only p orbital is deciding the orientations of hybrid orbitals then what's the purpose of mixing s orbital?

@happy I think that’s actually a very interesting question to ask; but the point is kind of lost amongst your main post, which still doesn’t make sense.

Your questions can be divided into the following two parts:

Since "$\ce {s}$" orbital is uniformly distributed throughout the space in a spherical form, how come your teacher told you that in the $\ce{sp^2}$ hybridization, it is due to the "$\ce {s}$" character in the hybrid orbitals that they can arrange themselves so that the molecule faces minimum repulsion {lone pair-lone pair, lone pair-bond pair and bond pair-bond pair}?
Why do the three $\ce{sp^2}$ hybridized orbitals involving mixing of "$\ce {p_z}$","$\ce {p_x}$", and "$\ce {s}$" orbitals arrange in the $xz$-plane only? Since the "$\ce {s}$" orbital with a spherical form points in all directions, why can't the $\ce{sp^2}$ hybridized orbitals point in other directions, e.g., in the $xy$-plane or the $yz$-plane? Are there any mathematical explanations for this?

What you are asking is basically that, since "$\ce {s}$" orbital is spherical, how can it dictate orientation? And if it can dictate orientation, why can't it cause the orientations to be out of the ones of the components?
Let's start by watching a video on interpreting $\ce{sp^2}$ hybridization based on Quantum Chemistry taken from the YouTube: Quantum Chemistry $\ce{sp^2}$ hybridization. To present a self-contained answer, we need to do a little mathematical homework to arrive at the expressions for the three $\ce{sp^2}$ hybridized orbitals shown in the video.
To that end, let's express the $\ce{sp^2}$ hybridized orbitals as linear combinations of the three original atomic orbitals as:
$$\psi_{sp^2_1}=a_1\psi_{2s}+b_1\psi_{2p_z}+c_1\psi_{2p_x}$$
$$\psi_{sp^2_2}=a_2\psi_{2s}+b_2\psi_{2p_z}+c_2\psi_{2p_x}$$
$$\psi_{sp^2_3}=a_3\psi_{2s}+b_3\psi_{2p_z}+c_3\psi_{2p_x}$$
where the three atomic orbitals are the three basis functions. The coefficients in the above-given equations can be obtained by using the normalization and orthogonalization of each atomic orbital and hybridized orbital, and the property that the "$\ce {s}$" orbital contributes $\frac{1}{3}$ and "$\ce {p}$" orbitals contribute $\frac{2}{3}$ to the integrated probability density. This is trivial and can be done on back of an envelope so I will just leave it to interested students.
Note that the solutions are not unique and depend on the coordinate system chosen. However, transformation between different coordinate systems does not change the electron densities of the quantum mechanical system. For convenience and the purpose of illustration, we can place the three atomic orbitals in the way as shown in the figure given below (left panel). As you pointed out, the "$\ce {s}$" orbital is spherical. In addition, the "$\ce {p_z}$" and "$\ce {p_x}$" orbitals point along the $z$ and $x$ Cartesian axes, respectively. Then you can understand that the "$\ce {p_y}$" orbital (not shown) should point along the $y$ Cartesian axis, which by the right-hand rule should point out of the screen. By aligning the atomic orbitals that way, we will obtain the solutions as shown in the video.
Putting the obtained coefficients back to the above-given equations, the three hybridized orbitals are expressed as linear combinations of the three original atomic orbitals in the following form:
$$\psi_{sp^2_1}=\sqrt{\frac{2}{6}}\psi_{2s}+\sqrt{\frac{4}{6}}\psi_{2p_z}+\sqrt{\frac{0}{6}}\psi_{2p_x}$$
$$\psi_{sp^2_2}=\sqrt{\frac{2}{6}}\psi_{2s}-\sqrt{\frac{1}{6}}\psi_{2p_z}+\sqrt{\frac{3}{6}}\psi_{2p_x}$$
$$\psi_{sp^2_3}=\sqrt{\frac{2}{6}}\psi_{2s}-\sqrt{\frac{1}{6}}\psi_{2p_z}-\sqrt{\frac{3}{6}}\psi_{2p_x}$$
As can be seen from the figure shown below (right panel), the three hybridized orbitals point to right, upper left, and lower left, respectively, with their corresponding labels shown. So you can see the key point here: the atomic orbitals are base vectors pointing along certain directions as defined by the coordinate system, and hybridized orbitals are also vectors with orientations determined by the coefficients and the base vectors in the linear combinations.
For the first part of your question, it would be better to say that it is due to the "participation" of the "$\ce {s}$" orbital in the hybridization, the hybridized orbitals can arrange themselves in the way they are, i.e., with certain directions and 120 degree apart. The "$\ce {s}$" orbital is spherical so it does not have any preferred direction. You can say that it does not dictate the orientation of the hybridized orbitals, but its participation in the hybridization causes the mixing of the three original basis functions and then the "cards" need to re-shuffled, so to speak, to give new orbitals with preferred directions.
For the second part of your question, as can be seen from the mathematical expressions of the three hybridized orbitals given above, there is no basis function coming from the "$\ce {p_y}$" orbital. This is the reason that the three hybridized orbitals do not point in the $xy$-plane or the $yz$-plane that involves the $y$ axis. For the same reason, the two $\ce{sp}$ hybridized orbitals only remain on the axis of the "$\ce {p}$" orbital, depending on whether it is "$\ce {p_x}$", "$\ce {p_y}$", or "$\ce {p_z}$" orbital. In contrast, the four $\ce{sp^3}$ hybridized orbitals will point in four different directions forming tetrahedral geometry. The difference between these different hybridizations can be understood as 1D, 2D, and 3D.
Here is a little bit more information regarding the 120 degree. Since you are a high school student, you can use geometry and algebra to prove that's the case by using the coefficients shown above. Basically, $\alpha = arctan(\frac{\sqrt{\frac{1}{6}}}{\sqrt{\frac{3}{6}}})=arctan(\sqrt{\frac{1}{3}})=30$ degree. Likewise, $\beta$ is also $30$ degree. Then it can be readily shown that the three hybridized orbitals are 120 degree apart. One more thing, not for now but when you enter college if you are still interested, you can practice to prove that each of these three hybridized orbitals is indeed normalized and they are orthogonal to each other.

I have down-voted this answer because the most crucial information is probably hidden in the video. Answers on this platform should be self-contained, so please include this information here and link to the video for context.

as you said that there is no py component in the hybrid orbitals that's why they don't have component towards y axis so it means that u are saying that the orientations are decided solely by p orbitals? then what's the purpose of mixing s orbital always?

like in ammonia why can't we mix all three p orbitals such that each hybrid have characters of all three and then rearrange for the position of minimum repulsion?

@Martin-マーチン, following your comment, I revised the answer. However, the derivation for the coefficients is very trivial and can be done on back of the envelope so I will just leave it to interested students.

@happy, in the example given above, the atomic orbitals are base vectors pointing along the x, y, and z axes, and hybridized orbitals are also vectors with orientations determined by the coefficients in the linear combinations. For ammonia, it is $\ce{sp^3}$ hybridization, where s has to participate, otherwise the three p orbitals cannot mix with each other since they are orthonormal. If you force the three p orbitals to mix without s, they will remain as the original degenerate p orbitals.

Apparently the most crucial part is not in the video. The reference system, borane, is planar. A planar system is best described with sp² orbitals, always provided you would need such a model in the first place. Since the choice of coordinate system is arbitrary, and in this example chosen to be Cartesian, there is always a choice where z is zero. And hence would be the coefficient for the orbital pointing along this axis. Is of course possible to use "all" p orbitals, but it will always reduce mathematically to two.

Also, I recommend looking up Bent's rule and the excellent Coulson's theorem. It elegantly describes how the angle between two spx orbitals is dependent on the "mixing" of the hybridisation coefficient x. We have some great posts on this site about this. (Always provided that the model then used is actually applicable.)

Figure is modified to have the directions of the atomic orbitals defined. The answer is more general with application to systems in addition to borane. I think it is sufficient to answer the question and will pause now.


text:
Ozone depletion with the OH radical

$\ce{OH*}$ radicals are known to deplete the ozone layer. Fill out the gaps below to complete the propagation stage:
\begin{align}
  \ce{? + O3 &-> ? + O2}\\
  \ce{? + O3 &-> ? + O2}
\end{align}

So I am an A-Level chemistry student in the UK. We have covered ozone depletion in the context of free radical substitution with examples as follows:
\begin{align}
  \ce{ClCH3    &->[\text{U.V.}] Cl* + *CH3}\\
  \ce{Cl* + O3 &-> ClO* + O2}\\
  \ce{ClO* + O &-> Cl* + O2}
\end{align}
I know that we need the overall equation $\ce{O3 + O -> 2O2}$. However, all the examples covered in class have been of the form: \begin{align}
  \ce{X + O3 &-> Y + O2}\\
  \ce{Y + O  &-> X + O2}
\end{align}
Clearly, the question provided at the top does not follow this form. We have never seen depletion reactions of the form $\ce{? + O3, ? + O3}$. Some internet research has lead to the idea that:
$$\ce{HO2, HO3, H2O3}$$
Are all chemicals that can occur in these atmospheric reactions. We have not studied any of them, but my best guess for an answer to the question is:
\begin{align}
  \ce{*OH + O3  &-> HO2* + O2}\\
  \ce{HO2* + O3 &-> HO3* + O2}
\end{align}
Is this right?

Yes, you are pretty right. This is $\ce{HO_x}$ chain reaction.

@NilayGhosh Thank you! Are there are other radicals which induce reactions not of the form ... + O$_3$, .... + O, do you know? It would be good to be prepared when something weird like this comes up

@NilayGhosh Actually, more importantly: what is the overall equation? In the in-class examples, we've ignored the radicals since they appear at the start and end, i.e. they have an overall spectator effect, but in this one the HO turns into HO3, so would it be: $$\ce{HO + 2O_3}\to\ce{HO3 + 2O_2}$$

Atmospheric radicals follow free radical reaction mechanism where the generated radical is used up in the next reaction cycle and with each cycle, ozone reduces. Here is a pictorial explanation. There are $\ce{NO_x}$ as well as $\ce{Cl}$ chain reaction 2. Trioxidane is unstable and decomposes to singlet oxygen and water. The singlet oxygen reacts with HO2 to oxygen. So, trioxidane don't stand as single species in the mechanism.

https://chemistry.stackexchange.com/q/9517/4945 seems to be relevant. Also the Nobel lecture should be freely accessible and have quite a bit discussion on this.

@Martin-マーチン I was just confused with the hydroxide radical in particular - I am not familiar with $\ce{HO_x}$ radicals and I had to guess the solution. This seemed like a good place to find someone to confirm or deny, as Nilay Ghosh so kindly did.

I understand that, and the question is clear, and needs a good answer. In any case I recommend reading the Nobel lecture, is here now https://www.nobelprize.org/prizes/chemistry/1995/rowland/lecture/ and I should update the other post.

Another likely correct answer that obeys the constraints of the equation:

$\ce{X + O3 → Y + O2}$

and forms likely stable radicals (as opposed to your suggested answer noting this reference "On the stability of the elusive HO3 radical") for two members of the chain associated with an ozone depletion cycle which results in the net reaction:

$\ce{2 O3 → 3 O2}$

is namely, the HOx ozone depletion cycle, where I claim that x may assume, at least functionally, the values of 0 and 1 as to oxygen presence. That is, possible acceptable answers could be $\ce{.H}$ and $\ce{.HO}$.
Support, per a source citing the HOx chain with x=1, namely: $\ce{.HO}$ discussed briefly here, presents the reaction chain for $\ce{.OH}$ as:

$\ce{.HO + O3 → .HO2 + O2}$

$\ce{.HO2 + O → .HO + O2}$

And, by extension for x=0 (no oxygen, just hydrogen):

$\ce{.H + O3 → .HO + O2}$

Source: "Theoretical study of the H + O3 ↔ OH + O2 ↔ O + HO2 system" where to quote:

The key features of the H+O3 potential energy surface are expected to be transferable to the X+O3 systems where X=Cl, OH, NO, and NH2.

And the associated next step reaction:

$\ce{.HO + O → .H + O2}$

Source: "O + OH → O2 + H : A key reaction for interstellar chemistry. New theoretical results and comparison with experiment".
which completes the cycling of $\ce{.H}$ as required.


text:
If the reaction is spontaneous, is it the ΔS surroundings that is always positive or ΔS universe?
If the reaction is spontaneous, is it the ΔS surroundings that are always positive or ΔS universe?
I tried researching this on Google. However, the answers that I have found vary with different explanation.

Hint: What would Delta G say about that? Consider spontaneous but endothermic processes.

It can be also said $$\Delta S_\mathrm{universe} \ge - \frac { \Delta G_\mathrm{r}} { T}$$ for constant T,p.

If a reaction is spontaneous, is decided by ΔG= ΔH-TΔS. The value needs to be negative. The second law of thermodynamics tells us that the entropy of the universe always stays the same or increases. Sometimes, for a chemical reaction with negative ΔG, ΔS is negative, but ΔH is as well, that means that the entropy of the system we are describing decreases. How is this compatible with the law I stated earlier? Well, the entropy of the system decreases, but the heat dissipated to the surroundings increases the surroundings’ entropy again, so that the second law of thermodynamics is valid. Otherwise, if a reaction has positive entropy change, it’s evident that is outweighs the (sometimes) positive enthalpy change. Hope that I could clear the doubts, if you have any questions feel free to ask.


text:
Does Sodium Chloride in the gas state meet the IUPAC definition of molecule?
Sodium chloride in the gas forms $\ce{NaCl}$ monomer and $\ce{Na2Cl2}$ dimer. So, no lattice, and those formulae show the number of atoms in the ionic compound, so not just an empirical formula that shows a ratio. So in that sense those are like molecular formulae.
Does sodium chloride in gas state meet the following definition of molecule  in the IUPAC gold book?

An electrically neutral entity consisting of more than one atom (n>1). Rigorously, a molecule, in which n>1 must correspond to a depression on the potential energy surface that is deep enough to confine at least one vibrational state.

Regarding the "electrically neutral entity" aspect of the definition, I understand that it means discounting any partial charges. Sodium chloride has no partial charges, only formal charges, and they balance, so it meets the definition. In the case of e.g. $\ce{H2O}$ , which has partial charges on hydrogen and oxygen then whether or not the partial charges cancel out,  they'd say that there would be an overall neutral charge on $\ce{H2O}$ because only formal charge is counted. And $\ce{H2O}$ is of course considered by all to be a molecule and thus considered an electrically neutral entity.  So Sodium Chloride meets the "electrically neutral" part of the definition.
And I understand that there is a concept of a zwitterion, which is composed of atoms that have a mix of charges but overall neutral charge, so it's not an ion, it's merely composed of ions, but since it has an overall neutral charge, it's a molecule. So, electrically neutral entity is meant to refer to overall charge as neutral, rather than each individual atom.
And Na+Cl- would be electrically neutral since it's an ionic compound,  compounds are electrically neutral, so it's not an ion / polyatomic ion,  it's ionic in the sense of composed of ions.
So that covers the electrically neutral aspect of the definition.
So then I wonder, does $\ce{Na+Cl-}$  meet the other part of the definition, i.e. the part that states "a molecule, in which n>1 must correspond to a depression on the potential energy surface that is deep enough to confine at least one vibrational state." ?

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

Certainly the electrostatic attraction between the ions can hold the $\ce{Na+Cl-}$ entity together when it vibrates, so yes, this does satisfy the vibrational stability definition of a molecule.
So we have a molecule. Given the ionic bonding I would call it an ion pair, which is a specific type of molecule.
Such ion-pair molecules are not as far-fetched as gaseous sodium chloride. We see them, in solvated form, for instance when magnesium sulfate is dissolved in water.

Thanks, though I notice on this question https://chemistry.stackexchange.com/questions/162004/what-is-the-structure-of-sodium-chloride-in-the-liquid-state   you commented  "I would not really call the gas-phase entities "molecules", or at least that is not the most specific term. The monomer is more lime an ion pair, and the dimer would be more like a larger cluster of ions. –
Oscar Lanzi
"   Whereas  in this answer to this question here,  you seem to take the view that not only they meet that definition of molecule, but you take the view that you consider it to be a molecule.

I wouldcsay "molecule" is not specific, so would prefer "ion pair".

Would even a    solid sodium chloride lattice,  fit the definition?

Looking at the definition
https://goldbook.iupac.org/terms/view/M04002

An electrically neutral entity consisting of more than one atom (n>1).
Rigorously, a molecule, in which n>1 must correspond to a depression
on the potential energy surface that is deep enough to confine at
least one vibrational state.

Looking at "in which n>1 must correspond to a depression on the potential energy surface that is deep enough to confine at least one vibrational state."
An ionic monomer and ionic dimer or even a large ionic crystal, would fit that part of the definition.
Speaking to a chemistry guy, they mentioned to me that they think that part of the definition is just to rule out very weak forces like van der waal forces.  So, the ionic bonds within an NaCl crystal or NaCl dimer or monomer, would meet that part of the defintion.  (I mention NaCl crystal for the sake of understanding the principle. Of course NaCl in gas form won't form anything big).
So now that makes a smaller question.
Does NaCl in gaseous state, meet this part of the definition

An electrically neutral entity consisting of more than one atom (n>1).

There's a very significant word that I overlooked when asking the question. The word "entity".
That's where an NaCl crystal gets ruled out. And even an NaCl dimer gets ruled out.  Since a crystal and I think a dimer too, has an entity NaCl. But wouldn't be considered to be an entity itself.
The word entity is a bit ambiguous 'cos it can mean "molecular entity", or it can include the term "formula unit"(which is a physical object pointed to by an empirical formula).
An ionic monomer is certainty meeting the defintion since it's an entity, even,  a molecular entity.. As it's an "ion pair"

Molecular Entity
https://goldbook.iupac.org/terms/view/M03986
Any constitutionally or isotopically distinct atom, molecule, ion, ion
pair, radical, radical ion, complex, conformer etc., identifiable as a
separately distinguishable entity. Molecular entity is used in this
Compendium as a general term for singular entities, irrespective of
their nature, while chemical species stands for sets or ensembles of
molecular entities...

As for a question in comment, what about an ionic NaCl crystal. (And maybe one could consider, if an ionic NaCl dimer is too).
In the case of an ionic crystal.  I'm not entirely sure.
And also with the term entity, do they mean just molecular entity. In their definition of "Chemical substance" https://goldbook.iupac.org/terms/view/C01039  they mention "formula unit" as an entity.
I did hear a suggestion that an ionic crystal is a species, but an expert I spoke to said it isn't. It might be possible to say that there is a species NaCl present in the crystal.  Another thing I heard which I don't think is correct, is that if it's a species it's not an entity, I don't think that's correct, 'cos that which is a molecular entity can be a species and vice versa, for n=1. So it comes down to is it an entity. If it has loads of atoms then it'd have an empirical formula, and the thing referred to by an empirical formula is known as a formula unit. A formula unit meets a definition of entity though not molecular entity.
If a crystal is considered to be a molecule then a dimer would be.
If it isn't, then maybe a dimer isn't either.
I will do some more looking into things re a crystal. And, likewise, for a dimer. (The situation for a dimer should be related in principle to the situation for a crystal).
But an ionic monomer is a molecule, it meets the "entity" definition along with the rest of the definition.


text:
Electrophilic substitution on disubstituted benzene, when deactivating groups are at para
As far as I have been able to understand this, the sites of electrophilic substitution are majorly decided on basis of the fact that "Where the electron density is more, electrophilic substitution occurs there".
This resonated with my intuition, but while looking at it for 4-nitrobenzaldehyde, my teacher told me that major site of substitution is meta with respect to aldehyde.

Pondering over it, I thought that since the reaction proceeds such that positive charge develops on the position adjacent to the site of substitution, thus it must be a location where electron density is more, i.e. meta to the nitro group. But this "ended" my intuition and I am doubtful whether such a reasoning is correct or not.

I did a quick search of Chem. Abst. for monohalogenation ofp-nitro benzaldehyde, benzoic acid and benzoic esters. The only example I found that may not involve a chelation effect with the carboxyl group is the bromination of p-nitrobenzoic acid with a quaternary ammonium perbromide to give 2-bromo-4-nitrobenzoic acid. I do not have access to the Slovenia journal. This result would support your contention.

According to Moodie & Willmer JCS Perk 2 (1992) 229-233 the major products of attempted nitration of 4-nitrocarboxaldehyde are oxidation to nitrobenzoic acids, the minor dinitrobenzaldehyde is 2,4-dinitrobenzaldehyde i.e. meta to the nitro group.


text:
How does cyclopentadienyl anion act as reducing agent?
I am a second year university student, so I'm afraid my chemistry isn't great. I am looking for an explanation of how the $\ce{Cp-}$ anion acts as a reducing agent.
To contextualise - in a simple synthesis of ferrocene, a hydrated salt of $\ce{FeCl3}$ (the hexahydrate) was mistakenly added to a dry solution of LiCp in THF, which had been prepared by reacting n-BuLi and cyclopentadiene under nitrogen. Clearly anhydrous $\ce{FeCl2}$ was what was supposed to be added.
So far I know that a re-protonation of $\ce{Cp-}$ would occur upon addition of a hydrate to regenerate CpH and thus ruin the reaction.
However, the reaction was allowed to continue and some ferrocene was collected by vacuum sublimation, and therefore I know that something present must have reduced the $\ce{Fe^3+}$ to $\ce{Fe^2+}$ in order for some ferrocene to form. I assume this was the $\ce{Cp-}$ anion as I know this is oxidised by the air. I would expect this to be oxidation under basic conditions because of the LiOH created in the re-protonation step.
I am looking for the half-equation for the oxidation of $\ce{Cp-}.$
What does cyclopentadienyl anion oxidise to? Why are the cyclopentadienyl salts air-sensitive?

Isn't it better to describe cyclopentadiene $\ce{<=> H+}$ + cyclopentadienyl anion as an simply an acid-base reaction (no change of electron count for cyclopentadiene/cyclopentadienyl anion). The anion stabilized (aromaticity), and -- contrasting to cyclopentadiene -- not so prone to dimerization. On the other side, the cyclopentadienyl cation would be a considerable uphill battle (because of its antiaromaticity).  And to be specific, ferrocene is an example of a well known, air stable compound, too (formally $\ce{Fe^{2+}}$ and two units of $\ce{C5H^-_5}$).

Huh? No, cyclopentadiene ($\ce{C5H6}$), dicyclopentadiene ($\ce{C10H12}$). There is a Wikipedia article about ferrocene via Grignard reaction as if the metal could dissociate (like famously for the anthracene/Mg/THF complex here). Here, this would lead to Li which could reduce Fe(III) to Fe(II). But (and thus hesitant to post this as an answer), the C-Li bonds are even more polarized than C-Mg; this sets the first reaction in question.

Cyclopentadienyl salts are not air-sensitive. Some complexes are, but it's more of a fault of the central atoms than the ligand afaik.

Could you give an explanation as to why something like TlCp is less air sensitive @Mithoron - as per a couple of sources I've found that say this is sometimes used. What is it about the metal cation

equally why was LiCp suggested for this lab and not NaCp or KCp - why is the lithium Cp used most often

@MatildaWillcox LiCp is commercially available (as a solution), though it isn't bread and butter in every ochem lab.  And in case it were not in stock, cyclopentadiene is easily deprotonated by commercially solutions of BuLi (which is a widely used chemical in many concentrations and solvents). Of course, one could cut sodium and use it as a base, too. But its reaction with water has more potential danger than nBuLi. I don't recall we had potassium metal in the lab class (for its oxides, and even more pronounced reactivity vs. water).

Matilda: If you are interested in a second opinion as to what could be occurring here, relating apparently to a growing body of recent research I am discussing elsewhere on the photo induced reactivity of aqueous FeCl3, you are invited to search the internet. Note: my suggestion photo path may explain the initiation process and may present clues as to forming organic radicals and products observed.

Search "ajkoer fecl3 photosensitive" for an extensive review of old and very recent science which may be at play here.

@AJKOER Like in the body of a question, or answer, it is possible to link directly to the doi of a publication. E.g., [primer cyclovoltammetry](https://doi.org/10.1021/acs.jchemed.7b00361) coalesces into primer cyclovoltammetry.

@Buttonwood:  Thanks for the tip.  I do admire your knowledge base and if continue to read my material and research in cross-disciplines, you actually may be filing patents before me on ideas that, frankly are somewhat obvious for someone with a cross-discipline perspective.  Or, perhaps not making big mistakes, as in the case of the billion dollar US navy expenditure on light Aluminum warships complete with a iron attached propulsion system now rapidly dissolving in polluted waters (just a bit of galvanic chemistry).

I'd like to remind everyone that the comment section should not be used for extended discussions. Please use [chat] vor this purpose.

Disclaimer: None of this constitutes proof; it's just an informed guess. Any corrections with reputable sources are welcome.

The simplest proposal seems to be single-electron transfer (SET) from $\ce{Cp-}$ to $\ce{Fe^3+}$. This forms the neutral cyclopentadienyl radical as well as $\ce{Fe^2+}$. This $\ce{Fe^2+}$ then reacts with any remaining unreduced $\ce{Cp-}$ to form ferrocene $\ce{FeCp2}$.
All in all for a balanced equation, we would need three equivalents of $\ce{Cp-}$ per equivalent of $\ce{Fe^3+}$. One equivalent for the redox reaction / SET:
$$\ce{Fe^3+ + Cp- -> Fe^2+ + Cp^.}$$
and two equivalents to form ferrocene:
$$\ce{Fe^2+ + 2Cp- ->FeCp2}$$
The radicals probably go off and dimerise, which leads to dihydrofulvalene $\ce{Cp-Cp}$. (Basically, fulvalene, but with the middle double bond changed to a single bond.)
There is some support for this hypothesis, namely the accidental synthesis of ferrocene from $\ce{FeCl3}$ and $\ce{CpMgBr}$ (see Wikipedia), which Buttonwood also mentioned in the comments. This follows almost the same scheme as outlined above. In the present case we have $\ce{CpLi}$ instead of $\ce{CpMgBr}$, and there are definitely differences between Grignards and organolithiums, but to a first approximation I'd expect similar behaviour.
Organolithiums are known to sometimes react as single-electron reducing agents, too (e.g. lithium–halogen exchange sometimes involves SET) so it's hardly implausible that $\ce{CpLi}$ might behave similarly.

Thank you for being so kind:) That's really interesting. It this how it would oxidise in air as well? I found this on Wikipedia - 'Lithium cyclopentadienide is a colourless solid, although samples often are pink owing to traces of oxidised impurities'

@MatildaWillcox This comment is be even more speculative than my actual post, but... potentially yes? I could see Cp- doing a SET to oxygen (forming superoxide anion). Only difference is that it might react again after that to form oxygenated products. I don't think the pink colour tells us much though. Could be anything.

Given the cited history published in the ACS (see my answer) of the light sensitivity of FeCl3, this answer is well deserving of the provided disclaimer.

I have 4 citations on the photo abilities ascribed to aqueous FeCl3. To deny its possible role here is a losing game. But please, attempt to prove me wrong by adding more air/light exposure to your FeCl3 mix.

Per this source: "The comparison of photocatalytic activity of Fe-salts, TiO2 and TiO2/FeCl3 during the sulfanilamide degradation process", to quote: "Experiments were carried out in the presence of homogeneous Fe-salts (FeCl3 , Fe(NO3)3 , Fe2(SO4)3 , FeSO4 ), heterogeneous compounds (TiO2 ) and a mixture of catalysts (TiO2 /FeCl3 ) under UV-A and solar irradiation. Based on the comparison of several factors...  economic and ecological aspects, FeCl3 is proposed as the optimal catalyst."

@AJKOER I don't doubt that there are iron-based photocatalysts. However I speculate the OP run the reaction in a glass reactor which would block much of light in the range of UV. Except for the cells to record an UV-vis spectrum, teaching labs I have seen so far have few (if any) quartz containers. (The blue LEDs for photoredox chemistry [McMillan group etc.] might be a different story.)


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Do molecules containing enough oxygen to burn themselves exist?
Is there a molecule with enough oxygen in it (for example, something like $\ce{HCOOOH}$) such that it can undergo complete combustion without requiring any additional oxygen? For example the above molecule could self-combust via:
$$\ce{HCO3H -> CO2 + H2O}$$

It depends on how narrowly you define "burn themselves". What you're describing is something that disproportionates into a more oxidized and a less oxidized molecules. Look up "monopropellant" rocket fuels for examples of this, but there isn't necessarily any "burning" going on in many cases.

Such compounds have usually explosive properties.

Why, many. Also, many compounds contain even more oxygen than that. It is just that when they burn, this isn't called burning.

Peroxo compounds are super oxygen rich and serves better as explosives. Have a look: https://digitalcommons.wayne.edu/oa_dissertations/1372/

Ammonium dichromate $\ce{(NH4)2Cr2O7}$ is well known for being able to burn with its own oxygen. If you dip a match into a crucible containing about $10\text{–}\pu{15 g}$ of ammonium dichromate, it will start to burn softly and throw out sparks like an active volcano, according to the equation $$\ce{(NH4)2Cr2O7 -> N2 + Cr2O3 + 4 H2O}$$ This operation is picturesque and not dangerous. In the 20th century, it was shown in high school classrooms. But dichromates are presently forbidden in many countries.

I remember seeing this. A small orange pile magically turns into a big green mound!

Then there is ammonium nitrate: Texas City, Oklahoma City and Beirut inter alia.

@user55119 Of course ammonium nitrate reacts in a similar way. But it is an explosive. It is hard to state that you see it burning

I remember doing this myself. Also, I don't think it is considered not dangerous by the modern standards. Then again, of all self-oxidizing reactions, this one is probably the likeliest to count as burning, for it does not look like anything remotely resembling an explosion.

As shown in the question, to fully oxidize one carbon and two hydrogen atoms you need three oxygens. Such a molecule, $\ce{HCOOOH}$, exists and it's called performic acid. It is used as a bleach and disinfectant and is explosive in high concentrations.

One of the best known examples is nitroglycerin, $\ce{C3H5N3O9}$, which has $18$ oxygen atoms in two molecules whereas only $17$ are required to oxidize all the carbon and hydrogen in those molecules. Not surprisingly:

In its undiluted form, nitroglycerin is a contact explosive, with physical shock causing it to explode. If it has not been adequately purified during manufacture it can degrade over time to even more unstable forms. This makes nitroglycerin highly dangerous to transport or use. In its undiluted form, it is one of the world's most powerful explosives, comparable to the more recently developed RDX and PETN.


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How can I use the MO diagram of hydrogen fluoride to demonstrate that the molecule is polar?

I am trying to explain that HF is a polar molecule using the MO diagram. Normally I say that the F atom has a higher electronegativity than the H atom, which causes a net positive dipole moment.
However now I am trying to use the MO diagram. My rough guess is that because the H atom has one of its electrons moving down to the energy level, it is effectively showing that it is less electronegative?

It's a bit more involved than that; after all, the fluorine atom also has one of its electrons "moving down".
The key is to look at the occupied MOs and to determine whether they are mainly hydrogen-based, or fluorine-based (i.e., which atom contributes more to the MO). This is actually not trivial to determine at all, and in general, the qualitative MO diagram you've attached isn't going to be sufficient.
In this case, however, the bonding MO (labelled $\sigma$) is only constructed from two constituent AOs: one from hydrogen ($\mathrm{1s}$) and one from fluorine ($2\mathrm{p}_z$). The fluorine $2\mathrm{p}_x$ and $2\mathrm{p}_y$ AOs are nonbonding.
Under these circumstances, it's possible to look at the energy levels to indirectly determine the contributions. In this case, the bonding MO is closer in energy to the fluorine AO than it is to the hydrogen AO. This suggests that the contribution of the fluorine AO is larger; often we say that the bonding MO is "centred on fluorine" or "localised on fluorine".* Effectively, the electrons in this MO would belong more to fluorine than they do to hydrogen.
Note that this can't be generalised to more complex molecules and MO diagrams. It only truly works in the specific case where two AOs make up two MOs. Even, for example, carbon monoxide is problematic: it is only formed from two atoms, but its MOs generally have contributions from more than two different orbitals (carbon 2s and 2p, and oxygen 2s and 2p). So this simplification falls apart.

* This can all be formalised properly using quantum mechanics and linear algebra, but I won't do it here. The treatment would be similar to that which I showed here, except that you would need to explicitly calculate the actual eigenvectors. It should follow from this that the lower-energy new eigenvector (bonding MO) will have a larger coefficient of the lower-energy old eigenvector (AO).

Since you've brought up carbon monoxide, I've actually done the qualitative approach you've described here. It's more involved, but still somewhat possible. The whole approach obviously falls apart, right before MO diagrams themselves stop being good visualisations or approximations. https://chemistry.stackexchange.com/a/67001/4945


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Are there any organic compounds with three or more oxygen atoms bonded together in a straight chain?
Recently, there was a question which enquired about some oxygen rich compounds that burn themselves completely and thus not requiring any external atmospheric oxygen. The author came up with the compound $\ce{HC(=O)OOOH}$* which I am not sure if it exists. We already have peracids ($\ce{RC(=O)OOH}$) but I am not sure if there is any acid that has longer oxygen chain than peracid and I am also not sure if it remains acid any longer. Internet does not give me valid result. I know that long oxygen chains are unstable and prone to detonation (as are all oxygen rich compounds) but OP's compound* made me curious.
We have had previous discussions before mainly discussion hydrogen polyoxides, $\ce{HO_nH}$ (which has been detected and produced experimentally), peroxide linkage vs disulfide and bond energy  comparison:

Why does sulfur, but not oxygen, catenate?
Are long oxygen molecules possible?
Why is an S-S bond stronger than an O-O bond?
Why are peroxides unstable but disulfide bridges considered stable? Why are esters stable but thiolesters are unstable?

However I am asking if there is any such corresponding organic compound that has been detected spectroscopically and that it exists even for some microseconds?

*the question was later edited to performic acid, thus its existence is already in question.
**When writing the formula, it looks like four oxygen atoms together in straight chain. However, one of the oxygen exist as carbonyl bond, thus compounds with three or more oxygen atoms in straight chain is in question.

Why don't you start with three in one chain? Or do I misunderstand the chain part?

What's the successor of Beilstein Crossfire?

@Karl Oh the old times of MDL's Crossfire. The GUI underwent multiple revisions since then and now doesn't discern as much as «back then» between Beilstein and Gmelin when starting as Reaxys(TM) by Elsevier. It becomes a rare occasion to search from the terminal (like in the old days via STN/télésystèms Questel) ...

Worth noting that peracids only contain two oxygens in a chain.

One could ask PubChem (e.g., for four oxygen atoms, for five oxyen atoms, etc.  However (my hunch), there are a bit too many properties labeled as computed property, e.g., for this one to think about them as stable/easy to handle in gram scale under ambient conditions.

Something like F3COOOOCF3 should be possible, per analogy to FOOOOF, which is known.

For cautionary and entertaining articles on oxygen "chains', read Derek Lowe: https://www.science.org/content/blog-post/things-i-won-t-work-peroxide-peroxides , https://www.science.org/content/blog-post/not-do-ether-peroxides and https://www.realclearscience.com/blog/2015/06/foof_the_chemical_most_chemists_avoid.html . You'll get a real bang out of these.

@Mithoron FOOOOF is known? What crazy lunatic made that (given how truly horrific FOOF already is)? ;-)

@matt_black Well, guys making FOOF couldn't help themselves and experimented some more, obviously ;)

"I am not sure if it remains acid any longer." It does (if it doesn't blow up), but the extra oxygen atom breaks the conjugation with the carbonyl group and so the peroxy acid will be weaker than the normal carboxylic acid.

@matt_black https://www.researchgate.net/publication/237850820_AN_IMPROVED_PREPARATION_ANALYSIS_AND_SOME_PROPERTIES_OF_TETRAOXYGEN_DIFLUORIDE.

Three oxygen atoms in a row is more common than one might think. In the first stage of the reaction between alkenes and ozone, the ozone is attached to both ends of the double bond to form an intermediate with a five-membered ring (a molozonide). This ring contains the three oxygen atoms still in a row, before it breaks up:

(image source)
In addition, ionic ozonides, $\ce{R^+O_3^-}$, are known with some organic cations $\ce{R^+}$. These are made by metathesis reactions from the ozonides of heavier alkali metals, as described in this answer for $\ce{(CH_3)_4N^+O_3^-}$.


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How do the odd proton-number elements originate?
I have a question about the origin of the odd proton-number elements in the periodic table, please.
As it is generally thought that the big bang produces hydrogen first, and then hydrogen combined to form helium. Helium combines into beryllium, and beryllium into carbon, etc... This means that all fusion atoms have even numbers of protons, does not not?
Then, if radioactive decay involving protons always result in a 2 protons decay, how then can we ever end up with atoms of odd number of protons, such as lithium, boron, nitrogen, etc?

This is question rather for nuclear physics and cosmology. Generally, they are created by the same processes as even-proton nuclei. BTW Li, Be and B are not produced by fusion. See also Nucleosynthesis. Search also for S-process, R-process and Spallation.

@Poutnik  According to the Wikipedia article you link, Li-7 can be produced by the fusion of tritium and He-4; and Be-7 can be produced by the fusion of He-3 and He-4.

@theorist I should say naturally and in majority. You cannot fuse 2 4He to 8Be, unless there is the 3rd 4He to form 12C. 7Be would quickly decay to 7Li.

@Poutnik  The article described these as the "chief nuclear reactions responsible for the relative abundances of light atomic nuclei observed throughout the universe", indicating these are the primary reactions that gave rise to these isotopes.  Yes, the Be-7 would quickly decay (half-life of 52 days), but Li-7 is stable.  I.e., it's saying the main origin of present-day Li is nuclear fusion.

@theorist I agree about the lithium, not about Be and B. The info I wrote may come from other source than the link and may be related  to higher than expected abundance of light elements on Earth, assuming the main their source us due Spallation. It is also supported by the periodic table chart in the link.

Correction: Stable isotopes of Be and B  were not created by nuclear fusion, but by fission by cosmic radiation called cosmic ray Spallation aka X-process.

More generally, it seems the asker may not be aware of beta decay (both minus and plus, but especially the former) as a major process capable of converting nuclei with an even number of protons into a nucleus with an odd number of protons.

@NicolauSakerNeto yes, thank you... i've forgotten about the decay that converts a proton to a neutron. So the odd numbers can be gotten by this beta decay. Thank you!

Even ones too. You cannot obtain stable iron nuclei by just adding alpha particles.

There are several different types of nuclear reaction that can create elements. Small, fresh stars do the one described in the question. But there are different routes to new nuclei that only occur in big stars when they become supernovae (these produce more heavy elements). Even heavier nuclei are thought to come from extreme events like neutron star collisions. Also, many of the elements common on earth are the result of the deaths of the first starts and are not present in the first generation of stars.

Note, in the English language, chemical names (including elements) are capitalized only at the beginning of a phrase.

One of the most significant sources of an odd-numbered element is the CNO cycle that operates to a minor extent in our Sun and more prominently in more massive stars like Sirius. There are actually several CNO cycles, but they all involve carbon, nitrogen and oxygen interchanging through the absorption of protons and the emission of alpha and positron beta particles. The net result is the same as direct hydrogen-to-helium fusion: four protons absorbed into the nuclei during this cycle are converted into one alpha particle and two positrons that are emitted, plus the same energy that we see with the direct fusion. This mechanism is responsible for nitrogen being the second most abundant odd-numbered element in the Universe, behind hydrogen but beating lithium and boron despite the lower atomic numbers of the latter two elements.


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Relationship between transmittance and color of the solution
As I understand, transmittance is light which passes through a material and it determines the color of material (for example green solution transmits green light).
Here’s the transmission spectrum of green mineral.
I have a question. This compound transmits in green (550 nm) and red (650-800) regions, so why is compound green? What’s my mistake?
Thanks in advance!

Try to make from this radiometric spectrum a photometric spectrum, involving correction to the eye spectral sensitivity.

@Poutnik unfortunately, I did not shoot the real spectrum of this mineral, this is a picture from the problem

That is, the main reason is that the eye is more sensitive to green light?

The source of data does not matter. The values can be picked from the chart.

Calculation of luminance of the RGB digital light source is 0.59G + 0.3R + 0.11*B.

@OscarLanzi I thought that transmittance in red is about equal to transmittance in green from plot (they’re on the same line), what’s my mistake?

@Poutnik excuse me, how to determine R, G, B, what’s it? And why can’t we determine color from original spectrum? Does this method involve correction to the eye sensitivity?

Forget RGB if you are not familiar with it. // Color is not objective parameter, but the subjective parameter of human eye+brain - see Photopic vision. Only for monochromatic ( or not too far from it ) light we can directly map the wavelength and color.


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Which ion is more stable in aqueous solution; Co2+, or Co3+?
We know $\ce{Co^{3+}}$ is more stable in complex compounds than $\ce{Co^{2+}}$ is in the same. I am told that for aqueous solution, $\ce{Co^{2+}}$ is more stable than $\ce{Co^{3+}}$, however in aqueous solution too, $\ce{Co^{2+}}$ will have to form complex compounds in water so $\ce{Co^{3+}}$ should still be more stable???

"Stable" is a vaque term with very different meanings, depending on context. The same thing can be very stable and very unstable.

I think stable here implies that they are not converted into other Oxidation state easily.

Then better to express it particularly as strong/weak oxidant/reductant. As another meanings can be thermodynamic or kinetic stability toward hydrolysis.

Welcome to the site. Note, chemical information may be advantageously formatted using on ChemSE with mhchem. Take moment to familiarize with this. You are encouraged to use it in the body of questions, answers, and comments. Because it is something special not all web browsers understand well, do not use it in the title of questions or answers.

Just because a metal can form complexes in different oxidation states with a given ligand, it does not follow that the complexes are equally stabilized in the different oxidation states by the same ligand. We often find that the relative stability of different oxidation states depends on the ligand environment. Copper, for instance, is stable only as copper(II) when solvated by water alone, but copper(I) becomes a stable oxidation state with a softer-base ligand such as thiourea or even chloride ion (see here).
In the case of cobalt, ammonia is especially suitable for stabilizing cobalt(III) but water is not so good, so we find that with water solvation cobalt(III) is highly oxidizing and cobalt(II) is favored instead, whereas with ammonia complexation cobalt(III) is more stable relative to cobalt(II).


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What is the difference between Cholestanol 5alpha and 5beta?
What is the isomeric relation between 5Alpha Cholestanol and 5Beta Cholestanol. I know that those forms appear after Hydrogenation of Cholesterol but unfortunately I cannot find any article or structural molecule of those, I think isomers, in internet, so if anyone can share a structural molecule, show the reaction mechanism or at least tell the isomeric relation I will be very thankful.
N.B: I assume that there is some kind of cis/trans isomerism but don't know exactly where it is. Am I right about that?

When cholesterol is drawn in the traditional way having ring A on the left and ring D on the right, hydrogenation of the 5,6-double bond from the top side produces C5-beta-cholestanol while hydrogenation form the bottom side affords C5-alpha-cholestanol. Beta is cis; alpha trans A/B-ring juncture. Mnemonic: top:bottom::beta:alpha. Words with more letters go together.

Ahha, ok thanks man, that was super helpful. I appreciate your answer:)

As someone who is new to ChemSE, you might want to familiarize yourself with some aspects of Comments.

@Nikimiskata if you got your answer, feel free to self answer it. Self answer are always appreciated and it also help future readers.


text:
Naming of an organic compound containing nitrogen and fluoride
I'm don't have a background in chemistry but I've been doing a project on computer aided molecular design and one of the resulting molecules was:

I'm having trouble naming this molecule let alone research if it is viable. I would appreciate any help in either naming or disqualifying it as infeasible.

Something along the lines of difluoro(methyl)amine. You also might find http://dx.doi.org/10.1021/acs.jpca.5b05157 interesting.

Maybe not so simple considering that the PIN for $\ce{CH3-NH2}$ is methanamine but the PIN for $\ce{CH3-NH-F}$ surprisingly is N-methylhypofluorous amide and not N-fluoromethanamine.

@Loong The again, is OP even asking about PIN, and what about "feasibility"? The question isn't clear.

@Mithoron Usually, we cannot be sure if an OP even knows what a PIN is, so we cannot expect that a nomenclature question will explicitly ask for the PIN. Generally, we recommend answering such questions with the PIN (see meta), but feel free to include other names in your answer.

@Loong thanks for the explanation. You were correct in guessing that I don't know what a PIN is. This was my first post and as I mentioned, I don't have a chemistry background. My question about feasibility was more of a general question. You could have looked at the molecule and said "that's not physically possible". That sort of thing. Thank you again.

The compound is definitely viable and has been subject to a study. The compound was made by reacting dinitrogen tetrafluoride with methyl iodide. (m.p. -114.8 °C, b.p. -16°C and density of 1.099 at -20°C). @Loong The compound was previously named Methyldifluoroamine before changing the name by considering it an amide derivative as per new IUPAC rules.

The IUPAC recommendations on this matter have been changed. Compounds such as $\ce{R{-}NH{-}Cl}$, $\ce{R{-}NH{-}NO}$, and $\ce{R{-}NH{-}NO2}$ are now named as derivatives of amides (see P-67.1.2.6).
The current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows:

P-62.4 N-SUBSTITUTION OF AMINES AND IMINES BY HETEROATOMS
Traditionally, substitution on the nitrogen atom of amines and imines was allowed for all characteristic groups cited as prefixes (see Table 5.1). This approach is maintained in these recommendations, unless a higher class is formed that must be named in accordance with the seniority of classes (see P-41).
This new rule is applied to prefixes such as Cl and other halogen atoms, $\ce{{-}BrO}$ and other acyl similar groups, $\ce{{-}NO}$, $\ce{{-}NO2}$, $\ce{{-}OR}$, $\ce{{-}SO2{-}R}$, $\ce{{-}SO{-}R}$, and even $\ce{{-}OH}$ groups and
chalcogen analogues.

Thus, the preferred name for the compound that is given in the question is N-fluoro-N-methylhypofluorous amide and not N,N-difluoromethanamine.
A similar example given in the Blue Book is $\ce{CH3{-}N{-}(NO2)2}$, which is named N-methyl-N-nitronitramide and not N,N-dinitromethanamine.

It could be that this is still not the correct answer, since there was a correction to the Blue Book concerning the N locants in 2021.


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Do mechanically recycled plastics decompose more rapidly?
Does it mean that mechanically recycled plastics decompose more rapidly in the environment? So even if mechanical recycling doesn't prevent plastics from going to landfills, just puts it off (incinerators aside), it ameliorates the situation by making them go away faster, doesn't it?

The authors suggest that the susceptibility of a polymer to shear-induced degradation is directly proportional to its chain length and degree of chain branching,  a  phenomenon confirmed by  Gooenie et al. and La Mantia et al. for PET. Thus, the length of polymer chain controls degradation kinetics (suggesting that it can autoaccelerate through repeated extrusion)  whereas thermo-oxidative processes are dictated by the both structure of and oxygen diffusion through the polymer matrix. Environmental oxygen reacts with shear-induced radicals and subsequent reactions produce peroxy radicals which propagate radical decomposition. Thus,  high oxygen permeability leads to increased thermo-oxidation rates within the material.

(from here)

Well, it would be very weird if opposite happened. Microplastics are still a problem, though.

@Mithoron Did I interpret that passage right, or did I overinterpret?


text:
Exothermic temperature rise on NaOH + Methanol
When mixing NaOH (w/w 0.05%) with methanol, there is an increase of temperature around 20 C
I wonder if we keep this ratio in large amount of methanol, will there be any spike in temperature increase or it will remain constant?
I am not quite good at math but i assume there must be constant?
But upon quick search, this guy
https://www.youtube.com/watch?v=6Wn-kD5YzqA
At 2:56 mentioning that:

The dissolution reaction is exothermic, meaning it produces off heat. At larger scale it can
even produce dangerously high levels of heat,
getting to above the point of boiling for methanol, which is 140 F

So is he correct or i do? thank you in advance

There would be increase even without NaOH, similarly as with water + ethanol. // Assuming you mix 0.05% w/w water solution of NaOH with methanol.

It's not constant at all, depending on many factors. I wager you're doing it wrong to let it heat up so much with so small concentration.


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What is the relationship between content of elemental magnesium and magnesium ʟ-threonate?
Looking at some supplements descriptions and comparing with recommendations of magnesium intake. From magnesium ʟ-threonate supplement description:

Recommended daily intake (4 capsules) contains 2000 mg of magnesium ʟ-threonate including 160 mg elemental magnesium (42% NRV*).
*Nutrient reference value

Why there is less elemental magnesium in magnesium ʟ-threonate? How to explain the relationship between both forms of magnesium for someone without a chemistry background?

Magnesium ʟ-threonate has the molecular formula $\ce{C8H14MgO10}$ with the molar mass $M = \pu{294.5 g mol^-1}$.
Magnesium molar mass is about  $\pu{24.9 g mol^-1}$.
Magnesium ʟ-threonate has then percental magnesium content
$$\displaystyle\frac{\pu{24.9 g mol^-1}}{\pu{294.5 g mol^-1}} \times 100\,\% = 8.46\,\%,$$
what gives for $\pu{2000 mg}$ of magnesium ʟ-threonate just $\pu{169 mg}$ of magnesium. The big rest belongs to carbon, hydrogen and oxygen content within ʟ-threonate.
The $\pu{160 mg}$ versus $\pu{169 mg}$ discrepancy may be due possible water content of the solid product due its high supposed water affinity.

Is there any central repository of such conversion ratios for FDA labelling or other such purposes?

Not sure, I have followed just general valid principles of chemistry, involving molar masses of compounds. It is nothing specific for FDA.


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Zn galvanic cell without electrolyte?

I performed a classical experiment with Zinc-Copper galvanic cell. The zinc electrode was made from pure zinc wire, the copper electrode was made from pure copper wire, the container was a glass beaker around 100cl. Water used for an experiment was pure, distilled ('demineralized') water you can buy at a gas station.
As a baseline test, I put both electrodes in that pure water without any electrolytes. I measured voltage between electrodes as well as current flowing between them.
To my surprise, the measured voltage and current were V = 0.8V I = 0.4mA, without any electrolytes in the water medium.
The classical explanation says that on the cathode hydrogen ions are taking electrons forming hydrogen molecules and that an electrolyte is needed (typically an acid). Moreover, it says that voltage will depend on the pH of the solution.
How to explain that this cell works even with pure water?
Additionally:

I check that copper is not even required, if I only kept zinc electrode in the water, the cathode could other metal, even barely sunk into water, still giving around 0.7~V potential difference.
Adding acid or salt to the water made higher current flowing (up to 2mA), made visible gas bubbles on zinc electrode (and that happens in acidic solution even without current flowing) but the voltage kept to be around 0.8V disregard of the presence of electrolyte or its concentration.

Water always contains ions,  from water auto-dissociation, from residual ions ( deionization is not 100% ) and from dissociation of acids from aerial CO2, H2SO4 or HNO3.

"the other could other metal" ?

My chemical intuition (As a chemist) would suggest the following aspects to consider: 1) The purity of a "demineralized" water source from a convenience store (demineralization is not the same process as distillation) and 2) as Poutnik commented, atmospheric CO2 can dissolve into the water to some extent, lowering the pH and providing some carbonate anions (But I don't know if that would happen very quickly as it could only enter through the water/air interface, and H2SO4 and HNO3 ought not be too significant unless you are in a very polluted/unhealthy environment, in my understanding).

Yes, of course. I am surprised there was no significant difference between "pure-maybe-not-so-clean" water and solution clearly acidic. I don't have pH meter here now, but I doubt there is no difference between pure water and vinegar in terms of its acidity or ph, right?
At first, I used tap water and the results were as above - my first thought was that it most likely has many contaminations so I went to a gas station for the clean water that is sold to be added for old-school car accumulators. Adding acetic acid, citrus acid or sodium chloride didn't change the voltage level.

It probably won't work on that voltage for long, though.

It might be interesting to put some of the "demineralized" water in a glass bowl and evaporate it to see how much residue there is when dry.

One could have a Zinc/Air battery! The water could be serving like the carbon in  said battery cell, here capturing O2 where the dissolved O2 is the cathodic zone. To test, take distilled water and shaking in a bottle with pure O2. Now, use this in the cell, better results? If yes, it is a form of the Zinc/Air battery discussed here https://en.wikipedia.org/wiki/Zinc%E2%80%93air_battery . Adding NaCl should improve (as was reported) but not in excess (salts out O2).

Note: The battery cell remains absence copper, so it is not the cathode here! Supports my metal-air battery answer. Further, I suspect, using Aluminum/Magnesium metal (pre-treated with HCl or NaOH, and washed in distilled water) will also produce similar results. Adding fine carbon to the water surface also may change results. Have fun!