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Nucleophilic strength of Fluorine vs iodine
Fluorine is less nucleophilic than iodine because it has greater electronegativity. However, in an aprotic solution, it is more nucleophilic than iodine. Why? I read that fluorine is less stable under an aprotic solvent (I don't quite understand why it is less stable; an explanation would be helpful), but anyways, it is STILL more electronegative than iodine, so shouldn't it still be less nucleophilic then iodine?
Another question- What makes something a good leaving group? does it have to do with its nucleophilic strength? Sorry for the three-part question, but I feel like these questions related to each other very well. Thank you


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Why is the relationship between vapour pressure and boiling point of water non-linear?
From my rudimentary understanding of chemistry, pressure should be directly proportional to temperature for a constant mass of gas under the same volume (Charles’s law).

However, in the diagram above, the vapour pressure seems to follow a non-linear relationship with the boiling point temperature of water.
Why is it the case that they are non-linear, and is there a mathematical relationship between the two quantities?

Evaporation of liquid is very different phenomena, compared to heating up a gas at constant volume.

See https://en.m.wikipedia.org/wiki/Vapour_pressure_of_water

It is linear if looked at differently. Check out the Clapeyron Equation in any P Chem textbook. As usual it involves the energy and entropy differences [ AKA Gibbs free energy] between the liquid and vapor.

By other words, there is not a constant mass for the curves above.

If you took out vapor at some moment and if you put it in a container of a fixed volume then its pressure would linearly grow with temperature, as in the Charles' law.
The point is, amount of vapor above liquid is not constant. It grows roughly exponentially with temperature and so does the pressure.
For a narrow enough temperature interval to consider the evaporation enthalpy approximately constant, the vapor pressure dependency on temperature is:
$$p_2=p_1 \cdot \exp {\left(\frac{\Delta H_\text{evap}}{R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)\right)}$$

source: Stack Exchange

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Will benzyl halides undergo SN1 or SN2 mechanism in a polar aprotic solvent?
If benzyl bromide reacts with KF in the presence of a polar aprotic solvent like DMF, will it undergo SN1 mechanism or SN2 mechanism ?
I understand that polar aprotic solvents favour SN2 mechanism but i came across a paragraph in my textbook which said that benzyl and allyl halides show high reactivity towards SN1 reaction (due to the stability of the carbocation formed) however, they failed to mention if it would take place in polar protic or polar aprotic solvents or both.
If it follows SN1, will it be more likely for the nucleophile to get attached to the benzene ring itself as compared to the side chain ?

Probably a mix of both processes but this will never result in nucleophile attack on the aromatic ring.

Why wouldn't it attack the ring ? The canonical structure in which the positive charge is delocalised in the ortho and para positions of the ring would be a secondary carbocation, which should be more stable than a side chain carbocation right ? @Waylander

It never occurs. Attack is always at the benzylic site. This demonstrates that the model of canonical structures is limited in its usefulness.

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Rotten Egg Gas Smell
Background context for question
When I was young, we'd go fishing. And Dad would put a camper on the back of the truck.  One time we came back in after fishing and the camper smelled like rotten eggs.  He used that as an opportunity to teach us how some gases are tasteless/odorless. And as a result that sort of rotten egg smell is added as a safety measure.
Question:
My question is this: What are those gases that are odorless that then have that rotten egg smell added to them?

Propane is a colorless, odorless gas. Ethyl mercaptan is added as a safety precaution as an odorant and is commonly called a "rotten egg" smell. wikipedia.org - propane. Similar is done for natural gas, containing mostly methane.

@Poutnik Thanks. There was just enough doubt in my mind that I thought I should ask and make sure. Thanks for giving me the details. I had thought that Propane might have been one of those, but didn't know that natural gas had an additive too.

tert-Butyl thiol is also used as an odorant for natural gas  https://en.wikipedia.org/wiki/Tert-Butylthiol

BTW, many natural gas suppliers have "Scratch & Sniff" cards to remind the user of the odor. You can request cards from your local supplier, or https://www.ugi.com/forms/request-scratch-sniff-brochures ,  https://www.dominionenergy.com/safety/natural-gas-safety/request-a-scratch-and-sniff and https://outreach.ngridsafety.com/product/ng-scratch-n-sniff-card-june-2019/ . Some state a generic "mercaptan"is used.

@DrMoishePippik Scratch and sniff odored cards. Do I give these to my friends or enemies?  (-:

@Epimanes, relabel and sell as Parfum de Pepé Le Pew, and make a mint... or Symplocarpus foetidus,

Rotten eggs happen much less frequently than 50-60 years ago because of better hen-consumer refrigeration. Most people today have not had the pleasure of smelling them. H2S and low molecular weight mercaptans [AKA thiols] have distinct more or less unpleasant odors tho at low concentrations they can smell sweet, and we can develop tolerances to many at higher concentrations when they can be deadly. H2S is supposedly the active ingredient in rotten eggs; Methylsulfide CH3SH the active gas in human feces, and thioglycolic acid in eau de skunk.
Manufactured gas, no longer used, was a mix of H2 and CO and is toxic. It was flavored with mercaptan at rather high levels; smelling gas then was almost a death sentence, we were trained to open windows and leave immediately. There was also the explosion hazard. Manufactured gas was replaced by "natural", really fossil fuel, gas that is mostly methane CH4. Methane is odorless, not toxic, but it can be soporific, and it can explode. A warning is still necessary tho it seems to me the odorant is now less noxious and less intense than before [It could be my sense of smell has waned.]. It is to protect from explosion, again evacuate and ventilate. Odorants are added to propane and LPG, I do not know about hydrogen. Perfumes are added to many products to overcome odors or enhance odors and Bitrex is added to some drugs to make them less palatable so it works both ways.
Gases in a laboratory or industrial, even a home, locations can be an unseen danger. They can be invisible and relatively odorless and gases such as nitrogen, carbon dioxide, argon, and low MW freons can displace oxygen in the atmosphere of confined spaces. A person entering such space can quickly become unconscious. Even oxygen can be a problem by making usually nonflammable materials flammable. This can be an under-appreciated aspect of Hazmat or fire situations.

This is some pretty fascinating context & details to add. Thanks.


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Point group of trimethyl borate $\ce{B(OCH_3)3}$
What's the point group of trimethyl borate $\ce{B(OCH_3)3}$.
The answer that my teacher gave is $D_{3h}$ but I'm struggling to see the n perpendicular $C_2$ axes and the horizontal plane of reflection.

It's c3h without the perpendicular c2 axis.

You can see some other $C_{3h}$ molecules as this site https://molecule-viewer.com/  where there are also many examples and also of most other point groups. You can add axes and planes etc to help identify.

Depends on what conformation you're talking about. It can be $C_1$, $C_\mathrm{s}$, $C_3$, $C_\mathrm{3h}$, or $C_\mathrm{3v}$. This particular conformation that you provided the image, with $\ce{H-C-O-B}$ angle of $180^\circ$, the point group is $C_\mathrm{3h}$.

The three C2 axes are along the three C-B bonds.
The horizontal plane of reflection is the plane defined by the boron and the three carbon atoms.

There are no C-B bonds here.


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Bioplastic from starch
I am doing research in bioplastic from starch. I have tried many quantities and concentrations of starch, vinegar and glycerol at different temperatures. But I am unable to get a certain result of the desired flexibility. I actually want to make bioplastic jewelry from it but either the products are too hard or not enough hard. I also tried putting agar, but the problem of contamination is seen. Can anyone suggest a suitable solution to the problem I am encountering?

Try starch, glycerol and water ! No vinegar.

tried that as well. Didnt work


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How to calculate oxidation number for silver hexafluorophosphate (AgPF6)?
I'm doing some exercise of assigning oxidation numbers and I am stuck with this compound:
$$\ce{AgPF6}$$
According to the guidelines fluorine is always -1. But how do I know how to split the +6 among the other to elements?

Consider it is a ionic compound, similar to LiPF6, the basis of electrolyte of lithium accumulators. And you must have knowledge about usual ONs of Ag and P.

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

Rules are not enough. There is needed to know at least basics of chemistry of involved elements. Otherwise, rules can lead to formally correct nonsense.

I just googled that Ag usually has an ON of +1, then it's clear that P has to have +5. I didn't have that information. Thank you for your input.

It is assumed you know that

It is probably assumed that you know P before you know Ag, too.

If we ignore the zerovalent forms, then silver has two oxidation states. The more common is +1 while we have the highly oxidizing +2 state which is sometimes used in nuclear reprocessing as an electrochemically generated oxidant.
We know that fluorine can accept one electron to reach a noble gas configeration, so we know that we have +6 changes which need to be shared between the Ag and P atoms. If we treat the substance as purely ionic for the electron counting process. Then
With silver(II) we would need to have P(IV), I am unsure if P(IV) has ever been reported. The common oxidation states for P include P(-III), P(0), P(III) and P(V). So this does not look great as an idea.
With silver(I) we would have five plus charges to deal with, as P(V) is common we can assume that the phosphorus is in the +5 oxidation state while the silver is in the +1 oxidation state.
I think that the latter is by far the most likely to be true.

There's also Ag(III) and lots of ox. states for P, not only IV, but even if Ag(I) somehow wouldn't oxidise lower P state, then Ag(II) or III certainly would.

I suspect the question doesn't demand speculation about the oxidation state of P in PF6, but assumes that the existence of hexafluorophosphate as a stable ion with a single net negative charge is already known.


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What is molecular "tumbling" in NMR?
In NMR textbooks, it is often said that small molecules tumble faster and large molecules tumble slower. What is "tumbling" exactly? When I searched for it, someone says that it is any molecular motions (translation, rotation, vibration), someone says that it is reorientation, and someone says that is is rotation.

Related: https://chemistry.stackexchange.com/q/105614

for 'tumbling' read rotational diffusion, see Wikipedia article.  You can estimate the diffusion coefficient as a sphere with $D=k_BT/6\eta V$ where $k_B$ is the Boltzmann constant, $\eta$ the solution viscosity (in Pa s) and $V$ the volume.

[OP]  When I searched for it, someone says that it is any molecular motions (translation, rotation, vibration), someone says that it is reorientation, and someone says that is is rotation.

The motions relevant for NMR are those that change the strength of the magnetic field an NMR-active nucleus experiences, i.e. motions that change the chemical shielding and the location of other NMR-active nuclei. When a molecule is rotated (as a rigid body), all nuclei are affected. When there are specific vibrations (or conformational change), nuclei in a specific part of the molecule are affected. Translation only makes a difference if there is a gradient in the external magnetic field (on purpose or because of experimental challenges).
Tumbling is a stochiastic reorientation of molecules over time. To make things easier, we will think of the molecule as a rigid body and exclude conformational change from here on. Stochiastic reorientation implies that if you pick a thousand molecules in your NMR tube that happen to have the same orientation at one time, they will have very different orientations - say - a second later. So the reorientation is uncorrelated, and this is important for solution state NMR.
To give you an analogy of correlated and uncorrelated reorientation, consider a washer and dryer. During the spin cycle in the washer, clothes rotate in a correlated way (on the axis of the drum). During tumble drying, on the other hand, items of clothes change their relative orientation because of the bevels in the drum, gravity and slow drum rotation speeds. This helps separate clothes and avoids wet spots.
Why are uncorrelated reorientations important in NMR? They pop up when explaining the chemical shifts you observe, and when considering T1 and T2 relaxation (how much time do spins need to go back to their equilibrium state, and what determines peak widths).
Chemical shifts are highly orientation-specific (the orientation of the "chemical environment" or lattice with respect to the orientation of the NMR magnet). Here is an example of two carbonyl signals:

From JMR 2002, 155:15-28, copied from slide 41 in a talk by Rienstra
Without tumbling, different orientations of the carbonyl show chemical shifts in a wide range (roughly 100 ppm to 250 ppm). The dotted lines are the average chemical shifts, and those are the ones you would observe for an isotropically tumbling molecule (if it tumbles faster than the timescale of individual scans in the experiment).
The influence of tumbling on relaxation phenomena is complicated but generally relies on a stochastically fluctuating magnetic field experienced by the nuclear spins. Suffice it to say that when molecules show fast tumbling rather than slow, solution-state NMR is easier to interpret and data are more useful.


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Van't Hoff Equation isolating for K2
So I wanted to find the theoretical Ka values of acetic acid from 25-65 degrees at 10 degree increments: 25, 35, 45, 55, 65.

I know that the dissociation of acetic acid is endothermic, and when I isolate for K(T2), and use -1.51kJ/mol as my ∆H, i somehow ended up with the following relationship: Ka decreases whilst Temperature increases... thats clearly not normal.

HOWEVER, when I add a negative sign in front of the 1.51, I end up with the following photo: Now this is normal, Ka increases while Temperature increases...

Can someone help solve this mystery?? I am probably tunnel-visioning so hard rn... Thank you in advance!!!

The value -1.51 kJ/mol  is the enthalpy of dissolution, not the enthalpy of dissociation, therefore cannot be used to determine Ka=f(T).

@Poutnik Ohh, thank you so much! Do you conveniently have the enthalpy of dissociation for acetic acid?

No, I don't. You have to Google it, or calculate it as the difference of neutralization enthalpies of acetic acid and a strong acid. The former consists formally from dissociation enthalpy and strong acid neutralization enthalpy.

@Poutnik Oh, Alright! Do you mind explaining it in slightly simpler terms? I am not quite familiar with these new enthalpies you mentioned. Thank you!


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Relationship between Gibbs free energy and surface energy
I am currently studying interface engineering, a field sometimes referred to as surface tension engineering. In the course, it has been mentioned that surface energy, also known as surface tension in the context of simple fluids, is considered a form of Gibbs free energy. I am attempting to grasp whether this classification is purely a matter of terminology or if there is a deeper connection between surface energy and Gibbs free energy. The Gibbs free energy (G) is expressed in various forms, and one common representation is:
G = U + PV − TS. My question pertains to whether surface energy could be conceptualized as a component of the PV term in the Gibbs free energy formula. Since surface energy represents the energy required to create a unit area of interface, is it plausible to interpret this as a form of work performed in establishing space for the material, thus linking it to the PV term?

I believe this is just a matter of terminology rather than a direct connection. The Gibbs free energy (G) is a thermodynamic potential that combines enthalpy (U), pressure-volume work (PV), and entropy (TS) to describe the maximum non-expansion work that can be extracted from a system at constant temperature and pressure. In simple words, it is a measure of the system's ability to do useful work.
On the other hand surface energy is the excess energy associated with the creation of a new surface or interface between two phases. It represents the energy required to increase the interfacial area by a unit amount. For example, in the case of a liquid-gas interface, surface tension is a measure of the surface energy per unit area.
While surface energy does involve work in the sense that energy is required to create an interface, it is not directly related to the PV term in the Gibbs free energy equation. The PV term in the Gibbs free energy equation typically refers to the mechanical work done by or on the system due to changes in volume under pressure. It does not specifically account for the work associated with the creation of an interface.
However, there is a connection between surface energy and the thermodynamic concept of excess free energy. Excess free energy (ΔG_excess) is a measure of the departure of the Gibbs free energy from its ideal behavior in a non-ideal solution. In the case of a binary liquid mixture, for example, the excess free energy can be related to the interfacial area and composition of the mixture.

One can happily add terms for any other energies to G. For example, magnetic terms are commonly needed for many element crystal structures. A surface energy term is quite appropriate.


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Are catalysts considered when assigning type of chemical equilibrium?
So I have very short doubt that whether catalysts are considered when we say that an equilibrium is homogeneous or heterogenous, like in an aqueous equilibrium with only aqueous reactants and products a solid catalyst is added, by intuition and the definition given in some books (i.e for an equilibrium to be homogenous the reactant and products must all be in the same phase) this should be an homogeneous equilibrium but I haven't seen a precise answer to this question anywhere or a reason behind it.

The catalyst changes neither the equilibrium constant expression nor the magnitude of the constant. So if all species (reactant and products) are in the same phase, it is a homogenous equilibrium. Adding a catalyst in (or as) a different phase does not change that.
I don't have a specific reference for this, but catalysts change the kinetics rather than the thermodynamics of a reaction. Thus, adding a certain type of catalyst is not expected to change the description of the thermodynamics.

With the little chemical kinetics I knew I gave similar arguement as well, but on discussion with some other people they argued that catalysts may react with the reactants, so in some sense the equilibrium is no longer homogenous. Could you point out what's wrong with their arguement.

The catalysis may be non-homogeneous, but the reaction is still homogenous, so its equilibrium is described like the non-catalyzed reaction. If you need a more specific answer, you should edit your question to make it more specific (which examples you are thinking of, which books you have consulted, etc). The catalyst always reacts with the reactants, so the reaction intermediates will be in a different phase, but the bulk of the reactants and products will not.

I think I have understood what you said and don't need more specifics, but just one thing, why have we in the first place segregated homogenous and heterogenous equilibrium? This question has been lingering in my mind since this doubt arised.

The concentration of pure liquids and solids don't change as the reaction goes on. Consequently, a heterogenous equilibrium might behave a bit differently than a homogeneous equilibrium. As you know, pure liquids and solids are treated differently in an equilibrium expression, which is related.

OK got it, Thanks!


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How to calculate theoretical value of mass output using electrolysis with different electrolyte concentrations?
I am writing a research paper on a zinc-copper electrolysis experiment observing the mass output of Zinc on a copper cathode, in which I am changing the concentration of the zincsulfate electrolytic solution in each trial.During the experiment, the controlled variables were duration of electrolysis and current. This experiment resulted in a positive correlation between mass output and the increase in electrolyte concentration. Now I am trying to calculate the theoretical mass output with the use of different concentrtions of electrolyte, but it doesn't seem possible using Faraday's laws. How can I calculate the theoretical value?

More diluted solution = more negative cathode potential = faster parallel reduction of hydronium = lower zinc mass for given charge.

Why is it not possible to use Faraday's laws? Nobody has ever thought that they are unusable.

@Maurice OP probably means it is not directly applicable if the same total charge leads to different masses of the product of the particular redox reaction. (as part of the charge is spent on a parallel reaction)


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Electron configuration in benzene chromium tricarbonyl
I am not sure what is the electron configuration in benzene chromium tricarbonyl?
From what I can work out, Cr is in 0 oxidation state (since all ligands are neutral). The complex is tetrahedral so likely high spin. But the CO ligand is strong field and so is more likely to cause low spin. There are three of these ligands and one weak field benzene ligand.
I'm confused by the conflicting information.

When there are strong pi interactions, a tetrahedral geometry no longer corresponds to weak splitting. With tetrahedral geometry the splitting is small between sigma orbitals but large between pi orbitals. So with the strong pi interactions from carbon monoxide and benzene that tetrahedral geometry can indeed impart strong splitting and a low-spin configuration. Add the fact that we have eighteen net valence electrons around the metal (six from the core atom, six from the benzene ring, two apiece from the three carbon monoxide ligands), which most strongly stabilizes the low-spin configuration, and low-spin becomes the natural choice. And the real one.

So this will leave one unpaired electron for the metal (two in $d_{x^2-y^2}$, two in $d_{z^2}$ and one in one of the $t_2$ orbitals), making it ever so slightly paramagnetic?

No, there are no unpaired electrons. When you meet the 18-electron rule as I describe,you are typically low-spin and the low-spin cilonfiguration has no unpaired electrons.

Chromium is $3d^5 4s^1$ so I'm confused how the electrons pair?  Does the 4s electron pair in a $t_2$ orbital with the fifth $d$ electron?

The s electron mixes in with the d. Then you have three pairs if electrobs in the mixed orbitals and six pairs from the ligands.


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Waste and Biproducts generated during eNRR
I am currently reviewing electrocatalytic methods of fixing nitrogen through eNRR and thereby looking up methods coming up in recent literature. However, I need to know what sort of waste products are usually generated from such electrocatalytic pathways. Most of the existing literature deals with the detection of the fixed product and doesn't elaborate on the details of side products and wastes generated.
Any literature quantifying (or even naming the waste products) will be appreciated.

I suspect the answer is quite dependent on the process used. The category of eNRR includes a wide variety of processes. Can you specify which one you are interested in?


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How to convert the uncertainty of my pH probe to the uncertainty of my Ka value
Literally the title. So basically I performed numerous titrations with ethanoic acid sodium hydroxide. I got the full pH curve, and subsequently found the Ka Value of the ethanoic acid. Now here is the issue, I have the uncertainty of the pH probe, which is ±0.1. At the half equivalence point, pKa=pH, so I can safely assume the uncertainty for pKa is also ±0.1.
However, I need to find the uncertainty of the Ka Value I experimentally determined. So I somehow need to convert that ±0.1pKa into the uncertainty for Ka.
Thank you in advance!

Just considering the resolution of the pH probe as the only source of error seems unlikely. What about the experiment error? How many times did you do the experiment? What were the values determined?

Following Metron's suggestion, suppose the $p$K$\ce{_a}$ of acetic acid is $4.75 ± 0.1 $. It means that both extreme values of $p$K$\ce{_a}$ are $4.65$ and $4.85$. Corresponding extreme values of K$\ce{_a}$ are $\ce{10^{-4.65} = 2.2 10^{-5}}$ and $\ce{10^{-4.85} = 1.4 10^{-5}}$. As the average of $1.4$ and $2.2$ is $1.8$, the final value for $p$K$\ce{_a}$ is equal to : $1.8 ± 0.4$

No you shouldn't average the two extremes to get thev estimated value. The estimate for the pKa is $10^{-4.75}$ which happens to be $\ce{1.778... x 10^{-5}}$ which in this case does round to $\ce{1.8 x 10^{-5}}$. My point however is that the difference between the +/- values and the expected value are not symmetric since the scale is logarithmic.


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Rearrangment during dehydration of 2,2,5-trimethylcyclopentanol

Here I thought doing the following mechanisms

I thought the product option through the first pathway (left) would be the major product because the intermediate cation is more stabilised due to more alpha-H. I was told rate determining step is formation of carbo-cation so the pathway where a more stable cation is formed would be faster be major product.
The answer given was obtained from 2nd pathway. Is there any specific factor behind it? My professor told that in this case TC product dominated KC product

The most substituted alkene is the more thermodynamically favoured. This is an equilibrium situation, the product alkenes can both protonate under the reaction conditions.

The name of the compound is 2,3,5-trimethylcyclopentanol. Not 2,2-dimethyl-4-methylcyclopentanol as you stated. I have corrected it for readers' benefit.

The major rearrangement of carbocation resulted from the acid-catalyzed dehydration of original 2,3,5-trimethylcyclopentanol is correctly tracked in your scheme. Keep in mind that the dehydration is done in thermodynamically control manner (in higher temperature) so that all rearrangements are in reversible condition (See Waylander's comment).
The initial secondary carbocation at C-1 (the top structure in your scheme) can rearrange either by a 1,2-hydride ion shift from C-5 to give a tertiary carbocation (structure 1.1) or by 1,2-methide ion shift from C-2 to give a different tertiary carbocation (structure 2.1). Both hydride and methide groups are available so that both rearrangements would undergo in different rates under given conditions. These two new tertiary carbocations together with initial secondary carbocation would account for most of the alkene products in corresponding percentages according to their thermodynamic stabilities. The structure 2.2 (1,2,3-trimethylcyclopentene) would be the major product since it contains the most substituted double bond.
For example, dehydration of 2,2,4-trimethyl-3-pentanol with acid has given in Ref.1 as a complex mixture of the alkenes with following percentages in parentheses: I) 2,3,4-trimethyl-2-pentene (29%); II) 2,4,4-trimethyl-1-pentene (24%); III) 3,3,4-trimethyl-1-pentene (2%); IV) 2,4,4-trimethyl-2-pentene (24%); V) 2,3,4-trimethyl-1-pentene (18%); and VI) 2-isopropyl-3-methyl-1-butene (3%).

References:

Robert J. Ouellette and J. David Rawn, In Organic Chemistry
Structure, Mechanism, & Synthesis; Second Edition, Academic Press: Cambridge, MA, 2018 (ISBN: 978-0-12-812838-1).

Thanks a lot. Just one more thing, is more substituted alkene, in general, major product of dehydration?

As a rule of thumb, more substituted alkene (Zaitsev’s Rule) would generally produced under thermodynamic conditions.


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What part of ascorbic acid is oxidized when it reacts with iodine?
When vitamin C reacts with iodine, it will lose two of its electrons (oxidation) which iodine will accept (reduction). What I do not understand is what part of the vitamin C exactly is losing the electrons? Where is the change in oxidation number happening when it is oxidized to dehydroascorbic acid?
Half equations:
$$\ce{C6H8O6  →  C6H6O6  + 2H+ + 2e-}$$
$$\ce{I2 + 2e- → 2I-}$$
I have looked for answers online, but I have not managed to find any, or at least none that I understand (high school student).

See a nice video of a related experiment here: Recreating the Iodine Clock Reaction at Home with Vitamin C - YouTube

A. M. Helmenstine states,"Triiodide oxidizes vitamin C to form dehydroascorbic acid." Wikipedia shows structural diagrams for the molecule in various states of hydrolysis.
Note that there is debate on representing the structure: "The actual structures of the various forms of dehydroascorbic acid have been known for at least a quarter of a century so that the continued use of the oversimplified tricarbonyl structure is hard to excuse."

Typically, when organic molecules get oxidized or reduced, the oxidation numbers of carbon atoms changes, while those of hydrogen (+1) and oxygen (typically -2) don't change.
The oxidation number is a formal number (not measurable), and does not imply certain charges on certain atoms. The atoms in organic molecules are held together by covalent bonds, so the electrons are not "assigned" to a given atom.
One easy way to figure out the oxidation number of a carbon (with four bonds and no formal charge) in a given Lewis structure is to start at zero, add one for each bond with oxygen, and subtract one for each bond with hydrogen.
The Lewis structure of the reduced form has two carbons at an oxidation state of +1, and they change to +2 in the oxidized forms shown below:

The image shows three forms of the oxidized vitamin C (with a nod to the other answer) but in all of them, the bottom two carbon atoms show two bonds to oxygen each (either two single bonds or one double bond) and no bonds to hydrogen.
The reactions that are shown, hydrolysis and hydration, don't change the oxidation state of the carbon atoms.

An oxidation number suggests that an atom is enriched or depleted in electron density relative to the free (unbonded) atom. This from the definition of the free form of an atom as the reference oxidation state, with zero oxidation number. If you inspect examples in the Wikipedia page linked above you'll note that increasing the number of bonds (or equivalently bond order) between carbon and oxygen increases the oxidation state of carbon, since oxygen is more electronegative. Formally, due to its high electronegativity, bonded oxygen is assigned a -2 state. As oxidation numbers of individual atoms have to add up to the charge of the molecule, this implies that the carbon OS increases with increasing CO bonds or bond order.

In the oxidation of ascorbic acid (structure above from the Wikipedia) electrons involved in the CC double bond are transferred to oxygen to form CO bonds or increase the CO bond order, implying that the carbons are oxidized. Electrons in a CC double bond are however not necessarily shared equally between carbon, meaning one carbon could be assigned a formal charge of +1 relative to the other with OS -1. Therefore one interpretation is that one carbon (that nearest to the carbonyl) is oxidized while the state of the other formally does not change.


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What will be the preferred reaction between phosphorous acid (H3PO3) and potassium hydroxide (KOH)
""To neutralize completely $\pu{20 mL}$ of $\pu{0.1 M}$ aqueous solution of phosphorus $\ce{H3PO3}$ acid the volume of $\pu{0.1 M}$ aqueous $\ce{KOH}$ solution required is: ____""
To solve this question a source used the reaction:
$$\ce{H3PO3 + 2 KOH -> K2HPO3 + 2 H2O}$$
While other source used the reaction
$$\ce{H3PO3 + 3 KOH -> K3PO3 + 3 H2O}$$
I know that $\ce{H3PO3}$ is dibasic. So I thought definitely two $\ce{H+}$ will get replaced by $\ce{K}$ atoms (As shown in the 1st reaction). But later I learnt that if $\ce{KOH}$ is in excess, then $\ce{H3PO3}$ will react with $\ce{KOH}$ to produce $\ce{K3PO3}$ and $\ce{H2O}$. However, if $\ce{KOH}$ is limited, then $\ce{H3PO3}$ will react with $\ce{KOH}$ to produce $\ce{K2HPO3}$ and $\ce{H2O}$.
So, my question that arose was do we take $\ce{KOH}$ as excess when we solve questions like this or do we take a limited value?
Edit:The 2nd reaction in the question does not take place as described in the accepted answer

@DrMoishePippik It is a phosphite. The last H is not acidic.

@Poutnik Sorry I clarified what I meant to ask.I think now one will be able to understand what I meant

Does this answer your question? Why is phosphorous acid diprotic and not triprotic?

Hello @Mithoron .Thanks for your reply but sorry my question was not why phosphorous acid is diprotic.My question was do we take NaOH in excess or in limited while we solve questions like that

Chem+Math formatting reference:
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Provide the source that talks about formation of $\ce{K3PO3}$.

@Poutnik Please see the production part in this page https://en.m.wikipedia.org/wiki/Tripotassium_phosphate. Edit Sorry I overlooked that it is tripotassium phosphate

Phosphorous acid $\ce{H3PO3}$ is not an ordinary acid. It has extremely unusual reactions.
First, although it has three $\ce{H}$  atoms, only two are related to oxygen atoms, and only these two $\ce{H}$ atoms can be replaced by metallic atoms like $\ce{K}$ or $\ce{Na}$. The third $\ce{H}$ atom is related directly to the central phosphorus atom, and cannot be replaced by alcaline atoms. The acid structure is probably better described by $\ce{HPO(OH)2}$ or $\ce{H2[HPO3]}$. So its neutralization by $\ce{NaOH}$ occurs through one of the two following reactions : $$\ce{H3PO3 + NaOH -> NaH2PO3 + H2O}$$ $$\ce{H3PO3 + 2 NaOH -> Na2HPO3 + 2 H2O}$$ These two sodium salts should be better described as :
$\ce{Na[H(PO)(OH)O]}$ or $\ce{NaH[HPO3]}$ instead of $\ce{NaH2PO3}$
$\ce{Na2[H(PO)O2]}$ or $\ce{Na2[HPO3]}$ instead of $\ce{Na2HPO3}$
The "expected" third salt $\ce{Na3PO3}$ is never produced by neutralization and does not exist.
Second, the acid and its salt have strong reducing properties. For example they reduce $\ce{Hg^{2+} and Ag+}$ ions into the corresponding metal : $$\ce{H3PO3 + 2 Ag+ + H2O -> 2 Ag + H3PO4 + 2 H+}$$
Phosphorous acid has the specific property of reducing sulfur dioxide in solution into elementary sulfur at room temperature : $$\ce{2H3PO3 + SO2 -> 2H3PO4 + S}$$

I learnt that if NaOH is in excess, then H3PO3 will react with NaOH to produce Na3PO3 and H2O.However, if NaOH is limited, then H3PO3 will react with NaOH to produce Na2HPO3 and H2O.Is this correct? @Maurice

No. The third $\ce{H}$ atom of $\ce{H3PO3}$ is not able to react with a hydroxyde, in excess or not, whatever the circumstances.  To be able to react with $\ce{NaOH}$ or $\ce{KOH}$, the $\ce{H}$ atom has to be bound to an oxygen atom. This third $\ce{H}$ atom is not bound to an oxygen atom. So it does not react as an acid. In the similar molecule $\ce{PH3}$, the $\ce{H}$ atoms don't react with hydroxides, even in great excess.

Please see the production part in this page https://en.m.wikipedia.org/wiki/Tripotassium_phosphate @Maurice

@Rajesh Paul. Your reference is related to tripotassium phosphate $\ce{K3PO4}$ which exists. But this well-known phosphate has nothing to do with derivates of phosphorous acid $\ce{H3PO3}$. There are $4$ oxygen atoms in tripotassium phosphate, and not $3$ like in derivates of phosphorous acid.  $\ce{H3PO3}$ ≠ $\ce{H3PO4}$ !

Sorry I overlooked that.You are absolutely correct

Technical note: do not use references to mobile versions of Wikipedia pages. If posting it from a phone, modify en.m.wikipedia.org to en.wikipedia.org


text:
Reason for positive enthalpy and volume change of mixing for mixtures showing large positive deviation from Raoult's law
At onset let me make it very clear that similar questions do exist already on this site, but none of them has a kind of answer I expect, hence I ought to ask a fresh question.
It may appear that this question is domain specific (to ChemE) which is not true, I had consulted peers from pure science background they confirmed that it's part of their curriculum too.
As per my understanding
\begin{align}
\Delta H_\mathrm{mix}
  &= -RT \sum\left(
     \frac{\delta \ln(a_{i})}{\delta \ln(T)}\right)_{P,{x_{i}}}\\
\Delta V_\mathrm{mix}
  &=\frac{RT}{P} \sum\left(
     \frac{\delta \ln(a_{i})}{\delta \ln(P)}\right)_{T,{x_{i}}}
\end{align}
Literature I am referring to reads

Positive deviation occurs when the interaction between unlike molecules is weaker than interaction between like molecules.The solutions of this type are characterized by positive Enthalpy of mixing and positive volume of mixing.

I want to know what is the mathematical reasoning as well as intuition behind this statement? How the equations (stated above) predict this behaviour?
Note:
A few comments had asked for derivations of those equations, I had learnt them from an Indian Book on Chemical engineering thermodynamics which may not be know outside my country. A proof for those equations is as follows:
$$ M = \Sigma x_iM_i° + \Delta M_{(mix)} $$
Where M is any property of solution $x_i$ mole fraction of component i, $M_i°$ molar property of component i, and the $\Delta M_{(mix)}$ property change occurs due to mixing of components in solution. By property of summability of partial molar properties of individual components in solution it is known that $M = \Sigma x_i \bar{M_i}$ where $\bar{M_i}$ is partial molar property of component i.
Rearranging the equation,
$\Delta M = \Sigma x_i (\bar{M_i} - M_i°) $
If M = Gibbs Free Energy
Eqn 1,
$$\Delta G = RT\Sigma x_i ln(a_i)$$
Where $a_i$ is activity of component i in solution,
(I have jumped the step, you may derive the equation using basic equation for G in terms of fugacity)
Eqn 2,
$$\bar V_i = (\frac{\delta \bar G_i}{\delta P})_T $$
Eqn 3,
$$(\frac{\delta (G_i°/T)}{\delta T})_P = \frac{-H_i°}{T^2}$$
Equation 2 follows from comparing partial differential equation for G(T,P,${N_i}$)with fundamental property relation for G
Equation 3 is from Gibbs Helmholtz Equation
Using equation 1,2 and 3 we can arrive at equations
\begin{align}
\Delta H_\mathrm{mix}
  &= -RT \sum\left(
     \frac{\delta \ln(a_{i})}{\delta \ln(T)}\right)_{P,{x_{i}}}\\
\Delta V_\mathrm{mix}
  &=\frac{RT}{P} \sum\left(
     \frac{\delta \ln(a_{i})}{\delta \ln(P)}\right)_{T,{x_{i}}}
\end{align}
If you are interested in detailed derivation then you may refer "Textbook of Chemical Engineering Thermodynamics" by KV Narayana 2nd Ed pg.no. 308- 310

Most of the Internet isn't localised to a single country. So maybe you could still just cite the literature you are quoting. That's just common practice. And Google might actually help you out with that. https://www.google.de/books/edition/Principles_of_Physical_Chemistry/SVBXXwAACAAJ?hl=en

Which version of the book ( "principles of physical chemistry" by Puri•sharma•pathania) are you quoting? (this is only to check the derivation). My version (46th) has a derivation (Eq 66 in chapter 21, solutions of non-electrolytes) very similar to the one you show, but is not the same.

@BuckThorn yes we have same edition, in equation 66 using the results from eqn 50 and taking ∆H on one side and rest of stuff on other and then Integration gives that same equation for ∆H,while I didn't went through the derivations from that book as I already had that covered in thermodynamics course for which reference was book that I have mentioned in last line of question, but equations are exactly same, just some rearrangement is needed. Really appreciate your efforts thanks.

I made an effort to provide a general explanation of what van't Hoff's equation says without dwelling into details of the equations. I also did not address the pressure response, only T response. You want an answer that makes a connection between the equations as derived and their deeper meaning, but the derivation is still not complete and one can see some potential confusing issues. First, when discussing mixing one distinguishes between ideal mixing and excess properties, which refer to changes that cannot be explained by the ideal term. That's what deviations from Raoult's law refer to.

For ideal mixing, $\Delta H _{mix} = 0$.

@BuckThorn sorry sir, but I can't agree that your answer is correct one, it isn't wrong but I was asking for different thing, also the equations are for general case you can play with them to use for ideal gases, thanks for taking interest.


text:
Carbon 13 NMR peaks - ketones
Why the shift of acetophenone's carbonyl carbon at C13 NMR spectrum is larger than the equivalent peak at trifluoroacetophenone ? Particularly , the first one is located at 197.94 ppm whereas the second at 179.59 ppm . Normally the presence of an EWG deshields the carbon more easily , right ?
It would be helpful if anyone could answer this .
Thank you very much !


text:
Why do t-butyl and isopropyl groups deshield the ipso carbon in C13 NMR more than a Me group?
I have a t-butyl group attached to an aromatic ring and I am confused about why it increases the shift of the ipso carbon about 20ppm. I know that C is slightly more electronegative than H, and this explains why Me increases the shift by about 8ppm, but why is the deshielding more for t butyl and isopropyl? Wouldn't extra hyperconjugation effects shield it rather than deshield?

Do you know what is hyper conjugation means? Methyl has only 3 hydrogens, but isopropyl has 6 and t-Bu has 9 (if I understand your question correctly). Put some more details to your question so9 that everybody'd understand it easily.

I understand the hyperconjugation but wouldn't that shield the nucleus which is the opposite of what i am seeing in my spectrum? Is it some kind of magnetic anisotropic effect which is deshielding the nucelus of the ipso carbon?


text:
Alkynes on ozonolysis
What are the products formed on ozonolysis of 1-hexyne followed by hydrolysis?
I am not getting whether Pentanal is formed or pentanoic acid is formed and sub products

According to libretexts the final product of ozonolysis of alkene will depend on what was along with ozone.
If we are using zinc along with ozone then we would get aldehyde.
If we are using Hydrogen peroxide along with ozone then we would get carboxylic acid.
In ozonolysis of alkynes we would get diketone or anhydride which could react further in presence of water (water would react with ozone to give hydrogen peroxide) to give two carboxylic acids.
In case of alkyne ozonolysis we would get no aldehyde.
In your example of ozonolysis of 1-hexyne we would get pentanoic acid and formic acid after cleavage of triple bond. Formic acid will get further oxidised to give carbon dioxide.
Source : wikipedia Youtube

Dimethyl sulfide is the more frequently used work up rather than zinc

@Waylander Yes you are right but zinc is more easily available and easier to remember from academics point of view. In my textbook we don't have dimethyl sulfide.

Get a better textbook. DMS is what is actually used by chemists who run ozonolysis

1-hexyne is not symmetric so we get two different products. Can you name the other product(s)?


text:
How would the regions of this graph showing percentage of ionic character, be classified?
Looking here at this website - Omnicalculator, at the section of the website for calculating percentage of ionic character. Omnicalculator - Percentage Ionic Character
They let you choose elements and it calculates percentage of ionic character.
And besides showing the electronegativity difference, and the percentage of ionic character
They show criteria for determining whether something is considered non polar covalent, or polar covalent or ionic

(No doubt the first row of that table  they mean non polar covalent)
And they show this graph

I think that table leaves some gaps e.g. what if the electronegativity is < 2, but the percentage ionic character is > 63.21%  (Then it's not in any of the three rows).
And they show this graph
Looking at that graph, the second intersection is just above 60. probably 63.21% (the figure they have in their table)
There are some regions of the graph i'm not sure about..'cos their table doesn't cover it.
Any idea how they'd be categorising/classifying sections i've labelled D/E/F?

@Poutnik Thanks.. So regarding the first two columns in the table.  Or regarding the graph.  Is it saying that there's no cases where ΔEN is < 2 and IC < 63.21% ?And if it's not saying that, then how would you classify a compound that has  ΔEN is < 2 and IC < 63.21% ?

https://chemistry.stackexchange.com/questions/9222/why-are-bonds-ionic-when-the-electronegativity-difference-between-bonded-atoms-i https://chemistry.stackexchange.com/questions/17064/is-kf-the-most-ionic-compound

The relation of the calculated ionic character and the Pauling electronegativity difference is the function of the ionic character model. There are no regions A-F in this context. There is just the blue curve.
For the measured ionic character, e.g. from the bond length and the bond dipole moment, the graph plays no role and there are just 3 regions nonpolar/polar/ionic according to experimental ionic character value.
Note that the threshold values are not sharply defined, they are rather fuzzy. But there is no reason not to follow the arbitrary values for the calculated values.
Note that the experimental values of ionic character and the difference of Pauling electronegativities are correlated, but not very strongly.

Thanks. And so besides 0.5<Δχ<2.0⟹6.06%<IC<63.21%,  Also,     6.06%<IC<63.21%  ⟹ 0.5<Δχ<2.0  (so that implication goes both ways). And so there are no holes in the table?

Do you see there any holes?

Not yet but I will check more combinations of elements.

They aren't truly dependent, rather correlated (not all that well).

I see what you mean re the mapping., I see there's a direct mapping between electronegativity difference and IC.  . So the folk that put the IC boundary for ionic at ~63%, use electronegativity difference of 2..  And the folk that put IC boundary for ionic at > 50%, use electronegativity difference at 1.7 or 1.8. https://www.calculatoratoz.com/en/percent-ionic-character-calculator/Calc-900

@Mithoron I do agree, it is measured versus calculated ionic character. I was speaking in context of the former. For the latter, the bond categorization does not obviously refer to electronegativity. for OP, see also https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_Chemistry_-The_Central_Science(Brown_et_al.)/08%3A_Basic_Concepts_of_Chemical_Bonding/8.04%3A_Bond_Polarity_and_Electronegativity

Hehe, I have turned it upside down in the last.by error. I have obviously talked about ionic character calculated from P. electronegativity, where those regions play no role. For measured ionic character, e.g. from bond length and dipole moment, the graph plays no role and there are 3 regions nonpolar/polar/ionic according to experimental ionic character value.


text:
Why is Benzaldehyde more reactive than propanone in nucleophilic addition, even though it has resonance?
I am not 100% sure that this is true but this was given in the JEE Main 2020 (03 Sep Shift 2).
Benzaldehyde has resonance and propanone has just +I effect why is then benzaldehyde more reactive in nucleophilic addition?

The carbonyl carbon exhibits reactivity towards nucleophilic addition reactions due to its partial positive charge resulting from oxygen.

Benzaldehyde's heightened reactivity, compared to propanone, can be attributed to the absence of electron-donating methyl groups. In propanone, two methyl groups act as electron donors, whereas benzaldehyde features an sp2 carbon attached to the carbonyl carbon, which, being more electronegative than sp3 carbons, does not exert a +I effect like methyl carbons.
Additionally, steric hindrance is induced by the presence of two methyl groups in propanone.
While benzene does contribute to reducing the positive charge through resonance, it is imperative to consider the impact of other influencing factors.


text:
Does Hess' law in standard conditions involve consuming all reactants?
The standard free energy change for equilibrium problems is relative to both reactants AND products starting at 1 molar.
Hess’s law takes the sum of the values for products minus the sum of the values for reactants. My interpretation is that the result of such a calculation is for changing from a state of all reactants to all products. So even if I use standard values the initial state does not involve 1 molar for BOTH products and reactants but instead involves an initial state of only reactants.therefore I can not equate $\Delta G = \Delta H-T \Delta S$ to standard $\Delta G= RT \ln K$ if the enthalpy I calculated came from using Hess' law. It would be like comparing apples to oranges. Is my logic correct?

The molar Gibbs free energy change of reaction refers to $\Delta G$ for conversion of a stoichiometric amount of reagents to products for one mole of reaction as written. For a reaction $\ce{aA + bB +...->pP +qQ +...}$ this means $a$ moles of $\ce{A}$ react with $b$ moles of $\ce{B}$ and so on to form $p$ moles of $\ce{P}$, etc, with reagents and products at the same T and p. An underscore "m", as in $\Delta_m G$, is used to indicate a molar change. An "r" subscript, as in $\Delta_r G$, is also commonly used. When the reaction is carried out under standard conditions, a circle or Plimsoll symbol, as in $\Delta_r G^\circ$, indicates standard states for reactants and products*.
It is correct to write
$$\Delta_r G^\circ = \Delta_r H^\circ -T \Delta_r S^\circ \\ = -RT \ln K \tag1$$
which follows from
$$\Delta G = \Delta H -T \Delta S \tag2$$
and
$$\Delta_r G =\Delta_r G^\circ + RT \ln Q \tag3$$
and $Q=K$ at equilibrium, when $\Delta_r G=0$.
In going from Eqs 2 and 3 to Eq 1 you are in fact also applying Hess' law by equating the result of two particular cycles between the reactants and products, as illustrated in the following figure:

One converts reactants to products in their standard states directly (Eq 2), while the other (derived from Eq 3) takes the reactants in their standard states to the equilibrium state, at which conversion to products occurs with $\Delta_r G = 0$, and then equilibrium products are converted to their standard states (hat tip to jimchmst).

*The definition of $\Delta_r G^\circ$ assumes all products and all reactants are ideal, which implies that these are isolated (not mixed). The 1M standard state is typically hypothetical  (an abstraction with thermodynamic properties determined through extrapolation from measurable states). The behavior of a real 1M solution is often non-ideal in the thermodynamic sense.

The standard free energy change is the change from the condition where all involved substances, reactants AND products, are in their standard states to equilibrium activities. If reversed it is the free energy from equilibrium to the standard states. Nonstandard conditions are evaluated by calculating the appropriate reaction quotient. The last paragraph makes no sense, calculation of state functions is independent of the path. Reactions run to equilibrium not completion.

ΔG=ΔG∘+RTlnQ
ΔG∘=−nRTlnK when Q is 1 which occurs when both reactants and products are 1moles/L.
Hess's law says ΔH∘f=ΔH∘f,products−ΔH∘f,reactant. and you can do that with gibbs free energy as well: ΔG∘f=ΔG∘f,products−ΔG∘f,reactant. However my interpretation of the math involved in Hess's law is that the reaction goes from a state of ONLY reactants(zero products) to ONLY products(zero reactants). So if I found ΔG∘f using Hess's law and then I wanted to find the equilibrium constant then I am saying it wouldn't make sense to equate it to −RTlnK which had products AND reactants start at 1M

@Alfred  Standard values are useful because they deal with pure substances or dilute solutions where conditions are well defined. In a reaction flask, say at equilibrium, conditions (concentrations in particular) can differ significantly from std states.

That is not what I said! Standard states are defined activities are 1.. The standard free energy change for a reaction is the free energy change from the condition of all involved components both products and reactants are in standard states to the condition of the equilibrium activities. Write out the equation and do the math: delta G -G[0] = RTlnKeq -RTlnQ. At equilbrium deltaG= O This is true when Q=1 and that is when both products and reactants have activity =1

@jimchmst Ok, I think I see what you were driving at. I included something along the lines of your comment into my answer. Hope it's clearer now.


text:
How to tell the difference between mineral spirits and isopropyl alcohol
They are both clear liquids, so is there a fairly simple test to determine if a container has one or the other? Maybe by mixing a sample with another chemical and looking for a color change? Thanks!

And why not just smell it?

I try to avoid inhaling mineral sprits as much as possible. Even breathing in small amounts often can add up.

Good luck avoiding vapor inhalation while making some solubility test :/

If by mineral spirits you mean low boiling petroleum distillate, simply add a small amount to an equal amount of water. It will float and not dissolve; 2-propanol dissolves in water. If you mean something else please be specific.

Ya that's what Poutnik had as an answer. I don't know what kind of mineral sprits the  liquid is composed of so I can't be more specific. Luckily the question was answered with a possible solution before this question was closed,

Shake well a small sample with similar volume of water and let it stand. If a single phase has been formed, it is isopropyl alcohol. If liquid has formed two liquid layers, it is a mineral spirit.
Isopropyl alcohol is, similarly as ethanol, miscible with water in any ratio. Additionally, its smell reminds ethanol.
Mineral spirits, similarly as mineral  oils, are not miscible with water and their mutual solubility is minimal. Additionally, their smell may remind gasoline or some paint thinners/solvents.

Thanks for sharing, I'm going to try this out when I return to the lab.


text:
Why does AsH3 have a higher boiling point than HBr?
Aside from the general trend of the boiling points of the hydrides, I noticed how the group 15 hydrides somehow "overtook" the group 17 ones in boiling point, which felt weird.

Notably, H-Br has an electronegativity difference of 0.7, which is enough to be considered polar, which to my understanding means HBr will exist with a permanent dipole. The As-H bond has an even smaller difference, so it can't experience these tiny permanent dipole attractions.
That probably means there's a big difference in London dispersion forces, but both molecules have 36 electrons, so they'd be similarly capable of forming instantaneous dipoles. Their molecule shape might affect their ability to induce dipoles, but I don't know where to start with that.
So if both AsH3 and HBr experience similar London forces, but HBr also experiences weak permanent dipole-dipole interactions, why does AsH3 ($\mathrm{-63^\circ C}$) have a slightly higher melting point than HBr ($\mathrm{-68^\circ C}$)? (The same could be said for SbH3 and HI, which has a slightly more significant difference in boiling point.)

What's the point of comparing apples with oranges? The molecules are simply different. But if you wonder about polarisability, AsH3 should have higher, simply because it's bigger.

The question behind the question is why do the hydrogen halides fall behind the other hydrogen compounds shown when we get to heavier periods?
A clue to the answer may be gleaned by completing each isoelectronic series with the noble gas: for example, $\ce{Kr}$ or $\ce{H0Kr}$ in period 4 and $\ce{Xe}$ or $\ce{H0Xe}$ in Period 5. In each case the noble gases are well below the other compounds in their respective series; in Period 4 for instance krypton weighs in at $-153°\mathrm{C}$ which is well below germane and the other Period 4 molecules.
As we move to bigger atoms with more electrons the predominant forces are dispersion forces and dipole-diploe interactions. The dipole-dipole interactions are absent for both the noble gases and the tetrahedral molecules produced by the Group 14 elements. Meanwhile the dispersion forces diminish as we move to later groups because more of the valence electrons are localized onto one atom in Group 17 and especially Group 18 (the noble gases) than in earlier groups.
As the dipole-dipole interactions fade in the heavier halides and they must rely more on dispersion forces, the lower dispersion forces compared with the more electron-delocalized polyhydrogen compounds (such as $\ce{HBr}$ versus $\ce{H2Te}$ and $\ce{H3As}$) lead to a drop is boiling point. This drop from Groups 15 and 16 to Group 17 is just the beginning of the huge drop going into the noble gases, where dipole-dipole interactions are completely gone and dispersion forces take a hit with all electrons stuck on one atom per gas particle.


text:
What's the mechanism in the formation of this quinone?
I need to solve this problem for homework. In an article, I saw a quinone formation, but I don't understand the mechanism. In these reactions, the authors used KI (potassium iodide) and AcOH (acetic acid) to generate an alcohol, but I do not understand why an alcohol is generated. Also, they used toluene and p-TsOH·H2O for the opening of the epoxide ring and the formation of a double bond, but I am not sure about this. Here I leave the name of the article (Maruo et al., 2013, "Biomimetic Synthesis of Zeylanone and Zeylanone Epoxide by Dimerization of 2‑Methyl-1,4-naphthoquinone"). I appreciate your help.

Although I'm not 100% positive, I assume the first step is for both conversion of the epoxide and removal of the MOM protecting groups. Acidic conditions are generally sufficient for removing MOM protecting groups, so I would expect the acidic conditions in the first step to remove the MOM groups; however, the iodide also plays a critical role  in converting the epoxide to the alkene.
In this case, conversion of an epoxide to an alkene is a net reduction. Therefore, iodide will act as a reducing agent. I assume the first step (A) to be nucleophilic attack by the iodine on the electrophilic epoxide, facilitated by the acidic conditions. We can then perform a net protodehalogentation by attacking the iodo subsitutent with another nucleophilic iodide (B). This is the key redox step. Iodide is oxidized to iodine and the substrate is reduced. Here, the only step left is to eliminate a molecule of water (C) to produce the alkene product. I assume this is the purpose of the second step of reflux in toluene with TsOH. Elimination of water generally goes better in nonpolar solvents like toluene as it can be removed directly with something like a Dean-Stark trap, shifting equilibrium towards the alkene.


text:
Which is more stable, a secondary allylic carbanion or a primary allylic carbanion?
I came across this picture in a polymer chemistry course, specifically on anionic polymerization of conjugated dienes (butadiene).

I dont understand why the secondary carbanion is more stable. Based on what I know, inductive effects should destabilize the secondary carbanion, so the primary carbanion will be preferred. Moreover, the hybridization and pi orbital orientation of the two ions seems to be the same.
For extra context, the carbanion could exist as an ion pair with Li+ as the counterion, although it should exist as a free ion in the most part since the solvent is polar. However, this also doesnt explain why the second structure is more stable.


text:
How does enthalpy of a reaction vary with pressure?
Kirchhoff's law gives us a way to find variation in enthalpy of a reaction with temperature
But is there some equivalent law for pressures?
Ie, given a reaction and it's enthalpy at a pressure P1, is it possible to calculate it's enthalpy at a pressure P2

These can be applied the Hess law for p shift instead of usual T shift.

Do you know how to find the effect of pressure on enthalpy for a pure species?

Yes, but I was wondering if there was a formula analogous to kirchoffs law

I guess writing down the enthalpy pressure corrections for reactants and products and combining them appropriately would be equivalent.  this would also take into account non-ideal effects.


text:
Is it not possible to balance the equation $\ce{I2 +O3 -> I4O9 +O2}$ in many ways?
The above reaction can be balanced in many ways. For example, the coefficient of $\ce{I2}$ is two when coefficient of $\ce{I4O9}$ is one. But the coefficient of $\ce{O3}$ and $\ce{O2}$ can be 5 and 3 respectively or 7 and 6 or 9 and 9. What will be correct balanced equation?

There is "formally correct" reaction equation and "reality reflecting" reaction equation.  Formally correct are all where are balanced charge and atom counts. They can be formally even any linear combination of several formally correct equations.  "Reality reflecting" correct equation must reflect observed stoichiometry, relating to observable chemistry and known reaction mechanisms. // All 4 provided reaction equations in Maurice's answer are formally correct, but only (1) is "reality reflecting" correct, if the assumed reaction mechanism is correct or if stoichiometry has been verified.

The basic equation of the ozone reaction on iodine is $$\ce{2 I2 + 9 O3 -> I4O9 + 9 O2\tag{1}}$$ because it is the sum of the following two equations: $$\ce{2 I2 + 9 O -> I4O9}$$ and $$\ce{9 O3 -> 9 O2 + 9 O}$$ It should be mentioned that the following reaction never happens : $$\ce{O3 -> 3 O}$$ But this original equation ($1$) may be modified by adding one or many times the following equation :$$\ce{2O3 -> 3 O2}$$ which produces one of the following possible equations : $$\ce{2 I2 + 11 O3 -> I4O9 + 12 O2}$$ $$\ce{2I2 + 13 O3 -> I4O9 + 15O2}$$ $$\ce{2 I2 + 15 O3 -> I4O9 + 18 O2}$$ etc. These equations are all correct.

But @Maurice, how can a single equation be balanced in multiple ways? Should not there be a single unique equation which is the correct one?

@HarikrishnanM, People write rules; the universe is not obligated to follow them. Consider nitrogen + oxygen, which forms various nitrogen oxides: https://en.wikipedia.org/wiki/Nitrogen_oxide

Yes @DrMoishePippik I am familiar with that example. But same equation with different stoichiometric coefficients?

I can form N2O5, NO2, N2O, N2O4 and so on... but I4O9 having many equations with different stoichiometric coefficients! Looks a bit new

This addition of reactions look like a similar situation in elementary mathematics. Suppose you have two equations $\ce{f_1(x)} = 0$ and  $\ce{f_2(x)} = 0$. You cannot prevent the following combinations to be true : $\ce{f_1(x) + f_2(x) = 0}$, or ; $\ce{3f_1(x) + 2f_2(x) = 0}$

You may want to use \tag{number} to enumerate or name equations. The way you're trying to do it breaks in many browsers, especially on mobile.

I disagree with this a bit. Just adding another reaction, which may happen at the same time, to the initial (or minimal) reaction is questionable. If these are indeed linked, there would be, given postcode conditions, still only one stoichiometry coming out. At least in that end, there answer is but very satisfying.

@HarikrishnanM https://chemistry.stackexchange.com/questions/58069/can-a-chemical-reaction-have-two-balanced-equations?noredirect=1&lq=1

It is not a good idea to gratuitously add in side reactions to an equation unless perhaps they are somehow linked to the reaction mechanism ie I atoms somehow hijack an O3 in the process of its decomposition to O2.


text:
Effects on pressure on reverse electrolysis
I am interested in the effects on pressure of the following reaction.
$$\ce{O2 + 4H+ + 4e- → 2H2O}$$
I have been trying to find data on this for a while, and I am sure that the information exists, but I haven't been able to find it. I need the math for the effects of pressure on the speed and energy output for this as the inverse of the electrolysis reaction. I can find data on electrolysis, but this seems to be more niche, if you know, or know an article that addresses this, can you please inform me if nothing exists? Is there some simple methodology that I am simply missing?

It's just a fuel cell, you likely didn't use the right keywords.

While it is related, it isn't explicitly about hydrogen fuel cells. Anyways, there are no tags relating to hydrogen fuel cells, to my knowledge.

$$E_\ce{O2/H2O} = E_\ce{O2/H2O}^\circ + \frac{RT}{4F} \ln{\left(p_\ce{O2} \cdot a_{\ce{H+}}^4\right)}$$

Follow up, can you confirm that this would say that the amount of energy released would decrease as the pressure increases?

No, I refute that statement. The potential of the oxygen cathode increases with oxygen pressure. As the used anode has probably lower potential (be aware it is a galvanic mode), the voltage of the cell increases with oxygen pressure. And provided energy for a given exchanged charge is proportional to this voltage.

You are writing the equation in a strange [at least to me] manner. Lets deconvolute. This seems to be a reaction of O2 and H2 to give water in a galvanic cell. At the anode, oxidation: H2 = 2H+ + 2e-; at the cathode, reduction: 1/2O2 +2H+ +2e- = H2O. Add the oxidation and reduction half equations to get:
H2 +1/2O2 +2H+ +2e- = H2O + 2H+ +2e-; H2 + 1/2O2 + 2H+ = H2O + 2H+.  The electrons have flown out the wire hopefully doing some work. H+ is a reactant at the cathode and a product at the anode.
The cell potential is the sum of the two half reaction potentials. These potentials are calculated from E[0] values [these can be looked up] modified for nonstandard conditions by the Nernst Equation. Detailed instructions can be found in any Physical Chemistry textbook. I left the H+ on both side of the equation because it is both a reactant and a product and since the cathode and anode are separated in space that means there is possibility of manipulation. H2 and O2 are reactants, an increase in either of their pressures should increase cell potentials and rates barring physical constraints in cell construction etc. Changes in H+ could be more complicated because cathode and anode mechanisms are different.

I do not see any issues with the way OP has written their equation, I do have to comment on the formatting of your reaction equations as they are a bit hard to read.

OP is conflating electrode reactions and overall chemistry in a manner that prevents OP from answering their own question. My equations are clear.


text:
Metal hydrolysis and equilbrium constants from stability constants
Problem: I want to create minimum solubility plots for a series of metal hydrolysis species against pH (below). But I cannot reconcile literature stability constants and equilibrium constants.
Context: Finding out a potential pH-range where certain metals (in mine wastewater) precipitate as hydroxides (adjustment with NaOH).

The plot above I created from equilibrium constants $log K$ found in Aquatic Chemistry: Chemical Equilibria and Rates in Natural Waters by Stumm and Morgan 1996.

But the literature values you find for these hydrolysis reactions are instead the stability constants $\beta_{p,q}$, for example here: https://www.cost-nectar.eu/pages/wg1_period.html that has compiled data from books books such as The Hydrolysis of Cations by Baes and Mesmer 1976 and Hydrolysis of Metal Ions by Brown and Ekberg 2016.
I know that the stability constant should the log-sum of the equilibrium constants leading up to that reaction. However I cannot seem to get the numbers right. Basically, my calculations of $\beta_n = \sum_{1}^{n}K_n$ in log-space for the tabulated values does not make sense. Below is a table of the equilibrium constants from Stumm and Morgan and the stability constants from Ekberg and Brown (with comma as decimal mark). I accept that there might be some numerical discrepancies since Ekberg & Brown are doing some quality assessment of previously reported values.

Note that aluminium is only used as an example here, since I had both data to create the plot from one source and the stability constants from another source. But I would like to create plots of 5-8 metals for which I have stability constants.
I feel I might be missing something really basic, but I have spent so much time reading the books and I thought perhaps someone could point it out easily.
Take $Al(OH)_{2}^{+}$ for example. From Stumm and Morgan 1996, the linear equation becomes:
$$log[Al(OH)_{2}^{+}] = -0.8 - pH$$
But calculating with the stability constants, I get:
$$log[Al(OH)_{2}^{+}] = -2.88 - pH$$
A fairly large discrepancy. The complete plot is:

What am I missing here? (I want to go from the stability constants $\beta$ to the linear equation systems)

The table containing data from both authors is confusing because it's unclear what the reactant is in two important rows. For the copy of the Stumm and Morgan data, the reactant is always solid aluminum hydroxide and the product varies. One product is aqueous aluminum hydroxide with a log K of –6.5, but this is mislabeled in the table as a solid. For Brown and Ekberg, what is the reactant? I thought for hydrolysis products, the bare metal ion would be the reactant, but if that's the case then the Al3+ row would be blank.


text:
Does reduction occur well above the reduction potential?
I was reading the voltammetry section of Analytical Chemistry 2.1 (Harvey), and it says that for the reduction of $\ce{[Fe(CN)^{III}6]^3-}$ to $\ce{[Fe^{II}(CN)6]^4-}$, which has a standard reduction potential of +0.356 V, if a potential of +0.530 V is applied to a solution of 0.100 mM $\ce{[Fe(CN)^{III}6]^3-}$ and no $\ce{[Fe^{II}(CN)6]^4-}$, there will be no faradaic current or change of concentrations at the surface of the electrode.
Using the Nernst equation, I would expect the concentration of $\ce{[Fe^{II}(CN)6]^4-}$ at the electrode surface to be 0.00114 mM. Is the textbook rounding this down to zero, or is there truly no reduction well above the reduction potential?

Shifting of electrode potential means shifting kinetic rate constants of reduction and oxidation. There is no shutting down threshold, just asymptotic convergence to zero.

You're right, the textbook statement is not entirely accurate. While applying a potential of +0.530 V, significantly higher than the standard reduction potential (+0.356 V), would thermodynamically drive the reduction of [Fe(CN)6]3 to [Fe(CN)6]4, some reduction will indeed occur. It's just that the rate of reduction might be very slow under those conditions.
Here's why:
Nernst Equation and Concentration: The Nernst equation tells us the equilibrium concentration of [Fe(CN)6]4 at the electrode surface for a given potential. In your case, calculating it using the Nernst equation gives 0.00114 mM. However, this value represents the equilibrium concentration, not the actual concentration at any given time.
Kinetics and Current: The actual rate of reduction depends on the kinetics of the electron transfer reaction. Even though the applied potential is thermodynamically favorable for reduction, the reaction might be kinetically slow. This means that the reduction will happen, but at a very slow rate, leading to a negligible faradaic current and a very slow change in the concentration of [Fe(CN)6]4.
Therefore, while the book took it zero might be a simplification for practical purposes, it's important to understand that some reduction will occur in reality, even if the rate is very slow and the change in concentration is negligible.


text:
Copper Blackening Biproducts
Apologies in advance for what is likely an incredibly basic question.
I am trying to form CuO from a Cu thin film using a bath of 1l DI water, 50g NaOH, 50g NaClO2.  I am wondering what the biproducts will be?
The reaction is described in this paper.  https://pubs.aip.org/aip/rsi/article/87/5/053119/982365/In-vacuum-scattered-light-reduction-with-black


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Translation of ferric and ferrous to German
I am not sure if this question belongs to the Chemistry or German forum.
My question is basically in the title. Are there good translations of ferric and ferrous to German which retain the meaning of Fe(III) vs. Fe(II)?
I.e. How would you translate something like:
The CAS(20,20) represents the smallest possible active space
  for this all-ferric model compound, with its 20 unpaired electrons [...]

to German.

German wikipedia pages for $\ce{FeCl2}$/$\ce{FeCl3}$ use the title names
Eisen(II)-chlorid resp. Eisen(III)-chlorid,
so it is almost identical to usual scheme element(ON) halogenide. (Eisen = iron).
So ferric = iron(III) = Eisen(III), ferrous = iron(II) = Eisen(II)}.
The model "element(ON)" is the preferred IUPAC way.
The alternative may be to use ferri-/ferro- prefixes.

This is definetely the right answer for single-core complexes. But if I have a multi-core complex and want to emphasize that all irons are Fe(III), then
"Eisen(III)-Modellverbindung" does not emphasize it in the same way as "all-ferric" does.
So I will wait a bit more before accepting it as answer.

Is "Alle(s)-Eisen(III)-Modellverbindung" applicable?

That sounds strange, at least to my ears. I guess I stick to "Eisen(III)-Modellverbindung" or "Eisen(III)-Modellkomplex". Thank you for the suggestions.


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Is the n factor/ valency factor of H3PO4, 2 or 3?
I'm very confused with search results I see on Google and different websites. I think the n-factor of H3PO4 is 3 because three H atoms are already bonded to three oxygen atoms. But some sources say it is 2, and I can't find out why.

If "these" sources say the $n$ factor is $2$, it may mean that they are speaking of an ion like $\ce{H2PO4^-}$. Can you mention the origin of these sources ?

It can be 1, if being neutralized to H2PO4-, 2 if being neutralized to HPO4^2- or 3 if being neutralized to PO4^3-.

Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking.

The n-factor of a molecule is not absolute. It can change from reaction to reaction.
If we are talking about H3PO4 then it can have a wide variety of n factors. The oxidation state of phosphorus in this compound is +5.
So it can change to only lower oxidation states(+3, 0, -3) (redox) or not change at all(neutralization reaction.
In the first case(redox) the n factor depends upon the change in  oxidation state. $$(+5\text{ to }+3):\text{ n factor is }2$$
In the second case that is in the neutralization reaction it depends on the basicity of phosphoric acid. $$\ce{ H3PO4\text{ to }H2PO4-}:\text{ n factor is }1.$$


text:
Ionic radii of cations and anions
My textbook mentions that across a period from left to right, the ionic radius of atoms decreases. I'm a bit puzzled because negative ions in the same period have an extra shell. I get the part about atomic radii decreasing from left to right due to the effective nuclear charge, but how is it that negative ions with an additional shell end up having a lower ionic radius than positive ions with a whole shell less?

TL;DR
Basically there are two possibilites:

Either the book is incorrect, or
The book is talking about isoelectronic species in a confusing manner.

The concept: The approach to studying atomic and ionic radii differs, as we focus on isoelectronic species rather than group or period trends. Ionic radii decreases as the charge decreases, while in a group, they increase. For example, the 2nd period of elements Li, Be, B, C, N, O, F, and Ne shows trends in ionic radii. The first three elements lose electrons and gain He configuration, while $\ce{C}$ gains electrons and becomes a negative ion, larger than the positive ions. The last few elements show a decrease in radii.

Glossary
Isoelectronic species: Species having the same number of electrons.
(e.g.) $\ce{N^3-, O^2-, F-, Ne, Na+, Mg^2+, Al^3+}$

Background note:
There is a difference in the approach when we generally talk about atomic and ionic radii.
We don't discuss group or period trends for ionic radii - rather, we talk about isoelectronic species and remember the fact that for isoelectronic species - as the charge decreases, the ionic radius decreases.
Obviously in a group the ionic radii will increase (as a shell is added.)

Now coming to your question:
What are the trends in ionic radii across a period?
Let us take the 2nd period.
Its elements are $\ce{Li, Be, B, C, N, O, F}$ and $\ce{Ne}$.
Now the first 3 would want to lose their electrons and gain He configuration. Compared to $\ce{H-}$ they are going to be smaller.
As we go across the period: $\ce{C}$ is like the cat on the wall. It could lose or gain 4 electrons, but both of them are quite unstable and thus it would share its electrons.
But here's the kicker: when it gains electrons and becomes a negative ion, it's larger than the positive ions before it. Why? The added electrons introduce some repulsion that slightly counteracts the nuclear pull.
And if $\ce{C}$ loses its electrons - it will become way smaller than $\ce{Li+}$, $\ce{Be^2+}$ and $\ce{B^3+}$.
As we go further, the last few elements of 2nd period - we can see a decrease in the radii. So the graph of ionic radii will look like:

Now THIS is the reason - this nitty-gritty trend - that we usually restrict ourselves (atleast at my level) to studying isoelectronic species.
The graph will look different if $\ce{C^4−}$ is taken. The peak will occur at $\ce{C}$ itself.

Sources for image: https://chart-studio.plotly.com/~hannahghutson/14.embed


text:
visualization problem of dilution of enthalpy $\Delta_{dil} H$ data with different initial concentrations
description
I have a question about the $m_f$ vs. $\Delta_{\text{dil}} H$  plot in some experimental studies.
For example, the dilution-of-enthalpy data at 300 °C with two initial concetration ($m_i$) 0.2470 and 0.5199 are reported as follows:

$m_i$: initial concentration
$m_f$: final concentration

then the author did the visualization by using $m_f$ vs. $\Delta_{\text{dil}} H$ like this:

however, when I try to do the same, I got:

it seems that the author explained a little under the diagram:

At each temperature, a constant was added to the
heats of dilution for the solution with the lower initial concentrations so that the zero value for these
solutions would correspond to the zero value of the solution having a higher initial concentration

but it is not very clear to me, could anybody explain a little bit about this? How a constant could be determined for all the enthalpies of solutions with lower initial concentrations?
reference
Oscarson, J. L., Gillespie, S. E., Chen, X., Schuck, P. C., & Izatt, R. M. (2001). Enthalpies of Dilution of Aqueous Solutions of HCl, MgCl2}, CaCl2, and BaCl2 at 300, 325, and 350° C. Journal of solution chemistry, 30, 31-53.


text:
What are the byproducts of ozonolysis?
I understand that ozonolysis of alkenes leads to formation of aldehydes or ketones or carboxylic acids depending on the type of work up, reductive or oxidative.
But when I tried searching online I couldn't find anything about the side products.
In summary my question is
"What are the by products when an alkene, say but-2-ene undergoes reductive ozonolysis(with Zn/H2O) and oxidative ozonolysis(H2O2) separately, in each case?"

In reductive ozonolysis, we get the Carbonyl compounds, and Zinc(II) oxide.
In oxidative ozonolysis (where we use water), we initially get the carbonyl compounds and Hydrogen peroxide. This hyrdogen Peroxide often reacts with aldehyde (if formed) to form the respective carboxylic acids and water.


text:
How to integrade a group of disconnected smiles descriptors from molecular into a single smiles descriptor?
I am new to cheminfo so this question may sound stupid.
I am trying to calculate the RDF(radial distribution function) of these two compounds below:
https://pubchem.ncbi.nlm.nih.gov/compound/155293646
https://pubchem.ncbi.nlm.nih.gov/compound/15182658
and I'm using http://www.scbdd.com/chemdes/start/ where it takes SMILES of the compound as an input.
However, I notice that RDF, due to its internal algorithm, can only calculate the descriptor for one single molecule per run, but since mine have several disconnected structures for each molecule, it seems impossible to run.
I wonder if there is some package/website/software for me to preprocess disconnected molecule like those so that I can get a single and continuous SMILES descriptor from it.

Look into "desalting" modules in rdkit.


text:
Is this a correct Lewis Dot structure for ozone?

I drew a Lewis dot structure of ozone but when I look online it is a bit different. Can I make the single covalent bonds like a triangle that connects all three atoms or is that not allowed? If so, what rule is that and you describe it and its name?

Dear OP, by the way, that 'trioxygen' is dubbed ozone.

For starters, the Lewis dot structure for ozone does not look like that. It is not a ring.
This is explained in this link: Can ozone have a triangular structure?
There it is also given (one of the simpler answers):

In order to form a 3-member ring your electron pairs would have to bend at 60 degrees; the energy required to push naturally repulsive electrons so close together would make the molecule very unstable.

The correct Lewis dot structures along with resonance structures, can be shown as below:

This is from my NCERT textbook. You can check out the bonding and the Lewis dot structure.

------------------------------------------------------------------


text:
Are buffer solution and acidic buffer different?
I have studied that acidic buffer is a type of buffer solution but the definition of buffer solution says that it's a mixture of a weak acid and its conjugate base or a weak base and its conjugate acid whereas the definition of acidic buffer says it's mixture of a weak acid and its SALT with a strong base. I don't understand this one buffer solution is an acid base mixture but acidic buffer is an acid salt mixture

Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking.

pH buffers are solutions with low value $\left|\dfrac{\text{d}(\text{pH})}{\text{d}n_\text{base/acid}}\right|$, i.e. their pH does not change much by addition of an acid or base.
How it is achieved is secondary. It is managed by two ways:

Having a conjugate pair (Broensted-Lawry) acid+base (or several ones). They are usually combination of

A weak acid and its salt with a strong base, like acetic acid/sodium acetate
A weak base and its salt with a strong acid, like ammonia/ammonium chloride
As acids/bases are often are used respective salts of multiprotic acids like phosphoric acid or citric acid.

Requirements for pH buffers with very low or high pH can be addressed solely by solution of strong acid or base, as high concentrations of $\ce{H+(aq)}$ or $\ce{OH-(aq)}$ have natively high buffering capacity.

Acidic/basic buffers may have several meanings:

Having acidic/basic pH.
Using a weak acid/weak base  and their salts as conjugate bases/acids.
Using strong acids/bases as the second buffer type.

So here the salt serves the same purpose that conjugate base does?

@yyzr The salt IS the conjugate base. More exactly, the anion of the salt - $\ce{CH3COO^}$ - / Or in the other example $\ce{NH4+}$ as the conjugate acid.


text:
How to chemically remove CaO from boron carbide?
Previously I asked (How to remove graphite substrate of boron carbide coating?) how to chemically remove a thick layer of graphite from a thin layer of boron carbide. It turned out that was a hard task. I hope the task of chemical removal of a thick (about 1.5 mm) layer of calcium oxide from a thin (about 150 microns) layer of boron carbide would be easier as CaO reacts with water. Any specific advice on how to do the removal? What rate of removal should I expect?

Water should quickly react it, hotter water will remove it faster.
Before you proceed, note that this reaction is exothermic and you need to have appropriate safety precautions. Additionally, this is toxic so keep your hands clean
It is highly likely that any calcium oxide has reacted to form calcium carbonate. To counter this, put it in water with some dilute nitric acid, which should react all the calcium carbonate.
Then, the calcium oxide should form calcium hydroxide, which is sparingly soluble. The calcium hydroxide will react with the nitric acid anyway, and any calcium nitrate is soluble so that should not be a concern.
You can expect a fairly quick reaction (depending on concentration of nitric acid as well as the heat of the reaction), but I would recommend trying the experiment at a lower temperature with proper PPE to assess it's efficacy


text:
Jahn-Teller Distortion: How to compare magnitudes (in Cu2+ complexes)?

Octahedral complex of copper(II) undergo structural distortion (Jahn-Teller). Which one of the given copper(II) complexes will who the maximum structural distortion? (en - ethylenediamine)

(A) $[Cu(H_2O)_6]SO_4$

(B) $[Cu(en)(H_2O)_4]SO_4$

(C) $cis-[Cu(en)_2Cl_2]$

(D) $trans-[Cu(en)_2Cl_2]$

The answer given is A.
How do I compare the magnitude of Jahn-Teller distortion on $e_g^3$ complexes?

I have subjectively understood the JT effect: $Cu^{2+}$ in its $d^9 -t_{2g}^6e_g^3$ configuration has an asymmetric configuration, so the octahedral complex breaks symmetry by distorting, usually by lengthening (z-out) along the $d_{z^2}$ orbital or shortening (z-in)
However in the question, all the complexes have the same configuration.
Is it something to do with the strength of the ligands?

This question is from the second shift of the JEE Main
exam held on July 29, 2022. There is a possibility the answer is wrong, however I would still like to know how to compare distortion.
I have referenced the following material:

JD Lee (Sudarshan Guha): does not compare magnitudes
Libretexts: says "stronger M-L interactions increase chance of observing JT" . That sounds too vague.

I would appreciate some insight.

Well until it defines "maximum structural distortion" I have no idea what it means. The en complexes will already be distorted from Oh before J-T is even considered, how are we measuring distortion on top of distortion?

@IanBush what's "Oh"? Is it a group theory thing?

Sigh. Yes. It is short for Octahedral, and represents the point group for that symmetry. It is where labels such as $t_{2g}$ come from.


text:
What is the priority of an alkyl substituent containing double or triple bonds, in comparison to the other functional groups?
If we have an alkyl substituent containing a double bond(s) or triple bonds(s), what is it's priority in comparison to other functional groups and substituents?
For example, take the compound given below

What would it's IUPAC name be?
We know that the cyclohexane ring will be the parent chain (as it contains two functional groups), and the ethenyl substituent as well as the $\ce{-Cl}$ will be prefixes, while the $\ce{-OH}$ will be the suffix. This means we start numbering from $\ce{C1}$ carbon atom, which the $\ce{-OH}$ is attached to.
But exactly how would we go about numbering the carbons? And what is the priority of the ethenyl group and alkenyl or alkynyl groups in general (so that we know how to number it such that the higher priority prefix receives the lower locant, in accordance with set rule)?

Related: Doubts about IUPAC nomenclature of 4-ethyl-1-fluoro-2-nitrobenzene

There is no priority order for functional groups that are expressed as prefixes. When assigning the lowest set of locants, they are all considered together.
The corresponding section in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows:

P-14.4 NUMBERING
When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:
(…)
(f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;
(g) lowest locants for the substituent cited first as a prefix in the name;
(…)

Note that Rule f takes precedence over Rule g.
Therefore, the example is named as 2-chloro-6-ethenylcyclohexan-1-ol rather than 6-chloro-2-ethenylcyclohexan-1-ol since ‘chloro’ is cited first as a prefix in the name.


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Largest molecules verified in a Miller Urey experiment
What are the largest molecules confirmed with confidence in a Miller Urey experiment?
I know the experiment can create amino acids (which are mostly <250Da) and nucleotides (which are mostly <350Da). However, I see some papers suggesting molecules even in the 800-1000Da range.
Is it true such large molecules can form in the experiment? And, if so, what types of molecules are they typically?
Thank you.

There might not be a rigorous upper bound. There's been many variations on the experiment, but the sludge they tend of form ("thiolin") often contains complicated polymeric molecules.

Thank you, Nicolau! So, in your opinion, it is not too rare for large >600Da (or even >800Da) molecules to form in the Miller Urey?

It's certainly possible. It's useful to have to keep in mind that you generally find what you are looking for, and it's typically easier to detect and characterize smaller molecules (especially ones we already know, or others similar to them). Larger molecules generated under these conditions are probably extremely varied and structurally chaotic, and even though their collective amount may add up to a substantial fraction of the sludge, each individual substance will be present in very low abundance, so they may end up ignored.

I suspect running high-resolution mass spec measurements on the sludge could give a so-called "forest" of weak signals at higher molecular weights, though I haven't read the articles to confirm it.


text:
Identifying analytes in Gas Chromatography
Since the retention time of a sample tested using GC will necessarily depend on the column parameters (stationary phase material, oven temperature, flow rate, etc), it follows that the chromatograph will also depend on these factors. How, then, can we compare our results to those published in databases online?
Either the parameters mentioned above should be standardized, or the instrument needs to be 'standardized' (or calibrated) by using a calibration sample and performing some post processing such that for this calibration sample we get identical chromatograph to what is found for the same sample in the data bases.
Which of these actually happen in practice?

You can use a retention index based on e.g. alkane standards. You run the alkane standards on your column, and compare their retention time to that of your sample to get a retention index. Then, you can compare it to published retention index data.

As a comment to Karsten, It can be added that the retention indexes depend on the polarity of the stationary phase and on the temperature. They do not depend on flow rate. Table exist that show the retention indexes R of a lot of substances at a given temperature, plus the rate of changing dR/dT. Using these parameters, you can probably calculate the retention times of all your compounds.

@Maurice so then I need to compare retention indices using same or similar stationary phase material?

Yes. You understand the technique. The only trouble is, by reading the tables,  that the same product may have sometimes different retention indexes (up to ± $5 - 10$ in retention indexes) for the same column at the same temperature. It only seems to depend on the author of the measurements. I don't know how to explain this strange discrepancy.


text:
How do you calculate the pH range for which one metal precipitates but the other stays in solution?
The problem I have been asked to solve is:

$\ce{Cd^2+}$ and $\ce{Fe^2+}$ (starting concentration $\pu{E-3 mol/L}$) should be separated by precipitation using $\ce{H2S}$. Does a pH range exist allowing to completely precipitate one metal (i.e. the remaining concentration of the precipitated ion has to be $\pu{E-5 mol/L}$ or less) whereas the other stays in solution at its initial concentration? Please calculate this $\pu{pH}$ range!

Solubility product: $\ce{(CdS)}$: $\pu{E-27}$; $\ce{(FeS)}$: $\pu{E-19}$; Acid constant $\ce{(H2S)}$: $\pu{E-20}$; concentration of $\ce{H2S}$ = $\pu{0.1M}$.

My approach to solving the problem was the following:
$\pu{K_{sp}}$ = [$\ce{M+}$][$\ce{S^2-}$] $\ce{->}$ $\pu{K_{sp}}$ for either metal can be defined in the following way:
$\ce{[S^2-]}$ = $\pu{K_{a}}$ $\frac{[\ce{H2S}]}{[\ce{H+}]}$ $\ce{->}$ the concentration of sulfide ions is given by this expression
[$\ce{M+}$] = $\frac{\pu{K_{sp}}\ce{[H+]}}{\pu{K_{a}[\ce{H2S}]}}$ $\ce{->}$ substituting the second equation into the first we get the concentration of metal ions in terms of $\pu{pH}$.
After this, we must find the $\pu{pH}$ range where the concentration of $\ce{Cd^2+}$ is less than $\pu{E-5 M}$ and the concentration of $\ce{Fe^2+}$ is greater than $\pu{E-3 M}$.

For $\ce{Cd^2+}$:
$\pu{E-5} =\frac{\pu{E-27} \times [\ce{H+}]}{\pu{E-20} \times 0.1}$
Solving for $\ce{H+}$ gives $\ce{H+}$ = $\pu{E-3 M}$, or $\pu{pH = 3}$.
For $\ce{Fe^2+}$:
$\pu{E-3} =\frac{\pu{E-19}\times[\ce{H+}]}{\pu{E-20} \times 0.1}$
Solving for $\ce{H+}$ gives $\ce{H+}$ = $\pu{E-5 M}$, or $\pu{pH = 5}$
Thus the $\pu{pH}$ range is $\pu{3-5}$

I realise that this is a homework question, but I hope I have shown that I have given this problem some thought. I am really uncertain, whether my approach is correct, especially the step where I found the $\pu{pH}$ at which the concentration of $\ce{Cd^2+}$ ions is less than $\pu{E-5 M}$ (because I could have also calculated the $\pu{pH}$ at which the concentration of $\ce{Fe^2+}$ ions is less than $\pu{E-5 M}$). Some help would be greatly appreciated.

Formatting guides for texts and formulas/equations/expressions.

Gosh, my retina is facing its apocalypse.

I'll edit the question.

@HarikrishnanM +1 for your heroic effort!

Actually I love problems like these as these are the exact kinds that come for my competitives.

@andrej you're right, no worries.

H2S had two constants. They probably mean  the product of both constants, skipping HS-. So K=Ka1.Ka2=[H+]^2[S^2-]/[H2S] . Be aware of the square.

@Poutnik - I would have thought, and it is certainly what I was taught, that there were 3 sulfur species too. However the Wikipedia article on hydrogen sulfide says that  $\ce{S^{2-}}$ does not exist in aqueous solution. I'm confused at this point how the sulfide ppt constants would work.

@MaxW I often read Ka2 is near 13.5-14., so I am not sure  But if it does not exist, all sulfide solubility products are questionable how they are defined.

@Poutnik - I was looking for the $\ce{K_{a2}}$ when I stumbled on the Wikipedia article. If I remember correctly some of the sulfide ppts are also slow to form. Precipitation also has some sort of kinetic factor since 1 molecule of $\ce{CdS}$ won't precipitate. A particle has to grow to a "large" size to ppt. This question is more interesting than it would seem.

I'll point out that the question also ignores hydroxide precipitates. The problem statement indicates that there is a $\ce{10^{-8}}$ difference in the sulfide ppts so I'd expect more than a 2 pH unit difference if hydroxide ppts are to be ignored.


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How does X! Tandem calculate mass from an MS2 spectrum?
I am very new to proteomics, and am using the X! Tandem tool to match MS2 spectra to a list of protein sequences. I understand that X! Tandem (and other similar tools) works by narrowing down the list of proteins using the mass of the precursor ion. I understand how the mass of the precursor can be calculated using isotopes in an MS1 spectrum, but Tandem only takes MS2 spectra as input. How are they calculating precursor mass?
Thanks in advance!


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Mn(OH)2 synthesis
I have a small problem while reading Inorganic Chemistry Reference book.
The book stated that "In laboratory, Mn(II) hydroxide was prepared by the reaction of Mn(II) salt with alkaline in hydrogen atmosphere"
I guess the hydrogen atmosphere was to avoid Mn(II) being oxidised to Mn(III). Could it be nitrogen gas or other inert gases?

Possibly yes: https://chemistry.stackexchange.com/a/165867/17368, https://chemistry.stackexchange.com/questions/165889/does-solid-manganese-reacts-with-atmospheric-air-when-heated

I was also thinking that hydrogen gas could act as reducing agent in this case. Nitrogen, on the other hand, is way more inert to act as reducing agent. I wonder whether this explanation contributing to the usage of H2 atmosphere.

Look up the Winkler method to measure dissolved O2 in water. In base MnII is oxidized by O2 to MnIII. The O2 could probably be removed by addition of a pinch of bicarbonate to an acidified solution, purging the CO2 with N2 or argon, and degassing the hydroxide.   Alternatively react Mn with HCl maintaining a hydrogen atmosphere and neutralize with degassed hydroxide.

The described method using hydrogen was published in Z. anorg. allg. Chem. 1937, 232(4), 369–381 and modified in J. Am. Chem. Soc. 1950, 72, 2, 856–866.

@jimchmst I'm pretty sure it's oxidised to MnO2 - Mn(IV) not III.

Nitrogen would work if it were pure. In the real case it may contain traces of air/oxygen, or pick up this impurity as it is being delivered, and Mn(II)  could be sensitive to that. Also the solution may carry air if it has be exposed to said air.
In steel processing applications a mixture of nitrogen with 5% hydrogen is often used, especially for heat treating, with the added hydrogen serving as a "buffer" against air or moisture impurity.