text:
Expression of the potential in the hydrogen atom while solving radial part of the wave function
I was reading about the derivation of wavefunction of Hydrogen atom from Atkins book. After the separation of variables and writing the  wave function, $\psi_{(r,\theta,\phi)}=R_{(r)}Y_{(\theta,\phi)}$, and further solving the equation, they reach the expression of potential as

$V_{eff}=-\frac{Ze^2}{4\pi\epsilon_0r}+\frac{l(l+1)\hbar^2}{2\mu r^2}$

The first term $(-\frac{Ze^2}{4\pi\epsilon_0r})$ represent the electrostatic potential between the electron and the proton. I am not able to understand the interpretation of the second term $(\frac{l(l+1)\hbar^2}{2\mu r^2})$. In book it is written that it arises from centrifugal effect.
Also why the second term is considered in the potential as potential is defined for a conservative force field? I know that second term in potential comes in picture as a result of the derivation. But the second term doesn't make sense to me as in Bohr's model,  in the expression of the potential we only consider the electrostatic potential. Please explain about the second term in potential.

Stop thinking of it as a potential for a moment. We separated the variables, and we got this; which step of it do you disagree with?

Basically the problem is that while doing the derivation, we got this. So the expression of the potential is mathematically consistent. But I wonder that second term in potential basically is $L^2/2I$, and while solving the hamiltonian of angular part of the wave function (taken under the different heading as "particle on sphere" in some books) $L^2/2I$ is included in the energy of it.

So why we again consider $L^2/2I$ in radial part of energy. Intuitively only radial kinetic energy and electrostatic potential energy should be considered in the total energy. I know that my intution is wrong as it doesn't met with the mathematical analysis of the radial Hamiltonian equation. But I am not able to find the reason for the same.

That's why I started like this: stop thinking of it as a potential. Once you've separated the variables, there are no energy, potential, or other words with direct physical meaning. There is just the mathematical derivation, and it is what it is. Follow the steps, check that the result is right. So it goes. Why people call it the effective potential is another story.

OK. So double the rotational Kinetic energy, $(1+1)L^2/2I$ (one from angular part and other from radial part) is included in the total energy ($E_{radial}+E_{angular} $).

There is only one energy and no innate way to split it into the radial and angular parts. You may do so, but that's a matter of convenience, not the law of nature.

OK, now I have understood my mistake. While separation of variables, the E included in the radial equation is the total energy not only the radial energy. So in books 2$L^2/2I$ is never taken.

The 'centrifugal' term appears when the coordinates are changed from $(x,y,z)$ to spherical $(r, \theta, \varphi)$. This is not to say that it is artificial, or a mathematical trick, but is a consequence of the fact that the electron has orbital angular momentum, and this is made clear by changing coordinate systems.
To summarize: the equation is $H\varphi(\pmb{r})=E\varphi(\pmb{r})$ where $\displaystyle H=-\frac{\hbar^2}{2\mu}\Delta +V(r)$. Changing to spherical coordinates produces,
$$\Delta =-\frac{1}{r}\frac{\partial ^2}{\partial r^2}+\frac{1}{r^2}\left(\frac{\partial ^2}{\partial \theta^2}+\frac{1}{\tan(\theta)}\frac{\partial}{\partial \theta}+\frac{1}{\sin^2(\theta)}\frac{\partial^2}{\partial \varphi^2} \right)$$
which means that eigenfunctions of $\varphi(\pmb{r})$ are variables in $r, \theta, \varphi$
The second term is almost the same as that for $\pmb{L^2}$ and substituting for this makes the Hamiltonian
$$\displaystyle H=-\frac{\hbar^2}{2\mu}\frac{1}{r}\frac{\partial ^2}{\partial r^2}r+ \frac{1}{2\mu r^2}\pmb{L^2}+V(r)$$
As $\pmb{L^2}\varphi(\pmb{r})=l(l+1)\hbar^2\varphi(\pmb{r})$  then
$$\displaystyle H=-\frac{\hbar^2}{2\mu}\frac{1}{r}\frac{\partial ^2}{\partial r^2}r + \frac{l(l+1)\hbar^2}{2\mu r^2}+V(r)$$
The potential is unchanged and still $V(r) = -e^2/r$ and the Hamiltonian is still a function of $r$ but now depends on variable $l$ thus the eigenvalues must depend on $l$.
The centrifugal term is so called because the corresponding force, the negative slope, always repels the electron from the centre and labelled a centrifugal potential or barrier. The term is always positive or zero and can change the shape of the potential the electron experiences and thus forms an effective potential written as  $\displaystyle V_{eff}=\frac{l(l+1)\hbar^2}{2\mu r^2}+V(r)$. When $l>0$ the centrifugal term dominates at small $r$ causing $V_{eff}$ to be repulsive. As larger $r$ the potential becomes attractive.


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The most acidic hydrogen among ethane, ethene, ethyne and allene
Among
$$\ce{CH3-CH3}\qquad\ce{CH2=CH2}\qquad\ce{HC≡CH}\qquad\ce{CH2=C=CH2}$$
I need to find out the one having highest acidic hydrogen.
As per my observations, there are two equivalent resonating structures possible in case of conjugate base of $\ce{CH2=C=CH2}.$ However, the correct answer is given as $\ce{HC≡CH}$ due to inductive effect of sp hybridization of carbon.
But the priority order in determining stability of a conjugate base is:

Aromaticity
Number of equivalent resonating structures
Resonance
Hyperconjugation
Inductive effect

As such, the correct answer should have been the allene. Where did I go wrong?

Does the downvoter have any gut to comment?

Related: Relative acidities of alkanes, alkenes, and alkynes.

I did not downvote, but likely some people will if they seem to see a homework problem with no/too little work input from the asker.  You might want to look at the link given be andselisk for some ideas, as well.

This "priority order" is hardly a law of nature. In my opinion, instead of trying to force everything into a fixed order, it's a better idea to get more experience with various functional group pKa's so that you have a better idea of how much each factor affects the pKa. That takes time. You might think that memorising some order is a quick shortcut to the top, and maybe it does work for quite a few cases, but I fear that this sort of approach comes with the risk of having to unlearn wrong things later on.

But that's just me. If you have to learn this sort of thing, it's ok, just make sure to take it with a pinch of salt. Especially because these sort of orders were worked out precisely by observing the pKa's of various compounds, so using them to predict pKa's is a little bit like circular logic. TLDR: Models only work in the contexts in which they were developed.

@OscarLanzi well, this isn't exactly a homework problem. I was trying this on my own and did my calculations and have also drawn conclusions accordingly as given in the question. It just gets irritating when you've genuinely asked a question and someone out of the blue votes it down without telling what to fix, you know -_-

I agree with @orthocresol. The stability order here would be $\mathrm{sp \gt sp^2 \gt sp^3}$. In allene, $\ce{H}$ is on $\mathrm{sp^2}$ carbon, so most stable carbanion would be that from acetylene.

@andselisk I know the order of acidity of different types of hybridization. Was just confused about the priority order :(

@orthocresol thanks, that way it works. But, I was wondering about the priority order. So asked this question. But, I was really disappointed when someone voted it down without citing a reason :(

@MathewMahindaratne Thank you so much. Seriously, I was just pissed of due to that downvote for this question. Sorry :(

To paraphrase @orthocresol, models work insofar as they reflect reality, but you may discover that a specific model that you frequently use does not cover a specific case, or use case or whatever. Models explain/predict some features in some cases, but that doesn't mean they are general or right or anything.

The inductive effect will always make the stronger acidic compound, especially with an sp hybridization as it will ultimately stabilize the electrophile best in solution. Though resonance will also contribute to the stability, a structure similar to allene will have more resonance as it will have more bonds focused in one area so close to the electrophilic site.
I think you over resonance, its an easy mistake, but my notes from Orgo I show that the difference in pKa from an sp2 to an sp hybridization are roughly a factor of 2.
TL:DR, Induction beats resonance.

But in most of the cases that I've seen in these types of problem, resonance is prioritized over induction :(

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Does Potassium Superoxide also absorb Carbon Monoxide?
Potassium Superoxide ($\ce{KO2}$) is used as an oxygen provider and carbon dioxide scrubber in life support systems
$$\ce{2KO2 + H2O -> 2KOH + O2} \\
\ce{2KOH + CO2 -> K2CO3 + H2O}$$
I encountered a question that asked if Potassium superoxide can also serve as a scrubber for $\ce{CO}$. I thought of a reaction where $\ce{KO2}$ oxidized $\ce{CO}$ to $\ce{CO2}$ and the $\ce{CO2}$ was absorbed as previously mentioned
$$\ce{2KO2 + 2CO -> K2O2 + 2CO2}$$
Another possible reaction would be reduction of $\ce{KO2}$ to $\ce{K2O}$
$$\ce{2KO2 + 3CO -> K2O + 3CO2}$$
I could not find these reactions or similar ones mentioned in any literature online. This article describes the working of $\ce{KO2}$ in detail but I could not find a mention of carbon monoxide in there.
Can $\ce{KO2}$ also be used to scrub $\ce{CO}$? If so, how?

CO is a rather stable substance. It will not react with any peroxides or superoxides. Anyway if a reaction were feasible, it would never produce $\ce{ K_2O}$ and $\ce{CO_2}$, because this substances would react to produce $\ce{K_2CO_3}$.

Alkali and alkaline earth peroxides and higher oxides need an acid to set off their oxidizing/oxygen-releasing power under ambient conditions, even if the "acid" is just water in some applications. Carbon monoxide, unlike carbon dioxide or water, is too weakly acidic to cooperate.

Let's assume the reaction is feasible. KO2 oxidizes CO to CO2 with KO2 getting consumed to form K2O2, K2O etc (simultaneously). As the reaction progresses, more and more KO2 is consumed in oxidizing CO to CO2. So there will be no free KO2 to absorb CO2 as all of it takes part in the reaction.

CO is difficult to remove because of its inertness. Because it has a low b.p and low critical temperature, it cannot be effectively adsorbed by any physical adsorbent at r.t. Also, its insolubility in various solvents adds to the difficulty in its physical adsorption. FWIW, CO is removed via catalytic oxidation using manganese dioxide and copper oxide.

As the other answer and comments imply, potassium superoxide is not a good reagent for absorbing carbon monoxide.  Cuprous chloride solution works better, as in this study[1] where the salt is dissolved in an ammonia-bearing solution.
Reference:

R. V. Gholap and R. V. Chaudhari, "Absorption of carbon monoxide with reversible reaction in cuprous chloride solutions", Ind. Eng. Chem. Res. 27, 11, 2105–2110 (1988).

No, the reactions involving KO2 do not directly react as you indicated, although there is a claimed reaction producing a curious analog to H2O2 with a potassium atom taking the place of one of the hydrogen atoms from the action potassium superoxide with water (water vapor):
$\ce{2 KO2 + H2O -> KOH + KHO2 + O2}$
But, still no reaction of CO with KHO2 forming CO2 expected.
Interestingly, in the presence of the hydroxyl radical, however, per this source, Eq (15), to quote:
$\ce{CO + .OH -> CO2 + .H}$
but maybe occurring only at low temperatures.
So, given a visible light, or UV photocatalyst like TiO2, in presence of water and carbon monoxide, at the appropriate temperature, perhaps a slow limited conversion of CO to CO2 is, after all, theoretically possible, but decidedly not practical.
Note: CO will burn in air/oxygen forming CO2:
$\ce{2 CO + O2 -> 2 CO2}$
Reference: See, for example, this ebook.

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Lewis acidity order of Aluminum Halides
There is a similar question already on StackExchange which was not satisfactorily answered. So I am asking it again. I am unable to find this information in any book.
Link to the other question:
Acidity of aluminium trihalides

What question do you refer to? What is unsatisfactory about the existing answer? Why the name of the scientist isn't capitalized, but the name of a chemical compound is? Also, please familiarize yourself with the homework policy.

I have provided the link @adnselisk

Does this answer your question? Acidity of aluminium trihalides

I just wanted an answer and so asked it again...as no one answered the previous one properly...

According to Ref.1: The order of Lewis acidity of gaseous halides of aluminum is established to be $\ce{AlF3(g) < AlCl3(g) < AlBr3(g) \approx AlI3(g)}$. The order of Lewis acidity of molecular and solid group 13 halides is $\ce{AlX3(g) > GaX3(g) > InX3(g)}$ (where $\ce{X = F, Cl, Br, I}$).
References:

Thoralf Krahl, Erhard Kemnitz, “The very strong solid Lewis acids aluminium chlorofluoride (ACF) and bromofluoride (ABF)—Synthesis, structure, and Lewis acidity,” Journal of Fluorine Chemistry 2006, 127(6), 663-678 (https://doi.org/10.1016/j.jfluchem.2006.02.015).

Does the solid state also have the same order?

Can't say without any data. Keep in mind that at solid state, these halides tend to make dimers, trimers, etc., thus different than gaseous phase.

Ok....does the same order apply for the catalytic action during Friedel Crafts reaction?

Should be. But most common one is $\ce{AlCl3}$ due to its covalent nature in non-polar solvents.


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Is chirality absolute or relative?
While going through some questions on chirality and optical isomerism, I came across this compound:

Note that I have marked 3 carbons as $C_1,C_2,C_3$
Its obvious that $C_1,C_3$ are chiral carbons.
$C_2$ is not a chiral carbon as it has the same group above and below it( With respect to the picture)
However a thought struck me that chiral carbon meant that 4 different groups are attached to it.
But what exactly does same group mean?
Does it mean that even the spatial arrangement of the groups should be different?
Or does it not matter?
So if spatial arrangement mattered, then the compound I have mentioned would have a chiral $C_2$ carbon for specific configurations around $C_1$ and $C_3$ carbons.

C1 and C3 are chiral atoms, and are of the variety R or S. As a consequence, RR and SS are optically active. But the combinations RS and SR are optically inactive. They are meso compounds.

Even in the meso compound, C2 is pseudochiral.

C2 is achirotopic but stereogenic with a configuration of r (lower case).

Yes, spatial configuration of attached groups matters. Meaning different stereo configuration of $C_1$ and $C_3$ may render $C_2$ either chiral or not:

Also because you have more than 1 stereo center you're not limited to just enantiomers, you can get diastereomers - the molecules will be different even though they're not mirror images of each other:

The compound must pass the sigma test, as in symmetry. If you imagine a mirror cross sectioning the molecule with more than one chiral carbon and the substituents of the chiral centers are cis, then the molecule is achiral. If the substituents are in a trans configuration, then the molecule is chiral.

Don't we use R and S for this instead of cis and trans? Thanks for the answer anyway!

R and S are to designate what direction the substituents of the enantiomers rotate in respect to the chiral center. Cis and Trans are used for denoting whether a compound with two or more chiral centers is chiral or not.


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Carbanion Stability order
Why is a 3° Carbanion unstable than 1° Carbanion? Carbon atom with negative charge will become little electropositive relative to a neutral carbon. So if a Carbanion is attached to three neutral carbons, essentially it should donate it's electrons distributing the negative charge of -1, thus stabilizing it. By this logic, more carbon atoms attached means more stability. But this does not happen. 1° Carbanion is most stable. Why?

You are half way through explaining the question yourself. You just need to realise that alkyl groups themselves are electron donating groups.

Why that is my question they should be more Electronegative than the negatively charged carbon, thereby accepting electrons.

To make it common knowledge, I am a high school student. For my reasoning I have considered only the Inductive effect which is basically electron density movement from less to more Electronegative atom. I have even not considered C-H bond polarity in alkyl group nor the hyperconjugation effect. Please tell me what I am missing particularly and why my logic is wrong?

Methyl group is EDG and exerts +I effect due to Hyperconjugation. Related: Why is the methyl group electron-donating via the inductive effect? and What is Hyperconjugation?

Does this answer your question? Why is the methyl group electron-donating via the inductive effect?

The above links explain why vicinal C-H bonds stabilize carbocations. I think the OP is looking for an explanation for carbanions that is essentally not just "Oh! It's the reverse order of carbocations".

Carbanions and Carbocations are perfectly opposite in order of stability because they are opposite in order of charge. The order is preferred of carbocations the same as acidity by Lewis' nucleophile/electrophile definition. Therefore, because a positive charge is stabilized better by a tertiary carbon, it is preferred to host a carbocation because the electron shells of its substituent carbons help stabilize the charge.
In the case of a carbanion, the negative charge is best stabilized when not in contact with neighboring nuclei that would resonate the negative charge away from the nucleophilic site. If you think about it like sterics, the further away the negative charge can be from other substituents, the better in the case of carbanions.


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Enamine Formation in a Diketone
I am trying to form an enamine, and I was trying to figure out if the following reaction is possible:

Would this reaction terminate after the formation of the shown product, or will the other carbonyl also react, producing a final product with two enamine groups? If so, is it possible under any conditions to stop the reaction after the formation of the first enamine?

You won't form a second enamine because the mono-enamine is a vinylogous amide. I suggest placing the diketone (dihydroresorcinol) in an inert solvent (THF?) with anhydrous MgSO4. Add excess condensed dimethylamine.  The reaction may have an exotherm.

Thank you so much! That makes a lot of sense now.

The MgSO4 is to absorb water. Filter and evaporate.


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How to make a sample in ppm? (not a specific compound, only the sample)
Is soaking 1000 mg tea powder in 0.1 L water for about 12 - 24 hours, can be considered as 10000 ppm tea sample?
Could anyone give me reference from a journal or something similar, about my question topic?
Really appreciate any help you can provide.

Why do you want this? What for? Who told you "10 000 ppm of tea" is what you need. Is this a school project?  ppm sounds scientific, but you need to use it properly!

Note that ppX units are discouraged from being used because of their vaqueness in many use cases. Also, 100 mL does NOT contain 1000 mg of tea extract, so it cannot be said it is 10000 ppm (w/V) tea extract.

@Poutnik ppm´s are only vague if used sloppily, but you don´t save sloppy scientific writing and bad reviews that easily. And the BIPM discourages ppb, ppt, because they are seemingly lost in translation, the old fodderheads. One might as well advocate writing up scientific works in "Simple English". ;-)

@Karl, of course they may be used precisely, but prevention is better than correction. I notice many sloppy ppx unit usage. Also, for non EN speaking countries, ppX units are less used. For ppb, ppt, there also come in the game the language habits, as not everywhere 1 billion is 1000 millions. There are historically the short French scaling system ( which US adopted ) and long English scaling system. US and UK did not agree here until 1974 when UK  accepted US(=French) meaning.  It would not confuse scientists, but it may confuse readers of popular science articles, or of old articles.

@Poutnik Writers of popular science articles are already too often so confused to begin with that it would make very little difference.

Anyway, if it is frequently seen to use ppx units in vaque or wrong way, it is better to discourage their usage. sometime there are used explicit variants as ppmw, ppmv, ppmn or ppm(V/V),ppm(w/w),ppm(w/V),ppm(n/n).

Let us continue this discussion in chat.

The main point is that your are not dealing with concentration. You are dealing with a recipe / procedure. And even not dissolving but perhaps extracting (tea powder could be a soluble form or the tea as it is). If something calls for that 10000 ppm, that it simply means what you wrote. If it is you making the protocol, better writing down "the tea was prepared using 1000 mg of xxx in 100 ml hot water" or so. Consider that a good portion of the water can leave during infusion. This is what you need / can do. It relates to the concentration of whatever component of the tea but

...it is not a concentration. Still it provides a clear protocol.

In layman's terms, ppm is similar to percentage, i.e., how many parts of X are there in a million. By that standard, 1000 mg tea / 0.1 L is indeed 10000 ppm tea "powder" in water. We are assuming that everything dissolves completely.
Scientifically, or in analytical chemistry, you question is meaningless. Tea powder is mixture so one cannot talk about the ppm of tea powder. This is why you will not find this in journals or books. Concentration is specified for a single component.
Tea powder is a very complex mixture. If we have a mixture of say K2SO4 and NaCl, it is not appropriate to mention the concentration of "salts" in water. It will be an ambiguous statement. A better way would be expressing the concentration of K2SO4 as x% and that of NaCl as y %. It is okay in analytical chemistry to mention "total carbonates" or "total acidity"; here the species being analytes are closely related.

Hm, I am not sure if concentration must be limited to single component, it makes sense even for soluble mixture. But the problem is, the amount of tea extract is undefined, so I agree it does not have sense in case of tea infusion. It would theoretically still make sense for the original powder, but I do not see much sense in using 10000 ppm instead of 10 g / L, especially as units as ppm, ppb are discouraged to be used for their vagueness.

It still makes sense to consider mass concentation of dissolved salts or more generally any matter.

E.g. in water treatment analysis there is determined residual mass after evaporation of the given water volume, expressed as mass concentration of dissolved salts. It is obvious it does not agree with true dissolved salt content because of bicarbonates.

ppm must be unitless. It can be µg/g, µL/L, apple/megaapple, but NEVER mg/L. That wouldn´t be "vague", but plain wrong.

It can be both vague and wrong. And it is frequently implicitly used for mg/L.

@Karl ppm is frequently approximated in the field of analytical chemistry as mg/L for aqueous solutions, because it is understood that 1 L waters weighs 1 kg. Yes, for physical chemistry work this is not acceptable.


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How do ions conduct electricity?
I mean to ask what do ions actually do while conducting electricity?
Do they simply take up the electrons and transport it?
If it is so, I do not understand how anions can take up more electrons if they are already negatively charged?
And if cations take up the electrons, won't they be converted to their elemental forms? Then they no longer remain an ion, so how will they keep conducting? Or is it a completely different mechanism?

This forum is probably more appropriate for the questin: https://physics.stackexchange.com/ But short answer - the movement of a charge (be it an ion or an electron) is electricity.

Forget electrons. Ions themselves are already charged, and the motion of charged particles is electric current.

So this question should be posted in Physics Stack exchange??

I recommend looking up the meaning of the various terms you are mentioning, for instance ion, electron, conductor, current. The Wikipedia or an online dictionary may be a useful resource.

@BlackFire I wouldn't cross-post or migrate your question to Physics.SE as it seems on-topic here, on Chemistry.SE. However, as others already pointed out, there seem to be some confusion on your side regarding terminology. Electrons are not the only charge carriers out there responsible for conductivity.

Ok so simply the movement of the ions is the conduction of electricity? It may or may not involve the transfer of electrons, right?

There are never neutral atoms in pure $\ce{NaCl}$ or its solutions. The pure $\ce{NaCl}$ crystal is made of piles of ions $\ce{Na^+}$ and $\ce{Cl-}$. When $\ce{NaCl}$ is dissolved into water, the ions $\ce{Na^+}$ and $\ce{Cl-}$ are separated. They become independent and can move in the water. If you dip in this solution two conducting pieces bound to the poles of a battery, a positive pole (anode) and a negative pole (cathode) are created in the solution. The positive pole (anode) attracts the negative ions (anions) like $\ce{Cl^-}$: they slowly migrate in the solution up to the anode, and when they touch the anode, they lose their charge and create a free electron which enters the anode and the external circuit. The anions become neutral atoms $\ce{Cl}$. So the anions do not take up electrons as you say. They already have electrons in excess, before the beginning of the electrolysis.
In the circuit "Negative pole of the battery - Electrode  - Solution - Electrode -  positive pole of the battery", the electric current is made of charge carriers which are electrons between the battery and electrodes, and ions in the solution between the electrodes. The nature of the electric current changes along the circuit.
To be correct I should also mention that positive ions like $\ce{Na^+}$ are also migrating in the solution, as they are attracted by the negative pole (cathode). So in the solution, there is one current of positive ions moving to the negative electrode. And there is also a current of negative ions moving to the positive electrode. As a result, the solution remains neutral, without any positive or negative electric charges in excess. But what happens to Na then? Bappie

In ionic compounds electron is transferred from one atom to another completely to form cation and anion, and resultantly, an ionic bond. To bring physics in this question, you have to understand that current is the flow of charge. All ions are charged. So positively charged ions move towards the negative electrode and anions move towards the positively charged electrode. As ions are moving, the charge is moving. This movement of charge produces current. However, Maurice has clearly answered your question and I would have upvoted, but my reputation doesn't permit that. My aim is to provide supplementary information regarding what ions do to conduct electricity as was your question.


text:
Make very accurate solutions without weighing
I am currently facing a hen-egg problem. For a LC-MS method development I need to prepare reference solutions which are used to calibrate the method. Since the method is to be used for quantification in the pg/mL range, the reference solutions should have the most accurate concentrations possible. Of course I know how to prepare exact solutions in volumetric flasks, but here substance quantities are used which can be weighed easily and also quite exactly. In my case, only a few milligrams of the substance are available to me, which is why I cannot weigh high masses. The precision balance that I have at my disposal has a lower weighing limit of 1 mg, which is about the same as the mass that I would have to weigh. However, since the lower limit is very error-prone, I do not like this method. Unfortunately, I cannot think of an alternative solution that would allow me to create an exact solution without weighing. One could of course dissolve any amount of substance in the desired solvent and then determine the concentration by LC-MS, but this is exactly where the hen-egg problem mentioned above occurs. I cannot quantify via LC-MS until I have a suitable method and an internal standard. Are there any other ideas how to get exact concentrations using small amounts of the desired substance?

Is what you are trying to weigh a salt? eg if a sodium salt then you could analyze the solution for sodium for which you could prepare standards by weighing a different salt and diluting.

No, unfortunately not.

Are you able to prepare a rather concentrated solution with maybe 100 mg and then dilute it 100 times so that to get your expected low concentration ?

@Maurice Unfortunately, the substances are quite expensive which is why I only have a few milligrams of them. Depending on the substance we're talking about 10 to 50 mg. Besides, I would love to avoid dissolving all I have. In solution, substances tend to decay faster, which is why it would be great to only use as little as possible. I could of course do it and then evaporate the higher concentrations. This would, however, be prone to contaminations of the substances and for the 10 mg, I wouldn't gain much. All the ideas that come to my mind have several disadvantages, it is despicable.

Can you find a fellow reasearch with a microbalance?

Is it possible for you to describe a little further about the nature of this reference substance? Molweight (roughly, say 100, 200, 500, 1000 g/mol)? What about mixing your compound with something down the road equally is diluted enough to not not interfere badly with your LC-MS, but initially eases the weight-in?  E.g., murexide (https://en.wikipedia.org/wiki/Murexide) is mixed 1:250 or even 1:500 with Na2SO4 and still then it is a powerful indicator for heavy metals.  Then an aliquote could be electrochemically analyzed (if reversibly responsive) by redox and cyclic voltammetry.

What is the nature of your sample (peptide / protein / oligomer / dye)?  In similar tune to @MaxW, researcher isolating natural products / determining LC50 values work with small quantities (e.g., doi 10.1039/C3NP20106F finally mentions a prep GC to yield just 2.22 mg of Caratol (MW < 200)).  Then, the folks in radiochemistry equally use tiny amounts of matter only.

@Buttonwood The reference substances are steroid hormones of an MW-range of 270 to 400 g/mol. As we have a common weighing room, the other labs won't have balances that are more precise and I even don't know anyone of my circle of acquaintances who has access to such a device. Your idea of mixing the compound could be an idea, I'll have a closer look into this. Thanks!

Are there any other ideas how to get exact concentrations using small amounts of the desired substance?

Every mass spectrometry lab should have microbalances to prepare samples and standards. A lab which can afford an LC-MS can also afford a semimicro balance or a microbalance. Such balances are meant to weigh 0.01 mg to 0.001 mg (with extreme caution).
Two simple options:
(i) Since you are apparently not the first user of LC-MS system in your group, ask how other senior researchers prepared standards in your lab?
(ii) Another (crude) option is that you try weighing by difference. You will tare your analytical balance to zero. By an analytical balance I mean a balance which reads to four decimal places in grams, this is the most commonly used in research labs. Place your standard container on the balance and read the mass. Carefully transfer approximate amounts to your desired container in which the standard will be prepared, and note the weight of the standard container again. The change in mass corresponds to the amount transferred. Practice it with some commonly available substance to see if this works and if your balance is sensitive enough or not. Balances age with time.

I wish you were right. I've now worked in several LC-MS laboratories and have never had access to a microbalance. Normally, an ordinary analytical balance as you describe it was sufficient, but this time I would need exactly that.
One more question about differential weighing: What's the difference between putting an Eppendorf tube on the scale and taring it, adding substance and reading off the mass, and doing what you describe? Are there concrete indications why one method is more sensitive than the other?

You did not address the simplest point. Can you ask someone in your lab how they prepared standards with small amounts? I can say exactly the opposite, every mass spectrometry facility I saw, had microbalances. I am talking about research labs and good universities. Microbalances are expensive (up to $ 30,000). The pros and cons of weighing by difference are described in standard textbooks. See https://www.google.com/search?tbm=bks&q=%22weighing+by+difference%22


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Discrepancy between preparation of yttrium aluminum garnet powder in a nonpreheated and preheated oven
I've made yttrium aluminum garnet (YAG) precursor powder and in my last batch, I placed one half of the powder to be annealed in a non-preheated oven (letting the powder heat up with the oven) and the other half of the powder, I placed it in the oven once the temperature reached my final desired annealing temperature, 1020ºC. Aside from this difference, everything about the annealing process was the same; Annealing for both were performed in air and in the oven for 2 hours.
The curious thing is, the XRD results showed that the sample placed in the preheated oven still showed traces of secondary phases (YAH and YAM), but sample that was placed in non-preheated oven (the one that was heated up as the oven was being heat up) showed phase pure YAG, which to me is counterintuitive. How can this be? Does it have to do with some sort of kinetic steps in the formation of YAG?


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Chemical Potential in the Fluorescent Molecule
I'm reading this paper, which states that the chemical potential $\mu$ is determined by the steady-state balance of up and down transitions in a fluorescent molecule. I am happy with this interpretation. But what I don't know how to verify is the next statement:
"For a high-quantum-efficiency fluorescent molecule illuminated at the earth surface, $\mu$ is usually $0.6$ or $0.7$ times $h\nu$ at the centre of the absorption band".
How would I show that $\mu$ is usually $0.6$ or $0.7$ times $h\nu$ at the centre of the absorption band?

This would seem to be due to the $T\Delta S$ caused by the Stokes shift that leads to heat not photons, but it's not clear why it has these values. You will have to test some values of frequency shift.

Apologies, I'm not quite sure I understand. So, a portion of the absorption leads to heat - but surely at high efficiency, this would be diminished?

There can still be a wavelength shift between absorption and emission even if the efficiency is high.  $v=0\to 1$ is different in wavelength $v=1\to 0$ due to relaxation in excited state vs ground state  due to differing interactions with solvent caused by changed electron distribution. The mean absorption and mean emission will always have a shift .


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Why we do not add salt water to the batteries instead of acid?
Actually I was just thinking that since we add acid to the batteries to increase the conductivity but the thought simply crossed my mind that salt are strong electrolytes so there should be a good conductor to and also they are quite cheap so why we do not add salt water to the batteries is electrolysis a possible reason please explain in simple terms.

We only add acid to lead/acid batteries. Here is an explanation of how they work  https://batteryuniversity.com/learn/article/lead_based_batteries

Which batteries ? Lithium ions ? Lead ? Cadmium-nickel ?

All depends on the used electrochemical system. The electrolyte is the part of this system, it is not just a passive medium to conduct the current.
Some systems work in acidic environments like acid lead cells, some use salts like the Daniell cell, some use alkaline solutions like nickel-cadmium or nickel-metalhydride cells, and some do not even use water in their electrolytes, like Li-Ion batteries ( using mostly LiPF6 in mixture of dialkylcarbonates ).

There is a very valid reason for using acid in batteries, where needed. You are right that salts are also good conductors but keep in mind they are not the best conductors. In aqueous solutions, proton is the best conductor known! Which ion is no. 2? Not surprisingly, the hydroxide ion. Look at the comparison, there is no match for proton in  various tables.


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Can more than one thermodynamic process occur at the same time?
Thermodynamic processes I know about are, Isothermal, Adiabatic, Isochoric and Isobaric (if you know any other, please mention).
Is it possible that any 2 or more of these four processes occur together in the same closed system?

It is also possible that some chemical reaction happens during any thermodynamic process.

There 4 are just 4 special cases of an abstract path a TD system  changes its state via a particular TD process.
In special cases, a TD system can have 1 or a combination of the above attributes, if there is enough degrees of freedom for the process to occur with multiple constant values of state variables.
But a general TD process may have neither of these  4 attributes. I.e. a TD process may be neither isothermal nor adiabatic nor isobaric nor isochoric.
E.g. chemical reaction can occur both isothermally+isobarically, or isothermally+isochorically.
Or, it may be neither isothermal as temperature changes, nor isobaric as pressure changes, nor isochoric as volume changes, nor adiabatic, as there is heat flow to surrounding space.


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Bleach in water only good for one day?
In this time of COVID-19 our establishment is being told by government sources to spray down surfaces with a 9:1 water bleach solution, and then told to throw it out within 24 hours. What makes it less effective when we have over the counter products with bleach in them that obviously sit on the shelves for months and years?

Commercial liquid chlorine bleaches, used in the laundry, are stabilized with sodium hydroxide at high pH. When you dilute the bleach in water, the alkali concentration also decreases. As a result, the active ion (hypochlorite $\ce{OCl-}$) slowly begins to decompose to the chloride ion (the common salt).

The Chlorox website says solution of 1 part of Chlorox Regular bleach in 24 parts of water is registered with the US EPA as a disinfectant. Its effectiveness is assured by making a fresh solution daily.
Chlorox says it should be stored between at about room temperature or cooler, and the undiluted commercial product has a shelf life of 6 to 12 months from the date of manufacture (which I assume can be obtained from the manufacturer using the dot code printed on the bottle). Undiluted bleach degrades at a rate of 20% per year until it is all gone. And a 1:10 bleach solution has a shelf life of 24 hours.
So, if you make up 1 part in 24 daily - perhaps this is good for an 8-10 hour shift, but if you want a whole 24 hours, make up a 10% solution. And fresh bleach is good for 6 months or 12, but no differentiation is made between 5% and 8% hypochlorite products (this concentrated version is also available). Well, even after a year, the bleach is 80% of its original strength, so you could continue to use it at a higher dose: 1.25 parts in 24, or 12.5% for 24 hours.
It all seems kind of vague, but you could analyze the bleach with H2O2 and collect the gas, or titrate with KI and then thiosulfate. But that seems labor-intensive. One thing to take away from the data is that bleach doesn't suddenly become ineffective, just less effective, so if you follow the guidelines set by authorities who have done the testing, you will kill all the bugs and be held blameless if one or two manage to escape (they will probably be weakened).


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Benzene - MO states
When constructing the different combined molecular orbital combinations (from the original 6pi orbitals) we normally say that p orbitals "combine" to produce the different molecular orbitals and antibonding orbitals etc.? I'll use just benzene as en example:

For the lowest energy molecular orbital (pi1, is it accurate to say it's because the electrons in particular are able to take "advantage" of the p orbitals of the carbon atoms and then can occupy the p orbitals equally, delocalize and share the p orbitals? (like in this manner they the p orbitals are "combining")? Because when I first saw that orbitals are combining, I thought it's inherently due to the electrons actually resulting in those orbitals.

Then how will the next two electrons "cause" to create a node in the next highest energy molecular orbital (pi 2 and pi 3)? I know that some of the p orbitals may be out of phase to create a node. But how does this conceptually happen with the electrons? Maybe the two electrons move, and one of the p orbitals, go out of phase when they realize the "pi 1" is full of electrons.

Sorry for the long worded question, yea I'm just conceptually looking for how these MOs form.

If the p orbitals can overlap one another it is possible for the electrons to lower their overall energy by forming  molecular orbitals. The delocalisation lowers the energy; think of a particle in a box, the bigger the box the lower the energy. All the orbitals are formed at the same time, the benzene symmetry determines where the nodes are going to be. The lowest energy has most delocalisation, the next lowest only half as much etc. The six pi electrons fill the lowest 3 orbitals with three remaining unfilled. Overall the energy is lowered. Look at the Huckel model for a simple calculation.

Let's stay with your example of benzene, $\ce{C6H6}$.  There are six carbon atoms bound in cyclic fashion with each other, and maintain each one bond to a hydrogen atom, too.  This is explained conceptually that for each carbon atom the outmost $s$ and three $p$ atom orbitals were hybridized to yield three $sp^2$ of equal energy, leaving one $p_z$ orbital «unaffected».
If you draw the molecular structure of benzene as flat, regular hexagon, the $p_z$ orbital may be assumed to point orthogonally out of the plane of drawing.  Joining the six carbon atoms will create the benzene molecule.  For $n$ (integer number) of atomic orbitals $p_z$, there will be $n$ molecular orbitals, and because they are not of rotational symmetric around the C-C bonds, they are called $\pi$-bonds.   In case of benzene, starting with 6 $p_z$ atomic orbitals, there will be 6 $\pi$ molecular orbitals ($\pi_1, \pi_2, \pi_3; \pi^*_4, \pi^*_5, \pi^*_6$).  Compared to the energy level of the atomic orbitals, the energy levels of the molecular orbitals either are lower, or higher and contribute to the molecular energy with either bonding or anti-bonding character. (Antibonding $\pi$ orbitals have this asterisk, too.)
You recall Hund's rules and Pauli principle; specicially:

if there are multiple orbitals of equal energy level, each one will fill first with one electron only
at maximum, one orbital may accomodate two electrons at maximum

An useful tool to sort these molecular $\pi$-orbitals is the Musulin Frost cycle.  Draw the benzene as a regular hexagon with a tip to the bottom.  Either keep in mind, or mark the corners as atoms, too:

(altered from doi 10.1063/1.1698970)
As you see, energy wise, $\pi_2$ and $\pi_3$ are degenerate but (as $\pi_1$ as well) below the level of the atom orbitals we started (the sketch symbolized this by 0).  The antibonding $\pi$ orbitals are higher than this level.  Again, there are two ($\pi^*_4, \pi^*_5$) which are degenerate because of the symmetry in benzene.  And you may fill the orbitals one by one, schematically by the added blue lines.  This representation equally helps you to identify HOMO and LUMO levels accessed, e.g. by UV-Vis spectroscopy.
As shown in the example below, it equally allows you to understand the energetic difference between cyclopropyl cation and anion, too:

(reference)

As for the name (cycle):  It may be easier to draw a cycle around the (cyclic) molecule in question with a centre at an energy level = 0 as in the example below and to quantify the energy differences, too:

(loc. cit.)

Thank you so much for the quick response! So in particular to how (like the process in which) the orbitals form, do the pi orbitals form sequentially or at the same time (pi 1, then pi 2, then pi 3, etc.). I imagine the electrons will be "in motion" from the Pz orbitals, forming the combined pi orbitals. So maybe 2 electrons form a pi2 orbital, but then since pi1 is of lower energy, they'd form actually go to the pi1 configuration, then pi 2 and pi 3 would form I imagine around the same time.


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Risks from overheated phenolphthalein in green/orange silica gel drying
Recently I tried to dry out some indicating silica gel. When I googled, I found advice on line saying that it should be heated to 150 C and that such was a standard procedure, so I popped it into the toaster oven set it to 300 F which is just over 150 C and it started to make a smell and turn black. I stopped of course, cooled it down, and aired out my kitchen. Futher research indicates that the temperature I used is applicable to blue indicating or non-incdicating gel, but the newer Orange/Green gels contains an organic indicator which turns out to be phenolphthalein, rather than Cobalt Chloride and as such should not be heated above 120 C. Basically I followed outdated advice for the older type of product, a classic internet fail.
So now the question is, what have I done? What did I breath? I was cautious, watched the process stopped it early and don't have any outward symptoms, but now I'm curious how dangerous this was.
I've done a lot of searching and found nothing beyond "Toxic oxides of carbon, acrid and irritating fumes" in MSDS sheets regarding what happens when phenolphthalein combusts let alone what might happen as it degrades inside a silca gel bead. MSDS for the beads are just as vague. The lack of info is slightly alarming considering the ease with which a mistake such as mine could be made. Does anyone know what happens when heating breaks down phenolphthalein? There was a smell so something's definitely been produced other than water vapor, and the blackening of the beads might indicate that some carbon probably was left behind.
I understand that a variety of chemicals might be released, but I wonder if anyone is aware of a characterization of the types and relative quantities, or if there are any specific concerns.

I believe you are trying to reactivate moisture absorbing silica gel with a indicator dye. However, you are blaming the internet for the mishap you have faced. Yet, the website you have directed is ScinceDirect, which provides subscription-based access to a large database of scientific and medical research (Producer: Elsevier). Therefore, I find they give wrongful information on scientific research (considering almost all of these papers are peer-reviewed). I think you made a mistake where considering the indicator dye in green/orange silica gel desiccants is phenolphthalein. My conclusion is based on the inventors' remark in their patent application (Ref.1), which states that color change is from orange (when the gel is dry) to creamy white (when the gel is saturated with moisture). My best guess is it contains something different. Even with something else as an indicator dye, it should be able to stand at $\pu{150 ^\circ C}$ temperature in order to be able to reactivated. What that suggest is, your silica gel might be absorbed unknown chemical without your knowledge, which must be thermally unstable.
According to the patent, there are three kind of materials they are suggesting as indicator dyes: (1) phenolphthalein (color change Orange-red to creamy white); (2) thymolphthalein (color change dark-brown to creamy white); and (3) mixture of phenolphthalein/thymolphthalein (color change Orange-is not provided). All three types are able to get reactivated under $\pu{150-250 ^\circ C}$ temperature range (Ref.1).
References:

Waclaw Lanowy, Krystian Lukaszczyk, Przemyslaw Kasprzycki, “Silica gel with moisture indicator, a method of obtaining them, and use of phthalein dyes,” International Patent 2006, WO2006/130029 A2 (PDF).

I found something regarding my specific brand, and it said 200-250F which sounds like maybe it's (3)... and anything containing phenolphthalein will of course be limited to temps that can be sustained by phenolphthalein. The pellets were placed directly from the distribution container into an polyethylene container next to 3d printer printing PLA, so the only likely contaminant would be PLA dust, but whole silica beads turned black and significant smoke was produced.

Hmm although your PDF does seem to say that the thymolphtaline is dark when dry, so I guess the source of greenish color is unknown and possibly more limiting than the orange... I've not found description of any other orange colored formulation though (not that I'm likely to have the resources for an exhaustive search though)

Hmm, well for some reason my google foo is better today. I just found something about Methyl-Violet-2B as a silica dye, so you're probably right that this is not phenolphthalein. This seems disheartening since the orange stuff is marketed as bad for the environment, and maybe it's not as not bad as cobalt blue, but it's still very toxic to fish and aquatic life http://datasheets.scbt.com/sc-206052.pdf

marketed as not bad


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Can elemental silicon react with water at room temperature and pressure?
Free silicon is not found in nature (in water, it might be present in the form Si(OH)4). So how is Si(OH)4 formed?
Is there any possibilities of direct reaction with water?

The bonds $\ce{Si-O-H}$ are formed when silica $\ce{SiO_2}$ is dissolved into hot NaOH, producing a solution containing silicate ions like $\ce{SiO_3^{2-}}$ and $\ce{SiO_4^{4-}}$. In the presence of acidic solution the $\ce{H^+}$ ions are attached to the oxygen atoms of these ions producing $\ce{Si-O-H}$ bonds

There is variable information in textbooks about the reaction of silicon with water. Unlike all common allotropic forms of carbon, silicon surface or dust can react with water at elevated temperatures to form oxide layer or orthosilic acid.

Oxidation of Silicon by High-Pressure Steam by  Joseph R. Ligenza Original Paper
ABSTRACT: The oxidation of silicon by high-pressure steam was studied
at temperatures between 773 K and 1123 K and pressures from 25 to 500
atm. Within a certain range of pressure and temperature a uniform film
of amorphous silicon dioxide forms. At moderate pressures the film
growth is linear in time and directly proportional to the steam
pressure. At higher pressures the kinetics for film formation are
complicated by the volatility of silica in steam, and, above a certain
large pressure for any temperature, no film will grow. In the linear
range, the rate depends on the orientation of the silicon surface and
is greater for silicon that is phosphorus doped near the degenerate
range than for high-purity material.

And steam oxidation is routinely used in silicon semiconductor manufacturing, and has been for 60+ years now. Although one does not get ultra-high purity steam from water, but instead from burning semiconductor grade hydrogen and oxygen.

I also read that steam vaporizes silicon. What goes on in that process?

Unless it is very high temperature (greater than 1000C), or unusual conditions, I’m not sure it does. It certainly is not an issue in silicon fabs.


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Do isotopes have the same ionization energy?
When learning about isotopes and IE in High School this question have come to my mind recently. I have asked the teacher, she told me that she read articles about it however could not come to a conclusion, given that she doesn't know English, I think her research may be shallow. Nevertheless to put it another words does the neutrons have any effect of the IE particularly the first one?

No. The ionization energies will be different. However the difference matters most for the light elements. If you think of the Bohr orbit, the orbit's radius depends on the center of mass.

For questions like these, always start with the simplest possible case. For chemists and physicists, it is deuterium and hydrogen atom. They have slightly different atomic spectrum. It means their ionization energies must be different too (slightly). See another answer https://chemistry.stackexchange.com/questions/67489/how-can-difference-in-neutron-number-cause-a-difference-in-ionisation-enthalpies

The difference is very litte. Between $\ce{H2}$ and $\ce{D2}$, the difference is $\approx$ 0.05%, and it is the largest. It is not caused by the electrical charge of the neutron, because it is neutral, but that the common center of mass of the deuterium atom is a little bit more close to the nucleus (in classical mechanical analogy).


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What is the link between the classical and quantum definition of the symmetry number?
From what I understand, the symmetry number for a molecule can be defined in 2 ways:
1. The quantum mechanical symmetry number corrects for overcounting the number of possible rotational states of a rotating molecule. Here's a source using this method.
2. The classical symmetry number corrects for including physical orientations of that molecule that appear the same because of its symmetrical physical structure. Here's a source using this method (a PDF file)
This source claims at the end of page 3, that the classical view is roughly based on the quantum view. I have a very hard time to grasp that.
How does the physical structure of a molecule have anything to do with nuclear spins being integer or half-integers, whether the total wavefunction needs to be symmetrical or antisymmetrical, and thus determining which rotational quantum states are allowed for a molecule?

Without looking at the actual links, the statement that the "classical view is based on the  quantum view" is non-sensical. The classical view for any system is derived from immediate observables. That's why it's classical. The quantum view requires an understanding of quantum mechanics, historically arriving much later. How could the earlier view be based on the latter view?

on page 3, $\sigma$ is constant, nothing to do with quantum or classical, but is 2 for homonuclear molecules because it is not possible to tell left from right and so extra counting is accounted for.

@Zhe That is what I am trying to understand. If you look at page 3 of this source, it says the following: The symmetry number is rigorously based on the nuclear spins. Here, the symmetry number corrects for the number of physical orientations of a molecule that are indistinguishable. How is this based on nuclear spins?

@porphyrin Please also see this end of page 6 of this source. Initially on page 3, it derived the symmetry number $\sigma$ based on nuclear spins and wavefunction symmetry to correct for the number of allowed quantum states with the factor $\frac{1}{\sigma}$ for the partition function. But on page 6 it says $\frac{1}{\sigma}$ is of classical mechanical origin, but must come out of QM

That appears to be extending classical statistical mechanics so that it's consistent with quantum mechanics.

in a more accurate derivation p 6 the factor of 2 appears naturally, (plus other terms) earlier on $\sigma$ is obtained classically by symmetry and as is classical should only apply at high temp. By that statement they mean that if  classical sigma is correct, then a full  QM derivation should produce same result. It is a clear derivation .

@porphyrin In that case, it makes me question how one deduced that structural symmetry has something to do with correcting for the number of allowed quantum states.  Is there a way to explain this?

this is shown in the source you quote towards the end of the derivations.

@porphyrin What I'm saying is, how does a correction based on the physical structure of a molecule determine which quantum states are allowed or not allowed semantically? I can see no link how the symmetry number is semantically connected to the total wavefunction being symmetrical/anti-symmetrical.


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Can the degree of dissociation of a weak acid be more than 1?
My common sense dictates that it can't be more than 1 because after all it is the number of moles dissociating per mole of a substance.
But I was reading about Ostwald's law of dilution and came across this formula:
a (degree of dissociation) = √K/√C
So if C is less than K by about 4 times, a becomes 2.
I'm sensing there's an extremely simple solution to this question but can't realise what it is.

No, not possible mathematically and physically. You have 100 molecules, how can 150 of them ionize?

Exactly. So is there any rule which prohibits C from being lesser than K?

Your very definition of alpha is wrong. You know the principle GIGO, garbage in garbage out. The equation is wrong, the output of the equation is also wrong :-)

The equation isn’t wrong. It is the form of Ostwald’s Law of Dilution...

It is wrong because it is an approximation. Remember the approximation a ~ 1-a, when a is very small for weak electrolytes.

The $\sqrt K/\sqrt C$ ratio is only an approximation.  Let's do something more exact.  (Why do I say "more exact" instead of "completely exact"?  Read on, nature has a nasty curveball in store for us!)
Let $C$ be the nominal concentration of the acid, assumed monoprotic ($\ce{HA}$) for simplicity, and $K$ be its dissociation constant.  So, recognizing all ions are of course solvated, we have
$[\ce{H^+}][\ce{A^-}]=K[\ce{HA}]$
$[\ce{HA}]=C-[\ce{H^+}]$
$[\ce{A^-}]=[\ce{H^+}]$
So
$[\ce{H^+}]^2=K(C-[\ce{H^+}])$
$[\ce{H^+}]^2+K[\ce{H^+}]-CK=0$
From the quadratic formula
$[\ce{H^+}]=\dfrac{-K+\sqrt{K^2+4CK}}{2}=\dfrac{2CK}{K+\sqrt{K^2+4CK}}$
And our dissociation is
$\dfrac{[\ce{H^+}]}{C}=\dfrac{2K}{K+\sqrt{K^2+4CK}}=\dfrac{\sqrt K}{\sqrt{K/4}+\sqrt{C+K/4}}$
Now, if $C$ is large compared with $K$ then the terms in the denominator with $K$ become small compared with $\sqrt C$ and we simplify to $\sqrt K/\sqrt C$.  But if we dilute the acid, then the terms with $K$ in the denominator become dominant and we find a  limiting value of
$\dfrac{\sqrt K}{\sqrt{K/4}+\sqrt{C+K/4}}<\dfrac{\sqrt K}{\sqrt{K/4}+\sqrt{K/4}}=1$.
Thereby the dissociation remains limited to no greater than 100%.
There is one little thing we forgot.  If we keep diluting the acid so its concentration approaches the ion concentrations from water autoionization, then we also have to count the hydrogen ions from the autoionization into the equilibrium.  This extra source of hydrogen ions cuts down on the acid dissociation, so the limiting amount of dissociation is less than 100%.  For acetic acid it's still about 99.4%, but if we pick a really weak acid like $\ce{HCN}$ with a $pK_a$ above $7$ at ambient temperature, we never get more than a small fraction to dissociate unless we force the issue by reacting with a base.

Also, shouldn’t it be [H+]/C = 2K/ (K+ sq rt.(K^2 + 4CK) ?


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Do Cyclohexane-1,3-diones Give Positive Haloform Reactions?
Recently a post appeared on ChemSE asking why cyclohexane-1,3-dione (dihydroresorcinol) 1a doesn’t give a positive iodoform test. Unfortunately, the post was subsequently taken down after several comments. A paper by Fuson and Bull[1] reviewed the haloform reaction and stated that the iodoform test was negative for 1a. However, dihydroresorcinol undergoes the bromoform reaction[2] to yield both glutaric acid 9a and bromoform 7 as does 5,5-dimethylcyclohexane-1,3-dione (dimedone)[3] 1b yield β,β-dimethyl glutaric acid 9b and bromoform. Moreover, dimedone 1b affords the same diacid 9b and chloroform when sodium hypochlorite is employed.[4]
Mechanistically, dibromination of 1,3-diones 1 affords dibromides 2, non-enolic β-diketones. In alkaline medium diketones 2 cleave to provide dibromides 4 via intermediate 3. Further bromination of dibromides 4 sets the stage for the bromoform reaction on 5 leading to bromoform 7 and the glutaric acids 9.

These reactions were found by searching Chem. Absts. but there was no sign of the iodoform reaction on these two diketones. So, the question is, why no iodoform reaction?
a) It is unimaginable that the reaction has never been conducted.
b) Perhaps more vigorous conditions are required.
c) The diiodo analog of dihalo diketone 2, if ever formed, may be formed reversibly by iodide or hydroxide reduction. Even so, the equilibrium should be shifted to the right.
d) If the diiodide is slow to fragment, might it undergo a Favorskii ring contraction instead?
Any insights are appreciated.

R. C. Fuson and B. A. Bull, Chem. Rev., 1934, 15(3), 275-309. DOI: 10.1021/cr60052a001. PDF link
D. Vorlander and M. Kohlmann, Chem. Ber., 1899, 32, 1878.
G. Komppa, Chem. Ber., 1899, 32, 1423.
W. S. Johnson and D. D. Cameron, Org. Syn., 1951, 31, 40. DOI:10.15227/orgsyn.031.0040

I wonder whether the di-iodo species forms under these conditions. This paper reports the selective mono-iodination of 1,3 dicarbonyl compounds including dimedone and claims no di-iodo products are formed. I would have expected to see some di-iodo product if di-iodination is possible  https://www.researchgate.net/publication/228528584_Selective_and_effective_oxone-catalysed_a-iodination_of_ketones_and_1_3-dicarbonyl_compounds_in_the_solid_state

@Waylander:I agree. I put my money on choice c). The diiodide never forms. If it did, cleavage is inevitable.

Are the diiodides of 1,3-dicarbonyl compounds known compounds?

@Waylander: They are known. HIO4/Al2O3 in dioxane forms diiodide of acetylacetone, AAE, diethyl malonate in high yield (Tet. Lett , 2006, 47, 3525). Conjugated alkynyl ketones with N-iodosuccinimide in aq. DMF (J. Org. Chem., 1998, 63, 4433). Some enterprising young chemist out there should prepare the diiodide by the first procedure and subject the product to the conditions of the iodoform rxn. The presence of iodoform will show that the diiodide is not prepared from the parent dione under haloform conditins.

There's a Tet. Lett in that. If only we knew of an enterprising young chemist!

@waylander: To be sure!

As per what I have learnt, methyl ketones, I.e., CH3-CO- and groups containing CH3-CH(OH)- show positive iodoform test. You may want to refer to this site for detailed explanation and mechanism
https://byjus.com/chemistry/iodoform-test/ So 1,3 cyclohexane-1,3-diones should not be able to give a positive iodoform test.
Hope it helps.

@Harikrishnan M: What is your concern? WIs it why the diiodo compound doesn't form? Other than that the second paragraph and the discussion between Waylander and me covers most of the issues and how to demonstrate that the diiodo doesn't form. Are you an "enterprising young chemist"?

@Waylander: FYI...See the Comment above.

There have been no answers to this question and this question was intriguing me hence the bounty. That is the point.

Btw answers in comments may risk deletion by the roomba here.


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Do reactions carried out in light always follow free radical mechanisms?
If light is used in a reaction, is it necessarily a free radical reaction? Or are there some other reactions as well which involve light but don't follow a free radical mechanism?

You mean, when light is the source of energy of reaction? Well, you can get excited singlet states, too.

Often, yes; but sometimes not, and generalisations like that often don't work in organic chemistry.
It's a better idea to understand why there's a correlation between "light" and "radical reactions". One way of generating radicals is by breaking bonds, and one way of breaking bonds is to use light. Of course, this only really works for relatively weak bonds, such as the Br–Br bond, or basically any kind of bond between Group 16 and/or Group 17 elements (O–O, N–Cl, ...). Many of the classic radical reactions involve this kind of pattern, and this is probably why you're drawing this link.* Examples include: free-radical halogenation, the Hofmann–Löffler–Freytag reaction, and the Barton reaction.
Alternatively, if there aren't any weak bonds to break, it could be that you're trying to use light to promote an electron to an excited state. Now, because the vast majority of organic chemicals are closed-shell molecules (all electrons are paired), when you excite something you're going to get a radical species. A simple example is the Norrish type II reaction. This radical generation can be indirect as well, occurring via excitation of a photocatalyst, which then oxidises / reduces an organic substrate to form an organic radical. This is the idea behind photoredox catalysis, which has been a very, very hot topic in recent years.
It is not wise to make a sweeping generalisation, though. You could excite a molecule and have it react immediately, without generating an intermediate radical species. This is the case in photochemical electrocyclic reactions, for example. And there are many photoisomerisations which (as far as I am aware) occur in a single step, without going through a true radical pathway. A very important one is the cis-trans photoisomerisation of retinal. The mechanisms of these reactions are not well-described by the curly arrows of organic chemistry: you'd need to learn some photochemistry to get a good idea of how these work.
And not every chemical relevant to organic chemistry is closed-shell. Dioxygen, O2, is a diradical species; and when you excite it, you get an excited state where the electrons are paired up, usually referred to as singlet oxygen.† Singlet oxygen reacts with other organic molecules in a decidedly non-radical fashion: for example, it does classic Diels–Alder reactions and ene reactions.

* Note that many of these can be also accomplished by simple heating, which can break weak bonds too.
† Usually dioxygen is not excited directly. A photosensitiser like rose bengal is used.


text:
What is meant by "rotation about an axis that passes through a lattice point"?
I am currently studying Introduction to Solid State Physics, 8th edition, by Charles Kittel. In the section Fundamental Types of Lattices of chapter 1, the author says the following:

Crystal lattices can be carried or mapped into themselves by the lattice translations $\mathbf{T}$ and by various other symmetry operations. A typical symmetry operation is that of rotation about an axis that passes through a lattice point.

It's unclear to me what is meant by "rotation about an axis that passes through a lattice point". In doing research, I found this page, but I'm not completely sure that this is referring to the same phenomenon. I would greatly appreciate it if people would please take the time to clarify this. I would especially appreciate pictures depicting this, as well as the mathematics itself.

yes normal rotation about an axis, say, by 90 or 120 degrees, etc. but with the restriction that any symmetry operations causes the structure to be indistinguishable from that before the operation occurred.

@porphyrin So exactly as is described here http://pd.chem.ucl.ac.uk/pdnn/symm1/rotate2.htm ?

yes, same is true of all symmetry operations

@porphyrin Ok, thanks for the clarification.


text:
How to plot the wave function after a Hartree-Fock calculation?
I have written a code in Mathematica and have obtained the final energy of the hydrogen molecule. I have used Gaussian basis functions for this calculation. Now I need to plot the ground state wave function. How can I do so? I just know that the ground state wave function is defined as
\begin{align}
  \psi_i &= \sum_{\mu=1}^k C_{\mu i}\phi_{\mu} &
  i &= 1, 2, 3, \dots, k,
\end{align}
where $\phi$ is a basis function. However I don't know how to obtain an expression for the ground state wave function.

Is not $\phi$ your basis function with coefficients C ? You have to know $\phi$ at xyz, (or equivalent in spherical coords)  for each quantum number. It should then straightforwards.

If you have calculated the final energy of the molecule, then you must have the final form of the wave function, in HF that should be a slater determinant. Given the general limitations of plotting a function with more the 3 parameters, you'll have to have a pick at what you want to plot.

@Martin - マーチン yeah you're right, but I don't know where can I find the final form? can you help me?

For that I would probably have to know (and understand) your code, but I don't speak Mathematica. However, others might. So (if it is not too long) you could include it here.

@Martin-マーチン Thanks a lot. My code is very long, so can you introduce me a source or a textbook where the final form of ground state wave function is defined after a HF calculation? In fact I wrote my code 3 years ago and that time I used Szabo's book but now I can't find what I need.

I would recommend that book. I think it is the most detailed and contains the best mathematics. I don't know of any other source more concise. But if you have understood it before, you'll understand it again. You might want to look for the definition of the density matrix, I remotely recall they called it like that... Good Luck!

@Martin - マーチン OK, thanks for your hints.


text:
Separating out iron from ferrous sulfate
I don't need to necessarily separate pure iron, but from a ferrous sulfate solution, I am trying to form a precipitate with iron. By getting a precipitation of an iron compound, I can find out the amount of iron in the solution. Any help would be appreciated. I'm completely new and tried finding information online but all of it was beyond my comprehension. I apologise for my ignorance, but help would really be great.

Are you a high school student, first semester, who asked you to do that? Please give some background to get a useful answer.

Anyway, ask your prof/teacher for a textbook on quantitative inorganic analysis, specifically gravimetry. You won´t get good results unless you do this by the book. The standard technique involves precipitation as hydoxide, filtration, and calcination to get the oxide, which you then weigh.

@Karl thankyou, yes I am a highschool student in Australia. My teacher doesn't know how to do it, but I'll look into the textbooks. I haven't been able to find any standard techniques anywhere yet but I'll keep trying.

@Karl and also, it's a depth study, so we chose our own experiment. I starting looking into doing something like extracting a substance from a product. Then came the idea of extracting (or finding the amount) of iron in ferrous sulfate supplements. I don't know enough about chemistry yet to know what works and what won't. The main aim of the assignment is to discuss percentage yield. I would like to learn why each step is done, but at the moment im just trying to find out how to actually do the experiment. Thankyou for the general procedure, I'll look into how to do both of them

Did I understand correctly: You would like to quantify the iron content of a ferrous sulfate solution?

Yes, thankyou for that experiment you linked to. I'll try it out next lab and see how it works

As an alternative to precipitation, you should look into titration. Iron forms very stable complexes with EDTA.

@Karl I was just wondering what the calcination was for? Once the hydroxide precipitate forms, can it be weighed and from there find the iron contents?

No, hydroxides are notoriously unstochiometric, i.e. full of water. You need to transform it into the pure oxide.

What a cute teacher. Instead of teaching, he leaves it to the pupils to find a suitable (university-level) textbook. wtf

@Karl thankyou, he is helpful with what he knows. I wasn't sure if calcination was needed. I did a rough run with just the hydroxides and got 74% yield (I understand this is likely innacurate due to not calcinating, and unreliable since I didn't do it thoroughly). You've been a great help in making it simple enough for me to understand

@Karl my teacher said that not doing the calcination and just discussing the unreliability of the hydroxides is enough. I would have liked to do the calcination to make it accurate, but I think this is the best I can. Is there a way I can contact you again for more resources/guidance. Thank you again for your help

If I were you I would have suggested a displacement reaction like putting a Mg rod in salt solution and finding out weight gain on the Mg rod, weigh the rod before and after reaction. It should be OK.

Decomposing ferrous sulfate to get iron(III) oxide should work:
$$\ce{2FeSO4 ->[\Delta] Fe2O3 + SO2 ^ + SO3 ^}$$
Iron content can be measured from iron(III) oxide. Commercially, iron is recovered from waste ferrous sulfate in the form of magnetite ($\ce{Fe2O3}$) by co-precipitation and magnetic separation$\ce{^{[1,2]}}$. Iron content is measured from magnetite. It was observed that after processing the magnetite, the grade of iron in magnetite concentrate increased from 62.05% to 65.58% and the recovery rate of iron decreased from 85.35% to 80.35%.
References

Wang YU, Ying-lin PENG, Ya-jie ZHENG Recovery of iron from waste
ferrous sulphate by co-precipitation and magnetic separation,
Transactions of Nonferrous Metals Society of China, Volume 27, Issue
1, January 2017, Pages 211-219, DOI: https://doi.org/10.1016/S1003-6326(17)60024-4

Yu, W., Peng, Y. & Zheng, Y. Effect of purification pretreatment on the recovery of magnetite from waste ferrous sulfate. Int J Miner Metall Mater 23, 891–897 (2016), DOI: https://doi.org/10.1007/s12613-016-1304-2

You don't really need step two if you only want to determine the iron content. This lab manual (pdf) oxidises first: $\ce{2Fe2+ + H2O2 + 2H+ -> 2Fe3+ + 2H2O}$ Then precipitates it $\ce{Fe3+(aq) + 3 OH- (aq) + x H2O(l) -> Fe(OH)3.xH2O (s)}$ Then probably heating it thoroughly: $\ce{Fe(OH)3.xH2O (s) ->[\Delta] Fe2O3(s)}$.

@Martin-マーチン This case is also possible. I edited the answer to address the issue. Shouldn't need pure iron to quantify.

Last week magnetite was Fe3O4, and hematite was Fe2O3. When did it change?

I would do the following to get pure iron powder without contamination:

Evaporate to dryness.

Roast the $\ce{FeSO4}$ in an oxygen enriched environment to form $\ce{Fe2O3}$. You will notice the change in color. This has to happen at high temperatures (minimum: $\pu{400°C}$ ideal: $\pu{600-700°C}$) and can take quite a while. The best thing to use is a quartz tube as you can see the change in color.

Hydrogen reduce the $\ce{Fe2O3}$ in an hydrogen atmosphere at about $\pu{400°C}$ to get the pure iron powder.

I have done this procedure - or at least a variation of this procedure with other metals and it works.
But be aware that this procedure will yield toxic gases and you have the risk of explosion when you work with hydrogen - especially when you do not know exactly what you do.

This is a very unorthodox procedure. Would you mind telling why exactly you did that?

The use of hydrogen $\ce{H2}$ is the best way for producing metallic iron out of iron compounds. @ Karl. I doubt Andreas will give you an answer more than three years after his remark.


text:
What is the differnce between energy and enthalpy?
I cannot understand the difference between thwe two although the internet suggests me there is a difference. Even asked my chemistry teacher but he could not give an explanation.

Have a look at this for intuitive picture of enthalpy.

You may try your luck with the enthalpy article on Wikipedia, where

... $\mathrm{d}H$ refers to the amount of heat absorbed in a process at constant pressure...

I will try to alternatively put it down in more layman terms:
Enthalpy is defined as $$H = U + p \cdot V,$$
where

$U$ is system internal energy,
$p$ is system pressure,
$V$ is the system volume.

If we consider constant pressure ( isobaric ) scenario, the above equation in the difference form is $$\Delta H = \Delta U + p \cdot \Delta V = \Delta U + W,$$ where $W$ is the volume work done by the system on the surrounding.
If we add $\pu{110 J}$ of thermal energy to isobaric system ( $\Delta H$ ),
which does $\pu{10 J}$ of volume work by expansion ( $W$ ), the change of its internal energy ( $\Delta U$ ) is $\pu{100 J}$. But those extra $\pu{10 J}$ is "stored" in the neighborhood and is released back if the reverse process is performed.
In an isobaric scenario, the change of enthalpy can be considered as the change of the system energy with the implied involved volume work. Formally, one can consider this work as as special way of energy storage of the system. Therefore, enthalpy is defined in such a way, that for isobaric systems, it is formally and effectively the energy of the system, as the reversible system work acts as a part of the system energy.
As majority of experiments are performed at isobaric, rather than isochoric conditions, thermodynamic data are collected in form of changes of enthalpy, rather then the internal energy.

That should be w, not $\Delta W$.   Work is not a state function -- each process has associated with it a certain amount of heat and work.  There's not a 'change in work.'  And the idea that  "formally...special way of energy storage" is (at least to my experience) unusual, and needs to be demonstrated in a  sufficiently precise and formal manner to show that it is correct (it's not clear if it's correct based on what you've written).

@theorist  I agree there should be W - fixed.  I also agree work is not generally a state function. But in a special case of isobaric scenario, it is. $W = p . \Delta V$, while in general case, like in thermal engines W obviously depends on path. About the rest, consider reversible isobaric thermal expansion of gases. I have also intentionally used wording close to the OP.

The most simplest explanation I can give is this: Enthalpy is the measure of energy contained in a system plus the energy required to assemble it.
So, The measure of energy contained in a system is it's internal energy (U) and the energy to assemble is work. So,
$$ H= U + W$$


text:
Does the electron displacement effect act only on sigma electrons?
Our teacher told us that the inductive effect or the electron displacement effect can only affect the sigma electrons. I did not understand why this should be.
I reasoned it can not affect the pi electrons, because they are "delocalized".
Also, for measuring the displacement, the C-H bond is taken as a reference. The C-H bond does not have a pi bond, and so we can not consider pi electrons as being affected
Is it true that the inductive effect or the electron displacement effect affects only sigma electrons? Why should it be so?

I am not asking if it operates through single bonds only, as asked in this question.

"I did not understand why this should be". Well, perhaps because it's not like that?

I guess you are too fixated on defining things as distinct "effects", while it may actually be more sensible to look at it from a perspective of unequal distribution of electron densities across bonds. In both the cases, sigma, and pi bonds, due to variations in electronegativity, electrostatic forces are set up due to formation of charges, and so the electron density tries to shift towards the negative end in both cases. The only real difference is that as the pi electrons are pretty "loosely" bonded as compared to sigma electrons, they can actually shift themselves in their entirety...

..so as to give rise to the so-called "mesomeric" effect, where the pi electrons delocalize over the system.Sigma electrons can't really move completely from one atom to another,as their bond strengths are relatively higher,and also,their geometry somewhat prevents them from delocalizing.So they have to make do with just shifting towards one end of the bond,which is called the "inductive" effect.So it is like a spectrum of ways of distribution of electron densities,and just because pi electrons manage to move completely themselves,doesn't mean that they can shift within bonds like the I effect

@YusufHasan Thank you,now I understand.Can you convert your comment into an answer?

@YusufHasan you said pi electrons completely shift themselves to more electronegative atom which would lead to development of complete charges rather than fractional charges but that doesn't happens

@kanchantolani When I say "complete shifting", I am talking in relative terms, and not a binary "yes" & "no". Yes, the ultimate resonance hybrid after conjugation will bear fractional charges on it's atoms, but it is contributed by canonical structures where the entire pi electrons are shifted to adjacent atoms. Compare that to sigma electrons, where you can't imagine "structures" where the sigma electrons are completely shifted on it's constituent atoms, and you can see why you can't have the "mesomeric" effect in sigma bonds, which relies on existence of such canonical structures.


text:
molecular orbital theory explaining molecular orbitals
i am a high school student,,and i am confused in molecular orbital theory,,,my confusion is that how bonding and non bonding orbitals are formed at the same time my teacher taught me that the two wave functions interfere constructively forming bonding orbital and if they interfere destructively the form anti bonding orbitale.g when we fill the electrons in say O2 molecule then why do we fill electrons in bonding as well as anti bonding at the same time,,,please explain this in a very simple language without going into mathematics???does really bonding and anti bonding orbitals are formed from constructive and destructive interference???

Your question is really hard to comprehend. Please use proper spelling and punctuation. For what I took away from it: in general terms (between two atomic centres) a bonding type molecular orbital will form by constructive interference. An anti-bonding orbital will form by destructive interference. When talking about more bonding partners, the definitions are not that straight anymore...

First of all, please keep in mind, that molecular orbital theory is just a model. Much like the Bohr model is valid yet incomplete, orbitals are not the whole truth. They are just closer to the truth than the Bohr model.
Orbitals are a way of representing how electrons are arranged in atoms and molecules. They give some idea of where the electrons are located in space and how much energy they have, meaning how easily they can be removed from the molecule.
As lower (more negative) energies mean the electron is harder to remove, those arrangements (orbitals) are preferred. As the name implies, bonding orbitals typically have lower energy than their corresponding anti-bonding orbitals and are therefore occupied first. If more electrons are available and need to be distributed in the molecule, energetically higher lying orbitals will be occupied.
The different available orbitals may be constructed as linear combinations of atomic orbitals (LCAO). In mathematical terms this is just a basis set expansion and nothing special in quantum mechanics. The choice of atomic orbitals as the basis set is arbitrary but valid, and physically motivated. The rule is: you can form as many molecular orbitals as you use atomic orbitals to form them. The simplest case is the one you described: two atomic orbitals will form one bonding molecular orbitals and one anti-bonding molecular orbital, with constructive and destructive interference respectively.
If enough electrons are available, both types of orbitals will be occupied at the same time. The anti-bonding orbitals will counter the effect of bonding orbitals, which is why $\ce{N2}$ is more stable than $\ce{O2}$ and $\ce{Ne2}$ will dissociate into $2\ \ce{Ne}$.

Related: How can antibonding orbitals be more antibonding than bonding orbitals are bonding? Please note that the terms you are using for bonding, non-bonding, and anti-bonding, are not congruent with how they are used within LCAO.

@Martin-マーチン what exactly do you mean?

I've interpreted your statement bonding orbitals typically have lower energy than anti-bonding or non-bonding orbitals and are therefore occupied first in terms of orbital eigenvalues,  I'm not sure you've meant it that way. Anti-bonding LCAO can be (for most diatomics are) occupied before bonding LCAO. The definition in terms of energy is not necessarily straightforward, and it depends on the knowledge of the AO eigenfunctions. You can also define bonding and anti-bonding from the mathematical description of the MO, i.e. (not) having a nodal surface perpendicular to the bonding interaction.

Do you mean situations like $2\sigma^*$ vs $3\sigma$? Does formed from different sets of AOs, therefore I did not consider them here.

Yes, but I only picked up this nuance when reading it again, after your comment. I might be worthwhile clarifying this more, since the OP's question style asks for 'very simple language'.

That is exactly why I did not want to go into too many details and special cases.

I reformulated this sentence a little bit: "bonding orbitals typically have lower energy than their corresponding anti-bonding orbitals"


text:
Trend in atomic volume across a period

"Atomic volume decreases along a period, reaches a minimum at the middle, and then increases for the rest of the period"

Why does the atomic volume, along a period, initially decrease, reach a minimum and then increase?
The atomic radius decreases along a period, so I would have thought that the atomic volume would also decrease along a period.

Could you please provide the source of the quote? It seems like a really dubious statement... As you said, atomic radius decreases across the period.

@Tan Yong Boon, I don't quite remember where exactly I had read it (it was online). But a quick Google search for the graph of atomic volume versus atomic number verifies this claim. I have added an image to the question.

Perhaps it's because of the way atomic volume is defined. Atomic volume is defined as the ratio of gram atomic mass to Density. That's probably why it has a different trend to atomic size. But I do not know why exactly does atomic volume follow this trend.

I believe molar volume is a more "macroscopic" property that depends on the structure that the element exists in. Covalent bonds tend to be relatively short between the atoms of non-metals while the interatomic distances are longer in metals. Hence, we see the greatest interatomic distances coming from group 1 metals and hence the greatest molar volumes from these elements.

@Tan Yong Boon, why would chlorine have a larger atomic volume than sulphur?

Also, I think you should not conflate "atomic volume" and "molar volume". The graph shows the trend in the latter, not the former.

Your graph suggests that sulfur has a larger molar volume than chlorine... It is likely that the measurement is taken as some form of mass and volume readings at room temperature and pressure, where chlorine is a gas and sulfur is a solid.

FYI, according to the database I got the data from, Cl has molar volume of 11,000 $\ce {cm^3/mol}$ while S has a value of 16.4 $\ce {cm^3/mol}$. The much larger molar volume of Cl is expected as chlorine exists as a gas at room temperature and pressure while sulfur exists as a solid.

Additionally, I would suggest you double check the source of the chart. It seems to me that the vertical axis values are very off, compared to the database I referenced in my answer.

Firstly, we need to clearly define "atomic volume" and "molar volume". Typically, one would understand "atomic volume" to be the volume of an atom of the element and it is directly related to the radius of the atom. Molar volume is a macroscopic property that is defined by Wikipedia as the ratio of the molar mass to the mass density of the substance at a certain temperature and pressure. As such, molar volume would not only be dependent on atomic radius but would also depend on the structure and bonding of the atoms.
With this understanding, it is easier to explain the trend that you have provided. The molar masses of the group 1 metals are considerably large but yet they have a relatively low mass density. As such, they are seen as spikes in the graph, with the highest molar volumes in their respective periods. Mass density generally increases from group 1 to group 13 resulting in the fall in molar volume across the period initially while for the transition metals, they stay roughly constant and take relatively large values. However, once we reach the non-metals, the trend seems to reverse. The density values for the non-metals seem to display some variability so I am not sure how the their molar volume trend can be explained. Obviously, their molar volume would be higher than that of the transition metals that come before them as those are significantly more dense.
Density data for the elements can be found here. More detailed molar volume data for the elements can also be found on another page at the same website.


text:
Why the electrolyte is made up of the same metal which is being electro refined?
For example for the electrorefining of Cu the electrolyte is CuSO4... WHY the salt of the same metal is used as an ELECTROLYTE..??

Think about this the other way around.
Let's say you got the impure copper anode and nickel sulphate electrolyte.
When copper is oxidized something has to deposit in the cathode, but it won't be copper if there is no copper electrolyte.
It's also worth notice that the concentration of copper sulphate remains the same, or close to, so you can re-use the solution.


text:
Characteristics of chemical reactions
Characteristics of chemical reactions are certain characteristics that tell us whether a chemical reaction has occurred or not.
Some important characteristics of chemical reactions are given below :
 Evolution of a gas
 Formation of a precipitate
 Change in color
 Change in temperature
 Change in state
I was wondering whether all chemical reactions can be identified by one or more of these characteristics. Since there is an extremely wide range of chemical reactions, I think that these are just the major characteristics of chemical reactions and there may exist some reactions that exhibit none of these characteristics.

Am I right about this? If yes, what are some examples of chemical reactions that show none of these characteristics?
Thanks!

No they can't. The majority of organic chemistry reactions show none of these characteristics and you don't know any reaction has occurred until you run some form of diagnostic on the reaction mixture.

@Waylander And how does it hold up with inorganic chemistry?

I'm biased as I'm an organic chemist. You are IMHO more likely to see one or more of the characteristics with an inorganic reaction.

I think this would not be possible because we signify or understand the reaction by these parameters if they does not change then we can't say the reaction took place or not, and you forgot to mention about pressure

'Change in Temperature'  is but one example of an energy manifestation, as in the formation of light. Also, transfers of electrons, as occurring in a battery cell, for example, may display no evident signs of a reaction occurring.

I think you are right about this judgment. There are certainly some outliers which show none of these characteristics.
One example I can think about is the catalytic reduction of nitrate to nitrite, by either transition metal catalysts or biocatalysts. And I think some reactions in liquid phase show none of the characteristics you mentioned above.
Still, I think the key here is that chemical reactions always involve the breaking-forming of chemical bonds and the change of electron cloud distribution. I think that's the essential characteristic of chemical reactions.
Thanks! Let me know if you have more questions.

While most chemical reactions can be classified by their characteristics, the overwhelming majority of reactions can be grouped into the following:
 Synthesis/Combination
 Decomposition
 Displacement
 Double displacement
 Reduction/Oxidation (Redox)
These reactions will exhibit their own telling signs of occurrence, however, you are right in your estimation that reactions exist which will not exhibit signs to the naked eye and will require indicators.

I edited your answer a bit, hope that's fine. You'll see it once it's accepted.

One characteristic is always present in chemical reactions. That is the change in energy. This is because in a chemical reaction, bonds are always broken and formed.
In endothermic reactions, temperature of the reacting mixture decreases. Or in cases where heat is not involved, it is to be noted that photochemical reactions actually absorb light and show quite noticeable characteristics as mentioned above.
In exothermic reactions, energy is produced, which can increase temperature, produce electricity and even produce light energy. In situations where chemical reactions show no noticeable change is because the change in temperature is either too small to measure or is affected by external factors like temperature, pressure and others.
Of course a chemical reaction always does not show noticeable characteristics; characteristics visible to the naked eye. Those characteristics may be too small to observe.
Hope this answer was useful.


text:
Can NaOH and NaH2PO2 exist together in aqueous solution?

Can $\ce{NaOH}$ and $\ce{NaH2PO2}$ exist together in aqueous solution?

My textbook denies it, but I feel that since $\ce{NaH2PO2}$ is acidic, it should react with $\ce{NaOH}$, because in my textbook it is written that salts of even weak acids can react with a strong Base (e.g., $\ce{NaHCO3}$ is mentioned to react with $\ce{NaOH}$).

$\ce{NaH2PO2}$ is not acidic and it is a common catch for less observant students. ( Or rather has extremely low, not mentioned acidity.), as the remaining hydrogens of hypophosphorous acid are bound directly to phosphorus.( $\ce{HO-PH2=O}$ ). Similarly, only 2 of 3 hydrogen atoms of phosphorous acid $\ce{H3PO3}$  are acidic.

Neither of substances really exists in aqueous solution. They're both fairly dissociated into ions.

Can you give the title of your textbook, and paraphrase a few lines?

@Karl , Its the study material provided by my coaching classes for IIT JEE prep. You won't really need to know the name.

Well, I can only advise you to take everything coming out of the IIT-JEE complex with the utmost caution. Just my opinion, based on the purely anecdotal evidence of questions I have seen here on chem.SE.

Your question: Can $\ce{NaOH}$ and $\ce{NaH2PO2}$ exist together in aqueous solution?

I assume what you meant is your textbook is saying when two separate solutions of $\ce{NaOH}$ and $\ce{NaH2PO2}$ are combined, there should be no reaction. If that is what you meant "exist together," then your textbook is correct because there won't be an acid-base reaction since $\ce{NaH2PO2}$ has no "acidic hydrogen" contrast to $\ce{NaH2PO4}$ and $\ce{Na2HPO4}$ as Poutnik mentioned in his comment (more descriptive presentation would be $\ce{Na^+^-OP(=O)H2}$). The following diagram summerizes this point:

As diagram shows $\ce{NaH2PO2}$ (sodium hypophosphite) does not have any acidic protons to react with strong bases like $\ce{NaOH}$. The two hydrogens in $\ce{NaH2PO2}$  have attach ed to phosphorous atom in covalent bonds. And also showing is sodium bicarbonate, which has one acidic proton to react with $\ce{NaOH}$.


text:
Why are bonded ions more stable than separate ions?
I was reading on the Born-Haber cycle when I encountered a problem. The textbook seemed to suggest that since energy must be supplied to separate an ionic compound into it's composite ions, then the ions are more stable when they are bonded than when they are separate. My question is why?

It is consequence of the general rules of electrostatics. The opposite charges have lower energy when closer each other. Directly related is the lattice energy of ionic compounds, where are alternating ions with opposite charge, so energy decrease due closiness of the opposite charges is bigger than energy increase due closiness of the same charges, as the the formaer are in average closer each other.

What's meant by stability is whether the compound "wants" to be in current state. If it wants to change the state (separated ions want to bond) then the system is less stable.
In your example you can simply imagine 2 spherical magnets placed close to each other - they would rush towards one another. They had potential energy that turned into kinetic energy. And once they met - they don't have potential energy anymore, thus they are at lower energy state (more stable) and don't want to separate. Then you'd need to apply force to separate them.


text:
Prevention of Electrophilic Aromatic Substitution
Actually, I want to prevent unwanted electrophilic aromatic substitution in the reaction without deactivating the phenyl ring. Is there  any way through which I can prevent E.A.S. I am thinking about certain solvents or other approaches.

I would be looking at blocking groups, depending on what chemistry you are doing installing Br, SiMe3 or SMe to block the site where you don't want reaction. They are all removable in high yield

Sulphonation/desulphonation can also be used to block the para position to obtain an ortho-substituted product (if that's what you want).

@AniruddhaDeb  that does deactivate the ring though

Instead of sulphonation, you may use tert-butyl group, which doesn't deactivate the ring, though para position would be more susceptible to electrophilic attack