text:
Why Calcium Oxalate is insoluble in acetic acid?

Calcium oxalate is insoluble in acetic acid, but not it's carbonate, oxide and hydroxide.

From internet, it is even considered as a test to distinguish calcium oxalate from it's carbonate and oxide. The reason from the resources which i had searched, only gave factual reasons such as

High Lattice Energy and
Low solvation energy of $Ca(C_2O_4)$

But intuitively the "short sized" compounds like $CaO$ and $Ca(OH)_2$ must have better packing and more lattice energy, then why much bigger calcium oxalate have higher lattice energy?
Also, i am clueless about how to compare solvation energy of different compounds. I think it has to do with polarity of the solute and the solvent. Any help will be appreciated.
Edit:
I had omitted the relative basicity of oxide, hydroxide and carbonate of calcium but i am more interested in knowing how to compare the lattice energy and solubility of the given compounds.

You have omitted to consider acid-basis relations, particularly the relative acidity/alkalinity of respective cunjugated acid/base pairs.

Yes, but what may be the reason for other facts, i am more interested in knowing how to compare properties like solubility and lattice energy

Such things are more related to solubility  product. For solubility itself, you have to involved the side acid-base equilibrium reactions. The play essential role.

Calcium oxide, hydroxide and carbonate react with acetic acid to form calcium acetate. Calcium oxalate does not react.
Edit: In real life, you'll of course will end up with a little oxalic acid and calcium acetate, as calcium oxalate is very slightly soluble and then you'll mix all available ions.

If you take a solution of calcium acetate and add oxalic acid, you will precipitate calcium oxalate, leaving acetic acid in solution. That's because oxalic acid is a stronger acid than acetic, as Poutnik suggested.

"That's because oxalic acid is a stronger acid than acetic". Probably doesn't work like this. Hydrochloric acid is stronger that oxalic acid, but calcium oxalate won't dissolve in it.
It's rather the solubility product of calcium oxalate that's driving the precipitation.

Reverse the comparison: Since oxalic acid is stronger than acetic acid, acetic acid is weaker than oxalic acid, and hence won't protonate oxalate (at least, not very much). Since acetic acid won't protonate oxalate, it can't bring calcium oxalate into solution. You are correct that the solubility product drives the precipitation, but to undrive it requires protonation of oxalate anion. CRC Handbook says calcium oxalate is soluble in acids; stronger acids will do better because a strong acid could protonate oxalate and drive down the concentration of oxalate anion as CaC2O4 dissolves.

You're right, James. There is the equilibrium between free oxalate and oxalic acid which will draw the calcium oxalate into solution at low pH. My bad.

James may I know edition of CRC Handbook says "calcium oxalate is soluble in acids; stronger acids will do better because a strong acid could protonate oxalate and drive down the concentration of oxalate anion as CaC2O4 dissolves". Thanks

@ yeni Sulastri: CRC 62nd edition, 1981-2. Should also be in earlier and later editions.


text:
Why was Nobel Prize given to Baeyer for his "faulty" Baeyer's Strain Theory?
Baeyer Strain Theory said that using the formula
$\frac{(109^ \circ 28' - \alpha)}{2}$, where $\alpha$ is the angle between 2 sides of the cycloalkane, the most stable cycloalkane could be found. Thus he proved cyclopentane to be the most stable.
But later using the value of heat evolved during combustion, cyclohexane was found to be the most stable.
Then why was the Nobel Prize given to Baeyer for a faulty theory, still resides beside his name?
For a theory which no longer could explain the maximum stability of cyclohexane?

Are you asking why Nobel prizes aren't rescinded? Or is this a question about the theory itself?

Yes. Why wasn't it rescinded?

Baeyer's Nobel was principally for his work on dyes. Windhaus received the Nobel Prize in chemistry in 1928 for his work on sterols even though his proposed structure for cholesterol was shown to be incorrect  a few years later. To my knowledge, the prize was not rescinded.

Well, that's more of a question for the Nobel committee. Not for Stack Exchange. You can send them a message at https://www.nobelprize.org/contact

That's not a question for this community.

Instead of being critics to questions, answers are more welcome. And about the sarcasm, I think I found my answer already, thank you very much

Scientists are human beings after all. Nobody gets a Nobel Prize for inventing a "formula". It is the idea that makes an impact and new ideas are respected. Scientists like normal human beings make honest mistakes, but try their best to be correct. The guy who got the Nobel Prize in polarography, Hervosky, in the 1950s made errors in other electrochemistry papers, does this mean the prize be snatched back from him as well? Max Planck when he derived his blackbody radiation law, he did not believe much in atomic/molecular theory (later on he changed his views). He also received a Nobel prize.
If you read the citation/biographer by the Nobel Prize committee, the award was not due to some cycloalkane strain angle but the prize motivation was "in recognition of his services in the advancement of organic chemistry and the chemical industry, through his work on organic dyes and hydroaromatic compounds."
The Nobel Prize clearly says "Von Baeyer’s work was at once pioneering and many-sided. With admirable penetration and extraordinary experimental skill he combined dogged perseverance and, even at 70 years old, a youthful buoyancy in his work. He was careful never to overestimate the value of a theory. While Kekulé sometimes approached Nature with preconceived opinions, von Baeyer would say: “I have never set up an experiment to see whether I was right, but to see how the materials behave”. Even in old age his views did not become fixed, and his mind remained open to new developments in chemical science." (https://www.nobelprize.org/prizes/chemistry/1905/baeyer/biographical/)
You can see here, (good) scientists are open to change and are willing to listen to others. I found very good scientists to be very humble (with a small fraction of outliers), but only the mediocre ones are at times rigid and arrogant.

Nice response! I believe the "hydroaromatic compounds" part of the citation was tied to his strain theory. He assumed that the rings were all planar. He argued that, if van't Hoff and LeBel were correct about the tetrahedral carbon, then there should be two cyclohexane carboxylic acids, "equatorial" and "axial". On Neptune perhaps, not on Earth.

Unfortunately, he did not deliver a Nobel Lecture, otherwise we could have verified the story behind hydroaromatic compounds.

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text:
Ozone in ozone layer
If ozone is a gas, and gases have the highest entropy, then how does the ozone gas stay within a few layers of the atmosphere, even though they span for kilometers?

Entropy and ozone at the top of the atmosphere are two unconnected things. Harmful UV rays from the Sun are unable to penetrate deep into the atmosphere because they encounter oxygen molecules. They absorb the UV and form ozone on the top. If ozone reached us, nobody could survive. It is a harmful thing. Read more about the https://en.wikipedia.org/wiki/Ozone%E2%80%93oxygen_cycle

related: https://chemistry.stackexchange.com/q/41261/7951 https://chemistry.stackexchange.com/q/81971/7951 https://chemistry.stackexchange.com/q/43498/7951

@M.Farooq - For clarification - there's plenty of ground level ozone around. It's not good for us, but it's not so bad at low concentrations that we don't survive. I suspect most people are familiar with the smell of ozone, even if they don't know what they're smelling.

Right, Andrew, ozone at part per million levels it is not that great! It immediately hurts the throat if you are sensitive. If you recall the smell of older photocopy machines, the smell is not that pleasant but fresh rain is of wonderful but that must be parts per billion conc.

Plus some rare highly reactive agens like peroxyacetylnitrate can be created, what reportedly acted in past during some Los Angeles smog situations as plant defoliant. Additionally, ozone is the starter of immission damage to conifer trees by SO2.

It does not have much to do with entropy, rather with the way and place of ozone creation.
Stratospheric ozone is produced typically at altitude 20-30 km by UVC radiation with $\lambda \lt \pu{280 nm}$:
$$\ce{O2 + \nu -> 2 O}$$
$$\ce{O + O2 -> O3}$$
See ozone cycle as courtesy of @M. Farooq.
Ozone in lower troposphere troposphere is created by UVB (280-320 nm)  (+UVA >320 nm ??) mostly due catalytic effects of nitrogen oxides that come typically of the oxidative smog of the Los Angeles type:
$$\ce{NO2 + \nu -> NO2^{*}}$$
$$\ce{NO2^{*} + O2  -> NO + O3}$$
$$\ce{2 NO + O2 -> 2 NO2}$$

Technically, the concentration and positioning of ozone in the atmosphere likely relates directly to both ozone formation, stabilization and a destruction cycle.
While formation creation cycles has already been outlined, I would like to add more detail on the chemistry based on this source:
Creation:
$\ce{O2 + ℎν → 2 O• }$
$\ce{O• + O2 → O3 }$
which is followed by the 'ozone–oxygen cycle' where the ozone molecules formed by the reaction above absorbs radiation in wavelengths between UV-C and UV-B.
$\ce{O3 + ℎν(240–310 nm) → O2 + O }$
forming atomic oxygen which can further react with dioxygen:
$\ce{O + O2 → O3 + kinetic energy }$
The cited net effect of the ozone–oxygen cycle is the conversion of  penetrating UV-B light into heat, with no net loss of O3. This cycle reputedly keeps the ozone layer stable and results in the protection of the lower atmosphere from harmful UV radiation.
Removal of Ozone:
$\ce{O3 + O· → 2 O2}$
$\ce{2 O· → O2 }$
Also, per a source, on the net effect of the two reactions, to quote:

$\ce{2 O3 → 3 O2 }$
This reaction proceeds more rapidly with increasing temperature.

So, higher colder altitudes are more conducive to the preservation of created ozone.
On its destruction cycles, to quote a source:

Ozone is a highly reactive molecule that easily reduces to the more stable oxygen form with the assistance of a catalyst. Cl and Br atoms destroy ozone molecules through a variety of catalytic cycles. In the simplest example of such a cycle,[11] a chlorine atom reacts with an ozone molecule (O3), taking an oxygen atom to form chlorine monoxide (ClO) and leaving an oxygen molecule (O2). The ClO can react with a second molecule of ozone, releasing the chlorine atom and yielding two molecules of oxygen. The chemical shorthand for these gas-phase reactions is:
$\ce{Cl· + O3 → ClO + O2}$
A chlorine atom removes an oxygen atom from an ozone molecule to make a ClO molecule
$\ce{ClO + O3 → Cl· + 2 O2}$
This ClO can also remove an oxygen atom from another ozone molecule; the chlorine is free to repeat this two-step cycle.

Now, pollutants in the atmosphere include chloro-(and bromo-) organics, which can, under UV photolysis, liberate a chlorine (or bromine) radical.
So, as one gets closer to the surface of the earth, radiation-induced ozone formation is abated, warmer temperatures are unfavorable and any diffused ozone entering the lower layers, could be further subject to attack by increasing concentration of photosensitized pollutants engaging in catalytic destruction cycles.
And, that is why the ozone layer resides at a high altitude, maintains itself and is not found closer to the surface of the earth.

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text:
Why is dinitrogen tetroxide a planar molecule?

I was wondering why $\ce{N2O4}$ is a planar species in spite of having a σ-bond in between the two $\ce{N}$ atoms for free rotation. Does it not form a conformer?

Because pi conjugation.

Why there would be delocalisation of pi electron cloud over N-N bond   as the RS formed would be highly unstable as it would have + charge on oxygen atom with  high electron deficiency

No, it wouldn't have + on oxygen.

The oxygen with the -charge is  in conjugation only with the oxygen present in the same N atom .It can't be in conjugation with the =O present in the other N atom . How there can be partial bond between 2 N atoms

Well, just like that. Each O with minus. Each N with plus. Double bond between N.

I have added the diagrams of RS

Good. Now see, there is one more resonance structure.

I think it is a very good question (why the down vote?) - by comparison the geometry that the isoelectronic oxalate ion adopts in crystals depends upon the counter cation (https://en.wikipedia.org/wiki/Oxalate) indicating that in that case at least the energy well keeping the ion planar is not that deep - so why is N2O4 planar? pi conjugation would suggest double bond character in the N-N bond, but as is well known that bond is weak and long, so I don't find that a very satisfying explanation.

I think that one would expect that a stable form of the molecule might be planar and have D2h symmetry.
Some support for the claim:

Early crystal structures suggested that the ONO bonds on either side
of the N-N bond are easily modeled by low level DFT
(doi.org/10.1039/C19660000082).
More recently, simulations of the trans and cis N2O4 structure show
planarity when optimized at B3LYP/ aug-cc-pVTZ (aka, pretty good not
amazing) See here.

The simulations also suggest that the barrier to isomerization is low, meaning that while the stable structures are planar, there are low barrier (low temperature) out of plane rotations possible.
Summary:

Just because something has sigma bonds doesn't imply that it will immediately be non-planar
yes the molecule can form a conformer


text:
Spraying difluoroethane into an enclosed space
I'm working on developing physics lab activities that my students can do at home during the covid quarantines. For an activity involving waves and thermodynamics, I came up with an experiment at home in which I blow over the top of a bottle (beer bottle or 2-liter soda bottle), measure the frequency, and then spray a can of "dust-off" (difluoroethane) into the bottle and measure the frequency again. The frequency gets lower because difluoroethane is much more dense than air.
Environmentally, it looks like difluoroethane is fairly benign. It's used as an alternative to substances that are worse for our planet's atmosphere. People can huff it, and if you do, it's dangerous. It's flammable, so you want to stay away from flames and sparks. All of those things seem like they're not huge problems for my purposes.
But the label also says, "Never spray into an enclosed space, such as a trash can or paper shredder." What's up with this? That's exactly what I need to do for this experiment. Why is this dangerous?
Would there be any safety or environmental advantage to using 2,3,3,3-tetrafluoropropene ("Opteon" or R-1234yf)? It doesn't seem as readily available in small packages.
[EDIT]
Following up a comment by @MaxW: Interesting, thanks. I found a video of the liquid exploding when it's poured into water. Also a demonstration that it's flammable in gas form if you collect it in an open-topped container. Burning produces toxic $\ce{HF}$ and $\ce{COF2}$. So I'm hypothesizing (tell me if I'm wrong) that it's safe for my application if you (1) stay away from any source of ignition, (2) do it in a well ventilated area, and (3) in case of ignition, get away and let the fumes dissipate thoroughly. I'd be interested in answers as to whether this seems sufficient.

Making an educated guess, I'd say that the danger is explosive limits in a container with an open top would be the worrisome problem. Explosive limits, vol% in air: 3.7-18

That 1,1,-difluoroethane (physically?) explodes in contact with water is rather surprising. Unfortunately the explanation given in the video is rather unsatisfying. Have you considering using other gasses and volatile liquids (saturating the air with vapour), like the butane in the video, or diethyl ether, or acetone? As you say, flammability isn't a huge issue for you, with the proper precautions. I do understand sourcing the material is a major issue, but I can't think of clear winner from a hardware store (e.g. dichloromethane would probably work really well, but it's more toxic, etc..).

I don't think the potential explosion of liquid difluoroethane is a risk if you are just working with the gas. And not using vessels filled with water.

@NicolauSakerNeto: I'd be happy to find an alternative material, but it has to be something cheap that we can buy and mail to students. The technique involves blowing over the top of the bottle, so I don't want to use acetone, etc. -- it has to be an inert gas.


text:
Boiling point of oxygen
I am studying chemistry from the book titled "Fundamental Chemistry for Cambridge O Level" written by Rosemarie Gallagher and Paul Ingram.
On page 17, it states that

oxygen boils at −219 °C and freezes at −183 °C.

Isn't this incorrect? Should it not be the other way round?

Yeah that looks like they have it mixed around. The boiling point will always be higher than the melting point measured at the same pressure. Wikipedia gives the same values, but in the correct order.

BTW, write (or have your instructor write) to the publisher, citing the error. You might get a "freebie" (e.g. a copy of the next edition, with corrections) and the satisfaction of being acknowledged, as well as performing a service for those who would otherwise have to "unlearn" the error.

I tried to find some sort of errata for the book and couldn't. According to Amazon the paperback  was published January 1, 2018 so it isn't an old book.

According to NIST Chemistry Webbook, for oxygen:
$$T_\mathrm{boil} = \pu{90.2\pm0.2 K}$$
$$T_\mathrm{fus} = \pu{54.8\pm0.2 K}$$
$$T_\mathrm{triple} = \pu{54.33\pm0.6 K}$$
The $T_\mathrm{boil}$ and $T_\mathrm{fus}$ are from Ref.1 while Brower and Thodos reported $T_\mathrm{triple}$ as $\pu{54.363 K}$ at pressure of $\pu{1.125 mmHg}$ (Ref.2). The uncertainties were given by Thermodynamics Research Center, NIST Boulder Laboratories, CO.
Accordingly, $T_\mathrm{boil} = \pu{-183 ^\circ C}$, $T_\mathrm{fus} = \pu{-218.4 ^\circ C}$, and $T_\mathrm{triple} = \pu{-218.8 ^\circ C}$. Therefore, as you correctly suspected, the values in the textbook is a misprint.
References:

Alex G. Streng, “Miscibility and compatibility of some liquefied and solidified gases at low temperatures,” J. Chem. Eng. Data 1971, 16(3), 357-359 (https://doi.org/10.1021/je60050a024).
Gerald T. Brower, George. Thodos, “Vapor pressures of liquid oxygen between the triple point and critical point,” J. Chem. Eng. Data 1968, 13(2), 262-264 (https://doi.org/10.1021/je60037a038).


text:
What is the correct pH value of weak and strong alkalis?
In the book titled fundamental chemistry for Cambridge O level, it states at page 149,

Ammonia solution with a pH of 11.1 is a weak alkali and that's why it shows much lower conductivity of 0.50 US/cm.

However at p. 147 it's written that

a solution with a pH of 10 is a strong alkali.

So either the correlation of pH with conductivity is incorrect  or the statement that ammonia solution with a pH of 11.1 is a weak Alkali is wrong.   What are the pH values for a weak and strong Alkali?
Is there anything wrong with the two diametrically opposite sentences/ statements given in the book?

Proper word is base and it's strength has nothing to do with pH!

There might be confusion as below:
strong/weak "Alkali"(1) - or rather an alkaline solution, which is related to the $\mathrm{pH}$ value. The conductivity and the strength increases with $\ce{pH}$, but conductivity can be obviously affected by presence of salts.
There is not a strict threshold between strong or weak alkaline $\ce{pH}$. It is often considered relatively and/or it depends on context. As a guide the threshold can be between 10-12, context dependent.
strong/weak "Alkali"(2) - or rather a base ( as an opposite to an acid ), which is related to the base dissociation constant. IF a base has too low dissociation constant, so there is significant fraction of not dissociated base ( like ammonia ), the base is considered a weak base. There is ongoing equilibrium $\ce{ BOH <=> B+ + OH-}$, resp. $\ce{ B + H2O <=> BH+ + OH-}$

E.g.
$\ce{NaOH}$ solution with $\mathrm{pH} = 10$ is weak "alkali"(1), but strong "alkali"(2).
$\ce{NH3}$ solution with $\mathrm{pH} = 11$ is stronger "alkali"(1) then above, but weak "alkali"(2).
Note: For historical reasons, "alkali" is reserved to hydroxides of alkalic metals, so it should not be used for ammonia. As the general term, a base is preferred, with alkalic/alkaline ( along with basic ) as adjective being accepted.

In the 19th century, ammonia solutions were called "volatile alkali" by druggists and keepers of hardware shops. And sometimes the word "volatile" was omitted by subsidiary people. This information was given to me a long time ago by my father who was a druggist in a hardware shop in the first half of the 20th century (I am an old man, born 1937). I am surprised to see that this word "alkali" is still used for ammonia solutions in a serious document... It should not.

I recently worked at a company with people who had been with the company 50 years(!), and when it started, the chemist was already old. The names toluol and xylol kept being used. BTW, 83 isn't old; you are just up to bismuth. You have another 35 to go.

I wouldn't tell my age (or other data) on the net without good reason. Also this is more of a comment on archaic wording then full answer.


text:
Entropy change of surroundings and system
The entropy change of the surroundings can be calculated by the equation $dS_{sur}=\frac{dq}{T_{sur}}$ regardless of the path (irreversible or reversible).
The argument is that because the surroundings may be approximated as either constant volume or constant pressure, the heat absorbed/released by it is equal to the internal energy or enthalpy change, respectively. As both are state functions, it does not matter if the path taken was reversible or not, the changes would be the same for both, so $dU_{sur,rev}=dU_{sur,irrev}$ and $dH_{sur,rev}=dH_{sur,irrev}$. However, $dH=dq$ for constant pressure, and as per definition $dS=\frac{dq_{rev}}{T}$, the initial equation follows.
This is somewhat the way my book derives the formula. It seems to be inconsistent to me because it equals $\frac{dq_{rev}}{T}=\frac{dq_{irrev}}{T}$, which some pages after the book says that is not the case and uses the fact that $dq_{rev}\geq dq_{irrev}$ to derive the Clausius inequality.
I do understand the Clausius inequality and its derivation, however I do not understand why is it that it is not used to calculate the entropy change of the surrounds, only the system. The arguments for its derivation seem to apply to the first case as well.
Similar question that has no answer: Entropy change of surroundings

In evaluating the entropy change of the surroundings, you need to first look exclusively at its initial state and its final state, before and following the irreversible process.  The only difference between the initial state and the final state is a difference in enthalpy of the surroundings, equal to amount of irreversible heat that flowed to the surroundings.  To get the entropy change for the surroundings, you next need to separate it entirely from the original system, and subject it to a new reversible process while in contact with its own new surroundings (i.e., a second set of surroundings).  No matter what new (alternative) reversible process you devise, you will always end up with the same enthalpy change and same beginning and ending pressures, and its entropy change will always be the same, and equal to the irreversible heat it received in the original process divided by its absolute temperature.
On the other hand, for the original system, if you follow the same procedure (separating it from the original surroundings and putting it into contact with a third surroundings), the reversible process you devise will give you a different value for the integral of $dq/T_{boundary}$ than the integral of $dq/T_{boundary}$ for the original irreversible process.

Could you elaborate on that separation part? I've never seen that.

That's too bad.  It has annoyed me for some time that the text books out there do such a bad job of explaining how to determine the entropy change for an irreversible process.  So I wrote a short primer of how to do it, which many people have found very useful:  https://www.physicsforums.com/insights/grandpa-chets-entropy-recipe/ Please feel free to ask follow-up questions.

Regarding separating them:  If the process was irreversible, to get the entropy change of the system and the entropy change of the surroundings, we need to devise a path that takes each of them from their initial thermodynamic equilibrium state to their final thermodynamic equilibrium state reversibly.  If the original process was irreversible, there is no single path that will take them both between their initial and final states (of the irreversible process) simultaneously.  So we need to separate them, and subject each of them separately to a reversible path between these states.

The most important concept to understand is that entropy is a state function.
Best explained by this image:

The change in entropy is just based on the final state and the initial state. Therefore we model all processes as a reversible process because it has a simple equation. The reversible process is the one in blue above. That allows us to use the equation for the change in entropy for reversible process:
$$\mathrm dS =\frac{\mathrm dq_\mathrm{rev}}{T}.\tag{1}$$
Hint: use the first law to get the equation for $q.$
So if I want to calculate change in entropy for the system I integrate from state 1 to state 2. It will give you the same answer if you want to calculate $ds$ for any process.
Nuance of Surroundings:
If I want to calculate the change in entropy of the surroundings it changes, slightly, because the temperature of the surroundings is constant. That is because the surrounding is a reservoir. So we end up with
$$\int^{2}_{1}\mathrm dS_\mathrm{surr} = \int^{2}_{1}\frac{\mathrm dq_\mathrm{rev}}{T_\mathrm{surr}},\tag{2}$$
which will simplify to
$$S_2 - S_1 = \frac{q_\mathrm{rev}}{T_\mathrm{surr}}.\tag{3}$$

"The nuance of surroundings" part is incorrect as the final result must be $\Delta S=\frac{ -q_{\text{irrevsible}}}{T}$

If what you're trying to point to is that a heat transfer to the surroundings is irreversible then yes. However I wanted to explain in a way where the reader could see that this is the same q.  Then they will see sign convention correctly. My point being it's the same base equation. Same heat transfer, always starting tith definition. Keeps reader focused on that then they can decide what the "system" is and sign vectors accordingly.  In other words... My solution is right, your just being it from a different basis system.


text:
1.0 g of AlCl3 is added to 100.0 ml of water. What would be the resulting pH?

$1.0\ \mathrm g$ of $\ce{AlCl3}$ is added to $100.0\ \mathrm{ml}$ of water. What would be the resulting pH?

I think the answer may be $4.022$ but I'm not certain.
The reaction wasn't provided but I think its this:
$$\ce{Al(H2O6)6^3+ <=> H^+ + Al(H2O)5(OH)^2+}$$
Assuming all $\ce{Al}$ dissolved in water:
Initial $[\ce{Al}]=7.49\times10^{-4}\ \mathrm M$
$$x^2/(7.49\times10^{-4}-x)=1.4\times10^{-5}$$
$$x=9.56\times10^{-5}$$
$$\mathrm{pH}=-\log(9.56\times10^{-5})=4.022$$

What happened to the Cl attached to the Al?!?

I have no idea, I'm so confused with this question

I think you have the right idea, but in an excess of water the overall reaction would be $$\ce{AlCl3 + 6H2O ->[aqueous] Al(H2O)5(OH)^{2+} + H+ + 3Cl-(aq)}$$

Oh okay. Do you know how to solve this

Assume volume of solution is the same as the volume of water and that the reaction I gave goes to 100% completion.

So assume all Al dissolved in water

Yes there will be an excess of water. 100ml of water is about 5.5 moles of water.

can you confirm if 4.022 is right

No. Add your work as an answer. I'll comment on it to help you get it right, then you can accept your answer and gain some site reputation. // Start with the chemical equation. You can edit your post to see the markup I did.

okay I've shown my work

Your first mistake is "$\pu{[inital] Al} = \pu{7.49\cdot10^{-4}}$. Please show how you got that value.

Not being given any other information assume the reaction goes to completion and that it is:
$$\ce{AlCl3 + 6H2O ->[excess H2O] Al(H2O)5(OH)^{2+} + H+ +3Cl-}$$
molecular mass $\ce{AlCl3 = \pu{133.34 g/mol}}$. So:
$$\ce{\pu{mol AlCl_{3}} = mol [Al(H2O)_5OH^{2+}] = mol [H+]} = \frac{\pu{1 g}}{\pu{133.34 g/mol}} = \pu{7.496\cdot 10^{-3} mol}$$
$$\pu{Molarity} = \dfrac{\pu{moles}}{\pu{volume}} = \dfrac{\pu{7.496\cdot 10^{-2} mol}}{\pu{0.100 L}} = \pu{0.07496 M}$$
pH = - log(0.07496) = 1.12

Now the truth is a bit more complicated. You can assume the reaction happens in two steps. The first step will go to completion.
$$\ce{AlCl3 + 6H2O ->[excess H2O] Al(H2O)6^{3+}  +3Cl-}$$
The second step won't go to completion since $\ce{Al(H2O)6^{3+}}$ is a weak acid. There will be an equilibrium favoring the left side. But with no value for $K_\mathrm{eq}$ you are stymied.
$$\ce{Al(H2O)6^{3+} ->[excess H2O] Al(H2O)5(OH)^{2+} + H+}$$
$$K_\mathrm{eq} =\dfrac{\ce{[H+][Al(H2O)5(OH)^{2+}]}}{\ce{[Al(H2O)6^{3+}]}}$$

oh ok? I'm very confused at this point

@nanxii - Sorry I fouled that up badly.

i'm kinda desperate at this point. I just need a simple explanation and answer

@nanxii - Since I messed up, I'll just give an answer. But this isn't a homework site. You've gotten a free ride on two questions. Don't expect more. See the site's homework policy.

Actually the OP has stated an equilibrium constant. The only problem i see is the miscalculation of the molarity. If we use the correct value you can calculate the pH assuming the first step goes to completition. (Wich is 3.0 considering the rounding for given significant digits.)


text:
Why does silver stick to the glass in Tollens' test?
Why does it not simply form a solid powder at the bottom of the flask?
Does silver also stick to the glass in other silver precipitating reactions?
EDIT:
My question is definitely not the same as the one linked (which btw does not have an answer and is inactive for three and a half years).
Normal precipitation of silver from silver ions results in a black powder and no mirror.
As pointed in the comments, kinetics seems to play a major role here, with slow precipitation being key to achieve a mirror.
So why is it that a slow reduction of silver ions make them deposit on the surface of the glass?

Well, only logical explanation is that the reaction happens on surface, not in the bulk of solution, I do wonder why it happens like that, though.

This has some information: https://www.compoundchem.com/2017/09/06/silver-mirror/. Stannous chloride solution is also commonly mentioned as a preparation step: the glass surface is rinsed with this reducing agent.

@EdV The important hint in there might be that the diammine complex is more stable than the aqua complex, and leaving out the ammonia does lead to a black suspension.

It is qualitative but whatever reason nucleation centers being at the glass surface isn't surprising. True is that the film forming is quite peculiar of the test.

@Karl I agree: it was the lower right corner that had a little tidbit of maybe helpful insight. There is certainly no problem making silver nanoparticles or growing a silver tree: I have done these as demos. And adding zinc metal to a silver nitrate or nitrite solution just makes silver powder. So there must be something about kinetically hindering the process to ensure slow and orderly deposition. I hope someone provides a correct answer for this one!

@EdV Me too! But I guess someone has to write a new question that already clears the more obvious points. OP: Please update, then we can vote to reopen!

@Karl Sorry, I just got distracted by a message to vote for moderators in the EE stack exchange. I see nothing that I can add to elucidate how the silver mirror forms. Maybe this could be studied using things like SERS, but I have nothing substantive to go on. I know people still make relatively large silver mirrors for fun, but presumably they are following prescriptions or old recipes known to work. In any event, an interesting question that does not directly involve honey or homework. ;)

@EdV Silvered (Dewar) lab glassware is still made this way, and amateurs grinding large telescope mirrors certainly do the same. I once tried to find scientific work on how to control the thickness of such silver films, wanting a dewar with specific a.c. electrical properties. Did´t find anything really useful, however. Very mysterious. ;-)

@Karl   Just found a scan of a yellowed 1931 article from the old N.B.S. Journal of Research: https://nvlpubs.nist.gov/nistpubs/jres/7/jresv7n5p935_a2b.pdf. This must have been studied before 1931, but the available techniques back then would have been problematic. Even pH meters date from around 1931.

@Karl    Another paper from 1911: http://articles.adsabs.harvard.edu//full/1911PA.....19..327C/0000329.000.html. And it references even older work more than half a century before 1911. But no luck finding out why this process, which appears almost alchemical, works. There is a sentence in German on the first page that my college German is too long gone to translate.

@EdV Quincke says that all those methods for silvering occasionally fail, or give substandard mirrors, and one is unable to find out a proper reason for that. Alchemy. ;-)


text:
To what degree can the Rare Earth Elements be interchanged in its workaday applications?
In their humdrum pedestrian "high-tech applications due to their unique magnetic properties", how interchangeable are the Rare-Earth Elements (REEs) if, e.g., the US lacks one of them?
RRautamaa comments on Pentagon in talks with Australia on rare earths plant.

Rare earths are not rare, but they're chemically very similar. (This is because when you go to higher atomic numbers, they start filling the $\mathrm{f}$-shell, but the $\mathrm{f}$-shell is lower in energy than the $\mathrm{d}$-shell, so it's "hidden" by the $\mathrm{d}$-shell. The $\mathrm{d}$-shell is the valence shell and is the same in all of them, so their chemistry is almost the same.)
Separating two chemically similar substances is expensive. Separating 15 of them is much worse. This means that the mining part is easy, but the processing plant on the site costs about $500 million. This is why rare earth mines are so few in number.

This question was spurred by the Pentagon's halt on May 22 2020 on rare earths funding program pending 'further research'. "Of course, the problem with rare earth elements has never been scarcity. They're actually pretty common. The problem is just that the cost of mining them exceeded their market value everywhere but China, so China ended up with a monopoly and the ability to drive up the price. So it's not clear that this changes much.

It depends on the particular use of the "rare earth". Mischmetal, containing an assortment of mixed lanthanides, can be used as-is for lighter "flints" or fire-starters. On the other hand, ultra-high purity lanthanide elements are needed to get specific spectral lines for lasers, for example.
Other uses, such as in magnets and in high-temperature superconductors, may be somewhat less demanding of purity. Didymium glass, for example, used in glasswork to block yellow sodium light, is a mix containing primarily samarium, praseodymium and neodymium.
Another expense involved in lanthanide production is separation and disposal of the radioactive isotopes from the ore. At one time, for example, gas lantern mantles contained mildly radioactive $\ce{ThO2}$, but those are no longer available in many countries due to concern about the alpha emitting thorium and the alpha and beta emissions from daughters.


text:
What is the difference between primitive Hexagonal unit cell and Hexagonal closed packing HCP
There are 14 types of Bravais crystal lattices (Primitive + Centered) in 3D. Every solid in this universe is made by the unit cells of these lattices.

With the help of these Bravais lattice can HCP be formed?.
Is there any relation between Hexagonal unit cell and Hexagonal close packing
As we see that Hexagonal unit cell exists only in primitive AAA form then how can HCP ABA be formed from it .

Primitive Hexagonal unit cell

Hexagonal closed packing HCP

If there is a clear distinction then Should HCP also be placed among 14 Bravais lattice.

With the help of these Bravais lattice can HCP be formed?

Yes. One choice of unit cell for HCP is a primitive hexagonal cell with two atoms per unit cell. If you place the first atom on the origin, the second will be at (1/3 2/3 1/2). Here is a 3D view of the situation and some more information: http://lampx.tugraz.at/~hadley/ss1/crystalstructure/structures/hcp/hcp.php

Should HCP also be placed among 14 Bravais lattice?

No, because it is already in there, just hiding a bit. If you look further into it, the space group is #194, P$6_3$/mmc, see http://img.chem.ucl.ac.uk/sgp/large/194az1.htm


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Optical activity in cyclohexane
In case of checking for the optical activity of cyclohexane or its derivative why we check for POS and COS in its cyclic 6 membered planar eclipsed state but not by its real conformation it forms chair, boat,twist boat etc)  .

Are you asking why we determine if the compound is chiral from the simplified planar hexagonal drawing, rather than its true conformation drawing?

Yes I am asking this

Why might you think? Which drawing is easy to show and visualise planes of symmetry? Remember the drawings are saying the same thing - the chair conformation is implied in the drawing of a hexagon.

It is not about being visualised easily but it is about the real structures which should be seen for the molecule.The complete planar form doenot exist in any of its conformation.Check out for finding the plane of symmetry of a simple molecule 1,2 cis dimethyl cyclo hexane which has POS according to planar structure,If there is no problem with this then the actual conformer should also have POS ,but where it is

So with the specific example of cis - 1,2 - dimethylcyclohexane, the molecule is achiral not because it has a plane of symmetry. Each of the conformations of the ring are chiral, but they interconvert rapidly at room temperature and so the overall compound is said to be achiral.

The idea of a compound being achiral doesn't mean it has to contain a plane of symmetry.

Can you explain it further

So each of the conformation is chiral because they have no plane of symmetry like you suggested. However the ring is constant flipping between the two chiral conformations, which are enantiomers of each other. So at room temp you moving between each enantiomer so fast you can call the overall compound achiral. But at very low temp when there isn’t the energy to ring flip it is technically a chiral compound as you stuck in one conformation

You can add this one in the answer section


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Finding heat of formation
The heats of combustion of $\ce{C_xH_y}$, $\ce{C}$, and $\ce{H2}$ are $a, b,\text{and } c$ cal respectively. The heat of formation of $\ce{C_xH_y}$ will be:
The answer is
$-\left(x b+\frac{y c}{2}-a\right)$ cal
But I am getting negative of it by adding equations:
$$\ce{C_xH_y} + (x+\frac{y}{4})\ce{O2->}x\ce{CO2}+\frac{y}{2}\ce{H2O} \text{ , }\Delta H=a \tag{1}\label{1}$$
$$\ce{C} + \ce{O2->}\ce{CO2}\text{ , }\Delta H=b \tag{2}\label{2}$$
$$\ce{H_2} + \frac{1}{2}\ce{O2->}\ce{H2O} \text{ , }\Delta H=c \tag{3}\label{3}$$
$\Delta H_f=-1\times\eqref{1}+x\times\eqref{2}+\frac{y}{2}\times\eqref{3}=-a+bx+\frac{cy}{2}$

What are your x and y?

The same x and y which is in question

Using photos of text instead of text itself is strongly discouraged for multiple reasons. 3 main reasons are  1/  Impossible to search for it 2/ Impossible to be reused in answers 3/ Laziness of writing puts extra burden on shoulders of responders while reading and evaluating it.

Be aware combustion and formation heat have switched products and reactants when the value is evaluated from component contributions.

All should be in units like cal/mol or cal/kg, not just cal.

You didn't do anything wrong excep for not applying correct sign. All values of $a,b,c$ should be negative values because for each equation, heat gives away. It seems, the values given in question is just the numerical value without a sign. Asker expects you to apply correct sign when you do your calculations. Fault is on who wrote the question.

it seems to me your assessment is correct, and the given answer is incorrect.

The question: The heats of combustion of $\ce{C_xH_y}$, $\ce{C}$, and $\ce{H2}$ are $a, b,$ and $c$ $\pu{cal mol-1}$, respectively. What will be the heat of formation of $\ce{C_xH_y}$?

We have three combustion reactions to consider:
$$\ce{C + O2 -> CO2} \quad \Delta H_{\mathrm{comp}(\ce{CO2})} = b \tag{1}$$
$$\ce{H2 + 1/2 O2 -> H2O}  \quad \Delta H_{\mathrm{comp}(\ce{H2})} = c \tag{2}$$
$$\ce{C_xH_y + ($x$ +y/4) O2 -> x CO2 + y/2 H2O}  \quad \Delta H_{\mathrm{comp}(\ce{C_xH_y})} = a \tag{3}$$
Assuming the given answer is correct, I have to tell this straight: The wrong doing in this question is giving all three heat of combustion values as their numerical values and not mentioning it in the question (thus, solver has to assume it). That, I called being tricky at this level of education. For example, if the question is a multiple choice, then it may include both $-\left(x b+\frac{y c}{2}-a\right) \ \pu{cal mol-1}$ and $+\left(x b+\frac{y c}{2}-a\right) \ \pu{cal mol-1}$ as chices to pick as an answer. Too bad for students.
Let's assume $\Delta H_{\mathrm{comp}(\ce{CO2})}$ is $-b$ in the equation $(1)$, and so on. Now we can rewrite above equations as:
$$\ce{C + O2 -> CO2} \quad \Delta H_{\mathrm{comp}(\ce{CO2})} = -b \tag{4}$$
$$\ce{H2 + 1/2 O2 -> H2O}  \quad \Delta H_{\mathrm{comp}(\ce{H2})} = -c \tag{5}$$
$$\ce{C_xH_y + ($x$ +y/4) O2 -> x CO2 + y/2 H2O}  \quad \Delta H_{\mathrm{comp}(\ce{C_xH_y})} = -a \tag{6}$$
From the equation $(4)$:
$$\Delta H_{\mathrm{comp}(\ce{CO2})} = \Delta H_{f(\ce{CO2})} - \left(\Delta H_{f(\ce{C})}  + \Delta H_{f(\ce{O2})}  \right) = \Delta H_{f(\ce{CO2})} - \left(0  + 0  \right) = -b\\ \therefore \ \Delta H_{f(\ce{CO2})}  = -b \tag7$$
Simillarly, from the equation $(5)$:
$$\ \Delta H_{f(\ce{H2O})}  = -c, \tag8$$
and from the equation $(6)$ :
$$\Delta H_{\mathrm{comp}(\ce{C_xH_y})} = \left(x\Delta H_{f(\ce{CO2})}  + \frac{y}{2}\Delta H_{f(\ce{H2O})}  \right) - \Delta H_{f(\ce{C_xH_y})} = \left(-xb - \frac{yc}{2}\right) - \Delta H_{f(\ce{C_xH_y})} = -a$$
$$\ \Delta H_{f(\ce{C_xH_y})}  = +a - \left(xb + \frac{yc}{2}\right) \tag9$$
The equation $(9)$ is the correct answer.


text:
Why does DFT-D finds different optimized geometries for the same structure when using different input geometry files?
When I change c lattice parameter and optimize the structure with DFT-D, every time I get a significantly different geometry with different energy? Why is this happening with DFT-D?

In general, the end point of an optimization depends on the starting geometry. Starting from a different geometry can lead you to wind up in different minimum.

@RagonLoso, how different are these starting structures? Are they minor 0.1Å tweaks? What are the molecules? Finally, what is the magnitude of the difference in the energies?

I personally find this question but very specific. What kind of structure are we talking about? What kind of DFA? Convergence criteria, programs involved, algorithms, etc.. An important question would also be how you reduced the issue down to the dispersion correction (which one?)

Simply put, a geometry optimization finds a local minimum. Many such minima may exist for a given system and it is up to the user to ensure that a reasonable minimum is used for further calculations. Techniques may include using many handmade starting geometries, using a random mutation algorithm or starting from known, experimental structures.


text:
In what way trans alkenes are more symmetric than cis alkenes?
While providing the reason for the higher melting point of trans molecules, everyone says that trans molecules are more symmetric than cis molecules. An example can be found over here, where the OP states that:

trans-2-Butene has more symmetry than its cis isomer which results in better packing and hence higher melting point.

But to my (uneducated) eyes there doesn't seem any more symmetry (if any) in trans form than cis form of but-2-ene. I mean cis form is vertically and horizontally symmetric, whereas trans form is not.
Why do we call trans form more symmetric than cis form?

When comparing the melting points, the most important point is not the symmetry of the molecule. It is the symmetry of the spatial arrangement of the molecules in the solid state. l

"More symmetric" is not a thing at all. However, trans isomers tend to have central symmetry, which is not the case with cis.

Trans alkenes have a $C_\mathrm{2h} $ symmetry, identified as such because it has a $C_2$ rotational axis (you need to rotate 180° to have an identical molecule) and a mirror plane perpendicular to that axis $\sigma_\mathrm h$. You need to follow the symmetry flowchart to assign symmetry groups.
A $C_\mathrm{2h}$ group is characterised by having:

$E$: the identity operation
$C_2$: a twofold rotational symmetry axis
$i$: a center of inversion
$σ_\mathrm h$: a horizontal mirror plane

Cis alkanes have a $C_\mathrm{2v}$ symmetry; therefore they have:

$E$: the identity operation
$C_2$ :  a twofold rotational symmetry axis
$2σ_\mathrm v$: two mirror planes that include the rotation axis

As you see a $C_\mathrm{2h}$ group has more symmetry elements and therefore sometimes people say "more symmetry" but this is not a technical way of describing it. To say which is more symmetric you need to consider the possible symmetry operations you can do.
Easier answer:
A symmetric object is an object that looks the same if you rotate, reflect etc. The more operations you can do keeping the object identical, the higher the degree of symmetry. For example a sphere is "more symmetric" than a cube because you have much more symmetry operations you can do (you can have infinite rotation axis and infinite mirror planes etc.). In the case of alkenes the main difference between the trans and cis ones is the so-called inversion point (blue drawing in first figure) that only the trans alkene has and therefore is "more symmetric".

Please don't mix bold symbols with MatJax. When people see bold italic serif, they immediately recognize it as a vector. Also, <br> are unnecessary if you are writing in Markdown and pollute the code. Please visit this page, this page and this one on how to format your future posts better with MathJax and Markdown.


text:
Do we consider the spatial orientation of a molecular orbital while assigning plane of symmetry?
I am following the video lectures by Neeraj Saini sir and in this video$^{\dagger}$ he said that $\ce{CO2}$ has infinite planes of symmetry.
But I think that if we take into account the orientation of $π$ bonds (which I don't know if we have to or not) then $\ce {CO2}$ molecule must  have $2$ planes of symmetry, as per the following argument:
A $\ce {CO2}$ molecule has similar $\pi$ bond orientation as that of an allene.
Now if we consider the orientation of pi bonds then $\ce {CO2}$ must have only 2 planes of symmetry.
So, to check whether infinite planes of symmetry statement is correct or not, I have the following question:

Do we consider the spatial orientation of molecular orbitals while assigning plane of symmetry? And if not then why not?

$\dagger$ Note that he uses Hindi language for the purpose of conversation.

We don't.$\mathstrut$

@IvanNeretin: I agree that we don't. But why don't we? We learn about symmetry of molecular orbitals and molecules as a whole. But in a molecule, we don't consider the orientation of orbitals. Is this simply a convention?

In a way, orbitals themselves are a convention.

You can use symmetry to examine orbitals, for example those in benzene or butadiene, but to determine a point group the atoms are all that is needed. The allene belongs to d$_{2d}$ for example and the orbitals / bonds can be remove and the conclusion is the same.

The reason that orbitals are not considered in the symmetry of a molecule is that the orbitals cannot decrease the symmetry of the molecule. If we look at the canonical delocalized molecular orbitals of any molecule, everyone of them is symmetric or antisymmetric with respect to a symmetry operation of the molecule, with those symmetry operations determined only by the positions of the atoms, not the electron density.
Looking specifically at carbon dioxide, the simplistic picture of $\ce{CO2}$ as being like allene with lone pairs instead of the C-H bonds is not an accurate view of the electron density. If one looks at the delocalized pi-type orbitals, both span all three atoms in either the xz or yz planes (using the standard convention of the z axis being aligned with the molecular axis). Thus, both orbitals are symmetric with respect to the xy plane of symmetry of the molecule (which passes through the C).
Similarly, the lone pairs do not occupy separate sp2 orbitals that have specific alignment in a plane perpendicular to the p orbital involved in pi bonding. Instead, the create a combined electron density that can be deconvoluted into an sp-type orbital along the z-axis and partial contributions from the px and py orbitals, such that the total has no bias along the x or y axis.
If we were to look at the total electron density of $\ce{CO2}$, we would therefore see only a pill shape that does not vary as one rotates around the z-axis.
To visualize this, I recommend looking at all of the calculated delocalized molecular orbitals of $\ce{CO2}$. You can do this on the molcalc.org website if you do not have other software for that purpose.


text:
ortho/para-Selectivity in the Reimer-Tiemann Reaction
I can't figure out why the dichlorocarbene molecule produced in the Reimer-Tiemann reaction gets attached to the ortho position of phenol.
The ortho position is sterically slightly more hindered due to the oxygen, so why dosen't the carbene attach to the para position?

Does this answer your question? In Reimer-Tiemann reaction why does phenol attack the carbene from ortho position?

No, because this answer was a bit more simple and was straightforward

@EVO Answers can be simple and straightforward. If they answers your question, then yours is a duplicate of that.

I don't think this is a duplicate; the linked duplicate asks about C2 attack vs O attack, not C2 vs C4.

The Reimer–Tiemann reaction will give a mixture of products unless the formation of one product is highly disfavoured. The Kürti-Csakó notes in its introductory paragraph on the Reimer–Tiemann:[1]

the regioselectivity is not high, but ortho-formyl products tend to predominate;

The statement already hints at a product mixture being obtained. To further show that both ortho and para attacks are possible, the 1960 review published by Wynberg contains the following scheme as its very first scheme in the introduction:[2]

So while both products are obtained, a back of the envelope calculation already shows that the distribution is not a perfect $2:1$ ratio. This can apparently be explaned by the effect of positive counterions, as Hine and van der Veen report:[3]

The ratio of ortho to para product was found to be 2.21 [under high base concentrations], showing that the tendency towards o-substitution is indeed increased under conditions where ion-pair formation is encouraged. One factor that would certainly be expected to be present and that would tend to favor o-substitution is an electrostatic effect. When a dichloromethylene molecule attacks the o-position of a sodium phenoxide ion-pair to yield the probable initial product, there is less separation of unlike charges than when the analogous para product is formed.

So while the steric hindrance of a single oxygen atom is not much more than that of a hydrogen atom and thus does not play a huge role, the presence of positive counterions – although increasing steric congestion – may serve to enhance the formation of the ortho-product through favourable electronic effects.

References:
[1]: L. Kürti and B. Czakó: Strategic Applications of Named Reactions in Organic Synthesis. Background and Detailed Mechanisms, Elsevier Academic Press, Burlington, MA, USA, 2005, page 378.
[2]: H. Wynberg, Chem. Rev. 1960, 60, 169–184. DOI: 10.1021/cr60204a003.
[3]: J. Hine, J. M. van der Veen, J. Am. Chem. Soc. 1959, 81, 6446–6449. DOI: 10.1021/ja01533a028.

In Reimer-Tiemann Reaction, a mixture of ortho and para isomers is obtained in which the ortho isomer predominates (it is not the sole product).
If one of the ortho positions is occupied, the para-isomer is the major product.
The two isomers can be separated by fractional distillation, in which the unreacted phenol and the ortho-isomer distil over leaving behind the para-isomer.
Ortho product is major mainly due to 2 reasons-

Probability factor ( there are 2 ortho positions available Vs only 1 para position )
H- bonding in the final salisaldehyde.(There is a formation of 6 membered chelated ring which increases the stability of this product.)

Is the Reimer-Tiemann reaction thermodynamically controlled, such that the product stability determines the outcome? To me, it seems more likely that it's kinetically controlled, where transition state stability is the key.

Although it is run at refluxing chloroform conditions, I want to agree with orthocresol on this one and think that thermodynamic control is slightly unlikely.


text:
How do scientists know they have discovered an element?
I have a background in math not chemistry and perhaps this question is inappropriate but I will try. I read that after pitchblende was discovered Martin Klaproth in 1879 a latter date discovered that the pitchblende contained another element which he called uranium.
How did he know the uranium was an element and not just another compound that could not be broken down further?  I read Mendeleeve published the periodic table in 1869 but how about before this date? Somehow the first few elements had to be discovered without the benefit of that publication and scientists had to know the substance discoverd and in question was not a compound but an element.

The question in the title is likely a duplicate of How were the elements discovered before modern technology?. If your question is specifically about uranium, Wikipedia has a brief description of the chemistry allowing to determine that the new element was indeed new. And oh, it's 1789, not 1879. Would be nice if Klaproth lived a bit longer though:)

Mistakes were not uncommon. Compounds were mistaken for elements and vice versa.

History tends to remember those who got it right.

@Karl Except, of course, for Stigler’s Law of Eponymy. ;)

From @Buttonwood again, here is a bonanza of periodic tables: https://www.meta-synthesis.com/webbook/35_pt/pt_database.php?Button=1950-1999+Formulations. Have fun!

@Ed V nice link  that will come in handy !!!


text:
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:
Can pthalic acid or 1,2-Benzenedimethanol be possible products of Cannizzaro reaction of phthalaldehyde?
I know the main product that is ortho-hydroxymethyl benzoate ion but I think that phthalic acid and catechol must be formed also in very small amounts but I am unable to get any confirmation from the internet. Please help me with the products.

As far as intramolecular Cannizaro is concerned, I believe that one part oxidizes and the other gets reduced only. Can you shed some light on your intuition (like mechanism) ?

what I am thinking is that in the case of intramolecular Cannizzaro what you write is correct but take an example, if there are two molecules of phthalaldehyde and one molecule oxidizes and molecules reduce completely that is no intramolecular Cannizzaro then some other products will be formed.

You won't form catechol (aka, pyrocatechol). I believe you are thinking of the diol, 1,2-phenylenedimethanol.

https://www.nrcresearchpress.com/doi/abs/10.1139/v81-157#.XwI4UpOpHlw

@user55119 you are correct catechol is not formed.

The Cannizzaro reaction is the disproportionation of an aldehyde, which do not have an $\alpha$-hydrogen, to an equimolar mixture of corresponding primary alcohol and carboxylic acid salt. For example, original Cannizzaro's experiment:
$$\ce{2C6H5-CHO ->[aq NaOH] C6H5-CH2OH + C6H5-COO^-Na+}$$
Since there is no $\alpha$-hydrogen, the mixture does not undergo an aldol condensation, which is the characteristic of this reaction. The Cannizzaro reaction is an example of a typical redox reaction, which is considered to be one of the most important synthetic reactions of organic chemistry prior to the discovery of $\ce{LiAlH4}$ in 1946 (Ref.1).
The proposed reaction mechanism is as follows (Ref.1):

According to Ref.1, the hydride transfer must be from one molecule of benzaldehyde to the other molecule (when the reaction is performed in $\ce{D2O}$, no deuterated $\ce{CD2-O}$ is found in benzyl alcohol). The formation of benzyl benzoate was suggested and supported (e.g., Ref.2), evidence in Ref.1 doubts this proposal.
The Cannizzaro reaction of phthalaldehyde is studied in aqueous base ($\ce{NaOH}$) to obtain sodium o-hydroxymethylbenzoate as the product. The studies include the kinetics of the reaction, which have been measured over a range of $\ce{NaOH}$ concentration at $\pu{40 ^\circ C}$. This kinetic study proposed the hydration of phthalaldehyde to hemi-ketal monooxoanion is one of the first steps in the mechanism (hemi-ketal formation is depicted in the insert box of above image). The dioxoanion formation is depend on the concentration of $\ce{NaOH}$ solution (Ref.3). The hydride transfer from mono- or di-oxoanion follows, which is the rate determining step (determined by the isotope effect). The observed rate is based on following equation:
$$k_\mathrm{obs} = \frac{a[\ce{OH-}]}{1 + b[\ce{OH-}]}$$
where $a= \pu{(1.38 \pm 0.02)\times 10^{-3} M-1s-1}$ and $b= \pu{0.99 \pm 0.02 M-1}$ (the reaction follows this equation for $\pu{0.05-2.0 M}$ $\ce{NaOH}$). There are no other product has been observed.
I understand your concern of intra- versus inter-molecular hydride transfer. If intermolecular transfer occurs, at least 1,2-bis(hydroxymethyl)benzene and 1,2-benzenedicarboxylic acid should form as minor components. Since hydride transfer is the rate determining, intramolecular hydride transfer is much faster than that of intermolecular version. Therefore no chance of minor product formation is possible since all other steps are in equilibrium.
Note: I assume your suggestion of forming phthalic acid and catechol as byproduct is mistaken for 1,2-bis(hydroxymethyl)benzene and 1,2-benzenedicarboxylic acid (phthalic acid). Keep in mind that there is noway of breaking two aromatic-CHO bonds and making two aromatic-O bonds to form catechol molecule.
References:

C. Gardner Swain, Arnet L. Powell, William A. Sheppard, Charles R. Morgan, "Mechanism of the Cannizzaro reaction," J. Am. Chem. Soc. 1979, 101(13), 3576–3583 (https://doi.org/10.1021/ja00507a023).
Arthur Lachman, "The Benzyl Rearrangement v. Cannizzaro's reaction," J. Am. Chem. Soc. 1923, 45(10), 2356–2363 (https://doi.org/10.1021/ja01663a017).
Robert S. McDonald, Christine E. Sibley, "The intramolecular Cannizzaro reaction of phthalaldehyde," Canadian Journal of Chemistry 1981, 59(7), 1061-1067 (https://doi.org/10.1139/v81-157).


text:
How to predict the reason for color in a transition metal complex?
I know about two reasons for the color of transition metal complexes that are dd transition and charge transfer phenomenon but I am confused when a complex can show both of the above-mentioned phenomena then how to choose which can be the dominating reason for the color of the complex.

How to predict the reason for color in a transition metal complex?

Nobody can do it just by "looking" at the color of any compound. Spectroscopists have to do calculations to figure that out. There are at least 15 reasons for the "color" of chemical compounds.
Now qualitatively one can "predict" whether it is a d-d vs. it is charge transfer transition for transition metal compounds.
Keep in mind that d-d transitions are "forbidden" (Laporte selection rule). Therefore, majority of the d-d transitions correspond to low molar absorptivity. For example, Cu(II) solution is light blue, manganese (II) is almost invisible pink, chromium (III) is light green.
Charge transfer transitions are very intense! Add two three crystals of potassium permanganate to water, and it is intensely purple.


text:
Feasibility of phenyl migration
I am learning the fundamentals of organic chemistry,
here is the reaction which got me stumbling me for hours,
in the below reaction i could see that the amine group would get converted into diazonium salt, but i am clueless how to proceed further..

The compund A(solution) is given below, it seems like a phenyl group has migrated to the adjacent carbon, i have never seen something like this and it is also not clear for me why a carbonyl group is formed in the middle. Overall I dont understand the reaction...it would be helpful if someone points out what did i miss out in reading that made me unfit to solve this problem. Thank you!

It is a carbocation rearrangement, common in organic chemistry. Did you heard Pinacol Pinacolone Rearrangement? This is similar to that.

yeah i got it....but i forgot to mention in the question about the product's sterochemistry, hope u would add on it

This is similar to pinacol-pinacolone rearrangement. Only difference is in pinacol-pinacolone rearrangement, carbocation forms by dehydration while here, it forms through diazonium reaction. Following is the tentative mechanism of the rearrangement:

Since reaction progress through carbocation intermediate, the final product is a racemic mixture. $\ce{CH3-C+-CH3}$ (in pinacol) or $\ce{CH3-C+-H}$ (in given compound) can rotate about middle horizontal bond before alkyl or aryl group migration.

When phenyl is the migrating group there is also the possibility of an electrophilic aromatic substitution mechanism at the ipso position (i.e. forming a cyclopropane intermediate) followed by deprotonation of the O-H bond and breaking the C-O bond.


text:
Half life of first order reaction in case of variable volume
I have a doubt about the meaning of half-life of a first-order reaction. Earlier I used to think that it is the time in which concentration of reactant reduces to half of its initial but recently I got a question (I don't have that question right now) undergoing a first-order reaction with variable volume and then my teacher told that half-life is the time when moles of reactant reduces to half and as the volume is constant moles are similar as concentration.
I want to confirm this thought here.

Is volume constant or variable in your reaction?If it is constant, then comparing with number of moles is also correct.

For constant volume reactions, half the reactant moles being consumed = concentration getting halved.

the volume is variable and I have came across two questions where volume is variable and they used moles

@EashaanGodbole you are right my friend but I am confused in case of variable volume.

If the volume is varying with time, you will have to solve the first-order differential equation with $V(t)$ to obtain the rate equation.

For first order reactions (assuming the reaction $\ce{A -> B + C}$):
$$-\frac{\mathrm{d}[\ce{A}]}{\mathrm{d}t} = k[\ce{A}]$$
If the volume is $V(t)$, then $[\ce{A}] = \frac{n}{V(t)}$. You will have to substitute $[\ce{A}]$ in the differential equation and solve it in order to obtain the rate equation in this case.
As for finding the half-life, Wikipedia states that it is the time required for a quantity to reduce to half its initial value. If $V(t) = V_0$, then $[\ce{A}] \propto n$ and the half life for the quantity of substance is equal to the half-life of the concentration of substance. If $V(t)$ is not constant, then you'll have different half-life times for the quantity of the substance and for the concentration of the substance.

sir, I still have some problems as in the case of radioactive decay which is a first-order reaction the half-life is the time required for the decay of half of the number of nuclei.

First order reactions are always about the total count or particles, molar amount or mass.
$$\frac {\mathrm{d}n}{\mathrm{d}t} = -k \cdot n$$
$$\frac {\mathrm{d}m}{\mathrm{d}t} = -k \cdot m$$
The point of using concentration instead is, that it is proportional to them in homogenous, constant volume scenario.
$$c = \frac nV$$
$$c = \frac m{MV}$$
so it can be expressed in terms of $c$  as well:
$$\frac {\mathrm{d}c}{\mathrm{d}t} = -k \cdot c$$
This has great advantage as $c=\mathrm{f}(t)$ can be monitored much easier than $n=\mathrm{g}(t)$ or  $m=\mathrm{h}(t)$
For homogenous, variable volume scenario, concentration alone is not useful, unless multiplied by volume.
$$\frac {\mathrm{d}n}{\mathrm{d}t} = -k \cdot c \cdot V = -k \cdot n$$
$$\frac {\mathrm{d}m}{\mathrm{d}t} = -k \cdot c \cdot V \cdot M = -k \cdot m$$
For non homogenous concentration, concentration must be integrated over the volume
$$\frac {\mathrm{d}n}{\mathrm{d}t} = -k \int_V{c \cdot dV}$$
$$\frac {\mathrm{d}m}{\mathrm{d}t} = -k \cdot M \cdot \int_V{c \cdot dV}$$
or the known molar amount or mass has to be used.
With such approaches, the half-time is scenario independent:
$$t_{1/2} = \frac{ \ln{2}}k$$


text:
Beckmann rearrangement v/s dehydration of Oximes

In the above question, A is an oxime, which is then reacted with phosphorus pentoxide. Now, phosphorus pentoxide is both a dehydrating reagent and a reagent used for Beckmenn rearrangement, but which reaction will form the major product mostly, ie, which reaction will dominate? My teacher told to consider beckmenn rearrangement with Phosphorus pentoxide, but the answer of this question is A, that is, dehydration is favored . What I think from my experience is that aldoximes will favor dehydration, and ketoximes beckmenn rearrangement( dehydration obviously cannot  occur in them), but I am not sure. Please tell the conditions under which any reaction will dominate over other. Also what  would be the answer if phosphorus pentoxide is replaced by phosphorus pentachloride?

The aldoxime will tend to dehydrate with P2O5 or PCl5. No Beckmann.

Or, A is the Beckmann product with aldoximes.

See beckmann fragmentation.

Do you think that the mechanism of "Beckmann" and "dehydration" are different?
Answer is No. They're essentially the same reaction. Aldoximes undergoes Beckmann rearrangement to form nitriles, or you can say "they dehydrate" (as hinted by Ben Norris in comments)
The mechanism for Beckmann rearrangement is as follows,

(Source: Wikipedia, Beckmann rearrangement)
Note: Just replace $\ce{R1}$ with $\ce{H}$ and then, in the intermediate (upper one) after second step, the arrow will be from $\ce{N-R1}$ bond to $\ce{N}$. That's dehydration!
Also, if you replace $\ce{P2O5}$ by $\ce{PCl5}$ or $\ce{H2SO4}$, the answer would be same for all.

The middle arrow in the intermediate is going in the wrong direction. 2) If you want phosphate to deprotonate intramolecularly, then draw the anti-oxime. 3) This is a dehydration, not a rearrangement. Atoms are not being reconnected as in the Beckmann rearrangement.

@user55119: due to limited search via Google, I can't find a standard mechanism for dehydration of aldoximes. Hence, I've edited it with Beckmann rearrangement.

What about Beckmann Fragmentation/abnormal Beckmann?

The top two arrows in the second  structure need work. R1 with its electron pair migrates to nitrogen as water is lost. Then the electron pair on nitrogen can stabilize the carbocation.

@user55119: "...Then the electron pair on nitrogen can stabilize the carbocation." - the intermediate which you're talking about is the lower one in square brackets. No?

@Safdar: you can see here for fragmentation. I think that's a different reaction.

No, the structure preceding the brackets. What is in the brackets is fine.


text:
sp2 hybridisation of alkyl radicals causing formation of racemic mixture

What I know about $\ce{^{.}CR3}$ radical is that it has both $\ce{sp^2}$ and  $\ce{sp^3}$ character but the $\ce{sp^2}$ character dominates and so the radical is $\ce{sp^2}$ in nature.
So, suppose we have the above situation. In this, a radical is formed  at  the carbon of the 6 membered ring at which one $\ce{-CH3}$ is attached since the bromination is selective, but will it form enantiomers? That is, can the $\ce{^{.}Br}$ radical attack from both sides?
In this question, the answer given is (c), but I think that enantiomers can not form, as the electron of the radical would be present only on one side of the radical carbon, and so cannot migrate to other side.

In the first step, due to the selectivity of bromine in a free radical reaction, the tertiary carbon which has a $\ce{-CH3}$ group attached to it forms the radical.
Now, since the radical is a $\ce{^{.}CR3}$ radical, the carbon radical in the intermediate becomes $\ce{sp^2}$. This means that $\ce{^{.}Br}$ can attack from both the top and the bottom since this lone electron in the $\ce{^{.}CR3}$ radical is in a p-orbital. This is because the electron is equally likely to be found in both lobes of the orbital.
Now after the attack of the $\ce{^{.}Br}$ radical, you get two products. The first one shows the product formed when $\ce{^{.}Br}$ attacks from the bottom and the second one shows the product formed after the $\ce{^{.}Br}$ attacks the intermediate from the top.
You may have noticed that both products are enantiomers of each other and hence the answer would be (c).
Reference
Chem Libretexts


text:
Unusual hydrolysis of methyl 2,4,6‐trimethylbenzoate
Problem 223 from Conceptual Problems in Organic Chemsitry by Singh [1, p. 367]:

The given answer is b and I don't get why.
Usual ester hydrolysis would've given a.
I somehow managed to know that the mechanism is SN2, but why?
Also, if someone could provide the reaction mechanism, it would be wonderful.
Reference

Singh, D. K. Conceptual Problems in Organic Chemistry: For Engineering and Medical Entrance Examinations, 3rd ed.; Pearson India, 2013. ISBN 978-93-325-8207-1.

This is a known reaction mechanism for methyl esters with very good leaving groups (more commonly sulfonates, but benzoates as well). Here it's also favored by the steric hindrance of the carbonyl carbon. I'll try to dig up some refs when I have a few more minutes.

Here is a link to a paper that describes the reaction and various examples: http://folk.ntnu.no/audunfor/7.%20semester/Fysikalsk%20organisk%20kjemi/Artikler%20brukt%20i%20eksamensoppgaver/SN2%20Reactions%20with%20Carboxylic%20Esters..pdf

@Andrew you should make this an answer as you are undoubtly correct

The problem with this question is that you cannot be sure of the answer a priori. It would be a much better question to present the actual product and ask one to rationalize why that's the case.

In the typical reaction of hydroxide ion with a methyl ester, the hydroxide attacks the carbonyl carbon of the ester, resulting ultimately in elimination of methoxide. Given the basic conditions, the carboxylic acid formed immediately deprotonates to a carboxylate, and the methoxide is protonated to methanol. In older texts particularly, this reaction mechanism is referred to as BAc2, indicating "bimolecular base-catalyzed acyl-oxygen cleavage" (*) because the bond that is broken is between the acyl group and the oxygen of the methoxyl group [see for example Ingold, C.K. (1953) Structure and Mechanism in Organic Chemistry, Cornell Univ. Press].
An alternative mechanism is the BAl2 mechanism, designating "bimolecular base-catalyzed alkyl-oxygen cleavage". As the name suggests, in this mechanism, the bond between the alkyl group and the oxygen is broken. To do this, the hydroxide attacks the methyl carbon in an SN2 reaction, directly yielding the carboxylate and methanol products. This mechanism is also described in detail in Ingold's book and many other sources.
In the years since this mechanism was proposed, there has been some debate about how often it really occurs. Certainly for most esters, BAc2 is the dominant mechanism, and even when reaction by that mechanism is very slow, only a fraction of molecules, if any, react by the BAl2 option.
There seem to be two requirements for the BAl2 type reaction. First, the alkyl group must be methyl or a very ring-strained lactone. Other alkyl groups react much more slowly, as is observed in any SN2 reaction. Second, the carbonyl carbon must be sterically hindered to the point that the BAc2 mechanism is essentially impossible.
I am not certain that the tri-methyl benzoic acid in your question is hindered enough to react exclusively by the BAl2 mechanism, but it is certainly hindered enough that BAc2 would be substantially diminished.
You can find a more thorough discussion of these mechanisms in many textbooks, including the Ingold reference above. Another paper that may be useful is Mueller, P. and Siegfried, B. (1974) SN2 Reactions with Carboxylic Esters. Selective Cleavage of Methyl Esters. Helv Chim Acta 57:987.
*Note that since the net reaction in my example is $\ce{R-CO2Me + HO- -> R-CO2- + MeOH}$, the base is a stoichiometric reactant rather than a catalyst, so "base-catalyzed" is a misnomer. That terminology is used because the mechanism is the same if a different nucelophile is used, and the product is a non-acidic acyl derivative rather than a carboxylic acid. In that case, the base is indeed a catalyst.


text:
Why is 4-hydroxypyridine more acidic than benzoic acid?
4-hydroxypyridine (or 4-pyridone) has a $\mathrm{p}K_\mathrm{a}$ in the ballpark of $3$ to $3.33\:^1$ (in the pdf linked below, the $\mathrm{p}K_\mathrm{a}$ of 1-methyl-4-pyridone has been given to be $3.33$, however I suppose removing a methyl would only cause a slight decrease in the $\mathrm{p}K_\mathrm{a}$ value). This seems an unexpectedly low value to me. 4-pyridone seems more acidic than I expected it to be, since the $\mathrm{p}K_\mathrm{a}$ of other common acids are:$\:^2$
\begin{array}{lr}
\hline
\mathrm{Acid} & \mathrm{p}K_\mathrm{a} \\
\hline
\text{Benzoic acid} & 4.20 \\
\text{Formic acid} & 3.75 \\
\text{Acetic acid} & 4.75\\
\text{Phenol} & 10.0 \\
\text{Carbonic acid} & 6.37\rlap{\:^1} \\ \hline
\end{array}
Now I had expected 4-hydroxypyridine to be more acidic than phenol, because of the negative charge in the former's conjugate base being delocalised onto the electronegative nitrogen, yielding a more stable resonating structure than the latter's corresponding resonating structure. I thought that 4-pyridone's $\mathrm{p}K_\mathrm{a}$ might be somewhere around $\ce{H2CO3}$'s $\mathrm{p}K_\mathrm{a}$, but I was wrong. Also, I wasn't quite sure that it would be acidic enough to have a $\mathrm{p}K_\mathrm{a}$ value any close to other carboxylic acids. But it seems to me that it is more acidic than benzoic acid, formic acid and acetic acid.
Why is this so? I can't see any excetional characteristic in its structure which might impart it such a low $\mathrm{p}K_\mathrm{a}$. This does seem more of an exception to me, as of now, but it would be very helpful if someone clarified and explained such a trend.
References:
$[1]:$ http://gissmo.nmrfam.wisc.edu/static/pKa_compilation_R_Williams.pdf
$[2]:$ https://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/acidity2.htm

https://www.chemicalbook.com/ProductMSDSDetailCB6303828_EN.htm 4-hydroxypyridine has a pKa of 2.8 according to this

@Safdar However, this page from the same website shows a $\rm{pK_a}$ value of $3.2$, so I wan't really sure about the authenticity of the $\rm{pK_a}$ value mentioned there. However, my question still holds, in both the cases, the $\rm{pK_a}$ value is definitely lower than expected.

Why is it lower than expected? The conjugate base is quite stabilized.

@Zhe It is stabilised, but I don't see why it is so stabilised that 4-hydroxypyridine is so much more acidic than phenol.

The following are two resonance forms of 4-pyridone,

So, when 4-pyridone gets de-protonated, the negative charge on nitrogen gets delocalised on more electronegative atom i.e., oxygen, resulting in more stable resonance structure!
Note that, this explanation is identical to that one which is given when we compare $\mathrm pK_{\mathrm a}$ of benzoic acid and phenol.

Ref: Stability of 4-pyridone vs 4-pyridinol

I thought about the answer, and I am still wondering what makes the two equivalent resonating structures of the formate ions less stable then the conjugate base of 4-pyridone. In fact, oxygen is more electronegative than nitrogen, so having a negative charge delocalised on two oxygen seems like a much stable configuration than delocalisation on one oxygen and one nitrogen. Moreover, the resonating structures of formate are equivalent resonating structures, imparting an "extra" stability to the anion.

Can the delocalisation over mere 3 extra carbons justify the increased acidic strength of 4-pyridone?

@FakeMod, extended conjugation and aromaticity in 4-pyridone is what making it more stable than formate anion

But aromaticity isn't new to the 4-pyridone anion. It was there even before deprotonation, so I don't think it can be a driving factor here, however, extended conjugation does seem reasonable. Thanks!

Well, you too have wrong tautomer in there...

Hey @Mithoron, I've drawn resonance structures, they're not tautomers.

Resonance structures of the wrong tautomer... Just like second answer had. One can think this position of proton is no biggie but 4-hydroxypyridine and 4-pyridone are different compounds, even if usually in equilibrium, and OP is thinking primarily about the other one.

@Mithoron, you're right that 4-pyridone != 4-hydroxypyridine. But, the link OP had provided at starting shows 4-pyridone.

First of all let us examine the structural formula of  4-pyridone:

Now, as you can see, when it acts as an acid a negative charge forms on the nitrogen atom. This negative charge is well stabilized as it undergoes extensive resonance which includes oxygen as well. The fact that the negative charge is formed on nitrogen also contributes to the acidity.

"when the hydrogen is abstracted and negative charge comes , it acquires aromaticity": I don't think so. Wasn't it aromatic all along, even before deprotonation? Reference: https://chemistry.stackexchange.com/a/126146/91457

@FakeMod yes , thanku for the correction , it has been updated.

Wrong tautomer you have there...

@Safdar: I changed it back to tautomers. Also, followed  andselisk♦'s instruction to make similar fond in both text and image.

I thought about the answer, and I am still wondering what makes the two equivalent resonating structures of the formate ions less stable then the conjugate base of 4-pyridone. In fact, oxygen is more electronegative than nitrogen, so having a negative charge delocalised on two oxygen seems like a much stable configuration than delocalisation on one oxygen and one nitrogen. Moreover, the resonating structures of formate are equivalent resonating structures, imparting an "extra" stability to the anion.

Can the delocalisation over mere 3 extra carbons justify the increased acidic strength of 4-pyridone?

The pKa value that triggered the discussion must be incorrectly reported in this old document. It corresponds to the pKa of 4-hydroxypyridinium, i.e. expressing poor basicity of 4-hydroxypyridine. 4-hydroxypyridine is also poorly acidic (pKa near 11). The reason is that the 4-pyridone tautomer shown in other answers is highly prevalent in aqueous solution. This tautomer has neither acidic (phenol like) nor basic (pyridine-like) functionality. Correct values can be found in: A. Albert, J. N. Philipps, J. Chem. Soc. 1956, 1294-1304.


text:
What is the dimension of Avogadro's constant?
What is the dimension of Avogadro's constant ($N_\mathrm{A}$).
On Wikipedia it says it is dimensionless, but in Nigel Wheatley's article (pdf) On the dimensionality of the Avogadro constant and the definition of the mole it says it is $\mathsf{N}^{-1}$.
\begin{align}
\text{Number of particles} &= N_\mathrm{A}\times \text{Amount of substance}\\
[\text{Amount of substance}] &= \mathsf{N}\\
[\text{Number of particles}] &= \mathsf{1}
\end{align}
If $[N_\mathrm{A}] = \mathsf{N}^{-1}$ then the above equation follows, otherwise it doesn't. Is this the right value?

See https://chemistry.stackexchange.com/a/108506/79678

Mildly related question about the value of $N_\mathrm{A}$ (not the units): What is the correct value of the Avogadro constant? And how was it derived?

The dimension/unit of the Avogadro constant $N_\mathrm{A}$ is actually $1/\mathsf{N}$ or $\pu{mol-1}$. It is shown that way in your first reference as well, i.e.
$$ N_\mathrm{A} = \pu{6.02214076E23 mol-1}. $$
What is "dimensionless" is the Avogadro number or the number of atoms / molecules in a single $\pu{mol}$ of that substance, sometimes written as $N$ or $N_0$.
The numerical value is the same, but the concept is different. Basically the relationship between the two is:
$N_0 = (\pu{1 mol})\times N_\mathrm{A} = \pu{6.02214076E23}$
See also the answer discussed in the comment by Ed V, where the new definition of the constant is explained: Effects of Changing Avogadro's Constant.

The dimensions of Avogadro's Constant is $1/\mathsf{N}$. In SI units, this is $\pu{1/mol}$.
Your unit analysis is correct. I have seen this type of question before, I believe the confusion is caused by people normally saying particles per mole, and "particles" is a "phantom" unit (similar to radian). But either way, the SI units are definitely $\pu{1/mol}$.


text:
Can a chiral molecule be optically inactive?
Does there exist a chiral compound with zero optical activity? It seems possible that for some (chiral) arrangement of atoms, the optical activity of one section of the molecule could be exactly compensated by another section of the molecule, even if these two sections are not mirror images of each other. Although this is just a mathematical coincidence, does any chiral molecule have an optical activity of exactly zero?

Are you asking for compound with smallest specific rotation?

Edited it... I wanted to ask something else

It seems you're asking if "a molecule with chiral centres can have zero optical activity?". But, I'm not sure.

I very much doubt that there is a compound with exactly zero activity. Meso compounds are achiral (by definition) even though they contain stereo centers.

@RahulVerma yes, but not just limited to chiral molecules that have chiral centres alone. There exist molecules without chiral centres that are chiral.

Do you mean allenes? they do have a stereocenter.

@Safdar allenes, spiro compounds, and combinations of rings and double bonds are a few examples

The likelihood of exact cancellation of optical rotation in a chiral molecule is extremely low, so much so that the answer to this question is practically no.

With infinite measurement precision then no, but within the limit of experimental error then yes - see cryptochirality

The intent of the question appears to be to ask whether any chiral molecules could have exactly no optical activity. But this is problematic and the best we can do is test whether any compounds don't have observable optical activity.
As the Link by Nicolau Sater Neto to the concept of crypto chirality shows, on this criterion the answer is yes. The molecule 5-ethyl-5-propylundecane shown here has no observable optical rotation even when a pure enantiomer is isolated.

One can intuit that the nature of the compound is such that its rotation probably exists but is very very small.
it seems improbable from first principles that non trivial examples (eg not meso compounds) exist where the optical rotation is exactly zero.

If the reason for the chirality were far from the center of chirality I would think the effect would approach zero or become not measurable.

Yes, tertiary amines where all groups are different are optically inactive  (due to amine inversion) even though it's a chiral molecule. Eg. N(CH3)(C2H5)(H) or NHDT

While technically true this feels like a cheat as the reason why optical activity isn't observed is the lability of the molecules under normal conditions. At lower temperatures they might well show optical activity.


text:
Applying Bohr's model for Helium atom
Consider the following atomic model for Helium. Helium contains 2 electrons revolving in circular orbit around the nucleus,in such a way that there is minimum possible repulsion between them.Calculate the first ionisation energy of this Helium.Will there be any difference in the ionisation energies calculated and experimentally found?

Bohr's model is applicable only to monoelectron species. So it may work in He+ ion.

Now that's an interesting question. Why don't you go and compare these two numbers?

@IvanNeretin I have calculated the value and I got 29.62 eV but the experimentally found value is 24.6 eV. Could you explain me why that difference occurs?

Yes. It means the model is not applicable.

To be strict it doesn't mean the model is not applicable - it has just been applied after all. It means it is not very accurate.

Can you show some work on how you calculated the value? It's not very helpful to ask what is wrong when we don't know what you did.

Nothing is wrong. Even applying a more accurate model hardly will bring you to the experimental values. I personally known active researchers still calculating hydrogenoid systems :)

What this exercise suggests boils down to placing the two electrons on opposite sides of the nucleus. In this way, the repulsion is minimal because the distance between both electrons is maximal. We can safely assume there will be a difference between the energy calculated with this simplified extension to Bohr's model and the experimental value, otherwise Bohr or his peers would have adopted it. The inability to satisfactorily calculate the ionisation energy of Helium with any such model was a major driver for scientists in the early 20th century to adopt a new quantum mechanical description of the atom.
Bohr's theory tells us that the ionisation energy for one-electron atoms varies with the square of the nuclear charge, i.e. $E=Z^2$ in Rydberg units (1Ry = 13.6eV). So the ionisation energies of $\ce{H}$, $\ce{He+}$, $\ce{Li^2+}$ are respectively 1, 4, 9, ...
The two-electron model described here can be simplified to a one-electron problem because the particles are assumed to be neatly aligned at all times. To this end, we define an effective nuclear charge $Z_\rm{eff}$ which sums the charge of the protons in the nucleus (+2 for Helium) and that of the other electron (-1), weighed by the square of the distance : since the opposite electron is twice as far as the nucleus, its repulsive force is four times smaller than the attractive force of one proton in the nucleus. So we get $Z_\rm{eff} = 2-\frac{1}{2^2} = \frac{7}{4}$. The ionisation energy then becomes $E=Z_\rm{eff}^2 = \frac{49}{16}$ = 3.06 Ry (or 41.6 eV).
The experimental value is 1.81 (24.6 eV), which is significantly lower.
Apparently the electron-electron repulsion in the Helium atom is (much) stronger than this simple model suggests.

I think we should go by the Slater's rules?

The exercise at hand seems to be seeking a calculation for a thought experiment. Slater's empirical rules do not apply to that thought experiment.


text:
Is it possible to manipulate some specific atoms of a liquid metal by dint of the magnetic field, electric field, etc?
Let's deem, we have a liquid metal that is multicomponent, for example, type A, B, and C. Is it possible that we single out the type A and put them in a specific shape, for example, a circle?
Let me clarify it, I have a liquid metal, having three different elements, and I poured it in a square framework. I want to separate element C and then set them in several lines that are parallel with each other; so that, they play the role of reinforcement and the alloy of A and B is my matrix.

While Field Structured Composites are a thing, they use nano- to micro-scale particles in a polymer matrix, with particle alignment controlled/influenced by applying electric or magnetic fields during curing.
Trying to do this at the atomic level, as you ask about above, runs into several problems. These include:

Overcoming the desire of the liquid to mix, so you need your desired element to have a fairly well tuned positive heat of mixing to just balance the entropic contributions to the free energy. Or, in other words, just barely soluble.

Overcoming diffusion in the liquid would require strong interactions of your field with the specific element (and only that element). This runs up against the limits below.

In a metallic liquid, applying an electric field isn't going to specifically target any one atom type since the free electrons will shield the bulk from the field. Ions in a dielectric liquid would be a better choice (much like the field structure composites above), but that is just the reinvention of ion chromatography.

The magnetic moments of atoms are pretty darned small, so you won't get the ferromagnetic boost that can be achieved with particles. Applied magnetic fields will have to be enormous if they would even work at all.


text:
What is the hybridisation of the bridging oxygen atoms in the borax anion?
Below is a neat looking borax anion.

While exploring the molecule in a computer simulation (Gaussian 16, level of theory: HF/3-21G), I found that the oxygen atoms in the 8-membered outer ring had sp2 hybrid orbitals. My curiosity led me to question the presented data. Using Natural Bond Orbital analysis, I found out that:
BD(1)B1(O2)
1s(20.94%)p3.78(79.06%)
BD(1)O2(B1)
2s(10.48%)p8.54(89.52%)

BD(1)O2(B3)
2s(23.60%)p3.24(76.40%)
BD1(1)B3(O2)
3s(36.38%)p1.75(63.62%)

LP(1)O2
s(17.50%)p4.71(82.50%)
LP(2)O2
s(48.39%)p1.07(51.61%)

Here BD denotes a 2-centered bond with atom(i)atom(j) and LP(i) denotes the lone pair and the s,p characters follow. My question is as follows:
Given the chaotic data, how does one analyse or interpret the harmony and claim that the oxygen[2] is indeed ~sp2 hybridised?
What is the correct hybridisation of the oxygen atoms not connected to a hydrogen?

NBOA was done using Engine provided as Gaussian, theory: Hartree-Fock in basis set 3-21G

Oh, than that wouldn't be borax, because it's a salt of such anion and you made another mistake in simulation. not to say this title is particularly vague.

@Mithoron Your suggested title was misleading, this atleast is in direct relation in what I want to know. And yes, Borax anion...that has been changed

First, it's important to understand that the simple integer hybridization labels sp, sp2 and sp3 are gross approximations and only accurately describe the electron density of simple symmetric molecules like $\ce{BeH2}$ and $\ce{CH4}$. Atoms in more complex molecules (even something relatively simple like $\ce{H2O}$) cannot be described accurately using these labels. Instead, the contribution of atomic orbitals to localized bond orbitals must be described by non-integer combinations.
That said, the calculations you did are remarkably consistent with an sp2 hybridization of oxygen [2]. If I understand the notation correctly, it says that the bond from O[2] to B[1] is ~90% p (roughly a pure p orbital), while the other bond to B[3] is ~75% p (i.e. sp3). The two lone pairs on the oxygen average 67% p, perfectly sp2.
We are able to average the two lone pair orbitals because electrons are completely interchangeable and are not localized to a specific lone pair. Any hybrid orbital description that captures the combined electron density of the four electrons in the two non-bonding orbitals is acceptable.
So the oxygen can be described as having two sp2 nonbonding orbitals, an sp3 bonding orbital and a nearly pure p bonding orbital. That's pretty close to sp2 hybridization, where we expect three sp2 orbitals and one pure p. The only deviation here is the shift of ~ 10% s from one of the sp2 orbitals to the p orbital, so that we end up with an sp3 and a not-quite-pure p instead.

Your interpretation was correct, Thanks for the explanation, can please explain how did you average?...say for lone pair, I did (5.78/2) ....and what you did was adding the percentages and obtaining a ratio. Also, last paragraph implies that as long as we donot have nearly 100% p character, we cannot say the bond is $sp^2$ am i correct?

@AnindyaPrithvi - The reason you cannot do 5.78/2 is that the 4.7 and 1.1 p contributions aren't normalized. They are only relative to an s value of 1. To give them equal weight, you need to add 4.7/(4.7+1) + 1.1/(1.1+1), where the denominator represents the total s + p contribution. These normalized values are the percentages shown in parentheses in your calculation. For your second question, I guess you are talking about the B-O bond that is sp3? Yes, it's sp3, but the original statement is that the oxygen atom is approximately sp2 hybridized, which is roughly true.

This answer is extrapolating from incomplete data on a very, very crude level of theory. It's probably also using an NBO version years deprecated and the data which is actually not meant to be interpreted. It's not even clear that the structural data is sound. Sure, this is one way or interpreting it, but in my opinion, this is a more misleading interpretation than just guessing hybridisation from the structure with Coulson's theorem (given a crystal structure of something).

@Martin-マーチン Supposedly as the website states, Gaussian 16 is used, and Gaussian 16 uses relatively newer version of NBO as indicated in the answer. Structural data is sound? (structure of NBO or the compound)....and Apply coulson's theorem on the most optimised geometry of Borax( the B.angle is 109), you get sp3 as opposed all over the internet

@Martin-マーチン - fair point about the theory. I was approaching the question as being specifically about why the calculated result did not match OP's expectation and simply pointing out that it does match. Whether or not the expectation and the calculation correctly describe the situation is a different question.

@pipiedipie You do not cite any details for the calculation. While Gaussian is certainly able to interface to newer versions of NBO (i.e. 6 or 7), it only has implemented version 3.1, which is deprecated for more than a decade now. Even if the older version works for this molecule, there is still the issue of the crude level of theory and whether you are actually interpreting data from a true (or even the global) minimum. On another note, I don't see much benefit in describing binding within the constraints of hybridisation for this molecule.

@Martin-マーチン yes right, should have cited. Also, if you meant least energy state by global minimum, it supposedly should be, because I tried tera-chem G.O. and OCRA G.O. before Gaussian (All on same theory).. And also as Andrew stated, i wanted to know how does this work, so might say a fault of the question