text:
Why is ionization always an endothermic process?
My textbook says removing an electron always requires an input of heat. That makes sense because of coulombic force between electron and nucleus. But it also says that the process is endothermic always.
Why is that? In the case of Na, after the input of "activation energy", isn't Na+ more stable than Na? So therefore it would be exothermic?
Furthermore, this is not always true for electron affinity like it is for ionization. Adding an electron to Helium is an endothermic process because He- is less stable than He.* But it's exothermic for Cl and Cl-.
*Not totally sure about that. Contradicting information on the internet. Some sources say electron affinity of He is 0. Others say its negative (thus endothermic).

You can't compare $\ce{Na}$ versus $\ce{Na+}$. The ion is missing an electron, and since $E=mc^{2}$, it makes no sense to compare things that don't have the same amount of matter. That also should hint at why it takes energy to remove an electron. Since $\ce{Na}$ and $\ce{Na+ + e-}$ are identical, except in the latter case, you put two charges farther apart.

Ionization is always an endothermic process as bound electrons have lower energy than free electrons. It is very analogical to astronomical bodies bound by gravity of the central body. Note that ionization can be a part of more complex overall process that can be ( even highly ) exothermic.

See https://chemistry.stackexchange.com/questions/146481/why-is-oxygen-more-stable-than-oxygen-dication/146497

My textbook says removing an electron always requires an input of heat. Better to say it needs input of energy, which can come e.g. from thermal motion, ongoing reaction or absorption of a photon.
Ionization (of neutral atoms or positive ions) is always an endothermic process as bound electrons have lower energy than free electrons. It is very analogical to astronomical bodies bound by gravity of the central body. Note that ionization can be a part of more complex overall process that can be ( even highly ) exothermic.
$\ce{Na+(g)}$ + $\ce{e-(g)}$ are not more stable than $\ce{Na(g)}$ by itself. In fact the opposite. $\ce{Na+}$ is more stable than $\ce{Na}$ if and only if there follows a process making all the operation thermodynamically favourable, like production of solid salt or hydrated $\ce{Na+}$ ion.
Imagine high temperature gas that consists of $\ce{Na}$ and $\ce{Cl}$ atoms. Chlorine afinity to electron is lower than sodium ionization energy. So reaction  $\ce{Na(g) + Cl(g) -> Na+(g) + Cl-(g)}$ is thermodynamically unfavoured. It becomes favoured and exothermic, if ions pack together to form ionic pairs and later solid ionic lattice. Then coulombic energy released by bounding ions together overcomes unfavoured process of forming independent ions in gaseous phase.
Helium electron affinity is negative, estimated byWikipedia: Electron affinity data page as $\pu{-0.5(2) eV}$ ($\pu{-0.5 \pm 0.2 eV}$). Note that (from the link):

Negative electron affinities can be used in those cases where electron capture requires energy, i.e. when capture can occur only if the impinging electron has a kinetic energy large enough to excite a resonance of the atom-plus-electron system. Conversely electron removal from the anion formed in this way releases energy, which is carried out by the freed electron as kinetic energy. Negative ions formed in these cases are always unstable. They may have lifetimes of the order of microseconds to milliseconds, and invariably autodetach after some time.

There are more elements with negative electron affinity than just noble gases. Most of them are metals/metaloids ($\ce{Be}$, $\mathrm{Mg}$, $\ce{Mn}$, $\ce{Zn}$, $\ce{Cd}$, $\ce{Yb}$, $\ce{Hg}$, $\ce{Pu}$), sharing some electron configuration patterns, but suprisingly also $\ce{N}$ ($\pu{-0.07 eV}$).

"Electron affinity can be defined in two equivalent ways. First, as the energy that is released by adding an electron to an isolated gaseous atom. The second (reverse) definition is that electron affinity is the energy required to remove an electron from a singly charged gaseous negative ion. The latter can be regarded as the ionization energy of the –1 ion or the zeroeth ionization energy. Either convention can be used."

https://goldbook.iupac.org/terms/view/E01977

@TomHardy That is known, but the former way is usually preferred.

If the latter definition was used, the electron affinity of N would've still been -0.07eV, right? Or would it change from -0.07eV to +0.07eV?

@TomHardy The same. It is like energy released by falling an item from the shelf on the floor versus energy required for lifting the item from the floor to the shelf. The same too.

All nonradioactive, isolated atoms are stable and do not emit electrons. If one is excited thermally by collision or photolytically by absorption of a photon with sufficient energy, the photoelectric effect, an electron can or will be removed from the atom to give a positive ion and either a negative ion or a free electron. In most chemical reactions the charge separation is not complete, and the electrons involved are attracted to both nuclei in a bonding orbital, a covalent bond. Removal of an electron from a neutral moiety always requires energy, the quantum mechanical bond energy and the simple energy in electrostatically separating a positive and negative charge.
The case of negatively charged ions is more complicated! An atom or molecule with a half empty orbital or an empty orbital of similar energy that is not shielded from the nucleus can add an electron to that orbital with a release of energy. This process is best examined in isolated atoms and can be extended to molecular ions and solids. [Careful here signs are confusing, positive Electron Affinities, EA, usually mean energy is released in forming the negative ion, some sources treat EA as an energy and give energy release a negative sign. Key is H and F release energy in forming H- and F-.] An ion with a release of energy in formation is endothermic in ionization. In the removal of an electron once the electron is out of the valence orbital the electrostatic force is electrons repelling an electron. This repulsion increases the KE of the fleeing electron increasing its energy. If the original binding energy to the nucleus is very low and a negative ion is formed it is possible that its decay could be exothermic. This is explained in the following link[that is also quoted in a previous answer to this question.]  Such a negative ion would have a short lifetime.
https://en.wikipedia.org/wiki/Electron_affinity_(data_page)
A second EA for an atom such as O- + e- = O= is invariably exothermic even tho the empty orbital is available because it involves bringing two negative charges together. This sets up a conundrum because oxide ions do exist in metallic oxides so there must be a stabilizing mechanism. These are usually said to be charge transfer between the negative ion and the positive ions surrounding. Compounds such as cesium oxide, Cs2O, have very low electron ionization potentials. There might be a situation where the loss of an electron might be exothermic in such a compound.


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Integration of chemical potential
I want to ask a question about chemical potential, $\mu$.
I dislike the use of integrals to describe quantities in thermodynamics. If we observe the definition of chemical potential for constant pressure:
$$\mu_i = \left(\frac{\partial G}{\partial N_i} \right)_{T, P, N_i \neq N_j}$$
which I can understand to mean the change in Gibbs free energy G for a small change caused by the number of molecules of species i.
However, using some high school mathematics, I wondered if the following interpretation was valid.
If we consider that if $N_i = 0$, $G = 0$ and logically if $N_i = N_i$, $G = G_i$, integration by separation of variables will yield:
$$\partial G = \mu_i \times \partial N_i$$
$$\int_{0}^{G_i} \partial G = \mu_i\int_{0}^{N_i}\partial N_i$$
such that
$$G = \mu_i \times N_i$$
and hence
$$\mu_i = \frac{G}{N_i}$$
where the chemical potential of a species $\mu_i$ can better be expressed as the Gibbs free energy of a species $G$ divided by the amount of substance $N_i$.
Is this a more reasonable yet valid understanding of the chemical potential?
EDIT a similar equation is derived in [1].
[1] Chen, L. (2019). Chemical potential and Gibbs free energy. MRS Bulletin, 44(7), 520-523. doi:10.1557/mrs.2019.162

Why should be G=0 if Ni = 0? In such a case there would have been G=0 for all matter, as there is always some substance missing. Also, chemical potential is not constant, but is a function of Ni.

No, it´s simply a mathematical formula, which you have derived under a number of assumptions. Please explain how and under what circumstances this formula can be used! What does it mean?

At the very least you have assumed mu is independent of N. If it is not your third formula is invalid - mu should be inside the integration

See updated reference

Your last equation is the definition of mu for a pure substance. Otherwise as Ian explains you can't perform the integration as written since mu is a function of composition.

source: Stack Exchange

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Why does this ligand form 6-coordinate complexes with Transition elements and 8-coordinate complexes with Inner transition elements?

The ligand given above shows different denticities with common transition and inner-transition metal ion i.e. 6 and 8.
The extra two denticities come from the two Nitrogen having 1 lone pair each. I want to know why is inner-transition metal ion able to somehow get nitrogen to donate these lone pairs while a common transition ion fails to do so?

I decided to do some more research for this question as I had faced it before. Turns out the proper name of this ligand is diethylenetriaminepentaacetate ($\ce{DTPA^5-}$). It's an expanded version of EDTA and is primarily used in MRI scanning where it forms complexes with the Gadolinium ion. It forms complexes with a coordination number of six and seven with transition metals simply because these complexes are more stable. Here's Fe(III)-DTPA as an example1:

Experimentally Fe(III)-DTPA has been observed to adopt seven- and six-coordinate
crystal structures. This agrees with our result where the 8-coordinate Fe(III)-DTPA is $\pu{23 kJ mol^-1}$ higher in energy relative to the most stable 7-coordinate structure

With inner transition metals, it forms complexes with a denticity of 8, with Pr-DTPA being an example2:

References:

Sillanpää, Atte J., et al. “Density Functional Complexation Study of Metal Ions with (Amino) Polycarboxylic Acid Ligands.” Phys. Chem. Chem. Phys., vol. 5, no. 16, 2003, pp. 3382–93. doi:10.1039/B303234P.
Fusaro, Luca, et al. “Insight into the Dynamics of Lanthanide-DTPA Complexes As Revealed by Oxygen-17 NMR.” Inorganic Chemistry, vol. 51, no. 15, 2012, pp. 8455–61. doi:10.1021/ic3010085.

So it is based on factual data or is there a way to see the structure of ligand and then decide?

inner transition metals prefer to form complexes with high CN as they have open f-orbitals. this ligand has 8 sites which can donate electron pairs, and it will prefer to coordinate with all those sites when it can. That's how I initially went about solving this problem without knowing the fact.

Okay. Can't transition metals show CN 8 with this ligand?

They can, but those compounds are unstable, as I mentioned in my answer

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Why are all 3 of propane, butane and isobutane used in deodorant sprays?

Here are the back ingredient labels of a few popular commercial alcohol-based deodorant sprays.
As one can clearly see, they all have many things in common, but the 3 which are of importance in lieu of this question are-

Butane
Isobutane
Propane

After a quick search on the internet, one can find all 3 have the same common function i.e. acting as a propellant to give a perfect aerosol spray.
My questions lies, why are all 3 of them used together when they have the same function? There are also many deodorant sprays that use only butane as the propellant. Is it something related to pressure factors or its ability to be used in different conditions(hot/cold weather)?
If there's no such strong reason behind it, and if cost effectiveness isn't harmed either, then why not just use one of the 3? And if so, what would be the best of the 3 alternatives?
Image source: Google Images 1
2
3
4

A mixture can easily be cheaper than any of its ingredients.

Well that's just propane-butane, or LPG! Nothing mysterious, just a fraction of petroleum... It's like ask why is gasoline not one compound.

@Mithoron It's nothing about being mysterious ;) I just wanted to know if there's any special reason for using 3 different things that do the same function, and if we can do it by just one of them, cause there are such deos that use butane only. If one asks why gasoline isn't one, one could ask why water isn't one either?!

Bu/iBu or Bu/iBu/Pr mixtures are easier/cheaper to get instead of pure components as Ivan says, as there is extra effort for their separation, not the extra effort for their mixing.
Butane(Bu)/isobutane(iBu)/propane(Pr) boiling points are $\pu{0 ^\circ C}$, $\pu{-11  ^\circ C}$ , $\pu{-33  ^\circ C}$  respectively. Real boiling points would be higher than boiling points of pure hydrocarbon mixture, as deodorants are solution of active components in volatile liquid hydrocarbons, what increases the boiling point.
Propane (and isobutane too to a lesser extent) shifts usability toward lower temperatures, similar as Pr/Bu mixture of liquefied gas fuel cartridges for outdoor usage. Butane itself does not work well around $\pu{0 ^\circ C}$ either as propelent either as fuel, as it has not enough pressure to overcome the external one.
The goal is to provide the vapour pressure within the given pressure range for the given operational temperature range. Pure compounds have usually too low vapour pressure at the minimal temperature or too high vapour pressure at the maximal temperature.
Another reason is the control of evaporation rate of the solvent when it is being sprayed out. Different compositions of deodorant matrix may prefer different rates of hydrocarbon evaporation and different related cooling effect.

I agree that "there is extra effort for their separation, not extra effort for their mixing", but how much does it convert to cost when we're talking about pricing? I knew one butane-propelled deo, that costed the same. Yes I mentioned that "hot/cold weather"  usability thinking about gas canisters.

Price is not the only parameter.

Exactly my point! I'd argue for hydrocarbon reserves than ease of processing (esp. when it isn't much of a big deal)


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Choosing polar aprotic solvents for substitution reaction

It is known that common polar aprotic solvents include acetone and DMSO. Since they do not have hydrogen bonded to highly electronegative atom, they cannot only form dipole-dipole interaction with nucleophile, instead of hydrogen bond & ion-dipole interaction.

Moreover, the electrons in 2 R-CH3 groups in acetone / DMSO are repelled by electrons of the nucleophile, hence increasing the distance of dipole-dipole interactions, further decreasing the solvating effect.

My question is: from (2), is it better to use reagents that have more bulky R groups linking to the partial positive atom like pentan-3-one & diethylsulphoxide instead? I know this will decrease the polarity of solvent, but won't these better decrease the dipole-dipole interaction hence further increasing the nucleophilicity?

Your alternatives are not available in bulk as solvents. The differences in properties from those readily available are minor and do not justify sourcing them.

@Waylander Thanks! So the dominant effect is just about the inability of forming strong ion-dipole interactions to the nucleophile but not so related to electron repulsions, hence those are not common as solvents?

Any part of a reaction that can be done by a wide variety of compounds would be done in the cheapest way possible at the same time minimizing the byproducts formed due to using a given compound. So, it's more about the economic benefit here is what I gather from @Waylander's answer


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What is the exact statement of Drago's rule?
While studying the topic of chemical bonding, I came across a statement given by my teacher:
Elements of the third period and above do not show hybridisation. An example, in the case of $\ce{PH_3}$, $99.9$% s-character exists in the lone pair due to $\ce{P-H}$ bonds being weaker.
Is this the exact statement of Drago's rule? On various other sources, I have found conflicting views, some say that hybridization actually exists and the actual statement is different, the one being taught to students is wrong.
What is Drago's rule? Does it really exist?
I had gone through this answer, it basically explains a quantum mechanical reason for explaining the observation. What I'm seeking is the exact statement of Drago's rule.
For example, in $\ce{PH3}$ does the lone pair exist in pure s-orbital or in a hybrid orbital that is not $\mathrm{sp^3}$ but some other orbital whose nature can be explained and generalized by Drago's rule?

Unfortunately, the idea that the lone pair on PH3 is nearly a pure s orbital has long ago been debunked. Linus Pauling popularized the idea in the 1930's, but modern molecular orbital theory showed it to be false.

@Andrew Can you please shed some more light on that? The pure s-character of the lone pair was told to me by a professor on another forum and he had made this statement with quite a conviction.

@Aniruddha Deb I had gone through this answer, it basically explains a quantum mechanical reason for explaining the observation. What I'm seeking is the exact statement of Drago's rule. For eg. In PH3 does the lone pair exist in pure s-orbital or in a hybrid orbital that is not sp3 but some other orbital whose nature can be explained and generalized by Drago's rule?

Drago's rule is :For any Group in general,the formula for calculating the number of hybrid orbitals is group number of central element + no.of atoms surrounding the central element(except oxygen)-charge on the central element(consider the charge including its sign) whole divided by 2. I pulled it from here. It seems to be in only Indian Textbooks and is an extension(albeit weak) using the assumptions of Bent's rule

A weird coincidence? Search "Drago's rule made by" on google (forgive grammar).

@Andrew I was not aware of that actually. Could you please post an answer with more details?

@AniruddhaDeb There are many discussions of this in other questions on this forum about PH3 and SH2 bond angles as well as about Bent's rule. I don't think another answer is needed.

Here's a quotation from Albright et al. Orbital Interactions in Chemistry 2nd Edition (2013) p. 144. "Occasionally somewhat peculiar hybridization arguments have been constructed for molecules like H2S and H2Se which have rather acute bond angles. Since they are close to 90, p atomic orbitals are used to form the A—H bonds leaving the lone pair in an unhybridized s orbital on A. It is clear from the shape of 1a1 and 2a1 that this is far removed from reality."

Thx for all the help, just a small query...@Andrew...so to conclude, what percent s-character would actually be present in PH3 lone pair, if not pure s.

Any "rule" that involves d-orbital hybridisation should be deprecated. There is a reason why you can't find a statement of Dragon's rule: it is not rooted in a solid enough theory.

@Martin-マーチン "There is a reason why you can't find a statement of Dragon's rule" Of course, we won't find a rule based on a mythical creature ;)

@NilayGhosh But... But... But... maybe we should though. It'll be an interesting one, I'm sure. And then we possibly can use Dragon's rule to burninate Drago's rule.


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Why is polonium tetrafluoride non-existent?
I learnt that the group 16 elements can form halides, but with the exception that polonium cannot form a tetrahalide with fluorine.
However, I thought that it was possible. This is because the electronegativity difference between polonium and fluorine is very high since polonium is a metal and fluorine is a highly electronegative atom and so should be able to form a salt.
Why is this wrong?

Polonium in it's +4 oxidation state would mainly form covalent compounds as removing the first 4 electrons (sum of first 4 ionization energies) would be very high. Thus, if $\ce{PoF4}$ were to exist, it would have to be a covalent halide.

but isn't polonium a metal, also the electronegative difference between Po and Fl is very high, so ionic character has to be high right?

It is hard to characterize polonium compounds due of its radioactivity.

@Anir is right - compounds on higher ox. states are molecular in spite of significant ionic component in bonding.

Before going into details, let me state Wikipedia:

$\ce{^{208}PoF6}$ was probably successfully synthesised via the same
reaction (passing fluorine gas over $\ce{^{208}Po}$) in 1960 where a
volatile polonium fluoride was produced, but it was not fully
characterized before it underwent radiolysis and decomposed to
polonium tetrafluoride.

Even attempts were made to dissolve polonium metal in hydrofluoric acid but no fluoride salts were isolated as a free salt as they radiolysed and decomposed. According to this 1961 paper:

Although no fluoride salts of $\ce{Po}$ have been isolated, $\ce{Po}$
is quite soluble in $\ce{HF}$. Solutions containing 2  curies of
$\ce{Po}$ per ml of $\ce{1 N HF}$ are reported to be unsaturated.
Attempts to prepare a volatile fluoride by reaction with $\ce{F2}$ gas or
with liquid $\ce{BF3}$ have been unsuccessful. However recent work with $\pu{1
 mg}$ of $\ce{^{208}Po}$ indicated the formation of a volatile $\ce{Po}$ compound,
which decomposed before it could be isolated.

Polonium tetrafluoride is very difficult to isolate as it is seen to form either a complex or oxidize/decompose radiolytically. Also, bringing analytical data on such compounds is hard as they are produced in minute amounts possible in the range of micrograms. Below is a text from this e-book 1

A white solid possibly polonium tetrafluoride is obtained by treating
polonium hydroxide/tetrachloride with aqueous dilute hydrofluoric
acid. Treating this solid suspension in sulfur dioxide yields a bluish
grey product (possibly $\ce{PoF2}$) which quickly reverts back to the
solid on standing owing to radiolytic oxidation. The solubility of
$\ce{Po(IV)}$ increases with the concentration of hydrofluoric acid
indicating a complex formation
There has been unsuccessful attempts to make volatile hexafluoride by
passing fluorine to $\ce{^{210}Po}$ but recently a fluoride has been
made with $\ce{^{208}Po}$ plated on platinum. This product appers to
be quite stable in vapor phase but on cooling, a nonvolatile compound
is formed possibly polonium tetrafluoride resulting from radioactive
decomposition. Analytical data are not recorded for any polonium
fluorides owing to difficulty of determining fluoride ion at microgram
level.

Reference

Advances in Inorganic Chemistry and Radiochemistry, Academic Press, 1962
https://www.osti.gov/servlets/purl/4367751/4367751.PDF

Also note PoCl4 is known - https://en.wikipedia.org/wiki/Polonium_tetrachloride With intensely radioactive elements especially one needs to carefully separate "not known due to some chemical instability" with "not known because nobody has observed it yet"


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Stability of carbocations with -M groups in beta position
First part
The original question was to arrange the following compounds in the order of their rate of dehydration with conc. $\ce{H2SO4}$:-

(A) $\ce{CH3-CH(OH)-CH2-CH=O}$
(B) $\ce{CH3-CH(OH)-CH2-C(=O)-CH3}$
(C) $\ce{CH3-CH(OH)-CH2-C#N}$
(D) $\ce{CH3-CH(OH)-CH2-NO2}$

In each of the cases, a carbocation will be formed (+ve charge on the $2$nd carbon via E1 mechanism). Since all the groups(not talking about $\ce{-OH}$, because that’s the leaving group) are electron-withdrawing, the weaker the electron-withdrawing group, the higher the rate.
So, the order according to me should be B > A > C > D. But the answer given is the exact opposite.
"Edit- My teacher who gave me this question now told me that my answer is correct when the reaction takes place in acidic medium and reverses in basic medium."
Second part
Here is a similar question,

Arrange the following compounds in the order of their rate of dehydration with conc. $\ce{H2SO4}$:-
(A) $\ce{CH3-CH2-CH(OH)-NO2}$
(B) $\ce{CH3-CH(NO2)-CH2-OH}$
(C) $\ce{NO2-CH2-CH2-CH2-OH}$

Here, the answer given is C > B > A which I agree with because the further the $\ce{-OH}$ group is from the nitro group, the more stable will be the carbocation. But I feel it contradicts the first question.
Is my answer for the first question correct? If not, what is the correct reasoning?

"the weaker the leaving group, the higher the rate" Huh?

@Zhe I’m sorry. That was not what I was thinking, question-writing mistake. Edited the question.

@Aniruddha Deb I actually just read this new answer-https://chemistry.stackexchange.com/questions/136970/rate-of-dehydration-of-alcohols-having-keto-group  It states that "the rate of dehydration isn't guarded by carbocation stability, but by CH- acidity". I think I got my answer here but I still don't know why is it guarded by CH- acidity.

That question was talking about why alpha EWG carbocations are not stable. Your answer is in the first part is correct. Also where was this question taken from? which book?

@Safdar That answer did somewhat answer my question in first part when it said “guarded by CH- acidity”. From that I can say if there is a stronger -M group then the H on beta carbon will be more acidic. My professor actually gave me this question in class, so I don’t really know.

You cannot answer a question with an assumption taken from the other. Atleast not here.. There, the reason for a being the answer was because in the end the answer was between a and c w.r.t the end product but c could not be formed. Here it is different.

Ok, so B>A>C>D is the correct answer?

The RDS for dehydration is the formation of carbocation. However, if the rate of this step is similar in all the compounds, then the acidity of the -CH group would come into play... I'm still not 100% sure of your answer so hold out till someone posts a thorough answer.

This is not at all a trivial task and I wouldn't expect a purely qualitative approach to yield the actually correct result (in terms of observation). In any case, an answer might contain a lot of hand wavy arguments and unprovable qualitative statements. For all practical purposes, this should be measured, or thoroughly investigated with means of Quantum MD. Protonation and deprotonation reactions in a catalytic cycle are notoriously hard to calculate. The RDS approximation will break down.

These would be the arguments I propose under the assumption that the RDS approximation is valid within the scope of the question:
The order (B > A > C > D)  was said to be correct for the conc. $\ce{H2SO4}$/ $\Delta$ case since there would be no effect of $\ce{CH-}$ acidity since the protonation of the $\ce{-OH}$ group takes place and so the carbocation stability takes precedence meaning that the stronger the -I group, the slower the rate since the transition barrier increases.
However, for the scenario where only heat is used in dehydration (I assume basic conditions here), the reaction proceeds via E1cB  since the two main requirements for E1cB is that there is a relatively acidic hydrogen and a poor leaving group. Here, the RDS of the reaction would be the $\ce{C-H}$ bond cleavage. This would be stabilised by the presence of a strong -I group since a carbanion is being generated here which is stabilised by the delocalisation of the electrons via the inductive effect.
So, for such a scenario the rate order would be D > C > A > B.
The answers depend upon the medium as you can see.
This entire argument seems hand wavy with the assumption taken but otherwise as Martin pointed out, this would not be trivial and would require quantum modelling:

Protonation and deprotonation reactions in a catalytic cycle are notoriously hard to calculate. The RDS approximation will break down.


text:
Any open-source to animate 2D materials in 3D. (e.g. van der Waals heterostructures, 2D crystal layers etc.)
I want to model different 2D nano-structures, layered materials in 3D (animated version). Is there any open source where I can build these structures. Like this:

This would be better suited on Matter Modelling SE

I agree, there are probably more experts over there. I'm hesitant though, because I don't think it is off topic here. If you choose to ask your question there, please delete it here. (It might have already been asked, too, so please check that before asking.) Cc @Safdar / To the matter at hand: I don't understand the animated aspect. The picture looks static to me.

Does this answer your question? https://mattermodeling.stackexchange.com/questions/467/which-are-the-freely-available-crystal-structure-visualization-softwares/470#470 Also: https://mattermodeling.stackexchange.com/questions/1391/what-is-the-most-user-friendly-software-for-frequency-visualization-rendering/1392#1392


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Solubility of salts in water
I came across the formula for enthalpy of solution, which is equal to the sum of lattice enthalpy and hydration enthalpy. Therefore, for salts that have endothermic dissolution, the lattice enthalpy has to be greater than hydration enthalpy. But if that is true, then how will the salt dissolve in water in the first place?

Spontaneity is not just based on enthalpy. Read up on Gibbs Free Energy and entropy. Also this should help with the problem at hand. https://opentextbc.ca/chemistry/chapter/11-1-the-dissolution-process/

your saying that since entropy increases in hydration(due to salt dissociating), the reaction is spontaneous?

yes.but you would need some energy to initiate the reaction.

that energy comes from hydration enthalpy?

In case of icepacks, you shake it to mix the components. so you would initiate the process

Does this answer your question? Why can dissolution reactions (e.g. dissolution of calcium chloride) be exothermic?


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Rate of dehydration of alcohols having keto group
The following question was asked in IITJEE-Screening 2000,

Which of the following will be most readily dehydrated in acidic conditions?

I thought that the answer would simply be (c), as its intermediate would be stabilized by resonance with keto group and it is a sec-carbocation (which is common in all the options, so deosn't matter).
But my answer was wrong as per answer key. Am I missing something?

The correct answer as per key was,

 (a)

The carbocation is not stabilised in c since the carbon connected to it is supposed to show a negative character. Alpha carbon to keto. Both (a) and (c) give the same resonance stabilised alkene at the end. so answer should be a

@Safdar, what do you mean by "negative char"? charge?

Alpha carbon to keto is electronegative in nature, an example of that would be in aldol condensation.

All resonance is not stabilizing resonance.

The reason that a) dehydrates faster than c) and d) is that a) dehydrates through the more substituted enol of the keto group. Effectively, the enol is now an allylic alcohol which, under acid conditions, loses water.

A carbonyl group normally destabilizes carbocations but stabilizes carbanions.
In this particular case, the rate of dehydration isn't guarded by carbocation stability, but by CH- acidity. In acidic conditions, OH group in alcohols is relatively easily protonated. Then, if there is a relatively acidic hydrogen in $\beta$-position it immediately dissociates, together with a water molecule.
Excercise: Try to draw a mechanism for dehydration of (a) in basic conditions. Hint: it does occur and produces the same product. It does require more heating, though.

Why is it guarded by "CH- acidity"? Just because found in experiments?

@RobinSingh ... I can't answer this question in simple terms. At best, I can reccomend to find a good org.chem.book and read chapter on elimination reaction mechanism?

The other answer by permeakra is sufficient and complete. But, I'd like to add an explanation on how

"A carbonyl group normally destabilizes carbocations but stabilizes carbanions"

For instance, someone can think that the carbocation would resonate with the carbonyl group as what we've learnt earlier, I'd also thought in that way, initially. But, from the following resonance forms,

We can see there is a positive charge on oxygen, which is highly unstable resonance form. So, in this way it destabilizes the intermediate (here carbocation).
Now, one can think what happens when we have a carbanion ;)

@Safdar It is possible and does happen on occasion. Electronegativity of an atom in a molecule and electronegativity of an atom on Polling scale are completely different things. For example, what is a better electron acceptor - carbon or chlorine? Probably chlorine. But one of the strongest electron-withdrawing groups in organic chemistry is $-CF_3$ with electron accepted by carbon. Similarly, $Mn$ in $KMnO_4$ Is a very strong oxidizer, capable of oxidizing chlorine in $HCl$

@Safdar It is and it happens. Carbocation with this structure WILL have electron on an pi orbital with contribution of p orbitals of two carbons.  Vacant valence active atomic orbitals remain completely vacant only in exceptional or very specific circumstances, and this isn't the case.


text:
Lewis structures and chemical-compound formulas
In the Lewis structures listed below, M and X represent various elements in the third period of the periodic table. Write the formula of each compound using the chemical symbols of each element:
a.
b.
c.
d.
What are the easiest ways to answer these questions?
I am working on these questions,meanwhile if any member of CSE knows the answers to these questions may answer these questions.
Solution:
Now, in the third period, following elements are included. Na, Mg, Al, Si, P,S, Cl, Ar.
For question (a),  NaCl is the answer$\big[Na\big]\big[:\ddot{Cl}:\big]^{2-}$. Note:-Cl has lone pair of $e^{-}$ below it.
In all the four questions X has 8 valence electrons. I want to know the meaning of $\pm$ charge inside the bracket containing M and $\pm$ charge as a superscript of the bracket containing X.
I don't know how to draw lewis strucures of element, compounds using LaTeX.
If any member knows it may reply to me.

No. Absolutely not. For (a), the idea is that you should combine an ion with charge +2 and an ion with charge -2. You should consult your notes on which elements are likely to give those ions. And for (b) for example, you are looking for an ion at +3 and an ion at -1 (of which there are 3 to balance the charge). And so forth...

I would recommend to insert the upper index ($1$- or $2$-) inside the brackets. In this case, a) is $\ce{MgO}$, and d) is $\ce{Al_2S_3}$. Try to find yourself b and c !

@Maurice, How to draw Lewis dot structures on Chemistry stack exchange? M and X represents various elements from the third period of periodic table.

@Maurice, O belongs to the second period of periodic table.

OK. Replace the oxygen atom by the sulfur atom in MgO that becomes MgS.

The elements in the third period of the periodic table has following electron configurations and they achieve their most stable ionic forms by gaining or losing indicated electrons (final electron configurations of each is also shown):

$\ce{Na}$: $\mathrm{[Ne]3s^1} \quad \ce{ ->[-e^-] Na+}: \mathrm{[Ne]}$
$\ce{Mg}$: $\mathrm{[Ne]3s^2} \quad \ce{ ->[-2e^-] Mg^2+}: \mathrm{[Ne]}$
$\ce{Al}$: $\mathrm{[Ne]3s^23p^1} \quad \ce{ ->[-3e^-] Al^3+}: \mathrm{[Ne]}$
$\ce{Si}$: $\mathrm{[Ne]3s^23p^2}$
$\ce{P}$: $\mathrm{[Ne]3s^23p^3} \quad \ce{ ->[+3e^-] P^3-}: \mathrm{[Ne]3s^23p^6}\ce{#[Ar]}$
$\ce{S}$: $\mathrm{[Ne]3s^23p^4} \quad \ce{ ->[+2e^-] S^2-}: \mathrm{[Ne]3s^23p^6}\ce{#[Ar]}$
$\ce{Cl}$: $\mathrm{[Ne]3s^23p^5} \quad \ce{ ->[+e^-] Cl-}: \mathrm{[Ne]3s^23p^6}\ce{#[Ar]}$
$\ce{Ar}$: $\mathrm{[Ne]3s^23p^6}\ce{#[Ar]}$

Briefly, the first element of 3rd period, $\ce{Na}$ can loose one electron to gain neon electron configuration (octet) than gaining 7 electrons to achieve argon electron configuration as shown in entry (1) above. Consequently, loosing one electron make it $\ce{Na+}$. Similarly, the seventh element of 3rd period, $\ce{Cl}$ can gain one electron to achieve argon electron configuration (also an octet) than loosing 7 electrons to achieve neon electron configuration as shown in entry (7) above. As a result of gaining one electron, $\ce{Cl}$ becomes $\ce{Cl-}$.
For $\ce{Si}$, it can gain 4 electrons or loose 4 electrons. However, either is very difficult to achieve so that its nature (along with that of carbon) is explained well using molecular orbital theory.
Your list contains compounds with three cations: $\ce{M+, M^2+,}$ and $\ce{M^3+}$. Those are $\ce{Na+, Mg^2+,}$ and $\ce{Al^3+}$, respectively. Even though it wasn't shown, all of these have octets in the last cell as explained above.
Also, your list contains compounds with only two anions: $\ce{X-}$ and $\ce{X^2-}$. Those are $\ce{Cl-}$ and $\ce{S^2-}$, respectively. Those also have octets in the last cell as explained above.
Since you know the cations and anions in the list, it's now easy to make the compounds.

So answer to the a, b, c, d questions are MgS(Magnesium Sulfide), $AlCl_3$Aluminum Chloride, $Na_2S$ Sodium Sulfide,$Al_2S_3$Aluminum Sulfide respectively.

Your answers are correct.


text:
What is the hybridisation in HCN molecule?
According to VSEPR theory hybridisation of the central atom should be sp. However, my teacher says that there exists a near-100% s-character in the carbon orbital that bonds with hydrogen due to some other reasons. I couldn't get why that would be the case. One can think of it as the nitrogen that polarizes the carbon atom, but shouldn't it lower the s-character of the C-H bond? What am I missing?
Is it based on quantum calculations (which are beyond my scope) , as a commentator points out?

Ok, no need for apologies. Now, s- and p-character is used to refer to orbitals, not atoms, and certainly not molecules; regardless of how the orbitals are hybridised, carbon always has one s and three p orbitals. The orbitals can have different amounts of s- and p-character.

I have rectified some inconsistencies and errors in my question, please check them out, sorry and thanks.

@orthocresol can you please shed some light on this

I'm sorry, but I'm not qualified to. I will say that I am slightly skeptical of this claim of your teacher (firstly, it is completely unsubstantiated, and secondly, why should there be a difference in the H–C bonding in H–C≡N and H–C≡CH? Nobody would dare to say the latter is not sp hybridised), but I dare not say for a fact it is wrong without some evidence, and I don't really have the time to do quantum chemical calculations now.

chem.libretexts.org says it is sp hybridized. But that's in acc. to VSEPR theory. If you're in high school, that should be enough (but not complete... I think, after @orthocresol 's comment).

Firstly, the thing you say about N polarizes carbon..you certainly mean N is more electronegative...If that is the case, the bonds along C--->N will have more p character, and to compensate on s character, C-->H ,s' character increases. Performing rudimentary calculations on Computational software, it is indeed the fact that s character increased at C-H bond...BUT, the claim about near 100% is wrong...(my calculations show near 55%s) which is nearly the same old sp hybrid.

Idk but we could bring in some concepts used in molecular orbital (molecular orbitals are like the s,p,d,f orbitals for molecules) theory. The truth is, CN- is a very interesting molecule and the hybridisation idea has a lot of limitations. If we really observe the molecular orbitals of CN-, the highest energy orbital is a p sigma orbital which would be the orbital that forms the H-C bond.

$\ce{HCN}$ and $\ce{HC#CH}$ are linear, triple bonded, with a $π$ system consisting of two perpendicular $π$ bonds. They would be symmetrical in $\ce{HC#CH}$, and slightly distorted in $\ce{HCN}$, and they leave two orbitals for the sigma system.
In $\ce{HCN}$, we hybridize/combine the two remaining orbitals on the carbon atom to form two bonding orbitals, one to the hydrogen, another to the atom on the other side of the carbon (a $\ce{C}$ or an $\ce{N}$). The natural first combination is a $50$–$50$ split to form two $\mathrm{sp}$ orbitals, one directed to the $\ce{H}$ and the other directed toward the $\ce{N}$.
That's enough most of the time, but if you get picky, you could point out that the electronegativities of $\ce{H}$ and $\ce{N}$ are quite different ($\ce{H}$ $2.1$, $\ce{C}$ $2.5$, $\ce{N}$ $3.0$), so the nitrogen will be pulling on its $\mathrm{sp}$ bond more than hydrogen pulls on its $\ce{sp}$ bond, so the $50$-$50$ split readjusts to maybe $70$-$30$ (did your teacher say $~100\%$-$0\%$?), meaning that the hydrogen gets less of the carbon orbital (i.e., less $\mathrm p$-character, more $\mathrm s$-character from the carbon orbitals), while the nitrogen gets more of carbon's $\mathrm p$-orbital.
If this is so, the $\ce{H}$ atom in $\ce{HCN}$ should be more easily removed than a H atom in $\ce{HC#CH}$. This appears to be correct: the $pK_\mathrm a$ of acetylene is $24$ (which is considered to be quite acidic for a hydrocarbon), whereas the $pK_\mathrm a$ of hydrocyanic acid is $9.21$, much more acidic (although a weak acid by any other measure).
Your teacher pointed out something interesting, but the exact split ratio remains to be calculated.


text:
Crystal Field Theory
When I learned about crystal field theory I had fallen on a paradox. When K4[Fe(CN)6] is formed then three d orbitals are occupied by electrons of Fe2+ pairly and the rest two d orbitals are occupied by electrons of lone pair electron of CN-. So when visible light falls on it, the d orbitals are splitted into two sets which are t2g and eg and the electrons of eg orbitals jump into t2g orbitals by absorbing a portion of this light. But my question is that how the electrons of eg orbitals jump into t2g orbitals when the d orbitas are already occupied by the electrons of Fe2+ and ligand during bonding.

They are jumping into an antibonding orbital.

In crystal field theory, the electrons of the ligands are not distributed among the Fe central atom. They stay outside. So the electrons from CN are not on d orbitals. The six d electrons of the ion $\ce{Fe^{2+}}$ are all paired on the three t2g levels, which are lower than the two eg levels. This splitting happens without any light. When light is adsorbed, one of these six electrons is excited and goes to one of the eg levels, which are unoccupied. Where is the problem ?


text:
Reason for no change in electrode potential in reference electrodes
Reference electrodes such as the saturated calomel electrode and silver chloride electrode have a constant potential because the $\ce{Cl-}$ concentration (or, to be more precise, the activity) is constant.
The semi-reaction for those electrodes are:
$$\ce{Hg2Cl2 + 2e- <=> 2Hg + 2Cl-}$$
$$\ce{AgCl + e- <=> Ag + Cl-}$$
Now, since chloride ions are produced (or consumed, when the electrode is an anode) from the half-reactions, how is it possible that their concentration remains constant?
Is the quantity of chloride produced or consumed is negligible compared to that coming from the $\ce{KCl}$ dissociation?

Current flow through a reference electrode is supposed to be negligible, i.e., pA or less. The chloride ion concentration is significantly high or there is excess solid (containing the chloride ions). So there is negligible change in chloride ion concentration when the reference electrode is used as designed.

Your last sentence answered your own question. If you want a simple example, assume a current of 1 pA flowing for 10000 seconds. Multiply those, divide by the Faraday constant, and the resulting tiny numerical value shows that chloride ion concentration change is negligible in any reasonable reference cell solution volume.


text:
Stability of free radicals
If given a question to compare stability of free radicals, can we always check it by same parameters that we use for stability of cations (since both are electron deficient) or is there any exception to this? By parameters I mean hyperconjugation, mesomeric, resonance, inductive effects. Is there any other effect that can be added to this list?

Stereoelectronic effects. Not to say your questions here are kinda "meta-homework" - this site is "chemistry" not "how homework is done".

The thing is I am still a High school student so yes  I just deal with the basics of chemistry, and its application and problem solving (and I am not into research). But if I am showing asking a question and have shown certain effort to solve it, then what's wrong in asking it? or if am learning some new concept by asking , I think it would be harmless, since probably some questions may help future readers. Also, this site has a lot of knowledgeable people(far better than me) , and it's great learning from them since I don't have many other sources in this lockdown(covid-19).

But still if there's some better way to ask, or community guidelines that I might had not abide by, do enlighten. :)

This is rather broad. I don't think it's easily answerable; my suggestion would be to look in a book. Most organic chemistry books will have sections dedicated to radicals. The short answer is partially yes, but radicals are also stabilised by electron-withdrawing groups, and if you have both electron-donating and electron-withdrawing groups, that's potentially even better.

Well in general, I would appreciate if people tried to make their questions less homworky - yes, most of HW could be transformed into more general and useful form.

Look at nitric oxide, a stable free radical at room STP. https://en.wikipedia.org See also https://en.wikipedia.org/wiki/Radical_%28chemistry%29#Stability_and_formation /wiki/Nitric_oxide


text:
Why does an mercury ore exist as HgS?
Cinnabar($\ce{HgS}$)  is one of the most common ores of mercury.
Why does it not exist at $\ce{HgO}$ or some other such compound?
Is there any reason for why it is so prevalent?
Does it react less with other compounds, or maybe due to some solid structure of HgS having some characteristic property giving it more stability?
I'm seeking an explanation without the mention of free energy, because it doesn't properly explain the phenomenon for me. I understand that cinnabar is abundant in the crust, and my question is why is $\ce{HgS}$ so common?

Because most of the mercury in the world is found in cinnabar.

I think my question does not convey my query, I'll edit it...

Because mercury has quite an affinity towards sulfur So, most of the mercury exist in its sulfide form.

This is kinda getting outside of scope of the site. For example HgSe could be even more "preferable", but there's very little Se in the world. Why? Probable because nucleosynthesis stuff, which I doubt s particularly on topic here.

One may also state that mercury sulfide is probably the most insoluble compound of all. Its solubility product is $\ce{10^{-52}}$. This corresponds to much less than $1$ atom per liter in solution.

@Maurice mercury sulfide is not the most insoluble compound ;-)

Why $\ce{HgS}$ and not $\ce{HgO}$?

Because of HSAB. F'x has already answered a previous question of why mercury has propensity towards thiols(or in general sulfur). Quoting the relevant information:

In Pearson's HSAB theory (hard and soft acids and bases), the reason
the $\ce{S-Hg}$ bond is be stronger than the $\ce{O-Hg}$ can be
explained because $\ce{S^2-}$ is a soft basis and $\ce{Hg(II)}$ is a
soft acid, making a good fit while $\ce{O^2-}$ is harder.

If this answer helps, please make sure to upvote F'xs answer.

I am afraid that no one knows why! The superficial rationalizations such as HSAB and ionic size etc are good for decorating general chemistry textbooks.
As scientists, we should accept the fact that all questions which a human mind can generate do not have answers. Unfortunately, as science students, we are trained that every question must have an answer at the back of the book.
There is a whole field of geology, called ore genesis which tries to think of the origin of minerals and ores Ore Genesis on Wikipedia. It is a remarkable wonder and miracle of Nature, as how it deposits ore in certain chemical forms.
More specifically, you would search the geochemistry literature, you would see how complex is the whole concept e.g. if you go through Quicksilver Deposits of Chile Quicksilver deposits you can get a feel.

+1 for the courage to simply state "...no one knows why!"

-1 because it isn't true that we have no idea why.

Mr. Matt black, illuminate us with your knowledge and confidence please. Where is your answer if you know the reason?


text:
Mechanism of three-membered lactone hydrolysis in base
I was reading up on the participation of neighboring functional groups and in the reaction below, I was wondering why hydroxide attacks the C–O bond instead of the carbonyl group of the lactone. Lactones tend to be more electrophilic compared to acyclic esters, as the lone pairs on the ether oxygen are not positioned correctly to participate in conjugation, so I would expect hydroxide to attack the carbonyl. Does it have to do with the angle of attack, given the small ring?

Apparently Mother Nature has decided otherwise.

"lone pairs on the ether oxygen are not positioned correctly to participate in conjugation" This statement is not true. Usually, the reactivity is ascribed to donation of the in-plane lone pair into the σ* orbital (or lack thereof for the lactone).

One thing that you cannot rule out here based on the stereochemical outcome is that the hydroxide does attack the carbonyl, but it is the preferred leaving group for stereoelectronic reasons, despite the ring strain. It would be an interesting exercise to devise a mechanistic experiment to test this claim.

@Zhe: The use of O18 hydroxide may leave O18 in the carboxylate of lactic acid if there is any addition to the carbonyl of the alpha lactone.

According to Organic Chemistry: Second Edition,[1, pg 934](from which I assume that the question was taken):

Enantiomerically pure (R)-2-bromopropanoic acid reacts with concentrated sodium hydroxide to give (S)-lactic acid. The reaction goes with inversion and is a typical SN2 reaction—and a good one too, since the reaction centre is adjacent to a carbonyl group. If, on the other hand, the reaction is run using $\ce{Ag2O}$ and a low concentration of sodium hydroxide, (R)-lactic acid is obtained — there is overall retention of stereochemistry.

Therefore this reaction that has been questioned takes place at a low concentration of $\ce{OH-}$ in the presence of $\ce{Ag2O}$.
How do we decide between competing mechanisms?
The next paragraph clearly outlines what we need to look for in order to proceed.

In the first, we have conditions suited to an SN2 reaction: a very good nucleophile ($\ce{OH-}$) and a good leaving group ($\ce{Br-}$). Improve the leaving group by adding $\ce{Ag+}$ ($\ce{Ag+}$ assists $\ce{Br-}$'s departure much as $\ce{H+}$ assists the departure of $\ce{OH-}$ by allowing it to leave as $\ce{H2O}$) and worsen the nucleophile ($\ce{H2O}$ instead of $\ce{OH-}$, of which there is now only a low concentration), and we have the sorts of conditions that would favour an SN1 reaction. The trouble is, without neighbouring group participation, the cation here would be rather unstable — right next to a carbonyl group. The carboxylate saves the day by participating in the departure of the $\ce{Br-}$ and forming the lactone. The key thing to remember is that a reaction always goes by the mechanism with the fastest rate.

From what I have understood the question lies in the step after the lactone is formed. What reacts, $\ce{C-1}$ or $\ce{C-2}$? Let's assume that both are possible and proceed with the two possible cases.
Firstly, the $\ce{OH-}$ attacks the carboxylic carbon($\ce{C-1}$). We get the possible reaction mechanism to be:

In the second case, let's assume that $\ce{OH-}$ attacks $\ce{C-2}$. The possible reaction mechanism would be:

So, now we are faced with the choice of which reaction could take place, attack on
$\ce{C-1}$ or attack on $\ce{C-2}$.
From the second quotation, we can say that if we are able to decide which is faster, that would be the reaction that takes place and so we reach our next question.
Which reaction is faster?
In pathway 1, we are talking about the base hydrolysis of esters. However, esters are not very reactive and the concentration of base in this case is low as well.
In pathway 2, we are talking about an SN2 reaction taking place on the $\ce{C-2}$ carbon which removes the ring strain that is seen in the lactone. Therefore, this would be faster than base hydrolysis that takes place at $\ce{C-1}$
Another way of checking the speed of the reaction would be to use the principle of least motion, (as pointed out by Yusuf Hassan).
According to Advances in Physical Organic Chemistry, the
principle of least motion is defined as:

The principle of least motion, states that those elementary reactions will be favored that involve the least change in atomic position and electronic configuration.

Using this, we can compare the relative rates of the two reactions.
During the base hydrolysis of esters, the hybridisation of the $\ce{C-1}$ carbon changes from $\mathrm{sp^2}$ to $\ce{sp^3}$. This leads to a big change in the atomic arrangements since you go from a trigonal planar configuration to a tetrahedral arrangement.
In case of the SN2 reaction on the $\ce{C-2}$ carbon, the configuration as a whole doesn't change. It remains $\mathrm{sp^3}$ throughout the reaction, so there is less rearrangement taking place.
This implies that according to the principle of least motion, the SN2 reaction is favoured over the the basic hydrolysis of the lactone formed via the anchimeric effect.
Hence, we can say that the reaction that would take place would be the attack on $\ce{C-2}$ and not the basic hydrolysis of the lactone.
Reference:

Organic Chemistry: Second Edition; Jonathan Clayden, Nick Greeves, Stuart Warren, 2014; ISBN 978-0198728719


text:
What is happening at the various boundaries of a phase diagram?
Hoping to clarify a few questions I have that the textbook doesn't address.
Lets just say I have a piston (at equilibrium) with 100 moles of gaseous compound that looks like this:

And here is the phase diagram for this compound:
Now lets say I start increasing the pressure on the piston so that the gas starts to approach the gas-liquid boundary.
1.) What would start to happen as you approach the boundary? For example: Would a few moles of gas begin to condense to liquid? Is this process exponential as you approach boundary?
2.) What is happening at the boundary? For example: Are the gas and liquid in perfect equilibrium (50 moles of each)? By that logic, at the triple point, are there 33 moles of each?
3.) Now let's say we approach the critical point along the liquid-vapor boundary. I read online that its when the gas and liquid become "indistinguishable." If so, that means it has a uniform density. Does that make it another state of matter? If not, why is it special?

Why the downvote? Let me know what I can do to clarify!

(1) No, the process is not exponential, but abrupt. (2) Yes they are in perfect equilibrium, and that's not what you think it is. The number of moles is irrelevant and can be anything. (3) Call it whatever you like.

Thanks for reply! Can you specifically describe what would be happening inside the piston at the boundaries then?

The gas will start condensing into a liquid. As you move the piston further, more gas will condense, until there is none of it left. Then you will be just compressing the liquid.

Why is saying that half the molecules would be in one phase (and the other half in the other) erroneous? I'm imaging a layer of liquid with a layer of gas on top.

That's not a matter of clarification - your thinking there is plain weird for a thing that's practically reverse of what you do all the time when boing  water! Another thing is the supercritical part is a different question that shouldn't be in the same post and was already covered in earlier posts.

Sure, at some point half the molecules would be in one phase and the other half in the other. Before that, it would be 0.9 and 0.1, and after that, it would be 0.1 and 0.9. The pressure in all these situations is the same. They all represent the same point on the phase diagram. So it goes.

Second question: heating liquid chlorine at constant volume (pressure and temperature increase at the same time) and then cooling it again: https://www.youtube.com/watch?v=Y6S7bZbx4-s

I got that you are trying to visualize what is actually happening. I have not done the compression myself, but this is what I think will be observed. It will be good if someone with real experience can confirm the answer.

At the beginning you are in the region marked 'vapour'. When you are approaching the liquid-vapour boundary from below, before you touch the boundary, you will still observe only vapour, no liquid at all. You are still in the region marked 'vapour'. Nothing will happen.

When you just reached the boundary, you will see some liquid droplets formed. It should look just like some dew formed at the wall. Regarding the relative amount of liquid and vapour at equilibrium, the pressure of vapour present must be equal to pressure exerted by you (or else the piston will move!) and equal to pressure on the boundary (determined by temperature). Say if you try to push the piston further inside, the volume occupied by sample will decrease, and more liquid formed, but you will still stay on the boundary. You will notice that the pressure needed to push piston inside further does not change, because the pressure of vapour does not change (push in, vapour pressure should increase, but some vapour condensed, lower back the pressure). Only when all vapour has been converted into liquid, you can depart from the boundary, move into the region marked 'liquid'. However, before all vapour turned into liquid, if you stop pushing in and hold your piston still, the liquid and vapour amount won't change. So there is no unique relative amount of vapour and liquid at equilibrium as written in your question. Above liquid-vapour boundary you have only liquid and the piston surface is resting on the liquid surface.

That state is known as supercritical fluid.

Interesting, so at the boundary, if I was to lock the piston in place. The relative ratio of vapour and liquid would become volume dependent?

Yes, that conforms to Le Chatelier principle.


text:
Is there an example of a bond with negative bond enthalpy?
My textbook says bond breakage is "generally endothermic." This implies there are some bonds out there that release energy when broken.
Is there a classic example? Perhaps diatomic Helium?

Does this answer your question? https://chemistry.stackexchange.com/questions/63908/why-is-bond-breaking-endothermic

This is actually a tricky thing there - while breaking of bond itself is always endothermic, its results can sometimes generate energy! https://chemistry.stackexchange.com/a/112114/9961

So I wager your book is talking about metastable molecules, which break down spontaneously with release of energy and He2 excimer may qualify.

Bond enthalpy is like a measure that tells us how strong/stable a particular bond is.
Definition:

Bond enthalpy (also known as bond energy) is defined as the amount of energy required to break one mole of the stated bond.

To break a bond, energy must be provided to the bond. How is that energy provided?

The reactants are either heated or they take up energy from the surroundings, that is, they take in energy in the form of heat from the surroundings, overcome their bond enthalpies and break the bond.
All bonds require energy to be broken.
A negative Bond enthalpy simply means that the bond is highly unstable and would not exist in the first place.
I highly doubt if any compound has negative bond enthalpy.{Not even Dimeric Helium}
From Wikipedia:

Based on molecular orbital theory, $\ce{He2}$ should not exist, and a chemical bond cannot form between the atoms. However, the van der Waals force exists between helium atoms as shown by the existence of liquid helium, and at a certain range of distances between atoms the attraction exceeds the repulsion. So a molecule composed of two helium atoms bound by the van der Waals force can exist.The existence of this molecule was proposed as early as 1930.


text:
How is the formula of mean activity coefficient derived?
The mean activity coefficient is defiend as follows:
$$\gamma_\pm = (\gamma_+\gamma_-)^{1/2}.\tag{1}$$
If Debye-Hückel equation
$$-\log\gamma_i = 0.5z_i^2\mu^{1/2}\tag{2}$$
is used, then the mean activity coefficient has the form
$$-\log\gamma_\pm = 0.5|z_+z_-|\mu^{1/2},\label{eqn:3}\tag{3}$$
where $\mu$ is the ionic strength, $z_i$ is the charge of species $i,$ $\gamma_i$ is the activity coefficient.
But I'm confused about the equation \eqref{eqn:3}. It can be found as equation \eqref{eqn:3-38} in Snoeyink and Jenkins' Water Chemistry [1]:

[…] developed for ionic strengths of less than approximately $\pu{5E-3}$ and can be stated as
$$-\log\gamma_i = 0.5Z_i^2\mu^{1/2}.\tag{3-34}$$
Because anions cannot be added to a solution without an equivalent number of cations (and vice versa). it is impossible to determine experimentally the activity coefficient of a single ion. Therefore, Eqs. 3-34, 3-35, and 3-36 cannot be verified directly. However, it is possible to define, and measure experimentally, a mean activity coefficient, $\gamma_\pm,$ as,
$$\gamma_\pm = (\gamma_+\gamma_-)^{1/2}.\tag{3-37}$$
The Debye-Hückel and Güntelberg relationships can be extended to the mean activity coefficient thus:
$$-\log\gamma_\pm = 0.5|Z_+Z_-|\mu^{1/2},\label{eqn:3-38}\tag{3-38}$$

Below is my derivation process:
$$
\begin{align}
-\log\gamma_+ &= 0.5z_+^2\mu^{1/2};\tag{4.1}\\
-\log\gamma_- &= 0.5z_-^2\mu^{1/2},\tag{4.2}
\end{align}
$$
so
$$
\begin{align}
-\log\gamma_\pm &= -\log[10^{-0.5z_+^2\mu^{1/2}}\times 10^{-0.5z_-^2\mu^{1/2}}]^{1/2} \tag{5.1}\\
  & = -\log[10^{-0.5\mu^{1/2}(z_+^2+z_-^2)}]^{1/2} \tag{5.2}\\
  & = 0.25\mu^{1/2}(z_+^2+z_-^2). \tag{5.3}
\end{align}
$$
This result is different from $0.5|z_+z_-|\mu^{1/2}.$ Can someone tell me which step is wrong in my process?
Reference

Snoeyink, V. L.; Jenkins, D. Water Chemistry; Wiley: New York, 1980. ISBN 978-0-471-05196-1.

The definition of $\gamma_{\pm} $ is the geometric mean $\sqrt{\gamma_+\gamma_-}$, you have used the sum so obtain a different answer.

@porphyrin I thought that I used the geometric mean since $\gamma_+\gamma_- = 10^{-0.5z_+^2\mu^{1/2}}*10^{-0.5z_-^2\mu^{1/2}}$. Then I put them into the $\sqrt{\gamma_+\gamma_-}$ and further into $-\log\gamma_\pm$. But the result is different.

Ah, I misread, my mistake.

This formula is used for Monovalent salts where the ratio of cations to anions is 1:1. In such a scenario, $z_+$ = $z_-$ so your formula and their formula are same.

The problem is you were starting with an expression for the mean activity coefficient for salts where both the cation and anion were monovalent, and then attempting to derive, from this, a general expression for salts of all empirical formulas. To correct this, you need to start with the general expression for the mean activity coefficient of salts of any emprical formula.
If the empirical formula of the salt is of the form $\ce{A_pB_q}$, where A is the cation and B is the anion, then the general formula for the mean activity coefficient is:
$$\gamma_{\pm} = \sqrt[^{p+q}]{\gamma^p_+\gamma^q_-}$$
[Adapted from: https://en.wikipedia.org/wiki/Activity_coefficient, where I've taken the Wikipedia expression and substituted $\gamma_+$ for $\gamma_A$, and $\gamma_-$ for $\gamma_B$, to correspond to the nomenclature you are using.]
Charge balance dictates that  $|z_+| = q$ and $|z_-| = p$. [E.g., in $\ce{A_2B_3}$, the charge on A must be 3+, and that on B must be 2- (or some integer multiple of those).]
Substituting, we have:
$$\gamma_{\pm} = \sqrt[^{|z_-|+|z_+|}]{\gamma^{|z_-|}_+\gamma^{|z_+|}_-}$$
If you use that in place of eqn. (3-37) in your post, and work through your calculations again, you will get the result show in eqn. (3-38), which is a general result for any values of p and q.

Thanks! I completed the calculations here: $\gamma_{\pm} = \sqrt[^{|z_-|+|z_+|}]{\gamma^{|z_-|}+\gamma^{|z+|}-} = \sqrt[^{|z-|+|z_+|}]{10^{-0.5\mu^{1/2}(z^2_+|z_-|+z^2_-|z_+|)}} = 10^{-0.5\mu^{1/2}(|z_+||z_-|(|z_-|+|z_+|))/(|z_-|+|z_+|)} = 10^{-0.5\mu^{1/2}(|z_+||z_-|)}$ Then take the logarithm of $\gamma_{\pm}$ and multiplied by -1, it becomes eqn. (3-38).

@TX Happy to help! I left the calculation for you to complete because (a) I figured you'd have no trouble completing it, since your original calculations were correct based on what you'd started with (you just started with the wrong expression), and (b) it would allow you to see for yourself that the calculation does give the right answer.


text:
How can I synthesize phenylpiracetam from styrene or benzaldehyde?
I am trying to design a proper synthesis route for phenylpiracetam.

I am thinking a good starting material would be styrene or benzaldehyde. Here is my proposed route.

Is there a better method? Please identify any mistakes I may have made in my synthesis and suggest improvements, even if you have a different synthesis route in mind.

This is a reasonably sound route. I have some observations:
The prep of nitrostyrene from benzaldehyde is well known, it is best to use freshly distilled benzaldehyde. Here are the conditions from org syn prep.
The sidechain can be added by treatment of the phenylpyrrolidine with $\ce{NaOMe/MeOH}$/Methyl chloroacetate then heating with $\ce{NH3/MeOH}$
(patent).
You are better off reducing the nitro group at the diester stage as it will spontaneously cyclise to the pyrrolidine. On hydrolysis it will then decarboxylate. In my experience, hydrogenation of alkyl nitro compounds does not give good yields. $\ce{Al/Hg}$ amalgam is better. Sodium hypophosphite /phosphinic acid with cat. 5% Pd/C in water/THF and ultrasound was recently reported here as an efficient method for reduction of alkyl nitro, and the paper contains an example of spontaneous cyclisation of the product amine onto a methyl ester.
A synthesis of phenylpiracetam is described here in this patent.

May I add the Al-Hg "as is", or should water and/or a solvent such as THF be present alongside Al-Hg before the amalgam is reacted with the nitro compound (after the Michael addition of dimethyl malonate)?

Additionally, after the cyclization, how would the H₃O⁺/Δ step only affect the ester but not the amide in the ring? If this would necessitate a lengthy answer on your end, I can start a new thread with this question. Also, isn't an application of heat (Δ) or water required for the spontaneous cyclization to occur? I did some research and apparently esters can react with amines in, e.g., water, heat, and NH4Cl/Δ (though I am not sure what the side products would be for the third example).

Al/Hg needs a protic solvent, EtOH is good. Cyclic amides are difficult to hydrolyse under acid conditions so I think you will achieve adequate selectivity over the methyl ester. Intramolecular cyclisations to form amides go readily, heating the reaction mixture after the Al/Hg reduction will make sure.

Would using Zn, Sn, or Fe with an acid (e.g., HCl), instead of Al/Hg amalgam, be a valid option too?

Zn/acid reduction of aliphatic nitro gives the hydroxylamine. I have not seen Sn or Fe used effectively. Pyridine zinc borohydride is reported as a good method, but I have not used it. https://onlinelibrary.wiley.com/doi/abs/10.1002/jccs.200300040. Sodium hypophosphite/phosphinic acid and 5%Pd/C in water/THF looks like a good method https://pubs.rsc.org/en/content/articlelanding/2013/gc/c3gc37024k#!divAbstract


text:
Comparing equilibrium constants between acetone and benzaldehyde in nucleophilic addition reaction
My teacher told me that the $K_\mathrm{eq}$ of acetone is greater than that of benzaldehyde in nucleophilic addition reaction.
But I am not able to understand how is this possible. Acetone has a +I group attached to it, so it will decrease the reactivity of carbonyl. If we talk about benzaldehyde, it has a −M group. Can someone tell what is the correct reason behind it?

The information given by your teacher is incorrect. The $K_\mathrm{eq}$ value of acetone for cyanohydrin formation (a nucleophilic addition reaction) is less than the $K_\mathrm{eq}$ shown by benzaldehyde
According to Organic Chemistry: Structure, Mechanism, Synthesis [1 , p. 595], the values of $K_\mathrm{eq}$ for cyanohydrin formation are as follows:
$$
\begin{array}{lr} \hline
\mathrm{Compound} & K_\mathrm{eq} \\ \hline
\text{Acetone} & 20 \\
\text{Ethanal} & 10^5 \\
\text{Benzaldehyde} & 210 \\
\textit{p}\text{-Methoxybenzaldehyde} & 30 \\
\text{Acetophenone} & 0.8 \\ \hline
\end{array}
$$
Hence the statement that the $K_\mathrm{eq}$ is greater for acetone is incorrect. This seems accurate considering that acetone is more sterically hindered than benzaldehyde, which is a dominating factor for nucleophilic addition reaction since they are kinetically stable reactions.
This document has a much larger cases of compounds that are compared for multiple reactions, in all of which benzaldehyde is more reactive than acetone.
However, there is one minor issue with your reasoning given in the question.

Benzaldehyde does not have a −M group in it. However if you add another functional group into the ring, then we say that the $\ce{-CHO}$ group acts as a −M group. Actually, $\ce{-CHO}$ has a +R effect from the phenyl ring attached to it which makes the ring in benzaldehyde deactivated. So, the carbon in benzaldehyde is less electrophilic than formaldehyde which has no such effects.
Reference

Ouellette, R. J.; Rawn, J. D. Organic Chemistry: Structure, Mechanism, and Synthesis, 1st ed.; Elsevier, 2014. ISBN: 978-0-12-800780-8 (Google Books Preview)

A quantitative justification using tables of actual values has already been provided in the other answer, where it is evident that the $K_\mathrm{eq}$ for nucleophilic addition for acetone will be lesser than the $K_\mathrm{eq}$ for benzaldehyde.
I will be talking about how on a qualitative basis this can be predicted. Acetone is an example of a ketone, while benzaldehyde has the aldehydic functional group. In a very general sense, ketones are more sterically hindered than aldehydes. It's true that conjugation with the phenyl ring can reduce electrophilicity of the carbonyl group for benzaldehyde, but that will simply lead to nucleophilic addition to benzaldehyde becoming slower with respect to other aldehydes like formaldehyde(nucleophilicity being a kinetic property) as the activation energy barrier for addition will be slightly increased.
Comparing $K_\mathrm{eq}$ is a question of the relative product distribution in the equilibrium. It is basically a thermodynamic consideration, which will largely be governed by judging the relative stability of the tetrahedral intermediates formed after the addition(the more stable counterpart will shift the reaction towards the intermediate's side in accordance with Le Chatelier's principle). Due to steric clashes caused by the two bulky alkyl side chains, the tetrahedral intermediate formed from the ketone will always be relatively less stable than that formed from the aldehyde(which has only one alkyl side chain, while the other substituent is a small hydrogen atom). Hence you get the $K_\mathrm{eq}$ for benzaldehyde to be greater than that of acetone


text:
Why is five electrons in d orbital more stable than seven electrons / eight electrons?
The half filled and full filled configuration is considered to be most the stable among the other configurations. But d7 or d8 configuration has a greater exchange energy mathematically. So why isn't it considered to be more stable than d5 configuration which is half filled configuration?

Stability is never defined in absolute terms and I don't think that a generalised statement can be made. So, I'm afraid you must be a lot more specific about what you are actually asking.


text:
Stoichiometric calculation of composition of a compound
I have a question from a high school review package. I don't need the answer, but one thing that is bugging me is that, I feel as if I do not understand the question at all. I don't know what it wants me to find.
Can anyone please tell me what the question is about?
The question is:

Copper oxide is a black powder. It can be decomposed by heating it with an excess of charcoal, a form of carbon.
The charcoal reacts with the copper oxide to produce copper and carbon dioxide. Any excess charcoal that was used can be separated from the copper by adding water. The charcoal will float on the water while the more dense copper will sink to the bottom. The charcoal can then be skimmed off.
Mass of copper produced = $\pu{2.76 g}$
Mass of copper oxide used = $\pu{3.45 g}$
a) Use this data to determine the simplest formula of copper oxide.
b) Predict the valence/charge of the copper in the copper oxide formed.

I basically got:
$$\frac{2.76}{63.546} = \pu{0.043 mol}$$
The balanced chemical equation that I deduced was $$\ce{2CuO + C -> 2Cu + CO2}$$
However, I am not sure that my understanding of the question is correct. Is there a mistake in my understanding.
Note: I was able to solve the question.

Why do you think your understanding is wrong? feels as if the question is asking for two things: the copper oxide formula and the valency of copper in the oxide.

@Safdar would the copper oxide formula be the balanced equation? I couldn't get how the formula for copper oxide could be simplified.. for the valency I think I get it.

They are asking you to find the value of x in $\ce{Cu_xO}$. Simple as that.

@Safdar OOHHH, thank you so much! I was thinking of xCuO but that didn't make sense, but this works perfectly. Thanks again!

Safdar, A minor correction. they are asking to determing Cu_xO_y. We cannot assume that there is only one oxygen, although your suggestion is right.

@M.Farooq well x could be a fraction.. A simple matter of LCM.. it would be simpler that way since you wouldn't have two variables. However this is more of elementary math than chem at this point.

You do not need to modify the question like that. Just state you solved it at the end. An accept normally means you got the answer.

This is an example of a classical method of determining chemical formulas by elemental analysis.
You are not supposed to know the chemical formula of copper oxide neither they are asking you to write a balanced equation. All they are telling you is that
(i) Mass of pure oxide of copper = 3.45 g
(ii) Mass of copper obtained after reduction (we do not need to know the chemical equation) = 2.76 g
Use mass balance, how many grams of oxygen atoms must be there in that oxide?
Now you know the masses of Copper and Oxygen.
Two options
(a) Use mol and find out molar ratios between the two elements
(b) Alternatively, if you do not need to invoke the mole concept ... old school way. Just for trying
Think about the fact that 16 (exactly 16) is the mass unit of 1 oxygen atom. What mass of copper atoms would be associated with 16 units of O if copper's atomic mass is 63.54 units. Use mass balance and some arithmetic.

Thank you so much for the help! :)
So I got 3.45-2.76 = 0.69 g as the mass of oxygen and 2.76 g as the mass of copper, and dividing both by their molar mass got me 0.043 moles. Would the formula for copper oxide by CuO this way then, or is there something I've missed??

Yes, right. Deduce the simplets ratios, which leads to 1:1


text:
Recognising a rate-determining step in E1
If there is a hydride/methyl shift possible in E1, then is the carbocation formation the rate-determining step (RDS), or the shift, or is it considered a cumulative step and happens all at once?

Anindya Prithvi has said everything you needed to hear in two sentences. It is true that rate determining step for E1 mechanism is the carbocation formation. Thus, I present here the detail energy diagram for reaction progress of acid-catalyzed dehydration:

The first step is the protonation of starting compound (SM), which is shown in the right hand top corner in blue box. The next step is the rate determining step, which has the biggest activation energy and progress through the transition state 2. The resultant intermediate is the relatively higher energy 2°-carbocation, which is quickly rearrange to lower energy 3°-carbocation by 1,2-hydride shift through the transition state 3. It is worth noting that this 1,2-hydride shift does not involve net bond-breaking or bond-forming (it is believed to be undergoing concomitant bond breaking-making mechanism involving orbital overlapping). Thus its transition state (transition state 3) would not be higher in energy than that of transition state of RDS. The 3°-carbocation would finally give away the expected product through the transition state 4. It doesn't matter if all other steps happen at once or separately. All those steps have to wait to happen until the formation of the 2°-carbocation. That's why it is the rate determining.

I think it is important to mention that the hydride shift occurs in an already-high-energy intermediate (i.e. the carbocation). Furthermore, the hydride shift does not involve net bond-breaking or bond-forming. Hence, the energetic requirement (i.e. the activation energy) for the hydride shift would not be higher than that of the carbocation formation.

@Tan Yong Boon: Good point. I thought my energy diagram itself would explain that. Nevertheless, I edited the text accordingly. Thanks for careful reading.

Why would the energy difference have a large effect on the kinetics of carbocation shifts, according to arhenius eqn, the k stands pretty close to an order of $10^{10}$

This is a catalysed reaction, at one point you have all species simultaneously. These are therefore not uncoupled reaction steps (the protonation will influence the whole mechanism). This is a rather simple reaction, but for more complex catalytic transformation, the rate determining step approximation is insufficient. But in a first order approximation, for a simple reaction like this, it works quite well; however, one should be aware that there are limitations.

The RDS for E1 mechanism is the formation of carbocation. The rearrangement steps are fast and are hence not considered in the RDS.

Didn't I said it in my first sentence? I didn't ask OP to accept my answer. I just give a overall view of the reaction to benefit other students in mind. It was OP's decision. OP can still change it back to where it was. However, keep in mind that answers should be descriptive. Should not look like a comment. This is for future perspectives.

The concept of rate determining step is a crude approximation and does not always lead to the right conclusions and often does not explain side products. A catalysed mechanism cannot directly be compared with an uncatalysed one. The energetics of all involved species determine the overall kinetics (including all possible steps and paths) of the reaction. While this comment answer might be correct (it better: sufficient) in a basic highschool textbook approach, it is by far too simple for most practical means.

There is a problem here. The concept of RDS is a crude approximation, like I mentioned. So the key point of the answer should be that the RDS Approximation ist insufficient. It's important to realise that for a catalytic mechanism all steps depend on each other, therefore you cannot separate them and therefore you cannot compare any rate constants, because they simply do not exist on their own. The catalytic cycle is described by one rate law, which is inturn characterised by the entire energetics of the cycle itself. So in conclusion: I say your answer is wrong.

The question is about a catalytic reaction, discussing it in any other terms is wrong. The kinetics of a reaction is entirely determine by energetics of the intermediates. The rds only works well for linear (consecutive) mechanisms with only elementary steps, and maybe not even there. Reaction mechanisms are usually not as simple as the textbook approaches suggest; discussing the energetics of the reaction is the only way to get close to understanding it. But I can't continue this discussion any further at this point.


text:
Am I getting the details of mechanism of adsorption right?
Particles on the surface of a substance have a net force towards the geometric center of the substance called residual force.
When molecules come close enough to the surface of a substance the residual force due to the substance pulls them on to the surface thus adsorbing them.
is this right?
if not then, give a detailed account on the mechanism.

The process of adsorption involves separation of a substance from one phase accompanied by its accumulation or concentration at the surface of another. The adsorbing phase is the adsorbent, and the material concentrated or adsorbed at the surface of that phase is the adsorbate.


text:
Calculating the viscosity of water using particles

There are two particles with radius 0.212 μm. The particles are in a thin liquid film and the microscope can only detect movement in the $x$- and $y$-axis. The temperature is 25 °C.
$$
\begin{array}{c|rrr}
\hline
t/s & 30 & 60 & 90 & 120 \\
\hline
x_1 & 4.4 & 10.7 & 11.0 & 12.4\\
y_1 & -3.1 & -5.2 & -10.8 & -9.3\\
x_2 & 3.9 & 2.0 & 8.3 & 5.8 \\
y_2 & 3.5 & 6.4 & 14.1 & 12.0 \\
\hline
\end{array}
$$
Use this information to estimate the viscosity of water.

I know that I should use the equations:
$$x^2 = 2Dt \tag{1}$$
$$D = \frac{kT}{6πηa} \tag{2}$$
Where if I plot $x^2$ against $t,$ I can calculate the viscosity from the slope $\displaystyle\frac{2kT}{6πηa}.$ However, what I don't understand is how to calculate $x^2.$
I tried taking the average of the two particles in direction $x$ and squaring the result, but that doesn't give me the right answer. I also tried taking the average of $x^2$ of the two particles, but once again I get the wrong answer.
How should I calculate $x^2?$

Suppose particle 1 starts its random walk at 0,0. After 30 s, its coordinates are 4.4 and -3.1. Square those and add them to get the x squared value for particle 1 at t = 30 s. Repeat for the next three times. Plot those x squared values vs. time. Repeat for particle 2.

The basic idea here is that the spherical particles, each of radius a, are being bounced around via Brownian motion. So the two relevant equations are
$$r(t)^2 = 2Dt \tag{1}$$
$$D = \frac{kT}{6πηa} \tag{2}$$
Note that the first equation uses r(t) rather than x, to avoid needless confusion with the x and y cartesian coordinates of the spherical particles.
Now consider particle 1. At t = 0 s, it is at the defined location 0,0, where all distances are in $\mu m$. At four subsequent times, the x and y coordinates are as given in the OP's table. The following figure shows particle 1's location at t = 0s and at t = 30 s.

At any given time,
$$r_1(t)^2 = x_1(t)^2 + y_1(t)^2 \tag{3}$$
To estimate D, an ordinary least squares fit is performed on the tabulated data for particle 1. The result is shown in the next figure:

Note that the origin is used: at t = 0 s, then, by definition, $x_1(0) = 0 \space µm$ and $y_1(0) = 0 \space µm$. So $D = 1.1486 \times 10^{-12} m^2 s^{-1}$. Rounding off to two significant figures will be performed at the end.
Solving equation (2) for $\eta $ yields:
$$\eta = \frac{kT}{6πDa} \tag{4}$$
where T = 298.15 K, $a = 2.12 \times 10^{-7} m$ and $k = 1.38064852 \times 10^{-23} J K^{-1} $. The result for $\eta $ is
$$\eta = 8.97 \times 10^{-4} Nt \space m^{-2} \space s \tag{5}$$
where the fact that $1 J = 1 Nt \times 1 m$ has been used. N.B. 1 Joule equals 1 Newton times 1 meter.
Other units for $\eta $: Since 1 Pascal of pressure is 1 Newton of force per square meter, i.e., $1 Pa = 1 Nt / m^2$, and since 1 centipoise (symbol: cP) is 1 milliPascal times 1 s, i.e., $1 cP = 1 mPa \space s$, then
$$\eta = 8.97 \times 10^{-4} Pa \space s = 0.897 \space mPa \space s = 0.897 cP\tag{6}$$
Finally, rounding to two significant figures, $\eta = $0.90 cP. The viscosity of water at 25 °C is reported to be 0.91 cP.


text:
Can a reaction be both exothermic and non spontaneous?
Can a reaction be both exothermic and nonspontaneous?

The spontaneity of a reaction is given by $\Delta G$
$\Delta G < 0$ = spontaneous reaction
$\Delta G > 0$ = nonspontaneous reaction
$\Delta G$ is quantitatively described by the formula
$$\Delta G = \Delta H - T \Delta S$$
$\Delta H$ is negative (exotermic reaction)
When $T \Delta S$ product is more negative (smaller) than $\Delta H$, we have $\Delta G > 0$ and the reaction is nonspontaneous

Is there an example of such a reaction though?

No, i agree with what you say. However it would be better if you could add an example of such a situation. Also, this would be temperature dependant. You might want to add that/

exothermic non spontaneous not endo thermic

You can add that to the answer (+1)

@Safdar another example would be conversion of diamond to graphite

@AniruddhaDeb Thanks.. I didn't think of that.

You can think of any combustion reaction


text:
Why does increasing availability of microstates compensate for increased energy of products in an endothermic spontaneous reaction?
I'm wrestling with the idea right now of processes that are both endothermic and spontaneous.I understand that there must have been an associated increase in entropy.
How does an increase in "microstates" compensate for the fact that a chemical reactions products have overall "more energy" than the reagants?
Edits: Clarified question
My attempt: Two closets have same volume, except one has shelving and the other doesn't. The closet with shelving is "preferred" because it is able to "handle" more shoes ("energy") than other closet because there are more ways to arrange those shoes (on the shelves).  The shelved closet with 10 shoes is more "organized" than the nonshelved one with 8 shoes. Which means, in my analogy, Gibbs is a measure of energy "organization."

Melting ice at above the freezing point is endothermic and spontaneous.

Every endothermic reaction you could observe qualifies and that's a whole type - virtually infinite.

Edited question! Thanks for feedback!

The shelved closet will have greater entropy because of all the many ways that you can arrange the shoes, i.e. particles can be placed many more ways into many energy levels than into a few levels level with the same overall energy. This means that as $G=H-TS$ and if $S$ increases then $G$ becomes smaller and more negative. It is the multiplicity of arrangements of particles (or shoes) in the energy levels that is important here.

Energy distributes itself among the available microstates. The greater the number of microstates, the greater is the statistical probability that the system will find itself in an arrangement corresponding to the macrostate—in this case, the endothermic product. Therefore, the spontaneity of the endothermic reaction is fundamentally driven by this statistical probability, which results from the nature of energy to distribute itself among all available microstates and the endothermic product possessing a preponderance of such microstates in the system.

@Blaise I think I understand. So you're saying that spontaneity is all about the possible arrangement of microstates. And if the majority of available microstates are higher energy, than the product will be statistically higher energy. So because a molecule has a ton more microstates available in the gas state, than the solid or liquid state, molecules will spontaneously evaporate. Blaise, I encourage you to write up an answer with examples so I can upvote!


text:
X - membered rings in formation of epoxide
https://youtu.be/d4RczD122QI?t=2025
In this lecture, I am learning to form epoxides from the intramolecular version of Williamson Ether synthesis. During that, the professor says the ring formation is favored when ring is 3/5/6-membered, and not favored in 4/>6-membered. I tried to think of: smaller bond angle increases ring strain, but this does not explain why 3-membered ring is favored.
I checked the answer to this question.
Why are 7-membered rings less likely to form than 5- and 6- membered rings?

It seems that 3-membered rings have a very positive ΔG, how come it is favorable?
In the last graph of the answer, overall rate of formation of 4 & 6-membered rings seems around the same (log k(intra)), which I really want to know the explanation of that. In general, what membered rings are actually favored?

That answer explains why three-membered rings are formed easily; it's because of the entropy factor.

Pretty much the same issue is discussed here: Dependence of rates of neighbouring group participation on length of alkyl chain


text:
Finding Equivalent representation of a molecule/crystal
Is there a website or online calculator to determine the reducible representation of a crystal/molecule using wyckoff positions without having to analyse the number of unmoved atoms/bonds for complicated structures?


text:
Why do Grignard reagents act as sources of nucleophilic R− and not Br−?
In reactions with Grignard reagents, I've always seen the alkyl group being the nucleophilic component, with the R–MgBr bond being broken. This is always rationalised by the fact that the alkyl carbon has a partial negative charge, whereas the electrophile (e.g. a carbonyl group) has a partial positive charge.
However, the halogen atom should also have a partial negative charge. Why doesn't it attack the electrophile instead, breaking the RMg–Br bond?

And if bromide did add to a carbonyl group? It is  reversible step.

but isn't addition of iodine the only halogen addition that is unstable?

You may be thinking about addition of iodine to a CC double bond. The species R2C(OH)X where X=halogen. Moreover, the species you propose has a negative charge on oxygen, which is a worse case scenario.  Although cyanohydrins are stable when formed, they are reversed by base.

In all likelihood the halide ion does attack the substrate.  But, the product such a reaction mode would form does not accumulate whereas the alkylated product does.
Recall that in a nucleophilic reaction halide ions are good leaving groups, at least if you stay away from fluorides which are rarely used for Grignard reagents.  So while the halide ion may add to the substrate, it may also be displaced if an alkyl-anion moiety from another Grignard molecule comes in, or possibly the halide ion may just launch on its own to attach to a magnesium atom (which, in a Grignard reagent, acts as an electrophilic center, for example by adding to the oxygen end of a carbonyl group).
In contrast, when the alkyl-anion moiety attacks, the carbon-carbon bond it forms is strongly favored thermodynamically and a lot harder to break; here we do not have a good leaving group.  So the alkylated product rather than the halogenated product is what accumulates.


text:
Catalysts and activation energy
Do catalysts increase rate of reaction only by decreasing activation energy? Or can they affect the Arrhenius factor or reaction orders? Like if one catalyst is more effective at increasing rate of reaction than another, does it mean that it's simply decreasing the activation energy more dramatically? Or are there other mechanisms at play?

They change mechanism completely, along with kinetics.

does changing the mechanism include changing the reaction order or Arrhenius factor (A) ?


text:
Does rate of auto-ionization of water change after adding a strong acid?
We know that a reaction could have different rates at equilibrium. For example, lets take the reaction:
$$\ce{A + B -> C }$$
At equilibrium, the rates of formation and dissociation will be equal:
$$k_\mathrm{r}\mathrm{[A][B]} = k_\mathrm{c}\mathrm{[C]}$$
But that means at various concentrations of $\mathrm{[A],[B]}$ and $\mathrm{[C]}$, the actual velocity (rate)  of the reaction at equilibrium will be different.
In other words, for the above example: If I were to put additional [A] into the reaction after its reached equilibrium once, the new equilibrium it would reach would have a faster rate. Also, this means that a fold change increase in $\mathrm{[A]}$ doesn't require an equivalent fold change decrease in $\mathrm{[B]}$ to reach the new equilibrium.
Now, the autoionization of water is unique because it is independent of the concentration of water: $$K_\mathrm{w} = \ce{[H+][OH-]}$$
This means that a fold change increase in $\ce{[H+]}$  requires an equivalent fold change decrease in $\ce{[OH-]}$
Thus, I am guessing the rate would be equivalent to some constant: $$k_\mathrm{r}\ce{[H+][OH-]} = constant$$
Does the rate of auto-ionization of water change when you add a strong acid or base? An explanation at the level of molecules would be appreciated.

The autoionization of water is written as $\ce{H2O + H2O -> H3O+ + OH-}$. As written, the forward direction does not depend on the acidity or basicity of the solution.

@ Zhe. Yes, it does. If the water contains some acid or some base in solution, the autoionization decreases. The autoioniozation of water is very low in pure water. It is even worse in presence or an acid or a base in solution. Don't forget that the product of the concentration of the ions H+ and OH- must be constant, whatever their origin. If one mole of acid or base is dissolved in 1 liter water, the auto ionization produces only $\ce{10^{-14}}$ moles H+ and OH- .

@Maurice I don't understand how that relates to the kinetics of autoionization. I suppose I am assuming (possibly incorrectly) that the equilibrium I wrote is the elementary reaction step. But if that's correct, the kinetics of the process are unaffected by pH. We're not talking about the degree of autoionization.

I don't think the kinetics of this reaction can be measured.

It’s complicated and hard to measure: https://www.ingentaconnect.com/content/scs/chimia/2012/00000066/00000004/art00008?crawler=true

Ideal solutions
If we pretend for a moment that all the solutions under consideration are ideal, and that we have already reach equilibrium, we can consider the following reaction:
$$\ce{H2O(l) + H2O(l) <=> HO-(aq) + H3O+(aq)}$$
As Zhe said in the comments, the concentration of the reactants don't change much (even if you have $\pu{1 M}$ sodium hydroxide or $\pu{1 M}$ hydrochloric acid), so the forward rate at equilibrium is the same at any pH (again, assuming an ideal solution). That means the reverse rate is also the same at any pH. Because the product of the concentrations of hydroxide and hydronium ions is constant, the reverse rate constant does not depend on the pH.
Considering ionic strength and activity coefficients
The above argument does not work in a real solution with high concentrations of dissolved ions. If the pH is far from neutral, the concentration of ions (and counter-ions is high), and equilibrium concentrations (and with that the rates) will depend on the solutes present in solution.
Molecular level
At the molecular level, the auto-ionization of water is a special case because hydronium and hydroxide ions seem to diffuse faster than other ions (Grotthuss mechanism). At neutral pH, both species are rare. At acidic pH, there are more hydronium ions, but fewer hydroxide ions, so collisions are also rare (and reverse for basic). A high ionic strength would make it easier for ion pairs to form and separate, while a low ionic strength would be advantageous for the combining of hydronium and hydroxide.


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Can you turn lead dioxide into lead?
I've read that you can reduce lead oxide to lead by heating it with carbon. Does this also work with lead dioxide?
I took apart an old lead-acid car battery and realised then, that there are almost equal parts of lead and lead dioxide.
Already having saved and distilled the acid, I started to think there was a lot more effort involved giving all the safety steps and time than buying the items separately.

Decomposition of lead(IV) oxide at different temperatures forms different types of oxides. At around $\pu{600 ^\circ C}$, lead(II) oxide forms. Wikipedia adopts the scheme of lead(IV) oxide decomposition in air summarized by White and Roy [1]:

$$\ce{PbO2 ->[\pu{293 ^\circ C}] Pb12O19 ->[\pu{351 ^\circ C}] Pb12O17 ->[\pu{375 ^\circ C}] Pb3O4 ->[\pu{605 ^\circ C}] PbO}$$

You can then reduce lead(II) oxide to lead using carbon or carbon monoxide at around $\pu{1200 ^\circ C}$. The effect of dissolved inorganic carbon is discussed by Wang et al. [2]
$$
\begin{align}
\ce{PbO + CO &->[\pu{1200 ^\circ C}] Pb + CO2}\\
\ce{2 PbO + C &->[\Delta] 2 Pb + CO2}
\end{align}
$$
Note that lead(II) oxide can also decompose to release toxic fumes of lead which can be harmful and also doesn't serves your purpose of getting solid lead.
References

White, W. B.; Roy, R. Phase Relations in the System Lead-Oxygen. J American Ceramic Society 1964, 47 (5), 242–249. DOI: 10.1111/j.1151-2916.1964.tb14404.x.
Wang, Y.; Xie, Y.; Li, W.; Wang, Z.; Giammar, D. E. Formation of Lead(IV) Oxides from Lead(II) Compounds. Environ. Sci. Technol. 2010, 44 (23), 8950–8956. DOI: 10.1021/es102318z.

Lead dioxide can be heated to 600 °C and it will thermally decompose into PbO and O2.  This likely creates lead fumes but this may be one way to accomplish the task.
Optionally, you could dissolve the lead dioxide in acid then reduce it from a salt.
Welcome to this community, but please use care.  Be careful with the DIY approach to doing chemistry when it comes to high temperatures or somewhat dangerous materials such as lead.

Thank you for your reply, I wear acid rated PPE, face shields and take extra precautions with a bicarb wash. Also outside. Using a acid to dissolve it to a salt, this sounds more dangerous then creating a old style coke forge to take it up to 1200 then collect the lead from the bottom once cooled. Then refine it.