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How to publish a review article on drugs in a reputed journal?
What are the things needed to publish a review article in a journal starting from scratch ?
Is it possible if i am not associated to any institution ?
How long does it take and what are all the guidelines that I should have in mind while writing an article ?

Why not try to write  some science articles for your local newspaper, many now have such columns,  rather than a full blown article, which as suggested in the answers is hard and time consuming even for an expert. That way you can get your name known and learn a lot  on the way.

You have no chance whatsoever to get a reputed journal to accept a review article. Doing research on your own is equally futile if you are just finishing your bachelor. Go and work on what you know, or should know. perfect grades in your final exams are really going to help you once this corona madness is over.

Better suited for Academia SE but as other said you are totally out of the track. It does not work like this.

I'm voting to close this question as off-topic because it's outside of the scope of the site.

The idea behind a review article is that it is an overview typically written by author(s) who is/are known to be experts in the field.  You need to be an expert because you are not merely publishing your own work, you are assessing and interpreting everyone else's work in this area. And it takes a lot of experience in the field to be able to do this -- to know what work is reliable, what work is questionable, what work is significant, and what work is of minor importance.  That's hard to do if you're starting from scratch.  If, on the other hand, you've been working in the field for 20 years, you know a lot of this to start.
For this reason, journals often invite known experts to write review articles.
The two exceptions I know of would be:
(1) a case in which a well-known expert is invited to write a review article, and then invites, say, a senior grad student to be a co-author.  But in this case, the senior expert is still providing guidance.  See, for instance:
https://www.asbmb.org/asbmb-today/careers/120111/writing-a-scientific-literature-review-article
(2) A review article written for a predatory journal.  If the journal is not legitimate, their standards are of course far lower! If you've been asked to write a review article for a predatory journal, or a predatory publisher, don't!  Here is a good list of predatory journals (though it may not be complete):
https://predatoryjournals.com/journals/
and of predatory publishers:
https://predatoryjournals.com/publishers/
You say you're interested in writing for a reputable journal.  Which one is it?  I would say that, before doing any work, you should contact them.  I expect, if it is a reputable journal, they would say the same thing I have.
In addition, is the journal in English?  If yes, then I can tell (even from your brief post) that you will need a good editor to clean up your English.

I am looking for ways to stregthen my profile in a short period for my masters application. I ve had a very limited research experience in my bachelors and I ve heard that students with publications have greater oppurtunity to get admitted. Are there any ways that you could suggest me to have any publication or to strenghthen my profile?

Usually the way students get published in peer-reviewed journals is by working for a research group, and being included on a paper as a coauthor.  Is there a university nearby where you could, say, do an internship in someone's lab?    The other option is to publish an opinion piece on a scientific subject of popular interest in, say, your local paper.   For instance, if you are interested in pharmaceuticals, you could publish an article about, say, their misuse, or their over-prescription, or about how doctors are being influenced by pharma companies to prescribe more expensive drugs, etc.

There are universities here  but due to the current situation of COVID-19, all I can do is with the computer and the application period starts in a month. I can think about articles in news papers. But i dont know why i couldnt lose the idea of having publication in a journal.

Having a publication as a bachelor just means that you were lucky, and/or had very good grades that helped you get into a project that resulted in a serious publication. It´s not going to impress anyone alone.

As you stated you are a bachelors student and wish to apply for a masters. You would like to improve your academic profile. This is perfectly fine. It will be very hard to write a full fledged review at this stage and it will consume a lot of your time. A good review by a field expert may take six months just to write a draft and read more than 250 publications in detail. You can write for science magazines like Resonance which is published by the Indian Academy of Sciences. It does not have an impact factor but they do general publish science articles for public outreach. I did publish one article for fun there on making a "monochromator" in a shoebox (from my mother) long long time ago. I still occasionally read it. It is a good science magazine. Having one magazine article also indicates that you do have an inclination to write and compose essays.
What is your area which you think you are really good at? Can you address some math problems in a simplified way? Can you simplify something in chemistry? Do you know any good trick to balance chemical equations? There are matrix based methods which most people are not familiar with. Look at the topics in Resonance and probably write something with your teacher.

Thanks a lot for the suggestion about magazines. I am afraid that I am pretty much on my own if I am gonna publish such an article but I will be really glad if you could tell me the ways i could learn on learning to write an article by my own. Where do I start ?

You start from reading and reading and reading your topic of interest. A well-read person can write very well. Explore some issues of Resonance and see if you get the feel for the topics covered there.


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Determine and draw planes and axes of symmetry of a molecule using software (VESTA or Mercury)
I designed the $\text{MoS}_2$ molecule (bilayer), with polytype 2H in the VESTA. My goal is to draw the plans and axes of symmetry of the bilayer set. According to the information in the literature, this molecule belongs to the $\text{D}_{3d}$ point group. How can I draw and determine the planes and axes of symmetry in VESTA?
These are the parameters I used in VESTA:

Mercury Attempt:
I exported the .vesta data to a .cif file.

Table with space and point groups for $\text{MoS}_{2}$ (taken from https://iopscience.iop.org/article/10.1088/0953-8984/28/35/353002/ampdf) (see 2H phase for $\text{MoS}_{2}$):

I do not know VESTA well enough to provide an answer for this program.
However because your data are about a crystal structure I suggest to export your model in the .cif format.  CCDC's freely available version of CCDC's Mercury is able to read this file type, and add to its visualizations the symmetry elements of the unit cell (Display -> Symmetry Elements).  There will be a new menu where you may choose among the symmetry elements the program is able to recognize, and how they are displayed:

For example only the centres of inversion:

The export options in Mercury include bitmap .png, ray-tracing PovRay .pov, and .vrml / .stl.

Given lattice constants and space group $P6_3/mmc$, do not forget to compare your model with the literature known models for example in the freely accessible COD about experimental data (entries #1010993, #1011286, #1531960, #9007660, #9009144); and consider contributing to its sibling TCOD about theoretical data and MPOD.

Thank you for your answer. I tried to follow your suggestion and do it on Mercury. The objective is to draw the axes and the planes to prove that it is a $\text{D}{3d}$ structure. To be a $\text{D}{3d}$ structure, it must have 3 $\text{C}_2$ axes of symmetry. I put what I got on Mercury, in the post. The problem is that I am getting more than 3 axes of type $\text{C}_2$, and I should only get 3 axes of that type. Why do more than 3 axes type $\text{C}_2$ appear to me, when this structure should only have 3 axes type $\text{C}_2$ and what are those 3 axes in the last figure I put in the post?

In addition, the molecule you took from the COD library must have a bug in Mercury. In $\text{MoS}_2$ with polytype 2H, there is a bilayer and only Mo-S bonds are allowed. In the drawing you have in the images there are Mo-Mo bonds, which should not appear.

@JoseMarin We are sharing the same page here.  Space group $P6_3/mmc$ (#194) is for unit cells of point group symmetry $D^4_6h$ (IT volume A1, p. 345, the 2006 edition). Yes, the diagram (e.g., http://img.chem.ucl.ac.uk/sgp/large/194az1.htm) indicates more than 3 $C_2$ axes perpendicular to $2/m$ and $6_3/m$ axes.

@JoseMarin Mercury is developed by CCDC; contrasting to ICSD with more inclination for organic, and metalorganic compounds. So its geometric approach draws lines between atoms if their distance is quite less than the sum of the van der Waals radii.  With S 1.80 A, Mo 1.90 A, d(Mo-Mo) = 3.15..3.16 A which is lesser than 2 x 1.90 A.  With the input given in your question, I was not able to reproduce the trigonal prisms of your question (vesta 3.4.8); indeed, pressing the «remove duplicate atoms» with default 1e-006 leaves only one Mo and one S (still no prisms).

@JoseMarin Loading one of the COD entries (#1010993, #1011286, #1531960, #9007660, and #9009144) as .cif into Vesta however yielded the prisms of the original question within a blink of an eye; each of them require only one Mo and one S for the model's definition.

Thank you for your response and explanation regarding the Mercury bug. I still don't understand why there are 24 elements of symmetry and more than 3 $\text{C}_2$ axes, when the material is a bilayer material (even number of layers). I saw an article that said that in these materials the point group depended on the number of layers (even or odd). I put the table in the post.

In the table it says that the space group is P$6_3$/mmc, but that for N=even (bilayer) the point group is $\text{D}_{3d}$ and in this case it should have only 3 $\text{C}2$ axes and 12 symmetry operations, which contradicts the Wikipedia table on space groups (https://en.wikipedia.org/wiki/List_of_space_groups) or the article you suggested to me (which says that this space group belongs to unit cell $\text{D}^4{6}h$).

Could you explain to me why there are 24 symmetry operations in this case instead of 12 and why the bilayer belongs to the group $\text{D}^4_{6}h$ instead of $\text{D}{3d}$, please? What does it mean to belong to the $\text{D}^4{6}h$ group (I can't find the character table for this point group)? What is the equivalent point group?

@JoseMarin I think, section 3.1 (p. 11) contains one key: "For example, the 2H-MoS 2 stacking arrangement has space group $D^4_{6h}$ and point group $D_{6h}$ for its bulk phase, but the point group is reduced to $D_{3h}$ or $D_{3d}$ for odd or even numbers of layers, respectively."  A symmetry $D_{3h}$ is found e.g., if there is only one layer of these $\ce{MoS2}$ trigonal bipyramides (like a film / 2D) instead of volume / bulk / 3D crystal modeled in $D^4_{6h}$ with maybe hundreds of (alternating) layers, p. 13: "Odd or even numbers of layers of a TMD can lead to different symmetry labels"

@JoseMarin As lay-out the symmetry operations in $P6_3/mmc$ or $D^4_{6h}$, this were worth a separate question (there already was such a question here, as far as I vaguely remember without pointing to examples of http://faculty.otterbein.edu/djohnston/sym/spacegroups/, though).

But in this case, I designed a bilayer so the point group is $\text{D}{3d}$ instead of $\text{D}^{4}{6h}$. Why does the Mercury software give me the elements of symmetry as if the structure were $\text{D}^{4}{6h}$ when, in reality, it is bilayer and belongs to $\text{D}{3d}$?

Why is the structure not P$\bar{3}$1m ($\text{D}_{3d}$ point group)?

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Are these different neutralization pathways?
When sodium oxide is added into water, it reacts with water to form $\ce{NaOH}$.
$$\ce{Na2O(s) + H2O(l) -> 2 NaOH(aq)}$$
This means that if we were to add an acid, it would be neutralized by the sodium hydroxide.
$$\ce{Na2O(s) + H2O(l) + 2 HCl(aq) -> 2  NaCl(aq) + 2 H2O}$$
Is the following equation a totally different reaction pathway, or is it simplified from the reaction pathways above?
$$\ce{Na2O(s) + 2 HCl(aq) -> 2 NaCl(aq) + H2O(l)}$$

Most likely, nobody actually knows the correct path. The question, does the oxide react with water or does it react with protons from HCl. Both must be extremely fast reactions. I think someone has to use to isotope labelling in Na2O and H2O and see what it is isotope of oxygen in NaOH.

It is surprising that there is no related paper on this topic.

First you have to decide what experiment you want to do. Add HCl to solid sodium oxide, or dissolve the sodium oxide in water first?

As water will present at about 55 molar, even in the presence of acid, the first reaction would seem to be be the most likely assuming that it does not exist as  Na$_2$O in solution.

And then perhaps give a thought to the concentration of your HCl solution.

Anyway, sodium oxide is rarely used in the lab. I would even be pleased to know how it is synthesized, because it is decomposed at about 400°C into peroxide and metal.

Both fast reactions happen in parallel, acting as "first comes, first served".
Concentration of water molecules ( $\pu{55 mol/L}$ in water) is much higher than $\ce{[H3O+]}$  for diluted acids, so oxide anions meet water molecules much more frequently.
If the oxide anion meets water molecule, it will not wait for hydronium.
Additionally, hydronium ions would be rather depleted by reaction with hydroxide ions next to the oxide surface.
Be aware that $\ce{Na+}$ ions are just bystander/spectator ions, so it is more illustrative to write equations as ionic equations:
The main reaction:
$$\ce{O^2− + H2O -> 2 OH−}\tag{1}$$
The secondary alternative reaction:
$$\ce{O^2− + H3O+ -> OH− + H2O}\tag{2}$$
The final reaction:
$$\ce{H3O+ + OH− -> 2 H2O}\tag{3}$$

Both are equally valid (different) pathways though indistinguishable in real life.
Sodium oxide is essentially an anhydrous version of sodium hydroxide. Your second pathway is valid as sodium oxide being a solid base would lose the oxide ion when in contact with an acid like HCl.
Your first pathway is equally valid, where the sodium oxide dissolves in water first then reacts with HCl.
However, these two both happen so fast that they are basically indistinguishable in normal settings. The only situation where I perceive only one happening would be in high temperatures where HCl is considered molten and not aqueous, then the oxide does not dissolve into water but reacts straightaway.

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Molecular orbital diagram for I₃⁺ cation
Do I need to go through the whole process of reducible representations/projection operator method to come up with the bent SALCs for the $\ce{[I3]+}$ molecule? Or is it simpler than that?

How is this unusual ion created ? Is it not $\ce{I_3^-}$ ? Because this ion $\ce{I_3^-}$ does exist. Never heard of $\ce{I_3^+}$ !!

@Maurice Apparently it is iodonium cation. https://www.sciencedirect.com/science/article/abs/pii/S0032386101004906

Yes I already found information on I3- but thank you. In a homework we are asked to compare the bond orders of I3- and I3+ after creating their MO diagrams. After much searching and trying to figure out the MO diagram for I3-, I found out that it's very easy and only involves the three 5pz orbitals in a "4 electron 3 center bond." But I don't know if I3+ is similar. I3+ is a different geometry, so I feel like there may be other bonding involved?

I suspect the intent of the exercise is for you to compare the linear and bent structures and determine which is more likely based on MO occupancy. One way to do this is to start with the linear I3- MOs and consider how they are perturbed by bending. Look up Walsh diagrams if you're not familiar with doing that sort of perturbation.

I do know about Walsh diagrams, which I've seen for water and BeH2. Linear vs bent for water wouldn't be the same for I3- and I3+ though, it's a different set of orbitals involved. And the homework asks me to construct and MO diagram for I3+, which I don't understand how to do.


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Effect of substituents on arene oxide-phenol rearrangement
When I was learning about the reactions of epoxides from my textbook, I came across the following mechanism for arene oxide-phenol rearrangement in the presence of acid:

NIH Shift - National Institute of Health Shift (c.f. Wikipedia)
In the above example, there is no substituents on the ring and hence the final product remains irrespective of which $\ce{C-O}$ bond in the arene oxide cleaves to open the ring. Later, the author presented the following reaction where the ring has an electron donating group - methyl group:

Here the final product depends on which of the two $\ce{C-O}$ bonds in the arene oxide is broken. Based on the general reaction mechanism provided by the textbook, I derived the following mechanism to form the product prescribed by the author:

Before writing the above mechanism, I first arrived at a different product by following the mechanism given below:

The reason I initially chose to cleave the left $\ce{C-O}$ bond in "Reaction Mechanism - 2" was, it gives rise to the carbocation 2(a) where the methyl group is at the para position. I've learnt that methyl group stabilizes a positive charge by hyperconjugation. Since, resonance effect is felt only at the ortho and para positions and not at the meta position, the carbocation 2(a) is more stable than 1(a) from "Reaction Mechanism - 1". Moreover, in the first mechanism, NIH shift involves a 1,2 Deuterium shift whereas in the second it's a 1,2 Hydrogen shift. I find a deuterium shift to be slightly energy expensive process due to its higher mass and stronger bond with carbon compared to a hydrogen shift.
The only reason why I think the first mechanism is better than the second is the product 1(b) is more stable than 2(b). But I don't understand why is the second reaction not preferred over the first even though it has two factors supporting (stability of 2(a) over 1(a) and hydrogen shift over deuterium shift) it.
I faced similar contradictory results while writing the mechanisms for the following reactions (from the same textbook):

My questions are:

Are the reactions provided by my textbook correct?
If yes, why is the cleavage of one $\ce{C-O}$ bond of the epoxide preferred over the other even though it produces a comparatively less stable carbocation? Are there any other factors which determine which bond cleaves?
Is arene-oxide phenol rearrangement "specific" or highly "selective"? Or in simpler terms, does the rearrangement exclusively form only one product or does it form some small amount of the other product too?

Image source: My own work :) | Made using Marvin Sketch.
Textbook: Organic Chemistry by Solomons, Fryhle, and Snyder | Chapter 10: Alcohols and Ethers | Section 10.14: Reactions of Epoxides

What your book says is correct, and the issue arose due to your premature judgement of the intermediates' stability without actually working it out.
First of all, [1,2] Wagner-Meerwin shifts are intramolecular rearrangements which are primarily driven by thermodynamic concerns. Therefore, if their is a possibility of a reasonable cyclic transition state being formed, then the presence of the deuterium isotope hardly makes much of a change for the anionotropic shift, especially since the $\ce{C-D}$ is only marginally stronger than the $\ce{C-H}$ bond as stated here. So, if their is enough motivation for the rearrangement to occur, it can happen via either of the pathways.
Now, if you actually try drawing out the resonance structures of 1(a) and 2(a), you will actually find that the cation on the former will receive stabilisation by the +H effect, and not the second. The positive charge will only be able to align suitably for hyperconjugation in 1(a) as shown below:

Hence, mechanism 1 should be the preferred pathway
I think the rule you are stating refers to the increment in electron density at the ortho and para positions of an aromatic ring by the virtue of a methyl substituent as shown below:

However, this thing is only valid when the methylation has been done at a completely aromatic ring and not at some broken fragment of the same.
Only when their is complete conjugation and the ring itself is aromatic does it even make sense to use ortho and para descriptors as per definition. When you are considering a part of the same with incomplete conjugation, then it's pretty much up to the bonding schematic of the current molecule to see where the positive charge is getting delocalised.
o-/p-/m- descriptors lose their value in such cases, and along with them, their delocalisation rules as well
Clarification
I guess the OP had some confusion regarding the mechanism after the formation of these intermediates as well. The competition of each pathway occurs stepwise:

First, the competition will be between the cleavage of either one of the two $\ce{C-O}$ bonds, one adjacent to the hydrogen and the other adjacent to the deuterium. As the intermediate 1(a) is more stable than 2(a) due to reasons stated above, the cleavage will occur as shown in the second step of Reaction Mechanism-1

Now, the intermediate 1(a) will try go for rearrangements that will generate species which are more energetically favorable. Due to the electron donating(+M effect) of the hydroxyl present adjacent to the cationic part in 1(a), a 1,2 deuterium shift will occur leading to the formation of 1(b) as shown in the third step of Reaction Mechanism-1

Note that as soon as the cleavage step occurred as described above, all considerations of Reaction Mechanism -2 and it's intermediates after 2(a) were rendered moot. 2(a) lost the stability battle against 1(a), and after that, none of the intermediates or products of Mechanism-2 would matter at all

Thanks for the answer! Now, I understand why is 1(a) more stable than 2(a). Similarly, we can say that, 1(b) is less stable than 2(b). If the presence of deuterium doesn't matter, how to decide which of the two pathways is feasible? For the labelled intermediates, relative potential energies follows this order: 2(b)<1(a)<1(b)<2(a). Product (b)s are more stable than (a)s due to the electron donating nature of hydroxyl group. I'm unable to figure out which reaction will be preferred from this data. Did you consider any other things to arrive at the conclusion - reaction 1 is the preferred one?

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Will a Grignard reagent react with an epoxide or episulfide first?
In this question, I am confused whether the Grignard reagent will attack the epoxide or the episulfide (thiirane) first. The solution says ring opening of the episulfide will occur due to greater strain.

I'm not convinced by the strain argument, but you could think about leaving group ability and which bond is easier to break...

This is kind of stupid set of answers. I mean any idiot can point to right answer by elimination.

There are quite a few reasons why the episulfide might open before the epoxide:

Sulphur is less electronegative than oxygen. Therefore, by Bent's rule , more atomic s-character will be directed by the ring towards sulphur in the episulfide as compared to the oxygen in the epoxide. An increased s-character will lead to a bigger increase in the $\ce{C-S-C}$ bond angle as compared to the $\ce{C-O-C}$ angle, and thereby increasing the angle strain in the episulfide relative to the epoxide
Sulphur will also tend to be a better leaving group, as basicity decreases down the group(in this case, group 16)
The carbon-sulphur linkage is weaker than the carbon-oxygen one

So, the mechanism should look like this:

Some points to note:

Sulphur compounds also tend to be pretty good nucleophiles due to their softness as per HSAB as elaborated here. Therefore, the $\ce{RS-}$ formed after ring opening will again attack on the epoxide to open the latter.
This will also lead to the formation of an  ethereal linkage in the final product, which will be quite favorable due to the high stability of $\ce{C-O}$ ether bonds

I'm not sure I buy the strain argument either. Is there enough difference between the reactivities that you would get selectivity? In the real world I think you would get a mixture, possible complex enough to be called a mess.
However according to this reference  here

Grignard reagents attack the least hindered carbon atom of thiiranes.

So the sequence is as follows: RMgX attacks the least hindered carbon of the thiirane in preference to the least hindered carbon of the epoxide as the thiirane is more reactive due to the ring strain. The resulting sulfide anion then attacks the epoxide at the alpha position giving a primary alkoxide and a thiirane, this step is reversible but as RS- is the better nucleophile the thiirane predominates.
When the MeI is added it methylates the primary alkoxide giving product c

I feel like I might have seen this kind of sulfur Payne rearrangement before, with a different (softer) nucleophile, maybe an alkoxide (I can't remember)... but am a bit busy now to look it up


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Conversion and reactor time
Lets say it takes my batch reactor 3 hours for 35% conversion.
If I wanted to find how long it would take for 15% then,
35-15 = 20,
then 3 hours/0.20 = 0.15 hours for 15% conversion.
I am not too sure if this is correct. Some guidance would be greatly appreciated

I am very curious how did you come to "then 3 hours/0.20 = 0.15 hours for 15% conversion."

For one, in chemical engineering, to describe the duration about how long a volume element remains in a reactor (e.g., batch reactor, continous flow reactor) is the residence time (e.g., https://en.wikipedia.org/wiki/Residence_time).  For two, knowing the conversion after a (mean / average) residence time is only one part of the equation; depending on the rate order, rates may depend on the time of reaction, too.  Have a look at the rate laws (including their integrated form); zero, first; second, fractional order.

The result very depends on the reaction stoichiometry, kinetic order and the ratio of reagents, as  "cheaper" reagents are often in excess.
Let suppose the conversion equation is $\ce{A + B -> C}$ and the reaction kinetics is of the second order.
Then
$$\frac{\mathrm{d}[C]}{\mathrm{d}t}=k \cdot \ce{[A][B]}$$
$$\frac{\mathrm{d}[A]}{\mathrm{d}t}=-k \cdot \ce{[A][B]}$$
$$\frac{\mathrm{d}[B]}{\mathrm{d}t}=-k \cdot \ce{[A][B]}$$
If the stoichiometric ration is managed, then  \ce{[A] = [B]}
$$\frac{\mathrm{d}[C]}{\mathrm{d}t}=k \cdot \ce{[A]}^2$$
$$\frac{\mathrm{d}[A]}{\mathrm{d}t}=-k \cdot \ce{[A]}^2$$
If we integrate the function over the time:
$$\int_{a_0}^{a}{\frac {\mathrm{d}[A]}{ -k \cdot \ce{[A]}^2 }}=\int_0^{T}{  \mathrm{d}t}$$
$$\frac {1}{k} \cdot [\frac 1A - \frac 1{a_0}]=T$$
$$\frac 1A =kT +  \frac 1{a_0}$$
$$A =\frac 1{kT +  \frac 1{a_0}}$$


text:
Why is propenoic acid more acidic than p-methoxybenzoic acid?
Why is propenoic acid more acidic than p-methoxybenzoic acid?
I looked online for their $\mathrm{p}K_\mathrm{a}$ values and found them to be $4.37$ and $4.25,$ respectively. The $\mathrm{p}K_\mathrm{a}$ of benzoic acid is $4.20.$ Why does the $\ce{-OMe}$ group make such a big effect on $\mathrm{p}K_\mathrm{a}?$

Correcting what is probably a typo, the pKa values I found for aqueous solution (Evans' pKa tables) are 4.25 for propenoic acid, 4.2 for benzoic acid, and 4.47 for p-methoxybenzoic acid. However, even with this increased pKa disparity, if anything it's interesting that the values are all so close. Acid strength can span over 80 orders of magnitude, so a 0.3 difference is almost a rounding error, all things considered.

Consider the electron donation of the OMe group

My question is how would you compare these pka values with centainity during an exam? The values are very close for benzoid and propenoic acid.

Usually, you are only comparing structures of the same class. For example, you would probably just compare the pKa's of propanoic vs. propenoic acid. It would not be reasonable to differentiate between compounds with such similar pKa's on an exam, and I would expect any questions regarding pKa to be more obvious than this.

Why is propenoic acid more acidic than p-methoxybenzoic acid?

This is propenoic acid, $\mathrm{pK_a}=$4.25

This is p-methoxybenzoic acid, $\mathrm{pK_a}=$4.47

An electron withdrawing group increases the acidity of a carboxylic acid. It disperses negative charge by inductive/resonance effect and stabilizes the carboxylate ion. Thus, propenoic acid (pKa 4.25) is more acidic than p-methoxybenzoic acid (pKa 4.47).
However the molecular structures are totally different. It seems to me to be comparing apples and oranges! To me such hand waving comparisons between molecules are only meaningful if the two molecules are similar.

The $\mathrm{p}K_\mathrm{a}$ of benzoic acid is $4.20.$ Why does the $\ce{-OMe}$ group make such a big effect on $\mathrm{p}K_\mathrm{a}?$

This is a more interesting question.
\begin{array}{|c|c|} \hline \hline
\text{acid} & \text{pka}  \\ \hline
\text{2-Methoxybenzoic acid} & 4.09   \\ \hline
\text{3-Methoxybenzoic acid} & 4.10  \\ \hline
\text{Benzoic acid} & 4.19  \\ \hline
\text{4-Methoxybenzoic acid} & 4.50  \\ \hline
\end{array}
So the phenyl group has about the same effect on the acidity of a carboxylic acid as observed with an ethylene group. However the substitution of other function groups, and their position on the ring can increase or decrease the pKa of the acid.
Data sources:
2-Methoxybenzoic acid

3-Methoxybenzoic acid

Benzoic acid

4-Methoxybenzoic acid

I agree on the first part, but are you planning to flesh out this second part? Right now you've just listed the pKa's...


text:
Which will be oxidised between water and Copper?
I have always read that the one having higher  discharge potential undergoes the reaction( oxidation or reduction). But In this example why is there also oxidation of water?

What were the alternatives? (Hint: there were none.)

But i dont get it. Why not just Cu2+ -->Cu.   Why is there the reaction H2O---> H2 if it have lower discharge potential ?

Why not? Because you can't have your hamburger for free. The electrons didn't just appear out of nowhere and reduce copper. They have to be coming from somewhere. Something is going to be losing these electrons. Something has to be oxidized. Just what could that be?

Yes I understand the oxidation of  water to oxygen. But will the reduction of Copper not suppress the reduction of water to Hydrogen gas( as given in solution)? As coppers discharge potential is higher? Thanks!

It will. That is, until all copper is reduced and we have nothing more to reduce but hydrogen.

Copper cannot be oxidized. It can be reduced. And it will be reduced. H2O could be reduced in H2 if the over voltage is high enough, which is difficult to decide, as it depends on the surface of the cathode..

The question is not fully complete (not your mistake but someone who wrote this exercise). I am not surprised that you are confused. The question is indeed silent about the electrodes. You should have asked the question: What is the cathode made of? What is the anode made of before starting the question. The second trick in your question in your question is, which I liked, is that it is an example of exhaustive electrolyte. You passed a huge current which consumed all free metal ions and the solvent can also be reduced or oxidized.
So let us look at your problem:
At the cathode you have two explicit choices: (i) Reduction of Cu ions which are floating in the solution, (ii) Reduction of the solvent, which is water in your case to hydrogen! There is nothing else which can be reduced here as per your question.
At the anode you still have two implicit (hidden) choices: (i) Oxidation of water and (ii) Oxidation of the anode material.
Your solved questions bluntly assumes that the electrode is inert, say it is made of carbon or Pt or gold, whatever but inert. So are left with only one choice. Oxidation of water. However, let us say they actually told you that the anode is made of Copper:
Oxidation potential of Cu
Cu(s) ---------> Cu$^{2+}$(aq) + 2e$^{-1}$    -0.34 V
Oxidation potential of water
2 H$_2$O(l)---------> O$_2$(g) +  4 H$^+$(aq) + 4e$^{-1}$    -1.23 V
Which one do you think will get oxidized? The Cu electrode instead of water!
This is the process and principle behind copper plating.
When your examiners are quiet about electrode materials, assume the solvent is going to oxidize, i.e., water.


text:
Why does the corrosion of a steel nail develop iron ions in the ends of the nail and hydroxide in the middle of the nail?
We did an experiment in school where we put a steel nail in a solution consisting of:

1 g agar;
100 ml water;
100 ml indicator solution (which gets blue whenever $\ce{Fe^2+}$ is created and red when $\ce{OH-}$ develops).

We let it rest for a few days and this is what happened to the nail:

My hypothesis was that the corrosion would create two ends of the nail where reduction takes place in one end and oxidation on the other end. Why isn't this the case?
My teacher gave us a hint and told us it could have something to do with both:

The length of the nail;
The ion concentration of solution.

Cool observation! The two ends of your nail are pretty much identical, chemically. If anything, the middle is different from the ends. As you have seen.  Can you imagine what that might be? Think about how (in what steps) a nail is made!

Is the nail surface uniform? Before putting it in the agar, did you completely remove contaminants from the nail's surface?

@Karl They're the same right? What am I missing here?

@DrMoishePippik no we did not

Very good question and thought provoking too. It appears that the nail somehow knows or is it is able to decide which part will be a cathode and which part will be an anode! Why is that so? Why one end was not a cathode and the other end an anode, just the one shown in the cartoon taken from here Imaging metal corrosion

So the fact you should keep in mind is that corrosion is superficially a random process just like cancer. Corrosion comes from corrodere "to gnaw to bits, wear away," it can start from any defect on the metal (scratch, impurity, etc.). However once it starts, just like cancer, it decides its propagation (i.e., the flow of electrons), hence the direction of cathode or an anode. It is likely that in your case the nail tip and the head had more defects than the center, so corrosion started at the tips. The pattern you are seeing is not universal or a fundamentally symmetric phenomenon.
Let us now bend a nail, create stress or defect in the center. Now the corrosion also starts in the center. Have a look at this gallery. Corrosion gallery


text:
Why is gas pressure measured in mmHg?
Pressure can be measured in $\pu{mmHg}$ which is equal to the pressure exerted by a “column of mercury” of a height expressed in $\pu{mm},$ at $\pu{0 °C}$ under Earth’s normal gravity.
Does the diameter of a tube that contains the mercury affect how high the mercury will rise? Why isn’t pressure measured in $\pu{mmHg^3}$ or some other unit of volume of mercury?

The diameter does not matter, which also answers your second question. That's just how pressure works.

Mercury barometersare typically scaled in millimeters, mm, which refers to the height of the column of mercury. The height of the column not only depends on the air pressure but also gravity. Here in US it is common for the barometric pressure to be reported in inches. Old timers like me also know "mm Hg"  as the unit "torr." The pascal (symbol: Pa) is the SI derived unit of pressure which is now used by scientists.

Indeed, the diameter and the geometry of the tube are irrelevant. You might want to read about hydrostatic pressure. And this isn't really a chemistry question.

To go off on a tangent, it would absolutely be possible to have the scale on a
mercury barometer to be marked in pascals. However I'd guess that
mercury barometers would be rarely produced now given concerns over the chemical hazard that mercury has. I'd guess that labs now have an aneroid barometer instead.

Pressure=Force/Area....thus, diameter doesn't matter.

@andselisk I suspect if there's no space between mm and Hg.

@Zenix Please double-check authoritative sources before editing. AMA Manual of Style leans toward "mm Hg". However, SI brochure, IUPAC Green Book and ACS Style Guide suggest to avoid space and use "mmHg". Since Chemistry.SE tends to adhere to the international standards, we go with the latter. And neither "$\pu{mm^3Hg}$" nor "$\mathrm{0°C}$" was correct, which you didn't address in your edit.

@Zenix Also, I don't think the [tag:atmospheric-chemistry] tag is relevant here. Hence, it was easier to rollback your edit and do the corrections from scratch.

Consider diving in the water. For every ten meters, pressure increases by about one atmosphere (~100 kPa). Would it make any difference in pressure if one were to dive in a swimming pool or in a large lake? Ignoring the greater density due to dissolved salts, what about in the Pacific Ocean?

Indeed, I see my misunderstanding. Makes a lot of sense.

@andselisk OP wrote mm^3Hg, not me. Also I didn't put any space between 0 and$^{\circ}\mathrm{C}$, not 0° C. I wasn't aware that mmHg is "more correct notation" I am used to mm Hg. Does it really matter to me, if you rolledback my edit or not?

@Zenix I just pointed out what else could be improved, not your faults, and no, the rollback is merely an event in editing history and doesn't affect your reputation in any way:)

History and because of how pressure works
The original reason why mm of Hg are used is because of how chemists first started to measure and understand atmospheric pressure. The idea of the vacuum and atmospheric pressure is surprisingly new. Otto von Guericke demonstrated the power of atmospheric pressure by showing that two horses could not separate a brass sphere containing a partial vacuum only in the 17th century (remember a vacuum doesn't suck the atmosphere pushes).
This kicked off a vast amount of experimentation. Around the same time scientists like Torricelli and Pascal realised that they could measure atmospheric pressure by filling a closed tube with mercury and inverting it in a bath of mercury (you can do this with any fluid but mercury is about 13 times denser than water so the tube only needs to be a meter or so long; with water it needs to be ~10m long which is an inconvenient length in an indoor laboratory). The height of the mercury in the tube is a measure of the external pressure as a vacuum is formed above it in the closed tube. This simple instrument was used by a friend of Pascal to show that atmospheric pressure drops with height when he took a barometer up France's Puy de Dome mountain.
The diameter of the tube is irrelevant as pressure is weight/area (do the calculations and tube diameter cancels out).
We now call a tube like this a barometer. And the height the mercury rises to is a convenient measure of atmospheric pressure and can be read off using a simple ruler. We could use mm of water, but would need a much longer tube and an unusually inconveniently sized ruler. Hence why mercury is the fluid and a convenient way of measuring pressure is mm of Hg.
We tend to use modern SI units for pressure these days, not least because the wide use of mercury has been banned and calculations in consistent nits are far easier.


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Where is Al2O3 in Al2(SO4)3 coming from?
let me know where is Al2O3 in Al2(SO4)3 containing from? as we know there is 17-18% Al2O3 in commonly Al2(SO4)3 product.
is it because low convertion of reaction Al2O3 + 3H2SO4 = Al2(SO4)3 + 3H2O, or something?
if i using Al(OH)3 to convert Al2(SO4)3, 2Al(OH)3 + 3H2SO4 = Al2(SO4)3 + 6H2O, is it still happen 17-18% containing Al2O3 in Al2(SO3)3 because dehydration reaction of 2Al(OH)3 = Al2O3 + 3H2O?

Writing "Al2O3 in Al2(SO4)3" makes no sense; a chemical formula is used to denote the exact stoichiometric composition of a given compound, and these are no matryoshka dolls. You might want to re-formulate the question referring to the content of $\ce{Al2O3}$ in a certain ore/clinker/schist/clay/mineral by adding corresponding details.

@andselisk, most likely this is being taught in some South Asian school/university. It is an old way of quoting or reporting analysis or assay of salts etc.

@Fahmi Reza  Contains is not the proper term. Rather is equivalent.  100g of $\ce{Al2(SO4)3}$ is equivalent about 17g of $\ce{Al2O3}$, in ratio of their respective molar masses.

You are referring to very old analytical methods and analysis. In older times, many analyses were reported as oxides. This practice comes from gravimetric analysis. This practice continues today in fertilizers analysis and in some other industries. Coming to your question, there is no Al$_2$O$_3$ unit per se however the purity of aluminum sulfate may be quoted as aluminum oxide. Just do a simple mole calculation using the relation to determine the actual purity of aluminium sulfate. Just keep in mind, quoting percentage of oxides is purely historical, but you can easily do calculations as follows.
1 mol Al$_2$O$_3$ $\equiv$ 1 mol Al$_2$SO$_4$ $\equiv$ 1 mol Al$_2$SO$_4$. $x$H$_2$O

should be $\pu{1 mole }\ce{Al2O3} \equiv  \pu{1 mole }\ce{Al2(SO4)3} \equiv \pu{1 mole }\ce{Al2(SO4)3} \cdot x \ce{H2O}$

Is there a difference between a kilogram and a kg? "mol" is a symbol for amount of substance as one the SI base quantities. https://physics.nist.gov/cuu/Units/units.html

In some complicated modern technologies, like cement, it gets even more distant from textbook chemistry. Tricalcium silicate, the major ingredient in cement, is Ca3SiO5 (alite), but it's usually written C3S, a shortcut. Then consider one of the products, ettringite: Ca6Al2(SO4)3(OH)12·26H2O https://en.wikipedia.org/wiki/Ettringite. We simplify writing it by using three "languages": C3A + 3 CaSO4 → ettringite (cement shorthand, chemical symbols, and English words), and even the chemical symbol isn't exactly accurate, because the CaSO4 is usually a hemihydrate plus dihydrate.


text:
How is the amino group basic?
According to the Khan Academy, the amino functional group is basic because it can remove $\ce{H+}$ from systems. Why is that?

Amino: Charged (forms $\ce{R-NH3+})$ at the pH of most biological systems. Since amino groups can remove $\ce{H+}$ from solution, they are considered basic.

My current understanding:

The $\ce{H+}$ atoms bond in a polar covalent way with $\ce{N-}$. So that means the amino acid does not release that many $\ce{H+}$ ions. I can see why this is not acidic.
$\ce{N-}$ also takes electrons from $\ce{R}$, which makes $\ce{R}$ become $\ce{R+}$. This should repel $\ce{H+}$ from elsewhere in solutions.

How does this functional group remove $\ce{H+}$ from its solutions?

R-NH2 + H+ -> [R-NH3]+

Right! Totally forgot that N requires 3 more electrons to be stable, not 2. Thanks! But what, then bonds the N to the R?

You have that on your own picture. Amin groups are bonded to R. There is equilibrium $\ce{R-NH2 + H2O <=> R-NH3+ + OH-}$ or $\ce{R-NH2 + H3O+ <=> R-NH3+ + H2O}$ It is the same as with ammonia.

Polarity and elementary electrostatics have nothing to do with acid-base definitions. Sometimes we try to find a reason, using our current understanding, and that ensues the danger of over-rationalization. At a truly fundamental level perhaps no chemist can tell you why an amine can accept another proton, may be someone can do quantum mechanical calculations and say why. So avoid over-rationalization... accept some observations as facts (temporarily until we develop a full understanding).
One can titrate an amine with an acid, by this experimentally observed fact, exactly just like we titrate NaOH with HCl. So an amine group is indeed a base or say it behaves like base. The acidic proton cannot annihilate itself during a titration, it has to go somewhere. The amine group gets protonated, in return.

Is there no prior concept at all in chemistry that explains why this is basic? I have no idea why, for example, some atoms are stable with 2 shells of electrons, but taking that as a given, I can see why an oxygen atom wants 2 more. What is the given in this case?

Perhaps a theoretical chemist can tell you a fundamental reason. I am not one of them. My understanding is basically from an experimental perspective. Amines can be titrated with acids, and some amines can turn red litmus blue.

According to Lewis acid-base concept, a molecule with central atom having lone pair can act as proton acceptor. In $\ce{R-NH2}$ there are lone pairs over $\ce{N}$ atom, which can combine with $\ce{H+}$ ions.
This can also be explained by Brønsted–Lowry concept, which states if an atom  accepts protons (like above), it is a base.

The latter came as an anhancement of the original Arrhenius model where acids release H+ and bases OH-.  But is does not apply to ammnia/amines, while the B-L model does.


text:
Isotopic profiling
If a molecule contains 12 carbon atoms and 2 nitrogens, what's the probability that it contains precisely 1 carbon-13 and precisely 1 nitrogen-15?
my working: 1.1(12)(0.36)(2) = 9.5 %
where 1.1 is the natural abundance (%) of carbon-13 and 0.36 is the natural abundance (%) of nitrogen-15
This seems too high. What is the correct probability calculation and why?

You can see the issue with your approach by considering what probability you would get of having exactly 1 carbon-13 in a chain of 100 carbon atoms. With you method, you would get over 100%, which tells you that your missing a piece that will keep your probability bounded between 0 and 1.

Yes, 9.5% is much to high a value.
Let's break the problem into two parts.

$p_\mathrm{C}$ will be the probability of getting exactly one $\ce{^{13}C}$ atom in a molecule with 12 carbon atoms.
$p_\mathrm{N}$ will be the probability of getting exactly one $\ce{^{15}N}$ atom in a molecule with 2 nitrogen atoms.
$p$, the probability sought will be $p = p_\mathrm{C}\times p_\mathrm{N}$

For carbon
The probability of getting one $\ce{^{13}C}$ atom when selecting the first carbon is 1.1% or 0.011.
The probability of not getting any $\ce{^{13}C}$ atom when selecting eleven additional carbon atoms is $(1-0.011)^{11} = 0.8854$.
So the probability of having the first carbon atom be a $\ce{^{13}C}$ atom and the next eleven atoms not being a $\ce{^{13}C}$ atom is $0.011\times 0.8854 = 0.009740$
But the $\ce{^{13}C}$ atom could have been chosen in any one of the 12 positions. So
$p_\mathrm{C} = 12\times 0.009740 = 0.1169 $
For nitrogen
The probability of getting one $\ce{^{15}N}$ atom when selecting the first nitrogen is 0.36 % or 0.0036.
The probability of not getting a $\ce{^{15}N}$ atom when selecting one additional carbon atoms is (1−0.0036)=0.9964
So the probability of having the first nitrogen atom be a $\ce{^{15}N}$
atom and the next atom not being a $\ce{^{15}N}$ atom is:
$0.0036\times 0.9964  = 0.003587$
But the $\ce{^{15}N}$ atom could be the first or second so the value needs to be multiplied by 2. So
$p_N =2 \times 0.003587=0.007174$
Finally $p = p_C \times p_N = 0.1169 \times 0.007174 = 0.0008386 = 0.084\%$

BONUS
Figuring out the number of combinations with just one $\ce{^{13}C}$ atom and the others being non-$\ce{^{13}C}$ atoms was easy. However listing all the carbon possibilities when there are  four $\ce{^{13}C}$ atoms and six non-$\ce{^{13}C}$ atoms can get tricky. Fortunately there a mathematical method call Combinatorics which allows the answer to be found easily.
$\large\binom{n}{m} = \small\dfrac{n!}{(n-m)!m!}$ is read as "n chose m."
So for for four $\ce{^{13}C}$ atoms and six non-$\ce{^{13}C}$ atoms we'd have:
$\large\binom{10}{4} = \small\dfrac{10!}{(10-4)!4!} = 210$ possible arrangements.

V clear answer, but you seem to have chosen 10 C atoms instead of the 12 in the question,

@porphyrin - Thanks for point out my stupid mistake!

no problem I make them all the time :)

I should vote to close this question as off-topic because it's really about mathematics :-)

@BuckThorn , maybe so, but a number of non-mathematical chemists will come to chemistry stackexchange and find this answer very useful as it's not only a common problem encountered in chemistry, but also explains the maths at a level that is easily understood by a chemist. maths stackexchange would likely either dismiss this as 'trivial' or give an overly complicated answer.


text:
Why isn't protium nucleus considered most stable nucleus?
My chemistry teacher told me that $\ce{^{56}Fe}$ is the most stable nucleus because of the highest binding energy per nucleon.
But if we consider a protium nucleus, it has only 1 proton so there is no chance of separation.
so why isn't protium considered the most stable nucleus?

Will answer properly later if nobody else does, but firstly the proton is theorised to decay https://en.wikipedia.org/wiki/Proton_decay, and secondly read https://en.wikipedia.org/wiki/Nuclear_binding_energy - nuclear particles bind together to form lower energy states in a not totally dissimilar way to how atoms bind together to make molecules. There's probably a duplicate here as well

When he made that statement he was certaintly thinking of nuclei with A > 1, making the interesting point that binding energy per nucleon peaks at Fe-56 (which is important for stellar evolution).  It's obviously not meaningful to talk about binding energy/nucleon for a proton.  As far as Fe56's vs. a proton's comparative stability, Ian's link shows the proton's estimated half-life is 10^34 years!  I don't know if the nuclear binding energy in Fe56 adds to the stability of the proton, nor whether there are any estimates of the half life of the Fe56 nucleus as a whole.

The stability is meant not as lack of self decay here, but disability to reach lower energy. Analogy:  Nuclei are like pits and nucleons are like tiny balls in a children game. The 56Fe is the deepest pit for the balls/nucleons, while a proton/ball lays on the flat surface. It is conditionally stable, until it falls into a pit/nuclei, where it is more stable ( not in all cases ), as it cannot jump back up.

Don't forget that "binding energy" is relative. Say the binding energy per nucleon of an isolated proton is 0, then the binding energy per nucleon of an iron nucleus is lower than zero (hence nuclear processes in stars can release energy by banging protons together to make heavier nuclei like iron (it is way more complex than this in reality but this view captures the essential point).


text:
Molecular Geometry of IF₄⁺
Recently I came across a question asking for the geometry of the aforementioned molecule.
The answer key claimed the shape as an irregular tetrahedron, but as per my knowledge, it should have been like a see-saw.
What am I missing here?

Yes it is seasaw. The charge plays a apart so that may be the missing link.

apply hyrbridisation theory to see geometry of IF4


text:
hydroxychloroquine (HCQ) Synthesis: Which mechanism?

Draw the product of the following reaction and provide an explanation for this reaction knowing that $\mathrm{ROH}$ or $\mathrm{RNH_2}$ +  $\mathrm{PhCl}$ do not react at all.

I have no doubt that the answer is hydroxychloroquine (HCQ), but how would it be formed if the reaction:
$\mathrm{RNH_2}$ +  $\mathrm{PhCl}$
does not happen? Or, is the question hinting that the reaction takes the $\mathrm{S_N1}$ pathway where the phenylcation gets formed? If so, why does the Cl atom para to the nitrogen leave more readily than the other Chloride?

The 2 and 4 positions of quinolines are similar to the 2 and 4 positions of pyridine in reactivity. 2 and 4 haloquinolines/pyridines are highly reactive to nucleophiles by SNAr. Indeed 4-Cl-pyridine has to be handled as the hydrochloride because of its propensity to self condense.
The 7 position, by contrast, is unactivated and similar in reactivity to benzenes, so the 7-Cl is not going to undergo SNAr without specialist catalysis.


text:
Is this galvanic cell representation right?
Question

13. In the galvanic cell:
$$\ce{Pt(s) | I2(g) | I-(aq) || Fe^3+(aq) | Fe^2+(aq) | Pt(s)}$$

Representation of anode is incorrect and cell will not work.
$[\ce{Fe^3+}] = [\ce{Fe^2+}] = [\ce{I-}] = \pu{1 M}$ is sufficient for $E_\mathrm{cell} = E_\mathrm{cell}^\circ.$
$\ce{I-}$ gets oxidized to $\ce{I2}$ and $\ce{Fe^3+}$ gets reduced to $\ce{Fe^2+}.$
None of these.

Given Answer

 3. $\ce{I-}$ gets oxidized to $\ce{I2}$ and $\ce{Fe^3+}$ gets reduced to $\ce{Fe^2+}.$

My Thoughts
The representation of anodic half confuses me. I am unsure whether it is right representation. According to me, it should be $\ce{I-} | \ce{I2}$ (anodic half). It seems to me that answer should be 1.

Both 1 and 3 seems correct..

But wouldn't both answers collide with each other?

The chapter 1 of the reference 1 describes the Line Notation for Galvanic Cells as follows:

A shorthand way of completely describing an electrochemical cell such as Galvanic cell is called line notation. It is a shorthand way of expressing the redox reaction progressing in the electrochemical cell. In this notation:

A slash (/) or a vertical line (|) represents a phase boundary, and a comma separates two components in the same phase. A double slash (//) or a double vertical line (||) represents a phase boundary whose potential is regarded as a negligible component of the overall cell potential (e.g., salt bridge).
When a gaseous phase is involved, it is written adjacent to its corresponding conducting element.
The constituents of the cathode on the right and the constituents of the anode on the left.
The phases of all reactive species are listed and their concentrations or pressures are given if those species are not in their standard states (i.e. $\pu{1 bar}$ for gasses and $\pu{1M}$ for solutions).

Thus, in the given cathode representation, both $\ce{Fe^3+ (aq)}$ and $\ce{Fe^2+ (aq)}$ are separated by a vertical line (|), which is an incorrect representation, according to the description (1) above. They should be actually separated by a comma (as $\ce{Fe^3+ (aq)}$, $\ce{Fe^2+ (aq)}$) because there is no phase separation between them (both are in same phase).
Also, according to the description (2) above, $\ce{I2 (g)}$ cannot be introduced into $\ce{Pt (s)}$ anode (as indicated in the representation) because $\ce{I2 (g)}$ is the product of the reaction, not a reactant.
Based on these, given answers (1) and (3) should be correct. Also, OP's representation of 'anode half' is also correct since there is phase separation.
References:

Allen J. Bard, Larry R. Faulkner, In Electrochemical Methods: Fundamentals and Applications; Second Edition, John Wiley & Sons, Inc.: New York, NY, 2001 (ISBN 0-471-04372-9).


text:
Do the concepts of Newtonian and non-Newtonian fluids still hold at a nanoscopic level?
Recently I've spent a great deal of time thinking about enzyme diffusion in cytoplasm. One thing that keeps me awake at night is the constant referral to cytoplasm as a non-Newtonian fluid (which it surely is when looking at it as a whole), and the implications this has for the diffusive properties of singular enzymes.
If I think about a single enzyme in the cytoplasm with its immediate surroundings, I would expect to find the protein with its solvation shells, and only a few non-water components. In this context I'm unable to understand how this single enzyme could be thought to be in a non-Newtonian medium.
Furthermore, when thinking about the diffusion of the enzyme, quite a few authors refer to the Scallop theorem, and define it in the context of non-Newtonian fluids. If the enzyme is mostly surrounded by water molecules, and only occasionally runs in to other proteins or peptides, shouldn't the protein be considered to reside in a Newtonian medium?
Shortly put: How far down can you go until the classification between Newtonian and non-Newtonian fluids starts to break down?

I'd go as far as to say that at a nanoscopic level there are no fluids, just atoms.

Generally Fick's laws of diffusion have been found to apply pretty well to molecular translational and rotational diffusion in solution. This means that we assume that the solvent is a continuous structureless fluid, which in practice means that the diffusing molecule should be far bigger than that of the solvent so that the solvent's molecular nature can be ignored. Surprisingly this works well with 'slip' and 'stick' boundary conditions down to situations where the solute is not that much bigger than the solvent, e.g.  rhodamine and similar dye molecules in solvents such as acetonitrile or ethanol. In your example of a protein in water then you might expect that it would behave 'normally' as just described.
However, in the cytoplasm the close packing of many similar sized and large molecules means that these can easily block the diffusion of any other (except for far, far smaller ones such as water) and so the situation is totally different and will be more like a percolation process where a slight reduction in packing can lead to a large, almost stepwise increase in diffusion coefficient. The idea of a single diffusion coefficient is therefore not helpful.


text:
How is the change in Gibb's free energy a condition for spontaneity for reactions not happening under constant pressure conditions?
According to what I have studied, the change in Gibb's free energy is only a condition for spontaneity for processes happening at constant pressure and temperature conditions.
Consider a reversible reaction happening in a closed vessel at a constant temperature.
We check the feasibility of the reaction at any instant using the following equation:
$\Delta G = \Delta G^{.} + RT\ln(Q)  $
As the reaction happens, the partial pressures of each gaseous component will keep on changing as well as the total pressure. Then how can we use Gibb's free energy for checking spontaneity?

Why would it be a problem for partial pressures to change?

What about Helmholtz energy A = U - TS ?

Gibbs energy have an another equation which is:-

So according to this equation spontaneity of a reaction depends both on the enthalpy change as well as entropy change.
We all know that a system attains stability as the system's energy is lowered that is it has to release energy. If in a reaction energy is released enthalpy change is negative.
We also know that a reaction is favoured in the direction of increased chaos or a positive entropy change.
So from this we can see that Enthalpy change is negative and entropy change is positive for a spontaneous reaction to take place and the overall equation of
becomes negative that is Gibbs free energy must always be negative for a spontaneous reaction.....Can be applied to any reactions.
Hope this helps.


text:
Writing a electrochemical cell representation in correct way
The answers to following question seem wrong to me. The books has plainly told correct answer is (3), but it seems incorrect to me. Please help.

The chemical reaction $\ce{2AgCl(s) + H2(g) -> 2 HCl(aq) + 2 Ag(s)}$ taking place in a galvanic cell (under standard condition) is represented by the notation
$$\ce{Pt(s) | H2(g), \pu{1 bar} |  1 M KCl(aq) | AgCl(s) | Ag(s)}\tag{1}$$
$$\ce{Pt(s) | H2(g), \pu{1 bar} |  1 M HCl(aq) | 1 M Ag+(aq) | Ag(s)}\tag{2}$$
$$\ce{Pt(s) | H2(g), \pu{1 bar} ||  1 M HCl(aq) | AgCl(s) | Ag(s)}\tag{3}$$
$$\ce{Pt(s) | H2(g), \pu{1 bar} |  1 M HCl(aq)  | Ag(s) | AgCl(s)}\tag{4}$$

Photo of problem: https://i.stack.imgur.com/Oflyg.jpg

3 seems blatantly wrong. Don’t think you would even need a salt bridge in this reaction.

It would be interesting to hear why it seems wrong to you. If there were a correct answer choice, this question would have been closed as homework.

You have developed not very welcome custom to post photos instead of elaborated text of the task question and your explicitly done effort. It may be considered as rather lazy part on your side, putting the burden on shoulders of responders to explicitly type the relevant text referring to you question. Additionally, the photo content cannot be indexed for site searching. It increases chances for the question to be closed for lack of the OP effort.

The chemical reaction is:
$$\ce{2AgCl(s) + H2(g) -> 2 HCl(aq) + 2 Ag(s)}$$
If we split this into half reactions, we get an oxidation half reaction at the anode of:
$$\ce{H2(g) -> 2 H+(aq) + 2 e-}$$
And a reduction half reaction at the cathode of:
$$\ce{2AgCl(s) + 2e- -> 2 Ag(s) + 2 Cl-(aq)}$$
If you want all the species at standard state, you need a partial pressure of $\pu{1 bar}$ of hydrogen gas, a $\mathrm{pH}$ of about 0, and a chloride concentration of $\pu{1 mol/L}$.
In cells where the reagents can't react directly, you don't need a salt bridge. In the reaction we are looking at, $\ce{AgCl}$ is confined to the cathode because it sticks to it, and hydrogen gas is confined to the anode because that's where we are releasing it, and it has low solubility in water.
The cell notation starts with the conductor of the anode half reaction, and ends with the conductor of the cathode half reaction. Different phases are separated by vertical lines. If there is more than one species in a phase, they are separated by commas.
$$\ce{Pt(s) | H2(g, \pu{1 bar}) |  H+(aq, \pu{1 mol/L}) , Cl- (aq, \pu{1 mol/L}) | AgCl(s) | Ag(s)}$$

The books has plainly told 3. to be correct.

The answer is (5) none of the above. Using the format of the answer key, the best answer would be:
$$\ce{Pt(s) | H2(g), \pu{1 bar} |  1 M HCl(aq) | AgCl(s) | Ag(s)}$$
You get this answer from answer 3 by changing the salt-bridge boundary ("||") to a phase boundary ("|"). Having a salt bridge between the hydrogen gas and the aqueous hydrogen ions would mean that you are generating hydrogen ions at the electrode where they are not in contact with $\pu{1 M}$ hydrogen ions.

Just one more doubt. If I represent it as : Pt(s) | 1M HCL(aq) |H2(g)| Agcl(s)|Ag(s) would it be right?

@Adityasuresh No, because the hydrogen gas is not in contact with the silver chloride at the other electrode. The hydrochloric acid, on the other hand, is in contact with both electrodes.

This question is very similar to the question I nave answered recently: Is this galvanic cell representation right?. I like the answer given by Karsten Theis elsewhere, but it seems OP still did not understand the line notation for electrochemical cells. Thus, I like to give the same description here to clear out OP's doubts.
The chapter 1 of the reference 1 describes the Line Notation for Galvanic Cells as follows:

A shorthand way of completely describing an electrochemical cell such as Galvanic cell is called line notation. It is a shorthand way of expressing the redox reaction progressing in the electrochemical cell. In this notation:

A slash (/) or a vertical line (|) represents a phase boundary, and a comma separates two components in the same phase. A double slash (//) or a double vertical line (||) represents a phase boundary whose potential is regarded as a negligible component of the overall cell potential (e.g., salt bridge).
When a gaseous phase is involved, it is written adjacent to its corresponding conducting element.
The constituents of the cathode on the right and the constituents of the anode on the left.
The phases of all reactive species are listed and their concentrations or pressures are given if those species are not in their standard states (i.e. $\pu{1 bar}$ for gasses and $\pu{1M}$ for solutions).

Thus, the line notation for the given Galvanic cell (under stranded conditions) for the reaction, $\ce{2AgCl(s) + H2(g) -> 2 HCl(aq) + 2 Ag(s)}$ (as correctly given by Karsten Theis) is:
$$\ce{Pt(s) | H2(g, \pu{1 bar}) |  H+(aq, \pu{1 mol/L}) , Cl- (aq, \pu{1 mol/L}) | AgCl(s) | Ag(s)}$$
Any changes to this notation other than combining $\ce{H+ (aq)}$ and $\ce{Cl- (aq)}$ to $\ce{HCl (aq)}$ would be given you a incorrect notation for the given cell reaction.
Note: Basically, all 4 answers are wrong in this measure.
References:

Allen J. Bard, Larry R. Faulkner, In Electrochemical Methods: Fundamentals and Applications; Second Edition, John Wiley & Sons, Inc.: New York, NY, 2001 (ISBN 0-471-04372-9).

So 1 atm as standard state is incorrect, and 1 bar is correct? Oh, I just looked it up, it changed from 1 atm to 1 bar in 1982 according to Wikipedia...

I'm sorry for my sloppyness. It should be $\pu{1 atm}$, according to the book I referenced (which is 2001 book).. I also confused sometimes on the difference between $\pu{atm}$ and $\pu{bar}$ because it is only 0.01! :-) I believe correct standard condition is $\pu{1 atm}$.

I think the book is behind the times. The update was from atm to bar, so they should have bar, at least in the second edition.

It was the second edition. Still have atm.

Clarification: Your first comment is correct. In 1982 it has been changed to $\pu{10^5 Pa}$ ($\pu{1 Bar}$) by IUPAC. However, the difference is so small that most scientist seemingly have ignored the  change. For example, according to the same Wikipedia article, NIST uses a temperature of $\pu{20 ^\circ C}$ ($\pu{293.15 K}$ or $\pu{68 ^\circ F}$) and an absolute pressure of  $\pu{1 atm}$. Yet,  $\pu{1 bar}$ is IUPAC accepted STP condition internationally.


text:
What is negative adsorption?
In adsorption, the concentration of adsorbate is greater on the surface. So, how is it possible that the solvent will be more on the surface and the solute (adsorbate) in the bulk… or does solute not always mean adsorbate?

Solute does not mean adsorbate always.An Adsorbate is a substance that is adsorbed  on a surface of another substance. In a dilute solution, negative adsorption occurs as solvent is adsorbed by the adsorbent and not solute. In positive adsorption, solute is adsorbed.

Per a source:

If the concentration of the adsorbate is less than its concentration in the bulk then it is called negative adsorption. Example: When a dilute solution of KCl is shaken with blood charcoal, it shows negative adsorption.

The reference link also provides an educational video discussing also positive adsorption and related topics.


text:
Crystal field splitting of other orbitals
According to crystal field theory (CFT), $\mathrm{d}$-orbitals of central metal atom splits due to non uniform repulsion of ligands around the metal atom. Why isn't $\mathrm{p}$-orbital splitting discussed?
Argument: As in case of square planar complexes $\mathrm{(dsp^2)},$ $\mathrm{p}$-orbitals are involved which are facing different conditions of repulsion. So, $\mathrm{p}_{xy}$ must have higher energy than $\mathrm{p}_{xz}$ or $\mathrm{p}_{yz}.$ This happens in case of trigonal pyramidal structure too.

@andselisk is there any page that tells what should be italicised, what not be?

@Zenix Sure there is. First, I already suggested authoritative sources in my previous comment. Second, have a look at the short version on Meta: Which symbols are written in roman (upright) font and which are italicized?

According to Crystal Field Theory, d-orbitals of central metal atom splits due to non uniform repulsion of ligands around the metal atom. Why isn't p-orbital splitting discussed?

To be blunt, the reason nobody mentions p-orbital splitting or non-splitting is that neither the teacher nor the learner has ever understood the crystal field theory properly in its true depth (including myself).  They just propagate what they saw in standard textbooks, repeat the same in class, the students pass never to see CFT again. The story ends. If that student ever becomes a teacher, the story continues.
You are mixing apples and oranges. Crystal field theory has nothing to do with hybridization. The original work of Hans Bethe, the man behind crystal field theory, wrote a 72-paged highly theoretical paper in Annalen der Physik in German. The translations are available and I quote the abstract where he does clarify what happens to p-orbitals. How many can claim that that they read this paper completely before teaching CFT. I cannot. The abstract clarifies your misconception.

The influence of an electric field of prescribed symmetry (crystalline field) on an atom is treated wave - mechanically . The terms of the atom split up in a way that depends on the symmetry of the field and on the angular momentum I (or J) of the atom . No splitting of s terms occurs , and p terms are not split up in fields of cubic symmetry . For the case in which the individual electrons of the atom can he treated separately ( interaction inside the atom turned off) the elgenfunctions of zeroth approximation are stated for every term in the crystal; from these there follows a concentration of the electron density along the symmetry axes of the crystal which is characteristic of the term. - The magnitude of the term splitting is of the order of some hundreds of cm-1. - For tetragonal symmetry, a quantitative measure of the departure from cubic symmetry can be defined, which determines uniquely the most stable arrangement of electrons In the crystal.

Nobody ever needs to read about CFT ever again! It is outdated and deprecated. It should not be taught anymore, especially not in the truncated form that is so popular. The successor, ligand field theory, is somewhat easier understood and more robust.

You keep telling me that I am outdated! All right, I am, and so are school textbooks. But these concepts have their importance too. My textbook has mentioned about LFT, but not explained it. Can you add a link that explains it sufficiently enough?

CFT is a part and parcel of most undergraduate chemistry curricula $around$ the world - Martin. Pray tell me which university has completely outcast it?

@B.Anshuman, There is no need to feel outdated. If a grade 1 students wants to become a doctor, nobody will hand him a medical textbook. Things have to be developed gradually in an educational setting and the knowledge of some previous theories or older ideas is essential. It prevents re-invention of the wheel.

This is true., can you put in a link to any explanation of LFT which explains it sufficiently enough?

I wasn't mixing oranges and apples. I just wanted to mention square planar has dsp2 configuration according to hybridisation, which in turn contains 2 p orbital. I just wanted to mention the use of p orbital there.

Ligand field theory $depends$ on crystal field theory or uses the knowledge of the principles developed in CFT. Focus on CFT first and then move on the ligand field theory. One thing at a time. I said mixing of apples and oranges, there is no need to mix two independent ideas of hybridization, with CFT. The person who developed it (Hans Bethe), did not need the ideas to invoke hybridization.

One more thing, Martin had commented on another question regarding my outdated concepts, that's why I said so.

Crystal field theory is to me a good rule of thumb for how orbitals are split in a complex.  When it is time to render precise predictions with this splitting then we go to its successores


text:
Measuring ions through color change during electrolysis
I have very little experience in chemistry, you might consider me a physicist, and I am mostly not even sure where to find the information I need. I will start off by describing my problem, I have a nutrient solution (for plants) and I want to measure the different nutrients using electrolysis based on the color change. I have read that if a Copper anode is used the solution will turn bluish and if ammonia is present it will turn deep blue.
Now this is great and all, but I have 18 other elements/compounds I want to be able to measure and I have no idea how to figure out what materials to use for the anode/cathode to get color shifts which are dependent upon the different compounds found in the nutrient solution. How do you chemists go about finding out such things, do I simply have to test a bunch of it for myself or do you have a secret lair of wisdom somewhere?
I'm mostly asking for guidance in information seeking here, but if anyone is interested in the specific ions present in the solution they can be found on Wikipedia: https://en.wikipedia.org/wiki/Plant_nutrition

The problem is that nutrient solutions for plants may contain a lot of different ions. But their concentrations is rather low. I am afraid the usual color tests for detecting copper ions or any other metallic ions will not be sensitive enough to react with very low concentrations. Or if they do, they are difficult to find for ordinary people. For example, for detecting Manganese, you need formaldoxime. To detect lead in low concentration, you need diphenylthiocarbazone. To detect Aluminum, you need Alizarine Red S. To detect Zinc, you need diéthyldithiocarbamate. How and where to get them ?

@Maurice Would boiling off the water increase the concentration to measurable levels or do the ions escape too? I am willing to skip measuring some nutrients (especially micro nutrients) if it's very difficult to get chemicals needed to detect them.

@Maurice Basically, what levels are detectable? I'm guessing that percentages are detectable, but how about say 100 parts per million?

In general colorimetric methods are not super specific to a particular element. A second problem will be the low levels of some of the analytes. // Are you doing the plant growth hydroponically? Plants would extract some the nutrients from soil without the nutrients being in "solution."

Frankly if I wanted to measure a bunch of elements in a liquid sample I'd look at something like Inductively Coupled Plasma-Optical Emission Spectrometry.

@Beacon of Wierd. Formaldoxime reacts with Mn2+ up to 2 microgram or 40 ppm. Only Cu interferes. Diphenylthiocarbazone (or dithizone) detects Pb up to 0.05 ppm. Zn,Cu interfer, but may be eliminated by complex formation. Aluminum may be detected up to 0.2 ppm with Sodium Alizarinesulfonate (Alizarine Red S).These informations, ans many others may be found in books like Fundamentals of Analytical Chemistry, 7th ed, by Douglas A, Skoog, F. James Holler Donald M. West, Saunders College, 1991

@MaxW How small concentrations is that type of spectrometer able to measure? I'm also constrained to a budget, this is a hobby project so I won't be spending 1000's of dollars on it, but playing around with high voltage and plasma does raise my interest :)


text:
How to determine which compound will get oxidised and which will be reduced in a cross cannizzaro reaction?
How to determine which compound will get oxidised and which will  be reduced in a cross cannizzaro reaction?
As a general rule of thumb, I have always been told that Formaldehyde is always oxidised, and the other is reduced.
But no general principle was ever given to me. What factors (eg electronic, steric, etc) affect this ??
There is this question already here, but the both the answers seems to be contradictory. @Jan says , the one with more hydrate forming tendancy oxidised , whereas
@RE60K says less Nucleophilic aldehyde oxidises (which I believe will have a lesser tendancy to form hydrates)

See this earlier answer, Crossed Cannizzaro reaction.

It correlates very strongly with ability to form a hydrate as @Jan's answer says, which is why formaldehyde is usually the sacrificial oxidised component. In general electron withdrawn aldehydes are more easily oxidised, electron rich aldehydes reduced.
This paper from J. Org. Chem. here compared several pairs of aldehydes and the conclusions were:

m-nitrobenzaldehyde is (relatively) extensively oxidised as contrasted to p-anisaldehyde, which is much less extensively oxidised - i.e., much more extensively reduced.  On the bases of this and prior investigations, the five aldehydes studied can be arranged (in order of diminishing susceptibility to oxidation): m-nitrobenzaldehyde, furfural, p-bromobenzaldehyde, benzaldehyde, and p-anisaldehyde. With reference to benzaldehyde and the substituted benzaldehydes, this order correlates with expected combined resonance and inductive influences of the substituent groups on the formyl group, as these should contribute to ease of a) hydroxide ion attack and b) hydranion transfer

The oxidation of aldehydes seems to go thru the hemiacetal. hemiacetal formation is enhanced by electron withdrawing. This explains the stability of vanillin that is a methoxyphenylaldehyde.

It is a kinetically controlled reaction so OH-  will attack that carbon which is having more Electrophilic character(more partaol positive charge) to yield the major product


text:
Equilibrium constant for the thermal dissociation of phosphorus pentachloride

Thermal decomposition of $\ce{PCl5}$ is studied in a closed vessel. Partial pressure of $\ce{PCl5}$ in gaseous equilibrium mixture is $\pu{1 atm}.$ $\pu{50 \%}$ $\ce{PCl5}$ dissociates. Find equilibrium constant.

I am aware of the formula of equilibrium constant, but I am finding difficult determining concentration of reactants and products from partial pressure and $\pu{50 \%}$ dissociation.
I know the formula
$$K_c = \frac{[\ce{PCl3}][\ce{Cl2}]}{[\ce{PCl5}]},$$
but how to calculate the concentration?

In the future, please try to come up with meaningful titles and avoid randomly using math mode. Please visit this page, this page and this one on how to format your future posts better with MathJax and Markdown. As for the question itself, I suggest to start with writing down balanced chemical equation and observe the relation between concentrations from there.

HINT For gases there are two equilibrium constants, $K_\mathrm{p}$ and $K_\mathrm{c}$.

Note the equation as follows: $$\ce{PCl5 <=> Cl2 + PCl3}$$
You may begin with an ICE table showing the initial and final pressures of the gases.
$$\begin{array}{|c|c|c|} \hline
 & \ce{PCl5}  & \ce{PCl3} &\ce{Cl2} \\ \hline
\text{Initial} & 2 & 0&0 \\ \hline
\text{Change} & -1 & +1 & +1 \\\hline
\text{Equilibrium} & 1 & 1 &1 \\ \hline
\end{array}$$
Then it is a simple matter to find $K_p$ with the equation $$K_p = \frac{[\ce{P_{PCl_3}}][\ce{P_{Cl_2}}]}{[\ce{P_{PCl_5}}]}$$
If you would like $K_c$ you would need the ideal gas law ($PV=nRT$) but I doubt that is what your question requires.


text:
Why is the molecular structure of water bent?
Water has the formula $\ce{H2O}$ and we can draw a Lewis structure with two lone pairs on the central oxygen. As a physics student and not a chemist, I think to myself, "Okay, there are two lone pairs, they will repel each other" and so we arrive at the conclusion of bent shape.
The issue in my mind is that I don't see why the lone electron pairs have to exist on the same side of the atom. Also, wouldn't the Schrödinger equation provide an equally plausible structure for water with the lone pairs on the opposite side of the oxygen from what we assume (imaging the electrons on the top or on the bottom of the oxygen in the Lewis structure)? If that were true, then there would be a resonance structure between the two states and we would get a linear geometry. Clearly I'm running around in circles here, please someone enlighten me!

The lone pairs repel each other, but you should also throw the bonding electrons of the O-H bonds into the mix if you're approaching the problem from just electron repulsion.

You must account for repulsion between bonding electrons and non bonding electrons too (Edit: just as @zhe said). Therefore, each pair are repelling every other pair. So, every pair of electrons must have farthest possible distance from each other, which results in tetrahedral shape. And because lone pair repulsions are greater, it is distorted tetrahedral.

You could have a look at the answers here https://chemistry.stackexchange.com/questions/14981/why-does-bond-angle-decrease-in-the-order-h2o-h2s-h2se/57973#57973 which deal with Walsh diagrams.

@B.Anshuman This view of the molecular structure of water is outdated. It also does not reflect the measurements, see Are the lone pairs in water equivalent?

@Martin, This is what our textbooks explained, thus the logic given by me. I thought this was relevant here, so commented. Thanks for updating me.

Does this answer your question? Are the lone pairs in water equivalent?

That's not a duplicate.

a minor side point on terminology - in chemistry the term "resonance structures" is not used to describe structures in which atoms occupy different spatial positions (such as your example of water molecules with H atoms on opposite sides from each other). Resonance structures are electronic states. Crucially, molecules do not alternate between resonance structures. All resonance structures simultaneously contribute to the single electronic structure of the molecule.

I mean, there is a time and place for VSEPR, and this is probably as good a time as any, because all beginning chemistry students go through it. The actual model has already been explained multiple times, so I will only briefly say that according to this theory, there are four pairs of electrons around the central oxygen. In order to minimise electron-electron repulsions, these pairs adopt a tetrahedral arrangement around the oxygen. It does not matter which two are lone pairs and which two are connected to hydrogen atoms; the resulting shape is always bent.
What's worth bearing in mind (and hasn't been explained very carefully so far) is that VSEPR is a model that chemists use to predict the shape of a molecule. The truth is that there is no real way to predict the shape of a molecule, apart from solving the Schrodinger equation, which is not analytically possible for water. Everything else is an approximation to the truth. Some of these approximations are pretty accurate, such as the use of density functional theory. Some of them are extremely crude, and VSEPR falls into this category: it essentially treats electrons as classical point charges, and seeks to minimise the electrostatic repulsion between these point charges. As a physics student you should know better than to do this. Thus, while it predicts the correct result in this case, it is more in spite of the model rather than because of the model. And you should not be surprised to hear that in some slightly more complicated cases, VSEPR can predict entirely wrong outcomes.
As you learn more chemistry you will find that there are increasingly sophisticated ways of explaining molecular geometry. Most revolve around molecular orbital theory. For a qualitative method, you have Walsh diagrams which have been explained at Why does bond angle decrease in the order H2O, H2S, H2Se?. For a more rigorous method you would likely have to run some quantum chemical computations, e.g. Are the lone pairs in water equivalent?. Of course, the drawback of this is that it becomes more and more difficult to extract true chemical understanding from the numbers. Although it should also be said that you cannot extract any true chemical understanding from the VSEPR model.
What interests me more is the followup question:

Also, wouldn't the Schrödinger equation provide an equally plausible structure for water with the lone pairs on the opposite side of the oxygen from what we assume (imaging the electrons on the top or on the bottom of the oxygen in the Lewis structure)?

Because the Hamiltonian of the water molecule is invariant upon rotation, this means that indeed, any orientation of the water molecule is equally likely. However, this only refers to the orientation of the water molecule as a whole. It does not say anything about the internal degrees of freedom, such as the bond angle.
In the absence of any external force, the molecule is free to bend in whichever direction it likes, and most water molecules indeed do do this as they float through space or swim in a lake. But it will always be bent.

If that were true, then there would be a resonance structure between the two states and we would get a linear geometry.

If you were to think of a single particle in a double-well potential, say something with
$$V = \begin{cases}
\infty & x < -b \\
0      & -b \le x \le -a \\
\infty & -a < x < a \\
0      & a \le x \le b \\
\infty & x > b
\end{cases}$$
then because of the symmetry of your system, in every eigenstate of your system, the expectation value of $x$ would be $\langle x \rangle = 0$. This is quite similar to your argument. In the case of water, let's set the oxygen nucleus to be at the origin. Because it can point either up or down, the expectation value of the hydrogen nucleus position along the up-down axis would be exactly level with the oxygen atom, i.e. 0. In fact, don't stop there: it can point to the left or the right, and to the front or the back. So the hydrogen nucleus has a position expectation value of exactly $(0, 0, 0)$, i.e. right inside the oxygen nucleus.
Does that mean it's actually there, though? In our contrived double-well system, it's patently impossible for the particle to be at $x = 0$, because $V = \infty$ there. If you were to measure its position, you would never find it at $x = 0$; you would only find it in the left-hand side $[-b, -a]$, or the right-hand side $[a,b]$. Just because the particle has an expectation value of $\langle x \rangle = 0$ does not mean that it is physically there, or that $x = 0$ is somehow its equilibrium state. You're confusing an expectation value with a genuine eigenstate (which is what a resonance structure is).
In exactly the same way, if you ever were to measure the properties of water (and bear in mind that practically every interaction with a water molecule is, in effect, a measurement), we would find that it is indeed always bent.

"Solving the Schrödinger equation" is of course also just an approximation to the truth. In a very basic sense one could even argue that its one that is not in principle any better than Lewis + VSEPR. It is in fact a misunderstanding to imply a truth behind a model. This is in my point of view the cause of almost all missunderstandings in Theiretical Chemistry. A model shall predict the phenomena. It does not make sense to say a chemical model (that is a" theory" really) is wrong. A theory is always correct. The question is only its range of validity.

@Rudi_Birnbaum "A theory is always correct."  -- I have a theory. My theory supposes that my theory is incorrect.

I like your placing of ideas into a spectrum of accuracy. However, your discussion of the "good" approximations are slightly off. DFT, as a theory, is exact. It is the approximation of the unknown universal functional that makes its implementation inexact. Also, "solving the Schrödinger equation" and DFT are typically associated with electronic structure theory, unless you are treating atoms quantum mechanically as well, in which case you can only predict the expectation value of position. Thus, a pin-point description of atomic locations necessitates an approximation (Born-Oppenheimer).

@LordStryker: That's not a theory in the sense of epistemology. It could be a hypotheses.

@jezzo thanks for your comment! I’m aware of your points, but am slightly loathe to include them because I feel like I already digressed too much. But the comment can and should stay as a useful reminder to other readers about technical accuracy :)

Lewis structures are dots around atoms on 2-dimensional paper. Although H-O-H is planar, when you throw in the lone pairs, you have to think three-dimensionally. How would oxygen be hybridized, in readiness to accept two incoming hydrogens, each with an electron?
Oxygen could hybridize as sp2p, with 120 degree angles between bonds to hydrogen and a 90 degree angle between the p2 lone pair and the sp2 lone pair. Or as sp3, with 109 degrees between all the bonds. In water, we have 104.5 degrees https://en.wikibooks.org/wiki/Structural_Biochemistry/Water. 104.5 is between 90 and 109, so there must be some accommodation between the repulsion of lone pairs, which would push toward sp3 hybridization, and perhaps some repulsion from the electrons in the O-H bonds, which would tend to spread the H-O-H angle toward 120 degrees  (sp2) and push one of the lone pairs more completely into the p2 orbital.
I just can't decide which is the more powerful force; perhaps the water molecule can't either, so it just compromises.

This reasoning is backwards. Hybridisation is a model of description, hence it always follows the molecular structure. For more see: Are the lone pairs in water equivalent? As an additional note, while the phrasing 'oxygen is hybridised' is unfortunately very common, it is incorrect. Atoms cannot be hybridised, only orbitals can be constructed as hybrid orbitals. The repulsion of these lone pairs is also a lot more involved than what is presented here.

Yet another way to think about the fact that $\ce{H2O}$ does not have the full symmetry theoretically possible, is that the number of electrons does not properly match the highest possible symmetry point group (which has the odd name $D_{\infty\mathrm{h}}$. In this sense it can be seen as simple example of symmetry breaking (its however not a simple first order Jahn-Teller distortion).

Why is $D_\mathrm{\infty h}$ an odd name? Please use the chemical construct $\ce{H2O}$ instead of the concoction H$_2$O which may have plenty of unwanted side effects. If you want to know more about MathJax, please have a look here and here.

I am writing to a Physicist which possibly never came in contact to the Schönflies notation. They usually prefer other ones like Hermann-Mauguin or more mathematically oriented ones like Coxeter.

I don't see why the lone electron pairs have to exist on the same side of the atom.

Nobody says they exist "on the same side of the atom". In the picture showing Lewis structures of waters, the lone pairs are shown on opposite sides in the left panel , and on the same side on the right.

These two structures are identical. A Lewis structure does not make a statement about the geometry of a molecule.
If you look at a 3D model of water where the lone pairs are shown according to sp3-hybridization in the valence bond view of things, it depends on the orientation of the molecule whether it looks like "on opposite sides" or "on the same side" (the elongated shape - bunny ears - of the lone pairs is exaggerated; they should add up to a roughly spherical electron density). (Source: https://www.biotopics.co.uk/jsmol/watersingle.html)

In the molecular orbital view (where the molecular orbitals share the symmetry of the molecule) the two lone pairs have distinct shapes (each panel shows one lone pair, orientation of the molecule distinct from figure above).

This was calculated using molcalc.org. As these two molecular orbitals are of similar energy, you could make linear combinations of them to arrive at orbitals similar to the valence bond picture.
So what is the real picture? Looking at hydrogen bond geometries, either one describes the directions from which hydrogen acceptors on other molecules would be located, so both models are consistent with experimental data.

There are eight valence electrons on the oxygen that have similar energies then there is a gap and two electrons very close to oxygen core follow. These eight electrons form four spin-pairs and those point in directions that more or less minimise the repulsion. These directions are called "tetrahedral" in Chemistry language, since they point from the center (= oxygen atom core) to the vertices of an tetrahedron. Now the two protons are connected with two of those "electron pairs" and two are "alone". Hence you obtain the bent shape of H$_2$O.
(This is kind of trivialized version of the VSEPR model.)

This kind of trivialised version of VSEPR is unfortunately as wrong as the VSEPR explanation for water itself. Water is indeed one of the most popular examples for when VSEPR breaks down.

@Martin-マーチン to add: I would make a distinction between when the VSEPR algorithm breaks down, and when the premises of the VSEPR model fail. The algorithm itself works surprisingly well on H2O: if one assumes that the model is correct (i.e. two equivalent LP + two equivalent BP + LP repel more strongly than BP), it predicts a bond angle slightly smaller than 109.5°, in line with experiment. The algorithm fails for H2S, even if one makes the necessary assumptions. The premises fail for H2O. But then again it is arguable as to whether the premises are ever true...

Not sure about which premises you are talking. Obviously VSEPR works fine, you can even predict a compression of the tetrahedral angle if you include the refined rule the the sp3 LP requires more space than the sp3 bond. The VSEPR model in that sense is one of chemistries most important models at all I would say, since it has a huge predictive power. In the sense of the amount of data you need and the amount of molecular structures you get proprerly described by that. Its paper and pencil and the next better one is the MO and that you have to buy with a huge complexity.


text:
What does H2SO4 do to YAS?
I have been reading VOGEL'S qualitative analysis and I came across the reagent Yellow ammonium sulphide. It's written here that it is used to distinguish Group 2B elements. But my question is that how will addition of H2so4 affect YAS itself?
Answer given is that it turns the cloudless solution into white /yellow but how does it exactly do this ?what is the reaction?

Ammonium sulphide is $\ce{(NH_4)_2S}$; it is colorless. Yellow ammonium sulphide contains some sulfur $\ce{S}$ dissolved in the preceding solution. So the anion  $\ce{ S^{2-}}$ is mixed with some disulphide $\ce{S_2^{2-}}$, which is yellow. If you add an acid to this solution, you produce the reaction $$\ce{S_2^{2-} + 2 H^+ -> H_2S_2}$$ or$$\ce{S_2^{2-} + 2 H_3O^+ -> H_2S_2  + H_2O}$$
But $\ce{H_2S_2}$ is thermodynamically unstable with respect to disproportionation. It gets slowly decomposed according to $$\ce{H_2S_2 ->  H_2S + S}$$The sulfur S produced here is not soluble into water, and will make the solution cloudy. If there is enough disulphide, the sulphur will make a precipitate of elementary sulphur, which may be pale yellow or white.
Ref.: N.N. Greenwood A. Earnshaw, Chemistry of the Elements, A. Wheaton & Co, Exeter, 1986, p.805 - 807.