text:
My Misconception of Entropy
It was recently brought to my attention that my understanding of entropy is wonky at best.
In my experience, entropy was introduced (superficially at best) during general chemistry/foundations of inorganic chem and described in terms of order/disorder of a system, usually followed by some messy room analogy. In one of my textbooks, the author, Gary Wulfsberg, was discussing how a reaction tends to favor the products "if it increases the dispersal of ions/molecules over a larger volume of space..."(Wulfsberg, G., 2018). Wulfsberg goes on to say that

The measure of this dispersal or disorder is known as the entropy (S) of the substance. A
positive entropy change for a reaction indicates increasing dispersal or disorder
(Wulfsberg, G., Foundations of Inorganic Chemistry; ch. 4, pg. 200)

which I interpreted it as him saying that all systems that increase in entropy have a corresponding increase in disorder.
Later in my undergraduate career, Boltzmann's entropy was introduced during thermodynamics and described by my professor as the amount of micro-states available for the particles within the system and that an increase in entropy corresponds to an increase in the states that the system can exist in.
All was fine and peachy until recently, when I came across an article that discusses the Shannon Measure of Information (SMI) and entropy. In an article by Arieh Ben-Naim, he talks about how the idea of order/disorder in relation to entropy is not necessarily correct, in fact it is a fallacy that does not hold true nor can order/disorder be measured definitely for all systems. He states,

It is true that many spontaneous processes may be viewed as proceeding from an ordered to a disordered state. However, there are two difficulties with this interpretation. First, the concept of order is not well-defined, and in many processes it is difficult, if not impossible, to decide which of the two states of the system is more or less ordered.
J. Chem. Educ. 2011, 88 (5), 594–596

Additionally, he talks about how some systems do have "order parameters", however, it is not in relation to entropy and that not every process where an increase in entropy is observed has a corresponding increase in disorder. He later goes on to describe the SMI treatment of entropy "as the number of binary questions one needs to ask to find the location of the particle." Hence, if the number of yes/no questions one needs to ask to find the location of particle increases, so has the thermodynamic entropy.
So here is my question
Is entropy described by Boltzmann/statistical mechanics the same as the entropy described by Shannon information theory?
Additionally, is there any validity in relating order/disorder in describing the change in entropy of chemical systems?

Thanks for raising this query. Hope we will see a lot of activity and various view points. The recent terminology in Atkins and others entropy as "dispersal of energy" is due to the efforts of Frank L. Lambert. Many textbooks adopted his terminology. When he was alive, he used to discuss this in a chemistry education forum. If I understood him correctly at that time, he meant to say that entropy is a measure in how many ways energy can be distributed hence the word dispersal of energy.

The biggest problem was created by Shannon himself by using the same word "Entropy" which really led to this "messy room" and "shuffled cards" stories.

@M.Farooq Absolutely, it bothered me that my understanding of such a fundamental concept was wrong. I could blame my teachers but it really is the individuals responsibility to question things and ensure the validity of ones claims. I appreciate your comments/information

Let's distinguish the "fundamental concept is wrong" from the "the often used, simplistic explanation for a fundamental concept is wrong/totally misleading". Two very different situations. Entropy has a very well done definition in statistical mechanics and thermodynamics, and these definitions hold well to chemistry, too. The problem is when people want to explain everything at a 5 years olds level: more often wrong than enlightening. The entropy as a disorder is a very unhelpful concept in chemical systems.

I think my comment was misinterpreted. That is meant as I was wrong, not the concept of entropy itself. That was a reflection of my own misunderstanding. Unfortunately I couldn’t edit the comment beyond five minutes after posting

This is by the way a huge discussion somehow ongoing. In my modest opinion is just the term disorder that is ill-defined. A big stadium with numbered places and supporters owning a ticket for a specific place can be see full of entropy but also very ordered. Yet the random selection of the place should increase both, if I am right. I am also on the line of @Greg.

Above it reads "should increase both entropy and disorder, if...."".

I'm with @Greg on this, originally it was described (by Boltzmann ?) as 'a measure of disorder' but the 'measure of' got lost on the way hence the confusion. Why not either think of it in the classical thermodynamic sense only as $\pm T\Delta S$  as related to heat or as the number of ways of placing particles into energy levels/boxes etc., which is basically Boltzmann's approach, when  an atomistic approach is needed.

@Alchimista There are several nice examples for entropic effects in chemistry, a chemical equilibrium. For example, a rotating methyl group in a solid crystal, which you can call disorder if you want. But entropic factors can come eg from high spin / orbital multiplicity of an electronic state or high symmetry of a structure. It is very counter-intuitive to label the high symmetry cases  as "disorder".

@dval98 This question has been flagged for needing more focus. wanted to let you know. IMO, this can be fixed as follows. Q2 and Q3, Q5 are opinion based and can be removed. However an answer to Q.5 can be asked as a different question using the tag [reference-request]. Merging Q.1 and Q.4 would make the question focused. However, this last idea may be wrong since I do not know enough to discern between the two. If they are not same, ask the two as two different question linking the two together. For Q.5: https://chemistry.stackexchange.com/questions/37303/resources-for-learning-chemistry

I also recommend reading this article: https://pubs.acs.org/doi/abs/10.1021/ed079p187

@Safdar I apologize for that. Please let me know if my edits will suffice.

This seems to be a very rampant bee in some highschool/undergrad teachers' bonnets. The explanation with order/disorder is qualitatively right and useful, and that's that. The Boltzman/thermodynamic/statistical interpretation is, beyond doubt, quantitatively correct.

@Greg I copy it and nothing in my comment is contrary to your last comment. But I can add more, the idea entropy - -> disorder seems also useful. It should be taken with a grain of salt. At the end we might say universe is going to be well ordered, in the sense that it is going to be empty and dark. Not to open a discussion on cosmology, but to put things to the extreme. What remains certain is that there will be no energy to extract.

I wonder why this question was closed. The OP has edited it to be more focused, and it seems a very legitimate question how a very common (and often taught) interpretation of a central thermodynamic concept holds up.

@Greg, the question was closed before the last revision, you can vote to reopen as the question is focused now..

The association of entropy with disorder is anthropocentric. "Disorder" is a concept derived from our experience of the world. The association with "disorder" is clearer once we explain what we mean by "order". The following definition among those provided by Merriam-Webster most closely fits the intended meaning:

a regular or harmonious arrangement

Since regularity always implies lower entropy, all else being equal, it is therefore fair to associate "order" with lower entropy.
The association with "order" or regularity also syncs with the concept of entropy according to Boltzmann's statistical mechanical definition ($S= k_\mathrm B \log \Omega$). $\Omega$, the number of microstates available to the system, can be quantified using the entropy. More possible unique microstates implies a higher entropy. Greater regularity implies more constraints regarding arrangement of the system and therefore less possible microstates. Solids are usually more regular and therefore have lower entropy than fluid states at the same T. Same when comparing gases and liquids. The possible microstates increases from solid to liquid to gas.
This also jives with the informational-content definition. You can use less information (use a more compact description) to describe an orderly (regular) system. You need more information to describe all possible arrangements of molecules in a gas or liquid than in a solid. Think of entropy as measuring the length of the recipe required to build all possible arrangements of the system.

How would we define regularity at molecular level? Is it periodicity? How would explain regularity in liquids vs. gases?

@M.Farooq For a gas you need to describe more possible arrangements, each unique, than in a liquid, to describe the system. Similarly on going from solid to liquid (where the difference is more striking). For a simple description of a solid all possible arrangements can be considered permutations of one template. Once you describe that one template and the way of permuting positions you are done. For liquids you are somewhere inbetween. You see this e.g. in g(r).

The topic deserves going through the examples in more detail, including those in the comments to the OP. I find the idea of dispersal of energy appealing at times but not entirely satisfying. For instance, in an isolated system (say a gas in an adiabatic box) what exactly do you mean by energy "dispersal"? Are you talking about the kinetic theory of gases? The Boltzmann distribution?

Yes, I am not fully satisfied with the wordings of "energy dispersal." Frank Lambert, who was behind this terminology indeed did a big service by getting rid of order-disorders mythology from many textbooks. On the contrary, he also introduced "energy dispersal" concept. Which is a bigger evil entropy or energy dispersal? Only time will tell.

Trying to read the original work of Clausius will require a year to understand where he was coming from and what was going on in his mind when he came up with this idea. His book is available in English.

@M.Farooq It seems you are channeling Kuhn and other philosophers. Granted it is a chore to read something written in those days. Clausius was apparently mathematically minded and attempted to derive thermodynamic principles on a sound mathematical basis. Therefore dense and tough going for non-mathematicians. It would take some time to become comfortable with his style but maybe not a whole year :-)


text:
Comparing stability of CF3+ and CD3+
This is a question from my text book.

Compare the stability of carbocation between $\ce{CF3^+}$ and $\ce{CD3+}$.

The answer is given $\ce{CF3+}$ due to backbonding.
But I see backbonding as a sort of resonance effect. And we know that for fluorine atom its inductive effect dominants its resonance effect, and so overall it is electron withdrawing. So, shouldn't the answer be $\ce{CF3+}$?

Do you know Bent's rule? That should simplify comparisons here: in the CF3 cation, most of the p-character will shift towards the fluorines, and hence the central carbon bearing the positive charge will be left with a lot of s character as compared to the carbon of the CD3 cation. For the same quantum number, s orbitals are usually lower in energy than p orbitals(and that is anyway true for 2s and 2p) so the CF3 cation will probably be experiencing a greater lowering of energy, and hence be more "stable"

Do not use HW tag, it specifically says so. Also, MathJax in the title reduces searchability. Use $\ce{}$ for chemical compounds.

for fluorine atom its inductive effect dominants its resonance effect - no it doesn't. Do you know how bonding in $\ce{BF3}$ works? Same thing here.

source: Stack Exchange

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

sort: votes


text:
Organic chem mechanism

I don’t understand how the reaction produces this product.
In the starting material, there is an $\ce{-OH}$ group attached to carbon 5 but where did the $\ce{-OH}$ group go in the product?

hydrolysis in step 1 produces acid, which condenses with -OH at C-5 to produce the lactone.

@AniruddhaDeb  not hydrolysis. Deprotonation of the -OH and intramolecular cyclisation by nucleophilic attack.

Just for the sake of completeness: the second step using TBAF is basically a deprotection of the silyl protective group. A tentative mechanism is that the fluoride will first add to the silicon to form a penta-coordinated intermediate, then the alkoxide departs, in the process yielding a Bu-SiF as a side product. The alcohol will be fianlly generated by quenching, as can be seen here

I suggest you to change the title in order to reflect the body of question.

This is straightforward enough: The $\ce{NaOMe}$ deprotonates the $\ce{OH}$ group on $\ce{C}$5, the alkoxide then does an intramolecular nucleophilic attack on the ester to give the lactone. This is favoured because it is intramolecular and forming a [6]-ring. It is possible that they may be some ester exchange of the $\ce{tBuO}$ group with the $\ce{MeO}$ group forming methyl ester, but either is more readily attacked by the internal nucleophile so the end result is the same.

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
Dipole moment of (1s,2s,3s)-1,2,3-trichlorocyclopropane

Which of the following compounds has a zero dipole moment?

Only one of the options is correct
According to me, it looks like both options (c) and (d) are right. Since the dipole moments of option (c) molecule seem to negate each other (shown in the image below), so is option (d). However, only option (d) is right. Why?

Note that the molecule (c) is not planar.

Look at c with a 3D model. All the chlorine’s are on one side of the molecule. There will be a dipole running through the middle of the 3 membered ring, with the negative end pointing towards face with the chlorine atoms.

Related: https://chemistry.stackexchange.com/questions/123121/link-between-centrosymmetry-dipole-moment-and-chirality/123129#123129

The issue stems from the fact that you may not have understood what each bond notation means.
Here, the solid wedge indicates that the bond is toward us and so a 3-D view of the same molecule would look as follows:

So as you can see the dipole moments – instead of cancelling, add up. So simply by elimination option (d) is the right answer
For trans-1,3-dichlorocyclobutane, the reason for the zero dipole moment is explained in ron's answer to the question "Does trans 1,3-dichlorocyclobutane have zero dipole moment?".

[...] Conformers A and B both have $C_$ symmetry (the only symmetry element is a plane that bisects the ring and contains the two cyclobutane carbons bearing the substituents) and therefore do have a dipole moment. However, the dipole moments of conformers A and B are equal and opposite, so when flipping is rapid the dipole moment averages out to zero.

Therefore (d) is the correct answer and not (c).

In my opinion, instead of trying to cancel out dipoles, you should fall back to first principles and just do what you are effectively doing: using symmetry.
The key concept is that the symmetries of the nuclei translate to symmetries of orbitals and electrons, and these together translate to symmetry of any properties derived from the positions of nuclei and electrons, for example, dipole moment.
Every symmetry that the molecule has will be reflected in the dipole moment:

A plane of symmetry implies that the dipole moment will be also in the plane. When you reflect the molecule and its dipole moment, both must be unchanged (since it's the same molecule with the same properties).

An proper axis of rotational symmetry implies that the dipole moment will be along the axis. Otherwise, rotating the structure would give back the same structure but with a different dipole.

This immediately leads to a simple corollary:
If a molecule has two rotational axes of symmetry (or a plane of symmetry and a rotational axis of symmetry perpendicular to it), then the molecule has zero dipole moment. Based on the two points above, the only way a for dipole to exist in this case (or cases) is to point in two different directions, so the dipole moment cannot exist.
So, in your example (d), can you find two symmetry elements?

 The two symmetry elements are a plane cutting through the chlorines and carbons they are connected to, and a rotational axis going NW-SE for the molecule as drawn. The axis is perpendicular to the plane, so there is no dipole moment.

(a) has just a plane of symmetry. (b) has a plane of symmetry with a rotational axis of symmetry in plane. (c) has three planes of symmetry and a rotational axis that is along the intersection of the three planes. All of these lead to situations where we will likely have dipole moment.

Symmetry is very useful in cases like this. The rules are that (a) any molecule with a centre of inversion ($(x,y,z)\to(-z,-y,-z)$ which makes the molecule  indistinguishable) cannot have a permanent dipole. In addition (b) a dipole cannot exist perpendicular to a mirror plane ($\sigma$) and (c) a dipole cannot exist perpendicular to a rotation axis ($C_n$).
Thus certain combinations of symmetry elements mean that a molecule cannot have a permanent dipole, for example a $C_n$ axis and a perpendicular $C_2$ axis or perpendicular mirror plane $\sigma_h$. The point groups that cannot have a dipole are therefore are,(1) any group with a centre of inversion (i), (2) any D group ($D_n, D_{nh}, D_{nd})$ and (3) the cubic groups $T_d,O_h, I_h$.
Only molecule (d) has a centre of inversion so the answer is d.


text:
When is Raoult's Law valid for solid-liquid solution?
Raoult's Law for solid-liquid solution is stated as such in my books that it is applicable for all non-volatile solutes.
However different solutes must have different attractive or repulsive forces and in such a case, all of them will probably hamper vaporisation in different ways.
Then how come can we simply define the law as:
$$P_\text{vapour}=P_\text{vapour}^0\chi_\text{solvent}$$
where
$$\chi_\text{solute}+\chi_\text{solvent}=1$$
Here the equations do not provide any information about the solute.

Is the law valid for only a class of solutes?

When did this form Raoult's law become applicable for all solutes? AFAIR, it was only for non-volatile solutes

It is valid for both volatile and nonvolatile solutes, as it refers to the partial vapour pressure, not total vapour pressure.
Saying that, note that  it applies only on mixtures with  ideal behaviour, as there are many more or less significant  positive and negative deviations from the law. These deviations relate to preference of intermolecular bonding to the same, or to the other molecules.
Typical examples are ethanol water mixture forming an azeotrop of the minimal boiling point,  or hydrochloric acid forming an azeotrop with the maximal boiling point.

So what you are saying is that we neglect these deviations same way as we neglect air resistance in projectile motion.

@Tony Stark I say sometimes we do because we know we can afford it. sometimes we do not because we know we cannot afford it. E.g. mixing 1 L of ethanol + 1 L of water leads to heating up the mixture. When it cools down back, it has volume 1.9 L. We cannot neglect this like it does not happen.

The law is for ideal solutions where the attraction b/w solute particles  and attraction b/w solvent particles is same as attraction b/w solute and solvent particles.
Consider the case when the solute is non-volatile:
The solute and solvent particles of ideal solution can be treated as same. But the solute particles will reduce concentration  of solvent particles in solution and on surface of solution too. This will cause fractional decrease in vaporization rate of solute and hence its vapor pressure( it is also total pressure as only sovent is contributing to pressure)
That's how you get $P_{solvent}= P°_{solvent}×x_{solvent}$    ( for non-volatile solutes)
Also here $P_{solvent} =P_{total}$
If the solute is also volatile, then it will affect the condensation rate of solvent particle in vapour form which should increase the partial pressure of solvent.


text:
Polar protic solvents with melting point between between 0 °C and 25 °C
Need a solvent for a freezing point depression experiment. Acetic acid is the only one I know of, but it is not ideal due to the low freezing point depression constant. Any suggestions?

does it have to be polar protic? Benzene is pretty good for FP depression experiments.

@AniruddhaDeb I want a solvent like water, what do u think?

What do you mean by telling acetic acid is not ideal due to the low freezing point depression constant? $K_f$ of acetic acid is larger than water.

Followings solvents have higher freezing point depression constants than that of water:
$$
\newcommand{\d}[2]{#1.&\hspace{-1em}#2}
\begin{array}{lrr}
\hline
\text{Solvent} & \text{m.p.}/\pu{°C} & K_\mathrm{f}/\pu{°C m-1}  \\
\hline
\text{Water} & 0 & -1.86 \\
\text{Nitrobenzene} & 5.67 & -7.00 \\
\text{Cyclohexane} & 6.54 & -20.2 \\
\text{Acetic acid} & 16.6 & -3.90 \\
tert\text{-Butanol} & 25.82 & -20.2\\
\text{Phenol} & 43.0 & -7.27 \\
\hline
\end{array}
$$
So far, the best value is that of cyclohexene. Even though its melting point (m.p.) is $\gt \pu{25 ^\circ C},$ I also included phenol, just for curiosity.

@Buck Thorn: I voted to close under "Openion-based" when it is up. Then I found OP's comment of acetic acid, which is not correct. So I put the data I found as an answer for benefit of readers including OP.

Between $5$°C and $10$°C, you may find the following freezing points : 4-chlorotoluene, nitrobenzene, ethylenediamine, cyclohexane, decanol, formic acid.
Between $10$° and $15$°C, there is : acrylic acid, cyclooctane, dichloracetic acid, dioxane, oleic acid, paraldéhyde, phenylglycol, safrol, p-xylene, m-cresol, ethanolamine.
Between $15$°C and $20$°C, there is : acetophenone, dimethylsulfoxide, glyoxal, lactic acid, methacrylic acid, undecan-1-ol, butane-1,4-diol.
Between $20$° and $25$°C, there is : cyclohexanol, dodecan-1-ol, glycerol, hexadecane, heptadecane

not all of these are protic though.

@AniruddhaDeb - so pick one that is? Lots of choices there.


text:
Why does Vapor Pressure decrease on addition of solute?
Consider a vessel with a volatile liquid in it. Let's say it has a vapor pressure $P^0$ when it is pure.
Now a non-volatile solute is added to it. Its vapor pressure decreases as $P^0\chi_{solvent}$.
Why does this happen at a molecular level?

The reason that I have come across says that less surface area is available for evaporation and hence this effect. But this does not make sense to me as we can have liquid in container of all shapes and sizes and will still get the same Vapor Pressure.

Some users have suggested me a link to a similar question. The answers in the link are remarkable but the explanation I was hoping for was more likely to involve forces (as in Physics) and maybe a sort of "free body diagram" explaining how these forces are acting(words rather than actual diagram will be fine too) rather than thermodynamic picture.

Linked thermodynamic explanation: https://chemistry.stackexchange.com/questions/22395/what-causes-the-lowering-of-vapour-pressure-in-volatile-nonvolatile-solvent-mixt?noredirect=1&lq=1

@Safdar The answer in the link is talking about the thermodynamic and macroscopic picture of the phenomenon but I am specifically interested in the microscopic interpretation.

For solvent vapour tension not to decrease with the molar fraction, tendency of solvent molecules to leave liquid would have to increase with decreasing it's molar fraction.

For solvent at temperature $T$, surface area $A$ and  having the molar fraction $x$, the rate of molecules leaving liquid is:
$$\frac {\mathrm{d}n}{\mathrm{d}t}=-C_1 \cdot A \cdot x \cdot f(T)$$
Lower molar fraction means lower surface density and lower evaporation rate per surface area.
Factor $C_1$ includes pure solvent molecular surface density and $f(T)$ includes probability of a molecule leaving liquid.
For vapour of the partial pressure $p$, the rate of molecules coming to liquid from vapour is:
$$\frac {\mathrm{d}n}{\mathrm{d}t}=C_2 \cdot A  \cdot p \cdot  g(T) $$
The $g(T)$ involves temperature dependent relation of pressure and collision rate with liquid surface. Higher temperature means higher kinetic energy and less collisions needed for the same pressure.
For equilibrium, the net rate must be zero, therefore:
$$p= \frac{C_1 \cdot x \cdot f(T)}{C_2 \cdot  g(T)}$$
Therefore the equilibrium pressure is proportional to the molar fraction.
It is quite analogical to the steady state of water level(=pressure), when it is simultaneously being filled up(=evaporation) and drained(=condensation). When you dilute a non-volatile solute, you decrease the filling up (= surface concentration of molecules is lower), so draining (=condensation), brings the water level(=pressure) to lower value to match draining rate with the filling.

Sir does that mean the explanation that I have come across is true though not properly depicted?

Secondly the way I am interpreting this is that as soon as non-volatile solute is added, the rate of evaporation decreases. But the rate of vaporisation is still high and hence some vapor comes back in the liquid in the net exchange. This continues until the pressure becomes smaller to an extent both rates again become equal.

The explanation is true, but you may have got it wrong. If some surface area contains 1000  solvent molecules,, the gross rate of evaporation is 2 time bigger than if there is just 500 molecules.

Your 2nd comment - you got it.

Imagine the converse situation:

A sealed chamber has an open beaker of the volatile solute.
After some time, equilibrium vapor  is established.
Now an open beaker of the non-volatile solvent is introduced into the chamber.

Some molecules of the solvent in the chamber's "atmosphere" will dissolve into the open beaker of solvent, and that process continues until all the solute evaporates. Since the solute dissolves into the solvent, it is absorbed from the atmosphere, lowering the pressure.
As an example, consider a desiccator containing $\ce{CaCl2}$. If you start with a container with some water in the desiccator, you end with a puddle of  $\ce{CaCl2. n H2O}$ and very little  $\ce{H2O}$ in the air.

If we add non volatile solute to solvent for example water,  we decrease the tendency of water to evaporate in gas phase because solute particle obstructs the evaporation of water in gas phase therefore vapour pressure decrease when  we add non volatile solute to solvent like water.  This image is from ncert:

That is the exact thing that does not make any sense to me.


text:
Weinreb ketone synthesis

For this problem, Can I use weinreb ketone synthesis? I found the product but not sure it is the right one
If the amide react with either R1-Mgbr or R2-Li,
then it produces the ketone?

Why does this addition happen to the amide and not the ketone?

The Weinreb ketone synthesis i know does not involve this amide but rather an O-methyl-N-methyacetamide.  https://en.wikipedia.org/wiki/Weinreb_ketone_synthesis

What is the source of the reaction shown?

All the reaction doesn't occur just in one pot. First, BuLi removes the tributylstannyl substituent from the compound 1 in nearly 0.5 h. Then, by adding the amide 3 the reaction pot, the anion 2 attacks to the carbonyl group only once to provide the ketone 4 without any tertiary alcohol, as expected by the Weinreb–Nahm reaction.

Isono, .N., & Mori, .M. (1996). A NOVEL CARBON-CARBON BOND FORMING REACTION USING BIS(TRIBUTYLSTANNYL)PROPANOL. Main Group Metal Chemistry, 19(5),

Is the selectivity of the amide over the ketone explained?

I think this is associated with the chelation of Li and the oxygen of the ketone. This looks some weird but if the above path is confirmed as a Weinreb–Nahm reaction, the ketone must play the same role as the hydroxylamine. Also, if the Weinreb–Nahm reaction could turn ketones to tertiary alcohols, it would primarily do this with amide groups. See the mechanism section of en.wikipedia.org/wiki/Weinreb_ketone_synthesis

Here is the reference cited above: DOI:10.1515/MGMC.1996.19.5.277 It has little to do with the reaction shown by the OP and Chem. Absts. doesn't know about the reaction.

@Reihani  I am very familiar with the Weinreb-Nahm and this does not look like it. I do not find this explanation convincing

@Waylander: Neither do I.


text:
Anode and cathode in electrolysis
Can the anode and cathode in electrolysis be the same material and still separate out the oxygen and the hydrogen? I could not find any information about if platinum makes a good cathode or anode, some sites said it was good for an anode, and other sites said it was good as a cathode.

Electrolysis of water using platinum electrodes..

Any conductor can potentially serve as a cathode or an anode.

@safdar, we don't have to be too rigid. It is clear that the OP did search websites. It is question based on a confusion. Why can't one reply in line, that potentially any inert conductor can work. It is certainly not a textbook or homework question.

@M.Farooq, that was just an auto comment. The fact is OP saw it can be good anode, good cathode but didn't think to piece them together. Also,  electrolysis of water is a very common electrolysis where the electrodes used are either graphite or platinum which is talked about in many cases. I replied with an example within 2 minutes and then waited for 30 mins before sending that comment

Water cannot be electroyzed alone. To electrolyze water, an electrolyte has to be added to water. The electrolyte can be an acid, a hydroxide or a salt. The electrode can be any sort of metal, provided it does not react with the electrolyte, and is not easily oxidized at the anode. Platinum does not react with usual electrolytes, and is not easily oxidized at the anode. So platinum is usually chosen for anodes and cathodes : it is convenient for both.

A good electrode for electrolysis is:

An electrical conductor.
Chemically stable in water and in oxydizing environment of the anode / reducing environment of the cathode.

Most metals are OK being cathode in electrolysis as they fulfill both conditions. Well, sodium and friends are not good - it is not stable in water in the first place. Some non-metal substances are also acceptable - e.g. graphite. Even metals that are somewhat unstable in water (e.g. iron) are stabilized when used as a cathode and the effect is sometimes used as a corrosion protection.
Anode is the harder task. A lot of metals that are more or less stable in water oxydize when used as anode and will not last long. They will also "eat" some electrical charge and you'll get less than expected oxygen (you may as well get none). Precious metals are OK, graphite in lower temperatures is OK, too. Some metals use a trick: nickel or lead (and some other metals) anode (in proper conditions) forms a thin oxide layer that is itself (somewhat) conductive and also stable in regard to further oxidation.

As a cathode is usable near any conductor, which does not dissolve in electrolyte.
Anodic material must withstand aggressive anodic oxidative processes without dissolving. Aside of platinum, glassy carbon is often used. Questionable can be stainless steel.
Electrodes should not  be vulnerable to desintegration by evolving gases. E.g. otherwise inert but porous carbon electrode from zinc cells ( my first attempt as a young chemist ) forms fine carbon dust during electrolysis.


text:
Why is atomic number is more or less the same as number of neutrons for light elements?
In periodic table, the atomic weight is more or less twice the proton number. Do neutrons naturally tend to be in similar quantity as protons? I know atomic weight weighted average of isotopes but why this more or less a trend if we move across the periodic table we find atomic weight is 2(protons) + $\epsilon$ where $\epsilon$ < 2

It is true for light elements only. E.g. 238U contains 92 protons and 238-92=146 neutrons. See also https://en.wikipedia.org/wiki/Semi-empirical_mass_formula  and https://en.wikipedia.org/wiki/Nuclear_shell_model

While an interesting question, you are basically asking about nuclear physics and how it works. Too many protons and the nucleus wants to fly apart - the neutrons help stabilize the nucleus. But, too many neutrons and one can beta emit and become a proton.

As mentioned, it's a physics question, but in brief, Maria Goeppert Mayer won a Nobel Prize for her work on a nuclear shell model. See https://en.wikipedia.org/wiki/Magic_number_(physics)  for a start.

@DrMoishePippik alright I can accept that an answer.

@Poutnik, yes, i did mean light elements. There is some order to light elements that I need to explore. I guess DrMoishe's reference might be great aid to understanding why is this the case.

@JonCuster I see. Thats a nice gist of whats happening. I guess its something to do with Magic numbers as DrMoishe mentioned here.

Your title is not the same as your question...


text:
Problem with epoxy silane (Random results in glass silanization). What are crucial points to look at?
I`m trying to immobilize an antibody onto the glass surface by means of GPTOMS (glycidoxypropyl trimetoxy silane). Using the same protocol (silanization in toluene) I get different results any time, what are key points to check?
Looking through literature on immobilization, I see a lot of articles where authors use APTES (aminopropyl triethoxy silane) and only a few using GPTOMS, why? Epoxy group seems so atractive...
For toluene:

Rince slides with water
Ultrasonicate them in isopropyl alcohol for 3 min, dry them
Put for 3 minutes into piranha solution
Rinse with distilled water
Treat with oxygen plasma for 3 min
Slowly put slides into 2 or 4 % silane solution in toluene, incubate for 3 min
Rince with toluene and isopropyl alcohol.
Put them on dry heater (max temp. of ours is 97 C, most protocols say 110 C), incubate an hour

For ethanol:

Rince slides with water
Ultrasonicate them in isopropyl alcohol for 3 min, dry them
Put for 3 minutes into piranha solution
Rinse with distilled water
Treat with oxygen plasma for 3 min
Put slides slowly into 2% silane in ethanol solution (silane + alcohol + acetic acid (pH 4,5-5,5)), incubate 3 minutes
Rince slides with ethanol
Put them on dry heater (max temp. of ours is 97 C, most protocols say 110 C), incubate an hour

At a guess, surface preparation and contamination before applying GPTOMS? Glass adsorbs monolayers from detergents, water and anything else around.

Please take a look at protocols - we actually clean the surface with everything avaliable...

That does not mean that there is no contamination! Heaters can give off VOC's, for example. Having worked in vapor deposition and in electron micraoscopy, I've seen how easily a surface can become contaminated. I never did get a perfect OsO4 stain...

Can you elaborate "I get different results any time, what are key points to check?" What is your criterion?

When Im trying to immobilize antibodies onto the surface, fluorescent-labeled or regular ones with further immobilization of the blood cells, I get results like "fluorescence"/"no flu..."/"only mild flu..." using the same protocol. With cells... they may cover all the surface, no matter if theres antibody immobilized or not. ANd may not cover at all.

The triethoxysilane group has to hydrolyze first in order to bond with a silanol group of glass surface. It seems that you are cleaning the slides very well. However, I think your temperature is too low and your time of exposure is too short. The hydrolysis of ethyoxysilane group is slow.
Secondly, I guess you are posting questions on this topic for almost a year without much success. In such cases, it is always good to go back and try a control experiment and think about a new approach. The control experiment is that you try just neat glass slides and see if your antibody sticks to glass or not. Glass itself is a good adsorber. Probably you might get similar results of "fluorescence, no fluorescence and mild fluorescence". This will confirm that your silane binding procedure is doing nothing!
Second control test: Epoxy silane coated slides are commercially available. Why don't you try using them and see if your immobilization chemistry works on commercial slides or not. Schott makes commercial slides by this name: Nexterion® Slide E
Let me tell you how silanes are bonded onto silica. The reaction time is typically 10 hours under reflux conditions of toluene and other added solvents and catalysts! Sometimes a very very small amount of water is added to created polymer like bonding. There are at least several hundred papers on this topic. The surface area is silica is quite high so it is not a direct comparison with glass slides but still 3 min and room temperature is doing nothing!

Could you please tell me, what is the correct temperature then?

I had mentioned reflux temperature around ~110 C of toluene.

From what I heard, its enough to prepare either acidified solution of silane in alcohol or toluene solution, sink slide into it, extract, rince, bake at about 110 degrees - and youll have coated surface...

Since you have already tried it and certainly enough it is not working. You will have to study multiple variables and optimize them.

Could you please tell me, how important humidity control and on what steps?

I do not have glass slides experience but I have done a a lot of bonding on silica particles. In those cases, water is avoided like plague or in modern terms like corona. There is no short cut for optimization. You will have to do a lot of experiments under different conditions and compare them with the commercial slides.

Is it on the step of silanization or on the step of protein reaction?

Silanization. Epoxides are also very reactive so you have make sure of the compatibility. Acids can open the rings.

There is no acid, just toluene and silane. We use carbonate-bicarbonate (pH 10,4) buffer solution for antibodies when immobilize them...


text:
Why can’t every element emit all the colours from the emission spectrum
If there is an infinite number of levels, this means that every electron can jump from as many variations as possible. Hence, why can’t every element emit every colour?
Please don’t use very complicated theory because I’ve just started my A levels.

I guess even you can distinguish infinity of real number set and infinity of Ry . (1/n1^2 - 1/n2^2), If we consider for simplicity a hydrogen atom. BTW, a potassium atom with n=600 has size about 0.1 mm, so practically,, it is not infinity.

Simply put, "infinitely many numbers" is not the same as "all numbers".

If there is an infinite number of levels

Yes, but the key point you are missing is that they are not equally spaced. The energy levels are not like stair cases (equally spaced). The electronic energy levels tend to converge. After that, you have basically provided enough energy to electron that it can overcome the attraction of the nucleus and leave the atom. If you recall, this is called ionization.
This figure illustrates the concept of convergence.
As an exercise, try to plot 1/$n^2$ as a function of n in Excel, where n=1 to 20. See how rapidly it converges to zero.


text:
Real life examples of chemical reactions
For my upcoming lab report, I want to look at exactly how a simple variable, such as heat or surface area, affects a reaction. But I'm having trouble finding a reaction to test this on, as my results need to have real life applications. What a few real life reactions that are easy to replicate in a school chemistry lab?

Litihum-halogen exchange at low temperature. If you don't keep you reaction temperatures below about -65C you start to get side reactions occuring that brings your yield of final product down significantly.

The reduction of an ester to an aldehyde with one equivalent of DIBAL (diisobuylaluminum hydride) is conducted at -78oC. At higher temperature, the intermediate collapses and the aldehyde is reduced to an alcohol.  Aldehyde reduction is faster than ester reduction. https://chemistryscore.com/reduction-aldehydes-dibal-hydride/

Do you own refrigerator? If so, why do you own this appliance?

First, a disclaimer: there is no shortage of opinion about what comprises physical or chemical changes.
This page is entitled Identifying Physical and Chemical Changes and is intended to help instructors guide STEM students on this topic.
You might find good ideas with low overhead in the Teaching Activities section such as these, quotes from the document:

Heating steel wool in air and collecting the black powder that results;

Dissolving sugar and salt in water and comparing what happens to the electrical conductivity of the two solutions;

Comparing boiling water with mixing vinegar and baking soda - both produce bubbles but what's the difference?;

Precipitation reactions, particularly ones that produce an obvious color change, such as Epsom salts and ammonia solution; and

Acid - base reactions can be illustrated through the use of indicators both natural (cabbage juice) and synthetic.

These are easy and safe for the classroom and have the advantage of illustrating what are considered physical vs chemical reactions. They also have real-life analogues: adding sugar to iced or hot coffee; taking an antacid for an upset stomach; the lime that shows up in your tea kettle after a while; and so on.
You can illustrate the effect of more or less heat on the extent and rate of dissolution of a solute in a solution; or combustion of steel wool under a partial vacuum. The former isolates the heat temperature variable; the latter, percent composition of oxygen in the surroundings.

Try to get samples of marble (pure calcium carbonate) in a marble mason workshop. Any piece of marble will do. Or try to find samples of ordinary calcareous stone (impure calcium carbonate), that can be found everywhere along the footpaths. Check if it is made of calcium carbonate by dipping them into some vinegar : it must produce bubbles if the stone is calcium carbonate. If it's a silicate, it will not produce any bubbles. Keep the only samples that react with vineggar.
Now hit and break half of your stones with a hammer, until the stone is transformed into smaller and smaller grains, like dust or powder. When this is done, half fill two beakers or two glasses with the same amount of vineggar. Then take a handful of stones in one hand, and a handful of the obtained powder in the other hand. Throw simultaneously the stones in one beaker and the powder in the other one. You will soon observe that the reaction is much stronger with the powder, because its surface in contact with the acid is much bigger than with the stones. The reaction produces a gas which is carbon dioxide $\ce{CO2}$
Chemically speaking vinegar is a solution of acetic acid $\ce{CH3COOH}$. . The equation of the reaction is $$\ce{CaCO3  +  2 CH3COOH ->  Ca(CH3COO)2 + H2O + CO2}$$If you don't like the formula of the vineggar, you can replace it by a a solution of hydrochloric acid $\ce{HCl}$ one molar. Both acids are reacting roughly the same. In that case, the equation will be easier to write : $$\ce{CaCO3 + 2 HCl -> CaCl2 + H2O + CO2}$$


text:
When is the activity coefficient greater than 1?
I am quoting a sentence from Non ideal solutions:

if the interactions (between solute and solvent) repel each other, the
activity coefficient is greater than 1.

How can interactions between molecules repel each other? Don't they always line up so that the negative end is in contact with positive (and vice versa)?

I’m voting to close this question because incomprehensible.

The classic case of having activity coefficient greater than 1 is the solution of very hydrophobic organic solvents in water. For example, if you have toluene or benzene in water, these molecules do not prefer to stay in water. This type of work was a popular research topic in the 50-70s.
Note that here we are talking about neutral solution so the concept of ordered lining up does not arise.
There are plenty of papers on this topic. Just search activity coefficient of benzene in aqueous solutions.

Note that the saturated solution of benzene in water has the same activity  as pure benzene ( or rather as benzene in saturated solution of water in benzene ). OTOH, the same concentration of benzene in toluene would have activity much lower.


text:
Hydrogen and Hydrogen ion
Is it correct that a hydrogen atom only has 1 electron and 1 proton since the atomic mass is 1?
Also, is the hydrogen ion composed of a single proton and no electrons?
In an ionization process we know that sometimes a hydrogen ion or H+ is taken out of a compound to be transferred to water, in what we know as an acid-base reaction. For example if we have HCl, which is attracted to water, the bond between the hydrogen and chlorine would be broken and the electron pair between them would have all gone to the Cl. Hence making the chlorine having a negative charge and the hydrogen a positive charge.
Does this hydrogen only have 1 proton making it an ion?

So, yes. A Hydrogen atom without an electron is a proton, and it is an ion, a positive ion.

A single hydrogen atom is just one proton and one electron. However, just to make clear, elemental hydrogen exists as $\ce{H2}$, which is a gas.
The atomic number gives the number of protons. The atomic mass is, approximately, the number of protons and neutrons.
When a hydrogen atom loses an electron, then it becomes the hydrogen ion ($\ce{H+}$) which is just a single proton.
As for dissolving HCl in water, we known that hydrogen and chlorine are bound covalently in the gas phase, however, dissolving in water causes the electrons to be totally given up by hydrogen to chlorine. This leaves  us with an $\ce{H+}$ and $\ce{Cl-}$.
Let me mention, though, that in aqueous solutions, that $\ce{H+}$ will form a hydronium ion with water, giving $\ce{H3O+}$.


text:
Retrosynthesis suggestion
Am I even close to solve this problem or it is a complete disaster?

For one thing, to make a diazonium salt with NaNO2/HCl you need an amino group on the aromatic ring. Perhaps only a partial disaster. ;) Oh yes, welcome to ChemSE!

@Frederico  I think it would help you to look up the reactions of 2-halopyridines with nucleophiles

Out of curiosity:  The scheme suggests a C-N coupling reaction between 1,8-diazanaphthalene and an aryldiazonium salt.  May someone indicate the name for this type of reaction or / and a literature example for such a transformation?

Photoarylation of pyridine with an aryl diazonium salt at all positions of pyridine has been reported: DOI:10.1021/acs.joc.9b0187 Can't speak to the suggestion by the OP. @Buttonwood

@user55119 Perhaps we are not on the same page, do you refer to Bartomoleu, 2019JOC10459 (doi 10.1021/acs.joc.9b01879) -- as the last figure was truncated?  Scheme 2 of the paper and the NMR of the 88 p. SI read like pyridines / quinolines / quinoxalines underwent C-H arylation at 2, or / and 4 position relative to N, but no NMR record evidence of a C-N bond formation with what was a pyridine-N.  But even if there was a misunderstanding here, good to know about this photo mediated approach, since one of rules of retrosynthesis is «the strategy is better if a similar reaction already was seen».

@Buttonwood: The reference was truncated. Sorry.  I searched pyridine and phenyldiazonium cation as reactants in Chem. Absts. No evidence for what the OP drew in the first step.

Here is a possible route:
Start from the commercially available 2,6-dichloro-5-fluoropyridine-3-carboxylic acid (1). Form the Weinreb amide (2) via the acid chloride. React this with vinyl magnesium bromide to form (3). React the product with 1 eq of aniline in the presence of non-nucleophilic base to produce (4). Isolate this product and react with excess piperazine for the final product (5).

The closest analogue found by ChemSpider is this one : http://www.chemspider.com/Chemical-Structure.26951527.html?rid=fa825ba9-6765-4789-9e1d-13d6495e6f3b ; I wonder if that keto ester could help with the synthesis and be removed at the end by hydrolysis and decarboxylation.

Yes, I think the hydrolysis and decarboxylation would deliver the desired structure. The difference would come down to the yields in installing the vinyl ketone v the methylene beta keto-ester

I was thinking more of installing that C-C moiety the other way around. Suppose you start from an ester of your (1), do the substitution of the 2-Cl with the aniline, then a Michael addition to an acrylate and an intramolecular acylation (Dieckmann?). Just to avoid harsh nucleophylic conditions on such an electrophylic molecule. But yes, your route seems more direct.

user6376297: The Dieckmann route has been applied in this series of cmpds. T. Miyamoto, H. Egawa and j. Matsumoto, Chem. Pharm. Bull., 1987, 35, 2280.

This discussion illustrates very clearly that there is rarely only a single potential route to a target

@user55119 : OK, thanks, good to know! So perhaps leaving the lab for the 'dark side' (chemoinformatics) did not make me forget all my organic chemistry...

BTW, computers are at it, too: ai3sd.org/aireact2020/abstracts ; postera.ai/demo ; https://askcos.mit.edu/ ...


text:
If you accumulate 100 identical spheres how many will end up on the surface?
So far I have asked this question in physics, engineering and math sections but have not gotten an answer in three years. If you have 100 identical spheres and group them into A sphere shape, how many will end up on the surface? They can be packed tight or loose in a hexagon pattern or cubicle pattern. They can be symmetrically or evenly spaced. I realize the answer can be off by one or two depending on how you group them, But in general it seems like something that can be answered. Is there a formula that could be applied to any number of spheres? For example what if you have 120 identical spheres, how many will be on the surface then? Imagine an atom with 120 nuclei in the nucleus. Thanks for your help.
EDIT:After seeing some of the responses it reminded me of my thought process a few years ago when I began to think about this. How do you define surface, what type of packing, etc. etc. I too tried the plastic bags and ended up using different colored magnetic balls. Experimenting has worked just fine and I asked the question because I truly thought a simple formula would be available. As for what the surface is I have found another question that actually comes closer to what I'm interested in. No matter how many spheres are accumulated into a spherical shape you can separate them into two categories. Spheres in the (interior) and spheres on the (surface or touching a sphere on the surface). This seems confusing at first but it actually simplifies the surface question by literally filling the gaps.  So really the question is how many spheres are not on the surface or touch a sphere on the Surface? Even still it can be difficult counting spheres when you transfer between hexagonal, cubical or symmetrically spherical shaped. That is why I have hoped for quicker solution.

Comments are not for extended discussion; this conversation has been moved to chat.

Colleagues, it would be a shame if this question gets closed.

Too short for an answer, but this might be helpful : https://en.wikipedia.org/wiki/Waterman_polyhedron

Experimental approach
Here are 100 marbles in a polyethylene bag:

If a marble touches the bag, it is counted as being on the surface. I did not count, and the bag was not perfectly round. A balloon might be better.
Theoretical approach
This works for large $N,$ and the result is as Ivan predicted (from scaling arguments in 2D and 3D, I would imagine). We neglect the packing defects in the bulk, and treat the surface as having no curvature and no defects. The maximal packing density of spheres is well-known by chemists,
$$P_\mathrm{3D} = \frac{π \cdot \sqrt{2}}{6} = 0.74046\ldots\tag{1}$$
The maximal packing density of circles is less well-known on this site, but you can look it up on Wikipedia:
$$P_\mathrm{2D} = \frac{π \cdot \sqrt{3}}{6} = 0.90687\ldots\tag{2}$$
If you have $N$ marbles with radius $r$, you can estimate the radius $R$ of the enclosing sphere, get its area ($4 \pi R^2$), and calculate how many circles (of area $\pi r^2$) fit in that area. The resulting expression scales with $N$ to the power of two-thirds, as Ivan said:
$$N_\mathrm{surface} = 4 \cdot P_{\mathrm{2D}} \cdot \left(\frac{N}{P_{\mathrm{3D}}}\right)^\frac{2}{3}\tag{3}$$
For example, for a million spheres, about four percent are on the surface. For a single sphere, all are on the surface. The expression for $N = 1$ yields $4.4$, pretty bad. For $N = 100,$ it yields $95.2$ That is a not a bad estimate; from the experimental approach, I see more than $30$ marbles on the surface, and the picture only shows one side.
Approximating a sphere with a cube
If you have 125 spheres that pack in a primitive cubic fashion, you would have a cube with 5 spheres in each direction. The top and bottom, right and left, front and back layer would be on the surface, leaving a 3 by 3 by 3 cube in the center. So 27 spheres would be on the inside and the remaining 98 would be on the surface. When we take those same 125 and pack them tighter and such that they fit in a minimal sphere, the surface area would get smaller, and the packing on the surface would get denser, leading to a similar fraction of spheres on the surface.

@Poutnik The bag is better than glue because it also helps to define which spheres are on the surface. Also, you can vary the number of spheres more easily, and adjust the packing.

The plastic bag seems to work fine. Last night I was using plastic wrap but the bag seems to work better. Thanks everyone for your input, this is very helpful.

I like the experimental approach. Would love to know the actual result. As for the theory it suffers from the same problem as Poutnik's answer: N is not very big, so the approximations fail. Consider my answer: the number of interior beads predicted by a numerical method is nearly the same as the number of surface beads according to the large N approximation. This problem was made for "experiments" (numerical or otherwise).

@Karsten Theis I did mention the bag/wrapping as well. I mentioned the glue to experimentally evaluate the particular sphere packing by counting, what is not available by bag wrapping.

@Poutnik - The 2D value (appox 0.90687...) comes from tiling the plane with hexagons, then inscribing a  circle in each hexagon. So the value is the ration of the area of an inscribed circle to the area of its circumscribed hexagon.

Karsten was right, I was supposed to be able to calculate triangle area correctly ;-). The ratio  is indeed pi / ( 2 . sqrt(3))= approx. 0.9069

@Karsten  J.D.Bernal would have been proud of you.

If you multiply the top and bottom of your $P_{3D}$ formula by $\sqrt{2}$ then it will look more like the $P_{2D}$ one.

Note that the classical idea of an atomic nucleus as a set of nucleon balls is very wrong, similarly as another classical idea of an atom as a planetary system of orbiting and spinning electron balls.

I suppose you have had bad luck until now as this task is in opinion of my non-math brain resistant to both general theoretical ( geometry, topology ) and experimental analysis.
I doubt there is a general formula ( or very complex at the best ), aside of the asymptotic formula for the group radius $R$ much bigger then the sphere radius $r$, wher $N_\mathrm{surf} = a \cdot N^{2/3}$.
The problem for small N as 100 and the related effective diameter is, there is collision between maintaining the given sphere arrangement and following the spherical shape.
Are there supposed centripetal forces corrupting sphere arrangement in favour of more spherical shape ?

If there is given the space filling factor $f$ ( a fraction of volume filed by spheres)  then for $R \gg r$ the sphere volume
$$V \simeq  \frac {4N}{3f} \cdot \pi \cdot r^3 \tag{1}$$
and
$$R=r \cdot {\left(\frac Nf\right)}^{1/3} \tag{2}$$
The outer layer of thickness $2r$ has volume $$V_\mathrm{s} = \frac 43 \cdot \pi \cdot  \left( R^3 -  \left(R-2r\right)^3 \right)= \frac 43 \cdot \pi \cdot    \left(  3R^2.2r - 3R.4r^2 + 8r^3 \right) \\= \frac 43 \cdot \pi \cdot r^3  \left(  6{(\frac Rr)}^2 - 12 \frac Rr + 8 \right) \\= \frac 43 \cdot \pi \cdot r^3  \left(  6  \cdot {(\frac Nf)}^{2/3} - 12 \cdot {(\frac Nf)}^{1/3} + 8 \right)\tag{3}$$
The volume of spheres in this layer is approximately $$V_\mathrm{s1} \simeq
V_\mathrm{s} \cdot f =  \frac 43 \cdot \pi \cdot r^3 \cdot f \cdot  \left(  6  \cdot {(\frac Nf)}^{2/3} - 12 \cdot {(\frac Nf)}^{1/3} + 8 \right) \tag{4}$$
The number of the surface spheres is then approximately
$$N_\mathrm{surf} \simeq \frac {V_\mathrm{s1}}{V_\mathrm{1 sphere}}= \frac {\frac 43 \cdot \pi \cdot r^3 \cdot f \cdot  \left(  6  \cdot {(\frac Nf)}^{2/3} - 12 \cdot {(\frac Nf)}^{1/3} + 8 \right) }{\frac 43 \cdot \pi \cdot r^3} \\
=f \cdot  \left(  6  \cdot {(\frac Nf)}^{2/3} - 12 \cdot {(\frac Nf)}^{1/3} + 8 \right) \tag{5}$$
For a very big N, we receive the previously posted:
$$N_\mathrm{surf} \simeq 6 \cdot f^{1/3} \cdot  N^{2/3} \tag{5a}$$
Note that this asymptotic solution for big N considers surface spheres as those touched by the thought wrapper foil around the object.
For the tightest arrangent is $f  \simeq 0.74 $, then
$$N_\mathrm{surf} \simeq 0.74 \cdot \left( 6 \cdot {(\frac {100}{0.74})}^{2/3} - 12 \cdot {(\frac {100}{0.74})}^{1/3} + 8  \right) \simeq 77 \tag{6}$$
The function $N_\mathrm{surf} = f(N,f)$ is not much sensitive to the space filling factor for small N. $N_\mathrm{surf} = f(100,0.6) \simeq 74$.

If we consider instead of that a loose rectangular arrangement as $\ce{NaCl(s)}$ crystal structure $5 \cdot 5 \cdot 4=100$, then the surface sphere number would be $100 - 3 \cdot 3 \cdot 2=82$.

For more information, see the table below.
\begin{array}{r|rrrr} \hline
&f&&& \\
N&0.74&0.7&0.65&0.6 \\ \hline
10&10&10&10&10 \\
20&19&19&19&19 \\
50&43&43&43&42 \\
100&77&76&75&74 \\
200&134&132&130&128 \\
500&270&266&261&256 \\
1000&450&444&435&425 \\
2000&744&732&717&701 \\
5000&1425&1402&1371&1339 \\
10000&2313&2275&2224&2170 \\
20000&3738&3674&3590&3502 \\
50000&7010&6887&6727&6559 \\
100000&11242&11044&10785&10511 \\
200000&17992&17672&17254&16813 \\
500000&33415&32815&32032&31207 \\
1000000&53294&52334&51079&49757 \\ \hline
\end{array}

As posed I really don't think you can get any better than the dimensional argument. To go any further the packing has to be much better described, for instance in terms of mathematical restrictions on the form the radial distribution function can take, and what constitutes "being on the surface" needs to be defined. For instance consider 1m balls and a 1 light year sphere. Is just 4 as a tetrahedron on the surface OK, with the rest tightly packed in the centre? Or can all 100 be on the 1 light year surface? From the vague description I don't think so, but I'm not sure.

@Ian Bush As surface I understand spheres touched by a foil if wrapped by it.

As surface I understand spheres touching other spheres for a ill defined "down, or below"...

The whole task is ill defined and rather self-purposed. I guess for big N I can assume the overall convex arrangement.

I tried evaluating $N_\mathrm{surf} \simeq \frac {V_\mathrm{s1}}{V_\mathrm{1 sphere}}$ but didn't come up with your result. What is the expression for $V_\mathrm{1 sphere}$?

@Buck Thorn I have fixed the error and added the correction for layer volume for  (NOT R >> r ) - the result 77 is very impressive, compared to 82 for the rectangular 5x5x4 case.

That's a nice result with the additional terms.

Here I used a numerical approach roughly equivalent to the analytic approach in Poutnik's answer. The procedure is to generate the smallest enclosing sphere containing within its radius 100 hexagonally close-packed beads, and then determine the number of beads in a second smaller concentric enclosing sphere. If r=1 is the radius of a bead, R=5.074 is the radius of the sphere enclosing 100 close-packed beads. Then a concentric enclosing sphere that is smaller by r/2 contains 69 beads, one that is smaller by r contains 52 beads, and one that is smaller by 2r contains 24 beads. Note that in order to place exactly 100 beads into a perfect sphere, the arrangement of beads is not symmetrical.
There are alternate (and better) ways of distinguishing between interior and surface beads, for instance to count the number of contacts. In an hcp arrangement this is a maximum of 12, with surface beads having less. Using such counting reveals that only 25 beads are not on the surface and make the max possible contacts. Like the previous estimation method, contact counting one does not check whether there are gaps etc. These clearly exist as shown in the following (interior beads are colored blue, surface ones black):

Since N=100 is not very big, some of the approximations used in the analytic approaches proposed in other answers are bound to fail. Consider my answer: the number of interior beads predicted by a numerical method is nearly the same as the number of surface beads according to the large N approximation. In fact the number of surface beads is more nearly that expected for a cube (see Poutnik's answer) rather than an ideal (large N) sphere. This problem was made to be solved by experiment (numerical or otherwise).


text:
Proof for second-order perturbation term of energy
In perturbation theory, if
\begin{align}{\hat {H}} &= {\hat {H}^{(0)}} +  λ{\hat {H}^{(1)}} \tag{1}\\
{ψ} &= {ψ^{(0)}} +  λ{ψ^{(1)}} +λ^2{ψ^{(2)}} + ... \tag{2}\\
{E} &= {E^{(0)}} +  λ{E^{(1)}} +λ^2{E^{(2)}} + ... \tag{3}\\
\end{align}
then, by applying these three equations on ground state $ψ_0$ and using the general time-independent ${\hat {H}{ψ} = Eψ}$ , it can be shown that:[1]
$$E_0^{(2)} = \sum_{n \neq 0}\frac{|H_{n0}|^2}{E_{0}^{{(0)}}-E_n^{(0)}} \tag{4}$$
The proof can be found in quoted reference; it is too long and not shown here. There is one step reads:

\begin{align}\int{ψ^{(0)*}_0}&\hat {H}^{(0)}{ψ^{(2)}_0} dτ + \int{ψ^{(0)*}_0}\hat {H}^{(1)}{ψ^{(1)}_0} dτ = \\ &E^{(2)}_0\int{ψ^{(0)*}_0}{ψ^{(0)}_0} dτ + E^{(1)}_0\int{ψ^{(0)*}_0}{ψ^{(1)}_0} dτ + E^{(0)}_0\int{ψ^{(0)*}_0}{ψ^{(2)}_0} dτ\tag{5}\end{align}
the first and last terms cancel.

I don't see why the first term (on LHS) is equal to the last term (on RHS). Based on the first three equations above, we can only conclude
$$\hat {H}^{(0)}{ψ^{(0)}_0} = {E}^{(0)}_0{ψ^{(0)}_0} \tag{6}$$
Why is $\hat {H}^{(0)}{ψ^{(2)}_0} = {E}^{(0)}_0{ψ^{(2)}_0} $? Also, why is ${ψ^{(2)}_0}$ also an eigenfunction of $\hat {H}^{(0)}$, with same eigenvalue as that of ${ψ^{(0)}_0}$? I believe it has something to do with linearity of quantum mechanics, but not able to prove it.

Reference
[1]: Peter Atkins, Julio de Paula. Physical Chemistry (8th Edition). OUP. 2006. Page 314-315.

As the unperturbed Hamiltonian is Hermitian it follows that
$$\int{ψ^{(0)*}_0}\hat {H}^{(0)}{ψ^{(2)}_0} dτ  = \left[\int{ψ^{(2)*}_0}\hat {H}^{(0)}{ψ^{(0)}_0} dτ\right]^*=\left[\int{ψ^{(2)*}_0}E^{(0)}_0{ψ^{(0)}_0} dτ\right]^*=E^{(0)}_0\int{ψ^{(0)*}_0}{ψ^{(2)}_0} dτ$$
by the definition of Hermiticity and the fact that all eigenvalues of a Hermitian operator are real.


text:
Is there an actual distinction between through-space charge transfer and excimer emission?
Variety of books and literature reviews consider excimer (or exciplex) emission something distinct from through-space charge transfer. Is this actually the case?
This question arises, because the more literature I read on this subject the less sure it seems to have any distinctive characteristics one from another. Mainly because the same ''evidence'' (e.g. the bathochromic shift of emission, lifetime increase, concentration studies) is used to argue for excimer or exiplex emission in one study, while other authors propose that such evidence supports CT emission.

In an excimer, e.g. classical example pyrene ($\ce{M + M^<=> MM^}$ ) the ground state is repulsive but the excited state is not and from which broad structureless emission occurs. In exciplexes the situation is similar except there has to be some charge transfer character as the two species are different. This makes the spectra etc sensitive to solvent polarity.  CT complexes are different as they are formed directly on photon absorption.

The featureles, broad and red shifted emission is observed for both excimers and CT bands. If we look at this from a chemistry stand point then excimers and exciplexes form from already associated ground state complexes or dimers (from pi-pi stacking). CT complexes can form between identical molecules (if these molecules are polarized enough) or different species (this is easier to achieve because one can easily choose electron rich and electron poor molecules). There are even examples of through space CT emission within a single molecule

The excimer / exciplex ground state is wholly repulsive, molecules $\ce{M  and M^*}$ diffuse together to form excimer $\ce{MM^*}$ the excimer or exciplex if formed from different species. The complex does not exist in the ground state. Emission is observed from both $M^*$ and $MM^*$. ($MM^*$ can also diffuse apart before fluorescence) The reason the excimer/exciplex emission is broad and structureless is because the ground state is repulsive. The exciplex is formed from different species so there has to be some charge transfer character, i.e. potential energy is sensitive to solvent polarity. Intramolecular excimers and exciplexes were first synthesised and studied in the late 1970's.
CT complexes are intimate in the ground state (via $\pi$ stacking etc.) and the CT absorption is generated upon absorption $(DA) + hv\to (D^+A^-)$. Any emission is then from this CT state plus normal fluorescence from either D* or A* should they be isolated in solution (as DA is in equilibrium with D and A) and then excited.
For details of kinetic see J.B. Birks, 'Photophysics of Organic Molecules' 1970, publ Wiley, and Organic Molecular Photophysics' 1975, publ Wiley. M. Gordon & W. Ware 'The Exciplex' 1975, Academic Press.

Indeed, this is what books tell us. Any thoughts on: https://pubs.rsc.org/en/Content/ArticleLanding/2020/TC/D0TC02562C#!divAbstract ? My main interest is in understanding the mechanism, because there are some instances where excimer emission provides a net PLQY increase and do not work as a general luminescence quenching route.

Time resolved experiments confirm this model in solution, but being in a crystal makes this very different, however, it must remain true that the ground state is repulsive  and the excited state MM* has a minimum otherwise it is going to be hard to call it an excimer.


text:
Total number of “hyperconjugable” hydrogens in 1-(3-ethyl-4-methylphenyl)-2-methylprop-1-ylium
How do I find the total number of “hyperconjugable” structures for 1-(3-ethyl-4-methylphenyl)-2-methylprop-1-ylium?

I read some sources stating that the total number of “hyperconjugable” structures is equal to the total number of α-hydrogen atoms. But which carbon is taken as the alpha carbon in a compound having no functional groups, just carbons?

How many hydrogens "contribute to hyperconjugation" is more understandable phrasing.

There is one obvious $\alpha$-position, which is the tertiary carbon atom adjacent to the carbon carrying positive charge. This contributes one hypercongujation structure. It looks like there are no more apparent structures, but the methyl group that is para to the cation can also participate in hyperconjugation

There are three $\ce{H}$-atoms on the methyl group, which give us three more structures. Thus, there are a total of 4 hyperconjugable hydrogens in this structure.
Interesting to note is that the ethyl group at the meta position does not participate in hyperconjugation. Can you find a reason why?


text:
The order of increasing covalent nature in group 12 (Zn, Cd and Hg)
Suppose $\ce{Zn}$, $\ce{Cd}$ and $\ce{Hg}$ were to form a bond with the same element; say $\ce{ZnCl2}$, $\ce{CdCl2}$ and $\ce{HgCl2}$ Then which of the compounds would be showing most covalent character
I feel like there are two schools of thought with regards to the covalent nature of group 12 elements.
One says according to Fajans' rule covalent character is greatest for smaller cations. Therefore the covalent bond character should be $\ce{Zn}$ > $\ce{Cd}$ > $\ce{Hg}$
The other says (my theory), due to weaker shielding effect of the $\mathrm d$-block there would be more effective nuclear charge for $\ce{Hg}$ than $\ce{Zn}$. So it should be able to cause more distortion and be able have stronger covalent bond. Therefore the order should be $\ce{Zn}$< $\ce{Cd}$< $\ce{Hg}$.
Which one is correct and why?

The ionic radii (VI coordination) increases from $\ce{Zn^2+}(\pu{0.74 Å)}$ to $\ce{Cd^2+}(\pu{0.95 Å)}$ to $\ce{Hg^2+}(\pu{1.02Å)}$. Applying Fajans' rule, the expected order of covalent character:
$$\ce{ZnCl2} > \ce{CdCl2} > \ce{HgCl2}$$
But consider the following properties:

Water solubility order (per $\pu{100 g}$ at $\pu{20 °C})$:
$$\ce{ZnCl2}~(\pu{432 g}) > \ce{CdCl2}~(\pu{120 g}) > \ce{HgCl2}~(\pu{7.4g})$$(Source: Wikipedia)

Colour intensity comparison due to cations:
$$\ce{ZnS}~\text{(white)} < \ce{CdS}~\text{(yellow)} < \ce{HgS}~\text{(black)}$$

Therefore, the correct order of covalent character must be
$$\ce{ZnCl2} < \ce{CdCl2} < \ce{HgCl2}$$
This may be due to increase of effective nuclear charge, for which Slater's rule also fails.

HgCl2 is a colourless/white solid, unless contaminated by metallic mercury.

They are all colorless crystals, not sure where yellow or black (!?) color comes from. It would be nice to see some references for physical properties and solubility data.

@andselisk Edited, also added references.

"(Source: Wikipedia)" You are not serious, are you?

@andselisk Yes, I am. Though, anyone can edit wikipedia, the information there is near perfect except few cases which can be with any source. Take this: According to "Lange's Handbook of Chemistry- 15E" the solubility data at room temp. are- $\ce{ZnCl2, 395g}$- page 554; $\ce{CdCl2, 120g}$- page 516; $\ce{HgCl2, 7.15g}$- page 533. You can see it is very close to what is in wikipedia. Don't put unnecessary sarcastic comments. You may know much, but should learn how to be polite with strangers!

@Apurvium Oh. I'm sorry you see this comment as sarcastic. Let's walk it through. First, there is no URL, just plain text which makes this addition useless. Second, Wikipedia is not a reliable source. Third, you ignored two comments regarding unrealistic colors and claimed you added references, which is not true. Fourth, "take this" part is what should've been added a reference in a first place. You may feel upset about critics, but others are equally upset about the quality  and transparency of your answer.

@Apurvium It's not a matter of who knows how much, it's the basic common sense of written communication, so it would be nice if you keep commands as to what I do or don't do to yourself.


text:
Transformed chemical potential of formation: something weird in Alberty 2003
I am trying to wrap my head around transformed chemical potentials, as described in Thermodynamics of biochemical reactions (R. Alberty, 2003), however, two equations seem to be contradictory in this book.
Lets consider a chemical species $AH_n$, which has n protons. The formation reaction of that species would then be
$$\ce{A^{n-} +nH+<=> AH_n}$$
According to equation 4.4-6 of Alberty 2003, $\Delta_f G$; the Gibbs energy of formation for $\ce{AH_n}$ (under constant temperature and pressure) would be

$$\Delta_f G = \Delta_f G^0 + RT \ln \{\ce{AH_n}\} \tag{4.4-6}$$

Where $\Delta_f G^0$ is the standard Gibbs energy of formation for $\ce{AH_n}$ and $\{\ce{AH_n}\}$ the activity of $\ce{AH_n}$.
According to equation 4.4-8 Alberty 2003, $\Delta_f G'$; the Legendre-transformed Gibbs energy of formation for $\ce{AH_n}$ (under constant temperature and pressure) would be

$$\Delta_f G' = \Delta_f G'^0 + RT \ln \{\ce{AH_n}\}\tag{4.4-8}$$

Where $\Delta_f G'^0$ is the Legendre-transformed standard Gibbs energy of formation of $\ce{AH_n}$, defined in equation 4.4-9 of Alberty's book as

$$\Delta_f G'^0 = \Delta_f G^0 - n(\Delta_f G^0_{\ce{H^+}} + RT \ln {10}^{-\mathrm{pH}})\tag{4.4-9}$$

that is

\begin{align}\Delta_f G'^0 &= \Delta_f G^0 - n(\Delta_f G^0_{\ce{H+}} - RT \ln (10) \cdot \mathrm{pH}) \\
\Delta_f G'^0 &= \Delta_f G^0 - n \Delta_f G^0_{\ce{H+}} + n \cdot RT \ln (10) \cdot \mathrm{pH}
\end{align}

At an ionic strength of 0; $\Delta_f G^0_{H^+} = 0$. So,

$$\Delta_f G'^0 = \Delta_f G^0 + n \cdot RT \ln (10) \cdot \mathrm{pH}$$

This development seems contradicted by what I can see in equation 4.3-9 of the same book; it states that the Legendre-transformed chemical potential of a pseudo isomer "1" is

$$\mu_1'^0 = \mu_1^0 - N_{H,1} \cdot RT \ln(10) \cdot \mathrm{pH} \tag{4.3-9}$$

where $\mu_1^0$ is the standard chemical potential of pseudo-isomer "$1$" and  $N_{H,1}$ is the number of hydrogen atoms in the pseudo-isomer "1", which is equivalent to the variable $n$ used above. I suppose that in conditions of constant temperature and pressure, chemical potentials are equivalent to Gibbs energies of formation. While this formula is given for a pseudo isomer, it should also be true out of this context, since it has to be true for a group consisting of a single pseudo isomer. What I do not understand is why the sign of the $- N_{\ce{H},1} \cdot RT \ln(10) \cdot \mathrm{pH}$ term is negative given the development of equation 4.4-9. Can someone tell me how those two equations (4.3-9 and 4.4-9) can be true at the same time?

What is equation 4.4-9?

What is the meaning of $\Delta_f G$, $\Delta_f G'$, $\Delta_f G^0$, $\Delta_f G'^0$, And is ln${AH_n}$ equal to ln$[AH_n]$ ?

I edited my question to clarify this

$- n(\Delta_f G^0_{\ce{H^+}} + RT \ln {10}^{-\mathrm{pH}})=- n\Delta_f G^0_{\ce{H^+}} - nRT \ln [\ce{H^+}])$. It therefore would seem to me that $\Delta_f G'^0$ is the standard free energy formation of $AH_n$ minus the standard free energy of formation of n times $H^+$ (plus an extra term ...)

The extra term appears to come from the expression for the reaction quotient. There appears to be terms for $A^{n-}$ missing. Otherwise the expressions are pretty normal.

I think your interpretation is right, but I still don't understand why there is this sign change in the expression of the transformed chemical potential, when compared to the expression of the transformed standard Gibbs energy change


text:
Why are Terminal Alkynes less reactive than internal alkynes?
I got in Paula Bruice under the section Mechanism for Mercuric ion catalysed Hydration of Alkyne

Terminal alkynes are less reactive than internal alkynes towards addition of water. Hydration will take place if Hg²+ is added to acidic mixture.

Why are Terminal Alkynes less reactive than internal alkynes?

The mechanism of hydration is an electrophilic addition and requires polarization of the carbon-carbon triple bond. This task is more difficult compared to case of alkene due to it having more %s character
In case of internal alkynes the carbocation intermediate formed ( since this mechanism is electrophilic addition the intermediate is a carbocation) is stabilized by the hyperconjugative effect and the inductive effect of the alpha carbons.
This support is taken away in case of terminal alkynes as their is no alpha-carbon. Now terminal alkynes can still be hydrated but only in presence of $\ce{Hg^2+}$ ion in acidic medium because of $\ce{Hg^2+}$'s high polarizing power(because it has a pseudo inert configuration). This polarization succeeds in catalyzing the reaction.
Ref.: Morrison and Boyd, sec 12.10 (hydration of alkynes)


text:
Finding the right order of carbocation stability
I should put these carbocations in decreasing order of stability.

Structure 1 is a primary carbocation.
Structure 2 is a secondary carbocation.
Structure 3 is a tertiary carbocation.
Structure 4 (a primary carbocation) and structure 5 (a primary carbocation) are less substituted, but one could form resonance structures.
Therefore I would say 5 > 4 > 3 > 2 > 1.
But I'm not sure why 5 should be more stable than 4. In both cases the positive charge is stabilized by the + I-effect through the sigma bond of the two methyl groups and both can be stabilized through the free electron pair of the heteroatoms.

I'm correcting myself: the right order should be 4 > 5 > 3 > 2 > 1. Nitrogen is less willing to give its free electron pair to the positive charge because it's more electronegative than bromine. So structure 4 should come before 5, would that be the right explanation?

5 will get electron donation from the N and the positive charge will locate on the nitrogen - so as drawn the cation is highly unstable

So, the right order should be:  3 > 2 > 1 >  4  > 5?  Structure 4 will be then unstable too. If it donates its free electron pair to the positive carbocation, the positive charge will be located on the bromine atom. I'm just a little bit confused by 4 and 5.

Br is nowhere near as much of an electron donor as N

@Waylander Don't confuse the OP! carl, he just meant that the mesomeric structure written there is less important then the other one, with double bond. As far as expected order for such question is concerned, your original order is correct.

https://chemistry.stackexchange.com/questions/37427/confused-about-carbocation-stability


text:
Effect on acidity of benzene due to electron withdrawing groups

With respect to the compounds I–V, choose the correct statement(s).

Compound II becomes more acidic when it has a $\ce{-NO2}$ substituent.
The acidity of compounds follows the order I > IV > V > II > III.
The conjugate base of compound IV is aromatic.
The acidity of compound I is due to delocalization in the conjugate base.

I am not able to decide about the first statement properly. I compared the acidity by considering the conjugate base, which is the same (benzyne) even if $\ce{NO2}$ group is added.
However, the solution says that it will increase acidity because of being an electron withdrawing group.

Do you really think conjugate base of II is benzyne?

Both the predictions are good. The $\ce{NO2}$ will actually be an effective electron withdrawing group, and the lost of that $\ce{H}$ will stabilize the compound (if there is and $\ce{NO2}$ on it). It is also important to rescue, that both effects will be stronger just on para or ortho position. You can make the different structures to check it.

It is talking about the acidity of the compound, so ortho para don't matter here, the acidity of the compound in general matters.

@PK Objection. The relative position of the nitro group to the deprotonated carbon influences the acidity of the compound.  Draw the deprotonated form and count the mesomeric formulae possible for each of them and you find o and p differ from the m-pattern.  It isn't limited to benzoic acids, although often the ortho effect is taught on them as example.

@Buttonwood The deprotonated form holds the extra electron (i.e. the negative charge) in an $\ce{sp^3}$ orbital which is parallel to the plane, whereas the pi system is perpendicular to the plane. So they can't interact with each other and conjugation is not a factor here. Simple inductive effect would predict ortho > meta >para in order of high to low stabilization of the anion.

Indeed... the answer is wrong (-1 from me) and @SRMaiti is absolutely spot on.


text:
Ring opening, nucleophilic substitution

Are the mechanism and the product for this given reaction right or is there anything wrong with it?
1,2 - Epoxybutane has originally the 2R configuration and the product got an R configuration as well.
I was not sure about, where the nucleophile attacks. Under acidic conditions steric hindrance will be ignored, right? In this case the hydrogen atom will be added to the less substituted carbon atom and the ring will be opened.

Aluminium binds to the oxygen first

Al loses hydride to form Alane, (AlH3 + H-). It binds to the oxygen as Alane freeing up the hydride to do the nucleophilic attack. Whether they fully separate or remain partially associated is not afaik fully established

Well, I wager that hydride transfer happens after coordination - hydride hardly ever is alone and Al can coordinate up to 6 groups.

...and LiAlH4 has aluminum impurities in it that can act as Lewis acids to get reduction going. This process frees up AlH3.

The free electron pairs of the oxirane may act as a Lewis base and coordinate with a Lewis acid like $\ce{Li+}$.  This form of activation facilitates the attack of the alanate $\ce{[AlH4]-}$ as a $\sigma$-nucleophile at the less substituted side of the epoxide.

Once opened, the nucleophilic oxygen attacks at the electron deficient centre of $\ce{AlH3}$ (Lewis acid).  But the $\ce{Al-O}$ bond hydrolyzes upon the workup, yielding the alcohol and the hydroxides of lithium and aluminum.


text:
How is equilibrium achieved in osmosis?
According to BRS Physiology book:

excessive NaCl intake will lead to an increase in the osmolarity of the Extracellular Fluid (ECF) compartment, and thus will lead to water shift from the Intracellular Fluid (ICF) compartment to the ECF so the ICF osmolarity increases until the osmolarities equal each other.

But if water shifts from the ICF to the ECF, wouldn't the osmolarity of the ECF decrease also as a result of the decreased concentration of the solutes, and this cycle continues?

What "decreased concentration of the solutes"?

Water is added so the concentration of the solutes in the fluid will decrease .

Well, there's an assumption then both salt and any water or whatever were put into ECF and it caused raising the concentration of NaCl otherwise the effect on ICF could wouldn't be like that. Also what kind of question is that? Obviously these concentrations are changing all the time and you didn't mention anything about water in post, one could eat, say, a spoon of salt..

According to this picture, as the water moves from the ICF to the ECF, the ICF osmolarity will increase, and the ECF osmolarity will decrease.  This will continue until they are equal (this simple picture ignores active transport).  Don't know why you are saying "the cycle continues" since, according to this picture, they meet somewhere in the middle, at which point the net flow of water stops.  Of course, actual physiology is far more complicated than this.

If the osmolarity of the ECF decreases, then water should shift from the ECF to the ICF, thus the ICF osmolarity will decrease, so water will shift from the ICF to the ECF, how could this cycle stop?  @theorist

@Positron12 What cycle ? It is very similar as if you put a hot stone in a cold water. Their T starts to converge each to other until become the same. Similarly, if osmolarity of ECF > ICF, water flows out of cells and osmotic pressure of ICF and ECF converge each to other. sure, it is more complicated as there is an active ion transport.

@Poutnik If water flows from the ICF to the ECF, that will decrease the osmolarity of the ECF and increase that of the ICF, so water will flow from the ECF to the ICF, increasing the osmolarity of the ECF and decreasing that of the ICF, and so on... How could equilibrium be achieved?

@Positron12  There is no  1 - 9 -> 9 - 1 -> 1 - 9  ....... there is 1 - 9 -> 3 - 7 -> 4 - 6 -> 5 - 5

Thank you! @Poutnik , I would accept it if you could put it as an answer.

If there is a difference in osmolarity and therefore in the osmotic pressure, water flows from solution with lower osmotic pressure to the higher one, decreasing this difference and therefore the flow itself as well.
The kinetic of this process follows the first order  kinetics. The flow rate is proportional to the osmotic pressure difference. Therefore, the water flow exponentially decreases with time, as the osmotic pressure difference decreases, reaching the equilibrium theoretically ( mathematically ) in infinite time.
$$ \frac {\mathrm{d}\Delta p}{\mathrm{d}t} = - k \cdot \Delta p$$
$$ \Delta p = \Delta p_{t=0} \cdot \exp {(-k \cdot t) }$$
The kinetic constant $k$ depends on the membrane permeability, water diffusion coefficient and geometry of the scenario.
Therefre, there is no cyclic oscillation of osmotic pressures like:

1 - 9 -> 9 - 1 -> 1 - 9

but there is an exponential convergence like

1 - 9 -> 3 - 7 -> 4 - 6 -> 5 - 5


text:
Synthesis cost of Fomepizole
Fomepizole is essential drug for methanol poisoning, I observed it cost too much like in 1000's of dollar for couple of ml.
Is cost due to patents, or due to synthesis cost,structure of molecule looks very small and easy, then why so much difficulty in synthesizing molecule ?

It's not a matter of synthesis, so it seems borderline off-topic as it's about economics of pharmacy.

its also about synthesis cost with purity enough for human use also. If someone make clear that it is relatively cheap to get high purity product, i guess one will get it in black market

Interesting that the cost in the US for the USDA-approved form is US$900 per 1.5 ml IV ampule. (Specific gravity is given as ~1, i.e. 1.5 g per ampule).
That chemical, 4-methylpyrazole, "For Research Use Only. Not Intended for Diagnostic or Therapeutic Use.", is available for US$74 per g .
So if standard use is 1.5 ml, then the research-grade chemical would cost ~US$130 per dose. Is the difference in cost due to difficulty in removing toxic substances from the research grade drug? Is the cost due to licensing fees, set to make a profit and to recover the cost of testing efficacy and safety?
As a side note, ethanol therapy is often used for alcohol dehydrogenase inhibition, to slow metabolism of methanol to formaldehyde, because, "fomepizole is more expensive than ethanol, and its use in pediatrics is off-label." Твоё здоровье!

How many companies hold a product licence for it? I suspect that the answer may be as low as 1. What are worldwide sales of the therapeutic material? I suspect not enough to support several companies holding a licence.

The WHO list (page 10) mentions it as sulfate (5 mg/mL á 20 mL) so it is not the pure stuff used (4 injections with 10 mg/kg body weight, apparently) while the German edition mentions as disadvantage the high price, 945 € for 5 × 20 mL). Among the 250 hits on EPO's esp@cenet with «fomepizole» as key, some try to improve the original formulation by Jazz Inc. (USA) already freezing around 20...24 C.

as described in patent WO2009117166A2, for example.


text:
Do higher orbitals have more energy or less energy?
I've recently learned that as an orbital gets larger, its energy gets closer to 0. Before this, I learned that when an electron moves down an orbital it releases the energy difference between those orbitals. This would suggest that higher orbitals have more energy, but apparently this isn't the case.
Can anyone identify the flaw in my understanding of orbitals? Also, is the energy of an orbital equal to the sum of the electrons in it? Thank you for your help.

Orbitals have negative energy, so closer to zero means higher energy. It is analogous to the energy of orbiting celestial objects.
Orbital energy is the energy of an electron in quantum state related to the orbital.


text:
Why isn't just a fully occupied valence s orbital stable?
Why arent elements like calcium with an electron configuration of $[\text{noble gas}]\ n\mathrm{s}^2$ stable, although all the populated orbitals are fuully filled? Why is it necessary to obtain an octet configuration instead? (Ignoring the trivial cases of hydrogen and helium)

Because it is an outer orbital that must be filled (although the value of "outer" is not intuitively obvious to the casual observer for transition metals).

I actually think this is a good question as it’s something that’s rarely discussed in introductory courses.

Elements with just 1 or 2 electrons in the most outer $\ce{s}$ orbitals, like $\ce{K}$ or $\ce{Ca}$, have very well screened off the central charge of the kernel. As consequence, the effective charge of the kernel and related energy needed to ionize the atom are relatively low.
OTOH, elements with just few missing electrons in the most outer $\ce{p}$ orbital, like $\ce{O}$ or $\ce{Cl}$, do not have well screened off the central kernel charge, which is relatively high, what leads to considerable electron affinity to some extra electrons.
Combining these 2 kinds of elements leads to energy release, as energy spent by ionization of the former is  compensated by energy release by capturing electrons by the latter and by a covalent contribution to the element bond.
In case of solid ionic compounds, there is additional large energy release du the ionic lattice energy.

No, it's not compensated. There's still covalent component needed for such reactions to happen.

Hmm, you may be right.


text:
Questions Regarding Osmosis
From  Costanzo's Physiology:

Osmosis of water is not diffusion of water: Osmosis occurs because of a pressure difference, whereas diffusion occurs because of a concentration (or activity) difference of water.

Also from Costanzo, describing two solutions 1, and 2 , seprated by a semipermeable membrane, and 1 having a higher concentration of solute:

The solute in Solution 1 produces an osmotic pressure and causes, by the interaction of solutes with pores in the membrane a reduction in hydrostatic pressure of Solution 1. The resulting hydrostatic pressure difference across the membrane then causes water to flow from Solution 2 into Solution 1.

My question is:
How does the solute produce an osomtic pressure, and what type of interactions with the membrane pores causes the reduction in hydrostatic pressure that ultimately drive water movement?

Osmotic pressure

is consequence of non zero net water diffusion,
which is consequence of non equal water activities on both sides of semipermeable membrane,
which is consequence of the fact dissolved solutes decrease activity of water.

Osmotic pressure of a free solution is formally an external pressure needed to be acting on this solution to keep diffusion equilibrium with pure solvent on the other side of a semipermeable membrane.
Applying external pressure then leads to the well known reverse osmosing, widely used in production of deionized water ( replacing expensive distillation ) from sweet water, or in desalination of sea water.
Osmosis continues until fully counteracted by extenal, e.g. atmospheric pressure. E.g. if there is osmotic pressure $\pu{1 atm}$, osmosis leads up to $\pu{10 m}$ high solution column ( eventually corrected by solution density ). Similarly, you would need to use external pressure $\pu{p > 1 atm}$ to start reverse osmosis.
If a book says the primary reason is hydrostatic pressure difference, it is wrong. We can have 2 solutions of the same density, but different osmotic pressures. Combining them on the opposite membrane sides would lead to disbalancing of solution levels, even if they would be initially the same and even if the hydrostatic pressure difference would be zero.

What about the hydrostatic pressure?

But the book states that the hydrostatic pressure difference is what drives water flow, how is the hydrostatic pressure difference generated at first?

@Positron12 The answer has been updated.

Thank you. Could you please give a look at this question https://chemistry.stackexchange.com/questions/140823/how-is-equilibrium-achieved-in-osmosis @Poutnik

In addition to Poutnik's answer, I would like to add some comments on the question:
The first statement 'Osmosis of water is not diffusion of water:....' is incorrect.
The cause of osmosis is simply diffusion: the solvent is able to diffuse through the semi-permeable membrane the solute is not. It is wrong to assume that diffusion occurs only in the direction of decreasing concentration it is rather the difference in chemical potential that drives this.
Statement 2 is also incorrect.  There is no special interaction with the solute  and the semi-permeable membrane that causes osmosis, other than that the solute is too big to pass through it but the solvent is not.
The pressure difference is not what causes osmosis but it is the equalisation of the chemical potential as equilibrium is reached. The pressure difference is a result of this process.
A solution made up with an involatile solute (say a protein) has a lower vapour pressure than that of the pure solvent due to the fact that some of the solution volume is now taken up with the solute. In thermodynamic terms this means that the chemical potential of the solution is less than that of the pure solvent (i.e Raoult's Law). To restore chemical equilibrium the chemical potentials have to become equal and this is achieved by solvent diffusing into the solution thereby diluting it (i.e. trying to make it pure solvent and so increase its chemical potential). The consequence is that the volume of the solution increases (e.g. as seen by pushing solution up a tube) and at a certain hydrostatic pressure, equilibrium is restored.

Regarding the first statement, the way I understood this -and correct me if I'm wrong- is that diffusion of water occurs when there's a difference in the amount of water between two solutions of constant concentration of solute, whereas osmosis occurs when the solute concentration is different, even if two solutions have the same amount of the solvent (water). @porphyrin

If you have 'constant concentration of solute' the two solutions are the same no matter what the volume. Osmosis occurs only via a membrane that is permeable only to solvent. Its not the amount of pure solvent that is important but it is the concentration of solute in solution on the other side of the membrane.  Diluting this tries to equalise the chemical potential.

"The diffusion model of osmosis is rendered untenable by the fact that osmosis can drive water across a membrane toward a higher concentration of water" - Wikipedia
What are your thoughts ? @porphyrin

Well, it is an experimental fact that the water does go into the higher concentration solution, call this diffusion or call this migration it makes no odds.  There is no other way it can occur. It is the equalisation of chemical potential that drives the process.

continued...  of course at equilibrium there is no net flow of water either way. Also as mentioned in my answer  'It is wrong to assume that diffusion occurs only in the direction of decreasing concentration it is rather the difference in chemical potential that drives this.'

How does the solute produce an osomtic pressure?

The solute creates a difference in the chemical potential of the solvent in two chambers separated by a selectively permeable membrane that allows passage only of the solvent. The system changes such as to reduce the imbalance in chemical potential and the only way for this to happen is by reducing the concentration of the solute in the high potential chamber through solvent diffusion from one chamber into the other.
Note that at the molecular level diffusion is a process that occurs all of the time in all fluid substances due to random molecular motion and irrespectively of chemical potential.

What type of interactions with the membrane pores causes the reduction in hydrostatic pressure that ultimately drive water movement?

In the case of osmosis, solvent movement is not caused by hydrostatic pressure. Rather the opposite. Increasing hydrostatic pressure in the more concentrated chamber increases the chemical potential of the solvent in the chamber, leading eventually to equilibrium.
The membrane is a selective filter.  Interactions of the solute with the membrane impairs solute passage (the pores allow only solvent to pass), sustaining the concentration gradient that drives osmosis.