text:
Chemistry 12: Calculating changes in enthalpy
I was wondering if I could please get some help with this:

In a coffee-cup calorimeter $\pu{100.0 mL}$ of $\pu{1.0 M}\ \ce{NaOH}$ and $\pu{100.0 mL}$ of $\pu{1.0 M}\ \ce{HCl}$ are mixed. Both solutions are originally at $\pu{24.6 ^\circ C}$. After the reaction, the temperature is $\pu{31.3 ^\circ C}$. Assuming all solutions have a density of $\pu{1.0 g/cm^3}$ and a specific heat capacity of $\pu{4.181 J/(g ^\circ C)}$. What is the enthalpy change for the neutralization of $\ce{HCl}$ by $\ce{NaOH}$?"

This is what I did, but I'm not sure if it's correct:
$$
\begin{align}
q(\text{surroundings}) &= m \cdot c \cdot \Delta T\\
&= \pu{200 g} \cdot \pu{4.184 J/(g ^\circ C)} \cdot (\pu{31.3 ^\circ C} - \pu{24.6 ^\circ C})\\
&= \pu{5.61 kJ}\\[2em]
q(\text{system}) - \pu{5.61 kJ} &= n \cdot \Delta H - \pu{5.61 kJ}\\
&= \pu{0.1 mol} \cdot \Delta H
\end{align}
$$
Thus: $\Delta H = \pu{-56.1 kJ/mol}$

In general, homework is not welcome here in the forum, which is probably why your question was voted down. However, since you have already given a solution here and only ask for a confirmation, I want to be so nice and reply to your question.
The short answer is: Yes!
In more detail, the following can be said: The heat from the neutralization added to the heat to warm the solution added to the heat to warm the calorimeter has to be equal to zero.
Thus: $Q(\text{neutralization}) + Q(\text{warm the solution}) + Q(\text{warm the calorimeter}) = 0$
In your example, it is assumed for simplicity that the calorimeter does not consume heat to heat up. Therefore $Q(\text{warm the calorimeter}) = 0$.
Thus:
$$
\begin{align}
Q(\text{neutralization}) + Q(\text{warm the solution}) &= 0\\
n \Delta H + m c \Delta T &= 0\\
n \Delta H &= - m c \Delta T\\
\Delta H &= -\frac{m c \Delta T}{n}\\
&= \frac{\pu{200 g} \cdot \pu{4.184 J/(g ^\circ C)} \cdot (\pu{31.3 ^\circ C} - \pu{24.6 ^\circ C})}{\pu{0.1 mol}}\\
&= \pu{-56.1 kJ/mol}
\end{align}
$$

"In general, homework is not welcome here in the forum, which is probably why your question was voted down." This is absolutely not true. The reasons for the downvotes could be many, sloppy formatting, a title without information, not explaining the thought process. The particular cases like this one, i.e. a please check my homework request, however, are not encouraged (and disliked) here, because they hardy ever lead to a good answer, and they are generally to specific to be of future use.


text:
Rate Expression with two products
Consider the two following cases
Case 1: Two Same Products
Given the following chemical reaction,
$$A + B \rightarrow C + C$$
where C are radicals, therefore I am not writing $2C$ on the right hand side.
If we are to write the rate equation for the above expression, which one of the following will be correct.
$$\frac{d[C]}{dt} = k \cdot [A][B]$$
or will it be
$$\frac{d[C]}{dt} = 2\cdot k \cdot [A][B]$$
Do we write the 2 given that there are two of the products on the right hand side?
Case 2: Two Same Reactants
Now considering the following,
$$2A + B \rightarrow C$$
Are the following correct?
$$\frac{d[A]}{dt} = -2k \cdot [A][B]$$
$$\frac{d[B]}{dt} = k \cdot [A][B]$$
$$\frac{d[C]}{dt} = 2k \cdot [A][B]$$
Do we multiply with the stoichiometric coefficient for each reactant? So if there were 2A and 2B then would we write $4[A][B]$? I find this part of kinetics to be a bit confusing, and it would be great to get some clarity on it.

The general relationship  is given in this answer https://chemistry.stackexchange.com/questions/103037/reaction-rate-coefficient-constant-multiplier-for-equation-with-multiple-atoms-c/103078#103078

@porphyrin Thank you for the link. It also helped out a lot.

In case 2, shouldn't $[A]$ be squared?

Case 2: Two Same Reactants
$$2A + B \rightarrow C$$
Are the following correct?
$$\frac{d[A]}{dt} = -2k \cdot [A][B]$$
$$\frac{d[B]}{dt} = -k \cdot [A][B]$$
$$\frac{d[C]}{dt} = 2k \cdot [A][B]$$

Let's call $k \cdot [A][B]$ the rate. Your equations simplify to:
$$\frac{d[A]}{dt} = -2 \cdot \mathrm{rate}$$
$$\frac{d[B]}{dt} = - \mathrm{rate}$$
$$\frac{d[C]}{dt} = 2 \cdot \mathrm{rate}$$
The first two equations are correct. They are consistent with the stoichiometry and they happen to be definitions of the rate. The last one is incorrect. If a certain amount of B reacts, the same amount of C is formed.

Do we multiply with the stoichiometric coefficient for each reactant? So if there were 2A and 2B then would we write $4[A][B]$?

No, just for the reactant whose concentration change you are considering on the left hand side.

Case 1: Two Same Products
$$A + B \rightarrow C + C$$
If we are to write the rate equation for the above expression, which one of the following will be correct.
$$\frac{d[C]}{dt} = k \cdot [A][B]$$
or will it be
$$\frac{d[C]}{dt} = 2\cdot k \cdot [A][B]$$

The lower expression is correct if the two Cs in the chemical equation refer to the same species. If not, they should have different names, like C1 and C2 or something. Your example reaction equation is kind of unusual, but something similar occurs when you have multiple steps in a reaction making the same species:
$$\ce{A + B -> C + D}$$
$$\ce{D + E -> C}$$
The net reaction would be (if there is sufficient E to use up all of D)
$$\ce{A + B + E -> 2C}$$
and you are not be able to distinguish species C made from the first or the second reaction step if everything happens in one phase ("one pot reaction").

@Karsten TheisCase 2: Two Same Reactants
$$\text{Is the following expression correct?}$$
$$\text{rate}=\frac{1}{2}\frac{d[A]}{dt} =\frac{d[B]}{dt} =\frac{d[C]}{dt}=-k \cdot [A][B]$$

@AdnanAL-Amleh Almost: rates are defined as positive, and when C is made, A and B are used up (decrease in concentration). So the correct expressions are:
$$\mathrm{rate} = -\frac{d[\ce{A}]}{2 dt} = -\frac{d[\ce{B}]}{dt} = \frac{d[\ce{C}]}{dt} = k \cdot [\ce{A}]^m [\ce{B}]^n$$ I added the exponents in the rate law to show that we don't now the order of reaction and would have to measure it.

Thanks If we call $k \cdot [A]^m[B]^n$ the rate. Can we simplify the following expression(second expression in case 2 in the above answer   )$$\frac{d[B]}{dt} = -k \cdot [A]^m[B]^n$$
to:
$$\frac{d[B]}{dt} = -k \cdot \mathrm{rate}$$ or
$$\frac{d[B]}{dt} = \mathrm{rate}$$

source: Stack Exchange

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text:
Preparation of standard Fe from FeSO4 • 7H2O
We were assigned to prepare a standard Fe for spectrophotometry determination of Fe. What we did is basically obtain a mass of FeSO4•7H2O, add a small a mount of distilled water before adding 18M of H2SO4 then dilute.
My question is what exactly is the purpose of H2SO4? I asked my prof about it and she said it was to prevent Fe2+ from oxidizing. I'm not really satisfied with that answer. Maybe because we did add hydoxylamine hydrochloride later to reduce the Fe3+ present so I don't see the need to do that addition aside from a different purpose. Can anyone provide the redox reaction of it if it really is a redox reaction happening? Thank you so much for any help much appreciated

Forget redox. Acid is intended to prevent hydrolysis.

And hydrolysis favours iron oxidation, as standard redox potential decreases for hydrolyzed forms. Mohr salt (NH4)2Fe(SO4)2  is preferred, AFAIK.

I think I already answered your question yesterday. Acid is not added to prevent oxidation (your teacher has a misconception). It is added to prevent hydrolysis and precipitation of iron hydroxides which happens over a period of time. The hydroxides are easily oxidized. In this sense the teacher is right. One could add any dilute acid (HCl, $\ce{HNO3}$) to achieve the same purpose. There is no redox reaction. Conc. sulfuric acid acts as an oxidizing agent at really high temperatures. Guess what, if sulfate ion were an oxidizing agent, it has to reduce itself as well to something? Right?

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


text:
Separation of methanol–water mix in a plate distillation column
I conducted an experiment to separate methanol and water. I am trying to determine the mole fraction of the distillate collected but having a little trouble with the calculation.
The flow rate was $\pu{55 ml/min}.$ When I collected my sample, I got a specific density of $0.815$ and a volume of $\pu{334 ml}.$ I tried to work out the mass, but I cannot use specific density as it is a dimensionless ratio.
Would I just assume the specific density is the density and set units to $\pu{g/cm3}$ so I can get the mass of the distillate and the concentration, and then determine the mole fraction from there, or does that not make sense?

How exactly you got specific density ? By areometer, or by pyknometry ? Are you sure your result is unitless as the true ratio ? Or, can it be the case the units are implied ? E.g areometers provide density in implied units. Pyknometry should provide results in units, considering temperature and water density. Sure, if you just took the mass ratio, you got massless specific gravity.

The density was measured with a hydrometer. Well the density value is a ratio so it must be unit less.

Well, the buoyancy is independent on water density.

Yes, you can do that.
Specific gravity is considered as unitless ratio of density related to water density.
Density as a specific mass is mass per volume, in $\pu{ g/cm^3}$ or $\pu{ kg/m^3}$
Regardless of country specific quantity naming convention:
Numerically, for 3 valid decimal digits,considering value uncertainty, the unitless quantity and quantity in $\pu{g/cm^3}$ are about interchangeable at $\pu{20 ^{°}C}$, as water density is near $\pu{0.9982 g/cm^3}$

Density should be mass/volume. Specific gravity is density of substance relative to density of water and thus unitless. Since the density of water is near 1, Density $\approx$ Specific gravity

@MaxW I have never heard the term specific gravity, only density ( relative specific mass ) and ( absolute) specific mass. But the truth is, it is translation to English. In English speaking countries, it may be different, so you may be right, also, consulting the Wikipedia. Generally, specific quantities are not unitless. like conductivity being specific conductance as conductance of 1m cube,, or specific heat capacity as heat capacity per mass.


text:
UK - Testing for Tap Water
Water is pooling in my garden next to my neighbour's fence. He has been harassing us in multiple ways for 8 years.
It is the only place on my property where this is happening. It has never happened like this before until last year.
Last year, in the driest summer we have had in years, we saw my neighbour had a hose up to the fence which he kept on which flooded our garden. This year we can see water trickling down from his garden, but due to rain cannot be absolutely sure as we cannot see the hose. The water turns on and off, but not relating to any rain. All winter with multiple weather conditions there was no pooling. It only happens during the summer when we try and use our garden.
This water pools a few inches deep, so it is easy to take a sample.
Is there any home way of testing this water to see if it is tap water or rain water? This test has to be something that can be purchased in the UK.
If not, is there any cheap way of getting someone else to test it?
His house is a good distance from the trickling area, so it is unlikely to be a leak in his garden as there will be no piping there.

I recommend striking up an innocent conversation about water prizes with him. Probably with a hint about how lucky you are living downhill and not having to water your garden so much.

I think the question is clear: to tell by testing if water is of rain or tap origin.

I think the question was clear, but have edited it to make it even easier to understand.

Tap water is treated with chlorine so a chlorine testing kit should do the trick.


text:
How to find a full electron basis set that approximate a given ECP basis set?
In computational chemistry of transition metals, there are many popular basis sets, from pople+LANL2DZ mixed basis sets, to Ahlrich and Karlsruhe basis sets like def-SVPD, def2-TZVPPD. Often the ECP versions of these basis sets were used to minimise computational expenses. However there are molecular properties such as topology analysis and energy decomposition analysis where a full electron basis set is not only recommended but also necessary.
One of the full electron basis sets commonly used in the literature is the ANO-RCC basis set. However, such a basis set is so large that it computationally infeasible in applications like the EDA2 calculation in serial qchem (where there are no parallelisation implemented) and barely feasible in generating wfx files within hours for analysis. Thus a slightly smaller full electron basis set (if any) is needed to be selected.
Say I start with some ECP basis set X, which I can read the list of all exponents and angular momentum functions used, what is the recommended approach to upgrade it into a full electron version of it so it retains most of the properties of the original basis set while making it full electron. Do I need to perform a benchmarking again and compare single point energies at optimised geometries to be sure?

Shouldn’t you benchmarking to the property which you want to study? It is not clear what you want to study.

I want to perform a energy decomposition analysis to elucidate steric effects. The problem is that ECP basis disallow the decomposition of the frozen interaction energy term into the pauli and dispersion term thus I cannot directly benchmark that. The only thing I can benchmark is the SCF energy but I am not sure if it is enough

How SCF energy is relevant in this case? Also, what is the discrepancy you would like to check in a benchmark?

I want to e.g. check whether e.g. def-SVPD-Rifit (the full electron counterpart of def-SVPD) is a good approximate for LANL2DZ or other double zeta ECP basis sets such that I will not get qualitatively different results thus allowing the def-SPVD-Rifit result to be comparable to the energy, TS barriers and geometry of the LANL2DZ basis. However, I am not sure what benchmark parameters I need to ensure the energies calculated using the two methods will be comparable to analyse together (e.g. SCF and Gibbs free energies for the calculations with def-SVPD and LANL2DZ vs EDA2

energies using def-SVPD-Rifit. That is, if the basis sets differ too much, I cannot draw any meaningful correlations between the EDA2 energies and the Gibbs free energy unlesss I recalculate the whole library of 1000+ molecules with def-SVPD-Rifit (all optimisations and vibrations), which is infeasibale in my project

This question may be better received on the new Materials Modeling SE: https://materials.stackexchange.com/


text:
Which pathway is better for robinson cycloaddition? 2-phenylacetaldehyde + 3-buten-2-one or 2-phenyl-2-propenal + acetone?
So this was a question on my orgII exam yesterday. The target molecule is 4-phenyl-2-cyclohexenone. By retrosynthetic analysis there is only one possible aldol condensation reaction, but two possible michael addition reactions afterwards. My "guess" was that the best one is 2-phenyl-2-propenal + acetone because in the other case the negative charge is delocalized throughout the benzene ring reducing the nucleophilic character of the a-carbon and also activating the ring in which case EAS could take place with the ring acting as the "Michael Donor".

The phenylacetaldehyde route is better because of the easy availability of the starting materials

Either route is subject to the difficulty that the double bond will isomerise into conjugation with phenyl ring giving 4-phenylcyclohex-3-enone

@Waylander I went and asked my teacher and it turns out the same exact exercise was solved in the textbook. The answer was what I said, but for very different reasons. Aldehydes undergo aldol condensation in base much faster than simple ketones. So, the issue with using the 2 - phenylacetaldehyde is aldol condensation with another same molecule as a side reaction.

This is where I take issue with your teacher. In the real world you would do the phenylacetaldehyde route (modified to reduce self-condensation) because of the easy availablility of the starting materials. 2-Phenyl-2-propenal is not widely available (ask your teacher how they would make it), you would have to make it first which means running and working up  two reactions rather than one.

To be honest I totally agree with you. I thought that using an excess of the ketone would probably greatly slow down the self-condensation process among other various procedures with the same goal. However, when she pointed that the exercise was identical to the one in the textbook I didn't continue the debate. Perhaps it's her mistake or perhaps she wanted an "easy" and already solved exercise to aid those who didn't study much. To my knowledge, there wasn't any other imprecise problem in the rest of test.

In my experience textbook answers rarely mirror the actual practice of chemistry. They are there to illustrate a principle.


text:
CuAAC "Click" reaction giving inconsistent results
I apologize in advance for the large amount of text and thank all those who take the time to read it.

The "Click" reaction after Fokin and Sharpless is known to achieve high yields without long reaction times or harsh reaction conditions. The reaction is used mainly because of its high selectivity. Many publications prove that the reaction is not significantly disturbed by other molecules and does not place any special demands on the choice of solvent. In the search for a suitable reaction to derivatize a molecule, which is shown in scheme 1, I remembered the CuAAC reaction, which was presented in a lecture.  So I ordered copper acetate and sodium ascorbate as well as 3-azidopropylamine and started with the reaction.

Scheme 1: Reaction of the alkyne with 3-azidopropylamine to the "click" product.
I did a first test with 100 ng of the alkyne. To this, 40 eq of azide as well as 40 eq of $\ce{Cu(OAc)2}$ and 80 eq of $\ce{NaAsc}$ were added. The alkyne was present as dry matter, the azide was dissolved in $\ce{MeCN}$, the copper acetate and the sodium ascorbate were present as aqueous solutions. The total volume of 20 µL was filled with a $1:1\,\ce{H2O}:\ce{THF}$ mixture. Incubation was performed at room temperature at a rotating wheel with a reaction time of one hour. The reaction mixture was then quenched with 80 µL of 0.2% formic acid and injected into an LC-MS/MS. A signal could be assigned to the calculated mass of the derivative, which confirms that the reaction was successful.
Since 100 ng is way too much for my purposes (I would have to be in the range of about 1-10 pg) the reaction was carried out with a dilution series (1 pg, 10 pg, 100 pg, 1 ng, 10 ng, 100 ng). Here it turned out that a usable signal can only be found with 100 ng.
In order to optimize the reaction, it was investigated whether $1:1\,\ce{H2O}:\ce{THF}$ is the correct solvent or whether $1:1\,\ce{H2O}:\ce{MeCN}$ does not work better. This could be refuted very well. The signals are significantly better in a $1:1\,\ce{H2O}:\ce{THF}$ mixture (Figure 1). The reaction was then stopped and analyzed after different reaction times. Here it was shown that the turnover is lower with increasing reaction time (Figure 1).

Figure 1: (a) Peak intensities for the "click" product using 10 ng, 50 ng and 100 ng of alkyne in a $1:1\,\ce{H2O}:\ce{MeCN}$ and a $1:1\,\ce{H2O}:\ce{THF}$ mixture. (b) Average peak intensities for the "click" product from three samples. In each sample 50 ng of alkyne were used. The reaction was stopped after 1, 2, 3, 4, 5 and 6 hours.

Now to my questions:
Do you have any idea what else I could try to do to get the reaction working in the quantity range I need?
Why is the turnover lower with increasing reaction time? Even if a longer reaction time does not bring any profit, I would not have expected a reduction of the intensity. As far as I am aware, 1,2,3-triazoles are pretty stable, even under harsh conditions. My starting molecule is also quite stable and should not decompose in the short time and under the mild reaction conditions.


text:
Liquid alternative to alcohol to lower density of water
I am trying to create a suspension of polypropylene particles in a water based liquid. I want the density of the liquid to be more than or equal to the density of polypropylene (855 - 946 kg/m3) so that the particles don't float or sink too easily. It doesn't have to precisely match, just be close. I had thought of using alcohol, but the problem there is that I need to house the suspension in a container made from either acrylic or polycarbonate, which are both not very resistant to alcohols.
What I'm trying to find is a safe and hopefully cheap liquid that can lower the density of water but not attack acrylic or polycarbonate.
Thanks

You mean less or equal. Or more or less equal? ;) Depending on the actual density of your polypropylene (.85 to .95 is quite a large margin), you get away with an ethanol concentration that doesnt harm your housing.

As you need decreasing of density, the only reasonable way is using organic solvents miscible with water, with alcohols being the least evil. I suppose methanol may be less aggressive, being closer to water, if health safety is not an issue. Alcohol-water mixtures will be much less aggressive than pure alcohols. Try long term testing of plastic sample with intended liquid. I probably would not like using 25% w/w ammonia with density 0.91 g/ccm. SE also https://www.organicdivision.org/wp-content/uploads/2016/12/organic_solvents.html


text:
Mathematical justification for Le Chatelier's principle
At equilibrium
$$
\begin{align}
K &= \exp\left(\cfrac{TΔS^\circ - ΔH^\circ}{RT}\right)\\
⇒ \frac{\mathrm d \ln K}{\mathrm dT} &= \frac{ΔH^\circ}{RT^2}
\end{align}
$$
If enthalpy change is positive, change in $\ln K$ w.r.t. $T$ is positive. Hence, $\ln K$ and therefore $K$ increase (position of equilibrium shifts to the right) as $T$ increases. And, $\ln K$ and therefore $K$ decrease (position of equilibrium shifts to the left) as $T$ decreases.
If enthalpy change is negative, change in $\ln K$ w.r.t. $T$ is negative. Hence, $\ln K$ and therefore $K$ increase (position of equilibrium shifts to the right) as $T$ decreases. And, $\ln K$ and therefore $K$ decrease (position of equilibrium shifts to the left) as $T$ increases.
Can you give a similarly mathematical reasoning for why a reaction
$$\ce{A(g) <=> 2 B(g)}$$
has its equilibrium position shifted to the left as the pressure of the surroundings increases?

You cannot use that sort of relation to rationalise changes in equilibrium position. The reason is that the van ‘t Hoff relation (the one with d ln K/dT) tells you how the equilibrium constant changes with temperature. When you change the pressure, the equilibrium constant K does not change.

The question is unclear. You have to carefully specify what is in the gas phase (just A and B, or inert gasses as well). $$$$The connection between temperature change and pressure change is that for both, the Gibbs energy of reaction is zero before and after, and equilibrium concentrations/partial pressures change. Of course, @orthocresol is right in saying that in one case, K changes and Q will change to match it again, and in the other case K does not change and Q will return to its starting value (but again, the concentrations change in either case).

I think that I might have an answer to my own question:
$$K=exp \left( \frac{TΔS^o-ΔH^o}{RT} \right) $$
implies that K is dependent on T, but not on total pressure of system, p (nor concentrations).
For
$$ A_{(g)}⇌2B_{(g)} $$
$$ K= \frac{{p_{B}}^2}{p_{A}} $$
Partial pressure of a gas = mole fraction (mf) x total pressure of system
If p is increased then K initially increases:
$$ K=\frac{(mf_B.p)^2}{mf_A.p} $$
$$ ⇒ K=\frac{{mf_B}^2.p}{mf_A} $$
But p doesn't affect K, yet K has increased. So the mole fraction of A must be increased by shifting the position of equilibrium to the left (thus decreasing the mole fraction of B), in order to re-establish the value of K to before p was increased.


text:
What is the Criteria for a Compound Being Non-aromatic vs. Anti-aromatic?
Anti-aromatic compounds are formed with a cyclic system of planar, overlapping p orbitals containing a non-Huckel number of electrons (i.e. not 4n+2 electrons -n being a natural number-). Non-aromatic compounds do not contain a continuously overlapping cyclic system of p orbitals. If a compound is antiaromatic, it is generally possible for the molecule to adopt a different conformation in order to disrupt the continual overlap of p orbitals. For example, 1,3,5,7-cyclooctatetraene, which would otherwise be antiaromatic, will adopt boat conformation to alleviate ring strain and antiaromaticity.  Some molecules containing a lone pair or empty p-orbital can hybridize in order to eliminate anti aromaticity. What I don't understand is when a molecule will hybridize its lone pair or adopt a different conformation to become non-aromatic. Are non-aromatic compounds more stable than antiaromatic compounds all of the time? Will a molecule that is capable of bending or hybridizing it orbitals in order to avoid anti-aromaticity always do so, or are there special circumstances? Also, can an empty p orbital sp3 hybridize to render a compound non-aromatic?

True antiaromaticity is very rare.  Most "antiaromatic" things find a way to become nonaromatic.

https://chemistry.stackexchange.com/questions/57678/is-a-compound-aromatic-if-it-also-has-anti-aromatic-rings


text:
Why do alkyl substituents stabilize alkenes?
My textbook states that because $sp^2-sp^3$ bonds are stronger than $sp^3-sp^3$ bonds, alkenes are more stable than alkanes. The rule given is that bonds with more $s$ character are stronger.
However, it then states that an alkene with more alkyl substituents at the C=C double bond is stronger than an alkene with just hydrogens attached around the double bonded carbons. Carbon $sp^2-1s$ Hydrogen bond is stronger than Carbon $sp^2-sp^3$ Carbon bond. So how can my textbook use bond strength to explain why the alkyl substituted alkene is more stable than the non-substituted alkene?

If your understand hyperconjugation in MO theory, you may want to see this: https://chemistry.stackexchange.com/q/20362/44877

@TanYongBoon thank you. yes this is perfect response! was confused how hyperconjugation applied to case of alkenes vs carbocations

@TanYongBoon Btw, your profile is very impressive for a HSer. do you have plans of US college and PhD in chemistry in future?

Welcome to chemistry.SE!. Since you are new here, I advise you to go through the policies of our community. I don't want to sound mean but comment section is not for friendly chatting. I suppose, you have already found what you are looking for. If you have any further doubts, feel free to comment or edit the question. Thank you.

@NilayGhosh Does stackexchange have a chat function?

Of course it has, check The periodic table

@ebehr He means https://chat.stackexchange.com/rooms/3229/the-periodic-table, but you'll need 20 rep to chat.


text:
How are orbitals displayed in computational Chemistry software?
In typical software like Gaussian or GAMESS, the molecular orbitals (or even the basis functions) can be displayed as surfaces.
I understand that the surface is probably constructed from a series of points that give a fixed value when substituted into the wavefunction. However, how are these points calculated? Since typical software usually use GTOs, I’d like a formula if possible using those.
Thanks!

This question would probably be better served on Matter Modeling SE

Only the program that you used to calculate the electronic structure knows all the idiosyncrasies of the computation to correctly evaluate the orbitals. Therefore, what most people do is to export the evaluation of the orbitals in a grid. Then, a plotting program performs an extraction of a polygonal mesh with a marching cubes (or similar) algorithm.

That is not really true. The question specifically refers to GTO codes (Gaussian and GAMESS). Here the molecular orbitals are well defined by the basis set and their coefficients, both may be found in the output file. Using for example the molden software, the Gaussian output can be opened directly and it will calculate and plot the grid points required for the 3D isosurface of each MO.

I would not say that it is not possible, but that it is not straightforward. You can not blindly rely on molden because there have been errors disclosed in the past. E.g. https://github.com/psi4/psi4/issues/60. In that case the programs were not writing on file a correct representation of the internal parameters. You have to be prepared to inmerse in others code to get an answer because the documentation may be lacking or is incorrect. The authoritative source is the code. Besides, in some cases the relevant information is not even written in a molden format.


text:
Which pressure is used for calculation of standard enthalpy change of the reaction with two or more gases?
Standard state conditions for standard Gibbs free energy change stipulate (among other conditions) that the partial pressure of each gas is $\pu{1 bar},$ e.g. for the reaction
$$\ce{2 A(g) + 3 B(g) <=> 4 C(g) + 5 D(g)}$$
for standard Gibbs free energy change calculations the partial pressures are equal:
$$p(\ce{A}) = p(\ce{B}) = p(\ce{C}) = p(\ce{D}) = \pu{1 bar}.$$
If we are calculating the standard enthalpy change for this reaction, are the partial pressures still $\pu{1 bar}$ for every gas? Or is the external pressure is constant and is $\pu{1 bar}?$
Because if the external pressure is $\pu{1 bar},$ then the partial pressure for every gas can't be $\pu{1 bar}$: if $\pu{2 mol}$ of $\ce{A}$ are at $\pu{1 bar},$ then $\pu{3 mol}$ of $\ce{B}$ will have a higher pressure if they are all present in the same closed container.

It isn't the partial pressure.  In the standard state, the species are taken to be pure, each at a total pressure of 1 atm.  They are not mixed in the standard state.

Or you refer to a mixture in standard state, in principle one could do that, and choose 1 bar as pressure, for instance. But mixture composition would be then a parameter, too.

See https://chemistry.stackexchange.com/questions/114673/how-to-measure-quantities-at-standard-state-when-this-state-is-a-hypothetical-on

@Alchemista The term “standard state” for a gas implies pure species.

External pressure is not relevant in a closed container.

@jimchmst External pressure is not relevant to a fixed volume closed container, where would be evaluated rather Helmholtz than Gibbs energy.

For this reaction, the standard state is a hypothetical one: All (gas) components are present as pure at standard pressure. Of course, you can create that state, but it would not react because each species is in its separate container.
You can still determine the standard Gibbs energy by extrapolation. The easiest way is to let a gas mixture come to equilibrium (so the Gibbs energy is zero), measure the concentrations, assume an ideal gas, and plug them into the equation below:
$$\Delta_\mathrm rG^⦵ = -RT\ln K \tag{1}$$
This may be derived from
$$\Delta_\mathrm rG = \Delta_\mathrm rG^⦵ + RT\ln Q \tag{2}$$
which is the "recipe" to calculate the Gibbs energy of reaction at any state from that of the standard state (assuming ideal gas or solution), combined with the fact that at equilibrium, $\Delta_\mathrm rG$ is equal to zero and $Q = K$.

[comment in the linked question] how can gases start out as pure component, they have to mix for the reaction to occur? Also, how can each of the products and reactants have 1 bar pressure, the pressure would depend on the no. Of moles of the different reactants and products(which are not the same for all reactions)?

The Gibbs energy of reaction is a state function. You can start with the standard state, mix the gases, let the reaction proceed for a while, and then separate the gases again. For each step, you have to measure the change in Gibbs energy, or calculate it. Then you add them all up to get the somewhat surprising net reaction of pure, separated substances reacting to make a bit more of the product, yielding pure, separated substances again. Curiously, you never actually have to do this experiment. The Gibbs energy of mixing is baked into the formulas above.

The standard Free energy is the free energy change from reactants and product each in their standard state and the activities at equilibrium. Mix them all up introduce a catalyst or initiator and measure the energy change. Alternatively establish equilibrium, measure the activities of products and reactants and determine Keq. Delta G[0] = -RTlnKeq. Again the energy difference between the states of the reaction quotient equal to the equilibrium concentrations and the state where all activities are 1 [ln 1= 0]. For gases the pressure[fugacity] is equal to 1 for each gas. the total pressure [fugacity] is the sum of the pressures of each gas [volume is constant].
There is a subtle problem with gases that I do not understand. Upon mixing there is an entropy [and possibly small energy] of mixing and the mole fractions are no longer unity. This can be removed by establishing an electrochemical cell such as in the electrolysis of water. The O2 and H2 can be separated physically, connected electrically, each at activities of 1 and mole fractions of 1 in contact with products, activity =1. Now the standard potential can be measured and the equilibrium constant calculated. [taking into account the differences in the anode and cathode chambers]
Standard conditions may or may not be physically attainable. The energies can be calculated from equilibrium data, electrode potentials, and various Born-Haber cycles using cascaded reactions.


text:
Can aroma compounds travel through metal boards and glass?
A friend from customs said "if drugs have been in a glass jar for more than a day, the aroma compounds travel through the lid and the dog can detect it, especially after 2-3 days. They can travel through almost any material given enough time"
If you decanted freezing perfume into a very cold jam-jar and washed it it with water and set it to react for a few days at 25C', would the lid permeate enough aroma compounds to be detected, i.e. by a police dog or a bloodhound?

I doubt your friend knows what he's talking about. Most likely either outside of the container has been contaminated or the seal between the jar and lid is not tight enough. It's not like he would know

The part about the dog is possibly true. As for the rest... wait, how did the compounds get into the jar in the first place? Did they have to travel through the glass to get there?

On another note, the word "aromatic" in chemistry came to be used in a different sense, quite totally unrelated to what you mean. For that reason, I took the liberty of editing your tags.

While as stated in my comment I don't believe this is the reason dogs can detect those substances, every material has a certain permeability for different molecules.
https://en.m.wikipedia.org/wiki/Permeation


text:
Why does concentration of solutions of alkali metals in liquid ammonia change the magnetic properties?
In "NCERT Chemistry for Class XI - Volume II", in chapter "s-block elements", under the topic "Chemical Properties of alkali metals", it is given:

(vi) Solutions in liquid ammonia: The alkali metals dissolve in liquid ammonia giving deep blue solutions which are conducting in nature.
$$\ce{M + $(x + y)$ NH3 -> [M(NH3)_x]+ + [e(NH3)_y]- }$$
The blue colour of the solution is due to the ammoniated electron which absorbs energy in the visible region of light and thus imparts blue colour to the solution. The solutions are paramagnetic and on standing slowly liberate hydrogen resulting in the formation of amide.
$$\ce{M+(am) + e- + NH3(l) -> MNH2(am) + 1/2 H2(g)}$$
(where ‘am’ denotes solution in ammonia.) In concentrated solution, the blue colour changes to bronze colour and becomes diamagnetic.

In the last set of lines, it is given that in concentrated solution, the blue colour changes to bronze colour and the paramagnetic solution becomes diamagnetic.
How does the change in concentration affect the magnetic properties of the solution?
Wikipedia has nothing to offer to clarify my doubt.

See this question and links therein

@NilayGhosh, As far as I read that, I find no answer to my question. But the source specified was useful.

I always suggest students to try Google Scholar when a simple Google search fails. I just searched three keywords : alkali metals ammonia solutions and the third result is highly relevant.
When your book talks about "in concentrated solution", it means more alkali metal in liquid ammonia. This paper, which you should search in Google Scholar, is titled A Revised Model for Ammonia Solutions of Alkali Metals, Journal of the American Society, Year 1962, volume 84, pg 2264). It talks about the formation of dimers of metals of the type $\ce{M2}$, because as you increase the metal conc. in ammonia solution conductivity decreases.
The authors write, raising objections to older models as completely wrong show the following evidence:

"We picture the » species as a quadrupolar ionic assembly of $\ce{e- + 2M+}$ in which there is little distortion of either the ammoniated electrons or ammoniated metal ions. Presumably the electrons and ions are held in a square or rhombic configuration."

In short the picture is not that simple and no wonder the authors of a grade 11 book never discuss more details. You can start from here and search more.


text:
Accuracy of Quantum Chemistry ML models
I am trying to compare the performance of few Quantum Chemistry property prediction ML models. I was looking at the following table from DOI:
10.1039/c7sc02664a

The problem is that it does not contain any units. A preprint (https://arxiv.org/pdf/1909.00259.pdf, Table 4) suggests that the energy units are Hartree. Among energy properties, HOMO, LUMO, and gap have reasonable values for Hartree unit. However, values reported for U0, U, H, and G look absurd. As a comparison, look at MAE for mean value of each property:

These are in [eV] units (1 Ha = 27.212 eV). In other words, for instance MPNN models MAE for U property is 55.78 [eV] vs 8.25 [eV] for the mean baseline. This does not make sense at all. [k cal/mol] sounds more reasonable but then why half the energies in one unit the other half in other unit, and then why there are no units in a paper published by Royal Society of Chemistry. Seems like I'm missing something. Thanks!

Yes, you're exactly right - multiple papers in chemistry ML drop the units.
There are even comparisons (usually by statistics, ML or comp. sci. researchers) where models are compared by "averaging" errors down a column like that. Of course that's meaningless, since you can't average electron volts or Hartree (energies), Debye (dipole moments), and volume (polarizabilities).
Worse, energies are often computed as atomization energies - so for large molecules, they can be enormous...
In my opinion, a more relevant and meaningful statistic would be the Mean Absolute Percent Error (MAPE) which is unitless and easier to understand (e.g., 1% error? 0.01% error?)
That said, the original QM9 paper gives the units Table 3:
Scientific Data (2014) 1, art. 140022

I'm adding an image of the data table - I believe this is under fair use, since one cannot copyright data - in this case the units of the QM9 set.

Thanks for the feedback @GeoffHutchison. I agree that it seems most reasonable to use datasets original units, but for atomization energies it doesn't seem to be right. I've decided to try recreate these results with deepchem. That seems to be the only way to find out.

Just to put it in perspective, the reason that I'm so suspicious is that if they are in fact using Ha units, then in a world that DFT error for U0 is 0.1 [eV] and people are already claiming to not only achieving this, but achieving chemical accuracy (0.04 [eV] I think), the Japanese paper is proudly reporting 1.35 [eV] and the benchmark method is 54.97 [eV]! So I believe that the Japanese paper is dead wrong (given that it's just on arxiv).

I have reviewed multiple manuscripts in the field that claim state-of-the-art accuracy when they're unaware of units or of other papers with better accuracy. I'm not sure what the "true" state is for U0 on QM9, but it's definitely < 1 kcal/mol

It is worth knowing the downsides of MAPE.


text:
Would graphene be a good efficient conductor of electricity in mobile phones
Gold is commonly used in mobile phones as an electrical conductor,as it is apparently better than copper at conducting electricity
Quoting from extremetech.com :
"The materials recovered are reused in new electronics parts and the gold and other precious metals are melted down and sold as ingots to jewelers and investors as well as back to manufacturers who use gold in the circuit boards of mobile phones because gold conducts electricity even better than copper."
(Some of the information is irrelevant but i was to lazy to take it out)
But when I actually look at this table it says that gold is worse than copper at conducting electricity and has a higher resistivity rate
Please follow the link to see what I'm talking about
Table of conductivity
So it's either that the information on the table is wrong or that the tech website was wrong (and it probably was)
Information found on investorintel.com
"Graphene is fantastically good at conducting electricity; its current density is 1,000,000 times greater than copper’s and its intrinsic mobility is 100 times greater than that of silicon. Electrons move through graphene with virtually no resistance and without mass. This means that graphene can carry electricity more efficiently, precisely and faster than any other material. It is anticipated that graphene will enable lithium batteries to have more than 10 times the electrical retention/capacity as anything presently in use."
But there are problems of it being to expensive to manufacture, but I do not understand why?
To clarify what I'm asking
1. Is gold better at conducting than copper
2. Is graphene better and more efficient at conducting energy (thermal and electrical) than any other known elements?

All kinds of materials, engineering, and manufacturing details are buried in such a comparison...

It's more like "too broad" though, if additional fluff won't get edited out.

If you trust the wikipedia (or whatever source it quotes) then copper is more conductive: https://en.wikipedia.org/wiki/Electrical_resistivity_and_conductivity#Resistivity_and_conductivity_of_various_materials


text:
Water replacement for pressure-testing
Due to circumstance, our Data Centre build has some delays and it currently looks like we're going to have to perform a hydrostatic test of our sprinkler system with live computers in it.
As I have an education as a chemist, but haven't done any serious chemistry in >30 years I thought of using another liquid than water to charge the sprinkler pipes and am looking for the following characteristics:

Same order of magnitude of compressibility as good old H2O
Non-corrosive to materials used in a Data Centre (Metal, plastic, ...)
Non-flammable
Non-conductive to electricity
Non-HAZMAT
Cheap (No liquid gold please, or the VP is going to blow a fuse) ;-)

I tried googling a bit but could not find anything easily, so turning to the experts that deal with this on a daily basis as I've sent off the idea to the sprinkler installers already mentioning my preference for CHCl3 and C2H5OH as a joke, but if they ask me which liquid I would propose, what would be the ideal solution to the above problem?
A citation to a PubChem or similar source would be appreciated

No, no, no... It has to be water. You can protect comps any way you wish, but using other liquid in water using one for test is just plain bad idea.

@Mithoron Could you elaborate why?

Chemists refer to water as "the unique solvent" for good reason. There is simply no other liquid with the range of properties you are looking for.

Water might be somewhat "unique", but it's just that the idea of testing if something works properly with improper liquid is bad. You might even break it.

What about testing in sections (say, per aisle), placing a bucket on a ladder -- just below such a sprinkler.  And if you have multiple of them, you probably find someones which are agile enough to hold such empty bucket for 5 min while standing on a ladder during such a test, too.

Water is the only liquid I know of that is not toxic or is considered hazardous waste when dumped in large quantities in a building (or anywhere else for that matter). As for a CPU, there is no liquid including water that I would dump on it.

We're procuring buckets, reverse umbrellas, tarpaulins, etc already but I was looking for something better and easier that we can do during the day instead of during the night shift.


text:
Separation of Al-Si alloys without electrolysis
Can pure (or roughly pure) $\ce{Al}$ be somehow thermally, chemically, or physically separated from the $\ce{Si}$ contained in an $\ce{Al-Si}$ alloy? How? Approximately what purity of Aluminum can be achieved in this way?

Have you checked the phase diagram?

Aluminium reacts with hydrochloric acid much more readily than silicon does so that might be an avenue to look into.


text:
SO3 in wood ash
I have seen several sources note that wood ash typically contains 3–4 % $\ce{SO3}$ by weight. My question is, why would ash have any $\ce{SO3}$? $\ce{SO3}$ boils at 45 °C, so it should have all boiled away by the time the wood turned to ash.
Some sources for wood ash composition:
Source 1
Source 2
Source 3

;/ It's oxide percent value, there's no SO3 there, but its salts.

Salts such as CaSO3 and Na2SO3? Why not Na2SO4?

No, Na2SO4 etc. - it's treated as Na2O + SO3

Great, thank you. That really clears this up for me.

Wouldn't the bulk of S in wood burn off as SO2 gas ?

Probably not. https://www.jstage.jst.go.jp/article/jcersj1950/94/1091/94_1091_615/_article says that Na2SO4 decomposes at 1450c. Since this is above the melting point(884c), not to mention is hotter than normal wood fire, substantial S is likely to remain in the ash

Generally, ashes of natural samples (e.g. plant material or soil) do not actually contain chemical compounds like $\ce{Na2O}$, $\ce{P2O5}$, $\ce{SO3}$, $\ce{K2O}$ etc. These are just the reference compounds for the calculation of results of the elemental analysis.


text:
Explosive reaction based on reaction rate
Question

We have the following chemical reactions:
$$
\begin{align}
\ce{A + B^* &-> C + D^*}\tag{1}\\
\ce{D^* &-> B^* + B^*}\tag{2}\\
\ce{B^* &-> B}\tag{3}
\end{align}
$$
If the initial concentrations are $[\ce{A}] = \pu{100e-6 mol cm-3}$ and $[\ce{B}] = \pu{1e-6 mol cm-3}$, is the system explosive? We know that $k_1 = \pu{1e6 cm3 mol-1 s-1},$ $k_2 = \pu{50 s-1}$ and $k_3 = \pu{25 s-1}.$

My solution
So I found the rate expressions for $\ce{A}$ and $\ce{B},$ by applying steady-state approximation (SSA) on $[\ce{D^*}].$ I get the following expressions:
$$
\begin{align}
\frac{\mathrm d[\ce{A}]}{\mathrm dt} &= -k_1 \cdot [\ce{A}] \cdot [\ce{B^*}]\\
\frac{\mathrm d[\ce{B}]}{\mathrm dt} &= (k_1 \cdot [\ce{A}] - k_3) \cdot [\ce{B^*}]
\end{align}
$$
Upon entering the numbers into the above expressions, I get the following numbers:
$$
\begin{align}
\frac{\mathrm d[\ce{A}]}{\mathrm dt} &= \pu{-\frac{1}{10000} mol cm-3 s-1}\\
\frac{\mathrm d[\ce{B}]}{\mathrm dt} &= \pu{\frac{3}{40000} mol cm-3 s-1}
\end{align}
$$
My conclusion was that the system is not explosive since the rate of formation of $\ce{B}$ is quite low. Since if the rate is too low, then the reaction can not be considered explosive.
But the problem with my explanation is that I would like to have a reference number. Is there a reference such as f.e. rate higher than $10$ is considered explosive, etc? I have tried looking everywhere but could not find any 'value.' Could anyone provide some insight into this:

Are my explanation and conclusion correct?
Is there a reference value for explosive reactions?

I have an additional question. What is the connection between an explosive system and the steady-state approximation? If the system is explosive, can I still use SSA or does it become invalid?

The steady-state approximation only requires that the concentration of intermediates is low. As long as that is the case, it is valid.

@Zhe I'm not sure that what you write is necessarily true. What is important is that the rate of change is low, i.e. assumed to be zero, rather than the amount being low.

Shouldn't be that if B• is formed faster than C + D than it is esplosive? Can't read the equation but my comment is general (and might be wrong).

Taking your equation $d[B]^/dt\cdots$ and integrating gives $[B^]=[B_0^*]\exp((k_1A-k_3)t)$  and if $k_1A \gt k_3$ the exponential will grow and then one might expect an explosion.

@porphyrin The steady state approximation argues that the rate of change is near zero by arguing that the concentration of intermediate is low. If it's always low, it can't be changing by much.

@Zhe I should have been more exact, so as a clarification the steady state holds if the intermediate's rate of formation equals its rate of loss which means that its concentration is small but only relative to that of the major species in the reaction and not necessarily in an absolute sense.

@porphyrin That is fair.


text:
Enthalpy change of water
A given amount of (liquid) water in an open vessel is heated from 20 degree celsius to 50 degree celsius. Suppose it absorbs x joule in the process. What is the enthalpy change in the process? Is it = x J ? But many authors define enthalpy change as heat absorbed or emitted at constant temperature and pressure. And here temperature is increasing.
What exactly is meant by constant T and P in enthalpy definition ? (in context of system and surroundings)

In which context did you find the "heat absorbed or emitted at constant temperature" part?

Enthalpy change is not defined as the heat added at constant temperature and pressure.  Please cite a reference where you think you saw that.
For a pure species, enthalpy is a function both of temperature and pressure, and also varies with change of phase at constant temperature and pressure.  It is defined as $$H=U+PV$$ where U is internal energy, P is pressure, and V is volume.  So the change in enthalpy for any process of a pure substance is given by $$\Delta H=\Delta U+\Delta (PV)$$  This applies to all processes including the one you described.  If the first law of thermodynamics is applied to the process you described at constant pressure, you have:
$$\Delta U=Q-P\Delta V$$From this, if follows that $$\Delta H=\Delta U+P\Delta V=Q$$So, for the particular process you described, the change in enthalpy turns out to be equal to the amount of heat added.


text:
Properties that probe electron kinetic energy
This post is inspired by a question regarding the meaning of off-diagonal elements of the KE matrix (in some AO basis). One answer suggests that a diagonalized KE matrix might not be very useful. I take the liberty of quoting:

I am not sure it really makes sense to give much of an interpretation to this, however, because what one is really interested in is diagonalizing the Fock matrix, of which the kinetic energy is only one part.

Indeed, we usually are interested in the wavefunctions (eigenfunctions of the Hamiltonian) whose eigenvalues are the energies of electrons inhabiting the orbitals. We like to think of electrons as existing primarily as standing waves with constant energy. Such states are central to computing electronic spectra and ionization energies, for instance.
This leads me to ask two related questions:
1) Is a diagonalized KE matrix, that is, an orbital representation in terms of kinetic energy (rather than full Hamiltonian) ever useful to chemists?
2) What if any spectroscopies probe the instantaneous value of the kinetic energy of electrons (or some portion of the KE spectrum)?


text:
The oxidation state of iodine in iodoxybenzoic acid
I am slightly puzzled by this seemingly trivial question. The idea behind assigning oxidation states is simply to treat every single bond made by the atoms as ionic bonds. That is to say, both the electrons of a covalent bond in a molecule would go to the more electronegative atom. For example, in the case of the $\ce {O-I}$ bond, the bond electrons would go to the oxygen atom, leaving the oxygen atom with a negative charge while the iodine atom incurs a positive charge. Based on this, we would say that the oxygen atom doubly bonded to the $\ce {I}$ atom has an oxidation state of -2.
However, I am slightly confused as to why various sources assign the oxidation state of $\ce {I}$ in IBX as +5, rather than +3. On the Pauling scale of electronegativity, it is shown that $\ce { I (2.66)}$ is indeed slightly more electronegative than $\ce {C (2.55)}$. Based on this premise, we would reasonably assign the $\ce {I}$ atom in IBX with a +3 oxidation state as the pair of electrons in $\ce {C-I}$ bond would go to $\ce {I}$ rather than to $\ce {C}$.

Image taken from here

Yeah, it should be +3.

@Mithoron If that is the case, why are so many authoritative sources claiming otherwise? For example, Zhdankin's Hypervalent Iodine Chemistry also states that it is +5. We can also extend this to another popular reagent, $\ce {PhI(OAc)2}$, why is it +3 and not +1?  It seems that the $\ce {C-I}$ bond is treated rather differently by these source.

Turned out @Mithoron was right after all.  I'm going over yo Space Exploration where things are more down to Earth.

This is a written reply from Professor Zhdankin*, author of Hypervalent Iodine Chemistry:

Indeed, many literature sources classify pentavalent iodine compounds as iodine in the oxidation state +5. It makes sense because the electronegativity of iodine is almost the same as carbon, or even less than carbon. According to different studies, electronegativity of iodine varies from 2.21 [A. L. Allred and E.G. Rochow, J. Inorg. Nucl. Chem., 5, 264 (1958).] to 2.66 [A. L. Allred , J. Inorg. Nucl. Chem., 17, 215 (1961)]. The original Pauling's value of iodine electronegativity is 2.5, exactly the same as carbon. This terminology reflects the similarity of iodine in IBX to iodine in $\ce {I2O5}$ or $\ce {HIO3}$ by chemical and structural properties.
Personally, I prefer not to use the "oxidation state" terminology regarding organoiodine compounds. Pentavalent iodine is a perfectly correct classification for IBX.

*He has given his permission for his reply to be shared here.

Perhaps hybridization plays a role, too? See https://web.ics.purdue.edu/~loudonm//pdf/Supplement_04.01.pdf


text:
Latimer diagram for a metal in basic solution

Write the four half cell reactions from the Latimer diagram for a metal $(\ce{M})$ in basic solution.
$$\ce{\overset{+6}{M}O4^2-(aq) ->[\pu{0.25 V}] \overset{+5}{M}O4^3-(aq) ->[\pu{0.55 V}] \overset{+3}{M}O(OH)(s) ->[\pu{0.42 V}] \overset{+2}{M}(OH)2(s) ->[\pu{-0.89 V}] \overset{0}{M}(s)}$$

I can only come up with the first half reaction, which I believe is:
$$\ce{MO4^2- + e- -> MO4^3-}$$
But for the others I'm having trouble with knowing whether the $\ce{MO4^3-}$ will react with water or hydroxide ion to create, for example, $\ce{MO(OH)}?$

I suspect that the metal is iron, whose known species line up with those given here.  The +5 entry $\ce{FeO4^{3-}}$ is uncommon but known as alkali metal salts.

Let's look at how you would add in the water and hydroxide ions for the $\ce{MO4^{3-}->MO(OH)}$ reaction.
Step 1:  Start with the given metal species, with the metal oxidation states included:
$\ce{M^VO4^{3-}->M^{III}O(OH)}$
Step 2:  Add the electrons according to the oxidation states.  Here the oxidation state on the metal drops by two, so there must be two electrons on the left.  Ignore the fact that the overall charges may be unbalanced, that comes next.
$\ce{M^VO4^{3-} + 2e^- ->M^{III}O(OH)}$
Step 3:  Now we are ready to attack the charge balance.  Add the hydroxide ions, which do not couple with the electron exchange, to balance the charges.  Counting out the charges we have going into this step we need five hydroxide ions on the right.
$\ce{M^VO4^{3-} + 2e^- ->M^{III}O(OH) + 5OH^-}$
Step 4:  Finally add water to balance hydrogen; if you rendered the oxidation states and did the math correctly, the oxygen balance falls into place simultaneously.  With six hydrogen atoms to balance we use three molecules of water to get the final answer for this reduction:
$\ce{M^VO4^{3-} + 3H2O + 2e^- ->M^{III}O(OH) + 5OH^-}$
Can you now do the remaining reductions?

I tried for the second one and got: MO(OH)(s) + e^- +H2O(l) = M(OH)2(s) + OH-(aq) Is this correct?

And for the third one: M(OH)2(s) + 2e- = M(s) + 2OH-(aq) ?

Got them both!  :-) .


text:
Electrolytic refining of precious metals — why the anode rather than the cathode?
In electrolytic refining, metals like platinum and gold are deposited as anode mud.
Why aren't they deposited on the cathode since they have higher reduction potential?

For these metals to deposit, they would have to be present in solution such that they can come into contact with the cathode and be reduced. However, this would first require them to be oxidized, so that they dissolve in solution. Unless you are actually electrolytically refining for gold, the voltage will not be enough to oxidize it.


text:
Understanding the structure of NCS⁻

Predict the most preferable structure of $\ce{NCS-}.$
$$\Large\underset{(\textbf{1})}{\ce{(:\!\!\overset{\huge.\!\!.}{N}=C=\overset{\huge.\!\!.}{\underset{\huge.\!\!.}{S}})-}}\qquad\underset{(\textbf{2})}{\ce{(\overset{\huge.\!\!.}{\underset{\huge.\!\!.}{C}}=S=\overset{\huge.\!\!.}{\underset{\huge.\!\!.}{N}})-}}$$

My Attempt
A chart for the formal charges:
$$
\begin{array}{l|rr}
\hline
\text{Atom} & (\textbf{1}) & (\textbf{2})\\
\hline
\ce{N} & -1 & -1 \\
\ce{C} & 0 & -2 \\
\ce{S} & 0 & +2 \\
\hline
\end{array}
$$
As a conclusion, in structure 2 the formal charges on $\ce{C}$ and $\ce{S}$ (i.g. bond becomes polar because of $+2$ and $-2$) make it weaker, whereas in structure 1 there are less charge separations. So, (2) must be more a accurate structure.
But according to my book, structure 2 is more stable. Why? Please explain. Is there any other approach to this question?
Edit

I got this statement in favor of structure #2 "is more stable because each atom has non-zero formal charge in the lowest energy state". But i can't understand this statement. Can anyone explain me?"

The name of your ion is the thiocyanate ion. Now,the word 'thio' basically refers to the replacement of a sulphur somewhere(usually an O) in a parent molecule to get the current one. The parent ion in this case will obviously be the cyanate ion with the molecular formula [OCN]- . Now,due to restrictions on the valency of O,you can't make it the central atom ,and therefore a structure like (2) will not be possible,and so a probable structure for [OCN]- would be something like [O=C=N]- . Now,simply replace O by S here to get the 'thio' derivative's structure,which will resemble (1)

I took a liberty to compact all lists of formal charges into one table for a better visual cue as well as corrected formatting. We prefer to use Markdown as a more lightweight markup alternative to plain HTML tags, and MathJax only if necessary. If you want to know more, pleasevisit this page, this page and this one on how to format your future posts better with MathJax and Markdown.

Did you book give any reasons? It is unlikely that statements are left unexplained in a chemistry textbook, although there may be a minority of such cases.

Did you mistype?  Looks like you meant the book favors #1.  They probably also said that structure has less formal charge separation, which you typically want with all-nonmetal molecules.

Just a note: it is an ill posed exercise. It should ask about stability not what is accurate. The fact that A is less stable than B it doesn't mske A inaccurate. The used terminology would be valid about mesomers, in which sometimes one is more closed to the the actual hybrid of resonance and as such can be called "a more accurate structure". Here we have different species so what? Perhaps is not the book fault but OP.

@OscarLanzi I rechecked it. #2 structure is more accurate.

@Alchimista yaa... By accurate my book means most preferable.Thanks for suggestion.I'll edit it.

@TanYongBoon it states that "structure #2 is more stable because each atom has non-zero formal charge in the lowest energy state". But i can't understand this statement. Can you explain me?

More stable not preferable!

This is a misprint. Here is a similar problem (OpenStax Chemistry, retrieved from https://opentextbc.ca/chemistry/chapter/7-4-formal-charges-and-resonance/) that makes sense and has a correct answer:

As another example, the thiocyanate ion, an ion formed from a carbon atom, a nitrogen atom, and a sulfur atom, could have three different molecular structures: CNS–, NCS–, or CSN–. The formal charges present in each of these molecular structures can help us pick the most likely arrangement of atoms. Possible Lewis structures and the formal charges for each of the three possible structures for the thiocyanate ion are shown here:

Note that the sum of the formal charges in each case is equal to the charge of the ion (–1). However, the first arrangement of atoms is preferred because it has the lowest number of atoms with nonzero formal charges (Guideline 2). Also, it places the least electronegative atom in the center, and the negative charge on the more electronegative element (Guideline 4).

The statement made rationalizing the incorrect structure is also flawed:

"is more stable because each atom has non-zero formal charge in the lowest energy state"

This is an ion. Some atom has to have a non-zero formal charge because the net charge should equal the sum of formal charges. Why having each atom have a non-zero formal charge should result in a more stable structure is not clear to me. Mentioning the lowest energy state is curious as well. Maybe they meant most relevant resonance structure.
One should also keep in mind that formal charges are mostly an accounting method and do not reflect actual charge distribution (see https://chemistry.stackexchange.com/a/119771).

Ah yes, the big issue between charges and formal charges. And the way we teach stupid, meaningless approximations to students. The reasoning, while fitting the interpretation, is incomplete at best. The the atom, linear molecules are a nightmare to describe it understand, if you're not considering resonance, you're having the wrong idea. The original question is bad, if it's not done at a proper background; and I guess this is high school, so it's even worse.

@Martin-マーチン Oops, trigger warning - formal charges and stability. I think formal charges have their place (teaching acids and bases, maybe), but guessing the stability of fictitious species is probably not the best use of time when teaching an introductory course.


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On diagonal terms in the Coulomb matrix
I have seen many machine learning algorithms for prediction of quantum chemistry properties that use Coulomb matrix as their input. Coulomb matrix is defined as,
$$\boldsymbol{M}_{i j}^{\mathrm{Coulomb}}=\left\{\begin{array}{cc}{0.5 Z_{i}^{2.4}} & {\text { for } i=j} \\ {\frac{Z_{i} Z_{j}}{R_{i j}}} & {\text { for } i \neq j}\end{array}\right.,$$
where the diagonal terms:

... can be seen as the interaction of an atom with itself and are
  essentially a polynomial fit of the atomic energies to the nuclear
  charge.

I was trying to read more about these diagonal terms and how exactly this equation is fitted. Essentially, something that says: "... and from there we get $0.5Z^{2.4}$." If these energies can be extended to bonds as well, as the off-diagonal term in Coulomb matrix represents repulsion forces rather than potentials, any references would be appreciated as well.

The model seems to stem from Rupp et al Phys. Rev. Lett.,108(5):058301, 2012, but I did not find much explanation for it there.

@Blade could you clarify what you mean by extending these energies to bonds. There are methods (e.g. bag of bonds) related to Coulomb matrices which use bond energies, rather than atom energies , to predict an overall atomization energy.

@Tyberius For instance, in molecular mechanics, we describe bond potentials using bond stretching interactions (either harmonic oscillator or Morse function). I was basically wondering how the empirical relation above was derived, so that I see if there are similar empirical relations for bond energies as well. I'm not sure if there is an answer to the second part of my question, just wondering if there is such a thing.

@Blade Coulomb matrix is not an empirical relationship, it is just a descriptor trying to mimic the Coulomb interactions between atoms and the atomic energies (interaction with itself).

In terms of how they got the relation for the diagonal elements, I believe it is relatively straightforward. Given a list of atomic energies (energy of atom relative to separated nucleus and electrons), one can try to fit these values to a function of nuclear charge. If you plot a few values, it suggests trying an exponential fit (I don't know why all the papers say polynomial fit, as this could not lead to a noninteger exponent). So, you can try to fit the values to a function of the form $aZ^b$ using, for example, least-squares regression. You should wind up with something pretty close to the equation given in your question, with some variation because I don't know what atomic energies they used to do the fit. (I'm also not sure why the function isn't negative since the atomic energies are negative. My best guess is the sign winds up not being important as a descriptor in the network).
As to why they choose to use these values along the diagonal, they describe the system in terms of the energies of the individual atoms, along with the energy of interactions between atoms given by the off-diagonal terms. The original Rupp et al. paper compared two Coulomb matrices by looking at the difference between their eigenvalues, with small differences implying similar molecules. These eigenvalues can be thought of as effective atomic energies within the molecule, that is, energies of the individual atoms with the influence of the surrounding atoms included. Similarly, the eigenvectors would describe fragments of the molecule or an effective atom which account for the influence of neighboring atoms.

Wouldn't the diagonalized matrix then be in a basis consisting of linear combinations of the individual atoms, so something closer to fragments?

@BuckThorn that's my understanding. I guess, based on the context of being used for atomization energies, these eigenvector/fragments can thought of as effective atoms with the influence of their environment included as part of their extent.

"I don't know why all the papers say polynomial fit, as this could not lead to a noninteger exponent" good catch! I would have repeated that somewhere and looked like a fool!

@Blade yeah unfortunately the original paper didn't explain things in much detail and later papers just seem to copy the limited and somewhat misleading explanation from that original paper without thinking about it.


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Genuine confusion about the equilibrium constant
I gather that the equilibrium constant K = 1 implies that, at equilibrium, neither the forward nor the backward reactions are thermodynamically favoured. But if pressure of the system increases and this shifts the position of equilibrium, the value of K remains to be 1.
How then can K = 1 imply that, at equilibrium, neither the forward nor the backward reactions are thermodynamically favoured?
Please can I have a simple and comprehensive explanation of what the value of K actually tells us?

If the pressure increases, the system will be out of equilibrium. What's unclear about that?

I gather that the equilibrium constant K = 1 implies that, at equilibrium, neither the forward nor the backward reactions are thermodynamically favoured.

There is nothing special about K = 1. At equilibrium, no matter the value of K, there is no net reaction. Forward and backward reactions proceed at the same speed.

But if pressure of the system increases and this shifts the position of equilibrium, the value of K remains to be 1.

Yes, even a reaction with K = 1 can react. Just imagine you start with some reactant and no product. This situation is not at equilibrium. The reaction will proceed in the forward direction until Q = K. Then, you can disturb the system, changing Q, and there will be a net reaction again, changing concentrations until Q reaches K again.

Please can I have a simple and comprehensive explanation of what the value of K actually tells us?

You said "please", so I will try for a short, simple and comprehensive answer. The value of K allows you to calculate equilibrium concentrations, no matter which concentrations you started with. For a lower K, the product equilibrium concentrations will be lower than for a higher K (same initial concentrations). The opposite is true for reactant equilibrium concentrations.
It should be clear that the equilibrium concentrations for a given K are not always the same. The simplest example is doubling initial concentrations (while keeping the volume constant). If the equilibrium concentrations were equal for both cases (initial concentrations doubled or not), it would go against the conservation of matter. (We doubled the amount of species present initially, so we can't have the "old" amount of equilibrium species).

The equilibrium constant normally written as $K_p$ for a gas phase reaction is independent of pressure but does depend on temperature. This means that if the pressure is changed  then the amount of dissociation of reactants and products changes (and so does their partial pressure) until equilibrium is restored. $K_p$ is defined as the ratio of the partial pressures at equilibrium, and the exact expression depends on the type of reaction, for example $A_2\to 2A$ produces $K_p=p_{a}^2/p_{A_2}$. In your example if $K_p=1$ then $p_{a}^2=p_{A_2}$ at all pressures and this can only be true if the amount of dissociation changes.
Any equilibrium constant $K_p$ is also related to the standard free energy of the reactants and products at $1$ atm because $\Delta G^\text{o}=-RT\ln(K_p)$.  If $\Delta G^\text{o}$ is positive then there will be more reactants than products and the reaction will not proceed to such an extent than if $\Delta G^\text{o}$ is negative i.e $K_p \gt 1$.  (Note that just because an equilibrium constant may be  large this does not imply that the reaction is fast: thermodynamics does not inform us about reaction kinetics only the final equilibrium state.)
If we know $\Delta G^\text{o}$ for reactants and products then it is possible by calculating the change (products minus reactants) to find the equilibrium constant for a reaction without having to try it experimentally.
Notes. Calculation of change in degree of dissociation for the reaction $A_2\to 2A$. Assume the degree of dissociation is $\alpha$ then $A_2$ changes by $(1-\alpha)$ and $2A$ by $2\alpha$ and the total change is $1+\alpha$. The partial pressure of $A_2$ is $\displaystyle P_{A_2}= \frac{1-\alpha}{1+\alpha}p$ with $p$ as total pressure, and for $A$ the partial pressure is $\displaystyle P_{A}= \frac{2\alpha}{1+\alpha}p$. As $\displaystyle K_p = \frac{P_A^2}{P_{A_2}}=\frac{4\alpha^2 }{1-\alpha^2}p$. As the pressure $p$ changes then so must $\alpha$ because $K_p$ is constant.

“If delta G is positive then there will be more reactants than products” is not always true. “If all species initially are at standard state and delta G is positive, the reaction goes in reverse to attain equilibrium and equilibrium concentrations of reactants will be higher than those of products” is a general statement.

I though that is what my statement said anyway, 'more reactants than products' vs 'concentrations of reactants will be higher than those of products' ?

Extremely small concentrations favor dissociation while extremely high concentrations favor association. Even in a situation where at standard state, association is favored ( or dimerization like in your example), at extremely low concentrations it won’t be. So the key difference in our statements is that I specify the initial conditions as standard state.

I think what you are saying is that (in my example) at  low pressures (concentration) to keep $K_p$ constant $4\alpha^2/(1-\alpha^2)$ is large as $p$ is small and vice versa.

Actually K tells u nothing but it implies the ratio of the concentrations of produts and reactants when state of equilibrium has been attained which as the name suggests is always constant for a given temperature...Whenever there is increse or decrease in pressure( or Volume) the reaction shifts as the concentration gets altered and the shift is to re-establish the equilibrium....thus there should be and there is no effect in the value of K....
Hope it helps!

Please avoid colloquialisms, salutations and excessive punctuation, and mind proper capitalization (you also might want to fix a couple of typos along the way). In general, this somewhat addresses the OP's question, but I'm not sure this is exactly the answer OP is looking for.


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Choice of solvent during thin layer chromatography
I understand most of the theory behind TLC and that different volatile solvents have different polarities and cause the 'spotted' substances to move at different rates, but what is the point of creating mixtures and then using them in different ratios? How could the results change if I changed the ratios?
For example, during our lab, we synthesized acetylglycine and made three spots on our TLC plates, one spot was pure glycine, one spot was glycine + acetylglycine, and the third spot was just acetylglycine. the chamber was equilibriated with Dichloromethane & Hexane in an 8:2 ratio.
Glycine didn't move, and the other two spots did move up.
How would the results have changed if I used only dichloromethane? only hexane?
What if I used a 7:3 ratio? 5:5 (1:1) ratio?
I have a feeling the answer is really obvious and i'm overthinking this but it's really not hitting me right now.
Thank you for reading!

Chromatography is really an empirical science. The fluid mechanical side and the adsorption phenomenon is somewhat well understood but choosing mobile phases is essentially based on trials, sadly to date. If someone has given you a ratio, say 8:2, it means someone must have tried it and they found that this mobile phase works the best. The main criterion for choosing the mobile phase is that all the mixture components must be soluble in it. Thus if glycine did not move, this means that the mobile phase has no affinity for this solute (it is perhaps insoluble in it) and it is a poor choice. One then starts to tune the polarity of the solvent mixture. Think about a petroleum sample, which literally has thousands and thousands of different molecules. Now think yourself: will they all dissolve in 100% hexane? You do need an additional solvent.
There is no theoretical restriction in chromatography to use multiple solvents or a single solvent. For example in gas chromatography, you always use a pure gas. The reason for using a mixture of solvents in liquid chromatography originates from the concept of "general elution problem of chromatography" which basically says that there is no single adsorbent or a single mobile phase which can separate everything. Not only separation scientists use a single mixture composition of mobile phase (isocratic), they can also continuously change the mobile phase composition as a function of time. This is called gradient elution. You cannot do that with thin layer chromatography without using fancy stuff.
People are trying to implement machine learning, so that one can enter the molecular structures contained in the mixture, available stationary phases, and with a lot of statistics and tons of experimental data, the computer can tell you to try a certain mobile phase composition.

This is not a nice answer but unfortunately it has been accepted. Solubility is absolutely required just because molecules cannot walk by themselves along the column. However has little to do with the mobility of compounds and thus, with their separation. This misleading idea is reiterated by " Thus if glycine did not move, this means that the mobile phase has no affinity for this solute (it is perhaps insoluble in it) an...".Technically, chromatography uses eluants, it is just contingent they are also solvents and we call them like that. It is true that eluant it is choose by feeling and...

..experience. The different compounds in a mixture must have enough different REPARTITION between the adsorbing phase and the liquid one, and solubility might, in principle, play no role at all still being a prerequisite. Mixing solvents allows for an eluant with the right polarity* for the best separation. In the given example using just dichloromethane would have push everything faster, resulting in no separation. A 7:3 mixture would probably have worked anyway, at least with an appropriate colum. *polarity to be taken broadly together with affinity and specific shape and chem interaction

I believe you are missing the punchline of $elementary$ chromatography principles- good for the beginning student. The partition coefficient of a given solute cannot be zero! It can only be non-zero, if the compound is soluble in it.  Polarity is really a poor term, IUPAC discourages it. I wish there were a better term for it. Organic chemists use it a lot, but polarity doesn't mean anything.

Which point? About polarity you made polar eluted by polar not me. I am quite for the opposite. Also note the * in my comment.

I am afraid of this misinterpretation of my point which was specific to the student's question regarding glycine spot not moving at all. "Polar eluted by polar" would be the worst way to teach separation science to anyone. Do you really think glycine is soluble in hexane-DCM system? The key point to begin a chromtography experiment, I reiterate, is having a non-zero partition coefficient, and partition coefficient is non-zero if and only if the solute is soluble in both phases. Don't take solubility in a literal sense. Having a non zero partition coef. does not mean a separation will occur.

Yes, your reference to insolubilty and the quoted sentence could have been the reason. I would separate that additional comment by comma to make clear that in the actual example glycine might even be insoluble.  like this is more understanble but still the fact that repartition is is a ratio, moreover dependent on a third term, wasn't clear. The most I got is that one need (some) solubility. Which of course is true, but has no role per se in the success of the procedure.

I can give a very pictorial answer that doesn't require the use of actual pictures nor typesetting and might useful for beginners.
It is all about the misconception, encountered even in chemists with a relative experience in successful experiments,  that solubility of the compounds in a mixture is a key factor for their effective chromatographic separation in thin films or columns.
Think of a large number of boats coming down a river just because of the water stream. Each boat has a boater aboard. This ensemble of boat/ers is the eluant.
Think of a number of berths and piers all along the banks. This is the stationary phase, the silica gel on plate in your example.
Now, you decide, or better someone wants you to do so, to go down the river as well. In order to use a boat, first of all the boat must no sink. This is an allegory for you being a molecule soluble in the eluant. If this prerequisite isn't met, you cannot even start your journey.
Say you find a boater that is friendly and the boat carries you, you start descend. However after some time you might find the trip tiring or boring, perhaps the boater is not the best companion you might think of.  So you decide to take a break and jump out of the boat at the first next free pier. You might well stay there and relax until another boat with a friendly boater accept you so you restart the trip.
The scenario is repeated until you arrive at the mouth of the river.
Now imagine myself having to do the same. Obviously some of the boaters with not sinking boats must accept me as well. However, we are different persons, and what you find a friendly boater might be unpleasant to me, and viceversa. Similarly, I can be more or less at comfort than you, while sitting on a pier, so I do my best to leave it watching for a friendly boater.
Also, we can be oblige to stay aboard longer between two stops, because the pier can be occupied by one of the descending boats we are using, the boaters stop, too.
Assuming our taste and behaviour is at least a bit different (at the end we are different compounds, so it is) even if we start the descent together we'll arrive at the river's mouth separately at different time.
Note that the reason for this delay is not in the fact that the boat we used weren't sinking, i.e. our solubility, but in the number of boaters we were happy to cruise with and the level of comfort we found on free piers.
In other words, the delay is not due to time that we have spent on the boats - which is identical and depends on the river length and stream velocity - but on the time we stayed stationary at the little harbours all along.
The separation between our arrival times

depends ultimately on our relative affinity towards the boat/ers and the berthing places, i.e. the repartition we have between the liquid eluant and the stationary phase;
it is shortened more often/longer the boats stop, i.e. by an increase in affinity between the eluant and the stationary phase.

Changing the eluant composition equal to change the number of friendly boaters and free piers. The trick is to find boaters that are very friendly, say, to you, while minimizing those friendly to me. Still I need some, if I have to arrive as well.
To complete the answer, is it true that similar goes with similar and thus, a compound is likely more soluble in a solvent for which it has affinity. This equals to say that all the boats that don't sink necessarily have a boater which is at least moderately friendly.
As with human beings character and behaviour, it is complicated to find rules in details. If the mixture to be separated consists of very different compounds then the eluant choice is straightforward, otherwise one must start from the basic rules and proceed by feeling and experience.
Mixing different solvents allows to prepare eluants of different polarities in chromatographic sense, as I tried to convey with the above allegory. Other than general polarity (in terms of electric dipole moment), specific shape related issues or the presence of particular functional groups play a role, and this is indeed similar to the situation found when describing solubility, which cannot be reduced to a polar non-polar drastic criterion.
What I want to stress by this already long answer is that even a polar eluent can push unpolar compounds fast down the column (up the plate) by occupying the stationary phase. The unpolar  compounds are forced to stay in the mobile phase in spite of their barely minimal solubility. They aren't separated exactly because they stay in solution. Chromatography is not a fractional crystallisation technique and passes huge amount of solvents over the sample, at the end.

In your case using just dichlorometane would have pushed the two faster togheter, using just hexane would have taken months, and for intermediate case you have done one to interpolate about.

The answer given by @M. Farooq may be interpreted as «the optimization of TLC depends on the choice of the eluents used, which till now (by broader advent of machine learning) was empirical».
In my opinion, this should not the impression to retain. Of course the optimization of TLC depends 1) on the eluent / eluent mixture, but should be seen in context of 2) the stationary phase available (e.g., silica, basic / neuter / acidic aluminum oxide, reverse-phase) and 3) the nature of the sample itself (e.g., molecules which may donate protons to / accept protons from the stationary phase / mobile phase, or not) as most common factors, too.
It is true instruction classes and research groups often stick to «typical» mixtures (e.g., petrol ether / ethyl acetate) to perform TLC for analysis and testing, often as preparation for eventual column chromatography on preparative scale, for «typical» samples.  Here and there, there are special recipies to address peculiar separation problems, too.
On the other hand, the optimization a TLC already is formalized algorithmically, for example by Glajch et al. 1982 (up to quaternery eluent systems, here), or with the PRISMA method by Nyiredy et al., back in 1985 (here, English abstract).  For the later approach, after an initial thought about the separation problem ahead

(source, open access)
multiple solvents, grouped for their properties, are tested as such (purely) as eluent.  The results of this initial screening then are further optimized in mixing solvents as eluents

(source, open access)
This formalized approach include computations which may be performed independent of a computer.  A not-too old review in 2000 (here) explicitly reflects again how to optimize TLC, including the PRISMA method.

A good watch for detail.  Like appreciating the about 2.5 cm difference of level, prior and after the Pont du Gard.

Very well explained, did you notice the interesting footnote of the PRISMA paper "Die Kunst des Chromatographierens besteht nach wie vor darin, die stationäre Phase auf die mobile Phase und beide auf die Trenntechnik abzustimmen.” which DeepL nicely translated as "The art of chromatography is still to tune the stationary phase to the mobile phase and both to the separation technique". The emphasis on art is mine. –


text:
If alkyl-substituted alkenes are more stable, how does one explain their increased addition reactivity?
My understanding is that in alkyl-substituted alkenes there is hyperconjugation between $(\ce{C−X})$ and $^∗(\ce{C=C})$. I also have read that more alkyl-substituted alkenes are many folds more reactive in addition reactions. I understand how hyperconjugation stabilizes alkyl-substituted carbocations.
However, based on the fact that alkyl-substituted alkenes are more "stable", and the fact that the hyperconjugation between their $(\ce{C−X})$ and $^∗(\ce{C=C})$ causing a delocalization of electrons at the $\ce{C=C}$ double bond, wouldn't the alkyl-substituted alkene be less reactive? Obviously stability comes into play, and additionally the loss of electron density at the $\ce{C=C}$ double bond should mean that the $$ bond is less of a nucleophile so would prevent attraction of electrophilic addition.
One way I came up to justify this is maybe that the stability from the hyperconjugation of the carbocation outweighs stability from hyperconjugation of alkene. Or that somehow the hyperconjugation between $σ(\ce{C-X})$ and   $π^∗(\ce{C=C})$ adds electron density to the $^∗$ making it more attractive to electrophiles that can break the $\ce{C=C}$ double bond. Of course, I am only using these as justifications to fill in misunderstandings.


text:
Bent's rule: electronegativity and s character
I'm very much confused about Bent's rule. What I perceive from it is that more electronegative element occupies a position which has less s character.
What I can infer from here is that s character is inversely proportional to electronegativity which is not true. Please do explain.


text:
Finding ionic radii from solubility data and density

$\pu{5.35 g}$ of a salt $\ce{ACl}$ is dissolved in $\pu{250 ml}$ of solution. The $\mathrm{pH}$ of the resultant solution was found to be $4.85.$ Find the ionic radius of $\ce{A+}$ and $\ce{Cl-}$ if density of $\ce{ACl}$ is $\pu{2.2 g cm-3}.$  Given
$$\frac{r_+}{r_-} = 0.73 \qquad K_\mathrm{b} = \pu{2E-5}$$

So first we can find $\log C$ by using the formula
$$\mathrm{pH} = \frac{1}{2}(\mathrm{p}K_\mathrm{w} - \mathrm{p}K_\mathrm{b} - \log C),$$
which gives $\log C = 4.3.$ Now for the density $ρ$
$$ρ = \frac{ZM}{N_0V},$$
which will give relation in $M$ and $V,$ which are lacking. But how to proceed further?

The pH and Kb are red herrings. What you need to consider is the density and the ionic ratio and thus what type of packing. See the radius ratio rule

Yup I know it is CsCl structure but can you give full solution


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Why does thallium hydroxide increase the yield of product in a Suzuki reaction?
I came across a research paper by Uenishi et al. [1] where relative rates of $\ce{KOH}$ and $\ce{TlOH}$ in Suzuki coupling reactions are compared.
Thallium hydroxide increased the yield of products, see Table I below. The reaction also occurred at room temperature as compared to $\ce{KOH}$.
What is the effect of thallium hydroxide?

References

Uenishi, J.; Beau, J. M.; Armstrong, R. W.; Kishi, Y. Dramatic Rate Enhancement of Suzuki Diene Synthesis. Its Application to Palytoxin Synthesis. J. Am. Chem. Soc. 1987, 109 (15), 4756–4758. https://doi.org/10/fhmqsr.

The Suzuki coupling reaction (also called Suzuki-Miyaura coupling reactions; Ref.1) is the coupling of an aryl or vinyl boronic acid with an aryl or vinyl halide or triflate using a palladium(0) catalyst similar to Heck reaction and Negishi reactions in mechanistic aspects. In particular, Negishi reaction uses organozinc reagents instead of organoboronic acids. Both reactions are powerful cross-coupling methods that allows the synthesis of conjugated olefins, styrenes, and biphenyls. The mechanism of Suzuki coupling reaction is depicted below:

The palladium catalyzed mechanism begins with oxidative addition of the organohalide ($\ce{R^1 -X}$) to the $\ce{Pd(0)}$ (e.g., $\ce{Pd(0)L_n}$) to form a $\ce{Pd(II)}$ complex (denoted in mechanism as $\ce{R^1 -Pd(II)L2X}$). It is worth nothing that this first step in both of the Suzuki and Negishi cross-coupling reactions is identical to that of the Heck reaction. This complex undergoes halide ligand exchange with the base present in the reaction mixture.
This base is present to serve one or two possible roles: Either to react with the organoboron reagent (denoted as $\ce{R^2 -B(OH)2}$) to form a trialkoxyboronate (denoted as $\ce{^-B(OH)3R^2}$), which then attacks the palladium(II) halide complex (denoted as $\ce{R^1 -Pd(II)L2X}$), and/or to convert of the palladium(II) halide complex to a palladium(II) oxo complex (denoted as $\ce{R^1 -Pd(II)L2OH}$) that reacts with the neutral organoboron reagent ($\ce{R^2 -B(OH)2}$) (Ref.1 & 2).
Transmetalation with the organoboron reagent then follows where the $\ce{R^2}$ group of $\ce{R^2 -B(OH)2}$ replaces the hydroxide anion on the palladium(II) complex to give $\ce{R^1R^2 -Pd(II)L2}$ and boric acid. Keep in mind that organoboron compounds are highly covalent in character, and may not undergo transmetallation readily in the absence of a base. Therefore, the reaction of the base with the organoboron reagent, $\ce{R^2 -B(OH)2}$ to form $\ce{^-B(OH)3R^2}$ is very likely. Also note that transmetalation step is highly depend on the base availability and the types of ligand on the catalyst (Ref 1 & 3).
Reductive elimination then follows to give the final coupled product regenerating the catalyst, hence the catalytic cycle can begin again.
Based on this mechanism, availability of a base is very important to reaction to proceed. It is clear that AP's question does not indicate the solvent used in the reaction. However, mentioned reference has been using THF as the solvent. Since solubility of $\ce{KOH}$ in THF is very low and hence its availability to the reaction. Thallium(I) hydroxide is strong base compatible with alkali bases such as $\ce{KOH}$ and key to its capability of enhancing the reaction here should be its readily availability in such solvents. This is evidence that in recent publication (Ref.4) where $\ce{KOH}$ and $\ce{TlOH}$ are equally effective in $\ce{Pd(0)}$  catalyzed coupling reaction, yield of which increases when solvent has been changed from THF to DMF.
References:

Norio Miyaura, Akira Suzuki, "Palladium-Catalyzed Cross-Coupling Reactions of Organoboron Compounds," Chem. Rev. 1995, 95(7), 2457-2483 (https://doi.org/10.1021/cr00039a007).
Karl Matos, John A. Soderquist, "Alkylboranes in the Suzuki−Miyaura Coupling:  Stereochemical and Mechanistic Studies," J. Org. Chem. 1998, 63(3), 461-470 (https://doi.org/10.1021/jo971681s).
Norio Miyaura, Kinji Yamada, Hiroshi Suginome, Akira Suzuki, "Novel and convenient method for the stereo- and regiospecific synthesis of conjugated alkadienes and alkenynes via the palladium-catalyzed cross-coupling reaction of 1-alkenylboranes with bromoalkenes and bromoalkynes," J. Am. Chem. Soc. 1985, 107(4), 972-980 (https://doi.org/10.1021/ja00290a037).
Hiroko Koyama, Hisashi Doi, Masaaki Suzuki, "Evaluation of $\ce{TlOH}$ Effect for $\ce{Pd^0}$-Mediated Cross-Coupling of Methyl Iodide and Excess Boronic Acid Ester toward Fabrication of $\ce{[^{11}C]CH3}$-Incorporated PET Tracer," Intl. J. Org. Chem. 2013, 3(3), 220-223 (DOI: 10.4236/ijoc.2013.33030).

Toxicity of thallium hydroxide. https://pubchem.ncbi.nlm.nih.gov/compound/Thallium-hydroxide#section=Safety-and-Hazards


text:
Enthalpy and heat relationship
They say that enthalpy of reaction is the heat when at constant temperature and pressure. So if enthalpy is
$$\mathrm dH = \mathrm dU + p\,\mathrm dV = T\,\mathrm dS - p\,\mathrm dV + p\,\mathrm dV + μ\,\mathrm dN = T\,\mathrm dS + μ\,\mathrm dN$$
and heat is $T\,\mathrm dS,$ then what happens to the chemical potential term if enthalpy is heat?

They say that enthalpy of reaction is the heat when at constant temperature and pressure.

They should not say that. At least not without saying in the absence of non-PV work. And in a closed system.

So if [...] heat is TdS [...]

You should not say that without specifying a reversible (near equilibrium) process in a closed system where dS refers to the surrounding:
$$q_\mathrm{rev} = T dS_\mathrm{sur}$$
The sign of $q_\mathrm{rev}$ is from the perspective of the system, i.e. if there is heat transfer from system to surrounding (positive value), the entropy of the surrounding will increase (we don't know what happens to the entropy of the system because there is more going on in the system whereas the only change in the surrounding is that it got heat transferred to it).
On the other hand, all the thermodynamic quantities in your formula refer to the system. By the way, the enthalpy and the entropy of a reaction are not linked, there is no general formula that lets you infer reaction enthalpy from reaction entropy without knowing other quantities.

Thanks, this makes some sense. So what you're saying is TdS is the reversible heat but is not the heat of reaction which is the enthalpy at constant T and P with no non-pv work and it being a closed system.

Yes, and dS refers to the entropy change of the surroundings.