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Data-processing pipeline for protein x-ray crystallography: uncertain of terminology and order of procedures
Background: I am trying to get an overview of the data-processing pipeline for solving protein structures by x-ray crystallography, and in a very simple way understand what mathematical procedures are conducted at what step of the processing pipeline. I find data-processing a bit confusing as both textbooks and information on the internet seem to either give very basic explanations or way to technical (i.e. mathematical) explanation (and crystallography also seems to be full of confusing jargon terminology). Or alternatively, the information is focused on “clicking” you way through technical programs to produce an electron density map (which is great!), but without really explaining what the main steps of data processing consists of i.e. why you are doing them.
I would appreciate some clarification/confirmation that I have understood the overall process correctly. I can "click my way through" programs such as XDS/ Aimless/ Pointless/ Ctruncate/ MolRep/ Buccaneer and then finally work in coot for model in the electron density map. However, I don’t really understand the steps (in the data processing), a lot of it is simply just "click this" then "click that".
Specification of question: Everything up to the point of getting a protein diffraction pattern is okay, and everything after the electron density map is generated - is also okay. It is specifically the part between diffraction pattern and electron density map that is a bit confusing (also in terms of terminally used to explain the processing).
My perhaps error-prone understanding so far:

The diffraction pattern (collected at 360 degrees) of a protein crystal represents the reciprocal lattice.

The reciprocal lattice is an abstract concept for mathematical calculations, and does not represent the real crystal which exists in what is referred to as “direct lattice”, “crystal lattice” or “real space” (they are all the same thing?). The reciprocal lattice exists in reciprocal space (which is also referred to as k-space in physics?).

We can index the spots of the reciprocal lattice to generate miller indices(hkl) for each spot.

By indexing the intensities of the spots on the reciprocal lattice, we can determine the amplitude of the waves that diffracted from the crystal, but they do not contain any information about the phases of the wave relative to each other. This is the part where braggs law is used (i.e. a part of indexing?)

We then scale and merge reflections

Merging is grouping together identical reflections which increases the multiplicity of the data set (i.e. a good thing for statistics). Scaling is a bit abstract, and I know it is done in programs such as XDS during indexing (or right after indexing) and it somehow places all the reflections on a common scale.

We need to determine the structure factors (F) that contain information of the phases (or is this done at an earlier step, or at several steps of the process?)

This has to be done my molecular replacement or other methods (i.e. to solve the phase problem)
Assuming we use molecular replacement to solve the phase problem, are we then using the Patterson method of solving the structure?

After determining the phases, we can generate an electron density map is in “direct space/real space” by performing an inverse Fourier transform from the frequency domain to the time domain (f(t))? (or is it opposite, i.e. “normal” Fourier transform?)

There is software to do all the analysis you mention, look for CCP4 and / or Coot and associated methods used. The first step is to estimate the spot intensity on the detector; not a trivial problem as it happens as spots can have incomplete shapes. Once this is done then FFT methods are used to work out what the crystal alignment was and so find space group and hkl.

Thanks, yes I know how to use coot and also how to generally go through XDS/Aimless/Pointless/Ctruncate/MolRep/Bucaneer and then coot. However, I dont really understand what I am doing and why, a lot of is is just "click this" then "click that". This is why I asked the question above. I will add this statement to my question.

@CuriousTree, Glad to see that someone is actually thinking rather than clicking buttons. It will be better if you talk to someone in person who is an expert in Xray domain (I mean not the users but actually people who know the practical aspects +theory).

Have a look at 'Crystallography made Crystal Clear' by G. Rhodes publ. Academic Press which covers most of what you are asking as it specifically covers macromolecular models. (btw, if you work things through note that eqns 5.21/22 seem to have have typos)

Thanks! I think I have seen that book around. Ill have a look.

re 5: phases are calculated from a model. The better model you have, the more accurate are the phases. If you use MR, you don't need Patterson.

Regarding books, I think "Biomolecular Crystallography" by B. Rupp is the most comprehensive one. I read only selected sections from it (and not about the data processing), but I really like and recommend it.

I am voting as "Needs more focus" because for each point, you should tell us what your understanding is and then what your specific question is (in separate posts).

There are two large misconceptions in the OPs summary of solving a structure: 1) The phases are continually improved as progress in solving the structure is made rather than determined in a single step. 2) The Fourier transform is not between time and frequency domain (like in Fourier-transform spectroscopy) but between length and reciprocal length (and in three dimensions). The comments have some misconceptions as well (@porphyrin You can index a diffraction image and find the symmetry without FFT).

@Karsten Theis, yes of course, but with hundreds if not thousands of images for each structure fast methods are necessary.


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Why are borates more stable than aluminates?
In the differences between Boron and Aluminium, it is given that boron forms stable Borates and Aluminium forms unstable aluminates .
Please explain the stability of the 2 compounds.

Firstly, in both borate and aluminate; oxygen is covanlently bounded to boron and aluminium. Oxygen being more electronagative than any element in group 13 from which boron and aluminum belong to will always have delta (-).
Now, about the stability of both, as you go down the group 3, size increase down the group but positive charge (+3) remain the same, This implies that charge density decrease down the group 3. Charge density is about charge of an ion divided by the square of the ionic radius. Therefore, it implies that the polarizing ability of the metal decreases down the group 3. (This also means that, the metallic potential of Boron is greater than that of aluminium** taking into account polarizing ability***). Therefore, comparing their oxyanion compounds (borates and aluminates )which is stabilize mainly by polarising ability of each of the elements ( boron and aluminium) then, Borates will be more stability than aluminate.

"This also means that, the metallic potential of Boron is greater than that of aluminium. " this is quite wrong. Al is a metal and B is almost a non-metal. I don't the answer but the reasoning does not make any sense whatsoever.

As used in that contest by my explanation. My argument and explanation centered around Charge density and polarizing ability of Boron which is better/stronger than that of Aluminium.

I am against the word stability, because stability itself does not mean anything at all. When someone asks A is more stable than B, what does it mean? Is it thermodynamically stable or is it kinetically stable? Stability itself has no definition, so the question is vague.

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What is it that makes this noncontact catalytic system unique in 2020?
Phys.org's Accelerating chemical reactions without direct contact with a catalyst discusses the new Science Advances paper Noncontact catalysis: Initiation of selective ethylbenzene oxidation by Au cluster-facilitated cyclooctene epoxidation. I expected to read about something really unusual, long range forces, virtual entangled photons...
But Figure 1 of the paper (shown below) helps to explain that the catalyst produces an intermediary species and it's this that initiates or catalyzes the desired effect.
Question: Is it possible to make clear to a non-chemist how new the published noncontact catalytic system reaction really is? What is it that makes this unique or novel in 2020? Did nobody think of this until now, or have people been trying for a long time but this is the first success?

Fig. 1 A noncontact catalytic system.
A catalyst (Cat) catalyzes reaction A (AR ➔ S ➔AP) in which the intermediate S is effective in either initiating or catalyzing reaction B (BR ➔ BP), although reaction B is not catalyzed by the catalyst.

I'm quite confident that an answer can be written that is fact-based.

As a general concept, this isn't new - biological enzyme cascades are one example. But I don't know of any industrial processes that use this approach, so if it gets scaled up to industrial scale, that might be a first? Since most industrial catalysts are proprietary compounds, it's hard to be sure that there isn't already one that works in this way.

@Andrew I hadn't thought about proprietary processes; that could make this more challenging to answer. Of course there are industrial process using bioreactors so if all else fails that could be kind-of an answer ;-)

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A sealed container contains 90 % hydrogen and 10 % oxygen. There is a spark in the container. What happens?
A sealed container contains 90 % hydrogen and 10 % oxygen. There is a spark in the container. What happens?
Assuming it is a 10-litre volume container like a fuel tank.

What do you think happens? Lets work this step by step. Would you happen to know the volume of the sealed container?

For volume, assume a 10 litre container/fuel tank.

Why Volume it is crucial here, given that you know the composition? Are you asking for absolute values like energy and damages? That would depends on the container walls and gases amount, too :)

Boom!

"The flammability limits based on the volume percent of hydrogen in air at 14.7 psia (1 atm, 101 kPa) are 4.0 and 75.0. The flammability limits based on the volume percent of hydrogen in oxygen at 14.7 psia (1 atm, 101 kPa) are 4.0 and 94.0."

Quoted from:
Gregory, Frederick D. (February 12, 1997). "Safety Standard for Hydrogen and Hydrogen Systems" (PDF). NASA. Retrieved 2008-05-09.


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Trends in solubility of group 2 nitrates
In my lab report, we are required to explain the trends in solubility of group 2 salts, going down the group. I had explained all of the trends except one, group 2 nitrates. The following is the data provided.
$\ce{Mg(NO3)2}$ – $\pu{0.49 mol}$ per $\pu{100 g}$ of water
$\ce{Ca(NO3)2}$ – $\pu{0.62 mol}$ per $\pu{100 g}$ of water
$\ce{Sr(NO3)2}$ – $\pu{0.16 mol}$ per $\pu{100 g}$ of water
$\ce{Ba(NO3)2}$ – $\pu{ 0.04 mol}$ per $\pu{100 g}$ of water
So as the data shown, the solubility first increases and then decreases. My speculation is that the general trend should decrease but $\ce{Ca(NO3)2}$ has some special properties which makes it more soluble or $\ce{Mg(NO3)2}$ has some special which makes it less soluble. I searched the internet and all I got is "All group 2 nitrates are soluble" with no explanation regarding the trend.
Please correct my speculations if they're wrong and also provide an explanation regarding this trend.

Solubility is one of the present scientific problems that nobody is able to explain  thoroughly. As you are speaking of Calcium compounds, look at the series of Calcium compounds made with halogens. $\ce{CaCl2, CaBr2, CaI2}$ are so soluble in water that they can be dissolved in less than their weight of water. $\ce{CaF2}$ should be similar. It is not. On the contrary, $\ce{CaF2}$ is one of the most insoluble compounds, because it forms the mineral called fluorite, which is one of the main sources of Fluorine atoms on Earth. If it would have been at least partly soluble, it would have been washed away by the rains a long time ago. Nobody is able to explain such a difference of solubility.
There are plenty of theories explaining the solubilities of some groups of substances using electronegativities, ionic or covalent radius, and other parameters, which work pretty well with a lot of substances. But there are always exceptions. The final explanation remains to be discovered.

That's what I find on some website like chemguide. But like I said, I need to give a "reasonable" explanantion for my lab report, so I use one of the most popular explanation which is enthalpy change of solution, enthalpy change of hydration and lattice change. It's suprising that there is no thorough explanation on solubility trends.


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Why does Calcium (specifically Calcium Carbonate) inhibit the Citrate buffer system?
Is it because citrate chelates calcium and then is unable to react with other chemicals being added? If so can someone explain this reaction to me, or if not provide what the reason is? If anyone has any studies they've found on it that would be incredibly helpful! Thank you!

"Inhibited" is not probably the right term, as there is no catalyst inhibition, affecting the reaction kinetics.
If we consider about neutral $\mathrm{pH}$ range, the citrate buffer keeps $\mathrm{pH}$ by equilibrium reactions
$$\ce{HA^2- <=> H+ + A^3-}$$
$$\ce{HA^2- + OH- <=> H2O + A^3-}$$
with $\mathrm{p}K_\mathrm{a3}=6.4 $
Carbonate as a base spends the buffer capacity
$$\ce{CO3^2- + 2 HA^2- <=> CO2(g) + H2O + 2 A^3-}$$
If it was sodium carbonate, sodium ions would be just bystander ions.
But calcium ions form chelate with citrate 3:2, decreasing concentration of free hydrogen citrate + citrate ions.
So the overall calcium carbonate effect is being base and chelating citrate.


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Why does a paracetamol tablet clean an iron
If you burn some clothes on an iron and some residue sticks to it, rubbing a paracetamol tablet over it (while it's hot) surprisingly removes the residue and cleans the iron. Please explain why this happens.

paracetamol tablet decomposes to p-aminophenol almost quantitatively at hih temperatures, not sure how it cleans the iron. Perhaps the molten p-aminophenol solubilizes the synthetic charred polymers which are sticking on the iron's surface.

Many tablets contain silica (silicon dioxide) to avoid formation of large particles of the active drug. Silica is abrasive, so I think that silica is actually doing the magic. For comparison, you might try another tablet with silica in its composition: it should work the same way


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How to tell if an acid disassociate completely or partially?
what is the difference between
 and

how can I tell which one is the stronger acid, base on the 2 equations?

You need to know the dissociation constants of your acids to solve your problem.

Well, writing acetic acid as CH3COOH would be good start, Knowing common strong acids and types of weak ones as next step would be enough to solve it.

The equations don't tell you the complete story, just as by looking at an equation one cannot predict the boiling point of HCl, in the same way we cannot say how many HCl or acetic acid molecules are dissociating in water.  In easy words, if you had 0.1 moles of HCl in 1 L of water, all of them will dissociate in water, however if you have acetic acid, then only 1.3% acetic acid molecules will dissociate and the rest will remain in molecular form. It is a dynamic process. To see how this is calculated see
https://www.chemteam.info/AcidBase/Calc-percent-dissoc-of-weak-acid.html
We should use double arrows when we show the dissociation of acetic acid.


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Electrical and Thermal conductivity of electrum alloys
I have tried to search for this to little effect - I am trying to find electrical and thermal conductivity values for Electrum (a gold and silver alloy). Most sites mention the values depend on the percentage of gold and silver in the alloy, but provide no actual figures.
Is there a table out there that shows approximate electrical and thermal conductivity for electrum under different amounts of component metals in the alloy?

This reference may be a good place to get started:
Thermoelectric power and thermal conductivity in the silver-gold alloy system from 3-300°K
https://doi.org/10.1080/14786437008228219

You should elaborate.


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Mechanism of Etard reaction
According to Wikipedia Etard reaction proceed via a ene reaction with chromyl chloride forming a Etard complex(Chromium complex) which is then decomposed by a [2,3] sigmatropic shift to give aldehyde.

But the reaction given in my textbook is as follows,

Mechanism of this reaction suggested by Wikipedia does not create the intermediate of this reaction given in my textbook.The Chromium complex is further hydrolysed to give benzaldehyde.
Some sources which I referred suggest that Etard reaction proceed via a radical mechanism.
https://pubs.acs.org/doi/suppl/10.1021/ic200877f/suppl_file/ic200877f_si_001.pdf
What is the correct mechanism of this reaction?Is the Chromium complex given in my textbook correct?

Exact copy of- https://chemistry.stackexchange.com/questions/58381/mechanism-for-an-%c3%89tard-reaction-chromium-complex

Which textbook are you using?

The answer is within the comments

Organic chemist can imagine a lot of $wrong$ things with their imaginary arrow pushing constructs. It is not necessary that if the arrows pushing story makes some sense like a typical movie, then it is real story. Only an experiment can draw a reasonable conclusion. Don't be impressed with DFTs or fancy names. Someone jokingly said in a meeting that hardcore theoreticians cannot distinguish between a cow and methane.  I am not an organic chemist, but as a student I wasted a lot of time in believing in those arrow pushing mechanisms and thinking that all reaction mechanisms were known.

Look for original papers from the oldest first. A lot of papers appear from 1960s and indeed it is clear that it is a very complex reaction. https://www.sciencedirect.com/science/article/abs/pii/S0040402001985722


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Reaction of dimethyl sulfate with Protocatechualdehyde
I know that $\ce{(CH3)2SO4}$ reacts with phenols to give methoxybenzene. But the question that came in INChO-2018 was:

When one equivalent of $\ce{NaOH}$ and one equivalent of dimethyl sulfate reacts with Protocatechualdehyde, K is formed in 87% yield. When carried out with excess of $\ce{NaOH}$ and only one equivalent of dimethyl sulfate, M is formed in 95% yield. Identify K and M.

How would I know that ether formation take place at p- to $\ce{-CHO}$ group or at m- to it?
This is Protocatechualdehyde

Which alkoxide is more stabilised by the -CHO group? Hint it is the one where you can delocalise the -ve charge to the -CHO

For K, think about which OH is more acidic. For M, which phenoxide is more nucleophilic.

When one equivalent of dimethyl sulfate reacts with 3,4-dihydroxybenzaldehyde (protocatechualdehyde), there are two nucleophilic sites in the molecule, namely 3- and 4-hydroxy-groups. If one equivalent of strong base such as $\ce{NaOH}$ is used, it will deprotonate the most acetic group between them, in this case, 4-hydroxy-group. For example, $\mathrm{p}K_\mathrm{a}$ of 4-hydroxybenzaldehyde is $7.61$ while $\mathrm{p}K_\mathrm{a}$ of 3-hydroxybenzaldehyde is $8.98$ at $\pu{25 ^\circ C}$ (Ref.1). Thus, it is safe to assume that product K is 3-hydroxy-4-methoxybenzaldehyde (isovanillin).
However, the regioselectivity of the synthesis (when two nucleophilic sites are present) can be controlled by the number of base equivalents in this kind of alkylation steps (Ref.2). Thus, in excess $\ce{NaOH}$, both phenolic hydroxy groups of 3,4-dihydroxybenzaldehyde would be deprotonated and methylation would occur at 3-phenolate, because it is the most reactive at this point between two (basically, negative charge of 4-phenolate should be largely delocalized within aromatic nucleus with benzaldehyde group than that of 3-phenolate without direct resonance effect from benzaldehyde group).  Therefore, the major product M in this case would be 4-hydroxy-3-methoxybenzaldehyde (vanillin). You can probably compare with methylation of ascorbic acid in different conditions.
Note: Some literature mentioned that when one equivalent of $\ce{NaOH}$ is used, reaction is must faster and tends to use both methyl groups of dimethyl sulfate. Thus, I won't surprised if the given answer for K is 3,4-dimethoxybenzaldehyde.
References:

E. P. Serjeant, Boyd Dempsey, Editors, IUPAC chemical data series no. 23: Ionization Constants of Organic Acids in Aqueous Solution; Pergamon: Oxford, United Kingdom, 1979 (Values extracted from PDF here).
Tobias Schneider, Vladimir Kubyshkin, Nediljko Budisa, "Synthesis of a Photo‐Caged DOPA Derivative by Selective Alkylation of 3,4‐Dihydroxybenzaldehyde," Eur. J. Org. Chem. 2018, (18), 2053-2063 (https://doi.org/10.1002/ejoc.201701749).

In my experience only one of the methyl groups of dimethyl sulfate (DMS) is utilized in alkylation. The formation of K involves one equivalent of NaOH and one of DMS. So K, formed in 87% yield, should be isovanillin as you argue and less likely that veratraldehyde, the dimethylated product unless there is evidence that both both methyl groups of DMS react. Otherwise, we are in agreement on K and M.


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Why is the number of covalent bounds typically formed by an element equal to 8 minus the group number?
Why would the number of covalent bounds typically formed by an element equal to 8 minus the group number?
E.g for C, we have 8-4 bonds, for Cl, 8-7 bonds

It isn't always.  Boron compounds, sulfur forming three or four delocalized bomds in SF4 and SF6, etc.

First of all, one has to adopt an outdated concept of group number for the concept mentioned in the question to work. The group number of carbon, silicon, germanium ... is 14, not 4, per IUPAC.
Reverting to the old definition (sometimes called main group), one notices that the old group number is equal to the valence electrons for the members of the group. In order to fulfill the octet rule$^{1}$, for example a nitrogen atom could engage in 3 bonds, following
$$
8 \quad \text{(due to octet rule)} \quad - \quad 5 \quad \text{(valence electrons = old group number)} \quad = \quad 3 \quad \text{(covalent bonds)}
$$
However, there are many, many exceptions to this concept: $\ce{ClO4-}$, $\ce{SF6}$, $\ce{PCl5}$, both polymeric and monomeric $\ce{BeCl2}$, almost all borohydrides beginning with $\ce{B2H6}$, "sub-valent" but stable carbenes (compounds in which at least one carbon has a free electron pair), and more.
So, overall, the concept from the question is a bit of a crutch$^2$ for the more useful valence electrons.

$^{1}$ which follows from quantum mechanics.
$^{2}$ though not a bad observation per se.

In a very general aspect we could think of the following, if $X$ was the more electronegative element, say group 14 - group 17 and $M$ was the less electronegative element, then if the compound $M_mX_x$ forms, $M$ would give its electrons to $X$ so $X$ can form an octett which is the most stable electron configuration.
To get this octett at $X$ you will require $8x$ electrons.
$M*e(M)+x*e(X)=8x$
Now let's asume there were covalent bonds between $M$ and $M$. Those would include electrons that cannot be shared with $X$, hence, we need to subtract $b(MM)$ for every bond. Same goes for lone-pairs at $M$ $e(M)$.
For $X$ we need to increase the amount of electrons for every $b(XX)$ and $e(X)$. Therefore we get
$8x = m[e(M)-b(MM)-E]+x[e(X)+b(XX)]$
or
$\frac{m*e(M)+x*e(X)}{x}=8+\frac{(m[b(MM)+E]-x*b(XX))}{x}$
The part on the left side is what we call the valence electron concentration (VEC), a useful property to understand many intermetallic phases and structures. Form this we can say
$b(XX)=8-VEC(X)+\frac{m}{x}[b(MM)+E]$
Now if we assume a pure element like sulphur for example:
$m=0$
$VEC(X)=e(X)=N$
$b(XX)=8-VEC(X)=8-N$ so for sulphur (N=6) we get $b(SS)=8-6=2$


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Effect of reduction potential on particle size of chemically precipitated silver particles
Is there no mathematical relationship between redox potential and particle size saying that an $X$ potential should yield a $Y$ particle size? Right now its general that large redox potential results in smaller particles, but I want to know if, for example, for silver particles I use $\ce{AgNO3}$ with glucose reducing agent and say the redox potential is $\pu{0.33V}$, what the ballpark for the particle size should be based on that potential (assuming $\mathrm{pH}$ does not change)?

"Is there no mathematical relationship between redox potential and particle size saying that an X potential should yield a Y particle size?" Electrode potential is a thermodynamic property, it can only say "yes, this reduction or oxidation is favorable" but particle size is a kinetic property. I feel it is very difficult to relate the two.

I do see a relation between the size of the attained nanoparticles and their electrochemical potential, of course. Still the main parameter should be the amount of current flown to the cell. A parallel would be precipitation, where fast means small. In case of a chemical reduction like in your case I would think of small particles with high reducer conc, though thinking of concentration suffices in this context.That the redox properties of the final particles are dictated by size should be out of doubt, as I said above, but I am not sure it relates to the question.

Here is an example of a study of fish oil nanoemulsions which examines, to quote:

The influence of surfactant-to-oil-ratio on particle size and physical stability was evaluated.

Here is yet another study to quote:

Although the role of surfactant in the synthesis of particles in the nanometer range has frequently been documented...The surface coverage per surfactant molecule could be related to the surface tension of the latex, thus providing a relationship between particle size and stability.

And, another more expansive work to quote:

It was reported that the formation process of these NPs could be divided into three stages: growth, incubation and Oatwald ripening stages. The higher boiling point of 300 °C of paraffin affords a broader range of reaction temperature and makes it possible to effectively control the size of silver NPs by varying the heating temperature alone without changing the solvent. Moreover, the size of the colloidal silver NPs could be regulated not only by changing the heating temperature, or the ripening time, but also by adjusting the ratio of oleylamine to the silver precursor.

While chemical reduction methods are connected to reduction potential, the multi-stage aspect of the creation process for NPs suggests only a partial correlation, so not a 'mathematical relationship', but perhaps a statistical one.


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Equation of water at triple point
I was thinking about what exactly happens at the triple point of water, and what could be the effect of adding solutes/solvents. I feel we can represent the triple point using the following equilibrium reaction:
$$\ce{H2O (s) <=> H2O (l) <=> H2O (g)}$$
Is this correct? How would this change if I add solutes?
Note: I know that adding solutes would increase the concentration of the solution, but how exactly will it shift the equilibrium?

Quibbles - (1) If the liquid isn't pure, then the system can't be at the triple point. (2)  There is also an equilibrium between the gas and the solid. That's why it is a triple point. Three equilibriums at once. // The salt would only exist in the solution. So ice would melt (have you ever made ice cream by hand?), the water gets colder, and the vapor pressure of the gas phase goes down.

@MaxW Are you saying that even if I prepared a ~5M solution of NaCl +water and the solution is homogenous, it would not have a triple point?

The triple point will no longer be a point. Ditto for the melting and boiling points, BTW.

The phase diagram in https://chemed.chem.purdue.edu/genchem/topicreview/bp/ch15/colligative.php shows a simple transition of the phase lines when a solute is added. Is this wrong?

It is an oversimplification.

At the triple point the water system is assumed to be adiabatic. The system has 3 phases, ice from pure water, liquid water, and pure water vapor. The various equilibria  are shown below.

Note that the ice floats on the water. So the ice is in contact with both the liquid phase and the gas phase. Furthermore The ice can't cover the whole surface like a frozen over pond. There must be some liquid water in contact with the gas phase too.
A subtlety here... You can't make a solid phase of ice that is homogeneously say 1 molar NaCl. If you take a 1 molar NaCl and try to freeze it you get relatively pure water first, then saltier and saltier water freezing as the solid phase gets colder and colder. Since the impure ice has doesn't have a homogeneous composition, it doesn't have a specific melting point.
So for the system you proposed there is pure water ice,  a 1 molar NaCl solution, and a gas phase of water vapor. Unfortunately humans can't make a perfectly adiabatic container, so as the ice melts, the 1 molar NaCl solution becomes more dilute. Thus there is no specific triple point.
In other words... if you just throw some ice into the 1 molar NaCl solution, the final temperature of the theoretical system will depend on the relative amounts of the three phases - How much 1 molar NaCl solution, how much ice, and the volume of the gas phase.

The triple point of water is an example of Thermodynamic Equilibrium, which is explained by Wikipedia as:

simultaneously in mutual thermal, mechanical, chemical, and radiative equilibria

Your equation only covers the chemical facet of this equilibrium and hence is not the correct way of representing total thermodynamic equilibrium. I am also not sure of the correct way of representing these kinds of equilibria which deal with both energy and physical states.
If this equilibrium is disturbed chemically i.e. by adding more water, according to Le Chatelier's principle, the equilibrium will shift towards the solid and gaseous states. As the enthalpy of vapourization is more than enthalpy of fusion, more ice will be formed as compared to the amount of steam evolved in order to balance the energies involved. Hence, it would tend to shift to the left.

No voodoo here. Ice from pure water, a salt solution and water vapor would be a well well studies system. The whole system depends on temperature. There has to be enough ice to melt which will cool the water and the gas phase.


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How to precipitate Chloroplatinic acid?
For the precipitation reaction, I have to react $\ce{H2PtCl6}$ with $\ce{(NH4)2CO3}$ while maintaining the $\ce{pH}$ in the range of 7-8 at 60 °C. But the problem is Chloroplatinic acid is not precipitating. Instead, the acid evaporated by making holes in the aluminium foil which were not present before.
Can you please point out what I'm doing wrong?
Theoretical reaction:
$$\ce{(NH4)2CO3 (aq) + H2PtCl6 (aq) -> (NH4)2PtCl6 (s) + H2O (aq) + CO2 (g)}$$
PS: Aluminium foil was just used for proper heating

What do you mean by "it got evaporated by making holes in the aluminum foil" ?Which aluminum foil ? Why do you use an aluminum foil ? What for ? Why  are there holes in it ?

Normal aluminum foil to prevent escaping of fumes.
Holes were not present on the foil.
Chloroplatanic acid "made the holes" for escaping rather than precipitating

Are you sure of the quality of ammonium carbonate ? It is a highly volatile substance. On the other hand, Ammonium chloroplatinate is weakly soluble in hot water : 33.7 g/L at 100°C

I used Ammonium carbonate of following specs: ACS reagent, ≥30.0% NH3 basis.

Possibly the precipitate is reacting with the foil, making the holes. Try using a glass lid to cover the apparatus.

I suspect that you could be forming an aqueous solution of $\ce{(NH4)2PtCl6}$.
I would like to make an observation as a person who has done platinum group metal chemistry, the anionic chloro complexes of PGMs like $\ce{PdCl4^2-}$ and $\ce{PtCl4^2-}$ are very toxic. They can induce a nasty allergy to PGMs. I would suggest that you do not work at home with these substances.
Also in my experience at least with $\ce{PdCl4^2-}$, I have always found that if you want to bring it out of solution then cation methathesis with a potassium salt can bring the substance out of solution as a brown solid. I suspect for the hobby chemist that low sodium salt would be the best reagent for doing this transformation.
For the lab chemist I would suggest using tetramethyl ammonium chloride or even tetraphenyl phosphonium chloride


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What is the precise definition of standard molar entropy?
I understand what the standard molar entropy is, and how to use it in calculations, but I'm interested in understanding exactly how it's defined and measured.
My recollection from a long time ago is that it goes roughly like this:

for each element, assume that its entropy goes to zero at 0 kelvin when it's in some standard state
then, for the compound of interest, calculate and/or measure the entropy difference between the compound and its constituent elements at 0 K.

However, I'm not even sure that's exactly correct, and there are a lot of details that need to be filled in if it is right. Most of the following issues will be covered by the exact definition, but to be more elaborate, these immediately come to my mind: what exactly is the standard state for each element? Do all elements form a perfect crystal at 0 K, and how can we prove this? Should I think of the third law (zero entropy at zero kelvin) as being a definition that sets an otherwise arbitrary baseline, or can it be proven that the quantum mechanical von Neumann entropy really is zero for pure elements at 0 K?
An internet search turned up plenty of information about the values of standard molar entropy for various compounds and how to use them, but I wasn't able to find a resource with a detailed description of its definition.
What is the precise definition (preferably with reference) of standard molar entropy?

That's kinda too broad. "Do all elements form a perfect crystal at 0K, and how can we prove this?" kinda doesn't make sense - they don't, as "perfect crystal" is a model.

Well yes, but if standard molar entropy is defined in terms of such a model then, in order to make it, we must know something about how reality diverges from the model.

Sure https://en.wikipedia.org/wiki/Residual_entropy

This is a great question. I can conceive of a few ways in standard molar entropies might be determined, the most obvious of which is to integrate from zero to an experimentally-accessible temperature using the the Debye $T^3$ law (which models the constant-volume heat capacity of solids as they approach absolute zero) and then measure the additional heat flow calorimetrically.  But that's just one possibility.  I did look at the CODATA tables for standard molar entropies, but wasn't able to find details on how they are determined.  I'd thus be interested to see an answer on this.

@Mithoron the question is about how that's taken into account when defining and measuring the standard molar entropy, as found in thermodynamic tables.

https://chemistry.stackexchange.com/questions/126803/what-is-the-precise-definition-of-standard-molar-entropy

Not quite an answer, but too long for a comment:
Here's a link to the official CODATA site showing the values:
codata.info (via the Internet Archive).
It references the following source:
Cox, J. D., Wagman, D. D., and Medvedev, V. A., CODATA Key Values for Thermodynamics, Hemisphere Publishing Corp., New York, 1989.
Unfortunately, that source isn't readily viewable online (all I found was what appears to be a German version, at Berichte der Bunsengesellschaft für physikalische Chemie 1990, 94 (1), 93–93 (It's behind a paywall.)  But if you could get your hands on it (interlibrary loan, say), it should contain a detailed description of how the values were actually calculated.
FWIW, here's a link showing how standard molar entropies might be calculated (but I don't know if the procedure described here is what is used to determine the official CODATA values): Standard Molar Entropy of Solid Aluminum Oxide, Comments to Tandy Grubbs, Stetson University (via the Internet Archive).
Essentially, it mentions using the Debye extrapolation I mentioned in an earlier comment up to $\approx\pu{15 K}$, and then using a differential scanning calorimeter to determine the reversible heat flow between $\pu{15 K}$ and $\pu{298.15 K}$
And here's a purely theoretical approach (unfortunately, everything except the abstract is behind a paywall):
Pascal, T. A.; Lin, S.; Goddard Iii, W. A. Thermodynamics of liquids: standard molar entropies and heat capacities of common solvents from 2PT molecular dynamics. Phys. Chem. Chem. Phys. 2011, 13 (1), 169–181. DOI: 10.1039/C0CP01549K.

Also, some general comments on the issue of the entropy of a substance at absolute zero:
It is possible the question of whether substances need to be perfect crystals to have zero entropy at $\pu{0 K}$ does not enter into the calculation of standard molar entropies.  Consider the following:

Any calculation of standard molar entropy requires determining the initial reversible heat flow starting at $\pu{0 K}$, at standard pressure:
$$
S^\circ(\pu{298.15 K}) =
\int_{0}^{\pu{298.15 K}}\frac{\mathrm{đ}q_\mathrm{rev}}{T}
$$
The initial part of this integral (starting at $\pu{0 K}$) can't be done experimentally, and thus has to be based upon a model.
We may not know how to model whether a substance being a perfect crystal at $\pu{0 K}$ will affect the amount of reversible heat flow needed to increase its temperature.  In addition, we may not know how to predict the actual state of substances at $\pu{0 K}$ (i.e., the extent to which they are not perfect crystals–and to the extent that they are not, what the distribution of their forms actually is).
Further, even if there were an effect on heat flow from not being a perfect crystal, and we did know how to model it, the effect may be small compared to the experimental errors that are present in determining standard molar entropies.
It's still an open question whether a substance needs to be a perfect crystal to have zero entropy at $\pu{0 K}$.  Some in the field argue it does, some argue it doesn't.  Those in the latter camp argue that, given that no other states are accessible at $\pu{0 K}$, the fact that the system is locked into only one microstate (even if it's not a perfect crystal) gives it zero entropy.

Please avoid editing your answer multiple times in short succession. It will cause an automatic flag eventually, which will unnecessarily have to be investigated. (Also, I am trying to improve this post now for about an hour now.)

@Martin-マーチン Understood, I'll leave it alone for now.  Though I'd argue the automatic flag is a problem with the system, not with the practice.  I.e., we're all volunteers, and if a volunteer is willing to spend extra time to repeatedly refine his/her post to improve it, that's to be encouraged, not discouraged.  Yes, one could argue the volunteer should find a work-around that avoids the flag, but that may be inconvenient to the volunteer (when reviewing, I like to see the entire post w/o the editing screen), and we don't want to create inconveniences for those wishing to optimize their ans.

@Martin-マーチン  Incidentally, I was thinking of flagging you in a comment to bring this question to your attention, since I know you have an interest in and knowledge of international standard-setting, and I thus suspected you might know the answer to this.   Also, wrt the automatic flagging: The SE sites are undergoing continuous improvement, and this is just my opinion on one way the site might be improved. I know the auto flag is there to prevent gaming the system, but it has its downsides.

The automatic flag is necessary to safeguard everybody from roll-back wars, vandalism, reputation gaming, etc.. This is all quite important. Draft versions will be saved, there is a live preview of the full text, so posting it doesn't have any benefit. I do not wish to discourage you from improving your answer, I am simply pointing out the things that are going to happen if you continue with this approach. I am also a volunteer here, but unfortunately I don't have that much time with the site anymore, so I prefer having no false positive flags.

@Martin-マーチン  I've found that, at least in some contexts, drafts aren't permanently saved:  A few weeks ago I created a draft question on Physics SE (I assume the system behavior is the same here), leaving it in draft form while I worked on it (I hadn't yet posted it).  But when I went back to it a few days later to continue working on it, the draft had disappeared.   Is the behavior (i.e., the saving of drafts) different for edits of questions or answers that have already been posted (as compared with drafts of those that have not yet been posted)?

We should rather be using [chat] for these discussions. The software is the same here and there. I don't know the exact mechanism on how drafts are stored. I think it is perfectly okay to save the improved post before closing the browser to avoid any loss; no question here that this has to be accomodated. Leaving the draft for a couple of days is probably not the best idea, but while editing an existing answer it is better to use the preview instead of directly posting.

This gets the bounty despite not quite being an answer, since the points have to go somewhere :)

Thanks @Nathaniel.  I haven't given up on this.   I'll either see if I can get my hands on Cox/Wagman/Medvedev's CODATA Key Values for Thermodynamics, or email NIST.


text:
Find pH and Kw for water with given values of conductivity

At $\pu{25 ^\circ C}$, $\lambda_\infty (\ce{H+}) = \pu{3.5 \times 10^{-2} S m2 mol-1}$ and $\lambda_\infty (\ce{OH-}) = \pu{2 \times 10^{-2} S m2 mol-1}$. Determine $\mathrm{pH}$ and $K_\mathrm{w}$.

Given: Specific conductance $= \pu{5.5 \times 10^{-6} Sm-1}$ for $\ce{H2O}$
My Attempt:
I know I need to find the concentration of $\ce{H+}$ or $\ce{OH-}$ for $\mathrm{pH}$. $K_\mathrm{w}$ can be found subsequently. However, I have no clue as to how do I find that with the given information.
I am also aware of the relation between conductivity, Specific conductance, and molarity. It is easy to find the molarity from the given information, but that would be of no use as we need the concentration of the individual ions rather than the molarity of the solution.
Any help would be appreciated.

Is your second value ∞ valid for OH- ?

Yes, I forgot to include that. My bad!

We know,
$$\lambda_c = \frac{(\text{Specific Conductivity})}{1000 \times c}$$
$$\alpha\lambda_\infty = \lambda_c$$
(where $c$ is concentration in moles per litres of the electrolyte and $\alpha$ is degree of dissociation of the electrolyte)
$\lambda_{\infty}$ of water (the electrolyte) is the sum of $\lambda_{\infty}$ of $\ce{H+}$ and $\ce{OH-}$.
From equations above, one can easily find  the the value of $c\alpha$ is $10^{-7}$, which is the concentration of $\ce{H+}$.

@Tony could you verify the answer calculated based on this answer and report us until we get a second opinion?

I do get the correct answer based on this explanation.


text:
Miller Indices and the case of a cubic crystal
My textbook, Solid-State Physics, Fluidics, and Analytical Techniques in Micro- and Nanotechnology, by Madou, presents the following image and explanation in a section on x-ray diffraction and Laue equations:

... Laue’s equations can also be interpreted as reflection from the $h,k,l$ planes. From Figure 2.25 it can be seen that the spacing between the ($hk$) planes, and by extension between ($hkl$) planes, is given as$^*$:
$d_{hkl} = \frac{\hat{\mathbf{n}}_{hkl} \cdot \mathbf{a}_1}{h} = \frac{\hat{\mathbf{n}}_{hkl} \cdot \mathbf{a}_2}{k} = \frac{\hat{\mathbf{n}}_{hkl} \cdot \mathbf{a}_3}{l} \tag{2.27}$

$^*$ Notice that in the case of a cubic crystal, Equation 2.27 can be simplified to give the distance between planes as in Equation 2.12.

Equation 2.12 referenced above is presented earlier in the textbook as follows:

Adjacent planes ($hkl$) in a simple cubic crystal are spaced a distance $d_{hkl}$ from each other, with $d_{hkl}$ given by:
$$d_{hkl} = \frac{a}{\sqrt{h^2 + k^2 + l^2}} \tag{2.12}$$
where $a$ is the lattice constant. Equation 2.12 provides the magnitude of $d_{hkl}$ and follows from simple analytic geometry.

The following comment is not clear to me:

$^*$ Notice that in the case of a cubic crystal, Equation 2.27 can be simplified to give the distance between planes as in Equation 2.12.

I was wondering if someone would please take the time to explain and show this.
EDIT:
In an attempt to provide further context, I am posting more of the textbook section surrounding this problem.

Constructive interference will occur in a direction such that contributions from each lattice point differ in phase by $2\pi$. This is illustrated for the scattering of an incident x-ray beam by a row of identical atoms with lattice spacing $\mathbf{a}_1$ in Figure 2.20. The direction of the incident beam is indicated by wave vector $\mathbf{k}_0$ or the angle $\alpha_0$, and the scattered beam is specified by the direction of $\mathbf{k}$ or the angle $\alpha$. Because we assume elastic scattering, the two wave vectors $\mathbf{k}_0$ and $\mathbf{k}$ have the same magnitude, i.e., $2\pi/\lambda$ but with differing direction. A plane wave $e^{i k \cdot r}$ is constant in a plane perpendicular to $\mathbf{k}$ and is periodic parallel to it, with a wavelength $\lambda = 2\pi/\mathbf{k}$ (see Appendix 2A). The path difference $A_1B − A_2C$ in Figure 2.20 must equal $e\lambda$ with $e = 0, 1, 2, 3, \dots$. For a fixed incident x-ray with wavelength $\lambda$ and direction $\mathbf{k}$, and an integer value of $e$, there is only one possible scattering angle $\alpha$ defining a cone of rays drawn about a line through the lattice points (see Figure 2.20).

Because crystals are periodic in three directions, the Laue equations in 3D are then:
$$\mathbf{a}_1 (\cos\alpha - \cos\alpha_0) = e \lambda$$
$$\mathbf{a}_2 (\cos\beta - \cos\beta_0) = f \lambda \tag{2.21}$$
$$\mathbf{a}_3 (\cos\gamma - \cos\gamma_0) = g \lambda$$
For constructive interference from a three-dimensional lattice to occur, the three equations above must all be satisfied simultaneously, i.e., six angles $\alpha$, $\beta$, $\gamma$, $\alpha_0$, $\beta_0$, and $\gamma_0$; three lattice lengths $\mathbf{a}_1$, $\mathbf{a}_2$, and $\mathbf{a}_3$; and three integers ($e$, $f$, and $g$) are fixed. Multiplying both sides of Equation 2.21 with $2\pi/\lambda$ and rewriting the expression in vector notation we obtain:
$$\mathbf{a}_1 \cdot (\mathbf{k} - \mathbf{k}_0) = 2 \pi e$$
$$\mathbf{a}_2 \cdot (\mathbf{k} - \mathbf{k}_0) = 2 \pi f \tag{2.22}$$
$$\mathbf{a}_3 \cdot (\mathbf{k} - \mathbf{k}_0) = 2 \pi g$$
with $\mathbf{a}_1$, $\mathbf{a}_2$, and $\mathbf{a}_3$ being the primitive vectors of the crystal lattice.
If we further define a vector $\Delta \mathbf{k} = \mathbf{k} − \mathbf{k}_0$, Equation 2.22 simplifies to
$$\mathbf{a}_1 \cdot \Delta \mathbf{k} = 2 \pi e$$
$$\mathbf{a}_2 \cdot \Delta \mathbf{k} = 2 \pi f \tag{2.23}$$
$$\mathbf{a}_3 \cdot \Delta \mathbf{k} = 2 \pi g$$
Dealing with 12 variables for each reflection simultaneously [six angles ($\alpha$, $\beta$, $\gamma$, $\alpha_0$, $\beta_0$, and $\gamma_0$), three lattice lengths ($\mathbf{a}_1$, $\mathbf{a}_2$ and $\mathbf{a}_3$), and three integers ($e$, $f$, and $g$)] is a handful; this is the main reason why the Laue equations are rarely referred to directly, and a simpler representation is used instead. The reflecting conditions can indeed be described more simply by the Bragg equation.
Further below we will learn that constructive interference of diffracted x-rays will occur provided that the change in wave vector, $\Delta \mathbf{k} = \mathbf{k} − \mathbf{k}_0$, is a vector of the reciprocal lattice.
Bragg’s law is equivalent to the Laue equations in one dimension as can be appreciated from an inspection of Figures 2.24 and 2.25, where we use a two-dimensional crystal for simplicity. Suppose that vector $\Delta \mathbf{k}$ in Figure 2.24 satisfies the Laue condition; because incident and scattered waves have the same magnitude (elastic scattering), it follows that incoming ($\mathbf{k}_0$) and reflected rays ($\mathbf{k}$) make the same angle $\theta$ with the plane perpendicular to $\Delta \mathbf{k}$.

The magnitude of vector $\Delta \mathbf{k}$, from Figure 2.24, is then given as:
$$|\Delta \mathbf{k}| = 2\mathbf{k}\sin(\theta)$$
We now derive the relation between the reflecting planes to which $\Delta \mathbf{k}$ is normal and the lattice planes with a spacing $d_{hkl}$ (see Figure 2.25 and Bragg’s law in Equation 2.20). The normal unit vector $\hat{\mathbf{n}}_{hk}$ and the interplanar spacing $d_{hk}$ in Figure 2.25 characterize the crystal planes ($hk$). From Equation 2.23 we deduce that the direction cosines of $\Delta \mathbf{k}$, with respect to the crystallographic axes, are proportional to $e/a_1$, $f/a_2$, and $g/a_3$ or:
$$e/a_1:f/a_2:g/a_3 \tag{2.25}$$
From the definition of the Miller indices, an ($hkl$) plane intersects the crystallographic axes at the points $a_1/h$, a_2/k, and a_3/l, and the unit vector $\hat{\mathbf{n}}_{hkl}$, normal to the ($hkl$) plane, has direction cosines proportional to:
$$h/a_1, k/a_2, \text{and} \ l/a_3 \tag{2.26}$$
Comparing Equations 2.25 and 2.26 we see that $\Delta \mathbf{k}$ and the unit normal vector $\hat{\mathbf{n}}_{hkl}$ have the same directions; all that is required is that $e = nh$, $f = nk$, and $g = nl$, where $n$ is a constant. The factor $n$ is the largest common factor of the integers $e$, $f$, and $g$ and is itself an integer. From the above, Laue’s equations can also be interpreted as reflection from the $h,k,l$ planes. From Figure 2.25 it can be seen that the spacing between the ($hk$) planes, and by extension between ($hkl$) planes, is given as:
$d_{hkl} = \dfrac{\hat{\mathbf{n}}_{hkl} \cdot \mathbf{a}_1}{h} = \dfrac{\hat{\mathbf{n}}_{hkl} \cdot \mathbf{a}_2}{k} = \dfrac{\hat{\mathbf{n}}_{hkl} \cdot \mathbf{a}_3}{l} \tag{2.27}$

Related: https://chemistry.stackexchange.com/questions/118953/what-is-the-origin-of-the-quadratic-factor-in-calculation-of-xrd-peaks

This answer uses vector algebra and the so-called reciprocal lattice to avoid trigonometry. There are probably other, more geometric ways to connect the Laue conditions to the lattice spacing.

$d_{hkl} = \dfrac{\hat{\mathbf{n}}_{hkl} \cdot \mathbf{a}_1}{h} = \dfrac{\hat{\mathbf{n}}_{hkl} \cdot \mathbf{a}_2}{k} = \dfrac{\hat{\mathbf{n}}_{hkl} \cdot \mathbf{a}_3}{l} \tag{2.27}$

We use the Laue conditions to define the reciprocal lattice vector $\mathbf{d}^*_{hkl}$ and reciprocal lattice unit vectors $\mathbf{a}^*_1, \mathbf{a}^*_2, \mathbf{a}^*_3$. The reciprocal lattice point is defined as:
$$\mathbf{d}^*_{hkl} = \dfrac{\hat{\mathbf{n}}_{hkl}}{d_{hkl}},$$ i.e. it is perpendicular to the reflecting planes and has a magnitude of one divided by the lattice spacing. We can rewrite the three Laue conditions as:
$$h = \mathbf{d}^*_{hkl} \cdot \mathbf{a}_1$$ $$ k = \mathbf{d}^*_{hkl} \cdot \mathbf{a}_2 $$ $$ l = \mathbf{d}^*_{hkl} \cdot \mathbf{a}_3 $$
and define
$$ \mathbf{a}^*_1 = \mathbf{d}^*_{100} $$
$$ \mathbf{a}^*_2 = \mathbf{d}^*_{010} $$
$$ \mathbf{a}^*_3 = \mathbf{d}^*_{001} $$
The reciprocal space unit vectors have magnitudes that are the reciprocals of the magnitude of the corresponding real space unit vectors (dot product equals to one), and are perpendicular to the other two real space unit vectors (dot product is zero), as you can see if you apply the Laue conditions to, say, $ \mathbf{a}^*_1 = \mathbf{d}^*_{100}$:
$$1 = \mathbf{d}^*_{100} \cdot \mathbf{a}_1$$ $$ 0 = \mathbf{d}^*_{100} \cdot \mathbf{a}_2 $$ $$ 0 = \mathbf{d}^*_{100} \cdot \mathbf{a}_3 $$
You can now express a reciprocal lattice point via the Miller indices and the reciprocal unit vectors:
$$\mathbf{d}^*_{hkl} = h \cdot \mathbf{a}^*_1 + k \cdot \mathbf{a}^*_2 + l \cdot \mathbf{a}^*_3$$
For the cubic case the three reciprocal unit cell vectors are mutually perpendicular just like the real space unit vectors, and they all have a magnitude of 1/a. We can calculate the magnitude of $\mathbf{d}^*_{hkl}$ as (using properties of the cartesion coordinate system, or Pythagoras' theorem):
$$|\mathbf{d}^*_{hkl}| = \frac{\sqrt{(h^2 + k^2 + l^2)}}{a}$$
To get $d_{hkl}$, you simply have to take the reciprocal.
Disclaimer
I called $d_{hkl}$ the lattice spacing. This is correct if $h, k, l$ have no common divisor. If they do, it is the lattice spacing divided by that common divisor. For example, the (1,2,3) reflection and the (10,20,30) reflection has the same reflecting plane, but a different diffraction angle.

Thanks for the answer! So we have that $$|\mathbf{d}^*_{hkl}| = \sqrt{ \left( \dfrac{h^2}{a^2} + \dfrac{k^2}{a^2} + \dfrac{l^2}{a^2} \right) } = \dfrac{\sqrt{(h^2 + k^2 + l^2)}}{a}$$?

@ThePointer Yes


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Geometry of Ni(acac)2
The complex $\ce{Ni(acac)2}$ is paramagnetic as expected , i.e it has two unpaired electrons giving rise to a magnetic moment of $ \sqrt{8}$ BM but doesn't have the tetrahedral geometry consistent with the CFT assumptions , why ?
well if the above thing is true then what is the actual structure of the complex ? please help !
source: Inorganic Chemistry: Principles of Structure and Reactivity, 4e

Good question, the OP missed a point. Is the Ni hydrated or is it an oligomer. I suggest it should stay.

It is a good question, it is a gemstone hiding in the mud.
I have searched the Cambridge database, all the bis-acetylacetonatonickel complexes which have four coordinate nickel centres are square planar. There are three common coordination geometries for nickel(II) which we need to consider.
Tetrahedral, square plannar and octahedral. Here is a diagram (not to scale) of the d oribital energies in the three types of complex.

Nickel(II) has eight d electrons, if we pack them into the octahedral or tetrahedral complex we will end up with two unpaired electrons. But if we stick the electrons into a square planar complex then if the difference in the energy between the two uppermost d orbitals in energy is sufficient then we will get a low spin $\mathrm d^8$ complex. To get spin-pairing, the energy difference between the orbitals must be greater than the energy required to spin pair two electrons.
There is a whole continuum of complexes between perfect octahedral and perfect square planar. If the distance between the metal and the two ligands on one axis is increased (tetragonal distortion of an octahedral complex) then the complex is part way between the two extremes. There is a good undergraduate lab experiment based on the early work of David Goodgame which is all about the nickel(II) complexes of N,N-diethylethylenediamine. I did this experiment as an undergraduate, I supervised students doing it as a PhD student and as an academic I have used it in a lab class. Trust me, it is good!
What the person asking the original question should have been thinking about is the question of square planar vs octahedral (or tetragonally distorted octahedral).
If I was to take an acac-type ligand with bulky groups in place of the methyls such as tert-butyls then I am sure that I would have an isolated nickel. This nickel could well be tetrahedral. But in fact it is square planar (F.A.Cotton, J.J.Wise, Inorganic Chemistry, 1966, 5, 1200, CCDC code DPIMNI)
If we look at a nickel(II) complex of an acac where the methyl groups have been replaced with phenyl groups, the closest oxygens to the nickel other than the four we expect are $\pu{4.4 Å}$ from the nickel. They are a long way in terms of angle away from the $z$ axis of the nickel. I think it is safe to say that they are not binding to the nickel. Here is a picture of the nickel complex. (D.V.Soldatov, A.T.Henegouwen, G.D.Enright, C.I.Ratcliffe, J.A.Ripmeester, Inorganic Chemistry, 2001, 40, 1626.)

Just in case anyone is not sure it is very much a square planar complex.

But if you look at the crystallographic literature you will see G.J.Bullen, R.Mason, P.Pauling, Inorganic Chemistry, 1965, 4, 456. Here is the answer: what happens is that the nickel acac complex forms a trimer and the bridging oxygens make all three nickels have an octahedral environment. The CCDC code is ACACNI01 for the best determination which was M.B.Hursthouse, M.A.Laffey, P.T.Moore, D.B.New, P.R.Raithby, P.Thornton, Dalton, 1982, 307.

Also look out for hydrated complexes such as AQACNI (R.L.Harlow, C.E.Pfluger, J. Coord. Chem., 1973, 2, 289).
By the way one of the lectures I had as an undergrad mentioned this problem of waters adding to the coordination sphere or oligomers forming, I think it was in a lecture by David Goodgame.
While I have never seen a 4-coordinate nickel bis-acac complex which is anything but square planar, if you look at the coordination chemistry of a related phosphorus(V) ligand (dithioimidodiphosphinate) you can see both tetrahedral and square planar nickels.
If we look at YUSFIB which is for the tetraphenyl ligand (P.Bhattacharyya, J.Novosad, J.Phillips, A.M.Z.Slawin, D.J.Williams, J.D.Woollins, Dalton, 1995, 1607) we can see that the nickel is tetrahedral. Also in the tetraisopropyl ligands complex (D.Cupertino, R.Keyte, A.M.Z.Slawin, D.J.Williams, J.D.Woollins, Inorganic Chemistry, 1996, 35, 2695) it is tetrahedral as can be seen in the following two pictures.

But the THF solvate of the dppaS2 complex of nickel is square plannar. (E.Simon-Manso, M.Valderrama, D.Boys, Inorganic Chemistry, 2001, 40, 3647.)


text:
Is Gibbs free energy change applicable to forward and reverse reactions at equilibrium?
Consider a reaction
$$\ce{A + B <=> C + D}. \tag{R1}$$
Now $\Delta G$ for the forward reaction is
$$\Delta G_\mathrm{fwd} = \Delta H - T\Delta S. \tag{1}$$
For the reverse reaction $\Delta H$ and $\Delta S$ are clearly the same in magnitude but opposite in sign. Therefore,
$$\Delta G_\text{rev} = -\Delta G_\text{fwd}, \tag{2}$$
which implies that if the forward reaction is feasible, then the reverse isn’t and vice versa. But both the reactions occur simultaneously in an equilibrium reaction. Can anyone please help me sort this discrepancy?

That was answered on this site alone so many times that I wouldn't dare finding them all. And "Related" column already makes a good start with that.

The Gibbs energy of reaction $\Delta_\mathrm{r} G$ determines in which direction equilibrium lies, i.e. in which direction there has to be a net reaction (with a change in concentrations) to reach equilibrium.
When equilibrium has been reached already, there is no net reaction (i.e. concentrations are constant). Nevertheless, at the molecular level, reactions in both directions are observed. They just are happening at the same rate, so there is no net change.
In general, thermodynamics makes statements about the bulk properties, not about individual particles. So if you look really closely, there are fluctuations in concentration, but they are negligible because an change of a single molecule in, say, a mole of molecules is a relatively small change (and not detectable by measuring bulk properties).

But both the reactions occur simultaneously in an equilibrium reaction. Can anyone please help me sort this discrepancy?

In a nutshell, reactions don't stop just because the system has reached equilibrium.

I feel the OP's question was if the forward reaction of a reversible reaction is spontaneous $\Delta G$<0 then according to him the reverse should be non-spontaneous $\Delta G$>0. So a reversible reaction shouldn't go in both directions at all. I think you can edit your answer accordingly as he seemed to know about how to treat equilibrium and that its dynamic. Sorry if I misunderstood anything.

@SirArthur7 Yeah, maybe I should add that the Gibbs energy of reaction is concentration-dependent.

I feel Koushal has not seen the difference between $\Delta G$ and $\Delta G°$, because $G$ of all reactants and products change during the reaction. $G$ and $G°$ are very different concepts.
$G°$ is the Gibbs energy of $1$ mole of any reactant and of any product in the pure state at 25°C and 1 atm. $G°$ is a constant of the substance, independent of the equation chosen, and independent of the concentrations.  $\Delta G°$ is obtained by calculating the sum of the Gibbs energies of the products in the pure state, and substracting the sum of the Gibbs energies of the reactants in the pure state, before the initial mixture is done.
During the reaction, the $G$ values are changing all the time, because they are proportional to the concentration of the reactant (or of the products) . The $G°$ values are never changing during the course of a reaction.
As soon as you mix the initial reactants, the Gibbs energy of each reactant decreases from $G°$ to a lower value $G$ < $G°$ even if no chemical reaction occurs. And the Gibbs energy of each reactant continues to decrease due to the chemical reaction. Simultaneously the Gibbs energy of the products are moving in the opposite direction. But they are is very low at the beginning of the chemical reaction. The reaction is finished when the sum of the Gibbs energies of the reactants is equal to the sum of the Gibbs energies of the products.

When we haven't yet reach equilibrium as you said how we are able to calculate Gibbs energies for both reactants and products? Isn't this variable and other thermodynamic potentials valid only at equilibrium?

First you considered the reaction reversible:
$$\ce{A + B <=> C + D}\tag{R1}$$
Then you considered each reaction irreversible to determine its $\Delta_\mathrm{r}G:$
\begin{align}
\ce{A + B &-> C + D} \tag{R2} \\
\ce{C + D &-> A + B} \tag{R3}
\end{align}
Its not allowed at all! This way the $\Delta H$ and $\Delta S$ of all the reversible reactions would come out zero which is totally vague. In a reversible, a forward reaction may occur with a greater/lesser spontaneity than the backward reaction giving us net $\Delta H$ and $\Delta G.$ So you cannot say that both reactions occur equally.
If $\Delta G$ is more negative for a reversible reactions at a specific concentration, forward reaction proceeds more than the backward reaction and likewise we can say that reaction is less spontaneous in the backward direction. In a reversible reaction $\Delta G$ can be determined as overall net effect taking place and not for separate reactions because we cannot consider both to be equally taking place.

Being a professor, the most difficult thing to instill in the mind of students is that thermodynamics don't consider microscopic changes.
Koushal, you did the same thing. In calculating the change in free energy, we should only see what is visible. At a particular instant when equilibrium is not reached, both  forward and backward reactions proceed with different rates and there is a net reaction in any particular direction.
Suppose the rate of forward reaction is assumed to be greater than the rate of the reverse one, so microscopically products are converting into reactants and reactants into products but macroscopically where it actually matters for thermodynamics reaction proceeds in the forward direction and hence change in free energy for the whole reversible reaction will be calculated to have a negative value only and only at this instant because rate is not constant since equilibrium is not reached.
What you did is that you split the reversible reaction into two independent irreversible reactions. Although that's true at microscopic level but not in thermodynamics.


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Effective nuclear charge and electron repulsion
To the best of my knowledge, the effective nuclear charge $Z_\mathrm{eff}$ is usually defined such that the potential energy of an electron in an atom, $\langle V \rangle$, can be expressed as
$$\langle V \rangle =
  -Z_\mathrm{eff}\langle\psi|\frac{1}{|r - r_a|}|\psi\rangle$$
The Hartree-Fock expression for $\langle V \rangle$, meanwhile, is
$$\langle V \rangle =
  - Z\langle\psi | \frac{1}{|r - r_a|} | \psi\rangle
  + \sum_i \langle\psi |\hat{J_i} |\psi\rangle
  - \sum_{i \in \sigma} \langle\psi | \hat{K}_i | \psi\rangle$$
Is there any relation between these expressions? It was taught in college Chemistry that $Z_\mathrm{eff}$ accounts for electron shielding, which is simply electron repulsion effects. Is the effective nuclear charge just a semiempirical value fitted to approximate the total potential energy of an electron in an orbital, or is there another meaning behind the number?
Based on what I know, Clementi et al’s values for $Z_\mathrm{eff}$ are not evaluated based on this principle. Could I inquire why this is not the case?

Hi @ANZGC, would you consider supporting the creation of a Stack Exchange exclusively for chemical/material modeling and quantum chemistry?  It would be appreciated if you could support the proposal here: Materials Modeling Stack Exchange

$Z_\mathrm{eff}$ is an empirically derived value, commonly estimated by Slater's Rules. In the Slater system, a shielding parameter $S$ is calculated for a given electron by adding up the shielding constants for all other electrons on the atom with $n$ equal or less than the electron of interest.
If the electron of interest is in an $s$ or $p$ orbital, the shielding constants for the other electrons are given as:
1.00 for any electron in a shell with $n$ less than or equal to 2 less than the electron of interest. (Example: For a 3p electron, the shielding constant of a 1s electron is 1, because $3-1\geq 2$)
0.85 for any electron in a shell with $n$ that is 1 less than the electron of interest.
0.35 for any electron in the same shell as the electron of interest (except 1s, which fits better to a value of 0.30)
If the electron of interest is in a d or f orbital, the shielding constants for the other electrons are given as:
1.00 for any electron in a lower shell
0.35 for any electron in the same shell as the electron of interest
To calculated the parameter $S$ for a given electron from these constants, one simply adds the constants for all other electrons on the atom. Then the effective charge is calculated as $$Z_\mathrm{eff} = Z_\mathrm{actual} - S.$$

Is the effective nuclear charge just a semiempirical value fitted to approximate the total potential energy of an electron in an orbital, or is there another meaning behind the number?

No, that's pretty much it. The Wikipedia entry has a partial periodic table with some $Z_\mathrm{eff}$ values. If you go further down to larger atoms, it becomes obvious that $Z_\mathrm{eff}$ decreases quite a lot towards the outer shells, so you could argue that the concept has its merits in capturing a quite significant effect.

Am I correct in saying that the “effective nuclear charge” is just a semiempirical value used to approximate electron repulsion effects in an atom?

Yes, I would say that's a fair statement.


text:
Why organic reaction mechanisms so popular
Why are organic reaction mechanisms so much more popular than inorganic mechanisms? When I come to think of it, what is a reaction mechanism? How can we use an arrow? Is it a rough approximation of the real picture i.e. the true quantum mechanical picture?

No, inorganic reaction mechanism are also popular: https://chemistry.stackexchange.com/questions/10959/inorganic-chemistry-reaction-mechanisms/15882?r=SearchResults#15882

https://chemistry.stackexchange.com/questions/125767/is-organic-chemistry-mechanisms-limited-to-organic-chemistry?r=SearchResults

https://chemistry.stackexchange.com/questions/65044/how-do-organic-chemistry-mechanisms-become-accepted?r=SearchResults

There is a simple answer: there are a lot more organic reactions because there are a lot more organic compounds than inorganic ones. This is a simple result of the enormous variety of molecules than can be built from carbon skeletons. Nothing else in the periodic table comes close.

If you have a look at the main types of reactions in inorganic chemistry (here I am only speaking of the common ones):

Redox reactions- These electron exchange reactions are often so fast, and so spontaneous, with K values of higher powers of 10. It hence becomes very difficult to have a 'mechanism' for them, as an important part of a mechanism is knowing what happened in the time the reaction was proceeding (the study of T.S., reaction intermediates, etc.). Thus such fast reactions we say occur in a single step and do not have a mechanism. The truth is, that there are just so many reactions of this type, that this becomes a common misconception.
Ligand substitution reactions- These do have proper mechanisms in place and various concepts to decide what happens first, who leaves better, whos substitutes better, like trans effect to name one such concept.
Organometallic reactions- Well as the name suggests, they, of course, do have mechanisms, and I won't go in great detail as to what those mechanisms are, you can see it here

Types of organometallic reactions:

Ligand dissociation / association
  Reductive elimination / oxidative addition
  Intramolecular insertion (e.g. CO migratory insertion) / elimination (e.g. β–hydride elimination)
  Metallacycle processes (e.g. σ-bond metathesis)
  Nucleophilic / electrophilic addition and abstraction are also possible, similar to organic chemistry.

Combination, displacement, double displacement- These are often so simple, that ion-exchange, etc. are the only things that happen in them. So they also most of the times do not have a mechanism.

Hence, the common misconception is that these reactions never have a mechanism, but as you can see, that is not entirely true, but in most curriculums of inorganic chemistry, you will not learn 2. and 3. at a basic level, hence it is fair to be having this misconception.

I believe a similar question had been asked before. Most inorganic reactions take place in aqueous media in their ionic form, and the substrate is simply not complex enough to warrant a reaction mechanism. As far as I know, mechanisms do exist for Coordination compound reactions.
You might want to check out this book which explains inorganic mechanisms using the arrow-pushing mechanism seen in organic chemistry.


text:
Do these lead plates have a protective coating? Is it possible to test?
I've just begun teaching chemistry and physics at a local school. We're just getting started on radioactivity and i was looking through the schools supplies when i discovered these soft heavy grey plates used for measuring decay through walls. I've measured the plates' density and IT matches that of lead.
The thing is that i know new sets of these have a coating that prevents lead exposure, i am however, unsure wether these plates have that coating. Sone have a greenish coating I recognize from my own time in school, but the others I am unsure about

L: Coated plate. R: plate in question

The right one looks like lead without any coating. If you want to be even more sure, check that it is soft enough to be scratched with a wooden toothpick. Check that it reveals a shiny metal surface which turns into the same dull grey in a matter of days.

Thanks! I've made a scratch test. Will check back in a couple of days

Agree, they certainly look like plain old lead. Never heard of any coating on lead other than paint.

Where I work we tend to paint our lead bricks

Spray it with a clear varnish. (I´d simply wash my hands after using them, but that might be regarded as frivolous in our paranoid century. ;))

Yes thats what I've been doing, but student safety regulation says that they can't work with pure lead. I've painted them all, so thats hopefully enough


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What do you mean by "periodic" in the periodic table?
Modern periodic law states:

“The physical and chemical properties of the elements are periodic
  functions of their atomic numbers”.

But I don't think this is so! (Forgive me for my stupidity. But please try to understand what I am saying and if you find any mistakes then please let me know).
If I consider the property of Li then I find that after 7 element there comes an element Na which have similar properties like Li. Now according to periodic law I must get another element which has similar properties to that of Li after 14 elements. And indeed I found one named K . Now again the same thing must happen (no?) ie, there must be an element which have similar properties to Li and is 21 elements away(Mn). But I don't find such element! Why?
The properties of Element is periodic so why then this "periodicity" doesn't hold in case of Li?

There is a pattern, but the period is not a constant.

@orthocresol but the word PERIODIC means : occurring or recurring at regular intervals [ https://www.merriam-webster.com/dictionary/periodic ]. Why then scientist are using such word which can led confusion in a law?

Well, https://www.lexico.com/en/definition/periodic suggests "appearing or occurring at intervals" as their definition, so not as specific. Yes, English can be a bit weird, but the fact remains that technical definitions and uses may diverge from standard definitions and usages. Language is not static.

I think most people understand that when someone says they go to the doctor periodically, that does not imply exactly equally-spaced intervals.

@JonCuster Science starts from words used in daily life and then make changes to it for it's own use. Indeed. But in this case a better word is "intermittent": .... stopping and starting repeatedly or with periods in between. [ https://www.google.com/amp/s/dictionary.cambridge.org/amp/english/intermittent ]

@Idk, As you can see in the answer, periodic concept is derived from mathematics. Hope you understood the answer. Periodic functions must be taught in grade 10,11 or 12 in the context of trigonometry. Intermittent has a very different context. Intermittent means "ceases for a time; coming at intervals; operating by fits and starts." The properties of elements don't have this fits and starts or ceasing of properties.

contd...One can have intermittent fever, which comes and goes but one cannot have periodic fever which will imply that fever comes at exactly the same time and goes away at exactly the same time, every day. Imagine a situation of fever starting at 8:00 am, ending at 8:00 pm, then starting at 8:00 am and then ending at 8:00 pm.

Science does not start with words from daily life. Science uses words which may be incorporated into daily life. Take "quantum mechanics." In daily life, no one knows what "quantum" is, and "mechanics" fix cars.

@Zhe This is even written in Einstein's book The evolution of Physics "But science must create its own language, its own con cepts, for its own use. Scientific concepts often begin with those used in ordinary language for the affairs, of everyday life, but they develop quite differently.
They are transformed and lose the ambiguity associated with them in ordinary language, gaining in rigorousness so that they may be applied to scientific thought. " https://b-ok.cc/book/2626393/ada93b  .

First you need to understand what a periodic function is in mathematics. For example, if you are standing at any point P on a circle, you start walking on the circumference and count the degrees you completed. Once you complete 360 degrees, and you will find yourself on exactly the same point P. Sines, cosines and many other functions are periodic. This term existed before the chemists started using the term periodic law in the early 1800s or perhaps well before that. If $f(x)$ is a function with period $T$, then $f(x+T)= f(x)$ for a periodic function in a defined range of $[a,b]$. The period $T$ may change to $T'$, when x is in a different range say, $[c,d]$. This is  an example of a discontinuous periodic function. As you will see, periodic properties of the elements form a discontinuous periodic pattern in a loose sense.
Now come to the periodic law: "The law that the chemical elements, when listed in order of their atomic numbers (originally, atomic weights), fall into recurring groups, so that elements with similar properties occur at regular intervals." [unabridged OED].
Instead of $x$ for a mathematical function, the independent variable is the atomic numbers Z (Zahl for a number in German) for the chemists. So this periodic law has a very loose connection with mathematical periodic function, because none of the values are actually repeated. Each element is unique, but in a loose way, their properties or even a better word is trends "repeat" themselves after a certain number of elements. See this chart for example. Atomic radii vs. Z, do you see a loose periodic pattern?

Once you have got the idea of periodic function in a loose way for the chemical elements, get ready to the see the physicists style of periodic table. It explains why some properties loosely repeat themselves.
Each element in the column, corresponds to filling of a certain orbital, for example look at the column $s^1$. All elements falling in that column correspond to the filling of $s^1$ electrons. Then move to the next column $s^2$, all elements listed there corresponding to filling of $s^2$. Can you see a pattern? In this sense, one calls it periodic function in a modern way, that certain elements repeatedly fall in $s^1$ column, some elements repeatedly fall in $s^2$ column.


text:
Which is more basic among N-ethylpiperidin and 1-azabicyclooctane?
Shouldn't A be more basic than A because in B the conjugate base formed in B is more unstable as compared to A and because a bridge head nitrogen is formed In B which is comparatively unstable than conjugate base formed in A?

According to me you can see stability , but how how are you comparing the stabilities, there is not much difference in stabilities of conjugate bases!

The basic nature of amines is mostly dependent on their ability to donate their lone pair rather than stability of the conjugate acid formed.
B, due to steric hindrance, does not undergo Amine Inversion. The lone pair is thus stationary and more prone to donation. Hence, B is more basic than A.


text:
Significant figures and accuracy
A sample was weighed using two different balances. The results were $\pu{(a) 4.929 g and (b) 5.0 g}$. How would the mass of the sample be reported?
Method 1
$$ \text{Average of the values} =\frac{4.929 + 5.0}{2}= \frac{9.9}{2}$$ (the result of addition can have only one digit after decimal as in 5.0) $$\pu{= 4.95= 5.0 g}$$ (the result of division can contain only one digit after decimal as in 9.9)
Hence, the mass of the sample is to be reported as $\pu{5.0g}$.
Method 2
Out of the two values, 4.929 is more accurate and will be reported as 4.9 after rounding off.
Hence, the mass of the sample is to be reported as $\pu{4.9 g}$.
Which of the above method is correct?

The first method. You might want to check out This Wiki, which highlights the difference between accuracy and precision.

Mods, this goes to phy SE right?

In method 1 the reported answer should be 5.0 g, not 5 g.

@HahaHahaha I feel such basic concepts are to be known by anyone who aspires to pursue science in their academic career in future. Actually there are few cases where the line between different branches of science seem to fade. I feel this can stay, rest is up to the moderators.

Both method 1 and method 2 are wrong. The concept of useless significant figures should be wiped out from school curricula and replaced with proper statistics. This significant figures and concept of estimation was good for the 18th century when eye-balling and manual measurements had to be made. The concept originated in the 16th century.
In method 1: If you are using an electronic balance and weighed your object, the display reads 4.929g, this is what you will record. No data should be rounded off until the end of the calculations. Data should be recorded as it is without manipulating it.
The "true" error of balance will never ever be estimated by the so-called significant figure rules. We will check the precision by repeatedly weighing the object and we will check the accuracy by using a known weight. The "constant" error in the balance if it existed, will be corrected either by subtraction or addition.
In method 2, I am not aware of any scientific balance which will measure only one decimal place, even if it does, an accurately calibrated balance will read the weight as 4.9 g not 5.0 g. If it is reading it as 5.0 g, there is a huge error in either balance 1 or balance 2.
You see we never needed to invoke significant figures here.

Interesting perspective. Why are significant figures still taught in school curricula in that case? And when do people realize that this method is not accurate? (I'm guessing when they do serious lab work, please confirm)

I think this is just traditional and perhaps to teach common sense that we should not report concentration to 9 decimal places. However, the problem arises when some teachers start teaching significant figures as scientific facts or a part of some holy book and "this is the only way". Nobody has taught them to think critically ---they simply pass it on to their poor students who just cannot make sense out of it.

The experimental way, is the way, most scientists will do it.


text:
Which hydrate isomer of [Cr(H2O)6]Cl3 have lowest electrical conductivity
I found this question in my book and the answer that was given is $\ce{[Cr(H_2O)_4 Cl_2]Cl.2H_2O}$ but I think it should be
$\ce{[Cr(H_2O)_5 Cl]Cl_2.H_2O}$ because in this case only one $\ce{H_2O}$ molecule is released.
Which one is correct and why?

So what if H2O is released? H2O is not what conducts current anyway.

If my edit is correct then I think OP is wrong, the increase in electrical conductivity here depends on the number of $\ce{Cl^-}$ ions that go into the solution when the hydrated complex is ionized, and not the $\ce{H_2O}$ because its already there in excess in the solution.

If possible do include the sources you cite as reference in future questions and answers. We would appreciate if you could also improve upon framing your question so that its easier for us to understand and help you better. Thank you :)

The substance having the lowest electricity conductivity is the substance producing the lowest number of ions in solution. The substance $[Cr(H_2O)Cl_2]Cl$ produces $1$ cation charged $1+$ and $1$ anion charged $1 -$. This is the lowest possible number of ions with complexes made with Chromium(III, water and Chlorine.
With $[Cr(H_2O)_6]Cl_3$, the dissolution in water produces 1 cation (with the charge +3) and 3 anions with the charge -1. This makes many more ions in solution.

Why isn't  $[CrCl_3(H_2O)_3]⋅3H_2O$ the right answer? No ion would have been produced then and hence no conductivity...

Alas ! The substance $\ce{[CrCl_3(H_2O)3]·3H_2O}$ does not exist.


text:
Why is Cadmium so carcinogenic compared to other heavy metals?
This free article as well as this article do a good job highlighting the carcinogenic effects of Cadmium and the mechanisms by which they occur. However, my question is why is Cadmium so carcinogenic compared to other heavy metals? What chemical property of Cadmium makes it more prone to these carcinogenic actions than, say, Mercury or Lead or some other heavy metal?
To give a more concrete example, the recommended safe levels of Lead in water is 50 ppb, whereas for Cadmium, it is 5 ppb. Why is there such a stark difference in their health effects?

Which standards for lead levels are you citing? The FDA cites 5 ppb in drinking water (Table 4)

Related: https://chemistry.stackexchange.com/questions/24229/are-there-any-good-examples-of-commonly-ingested-molecules-that-contain-particul

Because of its combination of charge and radius mostly, or generally its individual chemical properties. It can squeeze into a number of places where it shouldn´t.


text:
Is there an anisotropy factor (g factor) for TDDFT Abs and CD calculations?
Experimentally, anisotropy factor is calculated by dividing the CD spectra by the absorbance spectra multiplying by a factor of 32980 (in order to get a nondimensional quantity)
Theoretical calculations can be performed via TDDFT to obtain rotatory and oscillatory strengths, obtaining excitations for each case and adjusting, for each excitation, a curve, obtaining then the "theoretical absorbance/CD spectra".
How can then we define a theoretical anisotropy factor, does it have sense?
Should another calculation be performed?
Would you divide both adjusted curves or divide the corresponding excitations, adjusting a new set of curves to what you get?
Any clues?

Great question, unfortunately TDDFT is not my expertise, but it looks like you might be interested in supporting the launch of a stack exchange just for spectroscopy and materials modeling (DFT, MD, QM/MM). Could you please help us out?  https://area51.stackexchange.com/proposals/122958/materials-modeling?referrer=NDY4N2I5MzBmODViNGQwYWM2MTBmMzVkMzc0OWU0ZjQ5NDU2YTVhYzQyMmEyY2IwYjdjYzlmZDMwMWQyZmJhNGBP4m1cf3-vYfnwUjsCFf90E8N7syRuU35ETDRpKo-U0

One approach would be to divide the corresponding excitations as you mention. I would avoid adjusting curves to the absorbance and CD spectra first then dividing, as you would be introducing too many tunable parameters.
This being said, the more common approach in the literature is to calculate the gabs directly as the ratio of rotational strength to dipole strength at each transition, then fit a curve such as a sum of gaussians etc.
Rotatory strength is given by; $$ R_{0n} = Im\{\mathbf{\mu_{0n}.m_{0n}}\}$$
Here $ \mathbf{\mu_{0n}}$ and $m_{0n}$ are the electric and magnetic transition dipole moments. This is often called the Rosenfeld equation, and is valid for isotropic samples, which most calculations of CD are concerned with.
In most cases gabs is then calculated approximately as $g_{abs}= \frac{4R}{D}$, where D is the dipole strength ($|\mu_{0n}|^{2} + |m_{0n}|^{2}$).
For more theoretical details see Wakabayashi et al., J. Phys. Chem. A 2014, 118, 27, 5046–5057 or Molecular Light Scattering and Optical Activity by Laurence Barron.
Some quantum chemical calculation programs give you the rotatory strength tensor, for Gaussian 09 it is given amongst the output under  Excited states from <AA,BB:AA,BB> singles matrix: in the log file. Then there are a couple of ways of calculating R from this, which are given as R(velocity) and R(length), which are explained in Pederson and Hansen, Chem. Phys. Lett., 246 (1995) 1-8. The velocity method is, as I understand it, more suited to this problem.


text:
Oxidation of toluene to benzoic acid
On Wikipedia I found the reaction of the oxidation of toluene to benzoic acid:
$$\ce{5 C_6H_5CH_3 +6 KMnO_4 +9 H_2SO_4} \to \ce{5 C_6H_5COOH +14 H_2O +3 K_2SO_4 +6 MnSO_4}$$
which involves $35\, \ce{C},\, 58\, \ce{H}, \, 60\, \ce{O},\, 6\, \ce{K}, \, 6\, \ce{Mn}$.
But I managed to find a simpler version of the reaction:
$$\ce{C_6H_5CH_3 +6 KMnO_4 +5 H_2SO_4} \to \ce{C_6H_5COOH +6 H_2O +3 K_2SO_4 +6 MnSO_4}$$
which involves $7\, \ce{C},\, 18\, \ce{H}, \, 44\, \ce{O},\, 6\, \ce{K}, \, 6\, \ce{Mn}$.
Is one equation more correct than the other? Why/why not? Which reaction does actually occur when oxidizing toluene?

The second reaction isn't balanced. Count sulfur atoms. The first one from Wikipedia seems to be OK.

@andselisk Interestingly enough, OP counted all atoms of all elements except the number of sulfur atoms in either reaction;)

Alright, I absolutely forgot to count the sulfur atoms somehow...

Permanganate ion i acidic medium can have two reduction half reactions, $(1)$ and $(2)$:
$$\ce{MnO4- + 8H+ + 5e- -> Mn^2+ + 4H2O}\:\:\: E^\circ = \pu{1.50 V} \tag{1}$$
$$\ce{MnO4- + 4H+ + 3e- -> MnO2 + 2H2O}\:\:\: E^\circ = \pu{1.68 V} \tag{2}$$
Regardless of what oxidizing reagent has been used, the oxidation of toluene can be written as:
$$\ce{C6H5-CH3 + 2H2O -> C6H5-CO2H + 6H+ + 6e- } \:\:\: E^\circ = \pu{?? V} \tag{3}$$
I have no idea what would be the value of $E^\circ_\text{tol} $, but since the oxidation is not spontaneous, it should have been less than zero or close to zero (Note: More than a century years ago, Giacomo Ciamician, the pioneer of the modern day Green Chemistry, in collaboration with Paul Silber, has allowed toluene, o-xylene, p-xylene, and m-xylene to be influenced by oxygen and Sun light for about a year, and discovered the chemicals have been converted to the corresponding mono carboxylic acid as the main product; Ref.1 and 2). Yet toluene get oxidized by condition $(1)$, and hence, should get oxidized by condition $(2)$ as well, since $E^\circ_\text{(2)} \gt E^\circ_\text{(1)}$.
If you worked half-equations $(1)$ and $(3)$ you would get the redox reaction (disregarding $E^\circ$ values):
$$\ce{6MnO4- + 18H+ + 5C6H5-CH3 -> 5 C6H5-CO2H + 6Mn^2+ + 14H2O} \tag{4}$$
When you included counter ions $\ce{K+}$ and $\ce{SO4^2-}$, it gives your equation from Wikipedia:
$$\ce{6KMnO4 + 9H2SO4 + 5C6H5-CH3 -> 5 C6H5-CO2H + 6MnSO4 + 3 K2SO4 + 14H2O} \tag{5}$$
Thus, your first reaction is the correct one and, as everybody pointed out, the second reaction is incorrect solely because it is not correctly balanced.
Similarly, if you worked half-equations $(2)$ and $(3)$ you would get the redox reaction:
$$\ce{2MnO4- + 2H+ + C6H5-CH3 -> C6H5-CO2H + 2MnO2 + 2H2O} \tag{6}$$
But remember, under this conditions, $\ce{2MnO4- -> MnO2}$. Thus, it is not the same type. The major reason I gave this equation because you want to know what happens in this redox reaction. It is believe that "generally inorganic substance reduces acidic permanganate to $\ce{Mn^2+}$ state whereas organic substance
(containing oxidizable hydrogen) stop the reduction at the manganese dioxide $(\ce{MnO2})$ state (Ref.3).
References:

G. Ciamician, P. Silber, "Chemische Lichtwirkungen. XXII. Autooxydationen. I," Ber. dtsch. Chem. Ges. 1912, 45(1), 38-43 (https://doi.org/10.1002/cber.19120450107).
For review on G. Ciamician's pioneering work: Angelo Albini, Maurizio Fagnoni, "Green chemistry and photochemistry were born at the same time," Green Chem. 2004, 6(1), 1-6 (DOI: 10.1039/B309592D).
Mukul Chauhan, "Permanganate Oxidation mechanisms of Alkylarenes," IOSR Journal of Applied Chemistry 2014, 7(6), 16-27 (DOI: 10.9790/5736-07611627).

The long article by Mahindaraine is very interesting, but it contains the same mistake repeated through all its lines. It produces a neutral substance C5H5-CO2 that does not exist. It should be either either benzoic acid C6H5COOH or the benzoate ion C6H5COO-. And the equation (3) is not balanced. There is 7 H at the left-hand-side and 6 H at th right hand side. The equation (4) is also wrong, with 33 H at left-hand side and 28 at right.

Can't a person make a mistake? (e.g., you have a mistake in your comment as well) Should have corrected rather than complaining.

To the down voter, give me a good reason why you have down voted my answer. Is there anything wrong about this answer?

@MathewMahindaratne We can't control what people like right? That's up to their own discretion, maybe it was before your edit. But don't worry I upvoted your answer, it was really an informative one, but I feel the question as it's stated is not that intriguing.


text:
Applied Potential and Electrochemical Cell Potential

Electrolysis of molten lithium hydroxide at $\pu{750 K}$ to form lithium metal
$$\ce{4 LiOH -> 4 Li + 2 H2O + O2}.$$
Electrolysis will only proceed at an appreciable rate when the applied potential exceeds the electrochemical cell potential by $\pu{0.60 V}.$
(c) Calculate the minimum potential that should be applied.
$$ΔG^\circ = -nFE^\circ_\mathrm{cell}$$
$$E^\circ_\mathrm{cell} = \frac{ΔG^\circ}{nF} = \pu{-2.60 V}$$

Why should the difference of $\pu{0.60 V}$ be added onto the absolute value of $E^\circ_\mathrm{cell}~(+2.60)$ and not $-2.60$?
I understand  $E_\mathrm{cell}$ is the measure of the potential difference between two half-cells in an electrochemical cell, but what is applied potential in this case?  I thought electrochemical cells produce an electric potential difference (in this case $\pu{-2.60 V}),$ so why would they need an applied (positive) potential difference?

You are mixing two areas, electrode kinetics and electrode thermodynamics. If we have 3 apples and 2 oranges, this is not equal to 5 oranges or 5 apples but there are 5 fruits in total. Electrode thermodynamics tells you a "yes" or "no" answer in the sense that it is saying the reaction is spontaneous or not. Electrode kinetics tells you how fast the reaction will be.
When you wish to electrolyse something, it means that you want do a non-spontaneous reaction i.e., if you leave lithium in the presence of air and moisture, it will spontaneously form LiOH
$$\ce{4 LiOH <- 4 Li + 2 H2O + O2}.--(1)$$
Now, by forcing electric current through molten LiOH, you want to reverse this reaction and want the following to happen
$$\ce{4 LiOH -> 4 Li + 2 H2O + O2}.--(2)$$
The negative Ecell of -2.60 V only tells you that reaction (2) is not spontaneous at all. It has no other meaning whatsoever.  That is, if you wait forever, you will not see any Li spontaneously forming out of LiOH.
Now as you read in the question, they are saying that if you apply exactly +2.60 V potential difference between two electrodes the rate of reaction is very slow. Remember negative sign was just for saying that the reaction is non-spontaneous. However if you apply +2.60+0.60 = 3.2 V of potential difference between the electrodes, the rate of formation of Li from molten LiOH will be quite fast. They don't say how fast but you will not have to wait for a long time.
So +2.60 V is the apple = electrode potential
0.6 V is the orange, it is called the overpotential in formal language.
You can only add them (not subtract) and get the total voltage required to start the electrolysis at an appreciable rate.

But what if I have a galvanic cell like fuel cell? What's the difference between cell potential and applied potential (or bias) in that setting?


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Looking for color changing reaction to track snow layer depth
I'm trying to find a chemical to mark the position of a snow layer with for detection in future seasons.
The substance has to be:
1. Colorless
2. Not water soluble/solid
3. Stable for multiple years
4. Easily detectable in the field
I'm aware of pH tracers, but I am wondering if there are other reactions with one colorless solid/insoluble component that would produce a dramatic color change.

Have you searched Google Scholar https://scholar.google.com/? When you have a scientific idea in mind, you search the relevant literature and see what others have done in the field.

@M.Farooq Yes, unfortunately this research area (wet firn densification) is understudied, largely because of this issue. The method in the literature is either to use naturally deposited sediment layers (usually volcanic ash, problematic for multiple reasons), or water isotopes. It is also unfortunately impractical to remove our cores from the field. Coming up with a material that could more easily facilitate layer tracking would be a huge help both for this and a variety of other research areas in snow science/glaciology.