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Relationship of acids and bases with pH. (Proportional Relationship)
Can someone please clarify the relationship between acidic-basic strength and pH? I have come across conflicting information and am unsure about the correct understanding.
Some people including my chemistry teacher say that
$\text{Acidic Strength} ∝ \frac{1}{pH}
\\
\text{Basic Strength} ∝ pH$

According to my research, it's called proportional relationship.
Well, I'm asking whether this information is accurate or not. If not? show me the correct one.

Express the relations quantitatively and decide yourself. E.g. the simplified equation for weak acid is pH=0.5(pKa - log c), for weak base pH= 14 - 0.5(pKb - log c).

There is either a wrong terminology at play, or something lost in translation/definition.  pH is not how acidic/basic strength is quantified at all.

In pH p stands for "potenz" Which means taking negative logarithm of active mass of hydronium ion  $$ pH = -log(a) $$ where a is the active mass of hydronium ions.
As we know that the acidic strength increases with the increase of active mass of hydronium ions the negative logarithm of active mass i.e., pH should decrease with acidic strength.
Therefore,$$Acidic Strength ∝ 1/pH  And  Basic Strength ∝ pH$$

Chem+Math Expression formatting reference:
MathJax Basics / Chem+Math expressions/formulas/equations / Upright vs italic / Math SE Mathjax tutorial
// MathJax is preferred not to be used in CH SE Q titles.

Nothing written here is coherent or correct from the point of logic, formatting or facts, and does not address the main point (flaw) in OP's question at all.

The p for pH as $Potenz$ is nothing but a common myth.


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Why is micelle formation endothermic?
Most websites mention that it is endothermic and some mention that it is endothermic at low temp. and exothermic at high temp. Nor do I know the correct answer neither I can guess the reason for any of them. An explanation would be a great help.

Some references and quotes would help to attract good answers. // Review the guides Asking and How to ask to prevent misunderstanding,  need of clarification, objections, downvoting or closure.

Also, micelle formation starting from what? From monomers in solution? Write the full equation to which you are referring.

Do you mean enthalpy? The enthalpy of micelle formation (say SDS in water)  is normally small, the main effect is a huge entropy increase ($\sim 140$ J/K/mol) making $\Delta G$ negative. The enthalpy change is a mix of negative change moving the tails from water and a positive term of repulsion between charge on head groups.

source: Stack Exchange

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

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How does the concentration of a reactant change? Is it based on individual order or overall order?
Reaction $\ce{A + B ->[$k$] C + D}$ follows the rate law $$r = k [\ce{A}]^{1/2}[\ce{B}]^{1/2}$$ where $k$ is given. Starting with $1 \,\pu{M}$ of $\ce{A}$ and $\ce{B}$ each, what is the time taken for concentration of $\ce A$ to become $0.1 \,\pu{M}$? Do we use the integrated law expression of first order (overall order) or half order (order w.r.t. $\ce{A}$)?

In this special case, r=k.[A]

Why? What’s special in this particular rxn?

It is not about chemistry, but simple algebra. $x=y \implies x^{1/2}y^{1/2} = x$. // As $[\ce{A}] = [\ce{B}]$

*At High School Level
The order of a Reaction is the  Sum of powers of concentration terms in The Rate Law Expression. Therefore, acc. to the rate law expression provided in the question $$r = k[A]^{1/2}[B]^{1/2}$$ The order of Reaction is 1.  Let the Concentration of reactants at any point of time during reaction $$\ce{ A +B→C +D}$$ be $xM$  Now the rate law becomes $$ r = k x $$   Now you can integrate the rate law obtained as usual.

I think the correct transformation follows a similar patter as the second order reaction, i.e. $\frac{\mathrm{d} x}{\mathrm{d} t} = -k\sqrt{(A_0-x)(B_0 -x)}$, where $A_0$ and $B_0$ stands for the initial concentrations.

But it is given that bot reactants start witl $1M$ comcentration. And you would get the same results from you equation any my equation if the limits in integration are put properly.

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


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Is a Sodium Chlorite Product Safe for Car Odor-Removal?
I would like to ask the science community's opinion about the safety of sodium chlorite for use as an odor-eliminator in an automobile's interior.
While the ingredient in the product is sodium chlorite the technology listed is ClO2 (chlorine dioxide).
The product is Auto Shocker.  https://www.autoshocker.com/
It involves a tea bag like pouch of sodium chlorite reacting with water to form a gas that eliminates odors in a car:  https://www.youtube.com/watch?v=3EaZaPbth6k&t=4s
Does the gas turn into harmless saline in time?
Should there be any concern with reactions with other chemicals or materials present in a car?
Here is the msds for the product: https://www.autoshocker.com/wp-content/uploads/2020/01/msds.pdf
Thanks!

I wouldn't want it lying around. It's practically a bleach and could be dangerous if used improperly.

See fraud indictment for ClO2: https://arstechnica.com/health/2023/07/church-of-bleach-family-guilty-on-all-counts-plans-to-appeal/

Though personal medical questions are off-topic, here's a quote from a $\ce{ClO2}$ Safety Data Sheet, as opposed to that of the product, which you cite.

Severe respiratory irritant. May cause bronchospasm and pulmonary
edema, which may be delayed in onset. May also cause severe headaches.
All symptoms may be delayed and long lasting. Long term exposure may
cause chronic bronchitis.  May be harmful if absorbed through skin.
Exposure Limits: ACGIH 1992-93: TWA 0.1 ppm, STEL 0.3 ppm (0.9 mg/m3).

That said, how much $\ce{ClO2}$ is released by that product is another question. If it's strong enough to oxidize odoriferous substances, is it safe, and if it's safe, could it possibly do the job?


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9,10-dihydro-9,10-ethanoanthracene derivatives. Stereoisomers or different compounds? What are their chemical names?
What is the relationship between these compounds? Are they stereoisomers or some of them are different compounds? Additionally, what would be their names?
(For position 9, with the methyl substituent)
Green and Cyan -> S configuration
Orange and Purple -> R configuration
Trying to name them by having the antracene ring system as a core all of them would have a name starting as "2-Chloro-" and then R or S, but in this way, the green-cyan and purple-orange compounds would have the same name. Is there another way to name these?

Green and orange are enantiomers, as are purple and cyan.
The remaining relationships are all diastereomeric. I think your problem is that you arent considering the carbons at the bridgehead positions as stereocentres too (which they are).


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Are there materials or molecules where fast internal conversion is desirable?
For light-emitting devices fast internal conversion is typically considered as a downside as it competes with fluorescence and phosphorescence. A low internal conversion rate is thus typically desired for molecules in applications in light emitting devices, like OLED's, and molecules with high photoluminescence quantum yield are searched for.
I am wondering if there are materials or applications, where the opposite is the case. I.e. where one looks for molecules with fast internal conversion. Are there properties/materials that benefit from fast internal conversion of a molecule?

Sunscreen compounds need to have fast internal conversion if i understand correctly.

Intersystem crossing is more important if you want to form triplets, fast internal conversion can prevent forming triplets or fluorescence by competition. Fast internal conversion means that the molecule does nothing but heat up the surroundings.

The old joke is that only 10% of all substances fluoresce and they are the wrong 10%! In photoacoustic spectroscopy and related techniques, you can make use of the non-radiative pathways. But for purposes such as sensitive analytical detection of substances, fluorescence is usually highly desirable. Phosphorescence not so much. If you really need minimal fluorescence, you can usually quench fluorescence by adding a quencher: oxygen, xenon, nitromethane, various heavy metal ions, etc.

@porphyrin That is the common knowledge. I am wondering if there is anything else that can be achieved by internal conversion. I am not aware of particularly useful applications, but I'm not certain if there are truly none. It could be a simple lack of knowledge on my side.

One can use the OLED as a lumiphore as a continuous light source (e.g. ceiling of an office, wall of a staircase) where slow processes are not this much problematic. This contrasts to applications where short reaction times/high switching rates are desired (e.g. displays of TV sets, cellular phones). The design of HTL, ETL and emitter layer (for red, green, and blue pixels in one cell phone/TV screen, one uses different emitter materials each) of the two differ; e.g. for longevity of the device, TVs often stack two layers of red/green/blue pixels on top of each other.

@Buttonwood, Not so, high fluorescence yield is essential for dye lasers and intense (laser) pumping due to their short excited state lifetimes.

I would add that if you are using Raman spectroscopy a low fluorescence yield is most advantageous.  Photothermal methods are useful way of studying excited states, the heat generated distorts a probe beam etc.  see this paper Beddard & Yorke, Appl. Spectroscopy, 2016, vol 70,p 1292 , https://doi.org/10.1177/0003702816653927

@porphyrin Ok.  Makes sense as to keep the states differently populated than mere room temperature and its resulting Boltzmann distribution would predict.

When you create new tags, it would be much appreciated that you also add usage guidance for them.


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Is it a convention to divide rate of reaction by stoichiometric coefficient?
If we consider a reaction
$$ \ce{A + B -> 3 C} $$
The rate of reaction is given by
$$ R= -\dfrac{\Delta[A]}{\Delta t} = -\dfrac{\Delta[B]}{\Delta t} = \frac13\dfrac{\Delta[C]}{\Delta t} $$
I know we divide it by $3$ for equality, but we can also multiply $3$ in other two concentrations too, right? Then the result will be:
$$ R= -\dfrac{3\Delta[A]}{\Delta t} = -\dfrac{3\Delta[B]}{Δt}=\dfrac{\Delta[C]}{\Delta t}$$
So is it just a convention to divide instead of multiplication?

It is rather definition than convention. Also, search for related question on the site, there were several similar questions in recent history. // Take it as 1 mol of reaction formal operations, each taking 1 molecule A, 1 molecule B, forming 3 molecules C. So 1 mole of formal reactions consumes 1 mol A, 1 mol B, forming 3 mol C.

You may review also search results for site:stackexchange.com reaction rate stoichiometric coefficient

@Poutnik Thanks, for replying. My doubt is now cleared

Also, think of a more "complicated" reaction, such as $\ce{5A + 3B -> 4C + 2D}$. It is not entirely clear and unambiguous how your "method with multiplication" could be used in such cases, while the "method with division" is completely straight-forward :)

As @Poutnik indicates, it is a definition. The value of the reaction velocity does not depend on which reactant or product you select.

@PAEP Yes. Maybe we divide them with respective coefficients to fit any reaction into the general framework of what a reaction is!


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Problematic Menshutkin reaction
I'm quite new to organic chemistry and want to try menshutkin reaction. Like the reaction in an attachment.
I use that recipe:
I do everything in a 60ml beaker. I dissolve 18 mmol of pyridine in 35 ml of acetone. I add 115 mmol of 1,6-dibromohexane and stir for 30 minutes. Then I add 76.6 mmol KOH (in the form of crushed pre-flakes). Stir for the next 24 hours.
Then I put the whole thing into a solution of 50 ml of dichloromethane and 100 ml of water. I mix for 1 hour, then pour into a funnel (I have a small 25ml one, so I do it several times).
I evaporate the whole thing using a 60-80 degree water bath until all the dichloromethane evaporates. Then, after cooling down, I wash it a few times with pentane (it cleans the material from the residual dibromohexane, unfortunately I don't have a vacuum pump).
Unfortunately, I only got 5% yield and the product still look little reddish, what am I doing wrong? Could you help me on what to do better?
I can buy equipment but not so sure what to buy

Dichloromethane? That's a bad idea https://en.wikipedia.org/wiki/Menshutkin_reaction

I mean when the reaction end of course, to transfer product to organic phase, It's not my imagination it's one of the procedure from Reaxys base. What do you think will be better that boils low temperature?

Hmm, at this point it's probably indeed OK. Maybe adding pyridine drop-wise would help?

Last time I add 1,6-dibromohexane drop-wise, but can try with the amine drop-wise. The water-dichloromethane washing and pentane washing is ok? I thought maybe the reaction time is to long and another pirydine is brought to second bromine atom?

Yeah, you need to do what you can to avoid disubstitution - adding pyridine slowly during reaction would be good idea.

What will be the best solvent to make a reaction in? I had also toluene, pentane, DMF. After the reaction how much of DCM use? Can I use like 1g-15ml or need to use more? Will it be better to buy bigger separatory funnel? Could occuring the reaction in 25-30 C degrees be good?

Maybe wait if someone with experience in this reaction comes around.


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What are disadvantages of methane or ammonia compared to dihydrogen for metal ore reduction?
Metallurgy, such as steel-making processes, require reducing agents to get rid of the oxygen in metal ores. Using fossil coal for this is a large source of $\ce{CO2}$ globally. Recently, I have read a lot how green hydrogen is proposed as a replacement for the fossil coal as a reducing agent. I also believe that hydrogen has a few issues such as its adverse effect on some metals (embrittlement) and its difficult storage and transport.
Wouldn't other reducing agents that can be easily produced from renewable energies, such as $\ce{CH4}$ or $\ce{NH3}$, be much more useful replacements for coal in metallurgy? Or do these have some chemical disadvantages that make hydrogen a much better contender?
Exemplary references about direct iron ore reduction using methane:

Clean iron production through direct reduction of mineral iron carbonate with low-grade hydrogen sources; the effect of reduction feed gas composition on product and exit gas composition
Kinetics of Iron Ore Reduction by Methane for Chemical Looping Combustion

Challenges of technology, where the solution is missing for decades, are very seldom solved by a single Q/A

@Poutnik I am not solving any technological challenges but I am asking a simple chemistry question: "What are the disadvantages of CH4 and NH3 compared to H2 as a reducing agent in metal making?" If you feel deterred by the background I put in the question body, either ignore it or delete it, please. I edited the title to emphasize that question.

Aside from chemical disadvantages, using ammonia sounds like an energetically expensive solution, like using ethanol from corn as car fuel.

Aside from which chemical disadvantages? @BuckThorn Yes, using renewable energy surplus to form consumables endothermically is "expensive". The same applies to H2 formation, so this isnt really an arugment against CH4 or NH3 (unless of course the formation of the latter is much less efficient than H2 formation, which would alos be an answer btw). Using renewably generated fuel in cars is a rather nonsensical counter-argument because - indeed - much more efficient ways to propel cars from renewable energy do exist.

Well, it would help to define "chemical disadvantage". Ultimately you might reduce this to a question of cost (capital), which might include environmental costs, or complexity, which might include factors such as safety, time to build a plant, practical limits to the scale of the process etc.

https://cen.acs.org/environment/green-chemistry/Industrial-ammonia-production-emits-CO2/97/i24 // There might be advantages in terms of transportation of ammonia as opposed to H2.

@BuckThorn I am not after things like storage or transportation in this chemistry question. It is really about trying to reduce a piece of metal ore with CH4 or NH3 as opposed to C or H2. Does that work worse, i.e. requires higher temps or is slower, i.e. ultimately less economical?

I wouldn't use CH4 that has a GWP100 of 21-28 (various values found). I'd use NH3 that has a ODP of 0 and a GWP100 of 0 as well.

@ParaH2 The GWP is irrelevant because a) the substance would be cycled anyway. The only net emissions are water and CO2 (and N2). b) the question asks about the merit of these substances for the reaction itself.

Reactants in a solid state behave totally different from reactants in a gaseous state. And that is more than enough to limit the usefulness of hydrogen, or any hydrogen carrier such as methane or ammonia over carbon.
To start off easy; solid Carbon can be conveyored, dropped, siloed and charged with a hopper. It can be supplied with a dumper truck, shipped with an open bay vessel, and hoisted with a crane bucket. It does not constitute a mortal danger doing this. When it is inside the vessel you want it to be, it does not escape.
A solid substance has an activity (gamma) of 1, its behaviour is well understood and leads to an intrinsically stable operation. Gaseous reactants must be monitored and controlled carefully so as to maintain the right activity in the reaction zone. The vessel itself must perhaps be an autoclave for the right conditions. You could very well create Silicon by reduction with hydrogen, but unfortunately the thermodynamic equilibrium at atmospheric pressure is low yield; you would either need a two stage setup with gas transfer or an autoclave. For electrolytic reduction it is much the same. I wrote my thesis on titanium diboride anodes so as to get rid of the anthracite/graphite anodes; you could even gain efficiency by lowering the electrode separation distance substantially, but as se can see 20 years later it was and is still not economic to do so.
Not to mention the considerable added dangers when introducing hydrogen or ammonia to an already exciting situation.
It is usually, but not always, feasible. If it was easy or economic it would have been done already. Since we are talking about those that aren't, you can conclude that it is either a) impossible, b) so complicated it looks like option a or c) reduces yields or increases costs.
Finally even green hydrogen is no greener than the marginal green-ness of electricity. Which is very often a shade of absolutely pitch black.

The actual reduction agent in a classical blast furnace is CO. You're right with the rest of what you say. ;)

Would you consider removing the last two irrelevant paragraphs and maybe instead elaborate why solid reactants (I guess) are favorable in terms of reaction economics to gaseous ones?

@tobalt the second last paragraph is a direct answer to your why not, and the last one is my personal quest to educate the servile masses in using the word 'green' sparingly. But I will elucidate once I am back on my laptop, writing equations is too arduous on a mobile device.

@Stian Ok maybe "irrelevant" was the wrong term. What I mean is that these paragraphs contain no substance because they tell trivial truths. Thanks for now adding this substance :) I think, however, that you try to make a point in case of Carbon vs any of those replacements. But the question is about CH4/NH3 vs H2.

Chemists would not use either $\ce{CH4}$ either $\ce{NH3}$ in the first place.
Their true acting reducing agent is hydrogen again. But it would have to be formed first. The partial hydrogen pressure would be low, the reduction slow and inefficient.
$\ce{CH4}$ would still produce $\ce{CO2}$.
Additionally, iron and steel production count on carbon content in iron.

Why would they not use CH4 or NH3 is the main question. The partial pressure of "hydrogen" at the same gas pressure is higher in both CH4 and NH3 than in H2, having 4 and 3 hydrogen atoms per molecule, instead of 2 if I am not mistaken. And the production of CO2 from renewably generated CH4 is of course exactly a net zero. The C alloy content in these metals is not really an argument pro-hydrogen either is it ? In summary, I don't see how this answer adresses the question in its current state.

You are really mistaken. Review what the partial pressure is. Sure, C content is against the whole H2/CH4/NH3 agenda. All that would be energetically and technologically very challenging. I did say chemists would not use CH4 nor NH3 (but H2 or C or CO), as you have asked a "simple chemistry question".

Just to make sure I understand correctly: Does it mean that CH4 and NH3 aren't really reducing agents in molecular form, but only CO, C or H2 are? And that only the tiny amount of elemental H2 that thermally cracks from e.g. CH4 would work as an active reducing agent?

In a summary, you are correct. Additionally, NH3 and CH4 would still keep disadvantages of H2, regarding metallurgic aspects. For some other metals than iron, hydrogen is already used, especially if carbon is not applicable.

At temperatures where H2 effectively reduces iron oxide, NH3 falls apart into its base components. Which means there is no point in making NH3 in the first place.

Note that one of methods to produce hydrogen by easy means is letting water vapor to pass (red hot?) iron peelings.  $\ce{3 Fe + 4 H2O -> Fe3O4 + 4 H2}$, resp.   $\ce{ Fe + H2O -> FeO + H2}$.


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Discrepancies in basic strength of different amines
The order of basic strength of ethyl substituted amines is
$\ce{(C2H5)2NH}$ > $\ce{(C2H5)3N}$ > $\ce{(C2H5)NH2}$ > $\ce{NH3}$
Which means secondary amine is more basic than tertiary followed by primary and then ammonia.
But the order of basic strength of methyl substituted amines is
$\ce{(CH3)2NH}$ > $\ce{(CH3)NH2}$ > $\ce{(CH3)3N}$ > $\ce{NH3}$
Here the tertiary amine is less basic than primary amine.
What is the reason behind this discrepancy.

Review also search results for site:stackexchange.com OR site:libretexts.org relative basicity of alkyl substituted ammonia

These differences are so tiny, you can call all these compounds equally strong. Another thing is considering pKa values some kind of universal truth about acidity. All these values have uncertainties and are appropriate only for diluted aq. solutions.


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In hyperconjugation in carbocations, which atom has more positive charge?
If we consider a carbocation, let's say $\ce{C2H3^+}$
The hypothetical hyperconjugation structures will be like this

In the actual structure(i.e. the resonance hybrid, all the $3$ $\ce{H}$ atoms bonded to the $\ce{SP3}$ hybridised carbon atom will have partial positive charge. The rightmost $\ce{SP2}$ atom has partial positive charge too. I can't understand which one will be more positive? For that I need to know which structure will contribute more to the hybrid.
In the hypothetical and identical structures (b) , (c) and (d) there are 2 different "species" , one resembles an $\ce{H}^+$ ion and other is an alkene, so how can the rules to find which structure is more contributing be applied. (b) , (c) and (d) will have equal contributions but what about (a) ?

$\ce{C2H5^+}$ has a different structure, with a bridging hydride ion.

https://chemistry.stackexchange.com/questions/29000/which-carbocation-is-more-stable-the-ethyl-or-1-propyl-carbocation


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Power produced by expanding gas
Calculate power produced by CO2 gas pressure driven 100% efficient turboexpander. Inlet pressure is 48 megapascals and inlet temperature is 425 kelvin and inlet flow rate is 3886 kilograms per second and inlet enthalpy is 4900 kilojoules per kilogram. Outlet pressure is 10 megapascals and the outlet temperature is 295 kelvin and the outlet flow rate is 3886 kilograms per second and the outlet enthalpy is 15 kilojoules per kilogram.
To determine the solution I have down the following calculation which am not sure is correct because I am not sure the enthalpy drop accounts completely for the power of the expanding gas driving the turbine since the enthalpy tables I used are only provided at constant pressure and only a function of change in temperature. Please advise if my solution is correct and actually accounts for everything. Thanks
Enthalpy Drop = 4900 - 15 = 4885 kilojoules per kilogram Power = (3886 kg/s) * (4885 kjoules/kg) Power = 18,975,610 KJ/sec * 1000 w/KJ = 18,975,610,000 watts
Therefore, the power produced by the CO2 gas pressure-driven turboexpander is approximately 18,975,610,000 watts.
This doesnt seem right to me. Are my methodology, assumptions, and calc correct?

Do you insist on accuracy of 7 valid digits? // Good practice is starting with symbolic algebraic expressions and keeping it this way until all is ready to plug in literal numbers with units. It helps in focusing on principles, mistakes are easier to spot, orientation is improved, Q/A is reusable and has bigger  permanent value.
You may find useful  formatting mathematical/chemical expressions/formulas.

I have noticed you have reposted the closed question, what is against the site policy and this will be closed as well.
Address the reasons why the original one was closed and ask for reopening.


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How to calculate carbon to phosphorus stoichiometric ratios using results from ICP-MS and CHN analyzer?
I run a few chemostats to manipulate the carbon to phosphorus ratio (C:P) of algae, which is later feed to an invertebrate. I used Inductively coupled plasma mass spectrometry (ICP-MS) and Inductively coupled plasma atomic emission spectroscopy (ICP-OES) to run samples to measure P and I used CHN analyzer to measure C. Afterwards, I discussed the results with the techs that run those machines and they helped me calculate mol of P in the sample per unit dry weight. Yesterday I showed the C:P ratios of 6 algae samples to my advisor and she said that they were way too P-enriched that they would be toxic to the invertebrate - many of our invertebrates in our dietary experiments that have been feed this algae live throughout the whole experiments. She said that I likely had a calculation error and I should look at my work again. I want back and wrote down everything that I did with the techs and tried out a few things that I thought might fix the problems - none of them worked. So here I am, asking you all to provide me with help.
Before I go into the math, I want to make my preparation work is clear. I performed 2 acid digestions on 6 different algae samples with several replicates. Since we did not know the P concentration of the algae beforehand, we used a variety of weights during the first digestion and ran the samples on ICP-OES. Once we had an idea of the P concentration from the ICP-OES we decided we would run some of the same samples on the ICP-MS and also do another acid digestion, with some tweaks, and ran those samples on ICP-MS. To clarify, we ran 3 “High-P” algae samples and 3 “Low-P” algae samples. The preparation for the CHN analyzer was easy: 1) weighed algae, 2) put them into tin caps, and 3) tossed them into the analyzer.
Without further ado, I will now describe what calculations I have done.
ICP-MS calculations
We calculated the P concentration ppm of the algae by using the P (no gas) counts per second (CPS) value. First, we made a calibration “curve” (aka a linear regression) with the blank and standards. To make the calibration curve we had to make a scatterplot with the standards (10 ppb, 100 ppb, and 1000 ppb standards) and added a line of best fit. We then used the slope and intercept of that graph and input the sample’s P (no gas) CPS value in the equation [slope * CPS + or - intercept]. Then we calculated our dilution factor: we divided the algae sample weight but the final algae sample weight. To clarify, for the ICP-MS results we digested a small amount of algae (0.20 g) and digested them. Afterwards, we diluted this situation with deionized water until our sample weight reached 100 grams. The P concentration ppm was calculated as [calibration CPS value * dilution factor / 1000]; the values range from 300 to 2500. Here, I am unsure why the value was divided by 1000.
ICP-OES calculations
We calculated the P concentration ppm of the algae by using the P 185.878 value - if I recall we looked at some figures to determine which option was the best, but I cannot explain our choices anymore than that. Again first, we made a calibration “curve” (aka a linear regression) with the blank and standards. To make the calibration curve we had to make a “blank correction”, here we subtracted the blank from the 1, 5, and 10 ppb standards. We plotted those corrected values on a scatterplot and added a line of best fit. Then we used the slope and intercept of that graph and input the sample’s P 185.878 value in the equation [slope * P 185.878 value + or - intercept]. Then we calculated our dilution factor: we divided the algae sample weight but the final algae sample weight. For the ICP-OES results we digested a small range of algae (0.004 to 0.1 g) and digested them. Afterwards, we diluted this situation with deionized water until our sample weight reached 4 - 8 grams. The P concentration ppm was calculated as [calibration P 185.878 value * dilution factor]; the values range from 573 to 3306. Here, I would like to suggest that perhaps the division by the 1000 in the ICP-MS calculations was a correction for the 100 grams used. The ICP-MS solution was close to water.
Average of P mol/ g of sample calculations
Next, I wanted the mol of P in the sample per unit dry weight, so we multiplied the P concentration ppm by the atomic mass (31) and divided by 1000000 to convert from ug/mol to g/mol. I am confused about this conversion – these algae samples were in grams so how can this result be converted from ug/mol to g/mol? I then took the average of P mol / g of sample for each algae sample. A range of Low P algae was 0.01 to 0.025; the range of the High P algae was 0.054 to 0.075.
CHN analyzer calculations
We started with the % of C in the sample. We multiplied %C by the sample weight (micrograms) to receive micrograms of C. Then we divided micrograms of C by atomic mass (12.011) to receive the ug / mol of C. We took that value and divided it by the sample weight (micrograms) to cancel out the micro(grams) so we receive C mol / g of sample. Our end result was the average of C mol/ g of sample – we averaged the algae samples. The range of the samples were 1.79 to 3.09.
C:P calculations
We divided the average of C mol/ g of sample by average of P mol / g of sample to calculate C:P.
Those are all the calculations that we did! Please let me know if you need example charts. I also recognize that this was one of my first times digesting algae so I may have made a mistake with the digestions, but I want to make sure that I did not convert anything incorrectly. I am not a chemist so I could make very easy mistakes an that fact that I am confused by some calculations is probably telling. If you all do not have suggestions on how to calculate C:P from ICP-MS and CHN please let me know if you know of any video or pdfs that can help me with this! I want to be sure I did this right before I try to run these digestions again.

There's a greater chance you'll get some help if you avoid acronyms etc or include definitions. For instance cps=counts per second.

Why "slope * CPS + or - intercept"? Isn't it simply Cppm = (CPS-intercept)/slope ?

Since you supervisor (I suppose) quicky spotted it I guess is it order of magnitudes (x100, x1000) or is it double, triple the value. Do you know the "expected" value  ? This will greatly help you to spot in which part of the calculation you were wrong

@BuckThorn Thank you for your questions! The tech is the one that showed me how to account for standards - I assumed that CPS is the x in the equation for a straight line (y = mx +b). I added the "-" because sometimes the intercept is negative. What does Cppm stand for?

@Jojostack Thank you for your question! When I talked to my advisor she suggested that the data could be off by an order of magnitude. The High P algae are ~38 to ~49 and the Low P are ~83 to ~125. If we added a zero to at the end of those values they would be closer to C:P that my advisor expected.
I definitely think this is reasonable especially since we started the CHN analyzer samples at micrograms and the ICP samples at grams! However, I thought these errors were accounted for. Are there any conversions that stand out to you?

Cppm is the concentration of P in ppm units. A fit will provide y = mx + b (typically one treats x as the independent variable, in your calibration Cppm, and y the dependent variable, in your experiment cps, but you are choosing to do things in a different way). You write the fitted line as y=mx+b irrespective of the sign of b (keeping in mind of course that b is a signed quantity).

@bribina Do you have some standards (as in samples of known composition) for which you know in advance running the analysis what the results should look like? (Certified standards are more expensive, e.g. NIST presented by a veritassium video, perhaps for later.) Is application of the method on algae this well understood/mastered that there is a protocol typically followed by multiple labs (like interlaboratory tests about doping, ASTM standards e.g., about NMR spectroscopy)?

@Buttonwood We have P standards of known composition which we used in the MS, but did not digest them. My advisor sent a few grams of NIST spinach and apple leaves after these digestions unfortunately. The method we are using was offered to us by another lab at our university; this method is from the Milestone company. I used another method to prepare the samples for the microwave (G. Esslemont et al., 2000) that digested 4 species of plants (leaves) using a variety of procedures. We used nitric acid and hydrogen peroxide digestion in combination with a microwave.

Hello all! Thank you so much for your help. I finally got to discuss this with the 2 techs in person. The mistake was quite simple. The first C_result from the CHN analyzer was C% so what I was missing was simply multiplying that first result by 100.  I am thankful this turned out alright!


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Non-elementary reactions: Relationship between kinetic and equilibrium constant
As far as I understand, the equation for a kinetic equilibrium is always the same, and is given by:
$$K = \frac{\text{concentration of products}^{p}} {\text{concentration of reactants}^{r}}$$
Where $p$ and $r$ are the stoichiometric coefficients.
So, for elementary reactions, it is clear that the equilibrium constant is always the same (for the same temperature $T$), notwithstanding the initial concentration. For example, for the reaction:
$$ \ce{A <=> 2B}$$
If we assume dummy kinetic rates for forward and reverse reaction:
$$ r_f = 1·c_\text{A} = [\ce{A}]$$
$$ r_r = 3·c_\text{B}^2 = 3 [\ce{B}]^2 $$
Assuming the following: $[\ce{A}]_0 = \pu{10 kmol/m^3}$, $[\ce{B}]_0 = \pu{0 kmol/m^3}$, $\dot{v} = \pu{1 m^3/s}$. We know that the equilibrium is reached when the rate forward equals the reverse rate
$$r_f = r_r$$
$$\begin{equation}[\ce{A}] = 3[\ce{B}]^2 \qquad(1)\end{equation}$$
And if we consider the mass balances:
$$ [\ce{A}] = [\ce{A}]_0 - [\ce{B}]/2 \\ [\ce{B}] = 2 ( [\ce{A}] - [\ce{A}]_0)$$
We can substitute these in the rate expression $(1)$ and solve the equations for the roots:
$$  [\ce{A}]_0 - [\ce{B}]/2  = 3 [\ce{B}]^2 \\ [\ce{B}] = 1.74, [\ce{A}] = 9.13$$
And the equilibrium constant is therefore:
$$ K = \frac{1.74^2}{9.13} = 0.33$$
And if we change the initial concentration, to something like double: $[\ce{A}]_0 = \pu{20 kmol/m^3}$, the result of the quadratic equation changes to:
$$ [\ce{B}] = 2.5, [\ce{A}] = 18.75$$
And the equilibrium constant remains the same:
$$K = \frac{2.5^2}{18.75} = 0.33$$
So everything makes sense. But I understood that this behavior should continue even when the reactions are non-elementary. If we now have the same reaction, but the kinetics are given by:
$$ \ce{A <=> 2B}$$
$$ r_f = 1·[\ce{A}]^2$$
$$ r_r = 3·[\ce{B}]^2$$
Now it changes. If we develop the mass balance in equilibrium:
$$ [\ce{A}]^2 = 3 [\ce{B}]^2$$
$$ ( [\ce{A}]_0 - [\ce{B}]/2 )^2 = 3 [\ce{B}]^2$$
For an initial concentration of $[\ce{A}]_0 = 10$:
$$[\ce{A}]_0^2 - [\ce{A}]_0^\vphantom{} [\ce{B}] + [\ce{B}]^2/4 = 3 [\ce{B}]^2$$
Solving the quadratic equation and mass balance:
$$ [\ce{B}] = 4.48, [\ce{A}] = 7.76$$
And the equilibrium constant:
$$ K = \frac{4.48^2}{7.76} = 2.58$$
And when we repeat the calculations for $[\ce{A}]_0 = 20$:
$$ [\ce{B}] = 8.96, [\ce{A}] = 15.52$$
And now the equilibrium constant is:
$$ K = \frac{8.96^2}{15.52} = 5.17$$
Why does this happen? Am I calculating the equilibrium constant wrong? It should be constant for non-elementary reactions, right?

For non-elementary reactions, all bets are generally off. Kinetics of consuming reactants or producing products is not generally directly related to the powered multiplication of their concentrations (or equivalent quantities). There is multiple forward and backward reactions and their respective rate constants.

See e.g. https://chemistry.stackexchange.com/q/81063/35806

Nitpick: IUPAC's Green Book discerns number concentration $C_\mathrm{B} = N_\mathrm{B}/V$ in units like $\pu{m^{-3}}$ from amount concentration $c_\mathrm{B} = n_\mathrm{B}/V = [\ce{B}]$ in units like $\pu{mol m^{-3}}$ (though mol/L is more practical and hence more frequently seen).  And SI units typically are not set italics.  See mhchem you can use for the body of questions, answers and comments here on chemistry.se (don't use it on titles of questions).

I don't understand your notation, if $r_f$ is a rate then $k_f=C_A$ does not make sense, $k_f$ is, I assume,  a rate constant, with units 1/time for first order reaction, and $C_A$ is  concentration. The rate is $d[concentration]/dt=...$.

While the post now is set with mhchem, there obviously a hump in the second equation.  That is, (amount) concentrations run in dimensions like $\pu{mol/L}$, but lower case $k$, the rate constant has (at least) a time related contribution like $\ce{s^{-1}}$.  By dimension analysis, the use of equality signs however requires the left and the right of this sign to express the same number (perhaps differently expressed, as we have e.g. prefixes like kilo, milli), as well as the dimension. (I didn't introduce this by the use of mhchem.)

Maybe I put something wrong. What I intended to write was: "rforward = CA", "rreverse = 3CB^2". I have rewritten those, I hope it is clearer now

I'm sorry, but it still does not make sense to write $r_f=1.C_A=[A]$ etc. it is still dimensionally wrong. If, for simplicity, $A=B$ then you can write $d[A]/dt=-k_f[A]+k_r[B]$  where $k_f,;k_r$ are rate constants, and similarly $d[B]/dt=..$

I might be confused, but it is dimensionally sound, isn't it? $r_f[kmol/m^3s] = 1 [s^-1] C_A [kmol/m^3]$. If we omit the 1, cause multiplying for 1 doesn't change anything, we can write... $r_f = C_A = [A]$, can't we?

If you have a two-step reaction
$$\ce{A <=>[$k_1$][$k_{-1}$] I <=>[$k_2$][$k_{-2}$] 2B},$$
the rate of the forward reaction of the first reaction (with rate constant $k_1$) is not equal to the rate of the reverse reaction of the second reaction (with rate constant $k_{-2}$). Instead, for each elementary reaction, the forward and reverse rates are the same at equilibrium.
So the premise of your argument is incorrect.
It turns out that the forward rate of the first reaction is relatively easy to measure (just start with pure $\ce{A}$ and measure the initial rate), so it is easy to figure out the rate law. Similarly, you can figure out the rate law of the reverse direction of the second step. It is much harder to measure the rate laws of the intermediate reacting to form reactants, or to form products (you would have to start with pure intermediate, and where will you get that from).
However, if the steps are elementary, you know the rate law by definition. If you write down the conditions for equilibrium (all steps are at equilibrium), you will get to the equilibrium constant expression in a tedious but straight-forward way, e.g. see this answer.

But, isn't it possible that you have a power law reaction that reproduces correctly the experimental kinetic, and that it is not an elementary reaction? For example:
N2 + 3H2 <-> 2NH3
The elementary kinetic law would be "r = kPN2PH2^3 - kPNH3^2", but what if the exponents found through experiments are different?

@DanielV. You would expect them to be different. The reaction is typically catalyzed (getting all four reactants to collide in mid-air with sufficient kinetic energy is practically impossible). Nonetheless, when you do the tedious math, you can derive the equilibrium constant expression from the rate laws of elementary steps, and it looks the same no matter what the steps may be ("catalysts change the kinetics but not the equilibrium constant of a reaction").

And If I wanted just the number of the constant (Because I am going to fix the temperature), how would I obtain it, when it changes with the initial concentration? Because then the division changes as well, like in the example I wrote, right?

@DanielV. The value of the equilibrium constant does not change with initial concentration. It just looks like that because your strategy to calculate it has a flaw. For multi-step reactions, you could get the equilibrium constant by measuring equilibrium concentrations, or you could measure the standard Gibbs energy of reaction (via electrode potentials or enthalpy and entropy measurements) or calculate it from thermodynamic data (via Gibbs energy of formation).

I see. Then in an experimental setting, how would the "global" equilibrium constant of the example reaction be calculated? Or is it impossible without knowing the multi-step reactions? (I'm gonna mark your answer as the answer already, since you already helped me a lot and it is the correct answer, I'm sure, I'm just having difficulty grasping it)

@DanielV. You can't calculate it from the kinetic data provided unless you know the kinetic constants (four in my example) of the elementary steps (two in my example). However, you can also just plug in the equilibrium concentrations if you are able to measure them.

But when I plug the equilibrium concentrations (Calculated computationally from the invented kinetic, that is true...) the Kinetic constant changes as I show. If it was a real reaction, this would not happen?

Yes. You might want to ask a separate question or edit your question to ask for an example.

I cannot think of an actual example where I know that the kinetics are non elementary and come from a power law. I was just curious about how it would be calculated for this imaginary example, since at equilibrium, the relationship between concentrations does not hold constant


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Question about percentage purity and percentage yield
While doing questions on percentage yield, this one took my attention
As a result of the reaction of 20 grams of hematite with a purity of 60% with a sufficient amount of pure aluminum, how much impure iron with a purity of 70% will we get? (The percentage yield of this reaction is 40%)
$$\ce{2 Al + Fe2O3 -> 2 Fe + Al2O3}$$
My question is that why does it say "How much impure iron will we get " aren't all products of a reaction pure ?

They are not pure if mixed with something else.

For the body of questions, answers, and comments, chemistry.se offers you to use mhchem.  Don't apply it to the titles of questions, though (because it is an extension not every web browser processes equally well).

@Poutnik So do you mean when the reaction is taking place some molecules or atoms are forming their own reactions with our products ?

@Poutnik Wikipedia reports a density of solid $\ce{Al2O3}$ (though not sure which polymorph) of $\pu{3.987 g/cm³}$ (ref and for molten Fe of $\pu{6.98 g/cm³}$ (ref).  While I don't know if the two easily mix into each other during a thermite reaction (welding of train rails), maybe the lower part of the «bucket» the workers use to join rails to join is at least enriched in Fe, while portions richer in $\ce{Al2O3}$ yield a protective blanket to prevent Fe to oxidize again at these high temperatures.

@Buttonwood  If one mixes e.g. 7 g of pure iron with 3 g of mixture of original hematite impurities, remaining  Al, Al2O3 and not reacted hematite, one can easily get heterogenous mixture with 70% iron purity.(. Not sure if Al and Fe makes alloy).  Anyway, the task gives vibes of made up task for computation exercise, so I would not worry so much about how much it reflects reality.

To start let's assume that you're using some reaction to make some compound in the lab. There are two "products." First the products as given in the the chemical formula for the reaction and second there are the products that you collect. The products that you collect are never absolutely pure. 99.99% pure would be essentially pure since working to 4 significant figures in a typical chemistry lab is very good work. But that isn't absolutely pure.
Now to discuss the problem you posed. This is just a made up problem and to worry to much about the details will send you down the rabbit hole.
But let's think about the problem this way. Suppose you are going to do the reaction in a crucible and after the reaction is complete you crack the crucible with a hammer to get the "button" of metal as the "collected product."
Now the stated problem throws in some wrinkles to see if you understand chemistry and are not just solving a math problem.

First the problem states "the reaction of 20 grams of hematite" but then the problem backpedals and states that the hematite is only 60% pure. Who knows what the other 40% is. Make the assumption  that the 40% is inert. (The problem would have been better stated as the reaction of 20 grams of hematite ore which is 60% pure.)

The problem states that there is "a sufficient amount of pure aluminum" which means that there is at least enough aluminum to react with all the hematite.

Now, as in my contrived experiment, the problem states that the button of metal produced is only 70% iron. Let's assume that 12.0 grams of hematite should yield x grams of pure iron. Then the button would weigh x/0.70 grams since it is impure.

Thank you so much for your explanation  So when we say our product has percentage purity, we are not talking about the products given in the the chemical formula but the products that we collect ?!

And can this be another reason or not " when the reaction is taking place some molecules or atoms are forming their own reactions with our products "

@Shadowsparkle - Yes the chemical formula implies absolute purity, but the actual product collected will never be absolutely pure. // The problem statement doesn't give any indication as to what the impurity in the "hematite" is, nor what the 30% of the product that is not iron. This was just a "problem" not a lab exercise. Don't go down a rabbit hole trying to figure out either of those.


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Why does s and p orbitals dont contribute to para or ferromagnetism?
Like in metal Ag,contains unpaired electron in their s subshell and like Cl has unpaired electron in its p subshell,but they all show diamagnetism,what can be the reason for this?

Are you sure atomic Cl is diamagnetic? Cl2 obviously is. But even O2 is not.

Review the guides Asking and How to ask to prevent misunderstanding,  need of clarification, objections, downvoting or closure.

Sorry,I edited it,mistyped

All compounds have a more or less significant diamagnetic contribution. I will answer the case of ferromagnetism, the case of paramagnetism of molecules is often rare but occurs in molecules like $O_2$ where there is a degeneracy of the molecular orbital levels at the highest occupied molecular orbital (HOMO), the first Hund's rule is applied leading to a paramagnetic molecule, some free radicals are also paramagnetic. The paramagnetism of solids, especially metals, depends on the band structure not the atomic configuration : orbitals with large $n$ are delocalised and do not exhibit a signifiant paramagnetism it is often the case of $s$ and $p$ orbitals, but there is always a paramagnetic susceptibility.
A ferromagnetic order requires a solid for long-range coupling, a disordered system without significant translational symmetry often does not show ferromagnetic order. It is remarkable that all solids with $s$ and $p$ orbitals are not ferromagnets, it is due to the fact that for being ferromagnetic there are two conditions: localized orbitals, ($2s$, $2p$) can be considered as localized in diamond and boron, but does not satisfy the second condition: the density at the Fermi level should be high enough, it means a considerable number of electrons. These conditions are the Stoner criterion, summarized as the repulsion between two electrons should be very significant for a ferromagnetic order to occur, it is not the cause of the $s$ and $p$ orbitals because they are delocalized and have less repulsion.


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Are the inner circular rings in higher 's' sub shells part of inner 's' sub shells, as the diagrams suggests?
The diagram of, for example, 2s and 3s, is the innermost circle showing electron probability has to be 1s subshell. How could it be the part of 2s or 3s, since it is a complete circle around nucleus like 1s sub shell, and they obviously cannot overlap

Although I do not know the Schrödinger equation, I saw the graph of variation of wave function with distance, and graph of each higher 's' sub shell seems to add on to the previous one. What I mean is that the 2s looks like the continuation of 1s graph and 3 the continuation of 2s, and so on.

In short, my question is that are the inner rings of electron probabilities in for example 3s sub shell that of 1s and 2s sub shells?

It is good to refer sources of used images, for broader context and credit reasons. Using them without the credit often violates the usage rights.

A bit more advanced detail on this is Slater's rules for electron shielding https://en.wikipedia.org/wiki/Slater%27s_rules

Note that in spherical $\text{s}$ orbital context, the charts of $|\Psi|^2=f(r)$ are charts of probability density $\text{d}P=|\Psi(r)|^2 \cdot \text{d}V$, not of electron radial probability $\text{d}P=P(r)\text{d}V=4\pi r^2|\Psi(r)|^2 \cdot \text{d}r$. The former has at $r=0$ the local maximum, the latter is zero at $r=0$.

Errata: should be $\text{d}P=P(r).\text{d}r=4\pi r^2|\Psi(r)|^2\text{d}r$

Spherical nodes of higher s orbitals and related inner spheres are not lower s orbitals nor their parts. They just partially occupy the common place.
Different orbitals mean different states, not necessarily places. There is generally large overlap of orbitals. Lower orbitals occupy, in sense of significant probability regions, a subset of the space of upper orbitals.

The angular part of the $s$ orbital is always spherical, these pictures are misleading because they include radial nodes, only the radial functions are different. A wave function has no physical meaning, the overlap in this case is different from the overlap between the plotted functions. The plot of $\psi^2$ has no physical meaning a well, the radial probability density is $r^2\psi^*\psi$. The overlap gives the probability of two electrons occupying $1s$ and $2s$ states by $<1s|2s> = \int_{-\infty}^{0} \psi_{1s}(r) \psi_{2s}(r) r^2 sin(\theta)drd\theta d\phi =0$ (in spherical coordinates) since this probability is zero, $1s$ and $2s$ are orthogonal even though the probability density may overlap.
In the case of a simple molecular orbitals like $H_2$, the overlap in the plot is proportional to the overlap integral. There is another case of fully overlapping probability density without overlap : the case of two plane waves $\psi_1 = \frac{1}{V} e^{ikr}$, $\psi_2 = \frac{1}{V} e^{ik'r}$ confined in a volume $V$ (to be normalizable), the square modulus gives 1 for both inside the volume, so there is a complete overlap, but $\langle\psi_1|\psi_2\rangle=\delta_{k,k'}=0$ because $k\neq k'$. To summarize, there is $3s$ wave function overlapping with $1s$ and $2s$ in the plot, although an electron cannot have $1s$ and $2s$ states at the same time.


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Ionization and exchange energy
As we all know that 'Nitrogen' has higher 'Ionization Energy' than its neighboring elements both carbon and oxygen, opposite to the trend of increasing ionization energy across a period. The reason mentioned everywhere I have read is - the extra stability of half and full-filled orbitals due to exchange energy. Now let's discuss exchange energy. Many of you say that exchange energy is - The amount of energy released when electrons with the same spin swap positions in degenerate orbitals and since exchange is possible maximum in half and full-filled orbitals they are more stable. This statement raises a lot of questions in my mind -
1.What kind of energy is this? - electrical, magnetic, nuclear, or anything else that I have not come across yet
2.When we go across a period let's say from boron to carbon, a electron enters in carbon p orbital with the same spin as of another and that leads to release 'exchange' energy so from where did that exchange energy come to electron?
3.When we ionize hydrogen (no exchange energy in hydrogen) by supplying energy to it, electron use it to break the attraction between the nucleus and itself and comes out of the reach of the nucleus. But In nitrogen valence electron releases two kinds of energies when it enters in 2p orbital -
(I). Electric potential energy
(II). Exchange energy
So when we ionize nitrogen, the electron must gain all the energy that it has released before coming out and it gains electric potential energy by using the supplied energy through breaking the bond between the nucleus and itself but in what form and how it gains exchange energy. This means when an electron gets exchange energy does it stop exchanging its position or does anything else happens to the electron after getting Back exchange energy?
4.I also found someone https://chemistry.stackexchange.com/a/58628/129242 who is explaining exchange energy with coulomb repulsion between electrons but I think that's wrong because we already have used that repulsion factor when calculating effective nuclear charge and Nitrogen has less effective nuclear charge than oxygen https://en.m.wikipedia.org/w/index.php?title=Effective_nuclear_charge and according to that nitrogen will have lower ionization energy and we know that is not true so I think exchange energy is not related to electronic repulsion.Correct me if I am wrong
At the end please consider the fact that this question is asked by a 12 class student so please avoid higher mathmatical stuff.
Please comment if any point is not clear to you.

Be sure to cover broader but targeted search. Reviewing these search results may help site:stackexchange.com OR site:libretexts.org exchange energy

For your Q#1, it's not any precise specific energy. Electrons are very mobile and an uncertainty of motion and position means they have varying forms of energy. It is mutual electromagnetic force, then nuclear force of attraction, Pauli's exclusion principle also contribute keeping them apart.

I admit that this exchange energy stabilizing a half-full shell is very questionable especially when the spins are aligned and this interaction bring a positive energy contribution. If I am wrong, feel free to downvote or add a comment.
Exchange interaction
Without trying to do deeper, there are fundamental properties related to an associated interaction.  Most of the interactions can be described by a general potential of the form $\propto PP'\frac{1}{r}$, where $P$, $P'$ can be an electric charge, a magnetic charge, or a mass. A magnetic interaction is a relative dynamic effect of the electric interaction. An approximation will be from a classical rotational motion of a charge particle to deduce an intrinsic magnetic dipole generating a field rather than a static magnetic charge, but none of the classical representations will describe the symmetry of the intrinsic angular momentum generating a magnetic moment for a charge particle like an electron. Another approach will be to create a potential derived from the interactions of this intrinsic angular momentum (spin) : the exchange potential, which has the form of the Coulomb potential, but there is a permutation of the positions of the electric charges. The use of a permutation operator and the antisymmetry of the total WF after exchange gives the normal derivation of this contribution. Unlike the mass with always has a positive value, the charge and the spin can be either positive or negative, therefore the exchange interaction energy can be positive or negative, the positive coupling leads to an interaction that reduces the Coulomb repulsive interaction and promotes an overlap with antiparallel spins and a negative coupling avoids an overlap and creates parallel spins.

1.What kind of energy is this? - electrical, magnetic, nuclear, or anything else that I have not come across yet

The exchange energy in this case is an electrostatic or Coloumb energy but does not release any energy nor create a force. Instead, it couples spins by allowing two electrons with different spins to share a single orbital state (singlet state). To consider the exchange interaction as a true exchange of position is correct, an important case is an anyion, where a theoretical path of this exchange is important. That is all I can say about this interaction without using math.
Half-filled stability
The $s$ subshell is normally always an innermost subshell because of the orbital angular momentum $ \ell =0$. The orbitals with $\ell >0$ have a "centrifugal force" and are farther to the nucleus compared to $s$ orbitals. The solution of an electron SE in an effective potential gives orthogonal solutions, the orthogonality or two states ($|1\rangle$ and $|2\rangle$) implies that an electron measured in one state cannot be in another state, this means that the probability that an electron is in a state $|1\rangle$
position and after move to another state $|2\rangle$ is zero, there is no possible exchange between $|1\rangle$ and $|1\rangle$. Let's take the case of a half-filled $p$ orbitals, they are several configurations, the most stable configuration is the first Hund's rule, where the spins are aligned and each has one of the three $p$ orbitals as a state.
\begin{array}{|l|c|c|c|c|}\hline
m_l & +1 & 0 & -1\\\hline
    &  \uparrow & \uparrow & \uparrow \\\hline
\end{array}
This configuration avoids Coulomb repulsion when two electrons have the same orbital. The question is whether an electron can exchange a position from one $m_l$ orbital to another $m_{l'}$ in this configuration :  the answer is no because the orbitals are orthogonal and the probability that an electron goes from one orbital to another is zero, even the weak dipole interaction will avoid this transition.  As a consequence, I do not understand the stability due to an exchange interaction. If will consider an exchange, the coupling will be positive because the spins are aligned and the ionization energy will not change much because the electrons will always feel approximately the same potential before and after the exchange, this argument is not correct.
The stability of a half-filled subshell is mainly due to the symmetry of this configuration and quantitatively the orbital angular momentum. This process can be fully understood with a semi-classical approximation. Let's consider a half-filled $p$ subshell, the "centrifugal force" acts on the projections $m_l \neq 0$, as an orbital has $m_l = -1$ and another $m_l=1$, the centrifugal force is cancelled, therefore $L=0$, the geometry is spherical and similar to $s$ orbitals, the electrons are closer to the nucleus, feel a stronger potential of the nucleus, the ionization energy is higher.


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Can the strength of a covalent bond be compared to the stiffness of a spring?
I am modelling a solid with the density functional theory. The solid contains different types of atoms. The atoms are covalently bonded. I want to estimate the strength of each bond. How can I do it?
My idea is to shift the atom for which I want to estimate bond strengths a little bit along the direction of each bond, and measure the energy of the solid for each shift.
This is similar to a calculation of the stiffness of a spring. Stretch the spring a little bit, and measure the force. When the force is stronger, the spring is stronger (assuming the same stretch for different springs).
Is my thinking correct for a covalent bond? Can it be likened to a spring? Does it mean that the bond is stronger if the force on the atom is larger if I stretch the bond to a certain distance?

Yes you can use Hooke's law. This models diatomic molecules quite accurately for small displacements from equilibrium and so at low energy.

If you move one atom away from a neighbour you are presumably moving it toward atoms on the opposite side. In that case it is a collective motion involving multiple bonds. I believe normal mode analysis is the more general approach to this kind of problem.

The simplest model for a bond is a quadratic potential like a spring. Bonds are assigned force constants to explain the vibrational frequencies observed in infrared spectroscopy. In the quantum-mechanical theoretical treatment, one talks of a harmonic oscillator.
To model how bonds break, you have to add some complexity and consider the anharmonic oscillator, perhaps with a Morse potential.
Stronger bonds have higher force constants (and higher frequencies).

My idea is to shift the atom for which I want to estimate bond strengths a little bit along the direction of each bond, and measure the energy of the solid for each shift.

Yes, you can do that. You just ("just") have to make sure the shifts are large enough to see the effect while small enough to not break the bond.

This Hooke's law is accurate only for strong covalent bonds and not far from the equilibrium distance. Anharmonicity occurs at larger distances from equilibrium.

I am modelling a solid with the density functional theory. The solid contains different types of atoms.

When performing a DFT calculation, you should not stretch the bond too much, the resulting energy (anharmonic part) will not be important for your calculation and will certainly be inaccurate.


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Is fugacity the corrected pressure that should be used in thermodynamical models?
I am a little confused about fugacity and pressure of a gas. In ideal gases, pressure is same as fugacity. But for real gases this is not the case. Here is the link of the question that is the source of my confusion :

https://chemistry.stackexchange.com/questions/135002/what-is-fugacity

The answer by @AChem hints that the pressure gauge measures the pressure of the gas and not its fugacity.
My question is:
Is the value corrected with the help of fugacity coefficient ($f=$$\Phi$$P$,where $P$ is the pressure as measured by manometer and $\Phi$ is fugacity coefficient ) is the actual pressure felt by the tires?
Please leave reasons for your downvote, so that I can improve in writing questions.

Fugacity is not the actual pressure. It is used to account for the real behaviour of gasses and also to overcome the assumption that chemical potential for a real gas can be based on using that based on an ideal gas $\mu=\mu^\text{o}+RT\ln(P(atm)/1(atm))$ rather than starting over again with, say, the van der  Waals eqn.

So if fugacity is not the actual pressure, then what does $\Phi$$P$$=$$f$ tell us about? Isn't fugacity just corrected pressure ?

yes, but usually we use dimensionless activity $a$ instead of fugacity ratio $a=f/f^\text{o}=f/1$ such that  for a real gas $a=\gamma P$ with $\gamma$ as activity coefficient and so $\mu=\mu^\text{o}+RT\ln(a)$ or $\mu=\mu^\text{o}+RT\ln(P(atm)/1(atm))+RT\ln(\gamma)$

It could be said the fugacity is the pressure a real gas would have had if ( it had behaved ideally ) AND ( had had the same chemical potential as it has as a real gas).

@Poutnik, then fugacity is corrected pressure? Is it right to think like that?

It is not a corrected pressure. It is like a pressure at hypothetical conditions. It is not like the corrected pressure in the van der Waals equation, where $p$ is external, boundary, measured and perceived pressure, while $p + \frac{a }{{V_\text{m}}^2}$ is the internal, in bulk gas pressure, at least within the vdW model.

@Poutnik, I'm giving you the link to a document by Delaware university. If you can, please let me know about your opinion on points mentioned on 5th page of the pdf. Pdf: https://www.google.com/url?sa=t&source=web&rct=j&opi=89978449&url=https://www1.udel.edu/pchem/C443/Patel_Lectures_2016/Non-Ideality_Lectures.pdf&ved=2ahUKEwjJ_7Pm-6SAAxWG7zgGHbboAQ84HhAWegQICxAB&usg=AOvVaw0-AqWYhQK9F3m5njF7LlWU

For an ideal gas, the mechanical pressure (the one you can measure and that is relevant for the behavior of tires) and the fugacity is the same.
For a real gas, which quantity you use depends on the context. For mechanical questions (whether your reaction vessel can withstand a pressure difference, for example), you use the pressure. If you are interested in chemical reactions, the fugacity is the relevant quantity (e.g. in equilibrium calculations, or any other equilibrium thermodynamics like reduction potential or definition of standard state).
Wikipedia has a succinct definition that captures those two contexts:

In chemical thermodynamics, the fugacity of a real gas is an effective partial pressure which replaces the mechanical partial pressure in an accurate computation of the chemical equilibrium constant. It is equal to the pressure of an ideal gas which has the same temperature and molar Gibbs free energy as the real gas.

Other definitions spell out the "molar Gibbs energy" as the chemical potential $\mu$, which is at the core of many formulas (equilibrium constant, Nernst law, entropy of mixing) that describe changes when you change partial pressure or concentration.
The same concept (to deal with non-ideal solutions, in this case) exists for concentration, where concentration is replaced by thermodynamic activity $a$ to account for non-ideal behavior. However, activity is a dimensionless quantity while fugacity has the same dimensions as pressure.

[OP] Is the value corrected with the help of fugacity coefficient (f=ΦP,where P is the pressure as measured by manometer and Φ is fugacity coefficient ) is the actual pressure felt by the tires?

No, fugacity is the appropriate quantity for questions of chemical reactions or transport, i.e. "the actual pressure felt by a chemical reaction". For mechanical questions, the measured pressure (or partial pressure) is relevant. For example, in a gas mixture, individual fugacities do not add up to the total pressure of the mixture.

I see... so fugacity is relevant for chemical reactions whereas pressure is relevant for mechanical questions. But if fugacity considers all the non-idealness of real gases, then wouldn't thermodynamical models using pressure(instead of fugacity) lead to erroneous results? Thank you for answering!!


text:
Why is tritium radioactive?
The nucleus of tritium (hydrogen-3, $\ce{^3H})$ contains one proton and two neutrons. Nothing repulsive is acting on the proton that can make the nucleus unstable. Then why does it decay?
I suspect the answer is related to the internal structure of the protons and neutrons. This applies as well to the one with the greater number of neutrons than it should be. My question applies to those isotopes as well.

It is just 1 special case among hundreds other cases of radioactive beta decay. // Pay more attention to prior search before asking. // Review the guides Asking and How to ask to prevent misunderstanding,  need of clarification, objections, downvoting or closure.

Note that there is a semi-empirical rule that elements with odd proton number have maximally 2 stable isotopes. And another rule, if there are 2 isotopes of neighbor elements with the same nucleon number, maximally one is stable ( the one with lower energy). $\ce{^3He}$ has lower energy than $\ce{^3H}$.

The neutron is an inherently unstable subatomic particle. An isolated neutron decays with a half-life of about 10 minutes. The real surprise is that, in atomic nuclei, the neutron decay rate can be suppressed by 20 orders of magnitude or more.

@NicolauSakerNeto It is surprising that there are known $\ce{^4H}$,$\ce{^5H}$,$\ce{^6H}$,$\ce{^7H}$, even if all extremely short lived, as expected.Isotopes of hydrogen -Hydrogen-4

Tritium decay is spontaneous even if the binding energy of tritium is higher than the binding energy of 3He. Why?

Nothing repels the proton. That refers to the electrostatic/electromagnetic force. Radioactivity is driven by the internuclear forces.

https://physics.stackexchange.com/questions/303657/about-the-stability-of-hydrogen-isotopes


text:
electron - electron repulsion and shielding effect
What is difference between shielding and electron electron repulsion. When I read this article , I came to know that both are different. First Read this paragraph and see the graph of ionization energy.

Here we can see that author is writting that when we neglect electron electron repulsion we see that effective nuclear charge is increasing and ionization energy should increase but when we include this factor we find a drop in predicted ionization energy.
Now my questions is when we are talking about effective nuclear charge , haven't we already included the shielding factor which itself means repulsion  that valence shell electron feels due to all other electrons in atom. So how author can say that he neglected the electron - electron repulsion while predicting ionization energy? Is not shielding same to repulsion?
reference - B.J. Duke (1978) Educ. Chem., vol. 15, p. 186.

Everyone who is going to answer in one or two line , don't waste your time because I am not as Intelligent and Smart as you

I think you are smart enough to do this research and the answer should be in two lines.

Now my questions is when we are talking about effective nuclear charge , haven't we already included the shielding factor which itself means repulsion that valence shell electron feels due to all other electrons in atom. So how author can say that he neglected the electron - electron repulsion while predicting ionization energy? Is not shielding same to repulsion?

$z_{eff}$ includes only the repulsion of the core electrons, it is the potential experienced by the valence electrons, the e-e repulsion of the valence electrons is not included in the effective charge.

Just read the example section any of these  articles https://en.wikipedia.org/wiki/Slater%27s_rules  or https://chem.libretexts.org/Courses/Mount_Royal_University/Chem_1201/Unit_2._Periodic_Properties_of_the_Elements/2.06%3A_Slater's_Rules  and tell me you are right or wrong.

You should ask yourself: Why do we need an effective nuclear charge including all the electrons of an atom if an atom is neutral ?

Have you ever  calculated effective nuclear charge mathematically using slater's rule

@M06-2x In context of Slater's coefficients and Z_eff, influence of valence electrons is involved by coefficient 0.35(0.30 for 1s
), in contrary to the core electrons with the coeffiecient 1.00(0.85). While the mutual valence electron influence cannot be easily identified as being subtracted from nucleus charge via interaction of spherically symmetric charges(densities), they still contribute to the repulsion and Z_eff decrease, what is reflected by the 0.30 coefficient (which does not distinguish repulsions between and within orbitals).

TL;DR; version:
Nucleus shielding is electron–electron repulsion that fits the two conditions:

Shielding electrons as the group have spherically symmetric probability density around the nucleus.
Their probability density do not (significantly) overlap with the density of shielded electrons.

Electron–electron repulsion that does not fit those conditions is not called shielding. The boundary is not sharp, as seen for n-1 level electrons shielding ns, np electrons.
In context of Slater's rules, repulsion by other electron in the same (Slater's) orbital group is expressed as the shielding equivalent, like if the electron were shielded by an extra 0.35 electron in core orbitals instead.

Every nucleus shielding is electron-electron repulsion, but not every electron-electron repulsion is nucleus shielding. The shielding covers mostly radial repulsion component of inner electrons, effectively subtracting from nucleus charge due spherical charge density geometry.
Core electrons can be formally added to the nucleus to approximate the total electrostatic force of a single effective nucleus change.  Electrons of the same orbital groups have this component diminished ( 0.30 versus 0.85 for n-1 vs 1.0 for n-2 in Slater coefficients), as their radial repulsion is part-time neutral, part-time even reversed.
E-E repulsion within the same orbital or orbital group covers mostly the tangential repulsion part, increasing the electron potential energy as well.

Shielding and electron-electron repulsion are not different interactions, but the different interaction geometry. They are distinguished because of two different ways how they are evaluated and processed.
Objects with spherically symmetric distribution of their mass respectively charge, affected by a distant quadratic force(*), behave if they had been point objects. This goes  back to Newton, reportedly one of reasons for him to invent integral calculus.
It applies to a spherically symmetric probability density as well, assuming there is no or negligible overlapping. Fully filled p, d or f orbital sets have this spherical symmetry. s orbitals are spherically symmetric natively.
Core, filled orbital sets with spherically symmetric density can be viewed formally as if the nucleus had been surrounded by a big, spherically symmetric negatively charged object of the charge and size of the orbital set. It can be than evaluated as their charge being subtracted from the true nucleus charge.
This is addressed by the shielding coefficient 1.00 in Slater coefficients, the electron charge being fully subtracted (0.85 for partial overlapping).
OTOH, electrons in valence orbital mutually repulse each other in such a way that cannot be accounted as shielding in the above sense. They mutually repulse themselves from all angles. This increases electron potential energy and is addressed by the effective shielding coefficient 0.35 resp. 0.30 AS IF such a charge had been added to the shielding core.
Note that the Slater's rules/coefficients do not distinguish cases of valence electrons being in different or same orbital.

(*) gravitational and electrostatic forces with $F \propto \frac{1}{r^2}$

please provide some references that shielding is radial repulsion and E-E repulsion is tangential.

Well, such references should be in the first place the part of your homework within the question formulation. Shielding is explained in textbook chapters of classical electrostatics. And repulsion which is not shielding as is not radial....

Can a post a photo of such explanation because I couldn't find it

Review search results of queries like site:stackexchange.com OR site:libretexts.org  nucleus shielding electrons

Friend, you must have read this explanation somewhere,  just tell me where you read that  ?  Please don't play with me . I am mentally exhausted.

please  answer @Poutnik

I will not answer until I see the documented explicit effort of searching, reviewing and analyzing available info. Do not expect others to answer if you do not show significant attempts (not shown indistinguishable from not existent) to find the answer before asking, or at least after asking. It would be otherwise like spoon-feeding. See 3 of 5 closing votes. // Review the guides Asking and How to ask to prevent misunderstanding,  need of clarification, objections, downvoting or closure.


text:
What the book means for "percentage difference"?
From Atkins-De Paula, Physical Chemistry, Ninth Edition, page 485

Not sure what the books means as "percentage difference".
Btw, this is my attempt
Fundamental vibration, in wavenumber, generic formula:
$\varepsilon_{\nu=1}-\varepsilon_{\nu=0} = \bar{\omega}$
$\bar{\omega} = \dfrac{1}{2\pi c} \sqrt{\dfrac{k}{\mu}}$
$\text{let }1) \ce{^{23}Na}\ce{^{35}Cl} \,\,\, \mathrm{and} \,\,\, 2) \ce{^{23}Na}\ce{^{37}Cl}$
$\mu_1 = 2.31 \times 10^{-26} \, \mathrm{kg}$
$\mu_2 = 2.37 \times 10^{-26} \, \mathrm{kg}$
$\text{percentage difference} = 100 - \dfrac{\omega_2}{\omega_1} = 100 - \dfrac{\sqrt{1/\mu_2}}{\sqrt{1/\mu_1}}$
According to what is reported in the book, the solution is 1.089 percent. Is this procedure right?

Percentage difference is the ordinary relative difference, just expressed in percentage. The only problem is what to take as the reference 100%? Value A, value B or their average ?  // BTW if both omegas are the same the percentage difference is hardly 99% as it is 0%.

4 is by 20% smaller (percentage difference) than 5. 5 is by 25% bigger than 4. They are respectively (100/9)%  smaller/bigger than 4.5 as the average of 4 and 5.

I formatted the text as you requested. Now, is my reasoning right?

Read again the second part of my second comment.

By the last equation line you say -- by other words -- that 5 is by 99% bigger/smaller than 5. That is not true, is it?

So, is my reasoning wrong?

Is 5 by 99% bigger than 5?

as @Poutnik has pointed out it is actually the relative difference that is meant in the question. This has to be a ratio as you realise,  but of the difference in $\omega$ because you do not know $k$  it has to cancel out. Make this then multiply by 100.

The relative percentage difference (RPD) of $b$ with respect to $a$, where the value $a$ is given or taken as the reference 100% value(typically if $b$ is deviation from $a$), is:
$$\text{RPD} = \frac{b-a}{a} \cdot 100 \%.$$
In your case:
$$\text{RPD} = \frac{\bar{\omega}_
2-\bar{\omega}_
1}{\bar{\omega}_
1} \cdot 100 \%.$$
In case the reference 100% value is not given, is unclear, or is explicitly said so, as the reference can be taken the arithmetic average of both values(typically if both have equal, parallel status). Then:
$$\text{RPD} = \frac{2(b-a)}{a+b} \cdot 100 \%$$
In your case:
$$\text{RPD} = \frac{2(\bar{\omega}_
2-\bar{\omega}_
1)}{\bar{\omega}_
1+\bar{\omega}_
2} \cdot 100 \%.$$

Regarding how exactly it was meant by the book, I just refer to one of my canned comments:
A asking B "What does C mean by X?" may not provide a satisfactory nor reliable answer. A should ask C.
The hint can be which of 3 possible options for the reference fits the given result the best.


text:
If a crystal has multiple oxidation states of the same transition metal element, is there only one possible configuration?
Here is the crystal structure of soluble Prussian Blue $\ce{KFe[Fe(CN)6]}$. The Fe(II) and F(III) are spaced alternatively like NaCl with cyanides in between. You can see that the places of Fe(II) and Fe(III) are not equivalent because one is attached to carbon while the other attached to nitrogen. This leaves some ambiguities because the places of Fe(II) and Fe(III) can be swapped without causing charge imbalance. Historically it was believed that Prussian Blue and Turnbull’s Blue are two different materials of the same atomic structure, but more rigorous studies confirmed that they are actually the same thing.

My question is whether it applies to other crystals containing transition metal elements. My guess is that there are unlikely counter examples because electrons are not confined to single atoms but shared among multiple atoms in solids, which makes it very easy for electrons to move among atoms to reach the most stable configuration. In another word, there should be only one stable orbital configuration which is either configuration A or B or somewhere in between like the orbital resonance in organic chemistry.

It is the default norm for any crystal to only have one possible configuration. Then again, there are things like the non-stoichiometric $\rm Fe_{0.95}O$; does that count?

This leaves some ambiguities because the places of Fe(II) and Fe(III) can be swapped without causing charge imbalance.

It sounds like iron Fe(II) and Fe(III) are free chemical species, they are just emergent species in the most stable configuration. The oxidation state is sometimes relevant, but remain hypothetical, the fluctuation of charge of an atom depends mainly on the neighboring environment. The ground state of Prussian Blue has an iron atom between two carbons and therefore loses less charge $\ce{Fe^2+}$ and the other is between two nitrogens more electronegative and loses more charge $\ce{Fe^3+}$, if the electronic structure of this crystal is analyzed carefully, the exact charge at each iron atom will never be $+2$ or $+3$, swapping iron atoms will not affect much the crystal, the charge will remain the same as there is only one stable configuration.
Here is an academic example that has important magnetic properties: lanthanum manganite : $\ce{LaMnO3}$, the crystal structure is peroskite and the manganese has the oxidation state $+3$. When it is doped with Sr, the structure becomes $\ce{La_{1-x}Sr_xMnO3}$, the environment changes and the manganese has either an oxidation state $+3$ or $+4$ depending on the presence of Sr. The exchange of two manganese atoms even with the partial charge will not always converge to the same structure with the same charge configuration. There is a double exchange in $\ce{La_{1-x}Sr_xMnO3}$, a very important effect of these oxidation states.


text:
If a crystal has alternating layers of different atoms, will it display different properties depending on which layer is exposed?
Many metal diboride compounds like $\ce{MgB2}$, $\ce{TiB2}$, and $\ce{ReB2}$ have laminated crystal structures with alternating sheets of metal atoms and boron atoms. If we polish the surface to ensure that only one type of atoms are facing outwards, will the two surfaces display different physical (such as friction coefficient, water contact angle) or chemical properties?

It would be rather exfoliation, like with graphene, not polishing.

Not sure if the specific effect you want to see would be easy, but in general, different crystal faces will show different atoms on the surface and are known to have different chemical properties.

Simple examples are the passivation of metals by oxide layers or antireflection layers on optical surfaces or an oil slick on a still water. or cell membrane structure. The finesse is getting the layers right.

A good question, but not sure it's possible, at least by polishing the given structure: if the boride layer is tightly bonded, I'd imagine it would be hard to remove such a layer, leaving the isolated metal sitting in the interstices.
However, there are ways to create a surface with a specific monolayer atop, and monatomic layers do have strong effects on the chemical and electronic properties of materials. For example,see Novel monatomic layer clusters for advanced catalysis materials and Electronic Structures of Polymorphic Layers of Borophane.  If you have access, see also Monatomic Two-Dimensional Layers,  Monoatomic Layer Electronics Constructed by Graphene and Boron Nitride Nanoribbons,

Undoubtedly yes. Giamello et al.[1] give an example involving a very subtle effect of this nature on a magnesium oxide (100) surface (meaning a square array of magnesium and oxygen atoms lying along the faces of cubic unut cells).
Molecular oxygen is adsorbed onto the magnesium ions and reduced to $\ce{O2^-}$ (through electron donation from the strongly basic oxide ions). If the surface layer is perfect, with no defects, then the superoxide ion is only loosely held to one magnesium ion and relatively mobile on the surface. But if there is an oxygen anion vacancy, the $\ce{O2^-}$ ion will be joined to all five surrounding magnesium ions and anchored tightly into the vacancy. The two different modes appear as different adsorption rnergies for atmospheric oxygen.
Reference

E. Giamello, D. Murphy, E. Garrone, A. Zecchina (1993). "Formation of superoxide ions upon oxygen adsorption on magnesium-doped magnesium oxide: An EPR investigation".
Spectrochimica Acta Part A: Molecular Spectroscopy,
Volume 49, Issue 9, Pages 1323-1330,
ISSN 0584-8539,
https://doi.org/10.1016/0584-8539(93)80040-H.

If you are looking for a less subtle effect---imagine cleaving NaCl along the (110) plane two ways: 1) Leaving leaving the surface as Cl; 2) Leaving the surface as Na.  Of course, such surfaces would be stable in a vacuum,

Sure!
The example most surface scientists should be familiar with is that of alpha silicon carbide or $\ce{\alpha-SiC}$.

Alpha silicon carbide (α-SiC) is the most commonly encountered polymorph, and is formed at temperatures greater than 1700 °C and has a hexagonal crystal structure (similar to Wurtzite).

It has a hexagonal crystal structure, and while the $[0001]$ face is terminated with silicon atoms, the $[000 \bar{1}]$ face is terminated with carbon.
Further, if you anneal in vacuum you can generate a layer or bilayer of graphene from the carbon-terminated side and use that as a substrate for growing additional hexagonal 2D materials like TMDs.
If you have a c-axis polished $\ce{\alpha-SiC}$ crystal, the challenge is then to tell which side is the "silicon side" and which is the "carbon". There are several "tricks" and just for an example here's a patent Method for distinguishing Si surface from C surface of SiC (silicon carbide) wafer CN103630708A
Image from Role of structure of C-terminated 4H-SiC($000 \bar{1}$) surface in growth of graphene layers: Transmission electron microscopy and density functional theory studies (also DOI:10.1103/PhysRevB.85.045426)

is it impossible for a crystal to form with two of the same side? how does one side know what the other is doing?

@user253751 Yes that drives me crazy too! I am not a chemist but here's my thinking. If we call the vertical direction in the images the c-axis, the tetragonal carbon (four bonds) has one bond parallel to the c-axis and the other three almost sideways to adjacent silicon atoms. The carbon can tolerate one "free" bond (that will connect to hydrogen or water or whatever sticks to it as soon as you take it out of vacuum) but it will not tolerate three "free" bonds.

@user253751 Carbon atoms left in that state will move around, cluster, form graphene or do something else, but the silicon can't hang on to a carbon with one bond. So the ""building blocks are the almost flat layer of carbon and silicon triply bonded to each other, each with a single dangling bond in opposite directions. That unit seems inseparable.

that makes sense; it wouldn't apply to all crystals, just this specific one

@user253751 yes exactly; there will probably be a whole family of materials that do this, but they will probably be hexagonal and have this kind of strong in-plane, weaker out-of-plane, buckled kind of structure.


text:
Why were alchemists trying to get gold particularly from mercury?
This question is on chemistry/physics/history.
The first (and arguably the easiest) nuclear transmutation producing gold involved mercury, its neighbor in the periodic table.
Obtaining gold from other metals, without access to nuclear technologies, by purely chemical means, is equally hopeless, independently of which starting ingredients are chosen.
And yet, for hundreds if not thousands of years, it has been mercury who was typically considered as a major ingredient for such attempts. It it a mere coincidence? Or is it because mercury has such weird properties (liquid at room temperature, poisonous) that it kinda makes sense to use it as a starting point for obtaining other metals, including gold? Do mercury and gold have any similar chemical properties? (which could have been observed in ancient times, leading to a conclusion that such a transmutation is possible)

The question is interesting but you will need to add more personal search efforts as seeds for useful answers. Did you search on Google Scholar or consult any book on Chemical History like Partington's? Search this title on Internet Archive as well.

I agree with @AChem about doing some more searching. But two things: mercury was not known to be poisonous long ago and it was fascinating to alchemists and Isaac Newton that you could put mercury and yellow sulfur powder in a mortar and pestle and grind them together to produce brilliant red cinnabar. It must have seemed like magic.

You may find this interesting: https://hsm.stackexchange.com/questions/12248/was-the-wide-use-of-mercury-in-experiments-in-the-19th-century-related-to-alchem

Interestingly the connection between gold and mercury may have been indirect but true - they are both "strange" due to relativistic effects (gold's colour, mercury's low melting point, and the chemical inertness of both). So in a sense, alchemists accidentally got it right with absolutely no knowledge of the deep underlying physics of relativistic quantum mechanics. Pretty amusing to consider!

"mercury was not known to be poisonous" ahaha as I kid I though I would literally die if I break that precious thermometer.

@mavzolej I assume you were not a kid in the era of alchemists.

I think the question is overall well formulated but yes, probably a better fit at History of Science and Math SE. I'd add that obviously the coincidence you mention is a coincidence in the sense that, while there are certainly correlations in the properties that allow nuclear transmutation and induced people to attempt alchemical transmutation, the alchemists didn't know a nucleus from a hill of beans.

Were they really particular about that? I mostly heard about transmutation of lead.

Good question. Here is a SWAG. Well, maybe not so scientific.

Gold (Au) is exceedingly dense, ~19 g/cm3, as is mercury (Hg), ~14 g/cm3, and lead (Pb) floats on Hg.

Both Au and Hg are highly reflective, but Pb quickly dulls.

Au is soft as butter, and the alchemists knew how to make it into sheets 100 nm thin. (One can continue the process until the gold is a transparent greenish, useful for filters but not for gilding.) Mercury is not merely soft, but liquid.

What might have been the biggest red herring is that Hg dropped onto Au dissolves into it in seconds (as I found when I got some on a ring)! At first, the Hg leaves a silvery-colored stain... but in a matter of days, the stain disappears, leaving apparently unadulterated gold (as I also found... BTW, the amalgam is safe, not releasing Hg vapor).
Using Archimedes' method to check purity of Au can fairly easily show if Au is adulterated with less dense Ag, ~10 g/cm3, but the more-dense Hg could be added in small amounts, appearing to increase the weight of gold without a *perceptible8 decrease in density.
A mountebank would think, "Eureka! Drop some mercury on gold, wait a few days, and I've made gold from mercury"... until some wise-guy heats the gold, distilling off the Hg (and slowly poisoning all in the assay office).

Although there are some good answers, I will mainly focus on the historical aspect.
Elemental mercury was known to the ancient Greeks, Romans, Chinese and Hindus and in the theory of alchemy, mercury was considered as one of the three principal substance (other being sulfur and salt) and every metal was made of these three principal substance in different ratios. However, the alchemists considered gold as the most perfect metal and other metals as "sick". So, they wanted to "cure"(convert) every metal to gold particularly mercury. Below are some excerpts from some articles:

Alchemy is a pre-scientific field of study that was widespread up until the Middle Ages, and focused on the nature of materials and the reactions between them. Alchemy combined philosophical beliefs, like the belief held by Greek philosopher Aristotle, according to which every material in nature strives for perfection. The most perfect material, according to alchemists, was gold. Besides its unique color, gold can withstand other materials and environments, and it is easy to weld and shape. Alchemists thought, like Aristotle, that other materials, such as lead and iron, are inferior and "sick" They believed these could be "cured" and turned into the coveted gold in a process called "transmutation".
https://davidson.weizmann.ac.il/en/online/orderoutofchaos/can-lead-be-turned-gold

In the ancient art of alchemy, mercury, sulfur, and salt were the Earth’s three principle substances. The Hindu word for alchemy is "Rasasiddhi", meaning "knowledge of mercury". Believing that mercury was at the core of all metals, alchemists supposed that gold, silver, copper, tin, lead and iron were all mixtures of mercury and other substances. While alchemists in different cultures had different beliefs, one of the central themes to European alchemy was the belief that the correct combination of mercury and other ingredients would yield riches of gold.
https://sites.dartmouth.edu/toxmetal/mercury/mercury-element-of-the-ancients/

You can go also go through the below links articles for an indepth discussion on this topic.

Simioli, Carmen. "Alchemical Gold and the Pursuit of the Mercurial Elixir", Asian Medicine 8, 1 (2013): 41-74, DOI: 10.1163/15734218-12341289
https://timesofindia.indiatimes.com/home/sunday-times/deep-focus/the-curious-case-of-an-experiment-with-alchemy/articleshow/3089537.cms

I wonder they didn't put mercury on the pedestal instead of gold despite the fact that mercury can form amalgam with so many metals meaning it should be kind of superior in its ability to dissolve other metals including silver and gold.

I will give a naive and personal answer without any source, this question surely belongs to another SE site, I don't know much about alchemy, the use of mercury could be related to their ideology and a financial reason even if this attempt is totally absurd.
If the process of obtaining heavy elements in supernova nucleosynthesis would have required a significant amount of energy, some of them are really abundant on earth compared to light elements and are very cheap. There is a wiki table giving the Price and abundance of chemical elements (an approximate value can be interpolated for the price in the past). Lead is one of the cheapest and is an abondant element in the row containing gold. Pb is literally is inferior and a best choice to be transformed into gold. Nowadays the position of Pb itself three steps away from gold in the table might also be the reason of this choice. The transmutation of lead into gold was certainly a challenge. A wiser process will be to make a transmutation of another cheaper element very close to gold for an easier and historic transmutation.  If this is achieved, a series of transmutations can be made Pb $\rightarrow$ Tl $\rightarrow$ Hg $\rightarrow$ Au. It's the likely explanation for this choice.
There is certainly another way to say left to right, but Os, Ir and Platinum are rare and expensive, such a success wouldn't make much difference, which explains the right-to-left tendency (heavy to lighter elements) in the transmutation process. At least if they'd known about another similar quasi alchemical process : cold fusion, they might have tried it with light metals.

The alchemists long predated notions of a periodic table and had no notion that lead, mercury and gold were “neighbors”. They were just dense metals known from antiquity. Thallium was spectroscopically discovered much later. The nucleosynthesis stuff had no relevance.

Although your thoughts are appreciated but I will have agree with EdV that this answer does not address the original question at all.

Answers to historical questions may require reasonable efforts in a library. The OP has to make an effort given the fact that plenty of historical chemistry books are available on Internet Archive free of cost!