text:
Why is measurement in molarity preferred over molality?
NCERT Chemistry Part 1 states that:
"The osmotic pressure method has the advantage over other methods as pressure measurement is around the room temperature and the molarity of the solution is used over molality."
I do not understand how the second statement is really an advantage. Infact, if I had to guess, I would choose molality to be the easier of the two to measure as molarity might involve change in volume upon mixing.

"My textbook" says...., always cite the name of textbook. Otherwise it's considered as a case of plagiarism.

@Zenix I am pretty sure the authors have better jobs than to go around suing 17 year olds for posting a question online which quotes their book, but okay

I agree with you, but mentioning the book is $useful$ for other readers.

Molality is used for those people trying to understand how the melting point or the boiling point of water or any other substance, changes when they add a given amount of salt or any other impurity to one kilogram of this pure substance. They don't care about the volume of the obtained solution. They will not keep the solution once obtained. As an example, they just want to check if it is correct to say that adding $1$ mole of a solute to $1$ kg water lowers the freezing point by $1.86°$C, whatever the nature of the solute.
Molarity is used for another group of people. They don't mind about the process of dissolution of a substance in a solvant. They don't care about the volume of water to be used to make $1$ liter of the solution. They just want to know the composition of the solution they have to use, i.e. how many moles (or grams) of a solute are present in a given volume of this solution. If for a given use, they must have a solution containing exactly $9$ g NaCl per liter, they don't care about the way of dissolving the salt. All they need to know is whether this solution can be used for making an injection, or any other use.

when measuring concentration in molarity, do we only take the volume of the solvent before adding the solute? do we not care about any volume change which might occur upon adding the solute?

Also molarity changes with temperature, while molality don't. Text cited by OP seems to be ambiguous.

When preparing a solution with a molarity precisely known,  the solid stuff is introduced in a volumetric flask of say 100 mL or 250 mL. Some water is then introduced to about half fill the flask. The flask is stirred until the powder is dissolved. Then water is gradually added up to the graduation limit, so that the total volume of the solution  is exactly known. Nobody cares about the exact volume of water to be added. The only important thing is the final volume, which must be known with precision..


text:
Does storing apples alongside potatoes keep potatoes from sprouting?
I am looking for a chemical explanation which I wasn't able to find.
I have read in several sources such as this one that, in order to prevent potatoes from sprouting, it is sufficient to store a few apples alongside them. This is due to the fact that the apples produce ethylene.
But this is a bit odd, since the ethylene is typically a ripening agent. This reddit thread also discusses the same doubt, but doesn't dive much into the details. There is also this article and this Quora answer, none of which give a specific chemical reason about why and how this works.
Chemically speaking, do we actually know the exact reason and the necessary conditions to make this work?

Where I am the conventional wisdom is to not store apples and potatoes together (nor potatoes and onions) and ethylene is given as reason. But then, this conventional wisdom is not about storing potatoes for a few weeks but for months, so that you'd get from the first freezing in fall till early potatoes are ripe (they will be shriveled by then and also sprouting, but not rotting).

Ethylene only makes fruit that is very nearly ripe take the last step. The potatoe is not a fruit (the fruit grows above the earth and is slightly poisonous), so any possible effect would have to be sth different.

@Karl: there are also so-called ethylene-sensitive vegetables, e.g. cauliflower or red cabbage - neither of which is a fruit. Have a loo e.g. at https://unitedworldtransportation.com/ethylene-and-produce-sensitivity/ or if you read German: https://www.lebensmittellexikon.de/e0002410.php

I sometimes down-vote accidentally because the buttons are too small on my phone. Once you make that mistake, you can't undo it until the question gets edited. In any case, the majority thinks it is a good question, and you are getting credit for that. I wouldn't worry about down-votes.

@KarstenTheis : Thanks. I think they've implemented that hand-waving sign under the username for a reason. Most likely because they thought downvoting a new contributor's post without giving a single explanation is typically interpreted as being mean.

Ethylene has been used in storing potatoes for a while, see this article. In it, they address the puzzle of a ripening agent acting to prevent sprouting:

Sprout suppressant properties   of   apple   ‘vapours’   were   first   reported   in   1932   and   subsequently,  ethylene  and  apple  ‘vapours’  in  1933  (Elmer  and  Huelin,  respectively).    This  was  preceeded,  however,  by  work  demonstrating  a  ‘dormancy  breaking’  effect  of  ethylene  (Rosa,  1925).    These  apparently  conflicting  responses  to  ethylene  were  resolved  with  the  publication of work showing both effects, but under different ethylene exposure conditions.

The mechanism of the different effects are largely unknown. The cited article (by the British Potato Council) gives some references. Potatoes treated with ethylene do have slightly different properties (they mention the color after deep frying), and there is ongoing research to get an overview of all the differences.

[OP:] Chemically speaking, do we actually know the exact reason and the necessary conditions to make this work?

We don't know the exact reason, but there are companies who figured out how to make this work in a warehouse under controlled conditions. I think on your shelf would be more difficult because you have no way of monitoring the ethylene concentration, so the apples might influence the potatoes in etiher way.

So the answer to the last sentence in my question is NO?

@polfosol - I updated my answer. The answer is no for the first part (mechanism is largely unknown), and yes for the second (conditions are known at industrial scale). Thanks for asking the question - fascinating stuff!

Great, +1. Another substance that is used to inhibit sprouting in potatoes is S-carvone. So maybe we could try to put the caraway on the potatoes long before cooking ;-) see e.g. https://dx.doi.org/10.1007/s10847-006-9210-2 (they used both S-carvone and caraway seeds (and some other treatments), and also caraway seeds inhibited sprouting - applying S-carvone directly was more effective in one experiment, but lead to rot in another)

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text:
Hyperconjugation, stability of molecules, and heat of formation
I recently asked a question about why the heat of formation of organic radicals and positive ions decreases with their size and degree of branching at the radical or ionic site. The user "Buttonwood" made the comment that hyperconjugation is one of the forms stabilising cations and radicals.
This begs two questions:

What is the reasoning behind why hyperconjugation increases the stability of cations and radicals, and therefore decreases the heat of formation? Point 3. of this section of the Wikipedia article for hyperconjugation  says that "the heat of formation of molecules with hyperconjugation are greater than sum of their bond energies and the heats of hydrogenation per double bond are less than the heat of hydrogenation of ethylene."; does this not imply that hyperconjugation increases the heat of formation of molecules?
Why does an increase in stability imply a decrease in heat of formation? To a novice who is ignorant on this subject such as myself, it would intuitively seem that greater stability implies an increase in heat of formation, since the molecule is more "stable" in its current form and therefore it takes more energy to affect changes in its structure?

I would greatly appreciate it if people would please take the time to clarify these points.

Hyperconjugation is an electronic effect which is similar to resonance (one of the most fundamental concepts of organic chemistry). Here's how it works:
$$\ce{H3C-CH2+ <-> H+H2C=CH2}$$
You can write the above structure three times for each hydrogen replaced. Note that only α-hydrogens (on the adjacent sp3 carbon atom) can be replaced.

Now, answering your first question, hyperconjugation leads to stability because of the delocalisation of charge. The positive charge gets diffused away from the carbon atom, and starts to occupy a larger volume, decreasing the overall energy of the system caused by electronic repulsions. As a matter of fact, the greater the number of structures you can draw due to hyperconjugation, the greater the stabilisation. That explains why tertiary carbocations are much more stable than primary ones.
$$\ce{(H3C)3C^+ <-> H+H2C=C(CH3)2}$$
If you look closely there are nine replaceable hydrogens, so you can draw nine structures (explaining its stability).
The Wikipedia page says:

the heat of formation of molecules with hyperconjugation are greater than the sum of their bond energies

Here, heat of formation refers to the heat released on the formation of these molecules. So it is negative.
In the absence of hyperconjugation, you would expect that the heat released on formation is equal to the sum of all bond energies. But because of stabilisation due to hyperconjugation, the product molecule has lower energy than expected, so overall, more heat is released (higher magnitude enthalpy of formation).
It's important to understand that enthalpy of formation (including its sign) decreases on increasing the stability of the molecule no matter what. But in the case that the enthalpy of formation is positive (energy absorbed), a decrease in the magnitude of heat of formation is observed, while if the enthalpy of formation is negative (energy released), an increase in magnitude is observed. In your other post, you include a list of carbocations, and their heats of formation are seen to be positive, while those of the non-ionic species are negative. So all the ionic species need to absorb heat in order to form (indicating a loss in stability), but that loss in stability (and hence the heat of formation) would be less if they were stabilised due to hyperconjugation.

[...] the greater the number of structures you can draw due to hyperconjugation, the greater the stabilisation. That is only true for very well-behaved molecules and as a zeroth order approximation. The contribution of different structures can be quite different and the overall superposition of all of them will only produce a good approximation.

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If the activity of a solid is one why does adding salt to water increase the salt ion concentration?
Suppose salt dissolves in water according to the equation NaCl(s) -> Na(aq) + Cl(aq).  Since the activity of solid salt is 1, adding salt shouldn't affect sodium and chloride ions at equilibrium.  However, if I were to add salt to an unsaturated solution, the salt would dissolve and increase the ion concentrations.  How is it possible for the activity of solid salt to be 1 if it affects equilibrium in this way?  My only theory is that an unsaturated solution is not at equilibrium, but I'm not sure this is correct.

The activity of a solid stuff is 1, and it is independent on the presence of any more or less concentrated solution.

A system of unsaturated solution and solid salt is indeed out of equilibrium.

Salt always dissolves, no matter if the solution is saturated or not. When the solution is saturated, however, the rate of salt dissolving and salt precipitating is equal, so the concentration does not change anymore. In fact, when you add salt to a saturated solution, the rate of dissolution goes up, as does the rate of precipitation.

Yes, you are correct: an unsaturated solution is not in equilibrium. Equilibrium is the state when rate of dissolution is equal to rate of crystallisation. But in an unsaturated solution there is only dissolution of salt in the water and no crystallisation back to salt.

Not exactly. Dissolution is just faster than crystallisation,leading to the net dissolution.

That is too ideal for a practical idea like this. Going by idealism like this couldn't it also be said that activity of liquids is not unity.

You may forget about 2 principal things. 1/ Equilibriums are on molecular level dynamic, not static. There are always the opposite processes, that just at equilibrium happen ongoing at the same rate.2/ molecules have no clue if the system is in equilibrium or not. There are no traffic lights saying now you may  dissolve, or now you may crystallize. Similarly, the thermal energy goes always from colder to warmer object as well. Just at lower rater then in the opposite direction.


text:
Calculate the lattice energy of ammonium carbonate
The lattice energy is a measure of the stability of a compound and I need this data for the ammonium carbonate ($\ce{(NH4)2CO3}$) but I don't know how to calculate and I can't find it anywhere. I know that it can be calculated with the Born-Lande equation but I don't know how to apply this equation to solve my problem.
I wonder if anybody knows a database where I can find this value, I need it for an important chemistry assessment.
Thank you for your help

The lattice energy is not directly measurable, and it is difficult to calculate. In a paper about estimating lattice energies, the estimation for $\ce{Na2CO3}$ is 30% off from the reference value (which is from CRC Handbook of Chemistry and Physics). I mention sodium carbonate because ammonium carbonate is not one of their examples. The introduction explains why it is not possible to directly measure the lattice energy:

Lattice  potential  energy  (UPOT)  is  a  dominant  term  in  the thermodynamic analysis of the existence and stability of ionic solids.  Direct  experimental  determination  is  generally  not possible since, in practice, the crystalline solid dissociates into atoms  and not into  gaseous  ions,  as  is  required  in  the  lattice energy  evaluation.  Therefore,  its  indirect  experimental  determination, computation, or estimation is of considerable interest in modern materials science; indeed, whenever the energetics of condensed-state materials are studied, the chemical processes under consideration may be rationalized if the appropriate lattice energy steps can be incorporated into the thermochemical cycle.

They go one to discuss how you can obtain the lattice energy from experimental or theoretical data:

A  variety  of  estimation  methods  for  lattice  energies  is vavailable. These include the Born-Haber-Fajans thermochemical cycle (which requires ancillary thermodynamic data) and modern computational methods (which generally require knowledge of the lattice constants and the coordinates of the ions as well as needing an established force field). The computational methods  range  from  direct  energy  calculational  procedures through to programs that produce lattice energies in the course of their modeling of the solid. Quantum mechanical procedures are also available, but are highly computationally expensive and are generally applied only to the simpler systems.

Then they talk about cheaper methods that are less accurate. The paper is about a new quick estimation methods.

By contrast,  a  few  exceedingly  rapid  estimation  methods based on ionic radii or volume per formula unit (“molecular”volume) have  been  developed.  Originally  only  applicable  to the simpler binary ionic solids (such as MX (with a 1:1 charge ratio)), the volume-based approach has recently been extended by  us to  include  MX2(2:1)  and  M2X  (1:2))  salts.

So check the CRC Handbook of Chemistry and Physics.

Just to comment that in the 20 years since the paper computers and computational methods have come on considerably - it should be better to do a somewhat better job nowadays. This is easily in the range of ab initio packages, though I admit don;t know how to extract the lattice energy term from the calculated total energy.

@IanBush So you could calculate it with periodic boundary conditions, but if you use a spherical cut-off for the electrostatic term (assuming you don't do ab initio for the entire infinite lattice), you might end up with an answer that depends chaotically on the cutoff radius, see Madelung constant.

Yes, I know about Ewald summations (e.g.https://www.sciencedirect.com/science/article/pii/S0010465506001901) Any respectable condensed matter code will do this correctly, be it ab initio or not, as otherwise the answers are wrong - as you say.

The preview of this paper, DECOMPOSITION OF AMMONIUM CARBONATE AND AMMONIUM BICARBONATE AND PROTON AFFINITIES OF THE ANIONS, by J. E. House, Jr. cites the use of lattice energies for ammonium carbonate as an aid to determining proton affinities.
The full work may provide some data.


text:
Why do we write electrode potentials next to chemical reactions?
We might have a half cell consisting of the redox couple $\ce{Ag+}/\ce{Ag}$, which for example's sake might be fixed to be at the cathode (i.e. undergoing reduction).
The reaction at the cathode is then
$$\ce{Ag+ + e- -> Ag}$$
The standard electrode potential of the $\ce{Ag+}/\ce{Ag}$ couple is $+0.80\ \mathrm V$ with respect to SHE. I've seen the above reaction written as such:
$$\ce{Ag+ + e- <=> Ag} \quad E = +0.80\ \mathrm V$$
I'm confused as to why we write this electrode potential next to the reaction, considering that the electrode potential of a redox couple is reasonably disjoint to the concept of the chemical reaction. Although, at the end of the day, the chemical reaction results in the interfacial potential difference which causes the electrode potential, $E$ is a property of the electrode, not the chemical reaction.
Is it perhaps just because we want to write the equation of the reaction occurring at a certain electrode and the electrode potential of the electrode on the same line?
I'm aware that for associating potentials with written reactions, there exist two different conventions: European, where all reactions are labelled with the reduction potentials of the couple, and American, where oxidation reactions are labelled with the oxidation potentials of the redox couple and likewise reductions with reduction potentials. However, my question is why we need to attach the electrode potential to the chemical reaction in the first place!
That is, ultimately, aren't redox potentials associated with redox couples/electrodes, not reactions?

Your question is a valid question, and ignore downvotes. They don't mean anything.
Your understanding is very good and that you realized that the electrode potential is a property of the electrode and it really does not care how the reaction is written. However, a equation is $needed$ to keep track of the electrons lost or gained in the Nernst equation. So as a tradition and a matter of convenience, electrode potentials are quoted along with a balanced half-cell.
Another very important thing to remember (even some university teachers and PhDs lack this understanding), that the sign of the electrode potential in invariant. It does not care how the reaction is written. It is the electrostatic sign of the electrode. This part takes care of the European and American conventions. American convention is due to Gibbs and Latimer and no longer used. Most international electrochemists have agreed to stick to the European convention i.e., write all half reactions as reduction.

Thank you for your answer, this clears up my doubts. I also agree that the American convention for electrode potential is slightly nonsensical, since it makes little sense to label oxidation reactions with oxidation potentials - as you say, why not just stick with the electrostatic sign...! That is, even if the reaction proceeds as an oxidation at a particular electrode, we may as well just write the oxidation equation (as usual) and label it with a standard electrode potential, representing the electrostatic potential of that electrode - not its negative!

Having been taught the the American convention it does not seem nonsensical. It is simply based on the fact that if the sum of the the oxidation potential and the reduction potential is positive, then the reaction is spontaneous. It is easy to tell if the reaction is an oxidation or reaction based on which side the electrons are shown in the half-cell reaction. I'll point ot that even in the American convention that all the tables that I have seen are compiled for the reduction potentials. // Having said all of that there is logic in teaching/learning one convention.

The main difference in American and European convention is that Gibbs and Latimer treat the electrode potential as a thermodynamic quantity where one can flip flop signs, but the European convention stuck to the sign invariance because they took the plus minus as the sign of electrostatics. It is a very subtle difference but it took ages to settle.

@JamesWirth, If you don't mind, what grade are you in? I am surprised that modern students are aware of the American vs. European conventions. Are they still teaching them? Not that I am very old but I cannot recall seeing this in modern textbooks. I learned about the because of my own interest in chemical history.

I'm a senior but my textbook teaches the American convention. It caused quite a lot of confusion because, from learning basic circuit theory, I didn't understand why they were negating electric potentials. So I did some research and found out about the two conventions, which helped a little in terms of rationalising the intentions of the author. Of course, the two methods are algebraically equivalent, but IMO one is more intuitive than the other.

@MaxW I see your point, it is perhaps more convenient (even moreso if dealing with lots of half equations in one calculation) to sum reduction and oxidation potentials, as it could be argued there is less chance for error. But I think it should be viewed as a purely algebraic (but valid!) abstraction, since it doesn't have a physical interpretation. Unlike in the European convention where we're essentially finding the potential difference by calculating the difference in the electrode potentials!

@JamesWirth, Could you edit the question and share the exact part as to how the textbook explains the American convention?


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Why doesn't laser ablation decompose sample
I was reading up on LA-ICP-MS and MALDI. All the sources I read for LA-ICP-MS simply say that when the laser hits the sample immediately goes into the gas phase. Why doesn't the laser immediately break fragile molecules?
MALDI makes sense to me. You put the sample in a matrix that absorbs UV really well then hit the matrix-sample mix with a bunch of UV. The matrix explodes essentially, pushing the sample into the gas phase. The key here is that most of the energy of the laser doesn't go into the sample. It goes into the matrix. This is why the sample doesn't break down. LA-ICP-MS doesn't have the matrix to protect the sample so what gives?

Actually, the ICP will destroy any molecules that the laser ablation did not destroy. The LA just gets the sample to the awaiting ICP inferno.

Thank you! That helped a lot. What about LAESI? This technique still has the laser ablation but it doesnt seem like it completely destroys the molecule.

I have not used LAESI, but I would assume the LA in that is deliberately relatively gentle. Hopefully, you will get a solid answer here!


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Reaction in Zinc-Carbon-Ammonium Chloride battery
I made a dry cell battery from zinc (Zn), carbon (C) and ammonium chloride (NH4Cl) which produced a current of 0.4 amps and voltage of 1.1V. However when I could not find the redox reaction that was occurring.
What is the chemical reaction happening in the cell?

If you put litmus paper near the carbon electrode, when your voltaic cell is in operation, does the litmus paper turn blue? In other words, is the electrolyte basic near the carbon cathode?

Surprisingly this battery uses Zn without any oxidant like $\ce{MnO_2}$ and it  produces a voltage as high as $1.1$ V. In order to establish its characteristics, the two compartments should be studied independently.
First, the anode, with zinc. $\ce{Zn}$ should not produce simple $\ce{Zn^{2+}}$ ions, as there are ammonia and ammonium chloride in the solution. So the anode reaction is probably :  $$\ce{Zn  + 4 NH_4^+ ->  [Zn(NH_3)_4]^{2+}  +  4 H+  +  2 e-}\label{rxn:zn4}\tag{1}$$ The reduction potential for this anodic equation is not given in the tables. But it may be obtained from the next equation, whose standard reduction potential is known to be $\ce{E_2° = - 1.03 V}$.  $$\ce{Zn  + 4 NH3 ->  [Zn(NH_3)_4]^{2+}  +  2 e-}\label{rxn:zn}\tag{2}$$  In a $1$ M solution of $\ce{NH4Cl}$, $\ce{[NH_3]}$ = $2.4 ·10^{-5}$ M. So the corresponding reduction potential of ($2$) is : $$\ce{E_2 = - 1.03 \mathrm{V} + 0.0296\mathrm{V}· log (2.4 ·10^{-5})^4} = - 1.03~ \mathrm{V} - 0.54~ \mathrm{V}  = -1.57 ~\mathrm{V}\label{rxn:nh}\tag{3}$$
Now the cathodic reaction should be studied. Suppose the cathode reaction is  : $$\ce{2 H2O  + 2 e- -> 2 OH- + H2}\label{rxn:z}\tag{4}$$ with a redox potential equal to  $- 0.83$ V. Couping these two expressions ($2$) and ($3$) are never sufficient to get $-1.1$ V as the overall voltage. So, as the ammonium chloride solution is acidic, the cathode reaction must rather be : $$\ce{2 H+ + 2 e- ->  H2}\label{rxn:znn}\tag{5}$$ In a $1$ M  $\ce{NH4Cl}$ solution, the pH is $4.62$, so the corresponding potential of the cathode is E$\ce{_5}$ = - $0.059$ pH = - $0.059·4.62$ V = - $0.27$ V.
To summarize, the anode reaction is ($1$) or ($2$), at will, and the cathode rection is ($5$). As a consequence, the theoretical voltage of the cell is : $\ce{E_5}$ - $\ce{E_2}$ = - $0.27$ V + $1.57$ V = $1.3$ V, if the concentration of ammonium chloride is $1$ M.
Ref.: Standard electrode potentials have been taken from "Chemistry Data Book" from J. G. Stark and H. G. Wallace, John Murray, London, $1994$, Table $43$, p. $70$.

Per a source, the chemical reactions in a zinc–carbon battery can be detailed as follows, to quote:

In a zinc–carbon dry cell, the outer zinc container is the negatively charged terminal. The zinc is oxidised by the charge carrier, chloride (Cl−) via the following half reactions:
Anode (oxidation reaction, marked −)
$\ce{Zn + 2 Cl− → ZnCl2 + 2 e−}$
Cathode (reduction reaction, marked +)
$\ce{2 MnO2 + 2 NH4Cl + H2O + 2 e- → Mn2O3 + 2 NH4OH + 2 Cl−}$
Other side reactions are possible, but the overall reaction in a zinc–carbon cell can be represented as
$\ce{Zn + 2 MnO2 + 2 NH4Cl + H2O → ZnCl2 + Mn2O3 + 2 NH4OH}$

The battery cell is claimed to have an electromotive force (e.m.f.) of about 1.5 V. Note, personally above, I would write NH4OH as NH3(aq).

The cell that was made doesn't have MnO2 in it.

Anyway, $\ce{ NH_4OH}$ does not exist. And I would write the anodic reaction simply as $\ce{Zn -> Zn^{2+} + 2 e^-}$. Further more I would not introduce the molecule $\ce{NH_4Cl}$, but $\ce{NH_4^+}$ ion.

I just made this battery and then found this question here and wanted to add to the discussion.
My battery uses a stainless steel electrode rather than the carbon used by the original author.
The voltage and current both drop quickly in conjunction with the build up of hydrogen bubbles on the stainless steel electrode.
I found that voltage and current dropped when a small load was connected.  The current and voltage output (while loaded) was increased dramatically if I added a small amount of hydrogen peroxide to the electrolyte.
This seems to depolarize the stainless steel stopping the formation of the hydrogen bubbles.
Cheers
Edit Aug 8
The battery has been connected to some LEDs for the last 3 days. The output drops after a day so H2O2 has been topped up each day by adding 2 ml of 30% H2O2.
What is this white precipitate that is forming?  I am guessing ZnO?  I was expecting the Zn and the Cl from the NH4Cl to react which would have been soluble and that the H2O2 would be depolarizing the stainless electrode by preventing H2 gas bubbles from forming.  Now I don't have a clue what the reactions are in this battery.
Does anyone have any idea?


text:
Total spin for lithium atom
Why does the ground state of a Li atom have a total spin of S = 1/2? I get that the total z component of the spin has to be either Ms = 1/2 or Ms = -1/2 because of the Pauli exclusion principle, but that could also correspond to a total spin of S = 3/2, so why is it always S = 1/2?

If it was really $S = 3/2$, then there would be a $M_S = +3/2$ state, but there isn't.

But it doesn't have to, right? I mean, you could have a state with S = 3/2, Ms = +1/2 and it would be a valid state

It does have to. Anything with $S = 3/2$ will necessarily comprise four different states with $M_S = +3/2, +1/2, -1/2, -3/2$. You can't have some of them allowed and some of them forbidden by the Pauli exclusion principle.

I see. Is that because the spin wave function must be either symmetric or antisymmetric?

Not quite, that's just how angular momentum works, and it isn't peculiar to spin angular momentum. The symmetrisation thing (we might as well focus on "antisymmetric" only, since we're talking about electrons here, i.e. fermions) is basically the same thing as the Pauli principle.

I don't understand that. For example, you can see that, in converting a state from the uncoupled basis to the coupled basis via the Clebsch-Gordan cofficients, you end up with a linear combination of eigenstates with the same Ms but differents S, i.e. not all Ms appear like you imply

A state (ok, collection of states) with $S = 3/2$ is four states with $M_S = +3/2, +1/2, -1/2, -3/2$. If you're just transforming one arbitrary (uncoupled) state to the coupled representation, then you shouldn't expect to get a nonzero coefficient for every value of $M_S$.

Probably the easier way to explain it is to rely on the fact that if Li indeed had a ground state with $S = 3/2$, this would necessarily comprise four levels with $M_S = +3/2, +1/2, -1/2, -3/2$. This sort of splitting can be observed under a magnetic field, for example (the Zeeman effect).
Note that it's not possible for the Pauli principle to forbid $M_S = +3/2$ and $-3/2$ while simultaneously allowing $M_S = +1/2$ and $-1/2$. The Pauli principle doesn't mean that "this state can exist, but the electrons can't reach that state". It means that "this state simply cannot exist": because any state that violates the Pauli principle is equal to zero, which isn't a valid state. So it's not as if the $M_S = \pm 3/2$ states are there, and that the electrons simply choose to go into the $M_S = \pm 1/2$ states. The $\pm 3/2$ states are nonexistent. And consequently, because a $S = 3/2$ state necessitates the existence of $M_S = \pm 3/2$ states, it cannot exist either.
However, since that line of argument doesn't seem to fully convince you, you may want to go through the "exercise" of proving it. Unless I'm terribly wrong, this should be a completely foolproof (albeit terribly tedious) method. You can construct the full wavefunction for the Li atom as a Slater determinant (without loss of generality we choose $\alpha$ for the third spin):
$$\Psi = \frac{1}{\sqrt{6}} \begin{vmatrix}
1s(1)\alpha(1) & 1s(1)\beta(1) & 2s(1)\alpha(1) \\
1s(2)\alpha(2) & 1s(2)\beta(2) & 2s(2)\alpha(2) \\
1s(3)\alpha(3) & 1s(2)\beta(3) & 2s(2)\alpha(3) \\
\end{vmatrix}$$
so "all" we need to do is to apply $\hat{S}^2$ to this wavefunction and see what eigenvalue it returns. We know that
$$\hat{S}^2|\Psi\rangle = S(S+1)|\Psi\rangle$$
and so if $S = 1/2$ then the eigenvalue will be $3/4$; and if $S = 3/2$ the eigenvalue will be $15/4$. The $\hat{S}^2$ operator can be expanded as
$$\begin{align}
\hat{S}^2 &= (\hat{S}_1 + \hat{S}_2 + \hat{S}_3)^2 \\
&= \hat{S}_1^2 + \hat{S}_2^2 + \hat{S}_3^2 + 2(\hat{S}_1 \cdot \hat{S}_2) + 2(\hat{S}_1 \cdot \hat{S}_3) + 2(\hat{S}_2 \cdot \hat{S}_3) \\
&= \hat{S}_1^2 + \hat{S}_2^2 + \hat{S}_3^2 + 2(\hat{S}_{1x} \hat{S}_{2x} + \hat{S}_{1y}\hat{S}_{2y} + \hat{S}_{1z}\hat{S}_{2z}) \\
&\quad + 2(\hat{S}_{1x} \hat{S}_{3x} + \hat{S}_{1y}\hat{S}_{3y} + \hat{S}_{1z}\hat{S}_{3z}) + 2(\hat{S}_{2x} \hat{S}_{3x} + \hat{S}_{2y}\hat{S}_{3y} + \hat{S}_{2z}\hat{S}_{3z})
\end{align}$$
We already know what the individual operators $\hat{S}_i^2$, $\hat{S}_{ix}$, $\hat{S}_{iy}$, and $\hat{S}_{iz}$ do to the basis states $|\alpha(i)\rangle$ and $|\beta(i)\rangle$. Bearing in mind that operators don't act on different degrees of freedom, you can work out what this monster operator does to the Li wavefunction.
If you look online, there are examples of how to perform this calculation for a two-spin system (e.g. "prove that the singlet and triplet wavefunctions are eigenstates of $\hat{S}^2$"). The extension to a three-spin system is not difficult, but is extremely tedious, which is why I gave up on writing it out in full. But I am fairly certain that at the end of the day you will find the eigenvalue of $3/4$.

In terms of how the electrons are arranged*, the reason you get only spin $1/2$ is because the most stable configuration has two of the electrons in the same 1s orbital, which by the Pauli Exclusion forces them to have opposite spins.  This leaves only the third electron, sitting in the 2s orbital, to contribute a net spin.
You could get a spin of $3/2$ in an excited state where one of the electrons is promoted from the 1s orbital to a higher, otherwise empty one such as a 2p orbital.  Then with all three electrons you may have them all with the same direction of spin leading to a total spin of $3/2$.  However, in the case of lithium (and other alkali metals) the energy required to do this is much greater than that required to ionize the outer electron from the ground state, so you are more likely to get ionization rather than excitation to a state with spin $3/2$.
*I assume you mean electronic spin.  The nucleus also has it's own spin, but this is often ignored by chemists because nuclear spin does not couple into chemical processes.


text:
Will galvanic corrosion happen on a ring made of gold and bronze?
I have recently got a custom-made ring from a jeweler. It’s made from a byzantine-period bronze ring enveloped in a gold bangle. The bronze ring is “glued” onto the gold ring through some kind of resin, as far as he told me.
Would bimetalism with rhe corrosion of the bronze ring happen?
Thank you!

Moist air, rich in an electrolyte (salt particles) or human contact, providing both NaCl and H+ may supply the reagents needed for galvanic corrosion, with dissimilar metals in direct contact. This usually proceeds, albeit slowly, over time.
Note, exposure to fruit juices could be especially problematic, resulting in a matter of days of continuous contact in the presence of moisture and oxygen, in some visible signs of disfiguring pit corrosion.
Nevertheless, I recommend wearing the ring and enjoying it, but perhaps placing it in fresh distilled water (remove O2 exposure) at night after washing it with mildly alkaline soap (to clean off salts and neutralize body acids). If the vessel use is made of Aluminum, a very active anodic metal, the ring should do just fine.

I will try to place it in distilled water at night, as i really care about this ring. But why should the vessel be made of aluminum?

@DanteAndrei Well, you're worried about galvanic corrosion right? Aluminum is a metal too. Bronze loses to gold, but aluminum loses to bronze. It's a meat-shield for the bronze against the gold, so to speak. But aluminum develops a passivation layer which is why aluminum doesn't rust and passivation layers tend to do just that, prevent the material from reacting so I am not sure how effective it would be.

But I would actually recommend baby oil/mineral oil instead of distilled water Distilled water isn't conductive until it has solutes in it which your ring will introduce.

If it is maintained in air, no corrosion will happen. If it is maintained in water, and specially in salty water, the bronze part of the ring may be oxidized and will darken. Golden parts will not be modified.

So it will only darken? It won’t get literally eaten away?
If this is the case, then i’m really happy! Will keep it away from water as much as i can.

Yes. It will only darken with the time. And if you want to be sure that no darkening happens at all, you may cover it with a colorless varnish like the one used by pretty girls to make their nails brighter.

Haha! I’d like it darkened though, to get a patina. As long as it doesnt compromise the artifact.
But why wouldn’t it get corroded through bimetalism? Because i know there is a big potential difference between gold and bronze. It’s the first time i have heard of this phenomenon and i’d like to understand how it works


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Which is the best kind of filter papers to use to extract silicon nanoparticles after acid wash, hydeophobic or hydrophilic?
I'm looking for filter papers to extract silicon nanoparticles after acid etching. I found that much of my material is get wasted with my Whatman Grade 4 filter paper. I am thinking to use membrane filter papers to get the maximum amount of nanoparticles.  However, there are hydrophobic and hydrophilic varieties in the market. I'm confused. Also, is it possible to peel off the nanoparticle layer from the membrane surface?
Thank you

I never dealt with nanoparticles, but I'd expect extracting them out of solution would make them stick together.


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Are petroleum-based products prone to produce hard residues with black powder?
A common controversy about black powder is lubricant composition. Some people tell this is a bad idea to use any petroleum based products in such lube because it will produce hard, difficult to clean residues.
As far as I know, in the old days, gunners used mostly mutton tallow in order to lube their bullets, and most animal-based grease have been found an easy (available) and efficiant way for that use.
Now we can avoid the problems of stinking, liquifying or decaying animal greases, and some people tell they used petroleum jelly or lithium grease for decades with no particular problems. Unfortunetaly, other people disagree, and the exact recipes are so numerous nobody can give me rationnal and definitive answer.
Looking at classic lube recipes teach us there is no consensus on such questions :

Historical Black Powder Bullet Lubricants
Composition of Extensively Used Bullet Lubricants
(E.H. Harrison, American Rifleman, Jul. 1965)

U.S. Army 1855 - 1 beeswax, 3 tallow.
U.S. Army 1861 - 8 beeswax, 1 tallow.
U.S. Army 1873 - 8 bayberry wax, 1 graphite.
U.S. Army 1880 and thereafter - Japan wax.
Sharps Rifle Co., 1878 - 1 beeswax, 2 sperm oil.
Massachusetts Arms Co. (Maynard rifle), 1890 - 1 beeswax, 3 tallow.
Marlin Firearms Co., 1891 - 1 beeswax, 4 tallow.
Smith & Wesson, 1891 - tallow.
H.M. Pope, about 1900 - 3 mutton tallow, 2 bay wax, 1 beeswax, 1 steam cylinder oil, .2 of 1 acheson graphite. The bay wax could be
  omitted.
Automobile door latch stick lubricant, U.S. Patent 1,920,161
(1931) - 5 paraffin wax, 3 petroleum jelly, 2 oil.
A large police department, 1962 - 1 beeswax, 1 paraffin wax, 1 cosmoline. Notes: "Cosmolene" in this context refers to dark
  petrolatum with no anti-corrosion additives. Refined yellow petrolatum
  (petroleum jelly, Vaseline) may be substituted.
Any mixture containing paraffin wax must include a plasticizer, such as petrolatum. Microcrystalline petroleum waxes may be used
  as-is.
The 1:3 beeswax/tallow mixture (or any composition composed mainly of tallow) is probably the most traditional choice for "primitive"
  shooters. The 8:1 mixture is rather stiff, and better suited to
  conicals, paper cartridges, and the like. For paper-patched bullets,
  I'd be inclined to try the Sharps formula, substituting Dexron II/III
  automatic transmission fluid for the sperm oil.

Could someone explain or quote some good source explaining if and why petroleum based lubricant will produce such disagreement with black powder?

Can't give a detailed source, but oil-based hydrocarbons or natural fats will tend to result in the same sort of deposits when burned so there is unlikely to be a big difference between them.

Interesting, but a solid source would be decisive for me.

Anything containing aromatic hydrocarbons (see wikipedia) and related compounds tends to produce a lot of soot. Natural mineral oils tend to contain quite a lot of them, and some remain in diesel fuel. Fraction that becomes gasoline is free of aromatics, but is too volatile for use as a lubricant. Depending on source and processing, modern motor fuel might contain various amounts of aromatic hydrocarbons|| Motor oils are a very different beast, as modern day motor oils are synthetic and consist of paraffin with some additives. || No sources, but what I recited is a fairly common knowledge

Okay for soot-producing aromatic hydrocarbons but is there something specifically linked to black powder combustion? Because smokeless powder lubricants (like Alox) are generally forbidden with black powder as the resulting fouling would tend to become hard.

The answer may be more simple than some may expect, I start with Black powder, which contains, to quote:

Black powder is a granular mixture of

a nitrate, typically potassium nitrate (KNO3), which supplies oxygen for the reaction;

charcoal, which provides carbon and other fuel for the reaction, simplified as carbon (C);

sulfur (S), which, while also serving as a fuel, lowers the temperature required to ignite the mixture, thereby increasing the rate of combustion.

Next, the warming of an organic together with sulfur can produce a volatile sulfur-containing compound (see, for example, 'Reactions of sulfur with organic compounds XXI. The action of sulfur on 2-thienylchloromethane and di-2-thenyl sulfide').
Further, mixing of the new compound with carbon powder could account for the observed product especially after evaporation of all volatiles.
Note, per the comment: "problems of stinking, liquifying or decaying animal greases, and some people tell they used petroleum jelly or lithium grease for decades with no particular problems", the precise sulfur-based product, if any, created could account for the differences reported.
Also, this proposed explanation requires an element of heating, which likely accounts for the development of the reported problem over time.
So, to precisely answer the question: "Are petroleum-based products prone to produce hard residues with black powder? Yes, per my assessment.

Thanks. So, if I understand you well, some types of petroleum-based products are prone to produce hard residues because of the interaction with sulfur from the blackpowder. But are all of thoses petroleum subproducts equally prone to it? Or is this linked to their aromatics proportion (if I refer to @permeakra answer)?

I can confirm this with first hand experience from yesterday.  Petroleum based lubricants will form a tar when exposed to burning Sulfur.  Only took twelve shots to make the rifle unusable.
However the commonly available substitute uses Potassium Perchlorate and Sodium Benzoate, the only common ingredient being Potassium Nitrate.  Petroleum based lubricants work the same as they do with smokeless powder, and only the normal amount of residue is produced.
MSDS sheet for the black powder substitute: https://hodgdonpowderco.com/wp-content/uploads/2023/04/2023-pyrodex-sds-us-012423-1.pdf


text:
What is the hidden meaning of electrochemical stability window?
I don't really get the real meaning of the electrochemical stability window. I googled a lot, and the only thing I understand about the electrochemical stability window is that it's the voltage range that the substance is not oxidized and reduced. However, what does this term imply? I mean, I wonder that how we use this concept in practice.

I guess you are confused by the word window, just like a window has a limited opening [=in a loose sense an interval], electrochemical stability window means the range where the solvent does not get oxidized or reduced. As explained by CH3M, you cannot scan water past 1.2 V vs SHE. It will begin to decompose electrolytically. Of course you have to see the context as well.

I mean, I wonder that how we use this concept in practice.

In practice, you will look up the literature to find at what voltage a given solvent system begins to electrolyze. Basically you want your molecule to take part in the redox activity, not the medium in which your molecule of interest is present.

From your answer, can I conclude that electrode potential of the system should be within electrochemical window otherwise the electrolytes can be electrolytically decomposed?

Yes, this is partially correct. Not the electrolytes, but the medium which contains the redox species.

Do 'the redox species' that you mentioned mean those which are not my interest?

Redox species is the species is of your interest. The solvent or the medium is what we don't it to participate in reduction or oxidation.

Sorry for my poor understandings. So, you said that beyond the electrochemical window, not redox species, which are my interest, but other electrolytes which are not my interest ex) solvents can be decomposed?

Yes. Your understanding is correct now. Suppose you are doing an electrochemical experiment, you have a species A dissolved in water. A can undergo redox reactions and water can also undergo redox reactions. During an experiment you only want A to undergo oxidation or reduction and study its behavior, but you do not want water to decompose. So you will not use potentials outside the stability window of water- if that makes it more clear.

I'm really, really thankful to you. So far, there is no place that can give real meaning of electrochemical window. By the way, when it comes to Li+/Li potential(-3.04V), many people mentioned its advantage of the most negative potential. But in my opinion, although most negative potential can generate high discharge voltage, it might be difficult to find proper electrolytes so that electrode potential should be within the electrochemical window of electrolytes because of high voltage. I wonder it is right to think in this way.

You are welcome. I had guessed that the term window is creating problems in your mind. Electrochemical window is a jargon and it is not an officially accepted term. Yes, lithium is a good example, but its redox system is outside the stability range of water. So you have to try another medium to try reducing lithium.

The electrochemical stability window is most important when considering components of an electrochemical system that you do not want to be oxidized or reduced. This refers most often to the electrolyte or protective coatings.
For example, in lithium ion batteries it is highly desirable that the electrolyte does not react/change/degrade in any way as a result of battery charging/discharging or even in storage. A lithium ion battery typically has a voltage of 3.6 V. The electrolytes used are either lithium salts in organic solvents (ethylene carbonate, etc), solid ceramics, or polymerics electrolytes. If the electrolyte was water based, the electrochemical stability window would only be 1.2 V.
Why 1.2 V?
The standard reduction potential of water is 0.0 V (vs SHE)
The standard oxidation potential of water is 1.2 V (vs SHE)
You can not operate a 3.6 V lithium ion battery with an aqueous electrolyte because the electrolyte itself would react, creating dangerous hydrogen and oxygen gas. So you need an electrolyte with a larger stability window.

Thank you for answering my question. Can you check if I grasp the concept of electrochemical window after understanding your answer? As far as I know, electrochemical window can be calculated by subtracting reduction potential from the oxidation potential because the area between which reduction and oxidation potential has no reaction. (oxidation would happen when electrode potential is above reduction potential, while reduction would happen when electrode potential is below oxdiation potential)


text:
How can you build a model of tetrahedral coordination from objects found at home?
Our college is switching to teaching online mid-semester in hopes of slowing the spread of Covid-19. All the molecular model kits are still on campus. How could you build a model of tetrahedral coordination (say methane) from materials found at home?
I'm aware of computer visualizations (and will make those available), but I think having a physical model when first encountering three-dimensional structures adds value. I made a model from lawn toys (see below), but these are not common household items.

The review bot said "The question you're asking appears subjective and is likely to be closed". We'll see about that ;-)

You may also two pairs of compases

Just get (or make) a cubical box and put four balls (or whatever) on alternating corners, like here: https://chemistry.stackexchange.com/a/118202/79678.

Not exactly things you have at home, but my research group bought a set of these little magnetic rods and ball bearings and we make tetrahedra/octahedra and lattices all the time with them, they're very useful: image

Mini-marshmallows, toothpicks and food coloring.

@Kai Considering they've been sold as children's toys for years now under varying names (Magnetix for example), that's not a bad idea.

Inflate balloons, and tie them «at their stem» like a bouquet of flowers.  If you take four of them, not too much inflated, you well demonstrate a situation close to $sp^3$ hybridization. These models equally work well in larger lecture halls by the way, and intentionally using different colors allows many options.

(source)

(screen photo
You need some worked examples?  See videos like this or this.
You need a scientific paper? Well, there are as well, e.g., this (open access) expanding the picture to extended $\pi$-systems:

This is a great VSEPR demo because the balloons "find the solution on their own", i.e. you are just tying a couple balloons together, and they arrange to maximize bond angles.

It get's even better, as https://www.youtube.com/watch?v=Kb0mxAMHnfE shows $sp^2-sp^2$ to lead to $\pi$ bonds, too.  And his «VESPR» reminds a bit to «Vespa» (smaller motocycles) to ....

This one is inspired by ideas from Ed V and Todd Minehardt. The cool thing is that the angles are pretty accurate if the chop sticks reach into the opposite corners of the cube:

As a bonus, you can make an octahedral model if you make holes in the center of each square (or slightly offset so that the chop sticks can pass in the center of the cube).
The next one is just cardboard, but you have to get the 109.5 degrees right.

Etcetetra...

From left to right: Pipe cleaner, pins and glue gun blob, paper model (from Bob Hanson's Molecular Origami book, ISBN-13: 978-0935702309), carrot and toothpicks (inspired by the marchmallow and toothpick activity in Laura Frost's General, Organic and Biochemistry textbook, ISBN 13: 9780805381788).

@ToddMinehardt - Version 2 of your idea...

The cool thing is «is that the angles are pretty accurate if the chop sticks reach into the opposite corner» .AND. if the box is a cube. I recommend to state this second conditon to be met during the lecture.  An answer by Todd Minehardt (retracted by him) used a box of probably only $D_{4h}$ symmetry instead of $O_h$; but only boxes qualifying $O_h$ will do this trick where space diagonals coalesce with «space bisectrices» indicated by the chop sticks.

@Buttonwood I took two flaps of equal with, and used the width of one to measure off the lengths where I had to fold the other. So it is pretty much a cube (and the chop sticks did collide in the center). Thanks for the input (the lecture will have to wait).

Yeah, I retracted mine as it was more of an optical illusion than the real deal as @Buttonwood notes. Anyway...

@KarstenTheis The hint to the book (https://www.amazon.com/Molecular-Origami-Precision-Scale-Models/dp/093570230X) led to the corresponding «folding program» (https://www.stolaf.edu/depts/chemistry/courses/toolkits/123/mo/).  Maybe interesting as complementary or substitute to the anaglyph googles in crystallography is Sanii's presentation of «Augmented Reality» with Students' Phones in J. Chem. Educ.  2020, 97, 1, 253-257 (https://doi.org/10.1021/acs.jchemed.9b00577) for 3D objects in general, too.

With styrofoam balls and toothpicks:

Pick a central ball.  As an aid for the next step draw three mutually perpendicular great circles on this ball, dividing it into octants.
Pick any octant and the three octants catty-corner to it and place toothpicks through the center of each of these octants.  This should give a good approximation of the proper angles if your great circles from the first step are reasonably accurate.
Attach the coordinated balls on the exposed ends of the toothpicks.

Obviously the contents of your home are different to mine. But I imagine the downvote was because this is a different structure with five balls not four

Do we or do we not include the center ball?

@OscarLanzi In my question, I did not mean to imply specifics about how the model would be used, as model for methane, as model for sp3-hybrids, as VSEPR model etc. Depending what you use it for, a center ball would make sense or not. Instead of the styrofoam ball, lots of different food items will work (I tried an apple and ate it afterwards because while I never want to waste food, this is definitely not the time).

You may also use two pairs of compasses, and join them with their tips in opposite directions.
Or you can use your hands wide opened in front of your face. You then turn one of your hands at an angle of 90°, and joined them, without closing them.  The thumbs and the two forefingers represent the directions of the four bonds around a carbon atom in $CH_4$.


text:
What happens if the filter paper in vacuum filtration is BIGGER than the buchner funnel diameter?
I’m wondering what would happen if the filter paper in a vacuum filtration process is bigger than the buchner funnel diameter?
I mean would some differences occurs? Or would the whole experiment damage? If yes, may I know how and why?

Get some smaller filter paper, or trim the bigger ones with scissors.

If it is larger then it could make "waves" on the sides, and your product could leak. That's one of the reasons why you find scissors in labs :D

Place inverted funnel on the filter paper. Trace around circumference with a pencil. Cut inside the circle accounting for the thickness of the funnel wall.

I in fact made this mistake once while attempting to dry casein isolate from milk. If the filter paper is larger than the diameter of the büchner funnel, you will risk there being gaps on the side (or "waves" as Vinz called it in his comment), due by crumples in the paper caused by fitting the large paper into the small diameter of the funnel. The air will flow through these gaps/waves. Hence, the vacuum filtration will not work effectively, if at all, because the air is not being pulled through the paper but rather through the gaps on the side. We learn from our mistakes. As to how to fix this issue, the comment by MaxW should help you solve this.

The above photo shows two of my results when attempting to dry the casein isolate from milk. The contents of the left petri dish are predominantly a result of using a too large filter paper during vacuum filtration (among a few other errors incurred by attempting to fix this mistake), while the contents of the petri dish on the right are a result of using a well-cut filter paper that suited the diameter of the büchner funnel.

There will be leaks around the edge of the paper. If the size difference is not too large (a few mm at most) you can remedy the situation as follows: (1) turn on the vacuum with only the filter paper in the funnel, (2) take a spatula and run the tip of the spoon around the edge of the filter paper, creasing it into the corner between the flat bottom and the sides. This usually works for me.


text:
Most understandable notation for reduction and oxidation potentials
Nowadays a lot of electrochemistry is done using just electrode potentials (i.e. reduction potentials), so if in doubt the notation $E^o$ tends to imply a reduction potential. However, some chemists use the perfectly acceptable convention that oxidation reactions are accompanied with oxidation potentials.
I wondered whether anyone could inform me about the most unambiguous or widely used conventions for indicating which type of potential is being employed when written next to a half equation.
For instance, for the oxidation half equation $Ag \rightarrow Ag^+ + e^-$, I could write the associated potential in a number of ways:

$E_{Ag^+/Ag} = +0.80V$ - the electrode potential of the $Ag^+/Ag$ couple
$E_{red}(Ag^+/Ag) = +0.80V$ - the reduction potential of the $Ag^+/Ag$ couple
$E_{ox}(Ag^+/Ag) = -0.80V$ - the oxidation potential of the $Ag^+/Ag$ couple
Often, just $E$, or some other notation!

I think the notation in points 1 through 3 is fairly clear, with the first two applicable to the European convention and the third applicable to the American convention, when written next to the oxidation equation. However I was wondering if anyone more experienced could advise as to what the most common conventions in the field are!
Thank you!

I don't know why your textbook author is confusing and still teaching American & European conventions. It is obsolete now. I will show you how oxidation potentials were quoted by American electrochemists in the 1950s-60s. Have a look at the table Latimer's book: Oxidation States of the Elements and their Potentials in Aqueous Solutions pg 340. This was a very famous book of its time. Latimer and Gibbs had great influence on American science and thermodynamics.
I would say you should stick to only (1). Your number 3 is not correct because you are still writing the cell as reduction. The correct version would be
$E_{ox}(Ag/Ag^+) = -0.80V$ - the oxidation potential of the $Ag/Ag^+$ couple, so that you are consistent.
Also make sure your instructor is aware of these things (American European conventions, and that the sign of the electrode potential is an electrostatic one, and that electrode potential does not care how the equation is written. You seem to know more than an average PhD student). I will not be surprised, your instructor is not a trained electrochemist, he/she may not be aware of these subtleties.
Also note that the small "o" on E, indicates standard conditions.

I wonder, why is it that number (3) is incorrect? The oxidation potential is the emf of a cell where the redox couple in question is on the left side of the cell (undergoing oxidation), and is well defined independent of the choice of convention. I personally wouldn't use (3) (like you say, the convention of labelling reactions with oxidation potentials is rarely used anymore), but just for completeness' sake isn't it still a correct formulation?

@JamesWirth, I have edited the answer.

Ah, I thought a couple was an unordered pair of species (i.e. $/^+ \equiv ^+/$) which is just written in the form / for consistency. Because then, we could always speak of the electrode potential of the $^+/$ couple without needing to worry about whether the cell reaction is a reduction or oxidation. That is to say we can always refer to the $X/X^-$ electrode no matter the reaction occurring there.

So if $Ag^+/Ag$ represents a half cell which might perform two different half reactions, then its electrode potential is the emf if we connect that half cell on the right wrt SHE, the oxidation potential of that couple the emf if we connect that half cell on the left wrt SHE, etc. It actually appears important that the order of the couple does not imply a particular half equation.

"So if Ag+/Ag represents a half cell which might perform two $different$ half reactions", this does not make any electrochemical sense. It is one and only one half cell whether there is an oxidation or reduced. You can only say, Ag+/Ag  can either be an anode or a cathode.

I will give you an example. Let us say water boils at 100 C. Now if I write H2O (l) <=> H2O (g), and I write T_H2O = +100 C, now reverse the equation, H2O(g) <=> H2O(l), can we write, T_H2O = -100 C? The is true when people try to reverse the sign of electrode potentials.

I think the confusion is arising from abbreviating the half cell. The typical procedure is that if you consider the half cell as an anode, you write it as an oxidation i.e., Ag ->Ag^+ + e, but don't change the sign of the electrode potential. The sign change occurs automatically when you calculate the E_cell=E_cathode-E_anode

Sorry, I should have been more careful. I meant that $Ag^+/Ag$ is simply referring to a redox couple (an unordered pair of two species), and not a half equation in a particular direction. The  $Ag^+/Ag$ redox couple might be fixed at either the cathode or anode, depending on how we connect our circuit. What I’m saying is that even if the half reaction is an oxidation, the redox couple at that electrode is still $Ag^+/Ag$.

So in that sense, I see no need to reverse the ordering of the redox couple to $E_{ox} (Ag/Ag^+)$, since this is completely equivalent to the expression I wrote - because redox couples don’t equate to half equations!

Well, you are right that the electrode potentials do not care how the half cell is written, but there are norms in a given field. People do stick to writing certain conventions. For example if I invent a new symbol for integration, mathematicians will reject my proposal . The notation you are using is not canonical either. Today, $E$  implies a reduction half cell with respect to hydrogen. This has been settled by electrochemists and decided by IUPAC. I am not an electrochemist, but I was lucky to have a fine electrochemistry professor. I have personally studied these conventions for a decade.

If you are really interested in going deep into these sign conventions, their origins and the sources of confusion, search the Journal of Chemical Education on the sign of electrode potentials. There are many useful articles. Your university should have access to them. Don't search the web, search academic articles. The web is full of non-sense especially on historical matters.

Let us continue this discussion in chat.


text:
Resolving ambiguity around Lattice Enthalpy in Born-Haber Cycle
According to my textbook,

Lattice enthalpy is the enthalpy change that occurs when one mole of a solid ionic compound is separated into gaseous ions under standard conditions.$^1$

According to the same textbook, this is an endothermic process, which makes sense as a lattice is generally held together by strong ionic bonds and thus would require energy to separate the atoms. The following diagram is given for the born-haber cycle in the textbook:

The diagram supports the definition as the enthalpy of the individual gaseous atoms is greater than that of the lattice, i.e. the arrow moves up. However, as I was working through past papers by the International Baccalaureate, a question asked to draw the born-haber cycle for LiF.$^2$ This was the answer:

In this case, the arrow is pointing down for the lattice enthalpy. To me, this does not suite the definition of lattice enthalpy as separating a lattice into its ions would require energy and not release energy. Hence, should the arrow not be pointing upwards for $\ce{LiF_{(s)}}$ to $\ce{Li^+_{(g)} + Cl^-_{(g)}}$? Am I making a fundamental error, is the case for NaCl different to LiF or is the answer wrong?
Works Cited:

Pearson Baccalaureate: Higher Level Chemistry 2nd Edition. By Catrin Brown and Mike Ford
International Baccalaureate

The downward pointing arrow means that energy is released when the crystalline solid is formed from the gas phase separated ions. So it is the negative of lattice enthalpy. The diagram is just a schematic showing the enthalpy balance in the B-H cycle.

Ah, that makes sense, thank you. So, just to clarify, the arrow for the first image is for lattice enthalpy (endothermic) and therefore points upwards and the arrow for the second image is negative lattice enthalpy (exothermic) and therefore points downward?

Exactly correct!

Cheers for the help! :-)


text:
Assigning Peaks and Determining J Values for Mata-Disubstituted Benzene
I need to determine as many J values as I can for 3-methoxybenzaldehyde. The peaks overlap a lot, and I don't see the expected ddd or dt pattern that I would expect for the hydrogen on carbon 6. The HSQC I was given has the clearest peaks for the H NMR, so I attached that. Maybe someone with more NMR experience can help me tease apart the signals?

There's almost nothing you can do in this situation, unfortunately. Because of strong coupling and other issues, aromatic peaks often do not conform to the so-called $n+1$ rule, which makes it highly non-trivial to extract coupling constants. It is probably possible to "fit" coupling constants to the multiplet shape by simulating the spectrum for a given set of $J$'s and seeing which set best matches the observed spectrum.
The HSQC trace is unfortunately not useful in this regard. At first glance, it may look simpler than the 1D proton spectrum you undoubtedly have. But that is because HSQC spectra are invariably recorded with low resolution in the proton dimension. The main reason for that is because of the high-power 13C decoupling that is performed during the acquisition. Acquisition times therefore need to be kept to approximately 100 ms, or there is a serious risk of damaging the spectrometer. On the other hand, a 1D proton spectrum doesn't utilise 13C decoupling, and can have acquisition times of several seconds.
Assuming the spectral width is constant, a shorter acquisition time means poorer resolution (this is a property of the discrete Fourier transform).

The last point "Assuming the spectral width is constant, a shorter acquisition time means poorer resolution (this is a property of the discrete Fourier transform)." is very critical. The readers should keep in mind that the resolution with shorter acquisition time is poorer because of the frequency spacing with lesser number of data points. Larger the length of the data, smaller the frequency spacing on the x-axis.


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aqueous solution recipe calculations: Using oxy-bleach as surface sanitizer [NOT A PERSONAL HEALTH QUESTION]
NOTE: Please do not delete this question.  It is not a personal health question.  The question has broad utility, in cases where others find themselves in social-distancing or quarantine with need for surface sanitizer, and their only available ingredients are water and powdered oxy-bleach.
I need a quantity of surface sanitizer to decontaminate boxes delivered to the house during the Coronavirus19 social-distancing period.
CDC and WHO say 0.5% hydrogen peroxide solution will destroy coronavirus19.  And powdered oxy-bleach a.k.a. Sodium percarbonate contains 32.5% by weight of hydrogen peroxide, I'll call it roughly a third.
So to make a water solution of hydrogen peroxide using powdered oxy-bleach, I'm thinking I need to add enough powdered oxy-bleach (by weight) to equal 3 x 0.5% of the weight of the water. I.e, for a gallon of solution, 1.5% of 8.3 pounds, thus 0.12 lb, or ~2 oz.
A friend checked this and agreed, saying:
"A 0.5% H2O2 solution in water will consist of (per gallon): 0.42 lbs of H2O2 and 7.9 lbs of H2O. To get 0.42 lbs of H2O2 out of Oxy-bleach (32.5% H2O2), divide 0.42 by 0.325, which = 0.128 lbs or 2.0529230769 ounces..."
Can anyone poke holes in our thinking?  Or refine it?  Or suggest ways to make it easier for the lay-person locked-in without access to other surface sanitizers?
Thank you in advance!

Tell them to wash their hands after handling the boxes for heavens sake. There is no way of telling if this disinfecting scheme of yours will work, but rubbing your hands for thirty seconds with soap is definitely effective.

I voted to close this medical que4stion. We are not qualified to give medical advice.

Your first paragraph says ."It is not a personal health question" and then you say it is a personal health question, using different words.

"Personal medical questions are off-topic on Chemistry. We can not safely answer questions for your specific situation and you should always consult a doctor for medical advice." is a general criterion on ChemSE.  About Covid-19, in particular, the site refrains from medical advice other than proper hygiene (e.g., efficient hand washing), too (https://chemistry.meta.stackexchange.com/questions/4668/regarding-coronavirus-questions).

You math looks fine but be careful if you intend to scale up this (use proper protective clothes and ventilation). see also: https://chemistry.meta.stackexchange.com/questions/4668/regarding-coronavirus-questions

Yes, do use household dilute H2O2 but, in general, do not use commercial percarbonates powders as they can contain an additive to convert the friendly percarbonate into more powerful Peracetic acid (PAA). The latter allows the otherwise weak bleaching power of the H2O2 to compete with chlorine-based bleaching. However, this comes at a price (safety). Hence, the dire warning labels, usually in at least two languages, on Oxi-Bleach and generic related products.
I quote myself on this subject:

Sodium percarbonate is chemically just an addition compound of Na2CO3 and Hydrogen peroxide.
Also, while Sodium carbonate and Sodium percarbonate (basically Na2CO3 and 3Na2CO3.2H2O2) are often listed as the main ingredients in typical chlorine free bleach, an additive, TAED, which acts as a key activator (see, for example) is not even listed on the label. TetraAcetylEthyleneDiamine is apparently commonly employed as a bleach activator in many laundry products (see this).  TAED reacts with H2O2 in alkaline conditions (referred to as perhydrolysis) creating, in part, Peracetic Acid. The latter PAA is actually preferred over sluggish H2O2 for bleaching. In other words, it is not the H2O2 as the active bleaching agent, but PAA created in situ.  So, other than possibly misleading implying that eco friendly hydrogen peroxide is the active agent, it is, in reality, PAA, a probematic compound with associated  health concerns (see Marquand, E. C.; et al. , 2007. "Asthma Caused by Peracetic Acid-Hydrogen Peroxide Mixture". J. Occup. Health. 49 (2): 155–158.). Here is a MSDS on an actual PAA mix ( 76-61% water, PAA 20-35%, 3% Acetic acid and 1% H2O2) available at which is far from friendly but does eventually totally decompose into harmless O2, CO2, H2O,...
As such anyone using a commercial percarbonate product should at least be aware of safety issues and that one is likely employing a Peracetic acid mix.

However, such more powerful PAA based mixtures do have their use and is recommended by the U.S. Military as a chemical/biological warfare decontamination protocol (see "Development of Bicarbonate-Activated Peroxide as a Chemical and Biological Warfare Agent Decontaminant"). So, it is up to the user to weigh the relative risks of the situation.


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Difference between docking and scoring?
I'm taking an undergraduate level class on computer based drug design.  The class is more for biology students than people with a strong understanding in physics and computer science.
With that said, I'm having trouble discerning the difference between what molecular docking and scoring functions are.
To me they seem like they are the same thing since they have the same goal, which is to figure out which ligands have the strongest binding affinity to a target protein. But what are the different ways each method accomplishes that and when would each method be used if (to me) each method does the same thing? Is a scoring function a way to check the validity of a docking simulation, if so how?

https://en.wikipedia.org/wiki/Scoring_functions_for_docking


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Adding hydrogens to multiple pdb files using Openbabel GUI software
I am  completely new to using OpenBabel software.
I have about 9000 pdb files to which, I want to add hydrogens. It will be a very hectic task to open all of them 1 by 1 and then convert each of them. Is there any way I could convert all of them simultaneously.
Also, can you please tell me how to correctly select the output file? It asks for a destination file to which, it appends/overwrites the output. Is there a way I could choose the destination where it would create another .pdb file and save the output in it? I guess I am missing something here while choosing the output file.
Thanks in advance. Also, I asked a similar question before which was closed. Considering I am new with OpenBabel, if any further information is required or if there is some issue, please tell me so that I may improve my question.

If you got an installation of openbabel's GUI, you equally have an installation of openbabel for the terminal (e.g., in Linuxes) / command line cmd.exe (in Windows).  There is nothing wrong using the GUI for an edit on one file or a few data -- your instructions set are right -- however using the CLI is closer to the engine, thus often more powerful.

Enter the directory containing the .pdb to work with from the command line.
The instruction babel *.pdb -opdb *_h.pdb -h will request babel to look for all *.pdb there and write new .pdb with hydrogens added.  Each file by name in pattern of example.pdb will have its edit recorded in a file named in pattern of example_h.pdb.  While the CLI will report the progress of this conversion, I recommend to slice the task into smaller batches than just one of 9k.  How many .pdb per batch depends on the computing power available on your machine.
The new files written are both discernable by name and time stamp from the retained input.  So you have two criteria to sort and discern the files.  Assuming you subsequently moved the new ones into their own folder, you may rename them en bloc back to the pattern of example.pdb.  If you work with Windows, a program like Ant Renamer may considerably accelerate this task.

Commands and grammar openbabel uses are documented in depth.  For commands accessible from the CLI, just a few are compiled here.  As perspective: note its use may be automated (scripted), for example with Python like the portable WinPython.

Is there an option to save the example_h.pdb files in another folder, in the same directory in which the folder containing the actual example.pdb files exist.

Also, can you please elaborate on Ant Renamer and how can it help accelerating the process. Thank you.

@TapasviBhatt For 1) There is if you know some CLI commands applicable for the operating system.  Inside the folder with the .pdb to work with, create a sub-folder, e.g. by mkdir stash.  Then redirect openbabel's output «with Windows syntax» to this sub-folder by babel *.pdb -opdb .\stash\*_h.pdb -h. Somewhere with Windows 7, «the Linux-like syntax» using forward-slashes became something Windows now equally accepts to denote a file path, which then equates to babel *.pdb -opdb ./stash/*_h.pdb -h.

@TapasviBhatt For 2) With said program you may select files, or complete folders of files to rename.  Here, you possibly would like to truncate two characters in the file names to convert any edited file named in pattern example_h.pdb into a file named example.pdb, and such a file name truncation is one of the actions offered by the program.  If you opt for a babel action like babel *.pdb -opdb ./stash/*.pdb -h you may well overwrite the original files lacking Hs with those now containing Hs, especially if you trace changes with a version control like git (e.g., curating a database).


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Is NH3 a strong ligand or weak?
Sometimes it produces high spin complex and sometimes low one. Is there any generalised result

$\ce{NH3}$ is roughly in the middle of the spectrochemical series as it is only a $\sigma$-donor, so it can kind of though as to create splitting energy that isn't large or small. But instead other factors will play more of a role in deciding the size of energy split such as oxidation state, coordination number and whether it is a 3d, 4d or 5d metal.


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Photosynthesis being a synthesis reaction
If a synthesis reaction is a reaction in which two or more substances react together to form just one product, then how is it that photosynthesis has two products; glucose and oxygen?

Where did you get that definition of synthesis? It is not one I recognise. There is always more than one product from a synthetic reaction.

Did you encounter the word «byproduct»?  Its use is not limited to chemistry (https://en.wikipedia.org/wiki/By-product).

The school definition of synthesis is pretty much never used outside of school.

Photosynthesis is a most complex set of reactions, all of which occur in proteins and in the chloroplast cell membrane and have taken at least 50 years to sort out by numerous scientists.  Look for 'primary processes' and 'light reactions' in photosynthesis and the Calvin cycle for the dark reactions.

I think you're being a little too strict with your definition of what a synthesis reaction is. A better way to describe it might be building more complex molecules from simpler ones.
Wouldn't you say $\ce{C6H12O6}$ is more complex than $\ce{CO2}$ and $\ce{H2O}$?


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Electron transfer rates in Marcus Theory
Electron transfer rates (k) as derived from Marcus theory assumes i) thermal equilibrium and ii) neglects quantum nature of low-frequency phonon modes. Can anyone explain what is at the heart of these two assumptions? Thanks a lot!
I read this in the context of the electron transfer process in organic semiconductors.

There is a description of Marcus theory here How does the inverted Marcus region explain chemiluminescence? . The quantum nature can be added into the theory as necessary, it leads to a asymmetrical plot of rate constant vs free energy with the inverted region (large -$\Delta G$) decaying away more slowly than the normal region.  The equilibrium is between the solvent reorganisation with reactants and then with products. The solvent plays a controlling role in the process, the proof is that experiment demonstrates that  the rate of electron transfer can be limited by solvent motion. See Harrison et al.  Chemical Physics, Volume 116, Issue 3, 15 September 1987, Pages 429-448, and Irvine et al. ibid.  Volume 104, Issue 2, 1 May 1986, Pages 315-324. The novel aspect of Marcus theory is the role of the  solvent and this was not generally accepted until the inverted region was observed.


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Heat transfer to an ideal gas
In thermodynamics, why is the heat that is transferred to an ideal gas considered as work that was done on the system for an isothermal process?

While you already got a good answer here, in general you should frame your questions to show what you have done to try to the solve the problem yourself.

I started to study thermodynamics, and we learned about ideal gas in class and while reviewing the material this question raised in my head and I couldn't find an understandable answer on the internet so I asked here. I will be more specific next time:)

Welcome to the Chemistry Stack Exchange!
In general, heat transferred to an ideal gas is not considered as work that was done on the system. I suspect you're referring to a process in which the internal energy doesn't change: $\Delta{U}=0$. As $\Delta{U} = q + w$ (where q is heat and w is work), in the case where $\Delta{U}=0$ you get $q=-w$. This holds for a general system (not just ideal gases).
For an ideal gas, U only depends on temperature. So for an isothermal process, the internal energy of the ideal gas will not change, and you'll once again get the result $q=-w$.

Ok. I understand know. Thank you very much for the help:)


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Does the order of a written redox couple matter?
My understanding is that a redox couple is an unordered pair of two conjugate species. So we could speak of the redox couple consisting of $Ag$ and $Ag^+$; depending on whether a given redox couple is undergoing oxidation or reduction, we could then write a half equation involving the species given in the redox couple.
In my textbook, it is given as a convention that all redox couples are expressed in the form Ox/Red. So for instance, it states "at anode containing the $Zn^{2+}/Zn$ redox couple, the oxidation half equation is $Zn \rightarrow Zn^{2+} + 2e^-$". This convention makes sense to me, since it doesn't seem that the couple itself refers to a half equation in a given direction. The order of the half cell line representation, however, does change the order to $Zn | Zn^{2+}$.
However, some texts reverse the order of the written redox couple, which I would interpret as either them writing the couple in the order of the current half equation (which seems wrong) or that they are ignoring the order because it is unimportant.
So I wondered whether anyone could clarify, is a redox couple just a pair of species which exist at a given half cell, or do we need to change the order of the written couple depending on the reaction occurring at that half cell?

At @EdV but specifically with regards to the notation for redox couples. As in I would say the written couple $Zn^{2+}/Zn$ doesn't correspond to either a reduction or oxidation half equation, it just represents the species present at that half cell. However others might say that the written couple $Zn^{2+}/Zn$ implies the half equation $Zn^{2+} + 2e^- \rightarrow Zn$ whilst $Zn/Zn^{2+}$ implies $Zn \rightarrow Zn^{2+} + 2e^-$.

My understanding is that a redox couple is an unordered pair of two
  conjugate species

This is conceptually perfect and there is no problem when we talk about electrode potentials of half cells because as I had mentioned in your earlier queries, the electrode potential value and its associated sign do not know nor care how you write the half cell.
What is creating confusion in your mind is that you are trying to abbreviate a half cell. Abbreviating a half cell is of no use, because it is creating a problem of ordering.
All we care about the order is when we write a full cell. For that, we must follow the IUPAC recommendation and rules in order to prevent chaos in the published work. There your half-cell is no longer an unordered pair.
I used to remember the IUPAC recommendation with a mnemonic: Write Reduction on the Right.

This has to be parsed as follows: Reduction is on the right, so you must order the copper half cell in such a way that it appears as reduction.
On the left, of course, we have an oxidation half cell and it must be written as such.
Hope that clarifies your confusion.

+1 I will delete my other two comments and hope this brings closure to this triptych of redox notation queries.

But I am glad that the student is thinking and searching historical problems. Hard to find such students these days.

I agree. But standardization can be extremely helpful: I once read of a situation involving optical effects in polarizers and such, and there were 4 parameters involved. Each could be defined as positive or negative, resulting in 16 possible total sign conventions. It turns out that various authors had used 15 of the 16! I guess the unused one fell out of the ugly tree face first ...

Thank you, this is perfect. To quote @EdV, many thanks for all your patience in answering this triptych of questions!

I notice in Bard and Faulkner (which I hear is the gold standard for electrochemistry...!) that they are one of the few sources that do order written redox couples depending on whether the reaction occurring at that electrode is oxidation or reduction (i.e. they speak of the O/R and R/O electrodes). I assume then that this is purely a matter of readability. Since most more modern publications and textbooks always refer to couples in the order O/R, no matter!

Bard and Faulkner are truly outstanding, but, as per my potentiostat answer and comment here, even they can make (or not catch) dinky mistakes. Electrochemistry, unfortunately, is an area where it is surprisingly easy to make little mistakes, so standardization is a real benefit, especially to those of us who have taught parts of it.

Bard has crossed mid 80s. So their book may not represent the latest. The authors are notorious for their difficult but respectable book. Anyway, O/R and R/O electrodes are not of our concern. I think R/O or O/R notation is not standard either. It is related to electrolytic cells.

An example might be the AgCl/Ag redox couple, which can be used as a reference electrode. If we use the "modern" convention, this couple should be denoted as AgCl/Ag. Yet the literature overwhelmingly refers to it as an Ag/AgCl electrode. It's little things like this which seem benign (and probably are benign) but can cause significant confusion to people like me who haven't fully grasped the conventions yet!

You will have to tolerate little bit of bending the rules. You have got a good grasp, far better than typical PhD students who may not have heard of European or American conventions. Keep in mind a field which is > 100 year old will not have consistent units or notation. Forgive those tiny things and focus on the concepts, as long as you know in the heart what is right.

Try to write the complete notation for silver chloride electrode with Standard hydrogen Electrode (SHE) on the $left$. This rationalizes the notation of Ag/AgCl.

You may ask why did I keep SHE on the left, it because IUPAC tells us to do so, in the definition of the electrode potential.

If a couple is an unordered pair, then $X/X^- \equiv X^-/X$, since they are not abbreviations of half cells / line representations, just pairs. In that case, $E^o(X^-/X) = E^o(X/X^-) = \text{standard electrode potential}$. The convention that redox pairs are written Ox/Red conventionally appears to be a distinct (but similar) convention to the American/European distinction.


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Would the hydrocarbon gas dissolve in gelled oil based drilling mud (OBM)
I am an oil well integrity engineer.  Need help in understanding gas solubility in oil-based drilling mud (OBM) when the mud is gelled.  Normally, oil based mud is an emulsion, where continuous phase is some type of oil (~70%), and remaining is dispersed water/brine droplets, various additives to get to desired drilling mud properties.  Gas is highly soluble in OBM, mostly due to base oil is it's main component.  However, drilling mud is thixotropic and becomes like a gel when static.  Don't know exact mechanism, but something like molecules connect to create long chain polymers, thus forming gel (?).
I am trying to understand, if gas is diffused into the mud after it has become like gel, would the gas dissolve in the same way as it would in thin mud (before it's gelled).  Considering, how non-polar molecules dissolve in non-polar solvents, I tend to think that not as much gas would dissolve, but I might be wrong.  Please, help me.

That is not how a "gas kick" type incident happens in a well ,if that is what you are after.

No, I am not interested in drilling at this point.  I am interested in very low rate gas entry (leak) into well annulus space during production.


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Generating electron density from protein coordinate (PDB)
Normally protein coordinates are generated from electron density, such as that from X-ray scattering experiment. In my case, I want the reverse - PDB to electron density mesh. I thought it was more trivial and a program should be available. But Google did not give me any luck.
Is there any existing python library/program which converts protein coordinate (pdb) into electron density mesh (custom resolution)? If not, how should one proceed? The orbital and atomistic electron density varies among atom types.
Your suggestion appreciated.
Edited:
The precision of electron density I am interested in is ~ 1-2A so hopefully DFT level might not be necessary in my case. The aim is to visually resolve sidechain atoms. Since there could be manipulation of the PDB upstream, some crystal structure information could be lost.
Indeed the raw electron density might provide such information but a unified treatment of PDB coordinate homogenizes structures, making a more fair comparison between them, specifically in my case.

If the electron density is derived from the coordinates only, using the independent atom approach, then the coordinates and the density carry equivalent information. It would be interesting to know whether you are comparing the calculated electron density to experimental data (such as derived from cryo-EM), are using this for teaching, or for some other purpose. Just to be clear, you could not use it for something like determination of partial charges or such.

I am using this to compare with X-ray crystallography experiment. Indeed with an independent atom approach, the partial charge cannot be deduced unless atomistic forcefield is used, like AMBER and CHARMM.

http://www.phenix-online.org/pipermail/phenixbb/2020-March/024559.html --> Generating electron density from protein coordinate (PDB)

For the more recent structures, you can view the density (based on measured diffraction data and the model, so-called 2Fo-Fc density) directly in the protein data bank, e.g. http://www.rcsb.org/3d-view/6QU9?preset=electronDensityMaps:

For a theoretical model or a model without deposited diffraction data, you would first have to generate structure factors, and then calculate the electron density. In the simplest case, the electron density would just reflect the density of isolated atoms (i.e. no deformation density). Program suites such as CCP4 (originally in Fortran) or Phenix (in python) are available to do these steps.
In these calculations, the electron density of individual atoms is encoded in so-called atomic form factors (see e.g. http://lampx.tugraz.at/~hadley/ss1/crystaldiffraction/atomicformfactors/formfactors.php), which are fourier transform coefficients of the model electron density for each atom.

Thank you Karsten. I am exploring Phenix and fmodel generates structure factor from pdb. However, I fail to find a way to turn structure factor into electron density. Although I can take its square norm of its fourier transform as electron density, my result will be off by some normalization constant. Do you happen to know the exact command/function?

@Simon I don't know Phenix, but it might be phenix.maps: http://www.phenix-online.org/pipermail/phenixbb/2010-February/014521.html. You would calculate an Fc map (something like 0 Fo + 1 Fc).

What you've described is the field of electronic structure theory. To obtain the electronic density from the cartesian coordinates of the atoms requires immense amount of computation (assuming you have a protein in your pdb file). If you have a small molecule, you could use any number of common electronic structure packages. ORCA is open-source (assuming academic), or you could check if you have access to Gaussian (another widely used one) through a university.
EDIT: Having re-read your question, I see it is indeed a protein you are trying to get the electronic density of. In this case, you will not be able to use even the most basic electronic structure methods on the entire protein. If you need the electron density of a small section (<100-500 atoms, with 500 atoms already heavily straining most computational setups) of the protein (say an active site), then you could do it. (Worth noting that you could go beyond 500 atoms with some advanced computational setups, see the paper Ian Bush mentioned in the comments).
However, if you just need an approximate charge distribution, you can assign point charges located at the center of each atom (water can be modeled in that way or with more complicated methods with the TIP models, where TIP3P is the simple "point charge on each atom" approach). The tleap command in Ambertools will do that for you. You could also use VMD's built-in support of CHARMM to add point charges centered on each atom. This is the psfgen plugin for VMD.

You seem to be overlooking the context here.  Computing (good enough) electron density from coordinate data is routine practice in the field of crystallography, from which most PDB data is drawn.  Yes, there is a non-trivial computation involved (a symmetry-accommodating Fourier transform of a weighted sum of atomic scattering factors) that scales with the size of the problem, but "requires an immense amount of computation" just isn't a good characterization of the kind of calculation the OP seems actually to want to perform.

@JohnBollinger I may indeed be. However, I haven't heard of a method like this, and would appreciate a link. As far as I know, computing an electronic density from only cartesian coordinates of atoms can only respectably be done through electronic structure methods. If you were to work backwards, using atomic scattering factors for isolated elements, you'd miss the effect of nearby atoms on the density.

I guess it depends on how "respectable" you want to be, and to whom.  Crystallographers know that the independent-atom model on which most crystallographic computations is based is an approximation, but it turns out to be quite a good enough one for the purpose, which is primarily to model the (three-dimensionally symmetric and periodic) atomic arrangements that give rise to specific diffraction patterns.  Electron density is involved as the physical property that gives rise to (x-ray or electron) diffraction in the first place.  This is the subject area of the question.

As for links, Wikipedia's overview article isn't too bad, but it doesn't go into diffraction theory to any significant degree.  Most web sites don't, other than to make substantially the same assertions I have already done, as it isn't really a subject for general audiences. You would probably need to turn to the literature, which is extensive, or to crystallography reference works, such as the International Tables for Crystallography, for a discussion of details.

@jezzo Gicen a capable enough code and a suitably powerful computer you can do somewhat more than 500 atoms, especially for open systems. As an example from my own work https://pubs.acs.org/doi/10.1021/acs.jpcc.9b06533 , which does full structural optimisations at the B3LYP level on periodic systems with 2788 atoms pr unit cell.

@JohnBollinger Thanks for introducing me to a new area, I'll do some reading on it. I would point out that OP didn't clarify to what accuracy the electron density was needed. Assuming the electron density would be used to predict properties of the protein (not dynamics, but localized excitations, chemical reactions of active site, etc), they should still use a method that accounts for electron-electron correlation.

@IanBush Thanks for the very interesting read! Worth noting, 500 and 2788 are just barely (but still are) on the same order of magnitude. I'll update to make the phrasing softer.

@jezzo for your suggestion. I have just updated my inquiry to specify my interest.


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Comparing Electron affinity and electron gain enthalpy at 0 K
In my book it is given that first electron gain enthalpy is greater than second for elements. Should we compare the magnitudes in such cases or the actual numbers with signs? Does the same comparision hold with Electron Affinity at 0K?
For example Be has positive Electron Affinity and F has negative Electron Affinity. So should I consider signs and say Be has more Electron Affinity than F or consider magnitude and say F has more Electron Affinity than Be?
How to compare their Electron gain enthalpies also at 0K?

In my book it is given that first electron gain enthalpy is greater than second for elements. Should we compare the magnitudes in such cases or the actual numbers with signs?

Successive Electron Gain Enthalpies
After the addition of one electron atom becomes negatively charged and second electron is to be added to a negatively charged ion. Thus, due to electrostatic repulsion energy is always required for addition of second electron. First electron gain enthalpy is always more negative than second electron gain enthalpy (which is essentially positive). Author made an ambiguous statement, as we can't just compare the magnitudes. For more clarity see:
$$\ce{S_{(g)} + e- -> S- _{(g)}}             \Delta_{eg} H^- _1 = \pu{-200kJ mol^{-1}}$$
$$\ce{S- _{(g)} + e- -> S^{2-}_{(g)}}             \Delta_{eg} H^- _2 = \pu{600kJ mol^{-1}}$$.

Does the same comparision hold with Electron Affinity at 0K?

Generally, if energy is released when an electron is added to an atom, electron affinity is taken as positive, contrary to thermodynamic convention. That is, if $\Delta_{eg} H^-$ is negative EA is positive (since temperature mentioned is absolute zero).
At absolute zero, $\Delta_{eg} H^- = \pu{-EA}$. Thus if $\Delta_{eg} H^- _1$ is negative, EA is positive and successive EAs will be increasingly negative. Here, you can say first electron affinity is greater than second (actual numbers with sign).

For example, Be has positive Electron Affinity and F has negative Electron Affinity. So should I consider signs and say Be has more Electron Affinity than F or consider magnitude and say F has more Electron Affinity than Be?

First, $\ce{Be}$ has negative electron affinity ($\pu{-0.5 eV atom^{-1}}$) and $\ce{F}$ has positive electron affinity ($\pu{+3.4 eV atom^{-1}}$). So, fluorine has more electron affinity than beryllium, consider the actual numbers with signs).

How to compare their Electron gain enthalpies also at 0K?

Since, at 0K, $\Delta_{eg} H^- = \pu{-EA}$. Thus, first electron enthalpy of fluorine is more negative than that of beryllium. Strictly speaking, it's value is less than that of beryllium.