text:
Diagram of the electric field potential around water
Is there diagram of the electric field around a water molecule.  I have seen some simplifications that treat hydrogen as a positive point source and oxygen as a negative point source. But I am looking for one that shows the how the shapes of the orbitals effect the field.

Complications of complex quantum chemistry calculation greatly surpass the gain, compared to 3 partial point charges system.

"potentional"? ;p It would be easier to to find electron density diagram, than of potential.

Spelling error fixed.  Though I also found out a new word "protention (philosophy) An anticipation of a future event"

You might be asking about an electrostatic potential surface/diagram... They illustrate the electrostatic charge around the molecule at some distance away from the atoms in the molecule (you basically get an idea of how positive or negative the region around the molecule is). The diagram is only valid for where the electrostatic potential calculations were run (the coordinates where the charge was evaluated) — in areas closer or further away from the molecule, your surface would be different (i.e. magnitudes of the charge would probably be larger), and the trends would change.
If you're looking for a cross section, as opposed to a surface, you might be asking about a charge density plot, which are essentially heat maps (sometimes contour plots) of the charges in certain regions around atoms/molecules.
Or, to be honest, you might just be looking for a simple 2D contour plot of the orbitals. They're easy to find, and will often separate the different orbitals to show each of their shapes.
That being said, while I was looking for the names of these diagrams, I came across an article$^1$ with several warnings about accepting the implications figures of contour plots and other orbital representations. Not sure what level of chemistry you're at (high school, undergraduate, PhD etc.), but might be worth checking it out.

Autschbach, J. Orbitals: Some Fiction and Some Facts. J. Chem. Educ. 2012, 89 (8), 1032-1040. DOI: 10.1021/ed200673w.

That is almost what I want. But rather then a surface around the shape I want a plane cutting through the atoms.

@QthePlatypus added to my answer. Does that help?


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How can I compare the UV/Vis absorption difference after metal ion chelating of crown ether-containing polymer?
Recently I have synthesized the polymer that contains 1-aza-18-crown-6 in the repeating unit. I want to know if there is any UV/Vis absorption difference after the chelation of metal ion. Considered metal ions are Cs, Co, Sr. In the other papers that compare the UV/Vis absorption difference, they use some metal salts that are soluble in some organic solvents like $\ce{CH_3CN}$, EtOH, or MeOH. And in this case the crown ether-containing compounds are also dissolved in $\ce{CH_3CN}$, EtOH, or MeOH. But the problem is, in my case, the synthesized polymer does not dissolve in $\ce{CH_3CN}$, EtOH, or MeOH. It dissolves in $\ce{CH3Cl}$, $\ce{CH2Cl2}$, toluene, chlorobenzene, THF, and dioxane, but does not dissolve in $\ce{CH3CN}$, EtOH, MeOH, acetone, hexane, ether, and EtOAc. I tried to find some Cs, Co, or Sr salts that are soluble in THF, but I couldn't find it. Some said that $\ce{Cs2CO3}$ and CsF dissolves in THF, but it failed as I tried. In this situation How can I compare UV/Vis absorption difference after metal ion chelating based on solution measurement?

I want to measure the difference for UV/Vis absorption, not other spectroscopy!

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About the salt: Instead of $\ce{CsCO3}$ I would try $\ce{CsCl}$ for which sciencemadness reports a solubility in ethanol of $\pu{0.76 g/100 mL}$ (at $\pu{25 ^\circ{}C}$).  Perhaps this little already is enough. If your institution is subscriber, Pabsch et al. report comparison of experimentally determined solubilities of inorganic salts (including $\ce{CsCl}$, $\ce{CsHCO3}$, and $\ce{Cs2CO3}$) in organic solvents methanol, ethanol, and N-methyl-2-pyrrolidone (NMP).  According to the freely available SI (Table S4), both $\ce{CsCl}$ and $\ce{CsHCO3}$ worked well enough for MeOH, EthOH, NMP. If your institution is subscriber to the Reaxys database, solubility of salts (expressed in Hill formulae) is an indexed property, too.
To the spectroscopic experiment:

Does / do your polymer(s) without the metal cation loading dissolve well enough in these three solvents (perhaps especially NMP)? If so, record their spectra once "empty", and once loaded.

Often (early level of teaching lab report), the UV spectra's ordinate is absorption; this helps to note if the data are recorded within the linear range of the detectors.  For the subsequent comparison, it can become more meaningful if substituted by $\log \varepsilon$ as function either over $\lambda$ [nm], or $E$ [eV] instead.  With this in mind, you don't need to record the UV-Vis spectrum in one sweep either; if suitable/necessary prepare e.g., one solution suitable for the interval of $200 \dots \pu{300 nm}$, and a second one for the range of $300 \dots \pu{700 nm}$ to both stay well within the dynamic range of the detectors.  But equally to take full advantage of the linear range of the detectors, too.

Third, with the data in hand, compare both the position (energy scale) and $\varepsilon{}$ of the absorption peaks for the unloaded vs the loaded ligand.  There will be some variation of the spectra's bands because of the solvent (solvatochromism, red / batho-, or blue / hypsochromic shift).  Hence compare the spectra pairwise (both in MeOH, or both in NMP) about the "empty" and "loaded" ligand.  You can facilitate their determination with spectra recorded with good resolution / small increments while sweeping the wavelength, and processing the collected data by derivative spectroscopy. Agilent's application note (link to a pdf) can serve as a primer to this useful technique.  Depending on the spectrometer software, this functionality is not promptly seen (invest some time to read the documentation).  Alternatively, Tom O'Haver's pragmatic introduction to signal processing dedicates a special section to this topic (with templates for spreadsheet Excel .xls / Open Office Calc .ods, and MATLAB/Octave scripts ready to use).

Pabsch, D.; Figiel, P.; Sadowski, G.; Held, C. Solubility of Electrolytes in Organic Solvents: Solvent-Specific Effects and Ion-Specific Effects. J. Chem. Eng. Data 2022, 67, 2706-2718. https://doi.org/10.1021/acs.jced.2c00203.

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How does an equilibrium move when increasing the concentration of water?
I have the following equilibrium given:
$$\ce{[Co(H2O)6]^2+(aq) + 4 Cl-(aq) <=> [CoCl4]^2-(aq) + 6 H2O(l)}$$
(I imagine the (l) for water is wrong, as the water is a part of the solution)
The $\ce{[Co(H2O)6]^2+}$ complex is pink, and the $\ce{[CoCl4]^2-}$ complex is blue.
I had to more or less answer the following question: "Would adding water shift the equilibrium?"
How would you describe the shift in the equilibrium by adding more water?
As I understand it, Le chatelier's principle would certainly shift the equilibrium towards causing a pink coloration. Two chemistry teachers I have spoken to have disregarded this and tell me that I am wrong, whilst a higher-level chemistry teacher, and multiple videos demonstrating the exercise agrees with it "causing a shift". Since a "dilution" often does not shift an equilibrium, they claim that only a little water could cause a shift. I do, however, think that even at a microscale, the water molecules would still cause a shift, even if we used an absurdly large amount of water. Realistically, you would use an amount equivalent to a beaker for example, which is used in the video below.
For those curious about empirical evidence:
Cobalt Chloride Equilibrium on Youtube video, RamZland channel
This question is grade deciding, as I got a split between two grades, so it would be nice to hear some professional input. The question was specifically  "What will happen to the equilibrium and what will we observe when we add water?" I specifically answered that it would shift towards the side that caused a pink coloration, as I think the equilibrium will always change at a microscale, although be it less at larger volumes of water. Using an infinite amount of water is not a realistic consideration, just like asking if $\pu{0.000001 g}$ of chloride would. Keep also in mind I got zero points for this answer, and that the solution just says the equilibrium does not shift at all. No water or solution concentration is given in the question.

Comments have been moved to chat; please do not continue the discussion here. Before posting a comment below this one, please review the purposes of comments. Comments that do not request clarification or suggest improvements usually belong as an answer, on [meta], or in [chat]. Comments continuing discussion may be removed.

I have the following equilibrium given:
$$\ce{[Co(H2O)6]^2+(aq) + 4 Cl-(aq) <=> [CoCl4]^2-(aq) + 6 H2O(l)}$$

This above equation may rather confuse you, for your purpose is better to write it just as:
$$\ce{Co^2+ + 4 Cl- <=> CoCl4^2-} \tag{1}$$
with the equilibrium constant:
$$K = \frac{a_{\ce{CoCl4^2-}}}{a_{\ce{Co^2+}} \cdot \left(a_{\ce{Cl-}}\right)^4} \tag{2}$$
Activities of ions in concentrated solutions ( as implied from scenario) is very difficult to estimate or to obtain as experimental data. I will therefore use for this educational illustrative purpose concentrations instead of activities, being fully aware of significant errors:
$$K_c = \frac{[\ce{CoCl4^2-}]}{[\ce{Co^2+}] \cdot [\ce{Cl-}]^4}\label{3}\tag{3}$$
Dilution of the solution by water  has similar effect as decreasing pressure ( and therefore partial component pressures) for this synthesis:
$$\ce{3 H2 + N2 <=> 2 NH3}\tag{4}$$
with the equilibrium constant
$$K_p = \frac{p^2_{\ce{NH3}}}{p_{\ce{N2}} \cdot p^3_{\ce{H2}}}\label{5}\tag{5}$$
In both cases, $\eqref{3}$ and $\eqref{5}$, dilution, respectively pressure decrease, causes increase of the value of the respective fraction, as denominators decrease in both cases faster then numerators.
From the equation below, dilution to twice as much volume leads to 16 times lower ratio of concentrations of chloro and aqua complexes.
$$\frac{[\ce{CoCl4^2-}]}{[\ce{Co^2+}]}\label{6}\tag{6} = K_c \cdot [\ce{Cl-}]^4$$
Therefore, dilution causes $\ce{CoCl4^2-}$  to gradually converted back to $\ce{Co^2+}$.  In reality, the effect is even stronger, as activity coefficients start to steeply raise for concentrated solutions.

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Reversibility for Entropy Changes in Chemical reaction
There is a problem in my textbook that asks to calculate the entropy change in surroundings for the Haber process, assuming that the enthalpy change is -92.4 kJ/mol. The solution to this problem is to use the formula ΔS_(surr) = - ΔH / T, for T = 25 degrees celsius. I understand the usage of this formula in general, but I am confused as to why this formula can be used if the problem never specified that the reaction was reversible? From my understanding, the problem should have specified this to be reversible, because only then does ΔS_(surr) = -Q_rev / T, and under a constant pressure situation, the formula becomes ΔS_(surr) = - ΔH / T.

Whenever entropy changes are calculated we must apply the condition of 'reversibility' giving $dS=dq_{rev}/T$ at constant temperature $T$ and where $q_{rev}$ is the thermal energy. There is, however, no reason for the actual process to be reversible, because entropy is a function of state and $\Delta S$ is the same whatever the route taken between the starting and ending points. We usually need to devise, on paper, a reversible route merely to perform the calculation.
(btw. By 'reversible' we mean that the process occurs at an infinitesimal rate so that the thermodynamic system is always effectively in equilibrium. A minute change in one direction can be reversed by a similar minute change in the opposite direction. Under these conditions the maximum amount of work is obtainable. As an example think of a tug-of-war contest between two well matched teams. The amount of work done depends on the opposing teams strength not their own inherent strength, and the maximum amount of work will be done when the teams are very evenly matched and so are working flat-out. The performance will then take a long time.)

so does that mean we can use ΔS_(surr) = - ΔH / T because entropy is calculated along a reversible route (irregardless of the actual process), and the heat along the reversible route will equal ΔH if it is conducted at constant pressure (which is true for many reactions), so - ΔH / T is essentially always correct to calculate the entropy change at constant pressure for the surroundings. I assume that would mean, by similar reasoning, we can use ΔS_(sys) =  ΔH / T also then?

You are correct that $\mathrm{d}S = \dfrac{đq}{T}$ only holds for a reversible process. However, the usual convention is to assume that the “surroundings” are so big that they never change temperature no matter how much heat you pump into (or out of) them. If the temperature is constant, we can take it outside the integral:
$$\Delta S_\text{surr.} = \int\mathrm{d}S_\text{surr.} = \int\dfrac{đq_\text{surr., rev.}}{T} = \frac{1}{T}\intđq_\text{surr., rev.} = \frac{\Delta H_\text{surr.}}{T} = - \frac{\Delta H}{T}
$$
You can see that we choose to take the integral over a reversible process (so that we know the right formula to use), but actually, because $S$ is a state function, the answer is the same regardless of whether or not the process followed in real life is reversible.
In practice it is very rare for chemical reactions to happen in a thermodynamically reversible way.

Hello Aant. We agree that this is corret. However, you seem to affirm that $dq_\pu{surr,rev} = dH$ in one of the equalities. This is true for a mechanically reversible process. But if the path where reactants are converted into products is not reversible, then, how do you prove that that equality still holds? At the end you recognize that no chemical reaction traverses along a reversible path.

Doesn't that simply follow from the definition of enthalpy and the assumption (admittedly tacet here) that the reaction proceeds at constant pressure? There's no requirement for reversibility here.


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when would copper material have free electrons
If you imagine a wire of copper(note that I am not talking about electricity at all). Just only a wire of copper or copper metal or whatever full of copper atoms.
We know that each copper atom has 1 electron in the most outer shell and is very loosely bound to the atom. but at that point of time, even though it is still loose, it is still bound.
At a room temperature, inside this wire, would these loosely bound electrons(1 for each atom) become free or not ? I checked and the binding energy for this electron is
7.72eV so at room temperature, I dont think any of these electrons become free. Do they or don’t ?

If monkeys are jumping between trees, it does not mean they are flying.

Related: https://chemistry.stackexchange.com/questions/85740/what-happens-when-an-electron-in-a-metal-is-excited/85743#85743

When we are talking about a conductor material (metal), we should consider the behavior in the bulk phase (ie. a lot of atoms together, generally in some kind of crystalline order). The electrons in this situation behave differently compared to a free atom - exactly because of the metallic bond between the atoms. The binding energy you cite is about how an electron behaves when you remove it from a single atom, hanging is space and leaving a positively charged ion behind.
In bulk metal when an electron leaves one atom it is not leaving a positive ion behind, other electrons come in. The leaving electron also spread around many other atoms, and not just hanging in vacuum. Imagine the situation as the electrons would be in a constantly moving crowd, where everyone is moving up and down and can step in to empty spaces. The situation is a little bit similar situation how delocalized electrons behave in organic molecules with two crucial differences:

the electrons are delocalized in the whole system, not just along one set of bonds,
energetically speaking there are more space than electrons, so electrons have some space to move.

If you look at electrons ie in benzene, some are delocalized in the aromatic ring, but they require significant energy to go to an excited state. If you look at the electrons inside the metal, the electrons with the most energy (around the Fermi level) can be excited with an infinity small energy, which can kick these electrons to different excited states and move them around.

If no current applied, why do valence electrons jump around ? They must have needed energy to absorb to become free electrons.

@chem They, aren't "free" just delocalised (and there's always thermal energy).


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Are there exceptions to the rate law for elementary reactions?
According to several authentic sources (for example, ChemLibreTexts), for elementary reactions, the order is equal to molecularity according to the rate law.

Example: For an elementary reaction $$\ce{aA +bB -> cC +dD}$$ the rate of reaction is given by $$R = k [\ce{A}]^{a} [\ce{B}]^{b},$$ where $k$ is the rate constant and $[\cdot]$ represents concentration.

I am aware that, in real solutions, concentrations are replaced by activities; however, the structure of the real rate law is still same as that obtained from the ideal rate law.
I wanted to know if there are any exceptions to this. Is there an elementary reaction for which the rate law is not obeyed, and the molecularity is not equal to order?

I think it would hold perfectly, if not for things like fast two-step vs one step ambiguity.

Your first equation tells us nothing about the way the species react other than $a$ moles of $[A]$ plus $b$ of $[B]$ react and so on. The example by Poutnik below for $H_2/Br_2$ clearly shows this where the rate $R$ is not of the form you quote even when $a=b=1$. Order is an experimental quantity only found with great difficulty. If you like, the connection between $R$ and your chemical equation is  fiction and based on assumptions about the mechanism as a way or trying to fit mechanism to experiment.

@porphyrin the rate law helps us in this regard; if it is known that a reaction is elementary, we can determine the order theoretically using just the stoichiometry of the reaction.

@anata , 'if is known that the reaction is elementary' this is just the point, you do not know this unless an experiment is done, even atom+atom.

It is true that, for elementary reactions, the reaction rate order is equal to molecularity.
The point is, for $a + b \gt 3$ ( and often for $a + b = 3$ or even $a + b = 2$ ), the reaction is not elementary.
Typical case is $\ce{Br2 + H2 -> 2 HBr}$. It looks like it may be an elementary reaction, but it is not. It has the loopbacked chain reaction schema. Its reaction rate is a rational function with some non-integer exponents:

$$r= \frac{k[\ce{H2}][\ce{Br2}]^{1/2}}{1 + j[\ce{HBr}]/[\ce{Br2}]}$$

( $k$ and $j$ are rate constants for particular reactions steps withing the reaction kinetic schema. )
Another exception are reactions of the zeroth order, with the rate not depending on the analyte concentration, typically photolytic reactions.

As far as I know, all zero order reactions are complex (and not elementary) reactions, these do not make an exception to the rate law.

Hmm, $\ce{Cl2 ->[h\nu] 2 Cl}$ does not seem complex. at least far from the steady state when recombination matches photolysis.

What is the rate expression for the reaction $\ce{Cl2 ->[h\nu] 2 Cl}$? Actually, even my first thought was photodissociation reactions. I wanted to confirm if this is the case.

R_forward = k. I[W/m2], net rate R = k.I - k2 . (p(Cl2))^2

Could you please cite this?

I cannot, I always fail in referencing particular neurons. The forward part is typical for photolysis, the recombination is the elementary reaction.

@Poutnik what is j?

The reaction $\ce{Br2 + H2 -> 2 HBr}$ seems to be a chain-reaction rather than an elementary reaction based on web search. This doesn't seem to be an exception to the rate law. I think this reference clarifies it.

I have not said it is an elementary reaction, I was just hinting it may look like one. As, OTOH, $\ce{I2 + H2 -> 2 HI}$ is not a chain reaction, but elementary reaction, forming $\ce{I2 \bond{...}H2}$ activated complex. //
j is rate constant for particular steps in the reaction schema.


text:
According to PV=nRT,if we increase temp,no of moles will decrease,but no of moles is dependent on mass,mass is constant,then how can moles change?
We know the ideal gas equation is PV=nRT,then, according to this,if we keep Pressure,volume to be constant,then on increasing temperature,no of moles will reduce,this would mean that mass of matter or particles of the gas would simply vanish????how????how is conservation of mass followed here?

How are you going to keep the pressure constant? And please refrain from improper terms like "temp", "no", "number of moles" etc. as well as illiterate punctuation.

Do not try to put as much info to the title as possible. Think about titles like magazine or newspaper headlines.

There are only 2 ways how to change the molar amount of gas ($n$) in the system:

We add/remove some gas, so the system is open and the mass conservation law does not apply on the system content.
There is ongoing chemical ( or nuclear ) reaction with different number of reactant and product entities and $n$ is changing due that.

Otherwise, $n$ is given as a fixed parameter. As $R \approx \pu{8.314 J K-1 mol-1}$ is a physical constant, we can reduce equation
$$pV = nRT$$
to
$$\frac{pV}{T} = a$$
where $a = nR$.
We can freely change any two of all three quantities ($p$, $V$, $T$), but the remaining third one is fully determined by values of the other two.
So $p=f(T,V)$, or $V=f(T,p)$ or $T=f(p,V)$.


text:
why chemical bond between Na and Cl happens
We know that if Cl and Na get too close, they produce ionic bonding.
Cl has 17 proton and 17 electrons and is considered stable.
Na has 11 protons and 11 electrons and is considered stable.
I understand that in Na, we have 1 valence electron and in Cl, we have 7 valence electrons. Now, if they get too close, 1 valence electron is transferred from Na to Cl. I don't understand why. Na was already stable. I get that part that Cl, since it has more protons, it would attract the electron from Na more than Na would do for itself due to less protons(11), but when ionic bond happens, Cl would end up having 17 proton and 18 electrons. Wouldn't Cl itself become unstable as there're more electrons than protons ? It seems like that before bonding, they both were stable, but after bonding, Cl ended up not stable.
I'm trying to understand it, but I get a point that explanation with quantum would be a waste of time as I have no knowledge in quantums. So would appreciate the explanation in terms of attraction forces and why after bonding, Cl doesn't become unstable. Note that octet rule and staff like this is something I don't really need to know as they're rules and not explanation in terms of attractions. I checked this why do atoms want to have a full outer shell but doesn't help.

Stable and not reactive are two different concepts.

Fluorine and hydrogen are perfectly stable, but if mixed, even at deep low temperature -200 deg C, they explode.

Na (neutral) and Cl (neutral) are stable when separated. They are just more stable should they bond together.

@Volpina The question is wrong on multiple levels and was already asked many times. Not to mention title lacks capital letter and question mark - these are reasons to downvote. https://chemistry.stackexchange.com/questions/21981/what-physical-properties-are-manifested-when-an-electron-leaves-na-and-moves-to https://chemistry.stackexchange.com/questions/47904/energetics-of-bond-formation https://chemistry.stackexchange.com/questions/32533/why-is-an-ionic-bond-a-chemical-and-not-a-physical-bond https://chemistry.stackexchange.com/questions/59072/why-is-the-formation-of-nacl-an-exothermic-process

Here is an easier explanation not relying on "stability" or quantum mechanics.
You start out with some sodium ions and some chloride ions in aqueous solution. Each sodium ion is surrounded by water, and each chloride ion is surrounded by water. Then, you let the water evaporate, the sodium and chloride ions come together and form a crystal, where each sodium ion is surrounded by six chloride ions, and each chloride ion is surrounded by six sodium ions.

We know that if Cl and Na get too close, they produce ionic bonding.

While ionic bonding is often explained as a transfer of electrons, it would be hard to find chlorine and sodium as elements. It is much more common that you already have sodium ions (as in sodium acetate or sodium hydroxide), and already have chloride ions (as in magnesium chloride) or make them from a molecular compound (as when hydrogen chloride gas comes in contact with water).
If you start with ions, you just have to figure out how they form an ionic solid under the right conditions (e.g. taking away water as a solvent).

Na has 11 protons and 11 electrons and is considered stable.

Elemental sodium is highly reactive. When it comes in contact with air or water, it will form ionic compounds or solvated ions. As you can read in the comments, "stable" becomes meaningful by saying stable with respect to some condition or in combination with which other substances.

I'm trying to understand it, but I get a point that explanation with quantum would be a waste of time as I have no knowledge in quantums.

If you want to have an explanation rather than a collection of rules and facts, you have to engage with quantum mechanics. All other explanations are incomplete. There is a perfect explanation of the strength of ionic bonds using pre-1900 physics, but to explain why the ions form in the first place, you need quantum mechanics.

Thanks so much. So when we have neutral sodium and on google, it says it's stable, it actually is stable in vacuum when only that sodium exists, but in respect to the world(putting it in the real world), it actually is not that stable anymore. Right ? I wondered why we can't explain it with attractions. when 2 neutral atoms come(sodium and chloride), maybe because chloride has more protons in it, its attraction for that valence electron is bigger than that valence electron is attracted by 11 protons in sodium, hence attraction dominates and Cl takes Na's electron. Is this wrong ?

"because chloride has more protons in it, its attraction for that valence electron is bigger than that valence electron is attracted by 11 protons in sodium, hence attraction dominates and Cl takes Na's electron. Is this wrong ?" That might work for NaCl, but does not work for NaF, right?

you're correct. my logic is wrong.


text:
Resazurin not turning pink
I want to add resazurin to my media as an anaerobic indicator.
To my understanding, the resazurin is blue when first added to the media, then turns pink after autoclaving, and finally it turns colorless when adding a reducing solution (sulfide in my case). However, I’m having trouble to make the media change from blue to pink. I’ve autoclaved and it’s still blue. My media contains KCl, NH4Cl, KH2PO4, Na2SO4, NaHCO3. I'll add sulfide and other metals after autoclaving.
Should I flush my media before autoclaving? what controls that first reduction step? Heat alone seems not enough to reduce the resazurin. Should I add another reductant in addition to sulfide?


text:
Is the activation energy for vaporization equal to the enthalpy of vaporization?
When ethanol, say, evaporates, the intermolecular forces between molecules need to be broken. As a vapour, there are no new forces being created, so it is a purely endothermic process, with no "hump" in the enthalpy profile, i.e. the potential energy maximum is the final state. Is this logic correct?
If so, then wouldn't the reverse process have no energy barrier? Presumably the only reason this wouldn't be spontaneous at any temperature would be because of entropy changes.

Please notice that evaporation and vaporisation are not same.

Yes, you are essentially correct. The vapour pressure above a liquid is give by $P=Ak_BT\exp(-\Delta E/(k_BT))$ where $A$ is a constant and $k_B$ the Boltzmann constant. Experiment shows that plotting $\ln(P/T)$ vs $1/T$ gives a straight line.

@Rafael Hello Rafael. I would just like to add, that there is a jump in the enthalpy profile when liquid vaporizes to gas. In fact, every thermodynamic property ($V$, $A$, $S$, $H$, $U$) at the liquid saturated state suffer an abrupt change into the gas saturated phase. The only exception is the Gibbs Energy $G$. Do you agree? If you don't believe me, you can check the steam tables.


text:
Color change when heating organic compound
I did just attend a lab class where I should synthesize  3,6-dibromo- N-propylcarbazole from N-propylcarbazole by bromination. Once, the bromination was over the color of the reaction content was yellowish and I then started recrystallising of what I hope was 3,6-dibromo- N-propylcarbazole.
After recrystallization, the solid product I achieved was white in its color (which it should also be as described in the manual). But, unfortunately I did not achieve a high enough yield, so I decided once more to make a recrystallization  of the reaction mixture (solvent) obtained upon the first recrystallization  because I could clearly see a lot of the product was still dissolved in the solvent.
Unfortunately, the amount of solvent (glacial acetic acid) was too high, so i needed to make a destillation to get rid of the excess glacial acetic acid. However, once heating/boiling the solvent mixture, it started to change color from yellowish (probably due to bromine) to green - consequently I did obtain green crystals instead of white ones. It did not happen during my first recrystallization, but why now?
I asked my tutors, the simply did not know or they came with arguments like: "do not trust change color in organic chemistry", "your product may have been broken down while heating it, "it may haven been oxidized". I made a TLC to confirm whether the green crystals were the same as the white ones and it seemed so. In order to clarify it more precisely I have now made a H-NMR but I am still waiting for the answer.
Do you have an idea why my reaction mixture changed color to green during the second recrystallization upon heating it? I simply do not believe it is due to the impurities.

And yes, in both recrystallization processes I did heat the mixture but during my second recrystallization I needed to remove the excess solvent which I did by boiling it away, but suddenly the color changed to green when the temperature increased.

And I must also say that no heat was required to synthesize the product

Was HOAc the solvent of reaction, or the solvent of recrystallization? Can you record the melting points of the crops of crystallization and compare them with each other / a literature value? You don't need much material for this (capillary/heating stage microscope) and it is cheaper than recording an NMR spectrum / comparison with (ATR) FT-IR of an authentic sample. The color of crystals can be indicative for a different polymorph, or potential inclusion of (solvent) molecules (alkyl chains on carbazoles can be used to tailor crystallinity / glass transition temperatures of org. electronics).

HOAc was both the solvent of the reaction and recrystallization. And also thank you for your answer and time.


text:
Which way Le Chatelier's principle
I have been pondering about the effect of pressure change in regard to Le Chatelier's principle on reactions. For this, I considered the following reaction:
$$\ce{CO(g) + 3H2(g) <=> CH4(g) + H2O(g)}$$
Le Chatelier's principle states that when the pressure on the whole reaction mixture increases, the equilibrium will shift in the direction where the number of moles of reactant(s)/product(s) is lesser as the partial pressure of a gaseous substance is directly proportional to it's mole fraction. This happens to oppose the effect of pressure being applied that tends to disturb the established equilibrium.
But here is my argument that, if external pressure applier has to be opposed, why can't it just shift backward instead? If it would shift in the backward direction where there are greater number of moles, will it not be more effective to oppose the external pressure?

What is the meaning of the first sentence in the last paragraph, that I show here : if external pressure applier has to be opposed ? Which applier ? Why oppose ?

Means external pressure applied on the reaction environment is to be opposed in order to revert it's effect on the reaction. That is, I think, what Le Chatelier's principle says.

External pressure can be applied anyway, I suppose. Like a piston or so?

Rather than “opposes”, say “partially reverses” the change imposed. So you increase the pressure, and the reaction goes (net) in the direction that reduces it.

Like when all the concentrations are minute?

Expert explanations here is always helpful in clarifying science concepts/problems. Perhaps following question could also shine some light along the way: this and this

This can get confusing. What you seem to have expressed is you want to reduce the effect of an increased external force by increasing the internal pressure. This can be done in such a situation of a cartridge primer where the firing pin initiates a chemical reaction to give an increase in pressure; that is not a system at equilibrium. A discussion of LeChatliers principle is about the effect on the equilibrium under the changing conditions. Please Simplify your question in your own mind.
At equilibrium the rates of the forward and reverse reactions are equal; the equilibrium constant is the ratio of the rate constants. This reduces to the reaction quotient. A simple increase in pressure will increase each factor in the reaction quotient by the same ratio. There are 4 factors one CO and 3H2 in the forward reaction and two in the reverse. Increasing the pressure accelerates the forward reaction more. In a quick pressure increase it could be possible to follow the reaction by monitoring the pressure.

You can use the equilibrium constant to help.
This is $\displaystyle K_e=\frac{P_{CH4}P_{H2O}}{P_{CO}(P_{H2})^3}$. We can express the partial pressures $P_{CH4}$ etc. in terms of the extent of reaction $\alpha$ and the total pressure $P_T$ by using an ICE table but instead of working this out let us write the partial pressures as a function of $\alpha$ as $P_{CH4}=f(\alpha)_{CH4}P_T$ etc. (and we shall not need to know what it is) and so get
$$\displaystyle K_e=\frac{f(\alpha)_{CH4}f(\alpha)_{H2O}}{f(\alpha)_{CO}f(\alpha)_{H2}^3}\frac{1}{P_T^2}$$
The equilibrium constant is constant as pressure changes so if $P_T$ increases, then the denominator, $f(\alpha)_{CO}f(\alpha)_{H2}^3$ must decrease and consequently $f(\alpha)_{CH4}f(\alpha)_{H2O}$ must increase, i.e more product is formed, just to keep $K_e$ constant.

Your approach to answer is quite insightful. I am curious to know the source of this kind of approach. Could you please share the source of your approach?

No source just me !

The shift of the equillibrium of the reaction doesnt fully cancel out the added pressure exterted on the mixture.We can model such a reaction as a system with negative feedback.And in negative feedback the output resists to changes of the input it doesnt completely cancel out any changes of the input.

So you say, if you kick a ball, the ball should not deform, but you should injure your toes instead. As this would be the principle reversal.
The principle is not limited to chemistry.
The principle means that external change has smaller impact on the system ( in sense of the values of changed parameter, typically temperature or pressure, or change of composition), than it would have if the system did not adapt to it.
If the system response is slow wrt the externally caused change then the system response partially reverses the change. If the system response is fast, it rather weakens the change.
When in doubts, put the qualitative principle aside and just quantitatively evaluate the equilibrium constant equation for partial pressures.


text:
How can entropy of universe or surroundings be a state function?
If a system goes from state A to B in a reversible process, then $(\Delta S)_{\text{universe}}=0$. However, my confusion starts if we take an irreversible route. For the system, even if we take the irreversible route, we can always calculate it using the reversible route from A to B, so the system entropy is a state function. However, the surroundings entropy change is not negative of the system's entropy change in an irreversible process, meaning that $(\Delta S)_{\text{universe}}\neq 0$, despite the process still being between states of the system A and B. This suggests to me that $(\Delta S)_{\text{universe}}$ is not a state function but depends on the path (reversible vs irreversible). I was under the impression that entropy is a state function.
Is this reasoning correct?

Paths are reversible only in the limit. So your premise is incorrect.

A system going through a closed irreversible path increases surrounding entropy, because returning the system to initial state is possible only by having surrounding in other than initial state, with higher entropy.

TL;DR: entropy, for reversible and irreversible processes, for system, surroundings and universe, is always a state function. You are not accurately considering the initial and final states of surroundings (and universe) for reversible and irreversible processes.
The details
In system going from state $A$ to $B$ reversibly, say the state of the surroundings changes from $C$ to $D$. In system going from state $A$ to $B$ irreversibly, considering the same initial state of the surroundings, $C$, the final state will differ from $D$ (say $D'$)

For the reversible process,
$$
\begin{align}
(\Delta S)_\text{uni} &= (\Delta S)_\text{sys} + (\Delta S)_\text{surr}\\
&= S_B-S_A + S_D-S_C \tag{1}
\end{align}
$$
For the irreversible process,
$$
\begin{align}
(\Delta S)_\text{uni} &= (\Delta S)_\text{sys} + (\Delta S)_\text{surr}\\
&= S_B-S_A + S_{D'}-S_C \tag{2}
\end{align}
$$
From equations (1) and (2), the difference between the entropy change of the universe when the system undergoes an irreversible and a reversible process is given by $S_{D'}-S_D > 0$, where the inequality is a consequence of the second law of thermodynamics.

There is no reversible route from A to B if those are the states of the universe.
If A and B are system states, and you take different paths, the surrounding state will be distinct for distinct paths. So in one case the entropy of the universe might increase by a certain amount, and in the other case it increases by a different amount.
However, if the entire universe is changed on a reversible path, there is nothing outside of the universe that can drive that. So either the universe is at equilibrium, with no more increase in entropy, or things are still happening and entropy is increasing.
Bottom line: there is no way to go reversible from a state A to a state B of the entire universe.

The universe can undergo a reversible change. As pointed out by OP, when a system undergoes a reversible change, so do the surroundings and, hence, so does the universe.

@ananta what would be an example of such a reversible process? And how would you change that process to make it irreversible?

A quasi-static expansion/compression of the system is one example of a reversible process. To make this process irreversible, carry out the expansion/compression within a finite time-period (fast).

@ananta I interpreted A and B as the state of the universe. Going from A to B is either reversible or not, there is no choice.

@Karsten Hello Karsten. System can go from $A$ to $B$ reversibly, but the surroundings can't go from $C$ to $D$ reversibly. Am I misunderstanding you? States can be referred to the system and the surroundings, separately, but we cannot talk of step $A$ of the "universe" and step $B$ of the "universe". Do you agree?

@MetalStorm I think so, yes. Again, I used A to designate the state of the entire universe, but it is much more understandable in the other answer, where system and surrounding get their own states with separate letters.

A State Function has a value for a given state of a system. This means that if something undergoes a process and eventually returns to the exact same conditions as it started, its state functions U, G, H, S, PV will be exactly the same. The energy terms U, H, G are conserved so these will remain constant in the universe. S is not conserved and must increase for anything that happens. Since S has not changed in our system it must have increased in the universe. The trick is to expand our immediate universe to determine where and how the entropy increase happens.
This can be difficult but if reduced to extremes is very simple: Everything on Earth depends on Energy from the Sun, Directed energy of higher energy photons of LOW entropy.  To maintain an energy steady state High entropy energy must be emitted. This is low energy infrared photons with higher entropy.
An example: A large Oak tree is measured to determine the energy etc. of everything down to the tilt of its leaves in the Sun and the rotational states of all its molecules.  The tree is cut down, all its roots removed and used for lumber furniture and fuel.  One of its acorns is planted in the exact same spot and with much TLC Grows to an exact clone of the tree down to the rotational states of the molecules. Its state functions are the same as the original tree! but the entropy increase of the universe needed to accomplish this is not only beyond all the supercomputers but beyond imagination. In the end one is content to simply have a new oak tree.

The Gibbs energy is not conserved because its entropy component is not conserved.


text:
Copper sulfate pentahydrate crystal structure
I am interested in what exactly the structure of copper sulfate pentahydrate crystal is like. In the wikipedia, it consists of copper and sulfate strand with 4 water attached to each of the copper atoms in forms of ligands. Some of the oxygen atoms on the sulfate compound seem to have negative charges. For the fifth water molecule, it seems that it is held only by hydrogen bonds.
But here is my question: isn't the whole structure supposed to be quite solid, strongly bonded and devoid of unbalanced charges? Why are there negative charges on the oxygen atoms and the water molecule unfixed in position?
(I'm not sure if it is really unfixed in position though)
P.S. please do not send pictures because my internet somehow cannot access them.

Solid and strongly bonded? Not necessarily; not all crystals are very stable, but this particular one is. Devoid of unbalanced charges? Yes, it absolutely should be and is.

Though simplifying, a crystal structure is a regular spatial arrangement of objects repeated (in an eventually predictable pattern) only by translations in one, or multiple directions. Similar to a stack of shoe boxes with some enclosed content, you copy-paste this whole parallelepiped again, and again. A single layer of soap molecules between water and air prefers a regular arrangement in this plane, hence builds a 2D crystal (similar to some floor tiles' layout and wallpapers). And crystal structures may undergo rapid changes (see liquid crystals in LCD displays of pocket calculators).

The original structure is in Beevers & Lipson Proc. Roy. Soc. A (London) 1934, v 146, p570. The Cu atom is 6 coordinated, four water O atoms in the plane and axially two O atoms each one from a different sulphate ion. The fifth water is held between water molecules attached to cations and O atoms of sulphate ions, i.e this water bridges between two Cu octahedra attached to two sulphate O atoms (on different Cu centres) and an O atom on one of the Cu. The Wikipedia structure does not show this v clearly.

https://chemistry.stackexchange.com/questions/122522/sulfate-and-carbonate-ions-as-ligands https://chemistry.stackexchange.com/questions/544/how-do-copper-sulphate-and-the-like-change-crystal-structure-on-adding-water-of


text:
Chemical equation of wood to charcoal pyrolysis
The equation for the perfect burning of wood is approximately:
$$\ce{C_xH_yO_z + VOCs + O2 -> CO2 + H2O + volatiles + heat}$$
Where $\ce{C_xH_yO_z}$ represents the cellulose and lignin and VOCs are the volatile organic compounds. How would this equation change in an oxygen starved environment aimed at producing charcoal?

You can as well use CxHyOz as the empirical summary formula for particular wood and it's history. It would vary greatly as lignin has less oxygen than polysaccharides and there is variable moisture content.

Rather for "equation"  for lack of oxygen.

And we will never know, we can use only some summary placeholders with empirical formulas.

If it is "perfect" burning, there would be no "VOCs are the volatile organic compounds", just CO2 and H2O (discounting any nitrogen or other byproducts, e.g., NOx).

Better understanding the pyrolysis reaction


text:
Chlorination of 2-phenylbutane

When the compound is chlorinated on the extreme carbons, there is a chiral carbon in both the resulting compounds, which gives a total of 4 stereo isomers. When compound is chlorinated on 2nd or 3rd carbon, it still has a chiral carbon in both resulting compounds. So the total stereo isomers add up to 8 according to me, but it doesn't match the options given.

First of all, welcome to ChemSE! I presume that you are a student, which I am not, and that this problem arises from a preparatory examination. Too many of these types of questions are ambiguous at best and this one is no exception. The sec-butylbenzene as drawn in question 33 must be assumed to be racemic because there is no indication that it is a single enantiomer. The products are presumed to be monochlorinated and that the fractionating column is achiral and infinite in plates. In the diagram below, the Products (P) are presumed to be constitutional and stereoisomeric while the Fractions are designated as (F).
For the racemate, five Products are formed along with their respective enantiomers. Thus, P=10 and F=5, which is not an option in the answers. Even if the secondary chlorides were deemed inseparable, then the values would only be P=8, F=4 and also not an option. If the reactant were a single enantiomer, either R or S, then the total number of Products is six (P=6) and Fractions five (F=5). In this instance, answer (a) would be correct.
BTW: It is advisable not to accept answers too quickly thereby giving others an opportunity to respond. Who knows? There may be an answer you like better.

I am confused by the config change in the benzyl carbon should it not remain R except for benzyl chlorination.  Also liking should have little to do with it. Your much more accurate answer is what should be of importance.

@jimchmst: Introduction of chlorine at the methylene carbon changes the priorities of the atoms attached to the benzylic carbon. The configuration at the benzylic site does not change. Benzylic chlorination causes racemization.

@user55119 It's a common mistake among students that they check chirality at the reacting carbon only.

This is what I feel...

When we chlorinate the two extreme -CH3, a change in the compound occurs only in those positions. So, although a chiral carbon does exist in the compound, it would not contribute to providing an isomer.
When we chlorinate the -CH2 and -CH, 2 isomers each would be possible as the -Cl atom can get attached in 2 ways.
Note that this is because it would follow SN2 mechanism.
Phenyl group wouldn't get chlorinated cause of greater stability.
Therefore, in total, 6 products would be formed.

Let me know if you need further clarity on any of the points.


text:
What would be the product if ozonolysis of an alkene is done in the presence of water?
I have gone across several articles that have mentioned the below:

Alkenes on reaction with O3 and Zn/H2O or any other reducing agent gives aldehyde/ketone.
Alkenes on reaction with O3 and H2O2 gives carboxylic acid/ketone.
Alkynes on reaction with O3 and Zn/H2O gives acid anhydride/diketones.
Alkynes on reaction with O3 and H2O gives carboxylic acid/diketones.

But, no book or article mentions what would happen if ozonolysis of alkenes occurs with water. Would it give aldehyde/ketone or carboxylic acid/ketone?

Presence of water does not really influence the outcome of ozonolysis

Formation of aldehydes, ketones and acids depend on the type of alkene. Let's start with the most basic alkene(ethene), on ozonolysis of ethene we get 2 equivalents of formaldehyde. On ozonolysis of 2-butene, we get 2 equivalents of acetaldehyde. To get ketone as product, we need to use a second degree alkene, let's say 2-methyl-propene. On ozonolysis of 2-methyl-propene, we get propan-2-one and formaldehyde, so you can see that we need different types of alkenes for different types of product. Now talking about carboxylic acid formation, it does not form simply by only ozonolysis. We have to add an extra oxidising agent for example per acid that will oxidise the aldehyde formed by ozonolysis to carboxylic acid. So with water, it will not yield carboxylic acid but can yield aldehydes and ketones. Hope this helps.


text:
How do we synthetis a peptoid from beta-Alanine?
I did a reaction that contains 4 steps described in this picture :

I understood that this 3 steps helped me to synthetis the NGlu part but I don't know why did we do alkylation ? It can't be a protection because otherwise why did we deprotected it before ? Can someone help me please to understand. I read somewhere that it is used as a linker but I don't understand how it can help to polymerize and have a longer chain

You have effectively added a glycine unit

@Waylander how is that a glycine ?

look at the unit at the right hand end - it is NH2-CH2-CO2Et - which is Gly-OEt

@Waylander Oh I understand so it's a way to make the peptoids longer ? what if I want to add another peptoid how should I do ? What is the role of OEt ?

I have olso a question about the protection of the carboxilic acid, does its protection makes it very stable ?

Which carboxylic acid? It is entirely possible to selectively hydrolyse either ester. The tBu ester can be hydrolysed by anhydrous acid (TFA/DCM does this well). Mild base hydroysis (LiOH/THF/water) will hydrolyse the Et ester selectively. Either one can be selectively cleaved and then extended by use of an amine + coupling agent. Many coupling agents based on carbodiimides are used for this.

@Waylander I talk about the reaction from Cbz-beta-Ala-OH to the Cbz-beta-Ala-OtBu

when you say "look at the unit at the right hand end - it is NH2-CH2-CO2Et - which is Gly-OEt ": Does it mean I have a pseudo peptide composed of 2 amino acid : Glu and Gly ?

Yes, you have a pseudopeptide of beta-Ala-Glu. The tBu ester of Cbz-beta-Ala is generally quite stable with well-defined conditions for removing the tBu group.


text:
Mechanism of Denigés' test?
According to Wikipedia, Denigés' test detects tertiary alcohols through their dehydration, marked with a colored precipitate. The reagent itself is some form of aqueous mercury(II) sulfate. My question is, what is the mechanism behind this reaction? I would assume there is some kind of electrophilic attack on the formed double bond as in oxymercuration, but I can't find any answers on what would occur from there or what the precipitate is.

Yeah, I guess it's oxymercuration, with organomercury compound precipitating out.


text:
HPLC Solubility vs polarity
According to this data collected from PubChem, polarity of nitrobenzene and benzonitrile should be similar since solubility is dependent on the polarity:
Nitrobenzene: Solubility in water, 0.2 g/100ml
Benzonitrile: Solubility in water, 0.1-0.5 g/100ml at 22 °C
p-Chlorophenol: Solubility in water, 2.7 g/100ml at 20 °C
But according to the simulator (https://www.multidlc.org/hplcsim/4_2_0/) p-chlorophenol and nitrobenzene are more similar in polarity aspect, they both give peak at 2,2 min.

What is the reason of this difference? Is it H-bonding and being a charged molecule cause more affinity to the polar mobile phase, resulting a difference between solubility and polarity?

Elution times are not matter of a single parameter. See e.g. solvent triangles in HPLC context.

There are a couple of common misconceptions here. First a common mistake is to call a chromatogram a spectrum. A spectrum has nothing to do with a chromatogram generated by a chromatograph. Second common misconception is that aqueous solubility has very little to do with the order of elution in a typical C18 HPLC separation. What is perhaps somewhat relevant is the octanol water partition coefficient. You are completely ignoring the stationary chemistry in your reasoning. Third issue is that polarity is an ill-defined term. The fourth reality is that nobody can predict the elution time for a given black-box column chemistry from scratch for a given set of compounds. The simulator is built by a nice team. However, the retention time is based on experimentally observed data. There is nothing fundamental going on. Most of the retention models are based on interpolation of two or three experimental sets of data.
So look up the log octanol partition coefficients for three compounds and see if you can see a trend.
If you are deeply interested, then there is something called as a hydrophobic subtraction retention model. Forget about polarity, it is an umbrella term with very little utility/.


text:
Magnetic moment of an electron not parallel to its angular momentum?
For the total angular momentum of an electron $\vec{J} = \vec{L} + \vec{S}$, we're often shown a picture like this

where I assume that the $\vec{L}$ and $\vec{S}$ cones are centred on the $J$ axis because it is $J$ that we measure, and know the projection, $J_z$, of onto the z axis, and thus the uncertainty is in the x and y components of $J$.  (Someone verify this for me?)
Both the spin and orbital angular momentum components have a corresponding contribution to the magnetic moment.  The orbital component is $\vec{\mu}_L = \dfrac{-e}{2mₑ}\vec{L}$, however the spin contributes more strongly, so that $\vec{\mu} = \dfrac{-e}{2mₑ}(\vec{L} + g\vec{S})$, where $g \approx 2$.
Looking at this picture, the vector $\vec{L} + 2\vec{S}$ is not parallel to $\vec{J} = \vec{L} + \vec{S}$, and so the magnetic moment is not parallel to the angular momentum.
So Q1.  Is this true that they're not parallel?
Furthermore, the torque $\vec{\mu} \times \vec{B} $ on the magnetic moment $\vec{\mu}$ causes the angular momentum to precess in a circle.  But $\vec{L} + 2\vec{S}$ is not parallel to $\vec{J}$, or even generally on the plane containing $\hat{z}$ and $\vec{J}$, so the torque is not perpendicular to the angular momentum and therefore it does not precess in a circle, or even predictably due to the different directions that $\vec{\mu}$ can point around $\vec{J}$.
Q2. What's going on with the precession?  How do we explain circular precession, or what do we expect to happen instead?
If $\vec{L}$ and $\vec{S}$ are correlated such that $\vec{\mu}$ is in the $\hat{z}-\vec{J}$ then at least the torque will be perpendicular anyone will still get circular precession, but I see no reason to suppose this.
In fact, if we allow $\vec{L}$ and $\vec{S}$ to take any value on those cones, then $\vec{J}$ can have different lengths depending on the relative phase $\vec{L}$ and $\vec{S}$ and whether they are more or less parallel, where it should be fixed $\lvert J \rvert = \sqrt{j(j+1)} \hbar$.
Q3. Are the motions of $\vec{L}$ and $\vec{S}$ correlated?  How do we explain the fixed length of their sum for a given electronic state?

Obviously there are not parallel, but slightly in the expectation values : https://en.m.wikipedia.org/wiki/Land%C3%A9_g-factor

I'm not sure what you mean by "but slightly in the expectation values".
If the magnetic moment $\vec{\mu}$ and angular momentum $\vec{J}$ are not parallel, then $\vec{J}$ will not precess in a circle.

*They're parallel considering the expectation values. Have you read this wiki page? the expectation values of the moment and the total angular momentum are parallel. In this case, the precession is described by a classical expression : the magnetic moment is proportional to the total angular momentum and the factor is  the gyromagnetic factor.


text:
+M or -I for chlorine? Major product in free radical chlorination and bromination of 2-chloropropane?
Is the +M or -I effect more dominant in when an alkyl (mono)halide is subject to free radical chlorination/bromination?
For example what will be the major product in case of free radical (mono)chlorination of 2-chloropropane?
Would it be different if we did bromination?
I have found a similar question: Stability of carbocations attached to halogen atoms which was marked as duplicate but the other question was only about fluorine. I want to know in case of chlorine.

TL;DR: Even though alpha radical of 2-chloropropane might be more stable, due to a higher number of hydrogens present at the beta position, 1,2-dichloropropane might be formed in majority.
Edit: Research indicates 2,2-dichloropropane is probably the major product of the reaction because the radical is stabilized by 6 hyperconjugative structures. The +R effect of $\ce{Cl}$ and +I effect of $\ce{Me}$ groups also makes the alpha hydrogen more labile. More details at the end of the answer :)

The details
Monochlorination of Propane
Firstly, (citing ChemLibreTexts) the monochlorination of propane yields 1-chloropropane in majority:

Dichlorination of Propane
If we were to isolate 2-chloropropane and carry out the reaction further, there are two factors to consider:

Number of hydrogens: only one hydrogen is present at the alpha position while 6 hydrogens are present at the beta position.
Stability of Free Radical: we must consider the stabilities of radical formation at alpha and beta positions.

ChemLibreTexts avoids this problem by simply stating that four isomers are formed in the dichlorination of propane:

I searched a few textbooks, including 10th edition of Organic Chemistry by Solomons, T. W. G. and Fryhle, C.B. and 7th edition of Organic Chemistry by Morrison R. T., Boyd R. N. and Bhattacharjee S. K., where no discussion on this reaction is provided. Why hasn't this been discussed by anyone?
Nonetheless, we are expecting the formation two isomers, 1,2-dichloropropane and 2,2-dichloropropane.
My thoughts
Even though I expect the chlorine atom to stabilize the free radical at alpha position, the beta hydrogens outnumber this and, chlorine radical not being very selective, 1,2-dichloropropane might form in majority. Bromine radical, being more selective, might form 2,2-dibromopropane, or 2-bromo-2-chloro-propane upon bromination of 2-chloropropane.

A Research Study
Since OP was not satisfied with my (very informed) intuition, I thought to go out and find references for this. Now, this took some time, and it has not been explored by a lot of researchers, but I was able to find an excellent research article, namely "Cl Atom Initiated Photo-oxidation of Mono-chlorinated Propanes To Form Carbonyl Compounds: A Kinetic and Mechanistic Approach" by Kumar A. and Rajakumar B. of Department of Chemistry, Indian Institute of Technology Madras, Chennai, India, that, besides many other things, explores the lability of $\ce{C-H}$ bonds in 2-chloropropane. The following is from the cited study.
Figure 1: Naming of transition states for the abstraction of hydrogen by chlorine radical from 2-chloropropane

Figure 2: Transition states involved in the reaction of chlorine radical and 2-chloropropane

Figure 3: Products of the reaction of chlorine radical and 2-chloropropane

Figure 4: Relative energies of the transition states and products in the reaction of chlorine radical and 2-chloropropane

To quote the authors:

For the reaction of 2-chloropropane with Cl atoms, it is clear from
the computational calculations that the abstraction of the H-atom will
be either from −CH3 (TS1 or TS2 or TS3) or from −CHCl (TS4). The
radical formed via TS1 is the primary radical, and it has one
hyperconjugation structure. The radical formed via TS4 is a secondary
radical, and it has six hyperconjugation structures. Thus, TS4 is more
stabilized due to secondary radicals and has more hyperconjugation
structures than TS1. TS4 is also the more kinetically favorable
pathway than other pathways (TS1 or TS2 or TS3) because of the
electron-donating resonance effect (+R) of the directly attached Cl
atom along with the +I effect of the two methyl groups attached to the
radical carbon, which increases the electron density and makes it more
kinetically labile.

Thus, 2,2-dichloropropane is probably (because I wasn't able to find a study that answers the question exactly) the major product of the reaction because the radical is stabilized by 6 hyperconjugative structures. The +R effect of $\ce{Cl}$ and +I effect of $\ce{Me}$ groups also makes the alpha hydrogen more labile.
Hopefully, this goes to show that researching upon one's intuition is extremely important as not only it may contradict our intuition but also provide further insights into the problem.

Why does Cl stabilise the radical formed if -I is more dominating?

@Valoruz because it has empty d-orbitals to accommodate the free-radical electron. -I effect acts on charged species, even if -I effect is acting, it will withdraw the free radical electron density towards the chlorine atom, stabilizing the system.  From your question, you shouldn't know which effect is dominating, so don't make any assumptions. Nevertheless, since you have asked, I will look into the stability of substituted free radicals.

@Valoruz hope the latest edits answer your question.

Thank you for the detailed answer. But I just want to know one thing. Why doesn't the -I effect of Cl destabilise TS4?

@Valoruz perhaps this article would be useful. I am also reading it as I write this comment.


text:
Why is the reduction potential of Cu$^{2+}$ less than Cu$^+$?
The reduction potential is the electrical potential of the species to be reduced. It's basically how positive it is (compared to a reference, since voltages are relative).
Tables list the following
$$\begin{align}\ce{Cu^{2+} + 2e- &-> Cu}\quad &&E^\circ=+0.340\ \mathrm V\\
\ce{Cu^{2+} + e- &-> Cu+}\quad &&E^\circ=+0.15\ \mathrm V\\
\ce{Cu+ + e- &-> Cu}\quad &&E^\circ=+0.522\ \mathrm V\end{align}$$
Why is an electron more attracted to the less positive charge?  And why is the two electron reduction potential not the sum of the individual steps? In fact, if both steps are exothermic, why is the both steps done sequentially, less exothermic than just the second step?
I am aware that $\ce{Cu^2+}$ has a higher enthalpy of hydration, and so $\ce{Cu+}$ will disproportionate in (aqueous) solution, but as I understand, that is not what is being measured here. These potentials should apply in the gas phase or in hexane.
____ Edit____
I realised that the two electron reduction is the average energy for both reductions, so it does indeed sum approximately to the other two.  But my underlying question of why the second step is more exothermic remains.  I'll leave the second part here in case anyone else with this question is following the same line of logic.

We should qualify this properly. Water-solvated copper(II) has a lower reduction potential than water-solvated copper(I). As will be mentioned below, copper -- alone among fourth-period transition elements -- has the property that this difference is small and readily reversible with suitable other ligands.
More is involved here than just ionization energy. It takes more energy to ionize one copper atom twice than two copper atoms once apiece, but the greater electrostatic attraction of copper(II) to hard-base ligands such as fluoride ions or the oxygen ends of water molecules overcomes this difference and makes the +2 oxidation state more favored in such environments.
However, copper(I) has greater polarizability than copper(II), which favors it when more covalent bonding with softer, more polariz-able bases is introduced. This answer discusses the effect of chloride as a softer, more polarizable and more covalently-bonding ligand on copper redox thermodynamics. By contrast, with all preceding transition metals in the fourth period (and most of those in later periods), the electrostatically-driven preference for a +2 or higher oxidation state is too large to be as easily reversed, and the +1 state is realized only is specialized nonaqueous environments.

TL;DR: Standard electrode potentials are indicative of the thermodynamics involved in the half-cell redox reactions. Electrode potentials cannot be directly added since these are intensive quantities and only extensive quantities can be added.
The Problem
The standard reduction potentials do not (directly) reflect the energy changes or heats involved; therefore, we cannot deduce which step is more or less exothermic from the reduction potentials.
The Solution
The following shows calculations of Gibbs free energy for each step:

It is clear that reduction of $\ce{Cu(II)}$ to $\ce{Cu}$ is more exothermic than that of $\ce{Cu(I)}$ to $\ce{Cu}$. The reduction of $\ce{Cu(I)}$ to $\ce{Cu}$ is more exothermic than $\ce{Cu(II)}$ to $\ce{Cu(I)}$ because we are going from a charged (unstable) moiety to a neutral (stable) moiety.
This diagram should be useful:

Caution
Note: Gibbs Potential is not an accurate predictor of the heat involved in a reaction. However, since pressure and volume change only negligibly during the reduction, heat of the reaction may be approximated by change in Gibbs potential.

Thanks for the reply.  I do understand that the energy for the two electron reaction is indeed the sum of the individual steps because the energy for that step is the $\Delta V$ multiplied by 2 electrons.  I'm sorry if that wasn't clear from the edit at the bottom.  My question is why is Cu$^+$ at a more positive potential than Cu$^{2+}$?  These voltages are relative to the same SHE reference potential.  The diagram suggests that the $\Delta V$ is between Cu$^+$ and Cu$^{2+}$.

@FurrierTransform as stated in the answer, since in going from $\ce{Cu+}$ to $\ce{Cu}$, we are making a neutral moiety from a charged one, and this is a favorable reaction.

@FurrierTransform I may be misunderstanding your question. My point is that standard reduction potentials should not be compared with each other; a more accurate understanding of the reactions is obtained from comparing the corresponding Gibbs potentials. However, I have saved this question and will be reading up more about the absolute values of the reduction potential and what can be inferred from these.

@ananta: In solution section does not make sense because there might be at least one typo in the section. Would you please go throuth paragraph and correct it?

@MathewMahindaratne thank you for pointing it out, I have edited the answer for typos. Hope this addresses your comment.


text:
Does chloride ion have its electron in hybrid orbitals?
Where are the electron valamce shell electron located in $\ce{Cl}^-$. Are they in pure $p$ and $s$ orbitals? Or are they in $sp^3$ hybrid orbital?
If the answer is that they are in pure orbitals i.e. they will not be symmetric anymore around the sphere of atom like it will in tetrahedral arrangement.

Hybridization occurs in our heads.

A full shell has spherical symmetry.

Atoms dont have hybrid orbitals.Only molecules have hybrid orbitals.So the added electron in the $Cl^{-}$ anion is in one of the p orbitals.

You know this because.....  And which orbital is this....


text:
Why does increasing the temperature (K) of a 1M Cu²⁺ and 0.001M Zn²⁺ galvanic cell increase the EMF?
I have a galvanic cell composed of a cathode with 1M copper sulphate solution, and an anode with 0.001M zinc sulphate solution. I am increasing the temperature and recording the resultant effect on the EMF. According to the Nernst equation, EMF should increase linearly with temperature in this scenario:
EMF = 0.0003T + 1.1  (I have rearranged the equation)
I don't know why this happens. I thought it was because of Ohm's law, which states voltage is directly proportional to resistance, and hence, by increasing the resistance of the galvanic cell by increasing temperature, the EMF also increased. But I read on other websites that increases in temperature can decreases the EMF too. So I am quite confused and don't know whether I made a mistake with the Nernst equation.
Is anyone able to explain why EMF, in this situation, increases alongside temperature? It would be lovely if there was some theory explaining why (like Ohm's law or something similar). Sorry if this question isn't formatted properly, this is my first one.

There is the explicit dependence in RT/nF .ln(a) terms and there is implicit T dependence of E°. And there is possible temperature dependent kinetic factor of deviation from expectation.

Exactly. RT/nF ln(a) should scale with temperature, but so does E0 aswell. Difficult to predict

How did you rearrange the equation to get EMF = 0.0003 T + 1.1 ?

EMF is proportional to Gibbs free energy, whose partial derivative with respect to temperature is an entropy term.

@Karsten That assumes dH/dT = dS/dT = 0

According to the Nernst equation, EMF should increase linearly with temperature in this scenario.

The statement and the equation you have written are fine. Considering the comments, I will try to extend the equation taking into account the change of the standard cell potential with temperature, in order to see if it has an impact or not. After this I will address your questions.
The system is the Daniel's cell, where the half-reactions and the global reaction, are
\begin{align}
  \ce{Cu^2+ (aq)} + 2e^- &\rightarrow \ce{Cu(s)} \\
  \ce{Zn^2+ (aq)} + 2e^- &\rightarrow \ce{Zn(s)} \\
  \ce{Cu^2+ (aq)} + \ce{Zn(s)} &\rightarrow \ce{Cu(s)} + \ce{Zn^2+ (aq)}
  \tag{1}
\end{align}
Eq. (1), written this way, is spontaneous.
As temperature changes, the standard Gibbs energy of reaction will be modified. We find its dependence with temperature integrating the Gibbs-Helmholtz equation, at constant pressure and concentration of species, and we put in parenthesis whether a property is evaluated in $T$ or $T^\Theta$
\begin{align}
  \frac{d(\Delta_r G^\Theta/T)}{dT} &= -\frac{\Delta_r H^\Theta}{T^2} \\
  \frac{d(\Delta_r G^\Theta/T)}{dT} &= -\frac{\Delta_r H^\Theta(T^\Theta) + \Delta_r C_p(T - T^\Theta)}{T^2} \\
  \int_{\Delta_r G^\Theta(T^\Theta)/T^\Theta}^{\Delta_r G^\Theta(T)/T}
  d(\Delta_r G^\Theta/T) &= \Delta_r H^\Theta(T^\Theta) \int_{T^\Theta}^T -\frac{dT}{T^2} - \Delta_r C_P \int_{T^\Theta}^T \left(\frac{1}{T} - \frac{T^\Theta}{T^2}\right) \; dT \\
  \frac{\Delta_r G(T)}{T} - \frac{\Delta_r G^\Theta(T^\Theta)}{T^\Theta} &=
  \Delta_r H^\Theta(T^\Theta) \left(\frac{1}{T} - \frac{1}{T^\Theta}\right)  \\
  &-\Delta_r C_P \left[\ln\left(\frac{T}{T^\Theta}\right) + \left(\frac{T^\Theta}{T} - 1\right)\right] \\
  \frac{\Delta_r G(T)}{T} &= \frac{\Delta_r G^\Theta(T^\Theta)}{T^\Theta} +
  \Delta_r H^\Theta(T^\Theta)\left(\frac{1}{T} - \frac{1}{T^\Theta}\right)
 \\ &-
  \Delta_r C_P \left[\ln\left(\frac{T}{T^\Theta}\right) + \left(\frac{T^\Theta}{T} - 1\right)\right] \\
  \Delta_r G(T) &= \left(\frac{T}{T^\Theta}\right)\Delta_r G^\Theta(T^\Theta) + \left(1 - \frac{T}{T^\Theta}\right) \Delta_r H^\Theta(T^\Theta) \\
   &- \Delta_r C_P \left[T\ln\left(\frac{T}{T^\Theta}\right) - (T - T^\Theta)\right] \hspace{3 cm} (2) \\
\end{align}
The relation between the standard cell potential and standard Gibbs energy of reaction, in this case, is $ \Delta E^\Theta(T) = -\Delta_r G^\Theta(T) / 2F $. Dividing Eq. (2) on both sides by $-2F$ we get
\begin{align}
  \Delta E^\Theta(T) &= \left(\frac{T}{T^\Theta}\right)\Delta E^\Theta(T^\Theta) - \left(1 - \frac{T}{T^\Theta}\right) \frac{\Delta_r H^\Theta(T^\Theta)}{2F} \\
   &+ \frac{\Delta_r C_P}{2F} \left[T\ln\frac{T}{T^\Theta} - (T - T^\Theta)\right] \hspace{3 cm} (3) \\
\end{align}
Eq. (3) is very general. The only restriction is that we assume the specific heat capacites independent of temperature. The following data (see references) is useful for evaluating Eq. (3):
\begin{array}{c| c c c c}
  \text{Substance/Property} & \ce{Cu^2+ (aq)} & \ce{Zn (s)} & \ce{Cu (s)} & \ce{Zn^2+ (aq)} \\ \hline
  \Delta_f H^\Theta \; \pu{(kJ/mol)} & 64.77 & 0 & 0 & -153.89 \\
  \Delta_f G^\Theta \; \pu{(kJ/mol)} & 65.49 & 0 & 0 & -147.06 \\
  C_P \; \pu{(J/mol·K)} & - & 25.40 & 24.44 & - \\
\end{array}
Thereby
\begin{align}
  \Delta_r H^\Theta(T^\Theta) &= \sum_j \nu_j \Delta_f H^\Theta(T^\Theta) =
  -218660 \; \pu{\frac{J}{mol}}  \\
  \Delta_r G^\Theta(T^\Theta) &= \sum_j \nu_j \Delta_f H^\Theta(T^\Theta) =
  -212550 \; \pu{\frac{J}{mol}}  \\
  \Delta_r C_P &= \sum_j \nu_j C_{P,j}(T^\Theta) =
  -0.96   \; \pu{\frac{J}{mol·K}}
\end{align}
Now we can evaluate Eqs. (2) and (3). The results are as follows:

As the reaction is exothermic, higher the temperature the less spontaneous it will be. As the Gibbs energy goes up, less will be the energy that it can be converted to electrical work to the surroundings. This is reflected in the decrease of the standard cell potential. However, to be honest, the changes are not that important considering this temperature range. As a conclusion, at least for the Daniell's cell, the change of $\Delta E^\Theta$ with temperature is not a determining factor.
Back to the question The Nernst equation is
\begin{align}
  \Delta E(T) = \Delta E^\Theta(T) - \frac{RT}{2F}
  \ln\left(\frac{\ce{[Zn^2+]}}{\ce{[Cu^2+]}}\right) \tag{4}
\end{align}
At fixed concentrations, Eq. (4) is just a function of temperature, but now both members are a function of it. Lets consider your equation in the post (in red color), and the Nernst equation taking into account the change of $\Delta E^\Theta$ with temperature (in blue color):

Note also that the slope of your equation will overestimate the performance of the cell.
For your case, you have an excess of $\ce{Cu^2+}$ with respect to $\ce{Zn^2+}$. While increasing the temperature, you increase the electrochemical potential of both $\ce{Cu^2+}$ and $\ce{Zn^2+}$, remembering that
$$ \tilde{\mu}_j = \mu_{j}^\Theta + RT\ln(C_j/C_j^\Theta) + z_jF\phi^S \tag{5} $$
where $\phi^S$ is the potentail of the liquid phase where the ions are. This translates into a higher tendency of $\ce{Cu^2+}$ of being reduced, and also of $\ce{Zn^2+}$ of being reduced. However, with the concentration difference, copper ions get the upperhand. This tendency translates into a higher potential difference between the terminals in the cell, so that the performance is increased at higher temperatures. In more simpler terms, you are making the job "easier" for $\ce{Cu^2+}$ to release two electrons in the electrolyte/solid interface by giving thermal energy. However, you need to give that energy from the surroundings.

I thought it was because of Ohm's law, which states voltage is directly proportional to resistance, and hence, by increasing the resistance of the galvanic cell by increasing temperature, the EMF also increased.

When considering the Nernst equation, the current is zero, because we are studying the electrochemical equilibrium. This is approximated when you measure the voltage with an "infinite" resistance in the voltmeter. If this wasn't the case, you would see the reaction take place, and the voltmeter reading changing over time. The resistance is always "infinite" in all your readings.
References
Atkins, P. and De Paula, J., "Physical Chemistry", 9th ed, W.H. Freeman and Company New York (2016).

This is hands down an excellent answer. There are just a couple of things I'd correct with respect to formatting. First, we have mhchem for chemical expressions enabled (not just symbols of elements). Second, the standard state is never denoted with the capital letter Theta. Instead, it's either plimsoll (^⦵, Unicode, works with XeLaTeX) or a circle (^\circ, works for pdfLaTeX). Third, textual labels and subscripts are usually upright. And my personal recommendation is to use $p$ for pressure since $P$ is usually reserved for power in PC.

@andselisk Hello andselisk. Thank you! I will make the changes afterwards, so it doesn't bother incoming questions in the feed, when night comes in South America. I didn't quite get the comment about mhchem, where should I employ it?

For example, in place of \ce{Cu^2+ (aq)} + 2e^- &\rightarrow \ce{Cu(s)} just use the \ce{…} macro once: $\ce{Cu^2+(aq) + 2 e^- -> Cu(s)}$.


text:
Why aren't Gibbs energies additive in the serial reduction of cupric ion?
Since Gibbs potential is a state function, it should be additive, regardless of the path chosen to go from a state ($\ce{Cu^{2+}}$) to another ($\ce{Cu}$).
While providing a solution to another question, I noticed that the Gibbs energies are not adding up. Consider the following:
$$
\begin{align}
\ce{Cu^{2+} + 2e- &-> Cu} \hspace{20pt} &\Delta G\\
\ce{Cu^{2+} + 2e- &-> Cu+} \hspace{20pt} &\Delta G_1\\
\ce{Cu^{2+} + 2e- &-> Cu} \hspace{20pt} &\Delta G_2
\end{align}
$$
I calculated the Gibbs potential changes:

I verified the reduction potentials from ChemLibreTexts, which seem close to the accurate values. What is causing this discrepancy?

I had actually misplaced a decimal point and, upon correcting for the accurate values, the discrepancy was resolved.

You may also want to consider significant figures. Since the potentials with copper(II) are given to the nearest 10 mV, the free energies are required to agree only to that level of accuracy; the remaining difference between 680 and 677 mV is not significant.

+3.4 V should catch your eyes, such strong oxidants are not even in fluorine chemistry. :-)


text:
What could be the origin of a persisting distorted water NMR signal?
I am currently running some solvent suppression experiment on water samples. I first run a one scan 1H experiment to get the ''O1P'' and then implement it in the solvent suppression experiment (''zgpurge'' pulse program). However the water signal appears distorted, it is asymmetrical and most likely unphased (other signals appear completly fine), which results in an inefficient solvent suppression experiment. I tried to overcome the problem by performing a 3D shimming, and adjusting the phasing parameters (Autophase during the lock and calibrated the autophase offset) but the problem persisted.
What could be the origin of the problem ?

You should repeat the solvent suppression experiment with different values of O1P, generally this needs to be optimised in order to get the best suppression — it's not necessarily just the frequency at the top of the water peak.

Thank you for your feedback, what bothers me the most is the odd shape of the signal. Also, It could be impractical to repeat the experiment because it is time-consuming. Best solution for me is to have a perfect symmetrical signal, then taking the O1P at top of the peak is enough.

I am not familiar with "zgpurge" specifically (you may want to explain in more detail what kind of suppression it involves).

In my experience, even if you achieve good suppression, the phase of the water peak can vary by quite a bit. Just google for 'water suppression nmr' and you'll see plenty of examples, this was the first one I found. I don't know exactly why this is the case, but radiation damping seems like a reasonable guess. In any case, I think it's quite normal to have an out-of-phase H2O peak, and I don't think that your proposed solution will reliably work.

(Btw, if you can sit at the spectrometer, it doesn't take very long to manually optimise one parameter (O1P) using the gs command. Optimising multiple parameters is a bit harder—as it turns out, I have a paper on this exact topic, which may or may not be interesting—but it might not be necessary at all depending on your use case and what the pulse sequence does. I agree with @BuckThorn that more details about the sequence would be useful; and I think Buck may know more about it than me anyway. ;))

I tried to obtain the optimal O1P using gs command (I took the O1P corresponding to the minimal FID). However, the obtained O1P was not as effective as when I directly choose it from the top of the signal (I compared S/N of both spectra). As for the efficiency of the suppression with regard to the signal shape, I noticed that the suppression was more effective when the signal was symmetrical, the latter was completely wiped out. However, when it is distorted, it just seems that a new signal is appearing after the suppression (like the rise of the C13 satellites of an organic solvent).

How can I send you the pulse program ? so that you can have all the details

Note this has been asked before here. You may edit the original post to include the pulseprogram but it suffices to include a description or reference

@BuckThorn It was me who asked the question in research gate! As a follow up I would like to know if it is possible to tell me how to check for the gradients (as this was the suggestion). I would like to replicate the experiments that have been done by the installation engineer, those are about, the ''gradient recovery'' and ''the gradient profile''. Is it possible to provide a step-by-step procedure on how to conduct those two experiments? btw, the link you provided is exactly about the pulse program I am using. Thank you in advance.

These are questions for the facility manager. It is a technical question who's answer involves a step-by-step protocol that should be tailored to your instrumentation.

You can however try editing your question to include what you wrote in the comments. There are various ways to check the gradient calibration, but this is something someone reasonably familiar with the software and equipment should do. Instructions to obtain a gradient profile and calibration should be included described in the instrument's documentation.

I have to say that, within the framework of quality control (equipment performance control) I redo most of what has been done by the installation engineer, this allows me to check the ''Health'' of the equipment (P90 pulse calibration, Sensitivity, Line shape,...). However, I have never rechecked the gradients since the equipment installation and I don't know how to perform it. As you suggested I should reach a specialized technician. Thank you very much for your feedback.

A very simple way to check the gradient strength is to run a diffusion experiment, if you have ever done that.

I have never done it before, but it shouldn't be too complicated. Thank you once again for your help :)


text:
Is neo-octane possible and if yes, what is its structure?
I know iso-octane is an exception and is 2,2,4-Trimethylpentane which intuitively contains both iso and Neo groups. But then what is the structure of neo-octane or even is it possible?

It is no use making up prefixes for all isomers of higher alkanes. That said, what would you call a 2,2-dimethylhexane?

Neo-octane is not an accepted name.

There is no fixed "neo-octane" to begin with. If you look at isomers of octane, you will see 4 of them have neo group out of which 2,2,4-Trimethylpentane has been commonly referred to as iso-octane although it contains both iso- and neo- group. Even if we eliminate isomers containing iso- group, we are left with 2,2-dimethylhexane and 2,2,3,3-dimethylbutane. Which one will you consider "neo-octane"?
These non-standard prefixes are not defined in IUPAC, so it is better to define them with their standard IUPAC names to avoid ambiguity.


text:
How to express the lattice parameter of NaCl in terms of ionic radii?
In the $\ce{NaCl}$ lattice, $\ce{Cl^-}$ takes the FCC configuration, and all the voids at the edge and body center are filled by $\ce{Na^+}.$ So, do we calculate the edge length of unit cell $a$ as
$$a = \frac{4\,R(\ce{Cl-})}{\sqrt2} \tag{1}$$
because of FCC structure, or as
$$a = 2\,R(\ce{Cl-}) + 2\,R(\ce{Na+}) \tag{2}$$
because the edge void is filled?

I forgot to mention that it is only the chemistry stuff that ought to be put within the braces of the \ce{}, the more mathematical stuff should just be written directly within the dollar signs. To write this math, you have to know some MathJax, for which there is a tutorial on this site: https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference  I have submitted an edit of your post just to potentially stop it from being closed before you learn some MathJax.

Unlike the normal FCC, where the corner and face centred atoms are touching each other, we use a*root(2)=4r, but here in the case of NaCl, the positioning is somewhat distorted because of more number of atoms in the same amount of place.
So the Cl atoms are not touching each other rather the Na and Cl atoms, which are on the edge, are touching each other, and hence we use a=2R(Cl)+2R(Na)


text:
What determines how easily a gas can diffuse into a crystal structure?
I have been told by a metallurgist that the reason a factory oxidizes the iron oxides in ilmenite before reducing them down to pure iron, is that the the products of this have a more spacious crystal structure, thus allowing the $\ce{CO}$ gas (which they use to reduce the iron) to diffuse more wholly and deeply into the crystal structure of the products. Now, what are the products? According to [1, p. 555], the reactions are as follows:
$$
\begin{align}
\ce{4 FeTiO3 + O2 &-> 2 Fe2O3 + 4 TiO2} \tag{R1} \\
\ce{6 FeTiO3 + O2 &-> 2 Fe3O4 + 6 TiO2} \tag{R2} \\
\ce{4 FeTiO3 + O2 &-> 2 Fe2TiO5 + 2 TiO2} \tag{R3}
\end{align}$$
My post can be one of two question. A general one, which if too hard/lengthy, means my post  defers to the specific case I am most interested in.
General question: How do I find out which of two given crystal structures any given gas would most wholly and deeply diffuse into?
Specific question: Would CO gas diffuse more wholly and deeply into the above reactions' products than it would their solid reactants?
Reference

Tathavadkar, V and Kari, Chandrakala and Rao, S Mohan (2006) Oxidation of Indian Ilmenite: Thermodynamics and Kinetics Considerations. In: Proceedings of the International Seminar on Mineral Processing Technology and Indo-Korean Workshop on Resource Recycling (MPT-2006), March 8-10, 2006, NML, Chennai. http://eprints.nmlindia.org/6302 (PDF)

Sounds like an urban legend to me.

Besides, I must say I admire your perseverance at making the formula look right. If not for one slanted O, I would never suspect it was made without the tools we use around here.

@IvanNeretin Ah, one slipped past me! Thanks for the edit. Also, I didn't know you use a different display code here. I will acquaint myself with mhchem.

The process of heating a crystal, transforming into a different chemical, would destroy the crystal structure, making it into many small crystals or an amorphous solid. Inevitably, this makes the substance porous, having nothing to do with the defunct crystal structure. Consider what would happen to the selenite crystals in Cueva de los cristales if the temperature varied or humidity dropped.
That said, there are some interesting materials that are surprisingly permeable to gases:

Not only are silicones permeable to oxygen, but oxygen also diffuses through metallic silver.
Palladium and other platinum metals, and nickel, tantalum and vanadium are also porous to it.

Interesting. Are you saying that compounds within the same phases transforming into distinct compounds leads to a very large number of small phases, thus causing the bulk material to be quite porous? That makes a lot of sense, but makes me wonder why the metallurgist was talking about the crystal structure.

I do not think your general question has any answer. It is oversimplification to state that the greater distance between atoms in crystal lattice makes the crystal to be more permeable by gases. It can be a contributing factor, but most likely not the only factor. Specific interactions between gas atoms/molecules and constituents of the lattice (atoms, ions, molecules) will be of great importance here. So, what is permeable for, say, oxygen can be almost non-permeable for carbon monoxide. The crystal degradation correctly described in another answer because of the incommensurability of lattices of the matrix phase and the nascent one results in formation of polycrystalline or amorphic phases. Also, considerable number of pores can be created. All that can result in forming much more permeable material. This is what the metallurgist could have meant (if this is the case then he was not exactly accurate while choosing terms to convey his idea, but the idea itself still makes sense).
As for specific question, the answer will depend on the conditions of the process. Should there be reasons to believe it to be an equilibrium process, you could calculate which one of the listed reactions leads to the largest energy gain (see any thermochemistry textbook). But I doubt that is the case. There are a number of factors in non-equilibrium processes that can favour this or that reaction. It will depend on specific values of thermodynamic and kinetic parameters. If you are really interested in this, I suggest you check the papers describing experiments.


text:
Side reaction of I− and acidity of HI
In acidic conditions, I− tends to get oxidised by atmospheric oxygen to give iodine. This is evident in iodine based titration, where upon leaving the setup, blue colour of starch reappears as iodine is formed again.
$$\ce{4I- + O2 + 4H+ -> 2I2 + 2H2O}$$
This makes me believe that I− has less tendency to exist as the anion(at least at moderate temperature), since ΔH of reaction is of the order -150 kJ.
So my question is: How is it that HI is a strong acid if in acidic medium I− has tendency to get oxidised by atmospheric oxygen, reducing the pH in the process.
References:

Chemistry Textbook of NCERT
J.D. Lee Concise Inorganic Chemistry, for JEE Main and Advanced, 2021

These two things are pretty much unrelated. Strong acid can be reducer as well as oxidant, eg. HNO3.

I was able to find $\ce{4I- + O2 + 4H+ -> 2I2 + 2H2O}$, but  the source is not very authentic. Could you provide a reference for the reaction you have mentioned?

@Mithoron although I doubt it, but if the above reaction is correct, then wouldn't the acidic strength of $\ce{HI}$ reduce?

@Mithoron sorry I failed to word my question well, but Ananta 's remark is precisely what I wished to ask, why does even HI exceed even a moderate acid, given iodide ions' reaction. (Think of H2CO3)

Also note that the reaction is entropically unfavored. $\Delta H$ is not a good indicator of spontaneity, especially when gaseous moieties are involved.

@ananta and OP. Again, some tendency to getting oxidised with air is hardly an issue  - solution can get a bit less acidic if bottle is left opened, and so what? Nitric acid somewhat decomposes and is it an issue? HCl vaporises, H2SO4 sucks H2O from air... all this decreases acidity of conc. solutions.

The dissociation equilibrium and the oxidation equilibrium are mutually independent. Therefore, acidic strength and reductive strength are mutually independent either.

H2CO3 is less acidic than CH3COOH in water despite the fact that the latter has a +I group. It is because H2CO3 decomposes in water (or so I was taught). Due to this I had begun to over-appreciate side reactions. I had just begun to understand kinetics, it didn't occur to me that decomposition of HI is kinetically unfavourable, while it is not so for decomposition of H2CO3.

TL;DR:

Gas phase reaction (the fastest way to react two substances) between $\ce{HI}$ and $\ce{O2}$ is rather slow and often requires a catalyst.
Low amounts of dissolved oxygen in the solution will reduce the rate of the reaction to negligible amounts.
$\ce{HI(aq)}$ is commercially available and is expected to be stable with a long shelf life.

The details
Although, according to various sources, including Wikipedia, $\ce{HI}$ does react with $\ce{O2}$ to form $\ce{I2}$ and $\ce{H2O}$, I am not able to find sources that claim such a reaction in aqueous state; however, several studies have been conducted using catalysts. For example, the reference$^1$ suggests the following reaction mechanism for the reaction between $\ce{HI}$ and $\ce{O2}$:

where $S$ is a surface site. The same article does mention a concurrent slow reaction between $\ce{HI}$ and $\ce{O2}$:

Do note that the reaction has been suggested as being rather slow. Thus, in the absence of a catalyst, the reaction between $\ce{HI}$ and $\ce{O2}$ is not very fast. This study considers very large amounts of oxygen, while oxygen only feebly dissolves in water. The reference-suggested (important) transition state for the reaction between $\ce{HI}$ and $\ce{O2}$ is:

In aqueous state, such a transition state wouldn't be possible since the ions are dissociated and solvated.
Further, since Merck and other commercial chemical manufacturers provide the hydrogen iodide solution, I expect it to be stable with a long shelf life.
References

Shum, L.G.S. and Benson, S.W. (1983), The oxidation of HI at low temperatures and the heat of formation of HO2. Int. J. Chem. Kinet., 15: 323-339. https://doi.org/10.1002/kin.550150403

Please do not use images (such as the first one) and typeset it in the style of the site, as it is (1) not searchable; and (2) not cited here in the body of your post.
You have not addressed the OP's query, which is how is $\ce{HI}$ considered a strong acid given the conditions they gave.

@ToddMinehardt I have referred the source of the images. These (cited) images are from a research article published in the International Journal of Chemical Kinetics. I have added appropriate image descriptions to make the image searchable. Thank you for the suggestion though, I will typeset it as soon as I find time.
I am not sure if you have been following this question, but me and OP have been discussing this question for a few days now, and this answers OP's question, which is why $\ce{HI}$ does not degrade due to presence of dissolved/atmospheric oxygen, remaining a strong acid.

"Example" here is not the proper citation, it's a hyperlinked word. I also second the Todd's remark on posting text as images. And you are doing it again here: answering the question not really posed by OP, but rather addressing a different loosely-related issue, and pleasing the OP who is seemingly happy not to see the forest for the trees.

@andselisk I couldn't find any studies answering the OP's question exactly (probably because this is a question about something that doesn't happen). In such cases, I refer to articles that are closely related and provide a good explanation. Again, the question is not why $\ce{HI}$ is a strong acid, but why it doesn't degrade into a weaker (more dilute) acid due to reaction with dissolved/atmospheric oxygen, which, I feel, I have answered to a satisfactory degree. I have also mentioned that I will typeset the images, and provide a proper reference, as soon as I find time.


text:
evaporating DMSO/Ethanol
I'm preparing to do a thiolclick reaction and to get a homogenous phase I found it's best to use DMSO/ethanol as solvent for the -ene and thiol reagents. I need the ethanol specifically for hole-capturing, and the DMSO seems to dissolve both reagents well.
To validate my results I want to use H-NMR however ethanol tends to give some messy peaks so I would like to evaporate it off at room temperature under a fume hood.
I was wondering if any of you know if DMSO/ethanol are problematic together when trying to evaporate them off.

DMSO boiling point  is 189 deg C. It has its implications toward ethanol and  dissolved organic matter

Ethanol is often used as a co-solvent to evaporate DMSO and similar high-b.p. solvents under vacuum at the temperatures below 25 °C. Are you planning to reconstitute your product (BTW which one, exactly?) in DMSO-d6 for NMR, or is your goal is to selectively remove ethanol and leaving DMSO intact? What do you mean by ethanol's "messy peaks" and are you absolutely sure it's EtOH causing the issue? Also, what "hole-capturing" are you referring to? Is there a photocatalytic process involved, or you are speaking of porogenic properties of ethanol/DMSO mix?

The boiling point of DMSO is about 462 kelvin and that of ethanol is about 352 kelvin, you should be able to evaporate ethanol with ease without losing any of the DMSO solvent. For fast evaporation, use low pressure evaporation at the boiling point of ethanol (at the lower pressure) while constantly stirring the solution.
Caution
However, so that you don't lose any of your product, you have to take into account the boiling point and vapor pressure of your product as well. If your product has boiling point significantly lower than that of ethanol, it is best to evaporate the product and collect it in a different vessel containing DMSO.

The kelvin as the unit of measurement is not capitalized.

@andselisk sorry but isn't the unit named after Lord Kelvin (William Thompson). Shouldn't it be capitalized? Actually, when writing temperatures on the absolute scale K/Kelvin/kelvin is not used at all, I only added it for clarity. Web search does agree with you; therefore, I have incorporated your suggestion.

In place of randomly searching the web and drawing a conclusion from the most popular results that your search engine spoon-feeds you with, I recommend to develop a habit of referring to the authoritative sources directly. Here, I suggest reading sections 2.3.1 Base units and 5.3 Unit names of the SI Brochure. Alternatively, just avoid *** and read good literature that follow conventions and you will automatically write correctly because you will visually memorize how the term or its symbol should be formatted for a suitable occasion.

@ananta Hello ananta. All the names in the SI units are not capitalized, it is kelvin (not Kelvin); it is ampere (not Ampere); even though these are surnames of brilliant people. However, you can say Celsius, Fahrenheit, or Rankine. Preferably, don't use "°" for absolute temperature scales like kelvin or Rankine, and use $^\circ F$ or $^\circ C$ for Fahrenheit and Celsius scales, which are relative temperature scales.