text:
What is the alloying composition of the Zinc-Platinum System?
My research team and I are working on a bimetallic alloy consisting of Zinc and Platinum. We have looked for papers regarding the composition of Zinc-Platinum alloy. But there aren't many papers out there.
So far, we have found only one paper:
Moser, Z. The Pt-Zn (Platinum-Zinc) system. JPE 1991, 12 (4), 439–443. DOI: 10.1007/BF02645964.
If we could get some good insights on the Zn-Pt system, it will help us a lot and accelerate our research work. And if there isn't any paper regarding it, what will be the expert opinions?

Did you complement your search e.g., in SpringerMaterials, specifically for the binary system Pt-Zn, there are 27 «element systems», including five entries about binary phase diagrams (e.g., 55-70 at% Zn, 573…1173 K; or 0-100 at% Zn, 273…1173 K).  You equally may work through for Gmelin (print stopped in the 1990s, now within Elsevier's Reaxys database).

Also, if you have access to the ASM Alloy Center it gathers references going way back. I’ll check my personal database…

There are a few papers on Zinc-Platinum Systems (e.g., Ref.1 and 2, which are also sited in Ref.3), but most of them are in German. However, there were few useful data published by Johnson and Dillon in their Research and Development Report (Ref.4):
$$
\bf{The \ structures \ of \ platinum-zinc \ intermetallic \ phases}\\
\begin{array}{l|l r r}
\hline
\text{Compound*} & \text{Crystal Class} & \text{Lattice Parameters, }\pu{pm} & \text{References} \\
\hline
\ce{PtZn} & \text{Cubic} & a = 1811.6 & 2 \\
\ce{PtZn8} & - & - & 2 \\
\ce{PtZn2} (\xi) & \text{Hexagonal} & a = 410, \ c = 274 & 2 \\
\ce{PtZn}\ (\nu) & \text{Tetragonal} & a = 403, \ c = 347 & 1 \ \& \ 2 \\
\ce{Pt3Zn} & \text{Cubic} & a = 389 & 2 \\
\hline
\end{array}\\
\text{*All phases have a range of composition.}$$
Nowotny et al. (Ref.2) determined the solubility of platinum in liquid zinc by thermal analysis. Their measurements cover the range of $\pu{688-869 ^\circ C}$ and are shown in following Table:
$$
\begin{array}{l r r r}
\hline
T, \ \pu{^\circ C} & \frac{1000}{T}, \ \pu{K-1}  & \ce{Pt}, \ \text{weigh percent} & \ce{Pt}, \ \text{atom percent} \\
\hline
869 & 0.876 & 30.1 & 12.6 \\
825 & 0.911 & 23.2 & 9.2 \\
762 & 0.966 & 18.1 & 6.9 \\
728 & 0.999 & 14.8 & 5.5 \\
688 & 1.0404 & 9.8 & 3.5 \\
\hline
\end{array}\\
\text{*All phases have a range of composition.}$$
The direct measurements may be represented by the following empirical equations over the range of $\pu{675-875 ^\circ C}$ (Ref.4):
$$\log \left(a/o \ \ce{Pt}\right) = 3.881 - 3176T^{-1} \tag1$$
$$\log \left(w/o \ \ce{Pt}\right) = 3.917 - 2780T^{-1} \tag1$$
where $\left(a/o \ \ce{Pt}\right)$ is the atom percent platinum and $\left(w/o \ \ce{Pt}\right)$ is the weight percent platinum.

References:

H. Nowotny, E. Bauer, and A. Stempfl, “Die Kristallstrukturen von PtZn, PtCd und PdCd (English Translation: Crystal Structures of PtZn, PtCd, and PdCd),” Monatshefte für Chemie 1950, 81(7), 1164-1164 (In German; DOI: https://doi.org/10.1007/BF00899370).
H. Nowotny, E. Bauer, A. Stempfl, and H. Bittner, “Über die Systeme: Platin-Zink und Platin-Kadmium (English Translation: On the System: Platinum-Zinc and Platinum-Cadmium),” Monatshefte für Chemie 1952, 83(1), 221–236 (In German; DOI: https://doi.org/10.1007/BF00897822).
Z. Moser, “The Pt-Zn (Platinum-Zinc) system,” Journal of Phase Equilibria 1991, 12, 439–443 (DOI: https://doi.org/10.1007/BF02645964).
Irving Johnson and Ira G. Dillon, “The Solubility of Metals in Liquid Zinc,” USAEC Research and Development Report (Argonne National Laboratory, Argonne, IL), ANL-7083, pp. 78–79 (1965).

The ASM Alloy Phase Diagram center shows your (2) and (3), and further lists Y. Khan et al., J. Less-Common Met. 21(3) p. 293 (1970). It focuses on the brass-like phases between 55 and 70 at.% Zn.

Thank you everyone for your help. I am grateful to you all. And a special thanks to Mr. Mathew Mahindaratne sir.


text:
Potential difference between metallic electrodes in water
If I dip two pieces of metal in tap water I measure a potential difference between the two electrodes with a common digital voltmeter. For example, for Cu and Al I measured 0.55 V and for Cu and Fe I measure 0.86 V. Is there any way to relate these potential differences to the standard electrochemical potential given in tables? (for Al i found -1.66 V, for Cu +0.34 V and for Fe, +0.77 V).

Unfortunately, the system under test is ill-defined and ill-controlled: tap water should be potable, but that is not at all chemically specific. If you stick copper and iron electrodes into a lemon, then there is a rich body of literature we could consult. ;-) Tap water, not so much.

What if it's a 5%solution of HCl in deionized water instead of tap water?

Then you need to start looking at the chemistry: aluminum and iron react with hydrochloric acid, but copper does not react with 5% HCl solution. So what are the possible redox half-reactions? Please note that this is a Q & A site and people are expected to do some investigation/research on the topic of their question. Just asking a question is generally not going to get a positive or useful response.

Anyway you cannot compare the standard potential and the potential of a metal dipped n water, where the concentration of the corresponding ion is unknown. The tabulated redox potential are always defined in  a one-molar solution of the ions.

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text:
How does optical activity work?
I have learnt that chiral molecules are able to show optical activity; that is, they are able to rotate plane polarized light by some angle.
But how does it actually happen? Won't the orientation of the molecule also depend on whether it causes clockwise or anticlockwise rotation?
Any help would be greatly appreciated

There are quite a few answers and relevant questions here, such as this one: https://chemistry.stackexchange.com/a/84136/79678. I suggest following the link and exploring related items on the right side.

Excellent one here: https://chemistry.stackexchange.com/a/6277/79678.

@Draculin https://physics.stackexchange.com/questions/15503/how-do-optically-active-compounds-rotate-plane-polarized-light?rq=1 This one as well.

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text:
Can the equilbrium constant ever be exactly 1?
According to my book,

In the real world [...] it is unlikely that any reaction would have exactly equal concentrations of products and reactants at equilibrium.

This was mentioned in context to the fact that it was an ideal situation for $K_\mathrm{eq}$, wherein $K_\mathrm{eq} = 1$
What confuses me is the term "unlikely", which I take to mean that it is uncommon yet possible.   Can you explain why having equal concentrations of products and reactants at equilibrium such that $K_\mathrm{eq} = 1$ is both possible yet uncommon?

Can you give some more context? In what context did your book mention that statement? And equilibrium does not mean there is equal concentration of product and reactant, there are many reactions like $\ce{2A +C ->B} where that isn't true. (If someone is very pedantic then yes, technically perfect equilibrium is impossible to achieve)

The book said it was an ideal situation for Keq, wherein Keq = 1

@ShoubhikRMaiti I assumed this meant ideal equilibrium

Related: https://chemistry.stackexchange.com/questions/57075/is-there-no-actual-equilibrium/57135#57135

When you specifically want a value, everything is unlikely. While the thread contains right comments and answers, it is hard to understand the meaning of ideal in the context "to the fact that it was an ideal situation....". One can place 2 instead of 1.

For a chemical reaction,
$$\Delta G = -R T \ln K_{eq}$$
$$\implies  K_{eq} = e^{\frac{-\Delta G}{RT}} $$
Thus, in order for a chemical reaction to have $K_{eq}=1$*, it is necessary to have $\Delta G = 0$
For some chemical reactions**, we can achieve this by adjusting the temperature and/or pressure. Thus it is possible. However, it's unlikely that the temperature and pressure at which a chemical reaction is taking place is coincidentally just that needed to achieve $\Delta G = 0$.
On the other hand, in finding phase transition temperatures (melting and boiling points), we are sweeping through a range of temperatures specifically to find the points at which the phases coexist, i.e., where their free energies are the same and thus where $\Delta G = 0$.  Thus these are equilibria for which $K_{eq}=1$.
*We can never exactly get $\Delta G = 0$, because we can never exactly adjust T and p; consequently, we can never exactly have $K_{eq}=1$.   Thus what I mean (and presumably what your textbook means) when it says "$K_{eq}=1$" is that it is so close to $1$ that it's within the size of the fluctuations about equilibrium.
**I say "for some chemical reactions" because in some cases the reactants or products might break down before the required temperature is reached.   E.g., suppose you were trying to find a reaction involving glucose at which $K_{eq} =1$, and did a theoretical extrapolation in which you found that this occurred at, say, 10,000 K.  Well, glucose doesn't exist at that temperature.
Note #1:  $\Delta G = -R T \ln K_{eq}$ applies to processes carred out at constant T and p. If the process were instead at, say, constant T and V, then the equilibrium condition would be $\Delta A = -R T \ln K_{eq}$.  [And don't be confused: Constant T and p doesn't mean a specific T and p.  T and p can be whatever you please, so long as the reactants and products can exist at those conditions.  Rather, constant T and p just means that, whatever T and p are, they stay fixed during the reaction.]
Note #2 (on time to attain equilibrium): The above has nothing to do with whether or not you can get to equilbrium, which is addressed in the question linked in Ivan's comment on your post. That question pertains to any equilibrium, regardless of the value of the equilibrium constant (whether it's 1 or 10.303 or 4.8 x 10^-22, or any other value).  Instead, the above is about whether or not the value of the equilibrium constant can be 1.
Note #3 (on enantiomers in an achiral environment): Andrew raises an interesting point that the reaction:
$$\ce{A_L <=> A_D},$$
where $\ce{A_L}$ and $\ce{A_D}$ are two different enantiomers of compound A, will have $\Delta G = 0$, and thus $K_{eq}=1$, in an achiral environment because, in an achiral environment, the two compounds are chemically identical.  And as they are chemically identical, this will obtain at all temperatures and pressures at which the compound exists.  I.e., in an achiral environment, chemically,
$$\ce{A_L <=> A_D}$$
reduces to:
$$\ce{A <=> A}$$.
But: While enantiomers are chemically identical, the weak nuclear force that exists within their nuclei can recognize handedness, and thus the restriction in the first paragraph still applies:  $K_{eq}$ won't exactly equal 1, but can be immeasurably (at least with current technology) close to 1.  [I haven't done the calculation to determine if the difference in energy caused by the weak nuclear force puts the deviation from 1 within the, say, the 90% fluctation range about equilibrium for a representative pair of enantiomers at room temperature.]

Another example is enantiomers in an achiral environment

@Andrew: Thanks for your comment. See my addition at the end.

Couldn't the same be said about any value of K or $\Delta G^\circ$?? I don't see the point in discussing K=1 as a special case.

You mean $\Delta G ^\circ = -RT\ln K_{eq}$? (note the superscript indicating std conditions)

Would you consider an identity reaction, let's say intramolecular proton transfer $$\ce{\color{red}{H}O\color{red}{C}H2\color{blue}{C}H2O\color{blue}{H} <=> \color{blue}{H}O\color{red}{C}H2\color{blue}{C}H2O\color{red}{H}},$$ as still being a valid "reaction"? Apologies for the grotesque colours. This one is a bit contrived, but I think you could still define a rate constant for it, and I'd suspect that there are valid reactions like this which are of genuine interest to someone.

@orthocresol  I like the colors! Plus they're useful.  I didn't even know you could make those in comments.  Anyways,  I would consider that a trivial case, since nothing is actually changing.   I was also wondering whether the enantiomer reaction in an achiral environment might similarly be considered trivial, but decided in that case that something actually is changing, but it doesn't make a difference.

@BuckThorn I was thinking about that very issue (should it be $\Delta G^{\circ}$?) yesterday. I explicitly don't want to use that notation, because it restricts us to standard pressure and (for gases) ideal gas behavior, while I wanted to allow for variation in pressure, as well as non-ideality. So I was trying to think of what alternative notation I might use, and hadn't yet come up with anything.

@BuckThorn   As far as this applying to any value of Keq: Well, partly, but achieving Keq = 1 seems somewhat special, because that's when all the reactants and products are thermodynamically "the same".  Ultimately the question of whether it's worth giving special consideration to Keq=1 is pedagogical rather than scientific, so it's a judgement call.   Anyways, that was the idea behind the question, so I was honoring that.   Also: I said  "partly" in the first sentence because Keq=1 also allows us to include phase transitions.

How about the concentration of a solute in two connected vessels (e.g. via dialysis membrane). At equilibrium, it should be equal (all other things being equal). Again, the equilibrium constant is 1.

@KarstenTheis Thats's an interesting example. Let's consider an actual dialysis, where the solute is $\ce{Na+}$, and assume it can pass freely through the membrane. I'd expect $K_{eq} \ne 1$, b/c of the Donnan effect, where anionic proteins confined to the blood side will create a charge imbalance such that equal concs. of $\ce{Na+}$ wouldn't have equal activities.  And even for uncharged solutes, the environments are so diff. on the two sides that equal concs. wouldn't give equal activities. If the environments were identical, then $K_{eq} = 1$, but that seems like a trivial result. Thoughts?

@theorist For the famous transport examples in biology, the equilibrium constant will be differrent from one because of differences in local charge (proton gradient in oxidative phosphorylation) or ionic strength (potassium and sodium in nerve impulse transmission). I wouldn't say when the equilibrium constant is 1 because of some symmetry is not trivial, just like when the reason is that all species are pure solids or liquids. In either case, there is a reason the equilibrium constant is 1 (rather than a coincidence).

@BuckThorn that is what I think, too. I have commented (unnoticed, it seems) below the question. Sometimes threads are puzzling.


text:
How to interpret the relation between CO stretch frequencies and backbonding strength for polycarbonyl compounds?
I am working on a series of octahedral complexes of the type $\ce{[M(CO)2(CH3)X3]-}$ where $\ce{M} = \ce{Co}, \ce{Rh}, \ce{Ir},$ and $\ce{X}$ is a range of ligands (halide, cyanide, nitrosyl, etc.). Now, the complexes all have either a mirror plane or a $C_2$ axis which makes the two $\ce{CO}$ ligands equivalent by symmetry (This is a crucial point).
I know the usual rule that the more electron density the metal has, the stronger the backbonding to the $\ce{CO}$ ligand will be, and this will make the $\ce{C#O}$ bond weaker and will therefore reduce the $\nu_\ce{CO}$ stretch frequency in the infrared region.
However, in this case there are more than two $\ce{CO}$ ligands which are symmetrically equivalent. This means there are two different $\ce{CO}$ stretch bands. I have encountered cases where changing the $\ce{X}$ ligand causes one band to go up and the other band to go down. How do I interpret the change in back-bonding in these cases?
My first thought was that, since they are equivalent, we are sort of taking the linear combination of the bond stretch vectors ($\vec{r}_1 +\vec{r}_2$ and $\vec{r}_1 -\vec{r}_2$) so the higher frequency band is the one we want to look at to see if overall the $\ce{CO}$ stretches have gone down. But this seems a bit unscientific and not rigorous, and the stretch frequency values are certainly not linear combinations, they are both near $\pu{2000 cm-1}$.
My question is—how should I interpret the $\ce{CO}$ stretch data and the backbonding strength when there are more than one $\ce{CO}$ ligand?

Example data ($\ce{M} = \ce{Rh}$):
$$\begin{array} {|r|r|}\hline  & \mathrm{CO\;stretch\;freq. (cm^{-1})} \\ \hline \ce{X} = \ce{F} & 1980,2067 \\ \hline \ce{X}= \ce{Cl} & 1992,2062 \\ \hline  \end{array}$$
Here one band goes up and the other goes down when changing $\ce{F}$ to $\ce{Cl}$. So what can I infer about the backbonding to $\ce{CO}$?
Note that these data (i.e. $\ce{CO}$ stretches, geometries) are all from computations. But I am not just looking for computational approaches; I am also curious as to how the $\ce{metal-CO}$ $\pi$-backbonding in a polycarbonyl compound will be analysed, if the $\ce{CO}$ stretches are coupled to each other due to symmetry (even in experimental IR spectrum).

Are you sure they are symmetrical? (You shouldn't start anything you write with "So,".)

@Martin-マーチン Yes, it is a meridional, trans-complex, so the two CO ligands are equivalent by mirror symmetry.

I've read your post, so I understood this already. But I see this as it being an assumption by you. Is there actual evidence that the molecules are symmetrical?

@Martin-マーチン These data (CO stretches) are all from calculations, and the optimized geometry does have a mirror image. (Optimization without symmetry gives the structure).

Then you should see that they are coupled. Why don't you use EDA for the analysis of bonding? NBO could provide insight, too. What about QTAIM? If you already have the calculations, there's plenty of more rigorous ways to analyse backbonding, than guessing it from the harmonic approximation to the potential well.

@Martin-マーチン This is actually an undergraduate lab project, so we are not allowed to use anything else besides bond lengths/frequencies/energies. Also, I don't know much about EDA or any of the other things you mentioned. Could you maybe give some clues about what type of calculation need to be done, which data need to be looked at from EDA/NBO etc.? Also, my questions wasn't specifically about computational approaches, I am also curious as to how one would analyse the data if the CO stretches are found to be coupled in the experimental IR spectrum.

This paper provides a thorough discussion of the coupling of stretches and a basic mathematical approach to estimating the frequencies based on group theory analysis. It's old enough that it does not assume any advanced computational capability.  https://doi.org/10.1063/1.1712327

And if you're unfamiliar with the math of normal mode analysis, here's a tutorial using the similar but simpler molecule CO2 as an example to estimate the symmetric and asymmetric stretch frequencies using basic assumptions: https://www.colby.edu/chemistry/PChem/notes/NormalModesText.pdf

The take-home message with regards to your specific question is that you can assume that the force constants of the two C=O groups are the same (consistent with symmetry) and still get two IR signals, and the degree of back-bonding will affect the force constants so it will affect both signals.

Disclaimer: This is not going to be an answer to the question. This is just a collection of thoughts on it, which hopefully helps to form a better understanding.
First and foremost, and this should go without saying: If you have experimental data, use it. If you do not have experimental data, go find it. No calculation is good enough to be used without calibration, especially when using DFT (see also DFT Functional Selection Criteria).
However, calculations can give you some insight into the molecule's behaviour.
The standard methodology to calculating the spectrum is to use the harmonic approximation for the force constants. That won't be good enough for anything else than a qualitative interpretation of an experimental spectrum; for example if you need to identify signals in your spectra. It won't be good enough for something as sensitive as the carbon monoxide stretch.
I wouldn't be surprised if you found a couple of different opposing trends when using different calculation techniques.
From your example data it seems quite obvious that the values for the two structures are too close together to get any meaningful interpretation.
I also have doubts that the molecule actually is symmetric. If you have a $C_3$ moiety in the mirror plane this is always going to be challenging. It might be symmetric enough to approximate this and to analyse it within these constraints, but that is a point that should be carefully checked. Especially when using calculation techniques, this is something that needs to be calibrated, too.
Assuming that the complexes are symmetric and therefore the carbonyl ligands are equivalent, one would indeed expect two frequencies: one where the stretches are symmetrical, one where they are asymmetrical. Denoting the equatorial plane with $\ce{M^*}$, this would be
\begin{gather}
  \ce{
    \overset{\leftarrow}{O}\bond{-}\overset{\rightarrow}{C}
    \bond{-}M^*\bond{-}
    \overset{\leftarrow}{C}\bond{-}\overset{\rightarrow}{O}
  }\tag{symmetrical}\\
  \ce{
    \overset{\rightarrow}{O}\bond{-}\overset{\leftarrow}{C}
    \bond{-}M^*\bond{-}
    \overset{\leftarrow}{C}\bond{-}\overset{\rightarrow}{O}
  }\tag{asymmetrical}\\
\end{gather}
The key to the relationship between the carbonyl stretch frequency and the backbonding is the force constant of this bond in comparison to "free" carbon monoxide. In simple and only phenomenological terms, the electron density of the metal pushes into the antibonding orbital of the ligand, hence weakening  the bond. There is probably some data that relates the frequencies with some other properties of the metal and some orbital interpretations of that, and linking that to backbonding.
Within this framework you can probably form some differential equations to find the force constant and doe similar maths and come to similar conclusions and you can possibly generalise them to form trends.
This is typically the same mathematical foundation with which you would go about analysing experimental spectra. I'm sorry to not be of more help here; but since you're clever enough to come up with this question, I'm confident you'll find a way to do the maths.
If you have the calculations already and you are confident enough that they describe your molecule, there are probably better methods to estimate backbonding.
From the top of my head, I can think of Energy Decomposition Analysis (EDA), possibly coupled with natural orbitals (search ETS-NOCV). Similar approaches come with Natural Bond Orbital (NBO) analyses. Backbonding is a description based on molecular orbital theory, so it would be quite odd to not be looking at orbitals when analysing it.
I wouldn't be surprised if there were models that employ the Quantum Theory of Atoms in Molecules (QTAIM), too. There is probably a way to find an option within Multiwfn to analyse backbonding. In these cases, you'll probably find a better audience at Matter Modeling SE.
Again apologies for that long post with little to no actual information.

Thanks! this is quite helpful. Your comment about differential equations reminded me what I was missing. Time to take out my math notebook!

@ShoubhikRMaiti Re. the DEs, I think you should look into normal modes of coupled oscillators; it's traditionally taught in physics so maybe even Physics.SE might give you an answer. I don't think I could do the maths myself, but you basically have five bodies attached by four springs (and I guess you could assume that the two M–C springs have the same force constant $k_1$ and the C–O springs $k_2$). The overall system is further subject to the constraint that the centre of gravity doesn't change. (i.e. only vibrations, no translations)


text:
What is Nascent Hydrogen?
I am currently studying preparation of Alkane from Alkyl Halides (By Reduction)
But the term  "Nascent Hydrogen" Confuses me after reading diffrent threads and stuff
Please help :)

Perhaps see here for a previous time this popped up: https://chemistry.stackexchange.com/q/131162/79678. Maybe a unicorn or maybe not.

Some dyes are reduced by a mixture of $\ce{Zn + HCl}$, and are not reduced by zinc alone or by $\ce{H2}$ alone. This reaction is attributed to  the hypothesis that the reducing agent is "nascent", or "atomic" hydrogen which could have been produced by a reaction like $$\ce{Zn + 2 H^+ -> Zn^{2+} + 2 H·}$$ which is usually followed by the dimerization $$\ce{2 H· -> H2}$$


text:
How to Build Hydrogen Atoms in MOE using SVL
I'm working on automating a process in MOE using SVL commands. I've got pretty much everything worked out, except I can't seem to find anything for building hydrogen atoms using an SVL command (equivalent to Edit->Build->Hydrogens->Add Hydrogens).
Is there a method to do this, and if so what is it?
Edit: MOE (Molecular Operating Environment), SVL (Scientific Vector Language)

Please remember that ChemSE enjoys a broad spectrum of readers, early and seasoned in the study of chemistry, including those with interest but without formal education of chemistry, too.  There is nothing wrong using abbreviations in a question, however consider that many readers might not yet encountered them.  Both as act of courtesy as well as to limit potential misunderstanding (because one abbreviation may have a different meaning to different readers) first explain the thing, then use the intended abbreviation.  Specific to your question: what «MOE» and «SVL» stand for?

While MOE and SVL are technically acronyms (Molecular Operating Environment and Scientific Vector Language respectively) MOE and SVL are much more common usage. I have no problem adding it, but to me adding those would be additional unhelpful information. Additionally if someone does not know what MOE and SVL are from context they are unlikely to be able to help with my question. Do you disagree?

I agree that the addition of the «full name» of these abbreviations will be common parlance for those «in the field» likely to help with your question, and other readers stay observers.  The potential benefit indeed is aimed for the readers not yet familiar with these, who might be intrigued if these may be beneficial for their work.  For their length, to keep reading your question facile, I footnote largely would suffice.

Short Answer: _EditH 'addH'
As I continued researching how to finish up my project, I came across a tutorial describing the behavior of Menus in MOE. (Tutorial download available from https://www.chemcomp.com/moe/help/2018/index_crs.html). All menus in MOE have their behavior defined by a file within the MOE installation directory. ("\$MOE/menu/moe-menus" where "\$MOE" is the MOE installation directory.)
By following the configuration file you can determine the exact SVL command used upon the selection of any menu item. You then can use that command in the SVL Commands window.


text:
What does pentamethylarsenic smell of?
Trimethylarsine smells of garlic. According to Wikipedia,  pentamethylarsenic smells like pentamethylantimony but I can't find the smell of that anywhere.
What does pentamethylarsenic smell of?

Well, it says it's "otherwise unique" - if that's accurate, than only the two compounds share the smell and it's different then any other.

@Mithoron I took it to mean that they only shared the smell and nothing else, i.e. was unique from pentamethylantimony in everything else.

Hmm, the cited book says: "characteristic odor which resembles the analogous antimony compound". Thing is it's not like smell is really quantifiable in a way color is. So would would you like you know? Some common smells that are somewhat similar?

@mithoron common smells would be great

Well, I didn't find anything, but "characteristic odor" probably means it stinks terribly in unusual way  ;)

Funny how “characteristic odor” is used to describe something no sane person would ever deliberately smell. So how would that rare insane person know it was characteristic and did not smell like something they had never smelled previously? In any event, “garlic” seems to be the default descriptor I have seen most often used for smelly stuff.

Related: https://en.m.wikipedia.org/wiki/Cacodyl. Bunsen almost died from inadvertently getting a whiff of cacodyl cyanide. Derek Lowe writes about the episode, but I cannot locate the link offhand. But everything Lowe writes is fabulous!


text:
Applying the Pauling's electronegativity when the ionisation enthalpy and electron gain enthalpy of an element are given
I came across the following problem while practicing some exercises on inorganic chemistry:

If the ionization enthalpy and electron gain enthalpy of an element are 275 and 86 $Kcal mol^{-1}$ respectively, then the electronegativity (in $KJ mol^{-1}$) of the element on the Pauling scale is?

I know this is rather, a formula based question but I was really confused about this. According to the formula for pauling's scale which is,
$|\chi_A - \chi_B|= 0.208\sqrt{\Delta E}$ :

The question doesn't even seem right as it only mentions a single element while the formula talks about difference of electronegativities and considers A and B as:
$\ce{A2 + B2 -> A-B}$ where $\chi_{A}, \chi_B$ are the electronegativities of A and B respectively.
How do I work this out?  I thought of simply taking $\Delta E$ as the difference of the given Ionisation enthalpy and the electron gain enthalpy. And that got me the answer. But thats just for the sake of getting the answer. I can't get the point behind this.

What about using Mulliken scale and then using the relation between Pauling and Mulliken scale?

@Rishi Can you please elaborate more on this and post an answer? I actually tried applying it but I haven't got the desired output and the different formulae on different sources just increase the confusion.

You are right, there was a relation in one of my textbooks but it is knowhere to be seen on the web, wiki, chemlibre etc. It was $$\mathrm{\chi_{Pauling} =\frac{IP+E.A.}{2.8}}$$ If I find a reliable reference I will post an answer.[Here EN(P)=EN(M)/1.4]

From Atkins' Physical Chemistry,

The electronegativity is a parameter introduced by Linus Pauling as a measure of the
power of an atom to attract electrons to itself when it is part of a compound. Pauling used valence-bond arguments to suggest that an appropriate numerical scale of electronegativities could be defined in terms of bond dissociation energies, $\Delta$, in electron volts and proposed that the difference in electronegativities could be expressed as
$$\mathrm{|\chi_A - \chi_B| = 0.102 ({\Delta_{A-B} -\frac{1}{2} .(\Delta_{A-A} + \Delta_{B-B}}))^{\frac{1}{2}} \tag{1}}$$
Electronegativities based on this definition are called Pauling electronegativities.

The spectroscopist Robert Mulliken proposed an alternative definition of electronegativity.
He argued that an element is likely to be highly electronegative if it has
a high ionization energy (so it will not release electrons readily) and a high electron affinity (so it is energetically favorable to acquire electrons). The Mulliken electronegativity scale is therefore based on the definition $$\mathrm{\chi_M = \frac{1}{2} (I+ E_{Ea})  \tag{2}}$$ where $\mathrm{I}$ is the ionization energy of the element and $\mathrm{E_{ea}}$ is its electron affinity (both in electron-volts).
The Mulliken and Pauling scales are approximately
in line with each other. A reasonably reliable conversion between the two is
$$\mathrm{\chi_P = 1.35{\chi_M}^{\frac{1}{2}}-1.37 \tag{3}}$$

Converting electron-volts to $\pu{kJ mol-1}$ Eq-(1) converts to
$$\mathrm{|\chi_A - \chi_B| = 0.0103 ({\Delta_{A-B} -\frac{1}{2} .(\Delta_{A-A} + \Delta_{B-B}}))^{\frac{1}{2}} \tag{4}}$$
and Eq-(2) converts to
$$\mathrm{\chi_M = \frac{1}{192.97} (I+ E_{Ea}) \tag{5}}$$
Reference
Atkins' Physical Chemistry: Eleventh Edition , Oxford University Press.


text:
Mesomeric effect in -NHCOR and -OR
I've read in Wikipedia about +M strength of groups which goes:

–O - > –NH2 > –NHCOR > –OR > –OCOR > –Ph > –CH3 > –F > –Cl > –Br > –I

Now in every other website states that in terms of +M strength:
 website->brainly

website->Chemistry libretexts
(And many more)
My question:
Can you please verify the series and explain the reason if -OR actually has less activity than -NHCOR?

Also since in -NH of -NHCOR there is an EWG(Electron withdrawing group) which is -COR which should decrease the +M strength of the group appreciably.

Just pointing out that you compared to order of -NHCOR with -OR in your reference from wiki.

@RavirajBhosale thank you for pointing out brother :)

What I think is +M of -OR must be greater. There will be a cross resonance when  -NHCOR is attached to a conjugated system. Meaning that the lone pair of N is already be in conjugation with the carbonyl group. Another thing is that there is, as you say, -I of carbonyl group.

@RavirajBhosale yes even I think the information provided in Wikipedia is misleading.

If you check out the $\mathrm{p}K_\mathrm{a}$ values of paracetamol (acetaminophen) , it is 9.38 while $\mathrm pK_\mathrm a$ of para-methoxybenzene is 10.1.
So, $\ce{-OR}$ should exert a greater mesomeric effect as compared to $\ce{-NHCOR}$ making the wikipedia wrong.
References:

https://pubchem.ncbi.nlm.nih.gov/compound/Acetaminophen

https://pubchem.ncbi.nlm.nih.gov/compound/4-Methoxyphenol#section=Ionicity

Hammett $\sigma_m$ and $\sigma_p$ constants are based upon the acid dissociation of benzoic acid and m- and p-substituted benzoic acids $(\ce{X-C6H4-CO2H})$ in water at $\pu{25 ^\circ C}$. They are correlated with  $K_\mathrm{a}$ values of these benzoic acids as shown in few examples in the following table:
$$
\begin{array}{c|cccc}
\hline
\text{Substituent }(\ce{X})  & \sigma_p & K_\mathrm{a} & \mathrm{p}K_\mathrm{a} & \text{property} \\
\hline
\ce{p-OCH3} & -0.27 & 3.5 \times 10^{-05} & 4.46 & \text{EDG} \\
\ce{p-CH3} & -0.17 & 4.3 \times 10^{-05} & 4.34 & \text{EDG} \\
\ce{H} & 0.00 & 6.46 \times 10^{-05} & 4.19 & \text{Standard} \\
\ce{p-Cl} & +0.23 & 1.10 \times 10^{-04} & 3.96 & \text{EWG} \\
\ce{p-NO2} & +0.78 & 3.90 \times 10^{-04} & 3.41 & \text{EWG} \\
\hline
\end{array}\\
\text{Source: http://research.cm.utexas.edu/nbauld/unit4.htm}
$$
It is safe to say that $\sigma_p$ constants are more reflected on mesomeric effect while $\sigma_m$ constants are more reflected on inductive effect (Ref.1).
Thus, let's look at some of $\sigma_p$ and $\sigma_m$ values:
$$
\begin{array}{c|cccc}
\hline
\text{Substituent }(\ce{X})  & \sigma_p & \sigma_m \\
\hline
\ce{N(CH3)2} & -0.83 & -0.15 \\
\ce{NH2} & -0.66 & -0.16 \\
\ce{OH} & -0.37 & +0.12 \\
\ce{OCH3} & -0.27 & +0.12 \\
\ce{CH3} & -0.17 & -0.07 \\
\ce{NHC(=O)CH3} & -0.15 & +0.07 \\
\ce{C6H5}* & -0.01 & +0.06 \\
\ce{H} & 0.00 & 0.00 \\
\ce{OC(=O)Ph} & +0.13 & +0.21 \\
\ce{Cl} & +0.23 & +0.37 \\
\ce{C(=O)CH3} & +0.50 & +0.38 \\
\ce{NO2} & +0.78 & +0.71 \\
\hline
\end{array}\\
\text{*: Indicates that the group is in the most sterically hindered conformation.}\\
\text{Source: Journal of Medicinal Chemistry Reference and others; see Ref.2 & 3 below}
$$
According to $\sigma_p$ values, $+M$ strength of groups are:
$$\ce{-NH2 \gt -OH \gt -OR \gt -CH3 \gt -NH(C=O)R \gt -Ph \gt -OC(=O)R}$$

References:

Corwin Hansch, Albert Leo, Robert W. Taft, "A survey of Hammett substituent constants and resonance and field parameters," Chem. Rev. 1991, 91(2), 165–195 (DOI: https://doi.org/10.1021/cr00002a004).
Corwin Hansch, Sharon D. Rockwell, Priscilla Y. C. Jow, Albert Leo, and Edward E. Steller, "Substituent Constants for Correlation Analysis," J. Med. Chem. 1977, 20(2), 304-306 (DOI: https://doi.org/10.1021/jm00212a024).
Corwin Hansch and Albert Leo, In Substituent Constants for Correlation Analysis in Chemistry and Biology; Wiley-Interscience: New York, NY, 1979 (ISBN-13: 978-0-471-05062-9).


text:
Why would a mixture of sucrose, propionic acid, sodium formate, and agar turn light blue? What reaction is happening?
I made 3 mixtures which had 1 M sucrose, 1% w/v agar, .68% v/v propionic acid, and EITHER 0%, 4%, or 13% w/v sodium formate salt.
It was NOT blue at first, but after a couple hours at RT, the 0% sodium formate mixture was clear, the 4% mixture was barely blue, and the 13% was light blue.
I'm wondering what reaction is likely happening here to make it blue?
Clearly the formate is key. Perhaps some acid-base reaction? Formic acid is stronger than propionic acid, if that helps.
I'm concerned about unwanted byproducts affecting my experiments. Thx!
EDIT to answer a question: the solution became a gel almost immediately, as desired. I'm actually not sure if it was blue at that point, but it was definately NOT blue before the agar was added.
EDIT: more context: I don't care about the quality of the gel – I don't need it for blotting or anything like that, just for feeding fruit flies. So I'm trying to make sure that the ingredients I want in it are still in it, and with no NEW ones.

Perhaps growing of living stuff?

After mixing the ingredients, you obtained a solution.  After x hours of additional waiting, you got a) still a solution, or b) a gel?  What is the difference of the pristine solution with the aged "blend", is do you observe Tyndall scattering like in the cell on the right here?  (It is a guess, because of the presence of agar-agar.)

It is known that the solidification of agar is pH dependent. The propionic acid is probably bringing it down too much. Needs to be above 5 typically to get a good gel. Then the formate salt is raising the pH back up slightly and getting you closer to the necessary pH for gel to form, so you're getting a bit of scattering as buttonwood suggested. Try repeating the experiment but adjust the pH of all three to 5.5 with NaOH and see if you get the same result.

Thanks for the comments so far! I added some clarifying edits above...Is it safe to say that the ingredients I added are probably still there, and with no NEW products?

Smaller fragments of agar-agar polymerized to form a gel / a network (think mousse au chocolat, pudding, or gel derived from tetramethoxysilane).  Establishing new chemical bonds equates formation of new compounds; thus there are new products in the container.

@Buttonwood that should be the answer. Is likely Tyndall effect and not the growing of some coloured bacteria or fungi :)) as I have first suggest as a remote possibility.

Thx @Buttonwood. Any idea what those new products might BE, specifically? Are you saying they're probably just agar-agar polymers?

Did you determine the pH of the media? Was any coloring agent added? Were all the ingredients used pure? What is their origin? Did you prepare the media?

@Alchimista Following your suggest, I edited the comment into an answer.

@Buttonwood In contrast to polyacrylamide gels, there is no covalent bond formation when agar gels. You can melt and gel agar multiple times, it is a physical process (with non-covalent bonds forming and breaking, of course). Here is a nice write-up.

@KarstenTheis So there isn't a change in the network of hydrogen bonds the answer below attributes to the formation of the gel?

@Buttonwood I classify hydrogen bonds as non-covalent, and they do change as I said in the parenthetical remark in my previous comment.

Agar-agar consists of polysaccharides; Wikipedia for example mentions agarose:

(credit)
As with many naturally occurring polymers, the degree of polymerization (i.e., the repeating number of building blocks per polymer chain) varies.  Any form of processing may increase the variation with shorter and longer chains of this polymer.
The hydroxyl groups may interact with hydroxyl groups of the same polymer chain, as well as with other polymer chains by hydrogen bonds (see, e.g., a previous answer here) which contributes to optimize the intermolecular interactions / lowering the overall energy.  Despite different in chemical nature, conceptually there are similarities to proteins: in addition to the primary structure (a chain of building blocks form a chain), there is conformational flexibility around the glycosidic bond to yield a secondary structure.
Similar to gelatin in the kitchen, bringing the polymer chains of agar-agar into water allows these chains to re-adjust their conformation.  The hydroxyl groups of agar-agar now equally interact with water molecules to minimize the overall energy of the system.  The formation of new loops and pockets is possible in which water molecules are trapped.  These cavities may be formed by one, or multiple polymer chains.  In an analogy, think about cooked spaghetti (the agar-agar chains) and the much smaller grains of carbonara (the trapped water molecules); except for the example of agar-agar, the number of water molecules is much larger, than the about the agar-agar molecules.
The then again solidified agar-agar is like a frozen state: the polymer chains are locked in each other, shorter and longer ones.  Thus the size of the pockets including water, as well of local volumes with some degree of order vary.  This order is like crystallinity; though unlikely as nicely ordered like in a crystal of NaCl, there is some regularity at the scale of visible light (about $400-\pu{750 nm}$) causing diffraction and scattering.  At this scale of length, it is equally known as Tyndall effect:

«[...which] is similar to Rayleigh scattering, in that the intensity of the scattered light is inversely proportional to the fourth power of the wavelength, so blue light is scattered much more strongly than red light.» (source, emphasis added.)

Thus the light blueish appearance, equally seen with gels of other chemical composition:

(credit)
For polymers, the analysis of the light scattering allows to infer e.g., particle size, molecular weight, or degree of crystallinity.

It turns out the medium in question is used to feed fruit flies, and sometimes a blue coloring agent is added to such media. This introduces ambiguity in the OPs post since the person has not answered in more detail what went into the growth medium. Good answer though!


text:
How are compounds within a complicated mixture detected?
I was reading an article about an independent pharmaceutical watchdog group called Valisure which recently released a report stating that they found Benzene, a toxic substance, in many bottles of consumer sunscreen. (Full report here: https://www.valisure.com/blog/valisure-news/valisure-detects-benzene-in-sunscreen/  )

Valisure LLC has tested and detected high levels of benzene, a known
human carcinogen, in several brands and batches of sunscreen, which
are considered drug products by the Food and Drug Administration
(FDA), as well as in after-sun care products, which are generally
regulated by FDA as cosmetics. Benzene is known to cause cancer in
humans according to the U.S. Centers for Disease Control and
Prevention, the U.S. Department of Health and Human Services, the
World Health Organization, and other regulatory agencies

As a layman, it made me marvel at how they are able to do this. What is the process by which a chemist is able to start with a complicated multi-ingredient substance like sunscreen, and determine that it contains a particular molecule, like benzene?
Is there some kind of litmus paper that ONLY reacts with something containing benzene, and not ANY other substance? Or do they mix in substances known to react ONLY with the other sunscreen ingredients, leaving benzene behind? This sounds like it would be incredibly complicated.
I'd love some more insight about how this works

Analysis is also context based and, in particular, starts before the steps that ultimately lead to identification and quantification of a species. In particular a complex matrix (the ensemble that as be collected via sampling, the latter being already a critical step) can be treated as such to became simpler. For instance one can divide inorganic salts and ions from organic molecules, divide water soluble and water insoluble components, etc. Afterwards specific techniques are applied. The answer below goes pointing to one  important example in analysing commodities, pharmaceuticals, etc.

In case you are interested in reading the details of how the analysis proceeded, it starts on page 6 of this document: https://www.valisure.com/wp-content/uploads/Valisure-Citizen-Petition-on-Benzene-in-Sunscreen-and-After-sun-Care-Products-v9.7.pdf  Note that there is a difference also between detect and quantify. Doing the latter accurately adds another layer of complexity.

I suggest you narrow your question to how this particular analysis was performed, since that is interesting enough and sufficiently narrow in scope to be answered in a reasonably short post. Your question as it stands is the subject of analytical chemistry and the answer could fill one of more libraries.

There are three common classes of methods:

Separate the components of the mixture, then detect the amounts of the substances.
Use a fancy method that can identify multiple substances in a mixture (such as training a dog to smell different substances, or nuclear magnetic resonance)
Use a method that detects only your substance of interest, even when in a mixture (such as enzymes that only react with your substance of interest).

The last class of methods is your "fancy litmus paper". Examples of actual methods are a pregnancy test looking for a specific hormone in urine (class 3), a doping test looking for various illicit substances in a blood sample (class 1) or a chemist monitoring a reaction mixture to test for completeness of the intended reaction and for possible side reactions (class 2).

I don't know if I'd say using a trained dog is "common" (on a par with NMR) but ok in general #2 is common. Also, there is more to fancy spectroscopy than just NMR. Some people can look at an IR spectrum and say what's in a mixture, which I find even more amazing than a dog detecting benzene. My guess is the method applied to the sunscreen was GC-MS since benzene is volatile and the matrix is a mess that would certainly be unkind to NMR. Some kind of separation would probably be necessary for NMR detection.

NMR cannot even detect benzene at parts per million level. Very insensitive to impurities. GC-MS is the way to go in the OP's case.

Most likely chromatography. Gas chromatography is often used for detecting volatile compounds. You vaporize the sample at the inlet of a very long, thin tube (the column). The sample then gets pushed along by a carrier gas (the mobile phase). The different components of the mixture interact differently with the stationary phase inside the column and so take different amounts of time to reach the other end. A detector at the other end registers how much stuff comes out as a function of time, and so you can tell how many dissimilar components you have in your mixture. It's common to use a mass spectrometer as the detector, in which case you can also look at the mass spectrum to identify each component, often by comparison to a library. For benzene this would be easy to match.
https://en.wikipedia.org/wiki/Gas_chromatography

Determination of residual solvents is a routine exercise in pharmaceutical industries and benzene is one of them. The technique is called gas chromatography. When you have a very complex mixture, such as sunscreen, one might use a technique called headspace gas chromatography. You would heat a sample in a very small tightly closed vial, at a desired temperature, and sample the "headspace" with the help of a syringe. The "air" in the syringe then injected into in a gas chromatograph, whose job is to separate and detect the components in the gas phase. Benzene is very volatile and easily detected by GC.
For legal purposes, a mass spectrometer is connected to the gas chromatograph or an HPLC , which confirms the exact identity of the molecule exiting the chromatography column.
Just keep in mind that, NMR is the last choice the detect impurities. It cannot tell you what components are present at parts per million or parts per billion level.
As others have hinted at, analytical chemists often do a targeted analysis. It will be a futile exercise to ask someone "Can you find all the components in this random sunscreen"? The answer is no. It will take a 6 months, to probably more to determine and identify all components from scratch and it will easily cost thousands of dollars.

So when presented with an unknown substance, for example a mysterious powder, chemists try multiple targeted analyses? How do they avoid running out of powder to test?

Analytical chemist will not avoid an unknown, it just how much time and money is available along with the instruments. Identifying plant extract components may take 5 years of PhD work. There are many ways which may be used for non-destructive testing.

For that you still very expensive spectra libraries which contain spectra of thousands and thousands of known compounds. Infrared spectrum is the first way, once the spectrum of an unknown is collected, software tries to match it with the library. Second way is Raman spectrum, and finally liquid chromatography mass spectrometry, gas chromatography mass spectrometry.

@CodyBugstein if you invent a situation or a challenge, but even in practice, that is a possibility. One can run out of sample even when knowing exactly what it is but need to perform other characterisation. Your question tend to mix practice and principles. In principles we can do a lot. In many cases a melting point determination can already tell if we have a mixture, an impure sample, a pure compound... And so on. Of course quantification of everything depend very much on the level needed.


text:
In a perfect electrolytic cell, would the anode and cathode change by the same amount
If I had a theoretically perfect electrolytic cell, such as one with Copper anodes and cathodes, and a copper nitrate solution, over a given time period with a voltage applied, would the change in mass of the cathode be identical to the change in mass of the anode?

Probably it is exactly what you have in mind by "perfect".

At the cathode, copper ions get reduced to copper metal, but the copper so produced might be poorly adherent, i.e., it might flake off the cathode and become “cathode sludge”. This electroplating example nicely illustrates something Yogi Berra might have said: “In theory, theory and practice are the same; in practice, they aren’t!” So the mass balance is correct, regardless of where the reduced copper ions end up.


text:
Is ΔG the maximum non expansion work for reversible processes only?
The derivation of $$\mathrm{d}G=\delta w_{non.exp. max.}$$
$$G=H-TS$$
$$ \mathrm{d}G=\mathrm{d}H-T\mathrm{d}S-S\mathrm{d}T$$
$$\mathrm{d}G=\mathrm{d}U+P\mathrm{d}V+V\mathrm{d}P-T\mathrm{d} S-S\mathrm{d} T$$
$$dG=\delta w_{exp}+\delta w_{non.exp}+\delta Q+P\mathrm{d}V+V\mathrm{d}P-T\mathrm{d}S-S\mathrm{d}T$$
Now the term $\delta Q$ and the term $-T\mathrm{d} S$ are cancelled, Note that this means process is reversible($ \mathrm{d} Q_{rev}=TdS$).Also the term $\delta w_{exp}$ and the term $P\mathrm{d}V$are cancelled.So the only terms remaining are$$dG=\delta w_{non.exp.}+V\mathrm{d}P-S\mathrm{d}T$$
when pressure and temperature are kept constant then
$$dG=\delta w_{non.exp.}$$
So looking at this implies that Gibbs free energy is equal to the non expansion work only when a process is reversible and when the pressure and temperature are constant.
Is this correct?
If no then what am I missing here?
If yes then what does Gibbs free energy imply(Any kind of work?) when process is irreversible?

Gibbs energy change at p,T=const is the maximum nonvolume work (negative as done by the system ), achievable at reversible conditions. For irreversible conditions, this available work is lower. Imagine e.g. an ideal reversible rechargable cell versus a real cell with Ohmic ( and some other ) loses.

Yes, you are correct.

Maybe my answer here would help.

@Poutnik ok got it now

We are assuming in this analysis that the system is in contact with a single constant temperature reservoir at temperature T, the same temperature as the initial temperature of the system, and a single external pressure  at pressure P, the same pressure as the initial pressure of the system (see Denbigh).  From the first law, for such a process, we have $$\Delta U=Q-P\Delta V -W_{nonPV}$$and, from the 2nd law, we have $$\Delta S=\frac{Q}{T}+\sigma$$where $\sigma$ is the entropy generated within the system due to irreversibility (an always-positive entity).  So, if we combine these two equations, we obtain: $$\Delta G=\Delta U+P\Delta V-T\Delta S=-W_{nonPV}-T\sigma$$or equivalently, $$W_{nonPV}=-\Delta G-T\sigma$$So, for a reversible path between the initial and final states of the system, the nonPV work is maximum and equal to $-\Delta G$ and, for an irreversible path between the same two end states, the nonPV work is diminished by T times the amount of entropy generated due to irreversibility (and $-\Delta G$ exceeds the amount of nonPV work done by the system).

Hey Chet! I wanted to ask you something. I read your answer here and was wondering if $\Delta A$ is also diminished in irreversible path? I applied your method you wrote here, and came to a conclusion that just like $\Delta G$, $\Delta A$ is also diminished for irreversible process. What do you think about it? Is it correct?

@NatashaJ Yes.   It is worked out in detail in Denbigh, Principles of Chemical Equilibrium.

Thank you for reply! I read in a book that though the work involved is different from maximum work,  yet $\Delta A$ has the same value in both, reversible and irreversible path. I think it may be a misprint. Thank you.

@Natasha Why do you think that?

@NatashaJ What Denbigh shows is that, if the initial and final thermodynamic equilibrium temperatures of a process in a closed system are the same value T, and the process takes place with all the heat transfer to the system solely involving an ideal constant temperature reservoir also at T, then the work done by the system is less than the reversible work equal to $-Delta A$ between the initial and final states.

So, $\ Delta A$ is same for reversible and irreversible process, the only difference is the work done?

@Natasha J Yes, if the two end states are the same, then delta A is the same.  Another difference is the heat transferred.

Okay, got it! Thank you. Also I wanted to ask you something else, may I? Here, https://www.quora.com/How-do-I-understand-entropy-Gibbs-free-energy-and-Helmholtz-free-energy-in-simple-terms-without-using-equations,  in answer by Mark John Fernee, he says enthalpy is sum of internal energy and thermal energy. But, isn't thermal energy already included in internal energy?

@NatashaJ It doesn't make sense to me.  As I've said many times in these forums, trying to ascribe a physical interpretation to enthalpy is, in my judgment, a waste of one's valuable time.  Enthalpy is just a convenient function to use in many situations.

You are right about physical interpretation. It is a waste of time and energy to find physical interpretation. Thank you for answering!!


text:
Preparation of Alcohols from Grignard Reagent
We were given this question in the class:

Answer given:

 Pentan-2-ol

I got 2-Methyl Butanol. Why doesn't the methyl attack the 2nd carbon atom from the oxygen atom and form 2-Methyl Butanol instead of attacking the 1st carbon atom from oxygen atom forming Pentan-2-ol. Is it just because of easy access or is there any other reason for this happening?

Why do you think the grignard should attack the second C? That carbon is not next to an oxygen so it's not electrophilic, and there is no leaving group. There is no way for the grignard reagent to attack the second C

The ring strain inherent in a [4] ring makes it more reactive

As @Shoubhik R Maiti said, Grignard reagent won't attack the second $\ce{C}$ as, being farther away from the electronegative atom $\ce{O}$, it has less partial polarity.
Now, Grignard being a strong nucleophile, would attack the carbon with highest partial positive charge, hence we are left with a $3°$ carbon and a $2°$ carbon adjacent to $\ce{O}$ where attack is possible. For the $3°$ carbon, however, due to +I effect of $2$ $\ce{Me}$ group, partial positive is less compared to $2°$ carbon.
Also, $2°$ carbon is less sterically hindered. Hence attack is most preferred there.

TL;DR The Grignard reagent reacts via the SN2 mechanism and attacks the less sterically hindered carbon.

Now, this reaction is similar to the opening of an epoxide ring by a strong nucleophile, wherein the less sterically hindered position is favored for attack by the nucleophile.
According to SN2 type Reactions of Epoxides from Organic Chemistry, 5th edition by Francis Carey,

Scenario 2:
The $\ce{Nu}$ attacks the least hindered end of the epoxide at $180^\circ$ to the $\ce{C-O}$ bond that breaks.
Experimental results show that scenario 2 is observed by these stronger nucleophiles such as Grignard reagents, $\ce{RMgBr}$.

Therefore a similar reaction is seen here leading to the formation of 2-pentanol.


text:
What is the product when cyclohexanone reacts with morpholine?
The question asked to find the final product of the reaction between cylohexanone and morpholine. I attempted the reaction which is shown below:

I have understood the heat symbol. It simply facilitates the conversion of the alcohol formed to an alkene. My question lies in the catalyst. What is the role of the catalyst and why is it used in traces? Also what is benzene doing there ?

Benzene is just the solvent. pTsOH protonates the carbonyl group to activate it to attack by the amine. It is used in trace amounts because it is regenerated at the end of the reaction.

Some thing wrong in your product's structure. Oxygen cannot eliminate under this condition.

@MathewMahindaratne please go through the following page and update your knowledge :
http://www.orgsyn.org/demo.aspx?prep=CV5P0808

@om joglekar: I don't have to update my knowledge. You start with morpholine and end up with pyrrole derivative. You should get your brain checked first before advising others about their brain.

@omjoglekar The link you posted agrees with Mathew. Why does he need to update anything?

TsOH is a relatively strong organic acid (Source: Wikipedia). It is used to protonate the carbonyl oxygen and make the carbon atom more electrophilic, thus facilitating nucleophilic addition of the amine.
Benzene acts as a solvent in this case. It can dissolve the less polar cyclohexanone. Also, since benzene is a non-polar solvent, strong mineral acids (like H2SO4) cannot be used as it will not dissolve in benzene. Hence TsOH is chosen.

It is worth noting that benzene forms an azeotrope with water so the water produced by the reaction can be removed e.g. by Dean-Stark reflux to drive the reaction to completion. Benzene however, is not used these days because of its high carcinogenicity, toluene is the safer choice.

My experience with this reaction is that a catalyst is not mandatory. While the Org. Syn. prep. (http://www.orgsyn.org/demo.aspx?prep=CV5P0808) employs p-TsOH, this one doesn’t (
https://prepchem.com/cyclohexanone-morpholine-enamine-ii/.)


text:
Is it possible to make Dipotassium Phosphate from Mono-potassium phosphate and potassium hydroxide?
Having fertilizer formulated and asked for 100% Dipotassium phosphate,  the formulator says its reacting mono-potassium phosphate with potassium hydroxide to make it.  Just wanted to double check and be sure this made sense.

Yes, it makes sense.

yes, entirely possible

Yes,it's possible

Yeah, why not?
$$\ce{KH2PO4 + KOH -> K2HPO4 + H2O}$$


text:
Why does iodine stay in the aqueous layer of a hexane/salt water mixture?
All the procedures take place in test tubes. There is also stirring in each case.
$\ce{NaCl}$ and $\ce{I2}$ are simultaneously added to a water solution. Then we add hexane. There is no reaction between the $\ce{NaCl}$ and $\ce{I2}$. The water is colored yellow quite intensely, but the hexane stays colorless and transparent. Why the organic phase created is not colored, in spite the fact that we have $\ce{I2}$ in the solution?
A similar experiment, with a water solution of only $\ce{I2}$ added to hexane, showed, after a sudden stirring, that the hexane phase was indeed colored purple, as expected, and the water phase yellow. The colors were not as intense as previously.
Another experiment of $\ce{NaI}$ and $\ce{Cl2}$ in water solution added to hexane had as a result the coloring of the hexane phase to purple. Water phase was yellow.

Comments are not for extended discussion; this conversation has been moved to chat.

Separating iodine from aqueous NaCl is a standard teaching experiment. That is how people would isolate iodine from brines/ or seawater after oxidizing iodide ions.
I am unsure if your experiment has some information missing or if it needs to be repeated. Sometimes overthinking causes more problems.

People have studied the interactions of molecular iodine and seawater. It takes several days to observe disproportionation or chemical reduction. It is very likely there is an experimental mistake in your observation.


text:
Why is it preferred to sulfonate nitrogen instead of carbon?
In this question, sulfonation of aniline is discussed and the answer by Waylander suggests that the sulfonation occurs on the nitrogen instead first.
The above is clear, but I can't make sense of why sulfonation of nitrogen is preferred over the sulfonation of carbon?
Here is a picture of sulfonation of the carbons in benzene, it is clear that there is a high preference to carbons in the benzene ring here:

Picture from David R Klein's Organic Chemistry.

I believe Nu addition from lone pair of N would be preferred over EAS as aromaticity is not lost in the former. This is perhaps similar to why Friedel-Crafts acylation doesn't occur in aniline, instead N undergoes acylation.

In the reaction scheme shown, there neither is an aniline, nor any nitrogen atom.  Did you miss a line in the drawing?  In the case of benzene, you likely do not even need fuming sulfuric acid to install the group.

also, to add to @Ritam_Dasgupta, pyridine is a very good nitrogen nucleophile because it has low steric bulk around the nitrogen lone pair

That's what is said in my book, I have shown the reaciton for benzene with the sulphuric acid to show that it is indeed reactive with it. So, I can't understand why it is reactive w/ the nitrogen as well?

Aniline can be protonated by strong acid. However, the site of protonation was a subject of debate over decades. It has been confirmed after numerous experimentation that protonation is more favored at the nitrogen atom that the carbon atom at the para- position. See this previous chem.SE discussion to find out the key factors resulting in the N-attack more than C-attack. Alternatively, refer 1 and 2. Thus, we can conclude that direct addition of sulfonyl group at para position is not possible. The reaction starts by the nitrogen atoms getting attacked to form anilinium cation. This cation then reacts with the rest of the sulfuric acid and form anilinium hydrogensulfate which at higher temperatures decomposes to a stable isomer. But this decomposition is not a direct reaction. It goes through a number of intermediates before arranging to the stable para isomer:

However, this mechanism is different for different acid concentration. Using only sulfuric acid will lead to the para-isomer but usage of fuming sulfuring acid(FSA) will lead to the meta-isomer. It has been proposed that if the acid concentration is increased, then more and more aniline gets protonated to anilinium which is meta directing and hence meta-isomer prevails when FSA was used. The paper used dimethyl aniline for the demonstration:

You can find more details on the major products of the reactions and their yields in ref. 4.
Notes and References

The Site of Protonation of Aniline Rebecca L. Smith, Leonard J.
Chyall, Brandon J. Beasley, and Hilkka I. Kenttamaa Journal of the
American Chemical Society 1995 117 (30), 7971-7973 DOI:
10.1021/ja00135a016

Site of Protonation in Aniline and Substituted Anilines in the Gas
Phase: A Study via the Local Hard and Soft Acids and Bases Concept
R. K. Roy, F. de Proft, and P. Geerlings The Journal of Physical
Chemistry A 1998 102 (35), 7035-7040 DOI: 10.1021/jp9815661

Mechanism of the Sulfonation of Aromatic Amines by Elliot R. Alexander:

I. Sulfonation with Fuming Sulfuric Acid, DOI: 10.1021/ja01210a018
II. Sulfonation at Elevated Temperatures with Sulfuric Acid, DOI: 10.1021/ja01199a009

Aromatic sulphonation. Part 59. Sulphonation of aniline in concentrated aqueous and fuming sulphuric acid, Paul K. Maarsen  and  Hans Cerfontain, J. Chem. Soc., Perkin Trans. 2, 1977, 1008-1010, DOI: 10.1039/P29770001008

Why III is an intermediate?: Why does sulfonation of aniline occur para rather than ortho?

Even if there are multiple amide groups present (for instance sulfanilamide), only the amide group with the carbon-nitrogen bond will be protonated: Is sulphonamide group prone to undergo diazotization?

Is reverse sulfonation of aniline possible?


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Liquid Fire and Unclogging Bathtub - lingering residue
So my mom needed help with a bathtub drain that wasn't performing all that well. She sent me out to the hardware store to get some Drain-o, but they didn't carry that brand.  Instead, the young man helping me suggested this brand of drain opener called Liquid Fire.  Said that it will unclog the pipes real quick. I took his word for it and began to get checked out when the owner of the shop, an elder fellow who had been running the hardware store before I was even born, took one look at the product and gave one of those short whistles.  He said it was some potent stuff and to not be in the room after I poured it, to keep the door closed and window opened in the bathroom, and then to let it sit for about 10-15 mins before running hot water in the shower for another 15 mins.  Ok, can do.
I kept his directions in mind, took note that I was working with sulfuric acid, so wore gloves, even a respirator just in case. I poured the acid down the drain then closed the door behind me and waited.  Definitely got the rotten egg smell. After about 20 mins, I came back and there was this mass of brown residue all over the tub.  Not just around the drain, but it streaked all the way to the back, and even up the sides like the tub had been filled up about 6 inches. I didn't touch it, but decided to run the water and see what happened.  Came back after 15 mins, the rotten egg smell had dispersed some, but the residue was still there.  After running some more water to make sure all the acid had washed away, my mom had me throw comet over all the stain and we were able to scrub the residue off.
So my question is, does this normally happen with a tub drain and liquid fire? As in, the tub fills up and then drains back out leaving a residue. And if not, was this some kind of eruption and what caused it?  And what was this brown residue (debris, rust, etc.)?

A great deal of heat is generated when concentrated sulfuric acid is diluted, even more when it interacts with organic matter. This is certainly enough to boil the water in the pipework driving the brown residue back up the line.

Foam did this. You should have covered the drain with an upturned pot, then it wouldn't spread that wide. Then again, it's not a big deal.


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Acidity of alpha hydrogens in enolate formation

I am interested in why the most right alpha hydrogen is removed to form the carbanion prior to intermolecular aldol condensation (2nd row first molecule).
Why is it that the alpha hydrogen from the secondary and tertiary alkyl substituted carbon is not removed instead?
I would think that in deciding which alpha hydrogen is removed, the most acidic is removed. Considering that a more stable conjugate base would result in a more acidic alpha hydrogen, wouldn’t the tertiary and secondary substituted alpha carbons be more acidic then? Since the carbanion formed has better charge dispersal than the non substituted alpha carbon on the right (I carried over the idea of charge dispersal from acidity in halogens where a bigger volume can stabilise a charge).
also since the carbanion is not electronegative, the alkyl groups would not donate electrons towards the alpha carbanion to destabilise it.

Welcome to ChemSE! All of the alpha protons are removed continuously under reversible conditions. Under the conditions of the reaction every step should have reversible arrows except for the last one that gives structure A. There is a protic solvent here, presumably water.

I'm not posting it as an answer since I'm not completely sure, but I think that abstracting the least sterically hindered H-atom is favourable in this case, so that might be another reason besides the answers given by Waylander and Mathew

I'm interested in the OP's curiosity. I'd agree with Waylander about disfavoring to make 4-membered rings. Yet, it is possible to have two six-membered ring products. The first $\alpha$-proton abstraction followed by Michael addition resulted the first intermediate as OP suggested:

Now, this newly formed intermediate contain 4 types of eligible $\alpha$-protons, namely, $\ce{H_a, H_b, H_c,}$ and $\ce{H_d}$. Amongst them, only $\ce{H_a}$ is primary, while $\ce{H_c}$ is tertiary and the rest, $\ce{H_b}$ and $\ce{H_d}$, are secondary. The base would abstract each proton in rates according to the dissociation energy of each (e.g., the fastest rate with tertiary and the slowest with primary) but all abstractions are in fast equilibrium as shown in the scheme below:

$\ce{H_a}$ and  $\ce{H_d}$ proton abstractions resulted the formation of intermediates $\bf{I}$ and $\bf{II}$, which would abstract from a proton from water to give the corresponding aldols.  Intermediate $\bf{I}$ is the most stable among others and resulted aldol would undergo further dehydration to give the corresponding $\alpha,\beta$-unsaturated ketone as the major product (as correctly suggested by the OP). Meanwhile, formation of the intermediate $\bf{II}$ is slow compared to that of $\bf{I}$ due to greater ring strain (even though it is a six-membered ring formation, the resulting intermediate is a bicyclic product with a bridged-head). Nevertheless, some would form to give the corresponding aldol as a minor product, which would not undergo dehydration because $\alpha$-hydrogen is locked in a position that restrict the elimination.
On the other hand, although abstractions of $\ce{H_b}$ and  $\ce{H_c}$ protons are possible and in fast equilibrium,  the formation of intermediates $\bf{III}$ and $\bf{IV}$ are very minute depending on the corresponding activation energies. The rates of these formations are very slow or not exist compared to other two, mostly due to very unfavorable 4-membered ring formation.

to clarify, for intermediates III and IV would it be the activation energy leading to their formation or their stability relative to the other intermediates that hinders their formation? (i am leaning towards stability to be the main reason)
Is the reason why intermediate II cannot undergo elimination because the carbons containing the α-hydrogen and soon to be alcohol functional group do not have corresponding p-orbitals to form a pi bond?
for confirmation, do alkyl substituents aid in the deprotonation via charge dispersal? (my teacher insists that they have no effect so i am unsure)

@ImJustKawaii: It's true intermediate III and IV are unstable compared to other two intermediates, and hence contain higher energy. Therefore they need high energy to form and have higher activation energy to form. When formed, the reversed reaction need less energy to go back so eventually more stable intermediates win out. You may need to look at energy profile  with reaction progress to fully understand what's I'm saying.

The two deprotonated species are in equilibrium (along with the neutral parent molecule). The deprotonation you suggest will occur, but the forward reaction cannot because it would be forming a [4] ring which is dis-favoured. The reaction moves forward through the other anion even though it may be the minor component of the equilibrium mixture.

what about the secondary substituted alpha carbon on the most left? if it was deprotonated there, wouldn't it form a [6] ring making its forward reaction favourable?

it could, but the species it creates is subject to greater ring strain so will be disfavoured.

sorry but i do not see why the species created using the secondary substituted alpha carbon would experience a greater ring strain than the structure A. since both of them form 6 membered rings, wouldn’t they experience similar ring strain?


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Is Sulfur or Sulfur Dioxide harmful to the environment?
I'm studying the impact/effects of greenhouse gases/pollution on the environment. I found that sulfur dioxide has harmful effects , but I read here that "pollutants like sulfur actually keep our planet cooler." So, I wanted to know, what's the difference between the two? And which one is the true greenhouse gas? Thanks.

There is no difference and no two. Sulfur in the quotation is actually supposed to mean "sulfur dioxide". OK, to be more precise: "sulfur dioxide and a few other unimportant compounds".

In the absence of citations in your second link I would ignore information in it. Rely on more reliable sources, for instance EPA.

Expanding @Ivan's comment, the sulfur in that quotation pretty much meant other sulfur oxides ($\ce{SO_x}$). If it was to mean elemental sulfur, then it would interact with the environment in its in vapor state but sulfur vapor only exist at very high temperatures which is beyond even earth's recorded highest temperature.
While sulfur dioxide isn’t a direct greenhouse gas like carbon dioxide or methane, it is considered an indirect greenhouse gas because, when coupled with elemental carbon, it forms aerosols. Also, it is a no-brainer that $\ce{SO_x}$ is very deadly as it poses the threat of forming acid rain.

(+1) All the more amazing that people mine big blocks of sulfur at the active volcano named Kawah Ijen, in East Java, Indonesia. At night, the sulfur burns blue, making for an eerily beautiful but toxic spectacle. Wild and crazy planet we got here! ;-)


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Can ethanol and heat during distillation cause degradation/dissolution of vinyl tubing leading hot vapors of the distillate to the condenser?
I'm a chemist and a qa lab, and we occasionally test herbal remedy products from one of our customers for alcohol content using a distillation, then using the specific gravity of the distillate to estimate alcohol content as if the distillate were a binary solution of ethanol and water. Almost all of the distillations I do result in a clear, transparent distillate, however today I tested a sample, and the distillate appeared a cloudy white. The tube I have leading vapors to the condenser was glass, but due to it being worn and damaged I changed it today using what I had on hand here, which was vinyl tubing. The tubing unfortunately only says vinyl so I don't know exactly what kind of vinyl polymer it is, but during distillation the tubing went from transparent to white. The alcohol content of the samples was 18.1% which is lower than the expected 20 - 30%. I'm wondering if the ethanol and or heat (or another chemical in the sample) was causing the degradation or dissolution of the tubing??
Can anyone help me find out if our tubing was degraded or dissolved in the process of distillation?
Thanks for the help!

Even cold ethanol corrupts surface of many plastics. Using plastic tubing may not be a good idea, full glass apparatus is advised. Try to test the plastic behaviour with cold ethanol alone,  with heat alone and in hot ethanol.

A lot of standard "vinyl" tubing is plasticised PVC. It contains a lot of plasticiser (often phthalates) which is easy to displace from the tubing. Even when that tubing is used for inert gas only, this often accumulates in bends in the tubes over time. Hot gas passing through such tubes will extract a lot which is why you should never use such tubing anywhere where it will contact your reaction, solutions or distillate vapours. Your distillate will be heavily contaminated with plasticiser.

Both of these comments are very helpful. Thank you!


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Is the origin of non-integral atomic masses the difference in masses of the proton and neutron?
Yes, atomic masses are non-integral due to isotopes and binding energy. These are the two reasons I found on google and other answers on stackexchange.
But there is something else - isn't the most obvious reason for it the way atomic masses are calculated? We defined 1 amu as one twelfth the mass of a C-12 atom. Now C-12 has 6 protons and 6 neutrons (thus 1 amu isn't mass of 1 proton or 1 neutron). And mass of proton and neutron is different. And each element has a different ratio of protons to neutrons. So of course when we convert their atomic masses in terms of amu it is going to be non-integral.
But this reason isn't mentioned anywhere so now I am starting to doubt it. Is there a flaw in this logic?

Yes it will be non integral for example atomic mass of hydrogen is 1.0008 amu but  for calculations we round them off to nearest integer

1 amu = 1/12 ($\ce{6 x m_e^- + 6 x m_p^+ + 6 x m_n^o - (nuclear binding energy of C-12 + atomic binding energy of C-12))}$  When you apply this to determine the masses of other isotopes, you are essentially doing a similar calculation for them (effectively--in practice, these would typically obtained by measurement), and then dividing by the above.  So clearly you wouldn't expect atomic masses to be integral.  For an explanation of nuclear and atomic binding energies, see https://en.wikipedia.org/wiki/Binding_energy

Please note that the unit symbol "amu" is obsolete (in chemistry since 1961), and 1 amu was not "one twelfth the mass of a C-12 atom".

Does this answer your question? Why atomic masses aren't integers?

@Mithoron- No, actually that was the question I read before posting mine. But as I said, it doesn't include anything about differences in masses of proton and neutron. So my question is a little different

Yes this is a valid reason, albeit a small one, almost negligible. After all, 1amu and the masses of 1 proton and 1 neutron are pretty close to each other.
Then again, the binding energy would also be negligible, if not for oxygen's $15.999$ which catches the student's eye right away and begs for an explanation. And of course binding energy is the only possible explanation, as oxygen's other isotopes work in the wrong direction, and the ratio of protons and neutrons in $^{12}\rm C$ and $^{16}\rm O$ is the same.
The lack of a similar clear example in your case is probably the reason why your reason is never mentioned anywhere.
So it goes.

The mass difference between a bare proton and a bare neutron is 1293 keV, while the mass of an electron is 511 keV. That means that the effect you're talking about is often going to be on the same order of magnitude as the difference in mass between an atom and one of its ions. In something like liquid-phase chemistry, as opposed to mass spectroscopy, I doubt that an ion ever really even maintains its ionization state for long enough to make the electron-mass correction even meaningful, and if that's not meaningful then the correction you're talking about, which is on the same order of magnitude, is also not very meaningful.
When you want to start talking about corrections at this level, you probably need to define more carefully what purpose you're going to use the resulting numbers for. For example, if you're going to look at the infrared spectrum of the NaCl molecule in a gas, which comes from its end-over-end rotation, then it's got some moment of inertia that is mainly determined by the nuclei, but when the bond forms, one of the electrons shifts its mass around, and that will change the result.
Keep in mind also that, conceptually, the protons and neutrons in a nucleus aren't really the same creatures as the particles that exist in free space -- the "bare" proton and neutron I referred to above. A free neutron is a little ball of quarks and gluons, with pairs of particles appearing and disappearing spontaneously in a kind of quantum soup. The mass that we measure for the neutron is some kind of average or composite value that represents the masses and kinetic energies of all those sub-particles. When a proton or neutron is incorporated in a nucleus, it loses its individual identity, and it's no longer really a proton or neutron. It promiscuously shares all its quarks and gluons with its neighbors. It's only a kind of shorthand terminology when we refer to the nucleus as being made of protons and neutrons. So it's not even really totally obvious whether it makes sense to try to add the masses of all "the protons and neutrons" inside a nucleus. It does sort of work as an approximation (it's what's done in standard versions of the nuclear liquid drop model), but it's only an approximation.
Another thing to realize is that most chemistry is done with light elements, for which the proton number and neutron number are about the same. Since the atomic mass standard is based on 12C, which has equal neutron and proton numbers, you will only see any effect in a.m.u. masses when you go to heavy nuclei.

Wow that was a great explanation!

I understood most of it but can you explain the NaCl rotation part in more detail?

@DhruviKataria: The rotational band forms a series of energy levels with energies proportional to $J(J+1)/I$, where $I$ is the moment of inertia and $J$ is the angular momentum in units of $\hbar$. What you observe in spectroscopy is the differences among energy levels in this band, so those are $\propto 1/I$.

The NaCl infrared part is indeed not clear. What is meant by "when the bond forms, one of the electron shifts its mass around" in "For example, if you're going to look at the infrared spectrum of the NaCl molecule in a gas, which comes from its end-over-end rotation, then it's got some moment of inertia that is mainly determined by the nuclei, but when the bond forms, one of the electrons shifts its mass around, and that will change the result"?


text:
Behavior of amines as acids
I came across this question which asked us to find in which alternative does the amine behave as an acid the most:

In which of the following reactions does the amine behave as an acid
(a) $\ce{(C2H5)2NH}$ + $\ce{H2PtCl6}$
(b) $\ce{CH3NH2}$ + $\ce{H2O}$
(c) $\ce{((Me)2CH)2NH}$ + $\ce{n-C4H9Li}$
(d) $\ce{(C2H5)3N}$ + $\ce{BF3}$

Answer given:

 (c)

Now coming to my thought process:
So without considering the reagents (because I don't know any of these reactions!), I considered the acidic and basic nature of amines.

We know that secondary amines are the most basic while primary amines are less basic due to lesser $+I$ effect. So, my reasoning says that the answer should be (b), which is evidently wrong.
Any ideas about the reactions  and the solution?

It is a relative term. For instance, water also acts as an acid or base based on which reagent it is reacting with. Basically, anything with $\mathrm{p}K_\mathrm{a} \lt 7$, water acts as an base (e.g., acetic acid). Anything with $\mathrm{p}K_\mathrm{a} \gt 7$, water acts as an acid (e.g., ammonia).

Question does not ask which amine is most acidic, but in which reaction amine acts as acid!

Acid/base nature of any reagent is a relative term. For instance, water $(\mathrm{p}K_\mathrm{w} = 14)$ also acts as an acid or base based on which reagent it is reacting with. Basically, any reagent with $\mathrm{p}K_\mathrm{a} \lt 7$, water acts as an base (e.g., acetic acid; $\mathrm{p}K_\mathrm{a} = 4.75$):
$$\ce{CH3COOH + H2O <=> CH3COO- + H3O+}$$
Then again, if the conjugate acid of any reagent has $\mathrm{p}K_\mathrm{a} \gt 7$, water acts with that reagent as an acid (e.g., the conjugate acid of ammonia with $\mathrm{p}K_\mathrm{a} = 9.25$):
$$\ce{NH3 + H2O <=> NH4+ + OH-}$$
Now, let's look at the given reactions:
(a) $\ce{(C2H5)2NH}$ + $\ce{H2PtCl6}$: This is the though one among all four choices. The name of $\ce{H2PtCl6}$ is hexachloroplatinic acid, and thus, it is safe to assume that this acts like an acid in water meaning $\mathrm{p}K_\mathrm{a}$ of $\ce{H2PtCl6}$ is $ \lt 7$. Ammonia with water acts as a acid $(\mathrm{p}K_\mathrm{a} \gt 7)$ and hence the amine in this case must acts as a base.
(b) $\ce{CH3NH2}$ + $\ce{H2O}$:$\ce{CH3NH2}$ acts like ammonia with water because $\mathrm{p}K_\mathrm{a}$ of its conjugate acid is about $10.5$. Thus, $\ce{CH3NH2}$ acts like a base here.
(c) $\ce{((Me)2CH)2NH + n-C4H9Li}$:$\ce{n-C4H9Li}$ is a strong base with $\mathrm{p}K_\mathrm{a} \approx 50$ for conjugate acid of $\ce{n-C4H9-}$, $\ce{n-C4H10}$. Therefore, the amine acts here as an acid, giving away its lone proton.
(d) $\ce{(C2H5)3N + BF3}$:Here, $\ce{BF3}$ is a Lewis acid, $\ce{B}$ in which does not fulfil the octet. Thus, it is craving for a electron pair, $\ce{N}$ in $\ce{(C2H5)3N}$ would gladly donate. Therefore, amine acts as a base here.
Therefore, no doubt, the choice C is the answer.


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Bromochlorofluoromethane as a reagent in Riemer-Tiemann reaction
The reagent usually used in Riemer-Tiemann reaction is $\ce{CHCl_3 }$ however one of the variations I read used $\ce{CHFClBr}$ as a reagent. Now $\ce{CHCl_3}$ produces the $\ce{:CCl_2}$ radical using which the reaction usually proceeds. But what happens when we use $\ce{CHFClBr}$ what will be the radical formed when this is heated in presence of a base and why? Does it have to do with the individual bond strength or something else?

Where did you read that? Include the source. I'm pretty sure the intermediate radical of the reagent would be very, very unstable to the point where it would cease to exist and reaction won't occur. If you say variations of RT reaction, then it will the usage of carbon tetrachloride or bromoform.

@NilayGhosh was used as a reagent in abnormal Reimer-Tiemann reaction as a part of a problem but I didn't read it in any standard book though. Could be possible that it doesn't exist and it was given expecting us to know what radicle it would form theoretically.

The Wikipedia entry for bromochlorofluoromethane notes its relative instability to hydrolysis so I wonder if it would even produce a carbene on reaction with hydroxide. This may be a problem set by someone who did not adequately research the properties of the starting material.

I am inclined to agree with Waylander and Nilay's comments, but having solved such questions (purely in theory) before, I can attempt to provide an answer. Since the formation of a carbene is an elimination of $\ce{HX}$ ($\ce{X}$ is a halogen), the elimination proceeds fastest for the best leaving group. The best leaving group order in general is $\ce{I > Br > Cl > F}$, so if a methane has multiple halogen atoms, the best leaving group will be eliminated as $\ce{HX}$.
In your case, $\ce{Br}$ is the best leaving group, so the carbene you get is $\ce{:CFCl}$, with the elimination of $\ce{HBr}$.
Since you mentioned abnormal Reimer-Tiemann reaction, if the question involves formation of halobenzene from cyclopentadiene, it will lose another halogen atom in a subsequent step. Again the best leaving group is lost, so from $\ce{:CFCl}$, chlorine atom is lost, the final product thus being fluorobenzene.
Again, this is all purely in theory, practically I am sure Waylander and Nilay are correct.

The reason why $\ce{:CFCl}$ is the final radical is because it is more stable. adjacent lone pairs often stabilize radicals. HOMO-LUMO interactions that stabilize the radical are poorer for $\ce{Br}$ and $\ce{I}$  due to high energy level differences of interacting MOs which is why they are better leaving groups. ( Also $\ce{\cdot Br}$ and $\ce{\cdot I}$ radicals are more stable so there is no conflict here)

@napstablook I'm not learnt enough to be able to make complete sense of your remark, but I'm assuming you mean that the backbonding (like in $\ce{BF3}$ or $\ce{BCl3}$) will be stronger for carbenes with $\ce{Cl}$ and $\ce{F}$ attached to it, rather than $\ce{Br}$ or $\ce{I}$. Is that what you meant? Also, I've known that  $\ce{Br}$ and $\ce{I}$ are better leaving groups due to stability of larger anions that they form after leaving - and because the $\ce{C-X}$ bond strength decreases as halogen atom's size increases, so it becomes easier to break. Is that what you meant too?

I read it here: https://www.masterorganicchemistry.com/2013/08/02/3-factors-that-stabilize-free-radicals/  . This is a great article! I don't think there is any problem in your answer, just an addendum from me


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Whats the difference between unit cell and motif?
The definition in my material is that motif and unit cells are both a repeating unit, but whereas motif is with respect to 2D lattice the unit cell is with respect to 3D lattice. But I can't find anything online with respect to that.

Conceptually, the unit cell is the larger object, the (if relating to 3D crystals), the parallelepiped containing the crystallographic motif.  It is the unit cell which, by mere translation along $\vec{a}$, $\vec{b}$, and $\vec{c}$ builds the crystal.  In an analogy, unit cells stash like shoe boxes next to / on top of each other.
In addition to the outer shape and apparent symmetry of the unit cell (the Bravais classes; i.e., triclinic, monoclinic, orthorhombic, tetragonal, rhombohedral, hexagonal, or cubic) the unit cell contains symmetry operators in its inner (the proper rotation axes, mirror planes, centres of inversion, screw axes, rotoreflection axes, glide planes) which act on the crystallographic motif.  In an analogy, these symmetry operators are instructions how to copy and paste the motif within the unit cell.  The motif itself is a smallest representative design pattern within the elected unit cell.
As an example, the illustration below shows how the $c$-glide plane relates two molecules of maleic acid in respect of their position within the unit cell, as well as in respect to their orientation relative to each other:

(credit: Hanson, R. M.; Jmol – a paradigm shift in crystallographic visualization in J. Appl. Cryst. 43, 2010, 1250-1260; doi: 10.1107/S0021889810030256 (open access))
Depending on the the space group symmetry within the unit cell, the position and the symmetry of the molecules within the unit cell, you may encounter motifs which on first inspection might look like incomplete.  For example, if a centre of inversion (molecular level) coincides with a centre of inversion (unit cell level).  The molecule then still is fully described in its geometry, because of the exact completion by the symmetry operators.  As an example, see the following triptych:

(entry COD 2102215)
Here, the left image depicts the crystallographic motif alone.  In the centre, you recognize the triazine molecule, «complete» as if you would draw it in a reaction scheme; but it has been «completed» by applying the symmetry operator of the unit cell.  In the right image, using the same perspective (i.e., again the same direction of observation) you see the complete unit cell when all symmetry operators are applied.  It depends on the software used how (if possible all three of) these levels are visualized with ease.
Primary reference for the crystal structure: Fridman, N.; Kapon, M.; Sheynin, Y.; Kaftory, M.; Different packing in three polymorphs of 2,4,6-trimethoxy-1,3,5-triazine in Acta Cryst. B60, 2004, 97-102; doi: 10.1107/S0108768103026284.

Thanks for the answer. I love how you went in detail to explain the relation between the two (in a depth much better than my textbook too), but could you add a short definition of the two as well in the beginning?

I can't seem to find the definition of a motif on the net that I can compare with the definition of a unit cell. Thanks once again.

Unlike unit cell, motif is not a thing that can be defined rigorously.

Unit cell is a volume that can build the entire crystal if only the lattice translations are applied. Asymmetric unit cell is a set of atoms that can build the entire crystal if all space group symmetries are applied. Motif usually means, as far as I have seen, a set of atoms that can build the entire crystal if only the lattice translations are applied. So I do not quite agree with what @Buttonwood has written. Maybe motif is not always used with the same meaning. It is not so common as the other two concepts.

Here is a reference that is consistent with the meaning of motif that I understand: https://www.doitpoms.ac.uk/tlplib/crystallography3/structure.php

@10ppb The help of an elder version of CCDC's Mercury program, archived e.g., about asymmetric unit as the smaller, and unit cell as the larger object.


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Finding out the composition of a sample in analytical chemistry without knowing what to test for
I am little new to analytical chemistry, but after reading up on it, I realised that most analytical procedures need you to know what to test for beforehand.
May I know, given a sample of material G, is it possible to know its composition without knowing beforehand what to test for? ie. For a sample G, is there a procedure that can be followed so I can know that G has:
30% of compound X
20% of compound Y
50% of compound Z
And that no other compounds exist in the sample.
(But I do not know X, Y, Z are the compounds to test for beforehand.)
Or are all analytical chemistry procedures limited in the sense that you need to know what to test for first?
If yes, what is the closest possible procedure to follow to get what I want as outlined above?
Lastly, are analytical chemistry procedures also limited in the sense that you can never know for sure if that "no other compounds exist"?

You never start from zero. Some questions I would ask you before planning an analysis are "where did you get the sample?" and "why do you want to know the composition?" (this helping me assess exactly what it is about the composition that you really want to know, and therefore narrowing the experiments that need to be performed)

Analytical chemists are extremely reluctant to start from zero. Like Buck Thorn said, they'll ask you all kinds of weird questions. If you keep the mystery but continue to insist on analysis, they'll throw your sample at you and kick you downstairs.

It depends on what you have in mind. In practice, see the comments above. In principle, yes you can. See also my comment to https://chemistry.stackexchange.com/questions/152296/how-are-compounds-within-a-complicated-mixture-detected as it is related. But I like the way VDVH summarised the idea. First ask questions.

It depends. Which instruments have you got (ISP-MS, IR, XRF, NMR or just pH metre)? What type of sample have you got (liquid, gas, powder, homogenous, heterogenous)? What results do you need (elemental composition, molecule composition)? What is the purpose of analysis? Do you need to analyse whole sample or just the surface.
Therefore you always have to ask questions and then find the appropriate available method. Generally, start with the cheap non-destructive methods and then decide what to do next.
I think you never know 100% composition. However does it really matter? E. g. the analyte is 99 % pure. Is it important to know the 1 % of impurities? Sometimes it is, sometimes it is not. Sometimes it is enough to say that there is 99 % of compound X and 1 % are impurities. Sometimes you need to say that there is 99 % of compoud X, 0.8 % of compound Y, 0.19 % of compound Z and 0.0009 % of compound A. The rest are impurities.
I would say that analytical chemistry answers questions. And the questions have to be asked first. And as you learn, you are able to answer more and more questions.


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Effect of distance on the mesomeric effect
I have always thought that mesomeric effect is somewhat dependent on distance.
But I  recently came across a question where I was asked to predict the order of stability of o-methoxyphenyl carbocation and p-methoxyphenyl carbocation. And since electron donation can be done more efficiently in ortho case, I thought it would be more stable. But in fact, the para carbocation is more stable.
Why is the para carbocation more stable than the ortho carbocation?

Mesomeric effect has no relation to distance. Here the distance dependence of the inductive effect plays the role.

According to the IUPAC goldbook, the definition of the mesomeric effect is as follows,

The effect (on reaction rates, ionization equilibria, etc.) attributed to a substituent due to overlap of its p- or π-orbitals with the p- or π-orbitals of the rest of the molecular entity. Delocalization is thereby introduced or extended, and electronic charge may flow to or from the substituent. The effect is symbolized by M. Strictly understood, the mesomeric effect operates in the ground electronic state of the molecule.

This effect is distance independent unlike the inductive effect which is defined as follows,

In strict definition, an experimentally observable effect (on rates of reaction, etc.) of the transmission of charge through a chain of atoms by electrostatic induction.

In your case, the cations we need to compare are p=methoxyphenyl carbocation and o-methoxyphenyl carbocation. Here we need to see two major effects, the mesomeric effect and the inductive effect.

These are the resonance structures of the methoxyphenol substrate, which shows that there is an increased electron density at the o- and p- positions. This means that a carbocation is more stabilized when it comes at these two positions.
Now, how do we compare these two positions? This is where the inductive effect comes in. Due to the higher electronegativity of oxygen, the electron density nearer to the oxygen would be concentrated near the oxygen. However, this effect reduces the farther away you move from the oxygen. This makes the o- position destabilized and hence less stable than the p= position.
Therefore p-methoxyphenyl carbocation is more stable compared to the o-methoxyphenyl carbocation.
Remember that inductive effect is a distance dependent effect whereas mesomeric effect is not.

The answer is correct for the problem that OP posted, however, it isn't true that resonance has no distance limit. It's just that it can act over a really long distance compared to the inductive effect.