text:
Total bond dissociation temperature
I understand that covalent bonds break at a dissociation energy $D_0$ that is a function of the particular bond.  And it looks like an upper bound on $D_0$ is known for possible covalent bonds.
It's unclear to me whether continuing to add energy to a mass will at some point guarantee that all covalent bonds are broken, or whether there are other physical processes that could continue to absorb energy so as to prevent total homolysis.
Is there some temperature above which the bonds cannot be maintained in any substantial way?  I.e., for any soup of atoms is there a temperature sufficiently high that no molecular compounds can exist?
How hot does thermal plasma need to be to eliminate virtually all chemical bonding?

Maybe this idea is not meant to generalize over entirety of bonds. Because bonds form in so radically different environments.

@Proscionexium if necessary, yes.  I'm assuming that covalent bonds are both the strongest and the most meaningful for chemical properties, but if there are other inter-atomic bonds that can survive energy sufficient to totally homolyze a mass, and that affect the chemo-physical behavior of a mass of atoms, that would be a helpful answer (though perhaps to a related question I should pose?).

How is this any different from your old question? https://chemistry.stackexchange.com/questions/34964/plasma-creation-and-condensation besides being more focused...

@Mithoron that one is about what happens if all bonds are broken and then the mass is cooled.  This one is whether heating actually will break all bonds (and if so, at what point).

sigh Heat stuff to million K and what we call chemistry is gone. There is no sharp threshold. What would you want, point when less than one ppb of CO remains undissociated?

@Mithoron Just an elaboration of that would be a helpful answer.  E.g., "There is no finite temperature at which all covalent bonds are broken, because ....?"  And maybe, "The dissociated fraction of a mass varies with what functional form of temperature?, or what other variables?"

There certainly are such temperatures, million K should be plenty enough. But yeah, an answer like that would be nice.

It's a duplicate of https://chemistry.stackexchange.com/questions/99110/at-what-temperature-can-molecules-no-longer-form though...

related https://chemistry.stackexchange.com/questions/61621/dissociation-of-diatomic-molecules

Does this answer your question? At what temperature can molecules no longer form?


text:
Is the change ΔG for an ideal gas unknown?
When one is introduced to the first law of thermodynamics, one learns that enthalpy is only a function of temperature, and calculating its changes is straightforward
\begin{equation}
   \Delta H^\pu{ig} = H^\pu{ig}(T_2) - H^\pu{ig}(T_1) =
   \int_{T_1}^{T_2} C_\mathrm{p}^\pu{ig}(T) \; dT \tag{1}
\end{equation}
When the second law is introduced, one also learns that the entropy is a function of pressure and temperature, and calculating its changes is also straightforward
\begin{equation}
   \Delta S^\pu{ig} = S^\pu{ig}(p_2,T_2) - S^\pu{ig}(p_1,T_1) =
   \int_{T_1}^{T_2} \frac{C_\mathrm{p}^\pu{ig}(T)}{T} \; dT -R \ln\left(\frac{p_2}{p_1}\right) \tag{2}
\end{equation}
Finally, when the Gibbs energy is introduced, one is presented with the fundamental equation for constant composition systems
$$ \mathrm{d}G = V\mathrm{d}p - S\mathrm{d}T \tag{3} $$
And Eq. (3) for an ideal gas, at constant temperature $T$, gives also a famous relationship to calculate the changes in pressure. We replace $V^\pu{ig} = RT/p$, integrate, and get
\begin{equation}
  \Delta G^\pu{ig} = G^\pu{ig}(p_2,T) - G^\pu{ig}(p_1,T) = RT \ln\left(\frac{p_2}{p_1}\right)
  \tag{4}
\end{equation}
Now, I am interested in calculating the changes in Gibbs energy when the temperature changes and the pressure is constant. However, one faces the funny situation of being unable to obtain an expression in this case. For a constant pressure process, integrating Eq. (3) is not particularly attractive since one gets the entropy in the right-hand side
\begin{equation}
  \left(\frac{\partial G}{\partial T}\right)_p = -S \tag{5}
\end{equation}
It is possible to manipulate Eq. (5) and obtain the Gibbs-Helmholtz equation to eliminate $S$ for $H$
\begin{equation}
  \left[\frac{\partial (G^\pu{ig}/T)}{\partial T}\right]_p = -\frac{H^\pu{ig}}{T^2} \tag{6}
\end{equation}
and, for example, try to establish a relationship in a simpler case, of a constant heat capacity gas. Eq. (1) integrated for a reference temperature $T_0$ automatically leads to
\begin{equation}
  \left[\frac{\partial (G^\pu{ig}/T)}{\partial T}\right]_p = -\frac{H^\pu{ig}(T_0) +
   C_\mathrm{p}^\pu{ig}(T - T_0)}{T^2} \tag{7}
\end{equation}
And no matter what the effort, I can never obtain a simple expression that only depends on the fundamental variables $p$ and $T$
$$ \Delta G^\pu{ig} = G^\pu{ig}(p,T_2) - G^\pu{ig}(p,T_2) = f(p,T_2) - f(p,T_1) $$
And thus, the changes of $\Delta G^\pu{ig}$ for a non-isothermal process in an ideal gas are impossible to calculate. A similar argument can be raised for calculating the changes in the Helmholtz energy $\Delta A^\pu{ig}$ when the temperature changes, and again, one cannot find an equation.
Is this correct? And if it is, why is so?
Edit Doing by definition of $\Delta G$ won't work because one faces an issue in the entropy term, impossible to resolve
\begin{align}
  \Delta G^\pu{ig} &=
  [H^\pu{ig}(T_2) - T_2S^\pu{ig}(p_2,T_2)] -
  [H^\pu{ig}(T_1) - T_1S^\pu{ig}(p_1,T_1)] \\ &=
  [H^\pu{ig}(T_2) - H^\pu{ig}(T_1)] - \color{red}{[T_2S^\pu{ig}(p_2,T_2) -
  T_1S^\pu{ig}(p_1,T_1)]}
\end{align}

It seems $\int_{T_1=0}^{T_2} {\frac{C_p \mathrm{d}T}{T}}$ is easier for real matter, as $\lim_{T_1 \to 0} {(\frac{C_p}{T})} = 0$ for it,  but $\lim_{T_1 \to 0} {(\frac{C_p}{T})} = \infty$ for ideal gas.

@Poutnik Hello Poutnik. In the first case you speak about solids, this is, the Debye model of solids that states $C_V \propto T^3$ and continuous at $T\rightarrow 0^+$? Is this right? And for the second case, how do you see that $T=0$ is making a problem in trying to find $\Delta G$?

YEst, I do mean the Debye model. Integral dT/T from T=0 K is infinite, in attempts to get S as integration for T=0 K for ideal gas, as C_p,ig, does not go to zero.

@Poutnik So, if we don't include vibrational modes of motion, then $C_\mathrm{p}^\pu{ig} \neq f(T)$ and thus $\nexists \lim_{a \rightarrow 0^+} \int_a^{T_2} \frac{C_\mathrm{p}^\pu{ig}}{T} ; \mathrm{d}T$. You smell that this is the source of the problem?

Imagine helium is an ideal gas, then it will still have Cp=3/2.R + R J/K/mol. / $\lim_{T \to 0+}{(\frac {dS}{dT})}=+\infty$ // I do not say it is the problem source, but rather additional problem. My TD is after leaving  university 34 years ago quite dusty.

Bad luck is it is infinity if integrated too.

Could you not use the Sakur-Tetrode eqn. to calculate S in eqn 5? The entropy of the ideal gas is just due to translational motion and this can be found via the partition function and so to the ST eqn.

@porphyrin Sorry for this late reply. I wouldn't like to mix statistical thermodynamics here, my idea is to do it classically. However, still the unknown value $S(T_1,V_1)$ for the Sackur-Tetrode value will appear, if you integrate. Do you coincide?

The (classical) entropy for the ideal gas is $S=C_p\ln(T)-nR\ln(p)+S_0$ at constant $C_p$ and $S_0$ is a constant, can you use this in eqn 5 ?

@porphyrin Although I don't like that indefinite integration we try. At constant pressure $p$ we would get $S(p,T) = C_\mathrm{p}\ln(T) + S_0$. Integrated that would lead to $G(p,T_2) - G(p,T_1) = C_\mathrm{p} [(T_2\ln(T_2) - T_2) - (T_1 \ln(T_1) - T_1)] + S_0 (T_2 - T_1)$. This again leads to an undefined $\Delta G$, because in its form of a change depends on the arbitrarily number you give to $S_0$.

Yes, there will always be some constant, but this is calculable from the ST eqn which is effectively the same as $S=\cdots$ I quoted earlier.

Yes, you are correct.  The value of $\Delta G$ you get is always going to depend on the value of S in the initial state or in a reference state.  But that does not limit the utility of the G function in establishing phase- and chemical equilibrium..

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text:
Oxidizing property of permanganate
If permanganate is a stronger and efficient oxidizing agent in acidic medium than alkaline medium.
Then why does it oxidise iodide(-1) to iodate(+5) in alkaline medium but iodide to iodine(0) in acidic medium ?

Iodate is a stronger oxidizing agent in acidic medium than alkaline medium.

Oh I see. Thanks for the clarification

The iodate ion needs $\ce{H+}$ to give a reducing half-equation with electrons, in an equation like : $$\ce{6 e- + IO3^- + 6 H+ -> I^- + 3 H2O}$$ If no $\ce{H+}$ are available, the iodate ion may react with $\ce{H2O}$ instead, but it will produce $\ce{OH-}$ ions according to the half-equation $$\ce{6 e- + IO3^- + 3 H2O -> I- + 6 OH-}$$ but the ion $\ce{IO3^-}$ cannot react with electrons in basic medium. This basic reaction would imply a collision of three negative species : $\ce{OH-}$, $\ce{IO3^-}$ and electrons. Three negative entities would repel one another : the whole process is hardly possible.

Wouldn't that imply permanganate oxidizes iodide to iodate in acidic medium?

No ! Permanganate oxidizes iodide ions $\ce{I-}$ to iodine $\ce{I2}$. It cannot oxidize the iodine atom to a higher oxidation number. If it would succeed, the following reaction would immediately take place and reduce iodate ions to iodine : $$\ce{2 IO3^- + 10 I^- + 12 H+ -> 6 I2 + 6 H2O}$$

Oh.! Got it. thanks.

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text:
polymerized Dimethylbutadiene or "methyl rubber"
"Methyl rubber", polymerized Dimethylbutadiene, is the first synthetic rubber, but it falls behind polyisoprene having inferior shear strength & oxidizes more easily.
I'm looking into the isomeric composition of these methyl rubber products. Particularly, I'm looking into varieties mostly composed of isomers cis-1,4 & trans-1,4. Typically, ones that have been polymerized with Anionic Chain Polymerization. Have the physical properties of these varieties more of cis-1,4 & trans-1,4 been characterized?


text:
Radioactivity in pegmatite
The site in question has pegmatite which includes uranium-bearing microlite. The age of the pegmatite is 1.3 billion years. I've measured the pegamtite at 4.2 mR/hr (see photo)
The chart below show the decay series for U-238. The first number is 4.468 * 10^9 years = 4,468,000,000.
What can be said about the composition of radioactive element(s) and the nature/strength of the radioactivity itself? (Edit: Can I assume it started with U-238? How much U-238 remains? What other of the decayed elements are there? What is the exposure danger to a human?)
Nuclide Half-Life   Radiation *
U-238   4.468*10^9 years    alpha
Th-234  24.1 days   beta
Pa-234m 1.17 minutes    beta
U-234   244,500 years   alpha
Th-230  77,000 years    alpha
Ra-226  1,600 years alpha
Rn-222  3.8235 days alpha
Po-218  3.05 minutes    alpha
Pb-214  26.8 minutes    beta
Bi-214  19.9 minutes    beta
Po-214  63.7 microseconds   alpha
Pb-210  22.26 years beta
Bi-210  5.013 days  beta
Po-210  138.378 days    alpha
Pb-206  stable  -

What do you want to know ? Which mass of pegmatite have you measured ? How far away is the counter from the pegmatite ? ? Does it detect alpha, beta and gamma rays ? What is the purity of the pegmatite ?

The instrument measures this reading from directly against the rock up to a couple of inches away. The measurement is of gamma rays. I'm not sure how to answer your questions about mass and purity. Added to question based on your comment.

"What can be said about the composition of radioactive element(s) and the nature/strength of the radioactivity itself?" Nothing.

Uranium from some areas has been depleted by formation of a natural reactor.
There are other radioactive elements in pegmatite that are not daughters of uranium decay, e.g. 232Th. Furthermore, the ratio of uranium to these other elements varies with with locale.

You'd need further tests, whether chemical or physical (e.g., measuring the types and energies of radiation), to get an idea of the composition of the rock.


text:
What is the molecular orbital diagram of hydroxide anion?
Context
I am a teacher, and, during a class, my student asked me to draw the molecular orbital (MO) diagram of the hydroxide anion ($\ce{HO-}$). First, I made the diagram on the left, and then, when my student pointed out that the bond order is two, I corrected it by exciting some of the electrons to the antibonding orbitals, making the diagram on the right.
Note: This is a genuine attempt at explaining MO diagrams to a high school student. Please help me explain the concept right.

I attempted to draw the molecular orbital (MO) diagram of the hydroxide anion.
I have shown the one with electron filling according to Aufbau principle and the one with a simple fix I applied to correct the bond order.
The bonding ($\sigma_1$ and $\sigma_2$) and anti-bonding ($\sigma^\ast_1$ and $\sigma^\ast_2$) orbitals have been highlighted:

Now, according to the Wikipedia article on Aufbau principle, in the ground state, electrons fill subshells of the lowest available energy first, and subsequently the subshells of higher energy are filled, which is what I have done in the left diagram, $\mathrm{BO=2}$, and I have shown an excited state, with $\mathrm{BO=1}$, on the right.

Comments

I am fully aware, as remarked by the student and in the comments, that $\ce{H}$ is not known to form a double bond.
There might be somethings which are wrong with the MO diagrams.
It has been pointed out that the number of MOs is more than the number of atomic orbitals (AOs), which is supposedly incorrect.
According to the comments, one of the bonding orbitals I have drawn is non-bonding, though I do not yet know of this.
Some comments have suggested reading a few textbooks, which I will, but it will take some time, and I will edit the question in due time, please be patient.

Question
Several comments have pointed out that this MO diagram is incorrect. What is wrong with this MO diagram and what is the correct MO diagram?

You better start your editing already - with such title you're lucky I can't closehammer it.

@Mithoron any suggestions?

Ditch your alternate diagram and ask what's wrong with the "double bonding" pic. - one "bonding pair" is actually non-bonding. (I don't feel like this question can be good even then, though).

@Mithoron which bonding pair is non-bonding?

Your initial case is already wrong. You made 6 MOs out of 5 AOs. I'd suggest reading a textbook to check the MO diagram for hydroxide ion (or HF; it's the same thing, and you're more likely to find that in a book).

@Mithoron there is nothing wrong with that, a single AO can interact with multiple AOs while forming MOs.

@orthocresol yes, $\ce{HF}$ is isoelectronic to $\ce{HO^-}$. Would you recommend a textbook?

Of course, there are a few here: https://chemistry.stackexchange.com/a/37305/16683 I think Housecroft, or Miessler, are the better ones for this. I presume both your comments were directed at me. Yes, more than 2 AOs can interact; but that doesn't mean you can generate extra MOs out of nowhere. If you have 3 AOs with the same symmetry, they form 3 MOs, not 4. You've made 4 MOs out of the oxygen 2s, oxygen 2pz, and hydrogen 1s.

@orthocresol I will read the relevant sections from the recommended textbooks and edit this question asap.

@ToddMinehardt I teach remotely, online.

@Mithoron does the diagram in the answer address all the concerns?

The following diagram is from Kinetics of the Reactions of Water, Hydroxide Ion and Sulfide Species with CO2, OCS and CS2: Frontier Molecular Orbital Considerations)$^1$ George W., Luther III (2004).

This diagram addresses the concerns raised in the comments and the question:

Number of atomic orbitals is equal to number of molecular orbitals.
One of the orbitals, shown wrongly as bonding, is actually non-bonding.
Bond order of $\ce{O}\text{-}\ce{H}$ bond in $\ce{HO-}$ is 1.

Note: The $\mathrm{1s}$ orbital of $\ce{O}$ is not shown since it is low in energy and essentially does not interact.
References

George W., Luther III. (2007). Kinetics of the Reactions of Water, Hydroxide Ion and Sulfide Species with CO2, OCS and CS2: Frontier Molecular Orbital Considerations. Aquatic Geochemsitry, 10, 81-97. 10.1023/B:AQUA.0000038957.18584.b0

Honestly, even this one is kind of dubious. The σ MOs are correct now (we could argue about the best labels to give them, although that's not so important). More crucially, the π MOs shouldn't be raised in energy relative to 2px/2py when there is no overlap going on.

I would say that you can shift the π MOs down and it would be correct. But I also don't really think you should have gone to a geochemistry journal. No offence, but I don't generally think that would be the best place if you were looking for something theoretical, like MO analysis. You could google for 'HF MO diagram', find any of the top hits, replace F with O- and call it a day — and it would be accurate.


text:
Can't the wavelength of an electron increase or decrease to "fit" the wave perfectly in any radius?
I was reading about how de Broglie's electron-wave theory explained the quantisation of energy in a Bohr atom. What I understand is as follows:

If the wave is to be arranged in the form of a circle so that it
attaches to itself, the waves can only occur if there is a whole
number of waves in the circle.

The standing waves on the left-hand side of the figure are allowed to form in the brown box,
because they fit perfectly inside the box. In contrast, the waves on
the right-hand side of the figure cannot form in the brown box because
they do not fit perfectly inside the box.

The left-hand standing wave fits in the electron cloud while the right-hand one does not.

(Source)
Bohr said that electrons could exist at specific "allowed" energy levels, but that they couldn't exist between those energy levels. And, apparently, these whole number wavelengths are supposed to explain that.
But, can't the wavelength of an electron increase or decrease to "fit" the wave perfectly in any radius? Or, are there some other conditions/constraints involved?

There corresponds a unique energy to every $\ce n^{th}$ shell. And unique energy means unique frequency which in turn means fixed wavelength.

You can search for other related posts. Look for "Bohr's postulates". See for instance https://chemistry.stackexchange.com/questions/121488/bohrs-postulate-of-quantisation-of-angular-momentum

A complete derivation: https://alpha.chem.umb.edu/chemistry/ch115/carter/files/103more/BohrEquations.pdf

If anyone saw the source, it has 7 sub-sections of the Schörodinger's equation. It doesn't present a single equation, because it is for higschool. Hard to grasp anything of this without math... I'd rather state that Bohr said that the angular momentum had "allowed" values, and that energy also does is a consequence of it (although he wanted the latter to explain the spectral series...)

They do adjust to 'fit' the box/ring as you put it, those on the right are the wavefunctions for a bigger box than shown in the picture. You can have any size box/ring you want but the restriction is that the wavefunctions have to fit exactly. (This is called a boundary condition). If this were not the case, as in the pic at the bottom right, there will be a place where the wavefunction can have two values at the same position which is not physically possible.  As the size of the box/ring changes so do the energy levels, you should have seen the equations that demonstrate this.


text:
Highest possible temperature
I know the lowest possible temperature in nature is 0 K although we have been able to only getting ever closer and closer to it but not at 0 K.
But on the opposite spectrum of temperature, what is the highest that is achievable naturally or artificially? Is chemistry still relevant at this temperature as we know it or is it senseless to think of any typical reaction to take place.
My best guess is plasma followed by nuclear fusion as it creates heavier elements but I am not certain.

This isnt really a chemistry question because it cannot be answered by the rules of chemistry,try asking it at Physics SE.

Totally irrelevant to chemistry, but interesting. It might be better to ask what is the highest T relevant to chemistry, but that is likely vague or opinion based and depends on definition of T, what you consider relevant to chemistry, and bond strengths.

Well bonds are broken much much lower than the highest possible temperature.After the first 10^-20 seconds in the early universe even when the electroweak symmetry was broken and the Higgs field gave mass to any particles around , the universe was just a soup of particles.The temperature was so high a bound state could not be formed.

@BuckThorn Thanks for appreciation. But you are right. I am afraid even with your suggested edit, it might not become a well received question.

@BuckThorn https://chemistry.stackexchange.com/questions/173880/total-bond-dissociation-temperature

You can infinite as well as negative temperatures, see this answer https://chemistry.stackexchange.com/questions/173760/schottky-anomaly/173764#173764

One can look for highest temperature of the universe in a singularity of a black hole where the energy density is infinite according to General Relativity but super high and finite according to the correct theory of quantum gravity.
At that point because the energy density  is so high electromagnetism is not anymore a fundamental force but is unified with the weak force to form the electroweak force.
Because the symmetry of the electroweak force is not broken,the Higgs field no longer gives mass to the particles and chemistry doesnt exist there at that point.

Except we don't know the "correct" theory of quantum gravity. What we have suggests there isn't any singularity at all.


text:
Why the common name of 1,2-dichloro-ethane is ethylene dichloride if it has no double bonds in it?
Why the common name of 1,2-dichloro-ethane is ethylene dichloride if it has no double bonds in it? The -ene in the common name suggests that it should have a double bond. I believe this is due to the fact that ethylene dichloride is a derivative of ethylene and synthesised from it but I was told that the presence of -ene is due to the two chlorine atoms? I don't think this is correct. What's the actual answer?

Related: 1, 2, 3

According to Wikipedia Article on 1,2-Dichloroethane, it is known by many names:

Ethylene dichloride
Ethylene chloride
1,2-DCE
DCE
Ethane dichloride
Dutch liquid
Dutch oil
Freon 150

The, misleading, name of ethylene dichloride originates from its method of production; ethylene is a common name for ethene, which upon reaction with dichlorine (a more accurate name for $\ce{Cl2}$, commonly known as chlorine) produces ethylene dichloride.

There aren't many sources to track down exactly when ethylene dichloride became popular. Another possible explanation is that it is derived from the name methylene (IUPAC name: methylidene, also called carbene or methene). Consider the following hypothetical reaction.

Unfortunately, accurate nomenclature was not the first priority of 18$^\text{th}$ and 19$^\text{th}$ century chemists and industrialists; moreover, according to Wikipedia article on IUPAC, it was only established in the early 20$^\text{th}$ century, and it took time to standardize nomenclature.
Google Trends shows that the name, ethylene dichloride is, more or less, consistently employed, especially in India, United Arab Emirates, and some parts of the USA:

However, Dutch oil is more common worldwide:

Ethylene is

the name for the bidental functional group $\ce{\bond{-}CH2-CH2\bond{-}}$

Similarly,  $\ce{\bond{-}CH2\bond{-}}$ is methylene and $\ce{CH2Cl2}$ can be called also as methylene dichloride, aside of IUPAC preferred name dichloromethane.
Therefore, $\ce{Cl-CH2-CH2-Cl}$ can be called also as ethylene dichloride

the older name for ethene $\ce{CH2=CH2}$.

Distinguishing is as usually context dependent.


text:
What is the osmosis equilibrium point of water beads (for plants)?
In theory, water beads for plants take in water when immersed in it to gradually release it to the roots of the plants. This process happens through osmosis. When the relative humidity is high enough, water beads take in water. They release the water in contact with the roots, as the roots have a stronger “osmotic gradient.”
I understand there is no regulated standard for water beads, but I guess their chemical composition is fairly regular, e.g. when compared with plant roots, which may range from desertic to aquatic. The point for the equilibrium will be higher than for the roots, which take the water from the beads, and lower than 100%, as they absorb water when immersed in it.
A nice and unlikely possibility would be the osmosis equilibrium point of the water beads being well below 60%, e.g., 40%, and the roots for any plant (e.g., succulents) lower than that, e.g., 20%. Then plants on beads would remove humidity from the air when above 40%, making for a nice dehumidifier. Many articles on the Internet claim that plants help dehumidify houses, but they are more likely to humidify them, as they need watering.
Is (a range for) the relative humidity of the osmotic equilibrium point something known or easy to find out?
The answer won't be a number, it will be a range. We find confusion and misinformation on the Internet in this area.

As the beads swell the activity of the water within changes and there is for each value of the activity a corresponding equilibrium external water vapour pressure. There isn't therefore a single equilibrium point. Rather there is an equilibrium curve of % water content in the beads versus vapour pressure.

How would that curve look like? My guess: it's a power law, e.g. the first 10% drop in vapor pressure causes a 90% drop in water content. Is this something known or easy to know?

Let me know if further clarification is needed.

Your intuition about how the beads should behave relative to the activity in the soil looks reasonable, but the question at present is not clear. You are asking how to determine "the relative humidity of the osmotic equilibrium point". You might want to edit that. Generating a limited curve (a few points)  of RH versus water content seems something a do-it-yourselfer can do with the right supplies. The curve is also known as a "moisture sorption isotherm". The change in activity typically accelerates with increasing water content so might follow a power-law depending on the system.


text:
Ortho- Para- Hydrogen Names and Benzene
I read that the Hydrogen atoms with nuclear spin in same direction are called Ortho- Hydrogen whereas those in opposite nuclear spin are Para- Hydrogen. But is there any resemblance of such a nomenclature with Ortho- Meta- Para- of Benzene, cause for benzene, these denote the relative position of substituent. Or is it just named so?

Coincidental naming.

There are ortho- and paracompact spaces in math (and meta-, too). No relation. We people are lazy bunch, we use the same word over and over again until it looks like a dirty rag, and then some.

And there is also ortho and meta phosphoric acid as yet different usage, describing different hydration levels of element oxoacids.

For "ortho-"; for "para-"; for "meta-".

There is no connection of ortho- para- organic nomenclature with the nuclear spin isomerism of hydrogen. However, that raises an interesting point, that which nuclear state should be called the ortho- vs. para- or what was the thought process among the early German discoverers?
[Subscription only access: Farkas, L. (1933). Über Para- und Orthowasserstoff. In: Ergebnisse der Exakten Naturwissenschaften, vol 12. Springer, Berlin, Heidelberg. https://doi.org/10.1007/BFb0111841]
"Die antiparallele Kernspinorientierung entspricht einem, "Kernspinsingulett" Zustand mit dem statistischen Gewicht I, die parallele Kernspinorientierung dem , Kerntriplett"-Zustand mit dem statistischen Gewicht 3."
Then a footnote reads on later pages:
"...Im allgemeinen werden immer die Zustände mit dem größeren statistischen Gewicht als Orthozustände bezeichnet."
which means that the antiparallel nuclear spin orientation corresponds to a nuclear singlet state with a statistical weight of 1, the parallel nuclear spin orientation corresponds to the nuclear triplet state with a statistical weight of 3. In general, the states with larger statistical weight are designated as ortho- states in nuclear spin isomers.
Perhaps this ortho- nomenclature is due to the other meaning of ortho which means,  ‘right, correct, proper’, so the majority (=higher statistical weight) of the hydrogen's nuclear isomer is in the correct form/proper form, hence this is ortho- nuclear spin isomer of hydrogen.

o-together, p-opposite there might be a connection.

@jimchmst, Are you suggesting that those hardcore physicists were inspired by organic nomenclature, that is most likely not the case :-) So far, I could not find any.


text:
Do Benzyl Halides undergo SN1 or SN2 with Ammonia?

Does this reaction show SN1 or SN2 mechanism? Our teacher told that it shows SN2. But wouldn't the stability of benzylic carbocation mean an inclination towards SN1?

It also increases the rate of SN2 substitution.

Why/How would it do so? @Zhe

Consult a textbook for more details, but neighboring group effects are known to affect SN2 rates, presumably due to the adjacent orbitals contributing to a more stabilized transition state.

Answer to Your Question$^\text{1}$
$\text{S}_\text{N}\text{1}$ and $\text{S}_\text{N}\text{2}$ reactivities between benzyl chloride and ammonia are solvent and substituent dependent.

Effects of Solvent
In liquid ammonia, there is strong evidence that it will follow an $\text{S}_\text{N}\text{2}$ mechanism, while in aqueous/binary medium, it is likely to show mixed $\text{S}_\text{N}\text{1}$ and $\text{S}_\text{N}\text{2}$ mechanism.

The rate of solvolysis in liquid ammonia is retarded dramatically when one of methylene hydrogen of benzyl chloride in the α-position is replaced by a methyl group, α-methyl benzyl chloride is solvolyzed about 130 times slower than benzyl chloride. This is strong evidence for solvolysis occurring by an $\text{S}_\text{N}\text{2}$ mechanism due to a more sterically hindered transition state.

Effects of Substituent
In liquid ammonia, substituents on the benzene ring only feebly affect the kinetics and reaction mechanism while strongly affecting both in case of aqueous/binary medium.

There is no clear consensus on the mechanism of hydrolysis of benzyl halides in water and aqueous binary solvents, although that for 4-methoxybenzyl halides is described as an $\text{S}_\text{N}\text{1}$ mechanism with the intermediate formation of a carbocation, and that for the 4-nitro derivative is claimed to be $\text{S}_\text{N}\text{2}$.

This reasons that reaction between unsubstituted benzyl chloride and ammonia (in aqueous/binary medium) might follow a mixed $\text{S}_\text{N}\text{1}$ and $\text{S}_\text{N}\text{2}$ reaction mechanism.
Applications of Benzylamine
Reaction between benzyl chloride and ammonia is of paramount importance, being one the major synthesis routes to the production of benzylamine, which is used within the coatings industry, for example, as a building block for additives and solvents, and it is a valuable intermediate for various applications and a building block for chemical synthesis like for the production of Active Pharma Ingredients (APIs) and crop protection agents.
I am surprised that a reaction, with industrial applications, has not been studied much. It would definitely be an interesting topic of research.
Reference

Pi, J. (2011), Kinetics and Mechanisms of Organic Reactions in Liquid Ammonia. Doctoral thesis, University of Huddersfield. Available as PDF.


text:
How does sodium react with methanol?
Strong acids of $\ce{AH}$ type readily dissociate to its constituent ions (considering the solvent is water): $\ce{A-}$ and $\ce{H+}$. But the mechanism of the reaction between sodium metal and methanol $(\ce{Na + MeOH})$ as taught by my teacher appears to contradict with this statement:
$$
\begin{align}
\ce{Na &-> Na+ + e-} \tag{R1} \\
\ce{MeOH &-> MeO- + H+} \tag{R2} \\
\hline
\ce{Na + MeOH &-> Na+MeO- + H^.} \tag{R3}
\end{align}
$$
$$\ce{2 H^. -> H2} \tag{R4}$$
$\mathrm{p}K_\mathrm{a}(\ce{MeOH}) = 15.5$ according to Wikipedia — Methanol and reference therein [1], which is an indication that methanol is a weak acid. But then how the hydrogen is evolved?
Secondly, how does sodium gives its electron freely in the water? I can't believe free electron could exist in solution. Does electron solvation take place?
Reference

Ballinger, P.; Long, F. A. Acid Ionization Constants of Alcohols. II. Acidities of Some Substituted Methanols and Related Compounds. J. Am. Chem. Soc. 1960, 82 (4), 795–798. DOI: 10.1021/ja01489a008.

Same reason it reacts with water, even if difference in speed is huge.

Ultrafast submilisecond camera can catch the blue colour of hydrated electrons in water. Solution of sodium in liquid ammonia keeps the dark color of solvated electrons relatively long, as ammonia is much weaker acid then water. Methanol is  much closer to water than to ammonia, being somewhat weaker acid than water.

related https://chemistry.stackexchange.com/questions/41669/what-is-the-reason-for-the-blue-color-of-an-ammoniated-electron

@Mithoron If I already knew that reason why would I ask here? And what do you mean by "It"?

@Poutnik but how would that explain that reaction takes place since methanol is still weak acid.

Water, methanol and liquid ammonia are all weak acids (ammonia many orders weaker than both others) and sodium reacts with weak acids. More exactly, solvated electrons react with weak acids.

It's not an acid base, but redox reaction.

@Poutnik could you please elaborate mechanism?

@Mithoron Yes But I never said it is acid base. I meant to say that the hydrogen dissociated from MeOH is in very low amount then how does reaction takes place since Na is in higher amount.

$\ce{e- + H^{(+)}-A^{(-)} <=> H^{(-)}-A^{(-)} -> H^{.} + A-}$

Caution
Sodium ($\ce{Na}$) reacts explosively with water ($\ce{H2O}$); I would recommend carrying out the experiment in absence water under inert gas atmosphere (to avoid moisture). Moreover, even the reaction of $\ce{Na}$ with methanol ($\ce{MeOH}$) is highly exothermic, and ignition is likely. Remember, alcohols and evolved dihydrogen ($\ce{H2}$) form good fuels. Thus, the reaction would have to be carried out at low temperatures or using very small amount of $\ce{Na}$ in inert atmosphere.
The Reaction
$\ce{MeOH}$ is acidic enough to react (exothermically) with Direct $\ce{Na}$. Reaction of $\ce{Na}$ and $\ce{MeOH}$, and subsequent vaporization of excess $\ce{MeOH}$ is a common route to the preparation of pure sodium methoxide $\ce{MeO-Na+}$ crystals.
$$
\begin{multline}
\ce{2Na(s) + 2ROH_\text{excess}(l) -> 2RO- Na+ (ROH) + H2 ^}\\
\ce{->[vacuum distillation][] 2RO- Na+ + ROH(l)}
\end{multline}
$$
In the article Synthesis and Characterization of Sodium Alkoxides$^\text{1}$ by Chandran K., Nithya K., Sankaran, K., and Gopalan K. (2006), the authors used $\pu{100 mL}$ of alcohol ($\ce{ROH}$) and reacted it with $\pu{500 mg}$~$\pu{1000 mg}$ of $\ce{Na}$ to produce corresponding sodium alkoxides ($\ce{RO- Na+}$). Upon subsequent vacuum distillation, pure $\ce{RO- Na+ (s)}$ could be isolated.
Solvation of Electrons
There is no possibility of solvation of (highly basic) electrons ($\ce{e-}$) in acidic $\ce{ROH}$ (except at very high pressures, order of 10-100 $\pu{Mbar}$, as pointed out in the comments).
$$
\ce{2ROH + 2e- -> 2RO- + H2}
$$
This would be an immediate reaction. However, solvation of electrons is possible; for example, in when $\ce{Na}$ is dissolved in (much less acidic) ammonia ($\ce{NH3}$), the characteristic blue color is obtained due to solvated electrons (video demonstration).
Reference

Chandran K., Nithya K., Sankaran, K., and Gopalan K. (2006). Synthesis and characterization of sodium alkoxides. Bull Mater Sci 29, 173–179. 10.1007/BF02704612

"People saying it is not possible should not stand in path of people doing it." (Chinese proverb). :-) // If it is possible to catch hydrated electrons in water, it would be surely possible doing so for methanol.  Search for "hydrated electrons", namely work of prof  Pavel Jungwirth.

Closely related Nature - Spectroscopic evidence for a gold-coloured metallic water solution

One of multiple his papers: ACS - Unraveling the Complex Nature of the Hydrated Electron

This isn't acid base reaction. You said MeOH is acidic enough to react but that doesn't explain mechanism. According to me it involves Single electron transfer mechanism.

Also you're too cautious in first section - reacting Na with methanol is a routine way to get rid of it. Fume hood is more than enough. Another issue is I don't get why you call it vacuum distillation, if it's the precipitate that gets purified.

@Mithoron this asnwer pretty much talk about it the same way, any thoughts on why I am receiving downvotes. I even provided a reference for the amounts of $\ce{Na}$ and $\ce{MeOH}$ to be used for a safe reaction and a procedure to obtain purified crystals. If someone pours MeOH and a big sodium sample, it might catch fire. Cautionary statements are important in practical chemistry.


text:
Crystal structure of tin tetraiodide
I was looking at the molecule of $\ce{SnI4}$ and how its structure looks like in its crystal lattice:

Wikipedia.de says its crystal lattice is of the fcc (ccp) type (for the iodine atoms), which means that there are eight tetrahedral sites, 1/8 of which are occupied by the tin. I understand the calculation behind it and how it leads to the formula of $\ce{SnI4}$, but I don't quite understand how one can deduce it is of the fcc type just by looking at its crystal lattice, as I'm quite bad at visualizing 3-dimensional structures. How can I visualize it properly?

https://pubs.acs.org/doi/pdf/10.1021/ja01657a014 has the atom positions listed. Since the iodine atoms are not all in equivalent positions I find it hard to believe Wikipedia's assertion.

You can't and arent't supposed to deduce the crystal system just by looking at the picture. Also, it is not really fcc.

Wiki says it’s cubically close packed.. is that the same as fcc in this case though?

How can I visualize it properly?

Here is a visualization showing the iodine atoms as large spheres, and the tin atoms as tiny transparent spheres. The colors for iodine are chosen by layers along the c-axis facing the viewers, with the top and bottom layers in black, and the other three layers in yellow, purple and green. In one view, you can see that iodine atoms are roughly arranged on a grid with four grid lengths fitting into the unit cell.

Coordinates from https://legacy.materialsproject.org/materials/mp-23182/#
When you tilt the unit cell and look at the center of the unit cell, you can see that the yellow, purple and green layers show an approximate FCC pattern with a "unit cell" that has half the length of the real unit cell. The symmetry is broken by the tin atoms, explaining why the iodine atoms are not exactly on the smaller lattice and why we can't choose a smaller unit cell (because the tin atoms would be in different positions of the 8 smaller unit cells).

Now this summarizes nicely why it is not a true fcc, and also why calling it so is not entirely untrue either.

Or, maybe it is fcc. My rendering identifies $D_{3d}$ symmetry, but that can be embedded into a cubic structure (your original reference identifies the crystal system as cubic rather than rhombohedral). The five green atoms, five yellow atoms and four purple atoms centered inside the box boundaries seem to form a face-centered cube.

@OscarLanzi They do form a face-centered cube, but only approximately. On the other hand, the cubic symmetry of the actual unit cell is exact, as it should be.

Whether the structure is fcc is disputed in the comments, but the picture shown gives a hint of at least rhombohedral ($D_{3d}$) symmetry. The two molecules at the corners of the parallelopiped are arranged in a way that shows improper sixfold rotational symmetry (they are invariant under a 120° rotation and also an inversion through the center), and the count of six remaining molecules also conforms with this symmetry.

Well wiki says that’s it’s cubically closed packed.. I always understood it as a synonym to fcc. Is that wrong?

I wouldn't even call this an error. A metaphor, maybe.

@Mäßige I updated the German Wikipedia article to say "Dies entspricht ungefähr einer kubisch-dichtesten Kugelpackung von Iod-Atomen, in der 1/8 aller tetraedrischen Lücken mit Zinn-Atomen besetzt sind."

Okay thank you. However I am unsure if fcc is really the same as cubic close packed because here it is said that fcc is only =cubic close packed if it is regarding metals; non-metals (for example diamond) form fcc, but are not cubic close packed. Maybe you can spot why but I also have trouble understanding it: https://en.wikipedia.org/wiki/Cubic_crystal_system

@Mäßige Quoted from Diamond article in Wikipedia: "The diamond lattice can be viewed as a pair of intersecting face-centered cubic lattices, with each separated by 0.25 of the width of the unit cell in each dimension." The space group is #227, whereas the space group for closest packing is #224. The coordination number for diamond is 4, while it is 12 for fcc, so they must be different.

Okay I see :] thank u


text:
Zinc organic-water solvent extraction (Hydrometallurgy)
Consider a biphasical system consisting of an aqueous phase and an organic phase.
Consider the following equation:

$\ce{Zn^2+(aq) + 2HR(org) <=> ZnR2(org) + 2H+(aq)}$

Let [ $Zn^{+2}$ ]$^{0}$$(aq)$ = 0,1 $\pu{mol L-1}$ be the initial concentration of ionic zinc in the aqueous phase.
Let $\ce{[HR]}$$^{0}$$(org)$ = 0,5 $\pu{mol L-1}$ be the initial concentration of phosphinic acid in the organic phase.
Consider that the volume of the organic phase $\ce{V2}$ is twice the volume of the aqueous phase $\ce{V1}$ ($\ce{V2} = \ce{2V1}$ )
Consider that the equilibrium constant of the equation above is $K = 10^{-3,259}$
Let $D = \frac{[Zn^{+2}]}{\ce{[ZnR2]}}$ be the distribution coefficient of the equilibrium above and $E = \frac{100D\ce{V2}}{D\ce{V2}+\ce{V1}}$ be the extraction coeficcient. (note that since $\ce{V2} = \ce{2V1}$ , $E = \frac{100D}{D+0.5}$ )

Given the information above,plot the graphic of E versus the ph of the aqueous phase.
Suggestion: Use $\ce{V1} = 1L$ and $\ce{V2} = 2L$

My attempt at it:
Assuming $\ce{V1} = 1L$ $\ce{V2} = 2L$ and assuming that $\ce{n0,n1}$ are the initial mols of ionic zinc and phosphinic acid respectively,we obtain that $\ce{n0} = 0.1 mol$ and $\ce{n1} = 1 mol$.Also assume that $x$ is the quantity (in mols) of $\ce{Zn^2+(aq)}$ that reacted.Noting that we are analysing the equilibrium for each $\ce{[H+]}$ ,after we draw the reaction table we obtain:
$K = \frac{2x\ce{[H+]^{2}}}{(0.1-x)(1-2x)^{2}}$ and $D =\frac{0.5x}{0.1-x}$ .If we consider $2x <<<< 1$ we obtain that $K \approx \frac{2x\ce{[H+]^{2}}}{(0.1-x)(1)^{2}}$ $=>$ $K \approx 4D\ce{[H+]^{2}}$ and with that we obtain a relation between D and the ph of the aqueous phase.After that I was able to plot the graphic of E versus the ph.
Did I answer it correctly?Any help/correction would be kindly appreciated.Thank you in advance.

Ph is probably pH. See also Upright vs italic
(element and unit symbol upright)  and $n = \pu{1 mol}$ as $n = \pu{1 mol}$ // See CH SE META post about policy for "Am I right?" questions


text:
Formation of tetrafluoroborate using VBT and hybridisation
How can the formation of $\ce{BF_4^-}$(Boron is $\ce{sp^3}$ hybridized) be explained using VBT and Hybridization? So far, I understood that one electron from s orbital gets excited and jumps into a p orbital, and then the orbitals undergo $\ce{sp^3}$ hybridization leaving us with four $\ce{sp^3}$  hybrid orbitals (three $\ce{sp^3}$ orbitals with one electron each and one $\ce{sp^3}$ hybridized orbital with no electron). What happens next?

Perhaps my answer deals mostly with hybridisation and less with VBT. Your question seems to be arising when you find that only 3 of them contain a lone electron each and the fourth is still empty despite the hybridisation being $\ce{sp^3}$ which means total four hybrid orbitals.
Well, see that you asked about fluoroborate ($\ce{BF_4^-}$) in which there is an excess charge on boron. What happens first is, as you pointed out, that an electron from s orbital jumps to a p orbital. There is still one empty p orbital which acquires an electron and thus a negative charge on B. Now what you have is four singly occupied $\ce{sp^3}$ orbitals of B that can each overlap with a lone electron in the p orbital of four F atoms to give $\ce{BF_4^-}$.
Note that the hybridisation to $\ce{sp^3}$ doesn't happen unless the remaining p orbital of boron gets at least one electron.

I didn't add this image to my answer as it is out of context of your question but could still be useful.

That was really well explained. Thank you.

No, please, no. This produces a completely incorrect image of what is happening. This only furthers the most common misconception. Hybridisation is only a mathematical description, it's not a cause and there certainly are no electrons jumping around. I have to down vote this.

@Martin-マーチン Since OP asked about hybridisation, I explained how it is sp³. Please let me know where I had been wrong. I would like to prevent such possible mishaps in future.

@Martin - マーチン please elaborate.

Wikipedia states that $\ce{BF_4^-}$ is isoelectronic to $\ce{NF_4^-}$ and $\ce{CF_4}$ and the latter itself is similar to $\ce{CH_4}$ and bonding in $\ce{CH_4}$ is explained similarly. This is why I did so.

It is important to understand that there are electron clouds and nuclei. They arrange in some sort of geometry. This is the basis. The elections don't know where they came from and they don't know where they are going. Hybridisation is a mathematical description of one way of forming a wavefunction. That is the basis of Valence bond theory. There is simply not enough space in the comments or in an answer to explain that. It needs a textbook and probably a good one at that

Just for good measure: Valence Bond Theory cannot explain the formation of a molecule (or ion). It can only describe the bonding within under the constraint that this geometry exists.

Probably hybridisation is taken often for granted in everyday life. I see your point now.


text:
Difference between ketone rings in wavelength using an IR device
It is said that the smaller the ring size, the greater the wavelength of the carbonyl group, but why does cyclobutanone have a higher wavelength than cyclopropanone?

Why does cyclobutanone have a higher wavelength than cyclopropanone?

Seemingly, you were informed wrong. According to MSU website, following wave numbers listed for different cyclic ketones:

The increase in frequency ranges from $30$ to $\pu{45 cm-1}$ for a 5-membered ring, to $50$ to $\pu{60 cm-1}$ for a 4-membered ring, and nearly $\pu{130 cm-1}$ for a 3-membered ring. According to this, the increase in stretching frequency of exo-carbonyl group is because small rings have more $\mathrm{p}$-character in their ring bonds, which means the bonds outside the ring (exo-bonds) have more $\mathrm{s}$-character and are stronger.

Edit: There is a relation between stretching frequency and corresponding wavelengths:
$$
\begin{array}{|c|c|}
\hline
\text{Stretching frequency, }\pu{cm-1} & \text{corresponding wavelength, }\pu{nm} \\
\hline
1716 & 5820 \\
1748 & 5720 \\
1783 & 5600 \\
1850 & 5400 \\
\hline
\end{array}
$$
Also, you can find this convertor here.

The questioner seems unsure about wavenumber[frequency] and wavelength


text:
Positive and negative voltage readings of chemical cells
In the experiment of chemical cells, I connected a magnesium electrode , a copper electrode and a multimeter in a complete circuit.
When magnesium electrode is connected to the negative terminal and the copper electrode is connected to positive terminal, the reading of the multimeter is positive.
However, when copper is connected to the negative electrode and magnesium is connected to the positive electrode, the reading is negative instead.
I wonder what “positive” and “negative” reading means. Is it just defined by scientists that a positive voltage reading mean a flow of electrons from negative electrode to positive electrode, or can it be explained with scientific knowledge?

One fundamental concept which is not taught clearly in electrochemistry is that the electrostatic sign of the electrode is invariant. Imagine you had a charge "sniffer" or charge sensor device, and if you were to touch a particular electrode in a electrochemical cell, you will always see the same electrostatic sign of the single electrodes, either positive or negative with respect to earth (UK, South/Asia) or ground (US). Imagine the only two possible scenarios- the electrode has excess of electrons then it has a negative electrostatic sign. If the electrode has a deficiency of electrons, it has a positive electrostatic sign. It will not matter, what you do. This is Nature's set-up.
Now when you are measuring the voltage using a multimeter, this time you are measuring potential differences,  and these differences are directional not in the sense of a conventional vectors/scalars, but the way meter is supposed to read the voltages. Imagine the old style galvanometers, where one could actually see the coils. The coils were wound in a certain direction and energized as such so that if the plus sign was connected to the plus sign of the direct current source the needle would deflect in a positive direction. If you swapped the connections, the needle moved in the opposite direction. This is what your multimeter is showing, but alas one cannot see anything in these modern black boxes. In one set-up, you see negative potential difference and in the other set up you see positive potential differences. The magnitude must be the same.

Hello AChem. I have always found very troublesome the concept of an "excess" or a "deficiency" of electrons in an electrochemical cell. The moment there is a reduction in the positive electrode, it will travel to the negative electrode and a compound will get oxidized. I think this accumulation or the opposite doesn't exist.

Do you mind if we discuss this thing in chat? It would be nice if I could clarify some points within this seemingly simple thing, and see if you think the same as me.

@MetalStorm, No the electrodes are indeed electrostatically charged (you really do not need many electrons) as long as the cell is alive. I do not know how to initiate a chat. Alternatively, you can compose a question.

In fact, this has very little/nothing to do with conventions of chemistry.
The same would happen if you measured potential differences between two points of electronic circuits, powered by a DC source plugged in the wall AC socket.
If you attached the positive voltmeter pin to more positive circuit potential and the negative/ground pin to more negative potential, you would read positive DC voltage. And vice versa.
For more, see the  term electric potential on Wikipedia.

... is defined as the amount of work energy needed to move a unit of electric charge from a reference point to the specific point in an electric field. More precisely, it is the energy per unit charge for a test charge that is so small that the disturbance of the field under consideration is negligible.

In early years, before an electron was discovered, there was conventionally decided what is positive, what is negative and that a  current flows from conventional positive to negative potential. When an electron had been discovered, it was realized it had negative charge and its flow had the opposite direction than the conventional current. But the convention is kept since then.


text:
How much can we extend the Finkelstein reaction?
Finkelstein reaction explicitly entails the conversion of an alkyl chloride or alkyl bromide to alkyl iodide by treatment with sodium iodide in acetone.
$$\ce{R−X + NaI→[acetone] R−I + NaX↓ }\\
\ce{(X$=$Cl, Br; R$=$alkyl group)}$$
What I think I understand
Acetone is a polar aprotic solvent so $\ce{NaBr and NaCl}$ formed, being mostly ionic, are insoluble while $\ce{NaI}$ have covalent character (in accordance with Fazans's rules) & organic solvents can dissolve covalent compounds. so, $\ce{I-}$ will be in solution while $\ce{Br- or Cl-}$ will not be there due to being precipitated.
Le Chatelier's principle will make the reaction to be towards the product. Therefore, I also understand how a weaker Nucleophile can substitute a stronger Nucleophile.

Nucleophilicity order of Halides in Polar Aprotic solvent: $\ce{F- > Cl- > Br- > I-}$

What I ponder

Why can't we do it on alkyl fluorides?
Why only acetone? Will the reaction be same in other polar aprotic solvents too?

Some specific reactions which seem to deny all this

$\ce{CH3-CH2-Cl →[NaF/DMF] CH3-CH2-F}$
$\ce{CH3-CH2-Cl →[NaBr/DMSO] No rxn}$
$\ce{CH3-CH2-Br\text{*} →[NaBr/DMSO] CH3-CH2-Br}$

where $\ce{Br\text{*}}$ is Isotopic bromine
Note-

Here I am talking about major organic products.
DMF & DMSO are polar aprotic solvents.

I would like to know the theoretical reasoning for the above reactions.

While this question is a subset of a previous question: https://chemistry.stackexchange.com/questions/6948/can-we-prepare-alkyl-fluoride-by-finkelstein-reaction, I haven't closed is as dupe since it touches upon solvents and all which is not discussed in other post. But please do check that post.

Could you please provide source for the "specific reactions which seems to deny all this"?

@ananta Well I have not read it on something. These were taught by our teacher whose credibility I can give if you ask.

@NilayGhosh Yeah, I have checked upon many questions before posting the question including the one provided by you and this (https://chemistry.stackexchange.com/questions/116396/why-does-finkelstein-reaction-only-work-with-r-cl-and-nai) questions 's answer is also good but could not be helpful.

Fluorine is a bad leaving group and forms impressively strong bonds with carbon, considerably stronger than other halides. 2. The solvent needs to dissolve reactants and allow removal of side products, but also must be a weak nucleaophile on its own. Acetone occupies a sweet spot satisfying most of requirements. I suggest you to take March's Organic Chemistry and read extensively on theory of SN2 mechanism

The Finkelstein doesn’t occur on alkyl fluorides because it is an equilibrium $S_N2$ process, and the C–F bond is very strong, so the reaction rate is reduced. You can also think of it as the transition-state of an $S_N2$ reaction requires thermal access to a vibrational mode that bends the system into a trigonal bipyramidal geometry, and the vibrational modes of C–F bonds (including bending modes) are high energy. They need something like a silane to activate them because the Si–F bond is of a competitive strength.

Yes, the Finkelstein can work in other solvents, especially with the correct activators. You can even do the Finkelstein on aromatic rings if you add catalytic copper (I) iodide and a diamine to activate it.

Acetone is used often because it is cheap, accessible, was used when this reaction was dicovered early in chemistry, and has the best solubility of sodium iodide relative to sodium bromide that is easy to achieve.

Now those reactions.

This is perfectly fine as a type of Finkelstein. The fluoride is highly nucleophilic, so it has a high forward rate for $S_N2$ and a slow backwards rate.

Sodium bromide is not very soluble in DMSO. Nor is sodium chloride, but it is the bromide ion that kickstarts the $S_N2$ reaction. Can’t start if there’s no ions in solution. Adding a catalytic amount of soluble bromine species where the bromine can dissociate could kickstart the reaction at least a little bit. Even then, chlorine is more nucleophilic, and since their sodium salts have similar solutibilities nucleophilicity will win.

NaBr and NaBr* have the same solubility and the same bromide nucleophilicity, meaning the rates of the forward and backward $S_N2$ are the same. If you start with 100% CH$_3$CH$_2$Br*, you will approach a 50-50 mix.

Follow-up doubts on the reactions: 1. But $\ce{NaF}$ should also not be very soluble in DMF, a polar aprotic solvent, so no $\ce{F-}$ ions to start the reaction OR you are saying with the same logic as in '2.' ie. Both $\ce{NaCl and NaF}$ are insoluble so nucleophilicity will factor out. 2. Sounds reasonable as for now. 3. So reaction will proceed if $\ce{Alkyl-Br}$* is reacted with $\ce{NaBr}$ & also if $\ce{Alkyl-Br}$ with $\ce{NaBr}$* in a polar aprotic solvent.

i mean there’s nothing to doubt, unless you just made them up. if they happen, they happen. DMF typically has trace water in it as it forms a very stable dihydrate—that’ll increase solubilities more. you could test that if you tried to run that reaction over molecular sieves. but in general, there’s too many factors going into these things to just brain your way through it, it’s a complex subject with lots of intricacies and special cases.

Finkelstein Iodination
The Wikipedia article on Finkelstein reaction mentions a halide exchange equilibrium:
$$
\ce{R-X +X'- <=> R-X' +X-}
$$
As you correctly mentioned, in acetone, $\ce{NaCl}$ and $\ce{NaBr}$ are not soluble, which drives the reaction forward in accordance with Le Chatelier's Principle.
$$
\ce{R-X +MX' ->[\text{dry acetone}] R-X' +MX}\\
\ce{MX' $=$ NaI, KI}
$$
Swarts Fluorination
Swarts reaction, not an extension of Finkelstein reaction, is the recommended method to substitute $\ce{Cl}$ and $\ce{Br}$ with $\ce{F}$, that is to directly heat chlorides and bromides with fluorides of antimony, silver, mercury, or copper. According to NCERT$^\text{1}$

The synthesis of alkyl fluorides is best accomplished by heating an
alkyl chloride/bromide in the presence of a metallic fluoride such as
$\ce{AgF}$, $\ce{Hg2F2}$, $\ce{CoF2}$ or $\ce{SbF3}$. The reaction is termed as Swarts reaction.

$$
\ce{R-X + \text{fluoride} ->[\Delta] R-F + X-}\\
\ce{\text{fluoride} $=$ SbF3, AgF, Hg2F2, CoF2}
$$
Further Reading
You may like to read the article by Evano, G., Nitelet, A., Thilmany, P., and Dewez, D. F. (2018)$^\text{2}$ for a thorough review. From the same article:

References

Chemistry Part II: Textbook for Class XII. Chapter 6: Haloalkanes and Haloarenes. Available as PDF.
Evano, G., Nitelet, A., Thilmany, P., and Dewez, D. F. (2018). Metal-Mediated Halogen Exchange in Aryl and Vinyl Halides: A Review. Front. Chem., 6. 10.3389/fchem.2018.00114.

Thanks for your answer but It doesn't answer my both main questions under "What I Ponder". It is all the same information about the Finkelstein reaction as in question and I do know about the existence of the Swartz reaction but I wanted to discuss the possible explanation of 1st rxn (regardless of whether it is a good way). However 2nd & 3rd rxn can't be explained by swartz rxn. Maybe it was an honest miss, it would be extremely helpful if you could answer only the above questions. Thanks!

Follow up question: https://chemistry.stackexchange.com/questions/174509/why-is-more-than-one-product-formed-in-swarts-reaction


text:
Polyatomic molecule vibrational spectra - Molecular dipole expansion in Taylor series
To give some context: I am studying the molecular spectra in the context of an introductory course to atomic and molecular physics. My professor is using the transition rate expression from an initial state $|a\rangle$ to a final state $|b\rangle$ in the dipole approximation:
\begin{equation}
W_{a\rightarrow b} = \frac{4\pi^{2}}{e\hbar} \, \alpha \, I(\omega_{ba}) \, |\vec{\varepsilon}\cdot\vec{D}_{ba}|^{2}
\end{equation}
to derive the selection rules for the vibrational (and rotational) spectra. $\vec{D}_{ba}=\langle b|\vec{D}|a\rangle$ is the matrix element of the molecular dipole between the two states, and for a generic molecule composed of $N$ nuclei of positions and atomic numbers $\vec{R}_{I}$, $Z_{I}$ and $n$ electrons of positions $\vec{r}_{i}$ the dipole is given by:
\begin{equation}
\vec{D} = e \left( \sum_{I=1}^{N} Z_{I} \vec{R}_{I} - \sum_{i=1}^{n} \vec{r}_{i} \right)
\end{equation}
For a diatomic molecule with relative distance between the nuclei 1 and 2 given by $R=|\vec{R}_{1}-\vec{R}_{2}|\,$, if $\vec{D}$ depends on $R$ one can expand it in series around the distance of equilibrium between the two nuclei $R_{\mathrm{eq}}$ ["physics of Atoms and Molecules", B.H. Bransden & C.J. Joachain]:
\begin{equation}
\vec{D}(R) = \vec{D}(R_{\mathrm{eq}}) + (R-R_{\mathrm{eq}}) \, \frac{\mathrm{d}\vec{D}(R)}{\mathrm{d}R} \, \bigg|_{R=R_{\mathrm{eq}}} + \cdots
\end{equation}
Which is logic enough, since the derivation of the vibrational and rotational eigenvalues has been made only for a diatomic molecule on the assumption that the nuclear motion is centred around $R_{\mathrm{eq}}\,$.
The problem arises when my professor wrote that expansion for a polyatomic molecule like this:
\begin{equation}
D_{\beta}(\vec{R}) \simeq D_{\beta}(\vec{R}_{\mathrm{eq}}) + \sum_{I=1}^{N} \ \sum_{\alpha=x,y,z} (R_{I,\alpha}-R_{I,\alpha}^{\mathrm{eq}}) \, \frac{\partial D_{\beta}(\vec{R})}{\partial R_{I,\alpha}} \bigg|_{\vec{R}=\vec{R}_{\mathrm{eq}}}
\end{equation}
What does it mean? Has anyone ever seen this expression? And if so, could you suggest me where to search?
It seems like he is expanding every coordinate of the dipole $\vec{D}$ around an equilibrium point $\vec{R}_{\mathrm{eq}}$ , but then there are the terms $R_{I,\alpha}^{\mathrm{eq}}$ which suggests that every nucleus $I$ has its own equilibrium position. So the vectors $\vec{R}$ and $\vec{R}_{\mathrm{eq}}$ are vectors of vectors, with every component corresponding respectively to a position and an equilibrium point of one of the nuclei...right? But does this make sense?
Has anyone ever seen this method to deal with the spectra of polyatomic molecules, and if so, could you suggest me where to search?
Every previous calculation has been made with diatomic molecules, so this jump to the polyatomic case it's very confusing.

The expression is perfectly reasonable and your interpretation is correct.
Its essentially just a convention for summing over indices of all relevant quantities. The Dipole vector has three dimensions and in a Cartesian coordinate system it is conventional to call these $x,y,z$, and the index $\beta$ is a place holder for them, $\beta \in {x,y,z}$.
Each nucleus also "lives" in a three dimensional Cartesian space, and in the mathematical treatment we use the direct sum of these spaces, i.e. for each nucleus we have a set of $[x,y,z]$ coordinates, which we label additionally with an index for the corresponding nucleus, for example $\vec R_1 = (R_{1x},R_{1y},R_{1z})$ for the first nucleus. The complete vector that describes the structure of the molecule is then
$$
\vec R_1\oplus \vec R_2 \oplus \dots \vec R_N :=\left ( \begin{matrix}R_{1x}\\
R_{1y}\\
R_{1z}\\
\vdots \\
R_{Iz} \end{matrix}\right )=\vec R
$$
In your example that index is $I\in {[1\dots N]}$. So summing over all nuclei coordinates requires summation over a combination of $[x,y,z]$ and $[1\dots N]$, which is done in your example using $\alpha $ and $I$.
Its also important to understand that each component of the dipole moment is a function of all these coordinates, i.e.
$$
\vec D = \left(\begin{matrix}D_x \\ D_y \\ D_z \end{matrix}\right)
$$
where $D_x=D_x(\vec R) =D_x(R_{1x},R_{1y}, R_{1z}, \dots R_{Iz} )$, likewise $D_y =D_y(R_{1x},R_{1y}, R_{1z}, \dots R_{Iz} )$, and $D_z =D_z(R_{1x},R_{1y}, R_{1z}, \dots R_{Iz} )$. Each component is function of all nuclei coordinates and each function is expanded around the equilibrium coordinates,
$$
\vec D^{Taylor} \approx \left(\begin{matrix}D_x(\vec R_{eq}) + \vec \nabla_RD_x|_{R_{eq} }\cdot (R-R_{eq})  \\D_y(\vec R_{eq}) + \vec \nabla_RD_y|_{R_{eq} }\cdot (R-R_{eq})  \\ D_z(\vec R_{eq}) + \vec \nabla_RD_z|_{R_{eq} }\cdot (R-R_{eq})  \end{matrix}\right)\\
$$
The last equation of yours is index notation for this expression.


text:
Must a strong reducing agent be a weak oxidising agent, and vice versa?
As titled, must a strong reducing agent be a weak oxidising agent, and must a strong oxidising agent be a weak reducing agent? For instance, fluorine is a very strong oxidising agent, and it cannot act as a reducing agent. Is this true for all chemicals? If not, what would be the explanation to why it isn't the case?

This is an obvious and pretty good rule of thumb, but a compound isn't an agent, it can only be used as one, in one way or another. Also redox potentials are hardly standard ones in reality. That means some substances can be used as both reducing agent and oxidising agent, depending on circumstances.

What does it mean for a substance to be strong oxidizer? Well, an oxidizer acquires electrons. For a substance to do a good job of ripping out electrons from most other materials, it needs to have unoccupied electronic states of very low energy in order to coax the electrons to move towards it. For a strong reductant, the whole picture is flipped around - the substance needs occupied electronic states of very high energy, such that the electrons will readily jump out to anything even slightly willing to accept them.
So what would a compound that is a strong oxidizer and a strong reductant look like, conceptually? Well, it would simultaneously need both a very low energy unoccupied state, and a very high energy occupied state. You can roughly imagine taking all of the allowed electronic states in a substance, ordering them by energy, and "filling in" the electronic states from the bottom (lowest energy) and going up until you've assigned all the electrons the substance has.
The problem is that for simple/small molecules in their most stable condition, you can't really do this "filling in" procedure and be left simultaneously with a hole near the bottom and an electron way up high. If this happens, the molecule is going to do whatever it can to make that high energy electron fall into the hole. It can do this either by a direct electron transfer (the molecule self-oxidizes and self-reduces at the same time), or more likely, the molecule will rearrange itself somehow such that it forms new electronic states where there is no deep hole and sky-high electron. So in general, it is true that a substance cannot be simultaneously be a strong oxidizer and a strong reductant; if it were, it would just react on its own and stop being at least one of them.
But there are ways around this.
If a molecule is not simple and small, it can be conceivably engineered and subdivided into regions which don't "communicate" electronically very well. In a sense, a strongly oxidizing end of the molecule may not "know" there is also a strongly reducing end. I don't really know of a realized example of this, but it would be a situation similar to intramolecular frustrated Lewis pairs, where molecules simultaneously contain strongly acidic and strongly basic segments which can't interact on their own, usually due to spatial constraints. An electrochemical version of this would likely be a more subtle matter, but is not physically impossible.
The real workaround though, is to realize that you don't always have to operate in the ground electronic state. For many perfectly ordinary substances, even ones with no significant oxidising or reducing power, it is possible to expose them to high energy photons (say, blue, violet or ultraviolet photons). A photon of the right energy can then excite an electron from a low energy state to a high energy one. Not only does the excited electron carry a lot more energy (potentially making it powerfully reducing), but it leaves a hole behind (which can be powerfully oxidizing) - the photon is forcefully breaking the "filling in" procedure. Now, it is possible for the excited molecule to be, in all respects, simultaneously a strong oxidizer and a strong reductant!
This is a delicate condition, which in most cases lasts around a nanosecond, but in certain situations can last substantially longer. Nevertheless, that is plenty of time to do chemical reactions, which generally happen within a few picoseconds. This is the field of photochemistry. Even here though, one typically chooses substances in order to explore only their increased oxidation or reduction power when exposed to light, not both simultaneously in a single substance.

It is interesting how recently this understanding was achieved. I think looking at a chem book from, say, 1920 is a real eye-opener. One important contributor was Conant, later prez of harvard.

If you have a long molecule with one reducing end and one oxidizing end, wouldn't those molecules just interact with each other, neutralizing itself?

@ReverentLapwing In principle yes, even if individual molecules of a substance were unable to self-neutralize, it could also be possible for multiple molecules of the same substance to neutralize each other. However, in principle this could also be suppressed with proper design of the molecular structure. As a very rough analogy, imagine if the strongly reducing end were at the bottom of a deep triangular hole, and the strongly oxidizing end were a square-shaped surface. Try as they might, they simply would not be able to reach each other.

If the same material is both a strong oxidizing agent and a strong reducing agent, it will react with itself unless some kinetic barrier exists. Some of the molecules will, at random, act as oxidizers and oxidize other "reducing agent" molecules. This turns the former molecules into products with lower oxidation states and the latter molecules into other products with higher oxidation states -- a classical disproportionation.
As noted above, we may be able to stop this kinetically. With hydrogen peroxide solutions, we add stabilizers and keep them rigorously free of transition-metal bearing catalysis like $\ce{MnO2}$ or heavy metal salts to raise this kinetic barrier.

Related: Disproportionation

The reciprocal strength of acids/bases or oxidants/reductants applies to respective conjugate acid/base or redox pairs.
If an entity is simultaneously acid and base like $\ce{H2PO4-}$, or oxidant and reductant, like $\ce{Cu^+(aq)}$ or $\ce{H2O2}$, the comparative strength of both is much less related. As the entity is not to itself the conjugate pair complement.


text:
A reaction based question from stoichiometry
A 110.0g sample of a mixture of CaCl2 and NaCl is treated with Na2CO3 to precipitate the Calcium as Calcium Carbonate. This CaCO3 is heated to convert all the Calcium into CaO and the final mass of CaO is 11.62 grams. The % by mass of the CaCl2 in the original mixture has to be found out..
i've tried out finding the no of moles of CaCO3 which which i got as 0.2075 and assuming that the CaCO3 and the no of moles of CaCl2 should be equal i tried finding out the % which came out to be 20.75% in this method and if i separately found the masses of Ca, Cl2 and summed them up to find the percentage over 110 the % turned out to be 20.09%
need some suggestions to get the % anywhere near 15.2%
in this process taking all of the compounds to be anhydrous

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

Assuming no waste during the reaction process, amount of $\ce{CaO}$ isolated in $\pu{mol}$ is same as amount of $\ce{CaCl2}$ in $\pu{mol}$ in original mixture. You also have to assume $\ce{CaCl2}$ in anhydrous form during the calculation.

There is no way to produce $11.62 \pu{g}$ of $\ce{CaO}$ from a $\ce{CaCl2 + NaCl}$ sample weighing $110.0 \pu{g}$ which is $15.2\%$ $\ce{CaCl2}$ by weight. Your calculations are correct.*
Assuming $100\%$ yield, and, as mentioned, no hydration whatsoever, you need at least $20.96\%$ $\ce{CaCl2}$ in a $110.0 \pu{g}$ $\ce{CaCl2 + NaCl}$ sample to produce $11.62 \pu{g}$ of $\ce{CaO}$.
*However, if the final product is hydrated, this is possible. Let me know if you need help with calculating percentage hydration.
Underlying Principle
This is an application of the well-known law of conservation of matter, also known as the principle of atom conservation (POAC). You may refer to R. C. Mukherjee (2004), Modern Approach to Chemical Calculations: An Introduction to the Mole Concept, 7$^\text{th}$ edition for a thorough reading with lots of practice problems.
Calculation
Your overall (contracted) reaction is:
$$
\ce{CaCl2 -> -> CaO}
$$
We simply apply POAC:
$$
\text{moles of }\ce{Ca}\text{ in }\ce{CaO} = \text{moles of }\ce{Ca}\text{ in }\ce{CaCl2} \tag{1}
$$
$$
\text{moles of }\ce{Ca}\text{ in }\ce{CaO} = \dfrac{m_\ce{CaO}}{M_\ce{CaO}}M_\ce{Ca} = \dfrac{11.62}{56.077}40.078 \pu{g} = 8.30\pu {g} \tag{2}
$$
$$
\text{moles of }\ce{Ca}\text{ in }\ce{CaCl2} = \dfrac{m_\ce{CaCl2}}{M_\ce{CaCl2}}M_\ce{Ca} = \dfrac{m_\ce{CaCl2}}{110.98}40.078 = 0.36m_\ce{CaCl2} \tag{3}
$$
where, in Equations (2) and (3), $M_i$ is the molar mass of $i$ and $m_i$ is the mass of $i$ in sample of.
Equating Equations (2) and (3) in accordance with Equation (1):
$$
8.30\pu {g} = 0.36m_\ce{CaCl2}
\implies m_\ce{CaCl2} = 23.06 \pu{g}
$$
Thus, if the original $\ce{CaCl2 + NaCl}$ sample weighed $110.0 \pu{g}$, $\ce{CaCl2}$ was present in $\dfrac{23.06}{110.0}\times100 \%$ weight ratio, which is $20.96\%$
Note: All calculations have been rounded off to the second decimal place.


text:
Why does N,N-dimethylethanamide have a higher boiling point than butanoic acid?
According to Wikipedia, N,N-dimethylethanamide has a boiling point of $\pu{165.1 °C},$ while butanoic acid has a boiling point of $\pu{163.75 °C}.$
From what I learned, butanoic acid should have a higher boiling point than N,N-dimethylethanamide since it can undergo hydrogen bonding and form dimer structures, while N,N-dimethylethanamide, as a tertiary amide, cannot undergo hydrogen bonding. Butanoic acid even has a higher molecular mass than N,N-dimethylethanamide which also means it has greater dispersion forces.
All these factors indicate that butanoic acid should have a (much) higher boiling point than N,N-dimethylethanamide. Can someone explain the discrepancy in my reasoning and other (maybe more advanced or niche) factors that I have missed that can explain these boiling points?

It's just that you can't expect some "hydrogen bond beats dipole" rule to be an oracle. If you need a reason here, then I think the temperature gets high enough to seriously hamper the hydrogen bonding.

@Mithoron Sorry, I don't really get the" temperature gets high enough to seriously hamper the hydrogen bonding" part. Wouldn't this apply to all intermolecular bonds since, as temperature increases, the molecules have more kinetic energy and start spreading out, reducing the attraction of all intermolecular bonds?

I knew it was a next thing to explain, but all in all, it rather needs rather long answer than comments and I don't feel like making it. https://pubs.acs.org/doi/10.1021/jp3062465; https://www.sciencedirect.com/science/article/abs/pii/S0378381214004105

The bp’s are so close that a comparison is improbable. It is the equivalent of asking, “Do you walk to school or carry your lunch?”.

@user55119 improbable doesn't mean impossible.

I dont think we can justify the rules of which substance out of 2 has the highest boiling temperature especially when the substances are so much different from each other...

 The dipole moment plays a key role in deciding the boiling points of organic liquids. The dipole moment of N,N-dimethylethanamide is 3.7D while that of butanoic acid is 1.65D.

Introductory Remarks
This is a tricky question. Yes, your intuition is correct. Compounds capable of hydrogen-bonding often show higher boiling points (BPs) than those that are not. However, BP depends on several other factors, including molecular weight ($M$), dipole moment ($\mu$), symmetry, effects of functional group, and, perhaps most importantly, molecular structure.
Preliminary Observations
Before I get into the analysis, since, Wikipedia is not a reputable source, let us first verify the BPs of the N,N-dimethylethanamide (more commonly known as dimehtylacetamide, DMA, and DMAc) and butanoic acid and also check for another important factor that plays a key role in the determination of boiling points. I obtained the following data from PubChem.

Property
N,N-Dimethylacetamide
Butanoic acid

Normal BP
165 to 166.1 °C
163.5 °C

Molecular weight
$\pu{87.12 u}$
$\pu{88.11 u}$

Remarks on Data
Thus, despite N,N-dimethylacetamide not showing hydrogen-bonding and having a lower $M$ than butanoic acid, it has a higher BP. The value quoted on Wikipedia seems to be the lower end of the range.
The next step was to check for $\mu$, which, according to Reference 1, is 3.7D and 1.65D for N,N-dimethylacetamide and butanoic acid, respectively.
At this point, I could tell you that the deciding factor for this BP anomaly is the large difference in the dipole moments. But, who am I to decide the deciding factors. Although, $\mu$ is probably going to be major factor in determining the BPs, for accuracy, we must consider other factors such as molecular structure as well. The problem is, even the most accurate methods show up to $10\%$ error, and our difference in BPs ($\approx \pu{1.5 ^\circ C}$) is about $0.9\%$ of the values. Thus, our best efforts might just not be fruitful.
QSPR Approach
Quantity-structure property relation (QSPR$^\text{2}$) is a broad field which aims to predict a physical/chemical quantity (property) by looking at the structure of the molecule. There are numerous, specific and generalized, QSPR methods, but I am going to create a simple model here. I will consider a series of amides and carboxylic acids/esters. Our response variable, which I am trying to predict, will be BP, where, if a range is provided, I have used the lower value. Our predictor variables, which we use for predicting the response variable, will be the following:

Molecular weight ($M$)
Dipole moment ($\mu$)
N-substitution ($N_s$)
O-substitution ($O_s$)

Variables 3 and 4 account for hydrogen bonding. I constructed a multilinear regression model. I have used Reference 1 for all the data; where data is missing, I have used PubChem.
Data

Visualization

The model I have used is:
$$
\begin{align}
\text{BP} =& + 2.03M + 108.63\mu-49.48N_s-90.84O_s \\
&-0.15M*\mu -9.01\mu^2 -141.16  \tag{1}
\end{align}
$$
$$
r^2 = 0.97
$$
$$
\text{BP}_\text{predicted}\text{(N,N-dimethylethanamide)} = \pu{166.97 ^\circ C}\\
\text{BP}_\text{predicted}\text{(butanoic acid)} = \pu{170.61 ^\circ C}
$$
Inferences
$r^2 = 0.97$ is not bad at all, especially for such a simple model. From Equation (1), we see that boiling points are dependent on all factors that we considered contribute to the BP of the compound. The coefficients for $\mu$, $N_s$, and $O_s$ are large because the values are smaller compared to values of $M$, for which the coefficient is small. However, this doesn't imply that any one factor is more decisive than the other, all factors must be considered for an accurate prediction. However, the following can be inferred from these results.

BP increases with $M$ and $\mu$.
BP decreases with $N_s$ and $O_s$.

Reason for the first inference is that intermolecular interactions increase with increasing $M$ and $\mu$ (the latter commonly known as dipole-dipole interactions).

Reason for the second inference is that hydrogen bonding decreases with increasing substitution on the $\ce{N}$ of amides and $\ce{O}$ of carboxylic acids; accordingly, BP decreases.

References

Haynes, W. M. (2017), CRC Handbook of Chemistry and Physics, 97$^\text{th}$ edition.
Roy, K, Kar, S., Das R. N. (2015). A Primer on QSAR/QSPR Modeling: Fundamental Concepts.

Interesting modelling, but a couple of comments: (1) $r^2 = 0.97$ for 7 features and a training set of 15 is not really very impressive. It would be more interesting to see how the model generalises to data points outside the training set. (2) You need to scale your features to have the same range before you can draw any conclusions about which one is more important.

@orthocresol that is true. Proper analysis would constitute a structure like those of research articles on machine learning, just pointing the OP to the right directions here. Afterall, it is just a simple model that I came up with, and it may not agree with other models out there. However, I will take your advice and modify the answer to include better features and a more comprehensive dataset with a test set as well. By the way, I thought I have only used four features. Do second order terms count as features?

PubChem isn't particularly reliable either - both are as good as their sources and these sources should be used instead.

@Mithoron what would you consider a reliable source for information? Especially one that is easy to query and contains extensive data. Thanks for the edits.

Didn't you see PubChem cites sources much like Wikipedia? You can check them, at least in theory.

I think people do call them features, see e.g. https://scikit-learn.org/stable/modules/generated/sklearn.preprocessing.PolynomialFeatures.html — although I should've been more accurate and said that your model has 7 parameters. The point is that these terms make your model more flexible. If you had 15 parameters you could fit the data perfectly. You have 7, it's easy to fit the data well. The model is fine, it's the data that is too small. Anyway, the scaling is a bigger issue. $N_s$ and $O_s$ are smaller numbers, so their coefficients will be larger. You can't draw conclusions from that.

Is there a way to predict the dipole moment of a molecule? It's not a concept I'm very familiar with (I don't have any background in physics) :/

@PenandPaper I suggest formulating this as another question. I will try my best to answer it over there. Do ping me once you have done so.


text:
Colour of CuO and CuCl2
Colour of $\ce{Cu^2+}$ is blue.
Then why is the colour of $\ce{CuCl2}$ green and $\ce{CuO}$ black? Because both of these compounds have $\ce{Cu^2+}$.

Maybe $\ce{CuO}$ and $\ce{CuCl2}$ are not made of ions. Anyway $\ce{CuCl2}$ is not green, but brownish yellow. Only its hydrate $\ce{CuCl2·2H2O}$ is green. How do you know that $\ce{Cu^{2+}}$ is blue ? It may be blue in solution, or in the presence of water.

$\ce{[CuCl4]^2-(aq)}$ is even yellow. $\ce{Cl-}$ ligands attached to $\ce{Cu^2+}$ shift color toward green and yellow.
Solids are special. They may follow color of solutions, if structure is not far different. Or, the structure and related color can be very different.
There is also big difference if the ion in solid is hydrated or not. Anhydrous copper(II) sulphate is near white while its pentahydrate is blue.
Why should copper(II) oxide be blue, if copper ions are not hydrated? Also, the oxide structure and nature of interaction of copper and oxygen in the oxide are very different to $\ce{Cu^2+(aq)}$?


text:
Do molecular orbitals in metals form bands like in wave interference?
For bond that's formed between two identical atoms, I naively understand that one molecular orbital is a bonding orbital and the other is an antibonding one, and I believe they have to do with the superposition of the atoms' wavefunctions to constructively or destructively interfere.
For metal lattices (without any impurities), there's a lot more than a combination of two atoms' wavefunctions. I've heard of the conduction and valence band, which I believe are molecular orbitals of constructively and deconstructively interfering wavefunctions. I'm wondering, if you look for "thinner" bands between these two, would they look like wave interference pattern? (like a two-slit experiment) If they do, could I treat the the valence band as a "principal maximima" and the conduction band as a "principal maxima"?
Is this even a coherent question, or is my current understanding of molecular orbitals in solids very very wrong?

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

I know that there's almost a 'continuum' of orbitals in a metal right by each other, and I thought that two particular 'dense' points were called the valence and conduction band for being dense in bonding and antibonding orbitals, respectively. I was wondering if there were bands that weren't as dense as these two, but more dense than the 'background' that was similar to what I might see in slit diffraction.

Got it. No, bands are not points, and there may not be particularly dense points at all, and when there are, they don't mean all that much.

The simplest model is that of a 'free electron gas', in Chemistry this is called the 'particle in a box', and produces a set of energy levels that are filled with electrons to the Fermi level. However, a metal has regularly arranged atoms and these produce a periodic potential which perturb the electrons' energy levels producing energy gaps in which energy levels are forbidden. This is sometimes called the 'nearly free electron model'. In this model we use the idea of Bragg reflection of electron waves in the crystal, which is related to but different from your idea of multiple slit interference. If you want more detailed information look at Kittel, 'Introduction to Solid State Physics', although published a while ago now the explanations are very clear.


text:
What are the differences between the different Rate Equations?
I currently am about to finish general chem 2 so my knowledge is still basic.
There are different kinetics and rate equations and im not quite sure I know the differences.

Breaking rate of a reactant: ${\frac{\ce{-[R]}}{Δt}}$.

Formation rate of a product: ${\frac{\ce{[P]}}{Δt}}$

$v$ = $\ce{-\frac{1}{a}}\frac{\ce{d}\ce{[A]}}{\ce{d}t}
= \ce{-\frac{1}{b}}\frac{\ce{d}\ce{[B]}}{\ce{d}t}=
\ce{\frac{1}{p}}\frac{\ce{d}\ce{[P]}}{\ce{d}t}
=\ce{\frac{1}{q}}\frac{\ce{d}\ce{[Q]}}{\ce{d}t}$

Rate = ${k\ce{[A]}^\ce{{n}}} \ce{{[B]}^\ce{{m}}}$.....

If I remember and understand correctly, equations 1 and 2 are in regards to the initial concentration, so they're the rate at the beginning of a chemical reaction... I think..
If so, does that mean equation 3 is for any given point in time, just insert the concentration at given time?
Can anyone please explain what are the differences and when to use what?
Thanks!

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You should mention that your expressions are only valid in an equation written $\pu{a \mathrm{A} + b \mathrm{B} -> p \mathrm{P} + q \mathrm{Q}}$

For reaction
$$\ce{a A + b B -> p P + q Q}\tag{1}$$
is
$$r_\textrm{A} = -  \frac{\mathrm{d}[\ce{A}]}{\mathrm{d}t}\tag{2}$$ resp. $$r_\textrm{P} =   \frac{\mathrm{d}[\ce{P}]}{\mathrm{d}t}\tag{3}$$
the consumption rate of the reactant $\ce{A}$, resp. the production rate of the product $\ce{P}$.

$$r = -\frac{1}{a}\frac{\mathrm{d}\ce{[A]}}{\mathrm{d}t} = -\frac{1}{b}\frac{\mathrm{d}\ce{[B]}}{\mathrm{d}t} = \frac{1}{p}\frac{\mathrm{d}\ce{[P]}}{\mathrm{d}t} = \frac{1}{q}\frac{\mathrm{d}\ce{[Q]}}{\mathrm{d}t} \\ = \frac{r_\textrm{A}}{a} = \frac{r_\textrm{B}}{b} = \frac{r_\textrm{P}}{p} = \frac{r_\textrm{Q}}{q}\tag{4}$$
is the reaction rate, being the absolute value of the reaction rate of any reactant or product, divided by the respective stoichiometric coefficient.

$$r = k[\ce{A}]^a[\ce{B}]^b\tag{5}$$
it the reaction rate expressed by the reaction kinetic constant and concentration of reactants. Note that this is true only for single step elementary reactions, otherwise things get complicated by reaction chains and loopbacks.
If there is equation $y = x$ and $y = z$, then $x = z$, because we can substitute $y$ in the latter by $x$ from the former.
Therefore:
$$r = -\frac{1}{a}\frac{\mathrm{d}\ce{[\ce{A}]}}{\mathrm{d}t} = -\frac{1}{b}\frac{\mathrm{d}\ce{[\ce{B}]}}{\mathrm{d}t} = \frac{1}{p}\frac{\mathrm{d}\ce{[\ce{P}]}}{\mathrm{d}t} = \frac{1}{q}\frac{\mathrm{d}\ce{[\ce{Q}]}}{\mathrm{d}t} = k[\ce{A}]^a[\ce{B}]^b \tag{6}$$

You can determine

the reaction rate from the reaction rates of reactants or products,
the reaction rate from the concentrations of reactants or products,
the reaction kinetic constant $k$ from both rates and concentrations.

Thanks, it was trivial mistake.

Thank you so much.
So just to make sure, the above equations are used to express the rate of a reaction by using different data, i.e.  stoichiometry or concentration for example.
So when im asked to calculate the initial rate, I can use equations 4 and 5, and when asked to calc concentrations at specific time, I can use equation 5 (or 4, but i'd need the difference in both time and concentration)
Thanks again!

It is like the equation z=x.y . If you know z, you know x.y.  If you know x and y, you know z. If you know z and ý, you know x.

The second equation is the definition of reaction rate. The symbol used for rate sometimes is $v$, or it is written out as rate. The third equation is the rate law, telling us how the rate changes with concentration (at a given temperature).
The first set of equations is unfamiliar to me, and don't look like a rate as given.

the first ones are the average rates of breaking and forming of components.
So as Poutnik commented, are the reaction rate and rate law equal?

The quantities "rate" and $v$ are equal. The rate law is a different equation than the equation defining rate.


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Magnesium chloride to citrate conversion (reaction/synthesis)
How would one prepare magnesium citrate from magnesium chloride and citric acid? Will it work if I just heat a mixture of magnesium chloride and citric acid in water? I'd expect HCl to vaporize off.
Does anyone have experience with this reaction? Yield, purity, concentrations, byproducts, ... So if I just try to make magnesium citrate directly by mixing the chloride and citric acid, will that work – or is the citrate going to degrade/decompose into organic byproducts?
Background
I have a large amount of high purity magnesium chloride on hand which is slowly deliquescing anyway. I'd like to convert it into magnesium citrate (molar ratio of magnesium to citrate isn't hugely important, aim between 1 and 1.5) which is easier to store – and I need it anyways (need magnesium solution chelated with citrate and bitartrate).
I figure I could buy citric acid food grade, which is cheap enough, to make magnesium citrate. Objective is to have a high yield, low energy usage if possible (not having to boil off loads and loads of water, not having many steps), and no byproducts / few and easy purification steps necessary (a high grade end result consisting of only citrate and magnesium).
Economic feasibility (answer)
Magnesium chloride hexahydrate costs about 2.5 €/mol and sodium carbonate 1.1 €/mol. The product, magnesium carbonate is worth about 2.6 €/mol. So converting to carbonate results in a loss.
Citric acid monohydrate costs about 2.6 €/mol. The product, magnesium citrate is worth about 12.3 €/mol. So this step is profitable (gain 12.3 at a cost of 2.6+2.6).
Converting magnesium chloride into its citrate directly (if it were at all possible) would be profitable (gain 12.3 at a cost of 2.6+2.5).

Eventual direct mixing of $\ce{MgCl2}$ and $\ce{H3Citr}$ to get $\ce{MgH-Citr}$ and $\ce{HCl}$ and eventual precipitation of the citrate would hit the stumble stone of acidification of solution, suppresion of dissociation of citric acid and stalled precipitation. You will probably end up wasting all.

You can mix solutions of magnesium chloride and sodium carbonate, filter out precipitated magnesium carbonate and dissolve it in citric acid solution to get magnesium citrate.
$$\begin{array}{|l|c|C}
\hline \text{Compound} & \text{Molar mass }\pu{[g mol-1]}(*) & \text{Solubility }(**) \\
\hline \ce{MgCl2 . 6 H2O} & 203.31 \\
\ce{MgCl2} & 95.21 & \text{54.3 g/(100 mL) (20 °C)(anh)} \\
\hline \ce{Na2CO3.10 H2O} &  286.14\\
\ce{Na2CO3} & 105.99 &\text{ 16.4 (15 °C) / 34.07 (27.8 °C)(anh)}\\
\hline \ce{MgCO3} & 84.31\\
\hline \ce{C6H8O7. H2O} & 210.14 &  \text{59.2% w/w (20 °C)}\\
\ce{C6H8O7} &  182.12\\
\hline \ce{C6H6MgO7} & 214.41 & \text{20 g/100ml}  \\
\hline
\end{array}$$
(*) (Rounded) molar masses have been taken from respective compound Wikipedia pages.
(**) Solubility data have been taken from respective compound Wikipedia pages.
Compounds react with each other in 1 : 1 molar ratio, so the respective masses from the table are to be taken.
E.g.
$\ce{\pu{203.31 g} MgCl2 . 6 H2O \text{and} \pu{286.14 g} Na2CO3.10 H2O}$
leads to $\ce{\pu{84.31 g} MgCO3}$.
Adding $\ce{\pu{210.14 g} C6H8O7. H2O}$
leads to $\pu{214.41 g}$ of magnesium citrate.

It is always advantageous to provide the purpose and context of questions, leading to better answers and avoiding negative side effects, if responders know circumstances. If they know, they can for some scenarios say *Do not do this" or "Do rather that".

But at any rate, your solution using carbonate first seems easy and cheap enough.

Oops I hadn't yet calculated the amount of sodium carbonate required. That is a lot of sodium carbonate I'm going to waste; I wonder if the step can be skipped.

So I did some economic calculations. Converting MgCl hexahydrate into the carbonate makes a huge loss. So I can better try to sell my stock. Making Mg citrate is profitable. (see what I appended to my question)

Say, what if I attempted to mix MgCl2 with citric acid directly, but added NaOH to overcome the acidification tumble stone?

Then you will have cocktail of sodium, magnesium, chloride, hydroxide and several forms of citrate ions, competing which combination reaches the solubility threshold as first.


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Enamine formation mechanism. Can't there be alternatives?
I'm studying organic chemistry, more specifically the imines and enamines. I've read about the mechanism of formation of enamine from a generic ketone and a question popped up in my mind. The mechanism proposed by my book (and everywhere else) goes from the ketone, to the iminium ion when the water molecule goes out, and then to the enamine by the the breaking of the C=N double bond. Now, I've studied in the past the reactions of alcohols, and I remembered that a tertiary alchol can undergo by dehydration as an elimination reaction (E1 or E2), forming as a consequence a C=C bond. Now going back to the enamine formation where the water molecule is present, can't the water go out with the direct formation of the C=C bond forming the enamine by conseguence? Is the choice of the mechanism determinated by some kind of kinetics or spectroscopic study that confirm the presence of the iminium ion?

During enamine formation, “water does not go out” but rather hydroxide does. The hydroxide deprotonates the immonium cation to form water and the enamine. The immonium cation is isoelectronic with a ketone, and it is more readily deprotonated.


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What is produced from chromium, HCL and NH3
I wanted to see if I could get a chromium salt from stainless steel, so I put the stainless steel in HCL, turning the liquid green. Then I mixed the result with ammonia, and a green solid precipitated out. After drying it out, it was brown. Does anyone know what the formulas were for each of the reactions described, and what the resulting brown stuff was?

Plenty of different stainless steel do exist. They all behave differently. But all these stainless steels contain a large proportion of Chromium $\ce{Cr}$ apart from Iron $\ce{Fe}$. Most stainless steels contain also Nickel $\ce{Ni}$. Read the long and detailed article "Stainless steel" in Wikipedia.
One of the most common stainless steels is the famous $18-8$, made of $18$% Chromium + $8$ % Nickel, and of course $74$% Iron. Dipped in concentrated hydrochloric acid, it produced some Hydrogen gas, and the solution contains the metallic ions $\ce{Fe^{2+}, Cr^{2+}, and Ni^{2+}}$ according to the following equations :$$\ce{Fe + 2 H+ -> Fe^{2+} + H2}$$ $$\ce{Cr + 2 H+ -> Cr^{2+} + H2}$$$$\ce{Ni + 2 H+ -> Ni^{2+} + H2}$$ If ammonia is added to such a solution, the excess of acid is neutralized first. Then the metallic ions are transformed into insoluble metallic hydroxides like $\ce{Fe(OH)2}$ (green), plus some $\ce{Cr(OH)2}$ (grayish) and $\ce{Ni(OH)2}$ (green) $$\ce{Fe^{2+} +  NH3 +  H2O -> Fe(OH)2 + NH4^+}$$ $$\ce{Cr^{2+} +  NH3 +  H2O -> Cr(OH)2 + NH4^+}$$ $$\ce{Ni^{2+} +  NH3 +  H2O -> Ni(OH)2 + NH4^+}$$ The obtained mixture looks greenish, because the most important hydroxide in the mixture is iron(II) hydroxide.
After drying out such a mixture, it turns brown, because the iron(II) and chromium (II) hydroxides quickly react with oxygen from air and produce iron(III) and chromium(III) hydroxides, according to the reactions :
$$\ce{4 Fe(OH)2 + O2 + 2 H2O -> 4 Fe(OH)3}$$ $$\ce{4 Cr(OH)2 + O2 + 2 H2O -> 4 Cr(OH)3}$$ As $\ce{Fe(OH)3}$ is dark brown, and $\ce{Cr(OH)3}$ is green, the mixture takes the color of the most abondant product, namely $\ce{Fe(OH)3}$. The green nickel(II) hydroxide is not modified by air. It does exist in the mixture. But as it is not very abundant, it does not modify the brown color of the final mixture.

I think Cr+2 is oxidized in acid solution to Cr+3; check the potentials. First step is removal of the oxide layer followed by attack on the metal. final product is possibly mostly chrome salt that could be isolated by its amphoteric character in strong base under a N2 atmosphere to minimize reaction with O2 to form chromate.

@jmchmst. Thank you for your message. Of course the oxide layer is first removed, but the weight of the salt so produced is negligible. I have not spoken of  the other elements which are also present in the steel, like Carbon, or other metals which are present in traces. Remember that the problem was about the color of the precipitates, green in the beginning that becomes brown at the end.