text:
Transition metal chalcogenides
According to wikipedia, a chalcogenide is a chemical compound consisting of at least one chalcogen anion and at least one more electropositive element. In the same sense does transition metal chalcogenides mean atleast one chalcogen and one transition metal. If I have three classes of materials
$\ce{CrSbX_3}$
$\ce{BaVX_3}$ and
$\ce{Ba_2MnX_3}$ where $\ce{(X}$ = $\ce{S, Se)}$
Can I address them under a common term transition metal chalcogenides?
If not what might be a good term to address them together?
Update:
The common factor between these compounds is that they are all quasi-one dimensional. The $\ce{CrSbX3}$ structure consist of infinite chains of edge sharing $\ce{CrX_6}$. $\ce{BaVX_3}$ is made up of infinite chains of face sharing $\ce{VX_6}$ octahedara. $\ce{Ba_2MnX_3}$ consists of infinte chains of corner sharing $\ce{MnX_4}$ tetrahedra. Also this review has considered $\ce{BaVX_3}$ and $\ce{Ba_2MnX_3}$ among transition metal chalcogenides.

Where did you encountered those compounds? Any reference papers? Or is it hypothetical?

I worked on these materials. I am looking for a proper title for my thesis

"Ternary metal chalcogenides".

Related $\ce{PbSnS3}$ is known: American Mineralogist 2000, 85, 1066-1075 ("Suredaite, $\ce{PbSnS3}$, a new mineral species, from the Pirquitas $\ce{Ag-Sn}$ deposit, NW-Argentina: Mineralogy and crystal structure").

Find $\ce{SbCrS3}$ here.

Seems like the umbrella "transition metal chalcogenides" is used to describe these systems, though generally the properties differ enough when the number of metals/chalcogens change that they are typically grouped more specifically (e.g. transition metal  dichalcogenides).

As said by Oscar Lanzi, they are called "ternary metal chalcogenides". However, the name depends on the type of metal present in the compound. From here:

Systems to be studied include the layered intercalates $\ce{AMX2}$ (A=alkali
metal; $\ce{X=S, Se, Te}$) which are potential ionic conductors suitable for
use in solid state batteries: Group 2 ternary chalcogenides of type
$\ce{AMX3}$ and $\ce{AM2X5}$ ($\ce{A=Mg, Ca, Sr, Ba}$ ; $\ce{X=S, Se, Te}$) - some of which are superconducting and ternary mixed transition metal chalcogenides $\ce{MM'X2}$ ($\ce{X=S, Se, Te}$) examples of which have a wide range of magnetic properties including colossal magneto resistance.

You'll find more information on chalogenides of formula $\ce{ABS3}$ in this paper:

The structural diversity of ABS3 compounds with d0 electronic
configuration for the B-cation, John A. Brehma, Joseph W. Bennett,
Michael Rutenberg Schoenberg, Ilya Grinberg, and Andrew M. Rappe, J.
Chem. Phys. 140, 224703 (2014); DOI: 10.1063/1.4879659 (PDF
link)

Also, metal chalcogenides of different composition has also been noted:

($\ce{AB2X4}$): Wold A., Dwight K. (1993) Ternary Transition Metal Chalcogenides AB2X4. In: Solid State Chemistry. Springer, Dordrecht. DOI: 10.1007/978-94-011-1476-9_12

Mixed transition-post transition metal chalogenides of form $\ce{ABS2}$ and $\ce{AB2S4}$

($\ce{CuInS2,AgInS2}$): Ternary Metal Chalcogenides: Into the Exciton
and Biexciton Dynamics, Tushar Debnath and Hirendra N. Ghosh,The Journal of Physical Chemistry Letters 2019 10 (20), 6227-6238, DOI: 10.1021/acs.jpclett.9b01596

($\ce{AgBiS2}$): Effect of different growth temperature on the formation of ternary metal chalcogenides AgBiS2, S.Sugarthia,G.Bakiyaraj, R.Abinayab, M.Navaneethana, J.Archanaa, M.Shimomurad, Materials Science in Semiconductor Processing
Volume 107, 1 March 2020, 104781, DOI: 10.1016/j.mssp.2019.104781

($\ce{ZnIn2S4}$): Ionothermal/hydrothermal synthesis of the ternary metal chalcogenide ZnIn2S4, Cuixia Li, Honghua Lia, Lijun Han, Chunshan Li, Suojiang Zhang, Materials Letters, Volume 65, Issues 15–16, August 2011, Pages 2537-2540, DOI: 10.1016/j.matlet.2011.05.052

Note: if the chalcogen is oxygen i.e $\ce{ABO3}$, then it is called Perovskite.

I would be hesitant to apply the term transition metal chalcogenide to $\ce{CrSbX3}$ – and for the others, it depends on what the actual structure on a molecular level is.
A term like transition metal chalcogenide (to me) invokes the idea of the transition metal and the chalcogen being in close contact. Basic examples would include iron sulphide (where there are only two types of atoms so the relation is trivial), but I would likewise include complexes that feature some kind of $\ce{[M_nS_m]^x-}$ anionic complex. Looking at $\ce{CrSbS3}$, my first assumption would be a thioantimonate anion and a chromium cation which would not fulfill the transition metal chalcogenide condition laid out above. Of course, I could be wrong and there fould be a completely different structure in which antimony is not closely bonded to sulphur – but then that would likely invalidate the idea of a transition metal chalcogenide even more.
The other two compounds you mention have a main group metal alongside a transition metal and the chalcogenide. Depending on the actual structure, transition metal chalcogenide may be a good term for these compounds – but as stated above it may also not be the case.
In my opinion, a clearly better choice is ternary metal chalcogenide, as this removes the requirement of proving that the transition metal and the chalcogenide are part of the same complex anion.

Pre-emptive note to potential commentators: I have not checked the individual structures of these compounds if known. I am basing my entire answer on a priori ideas.

The common factor between these compounds is that they are all quasi-one dimensional. The $\ce{CrSbX_3}$ structure consist of infinite chains of edge sharing $\ce{CrX_6}$. $\ce{BaVX_3}$ is made up of infinite chains of face sharing $\ce{VX_6}$ octahedara. $\ce{Ba_2MnX_3}$ consists of infinte chains of corner sharing $\ce{VX_4}$ tetrahedra.

This review has considered $\ce{BaVX_3}$ and $\ce{Ba_2MnX_3}$ among transition metal chalcogenides,


text:
Potential energy surface in transition state theory
I would like to study the typical reaction from the transition state theory:
$$\ce{AB + C <=> ABC^‡ -> A + BC}$$
Assuming $\ce{ABC^‡}$ is a colinear molecule, the potential energy surface of the system is pictured as in Fig 5.4 on University of Bristol — Potential Energy Surfaces for Triatomic Molecules:

Figure 5.4 - Contour plot PES for a triatomic system. Ligher colours represent greater energies

My questions are:

If I project the potential energy from the figure onto the axis $r(\ce{A-B}),$ I would actually plot the potential energy of the triatomic molecule against $r(\ce{A-B}),$ but not the potential energy of the diatomic molecule $\ce{AB}.$ Is that correct?
Is there a way to obtain potential curve of molecule $\ce{AB}$ and $\ce{BC}$ as well as the atomic energy level of $\ce{A}$ and $\ce{C}$ from Fig 5.4? (I was given the hint that WavePacket program package is able to do this, but I'm not sure how)

Look for LEPS potentials and you will be able to calculated these for yourself.

Thanks a lot for your comment. Can you provide like an example (paper/website)? Because i tried googling but haven't found what i want

For the potential this is the original paper:  Sato J Chem Phys v23 p592, there are many other papers, a random selection of which see  Polanyi et al Chem. Phys. v9, p403, 1975, Johnson & Winter, J. Chem. Phys. v66 1977, p4116, Hemphill & White J. Chem. Educ. v49, 1972, p121, Rosenman & Persky, Chem. Phys. v195, 1995, p291.

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How does work from other sources affect Gibbs free energy of the system?
In most physical chemistry textbooks, Gibbs free energy is given like this:
$$\Delta_\mathrm rG = \Delta_\mathrm rG^\circ + RT\ln Q\tag{1}$$
and Gibbs free energy is part of this three-way relationship:
$$\Delta_\mathrm rG^\circ = -RT\ln K = -nFE^\circ\tag{2}$$
which gives rise to the Nernst equation.
However, in many cases, $\Delta_\mathrm rG > 0,$ so it is not thermodynamically favourable. In these cases, there would be an external source of work.
First Question: In the case of an electrolytic cell, in the system, there is also a power source. How would this external voltage source affect the Gibbs free energy equation? Would it just be an extra $-VIt$ term or something else which would cause $\Delta_\mathrm rG < 0?$
Second Question: In the case of a biological cell, the work usually comes from the hydrolysis of ATP. How would this alter the equation and cause $\Delta_\mathrm rG < 0?$

I will treat the first and second question as if they were part one and part two of a single question. I think the OP wants to know how non-PV work plays a role when looking at the Gibbs energy of a reaction, and its interpretation.
To start off, the Gibbs energy for a given reaction is a state function, so it does not depend on how much work or how much heat are exchanged between system and surrounding. The total change in entropy, however, is affected by what is going on in the system and in the surrounding, so work vs heat is relevant for the second law of thermodynamics and the direction of the reaction.

First Question: In the case of an electrolytic cell, in the system, there is also a power source. How would this external voltage source affect the Gibbs free energy equation? Would it just be an extra $-VIt$ term or something else which would cause $\Delta_\mathrm rG < 0?$

The easiest way to deal with the power source is to place the system/surrounding boundary such that the power source is in the surrounding. This shows it does not affect the Gibbs energy of the reaction at all. The Gibbs energy is positive, but the reaction can go forward because it is offset by the work done on the system. As a result, the combined entropy increases, and we are not breaking the second law.
In general, the criterion is not that the Gibbs energy of reaction has to be negative, but that
$$\Delta_r G < w$$
When there is no work done on the system or by the system, this simplifies to the requirement that the Gibbs energy be negative.
As an aside, when the Gibbs energy of reaction is negative, we can not have the reaction do an arbitrary amount of work on the surrounding, but that amount is limited to the Gibbs energy. That's why the Gibbs energy is sometimes called the maximal work (this is a good term for cases when it is negative; when it is positive, it is the minimal work that needs to be done on the system for it to be possible for the reaction to go forward).

Second Question: In the case of a biological cell, the work usually comes from the hydrolysis of ATP. How would this alter the equation and cause $\Delta_\mathrm rG < 0?$

The hydrolysis of ATP usually does not do work on the surroundings. Instead, the uncatalyzed reaction has a positive Gibbs energy of reaction. To use up the reactants and make the products, the cell runs a different reaction that includes the hydrolysis of ATP. It is not sufficient that ATP be hydrolized somewhere while the desired reaction goes on somewhere else. Instead, the two reactions are coupled so that ATP hydrolysis can not occur without the other part of the reaction happening. Here is an example how that might look for the reaction $\ce{A + B -> C}$:
$$\ce{A + ATP -> A-P + ADP}\tag{1}$$
$$\ce{A-P + B + H2O -> C + P_i}\tag{2}$$
In this scheme, A-P refers to a phosphate group bound to A, and $\ce{P_i}$ refers to phosphate. Both steps can have a negative Gibbs energy of reaction, even thought the Gibbs free energy of reaction for $\ce{A + B -> C}$ is positive. As net result, A and B turn in to C, without work done on the system.
There are other cases where there is work done on the system or by the system. These include proton pumps, and myosin hydrolyzing ATP to contract muscle fibers. You could also discuss photosynthesis in terms of work done to drive the synthesis of ATP.

This is what I think. Please correct me if there is any wrong notion.

First Question: In the case of an electrolytic cell, in the system, there is also a power source. How would this external voltage source affect the Gibbs free energy equation? Would it just be an extra −VIt term or something else which would cause ΔrG<0?

At first ΔrG > 0. You plug in external voltage, and the process happens. That does not mean ΔrG somehow becomes < 0 though. You did the work, the process is still not spontaneous, it is your work that makes the process happen. So, the external voltage source has no direct effect on ΔrG (Of course you can say that voltage causes change in composition, and the composition will affect ΔrG but that is the effect of composition on ΔrG).

Second Question: In the case of a biological cell, the work usually comes from the hydrolysis of ATP. How would this alter the equation and cause ΔrG<0?

Same reasoning as above, no effect on ΔrG. In this case, the work from hydrolysis drives non-spontaneous process. Just because a non-spontaneous process happens does not mean it is spontaneous, you are doing work.

I disagree with the term work in the case of a coupled reaction. There is no mechanical, electrical or other work. Also, the ATP hydrolysis reaction would have to be running in the surrounding in order to do work on the system. In most cases, ATP is not hydrolysed in an actual reaction. Instead, it undergoes a multi-step reaction, with a net reaction that is the sum of ATP hydrolysis and the reaction you want to drive. I like the first part of the answer.

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text:
The main difference between using AlCl3 and FeCl3 in Friedel-Crafts alkylation

What is the main difference between using $\ce{AlCl3}$ and $\ce{FeCl3}$ in Friedel-Crafts alkylation?

Is that $\ce{FeCl3}$ disfavors carbocation rearrangement in Friedel-Crafts alkylation?

This question is similar to yours: https://chemistry.stackexchange.com/questions/135259/alkylation-of-benzene/135262#135262

AlCl3 polarises stronger than FeCl3

Does this answer your question? Alkylation of benzene


text:
How to calculate membrane potentials?
I have recently been reading about membrane potentials in the cell and how they are derived. The textbook (Guyton and Hall Textbook of Medical Physiology, 13th Ed) gave this equation:
$$EMF=\pm\frac{61}{z}\log\frac{C_i}{C_o}$$
Where $i$ refers to the interior of the cell, $o$ refers to the exterior of the cell, $z$ is the relative magnitude of the charge of the cation/anion, $EMF$ is the potential in the cell, taking the external potential as 0. Additionally, the sign is + for anions and - for cations.
I understand that this comes from the Nernst Equation:
$$E=E^o-\frac{RT}{\nu_eF}\ln Q$$
Where $\nu_e$ is the number of electrons transferred per mole of reaction.
I understand that $E^o=0$ because at equilibrium, the concentrations of ions on both sides of the membrane are equal and I also understand where the magnitude comes from: $$61=\frac{1000RT}{\log eF}$$
Where $\frac{1000}{\log e}$ is used to change the base of the logarithm and change the units to mV and T is taken to be the body temperature of 37 C, converted to Kelvins.
However, what confuzzles me is

The sign convention: Why are cations - and anions +.
How do we derive the $-\frac{RT}{\nu_eF}\ln Q$ for transfer of ions, since unlike the Nernst Equation, which is derived for electrons, ions are being transferred instead?
How do we measure the membrane potential? According to the text, they put one electrode in the extracellular fluid and pierce another electrode (a typical Ag/AgCl electrode) into the cell, in contact with the intracellular fluid, as shown below. Since there is no transfer of electrons, how does this apparatus actually work?

Your textbook description is correct. Recall the principle of potentiometry, the potential difference is measured when there is no current flowing in the system. This is true under ideal conditions. Do you recall the pH meter? What is the electrode made of? It is made of glass, do you feel that current can flow across glass when you are measuring pH of your solution?
BTW, the working of the glass electrode puzzled electrochemists for ages and the still do not know how it works exactly. So if you are confused, welcome to the club.
You should rather ask how do high impedance voltmeters work? Such voltmeters work when there is no current flow and you are solely interested in potential difference between the electrodes.

Since there is no transfer of electrons, how does this apparatus
actually work?

This is perhaps a misconception. Potential difference only develops when electrons are transferred (only once) but the is no continuous flow of electrons (which constitutes current). If you are combing your dry hair in winter, a huge potential difference develops between your hair and the comb. You comb is still "charged" when you move it away from your hair. The electrons were transferred but only at one instant. There is no current flow between your comb and the hairs.


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Solving for Concentration Using pseudo-First Order Kinetics

The following graphs represent a pseudo-first order bi-molecular reversible reaction with the formula $\ce{A + B <=> C}$:

The reaction product has an extinction coefficient of $\pu{50000 M-1 cm-1}$. Solve for the concentration of $\ce{A}$.

I think you are supposed to use Beer-Lambert's equation ($A=\epsilon cl$) with the assumption that cellular path, $l=\pu{1.00cm}$, and get a value for $A$ (absorption) from the graphs to then solve for $c$. I just do not know how to get a value for absorption from the graphs. Furthermore, after getting a concentration for $\ce{C}$, I do not know where to go from there to solve for the concentration of $\ce{A}$. I think it has something to do with the equation $y=2.0172+0.498x$ (the slope is equal to $k_1$ and the y-intercept is equal to $k_{-1}$) and messing around with the rate constants to solve for $\ce{A}$.

Keep in mind that time in first graph is in log scale.

In a reversible reaction $k_{obs}=k_1(A_e+B_e)+k_{-1}$. You know $B_e$, the equilibrium conc'n of B from the right hand graph,  calculate the value $A_e$ from the absorbance. Species B is in excess.


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Estimation of dipole moments in aromatic compounds
If it asks to compare dipole moment in aromatic compounds then do we see inductive or resonance effects or do we see a mix of both?
E.g. we have to compare dipole moment in pyrrole and furan then in one we have opp direction of dipole moment in both why? Like it's a specific question but I wish to understand the mechanism behind every such tricky question so that I can do on my own.
Or suppose we have ortho hydroxyphenol then what would I say the direction of dipole moment would be? Would it be inwards towards benzene or outwards towards $-\ce{OH}$ group. It's not because of convention doubt, but like whether it's the resonance effect that dominates which would result in majorly + on $-\ce{OH}$ group and - on $-\ce{Ph}$ group or is it the inductive effect dominating which would result in opposite sign.

Does this answer your question? What is the origin of the "positive-to-negative" convention concerning dipole moments in chemistry?

Then your question is too broad.

https://chemistry.stackexchange.com/questions/96923/how-to-compare-dipole-moment-in-thiophene-furan-and-pyrrole

No I think I have got what the OP wants to ask

I read up in my text book that dipole moment direction of phenol is towards OH group so I guess O being highly electronegative pulls the electrons more and participates less in resonance hence it has deta negative charge on it thus acc to convention dipole moment direction is from positive to negative hence direction is towards OH group.


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Without attractive forces, why would there be only ideal gases?
I've read this statement in several places. What is the importance of attractive forces for solids and liquids?

The importance of attractive forces in solids and liquids is that you exist as a human being made of flesh, bones and liquid. Otherwise, if intermolecular attractive forces did not exist, we could not have existed in the human form. Everything would have existed as a gas.

Attractive forces is the reason why solids and liquids, and real gases exists. Else everything is fundamentally, collection of atoms.

The title is omitting the other attribute of ideal gasses - negligible molecular volume. So even without attractive forces, gases would not be ideal yet.

Not having Intermolecular forces would have far reaching impacts beyond the scope of the field of chemistry alone.

If there were no attractive forces solids and liquids would not exist
The very existence of liquids and solids depends on attractive forces. two molecules that don't attract will never form a liquid or a solid. Liquids and solids only form because, under certain conditions of temperature and pressure, the attractive forces between their constituent molecules is strong enough to overcome the thermal energy trying to shake them apart.
A truly ideal gas has no attractive forces between the molecules and the molecules have no volume. Real near-ideal gases have negligible molecular volumes and attractive forces insignificant compared to thermal energy at that temperature and pressure. So, in the absence of any attractive forces, everything would be something close to an ideal gas unless compressed enough for the molecular size to become a factor in determining the volume of the gas (which is usually at extreme pressures).


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Why energy is released when an electron is added to a neutral atom?
Question :

Why energy is released when an electron is added to a neutral atom?

I read somewhere

“When electrons are added to an atom, the increased negative charge puts stress on the electrons already there, causing energy to be released.”

I didn't understand What is stress and how energy is released due to stress?

Please do not personify atoms and sub-atomic particles. Electrons and atoms are not human beings who have to vent their energy in order to de-stress. Whoever wrote it just ignore this statement because the writer does not know what he/she is writing. Sub-atomic particles have a world of their own which is very different from what we know and understand. There is no such thing as stress or congestion of having too many electrons.

@M. Farooq Ok I ignore that statement.

Search about electron affinities "electron affinities can be negative (energy is released when an electron is added), positive (energy must be added to the system to produce an anion), or zero (the process is energetically neutral)".  See https://www.google.com/search?q=why+is+electron+affinity+negative&oq=why+is+electron+aff&aqs=chrome.0.0l3j69i57j0l4.14311j0j7&sourceid=chrome&ie=UTF-8

The point is that energy is not always released.

@M. Farooq Note that stress has multiple meanings and only some of them are related to feelings as a high discomfort. Some others are objective, like the one used in mechanical engineering as a tension. It may be stretched being applied to quantum systems, but has nothing to do with human emotions.

@M. Farooq I know that electron affinities can be negative. But my question is “Why sometime energy is released?”

But the stress=repulsion is not reason of energy release.

@Poutnik Yes. That's why I am asking the reason.

If the electron statistical distribution around the atomic kernel had been perfectly spherically symmetric, and if the electron occurance distribution had not mutually overlapped, than by the Gauss law of electrostatics, the net force between a neutral atom and an electron would have been zero.
But as neither of above conditions is true, a kernel charge is not fully screened off by electrons, acting as having a residual, "effective charge", what allows releasing energy by bounding an extra electron. See Slater rules.
An extra electron puts among other electrons some extra stress=mutual repulsion, what somewhat decreases this released energy.
As effective kernel charge and electron mutual repulsion ( classical and Pauli ones ) are 2 major factors affecting energies of electron orbitals in multi-electron atoms.
When the former factor is stronger, energy is released by an extra electron bounding, like for fluorine. And vice versa, like for helium.

Can you define energy? I thought heat energy changes.

Who am I to define energy, if it is already defined and definition available on many places ? :-)

In my physics textbook, it is defined as "The capacity to do work".

See e.g.Wikipedia

Not every atom releases energy; it is all about the stability. For your understanding, stability is inversely proportional to energy, so as the stability increases atom try to de-excite itself. If you look at the periodic table, metals generally have no affinity towards the electron because they are e donor, so they release energy by losing an electron. In non-metals like chlorine, which has highest electron affinity, release much high energy and gain the stability. And why it happens and how it happens, you need something more than normal Bohr and other atomic model, which is not the real model of an atom.

No element releases energy by losing an electron.

@ Yuvraj. Sorry ! Your long comment is nearly unreadable, because it has no punctuation sign. No points, no commas !


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Covalent atomic radii: oxygen vs nitrogen
Many books state that $R_\ce{N} > R_\ce{O}$ which is in accordance with the general trend. However, some books say that $R_\ce{O} > R_\ce{N}$ because of repulsion caused by pairing of electrons.
Which one is correct and should be followed?
Older version of NCERT, cengage, arihant and JD Lee state the latter, while the newer version of NCERT, Wikipedia and many websites say the former. Why was the theory discarded?
Note: van der Waal's radii follow the general trend.

The radius of atoms decreases from left to right across the table. As the smallest radius per row will always be the noble gas, then opposite the noble has having the largest radius in the row.

The van der Waals radius of nitrogen is larger than that of oxygen, and has been calculated as such for quite a long time.
$$
\begin{array}{lll}
\hline
\text{Reference} & R_\ce{O} & R_\ce{N} \\
\hline
\text{Pauling, 1939} & 1.40 & 1.5 \\
\text{Bondi, 1964} & 1.52 & 1.55 \\
\text{Zefirov, 1974} & 1.29 & 1.50 \\
\text{Gavezzotti, 1983–1999} & 1.40 & 1.50 \\
\text{Batsanov, 1995} & 1.51 & \\
\text{Wieberg, 1995} & 1.5 & 1.6 \\
\text{Rowland, 1996} & 1.58 & 1.64 \\
\hline
\end{array}
$$
As is evident from the table, $R_\ce{N} > R_\ce{O}.$ An interactive webpage of this reference can be seen here, which has the van der Waals radii of quite a few other elements as well.
Edit:
After OP mentioned the abnormality in Covalent Radii, I checked the radii in Concise Inorganic Chemistry, by J.D. Lee. The Covalent radius of Nitrogen $R_N = \pu{0.77 Å}$, whereas that of Oxygen is the same; $R_O = \pu{0.77 Å}$. Lee mentions that the radii are taken from Tom Moeller's Inorganic Chemistry book (which is from 1952)
OP should also be aware of the method used for calculating covalent radii: The Covalent Radius is defined as half the length of a single bond between two uncharged atoms. According to a more recent reference2 (which is used by Wikipedia), $R_N = \pu{73.4 pm}$ and $R_O = \pu{70.2 pm}$
Even more recent references, such as (3) use a different method of calculating covalent radius:

The covalent radius for nitrogen was obtained from N–N bond
distances in substituted hydrazines (only H or C atoms bonded to
N) with three-coordinated nitrogen atoms, excluding all structures
with an R factor larger than 10% or presenting disorder or
errors. The resulting value, $\pu{0.706(13) Å}$, is the average of 2200
crystallographically independent data and was used throughout
with three decimal figures for calculating other radii from N–
element bond distances to minimize rounding-off errors.
The covalent radius for $\ce{O}$ was derived from a
sample of 10,000 acyclic $\ce{C–O}$ bond distances, in a search limited to
purely organic compounds with two-coordinate oxygen and four-coordinate carbon atoms with R ≤ 5%, and the value obtained was $\pu{0.676(28) Å}$

A comparision of covalent radii has also been provided by the authors:
$$
\begin{array}{lll}
\hline
\text{Author} & R_O\ (Å) & R_N\ (Å) \\
\hline
\text{Cordero et al.} & 0.706(13) & 0.661(19) \\
\text{Alcock} & 0.702 & 0.659 \\
\text{Mingos} & 0.74 & 0.72 \\
\text{Butler and Harrod} & 0.75 & 0.73 \\
\text{Wells} & 0.74 & 0.74 \\
\hline
\end{array}
$$
References

Batsanov, S. S. Van Der Waals Radii of Elements. Inorganic Materials 2001, 37 (9), 871–885. DOI: 10.1023/A:1011625728803. (PDF)
Sanderson, R. T. “Electronegativity and Bond Energy.” Journal of the American Chemical Society, vol. 105, no. 8, Apr. 1983, pp. 2259–61.  doi:10.1021/ja00346a026.
Cordero, Beatriz, et al. “Covalent Radii Revisited.” Dalton Transactions, no. 21, 2008, p. 2832. doi:10.1039/b801115j.

Indeed. But here we are not talking about van der Waal's radii.

Then what is it that we are talking about?

@Safdar the covalent radii!

My bad. I'll do it right away

Nice one! Can you also provide some info on the effect of repulsion of $e^{−}$s?

@DatBoi In this case, repulsion of electrons (more specifically interelectronic repulsion) is hand-waving at best and hopelessly incorrect at worst.

There's an (or more) excellent paper by Pekka Pyykkö about radii, which should not be missing from this answer. I can't look it up right now, but in the following list that I compiled you should be able to find it: https://github.com/polyluxus/rwth-tools/blob/master/misc/data-sheets/radii.md


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Why does more s-character stabilise Carbanions?
Since carbanions carry a negative charge, more the distribution of electron density, more is the stability.
What does this have to do with more s character?
Does the distribution of electron density in a spherical fashion more favourable than p orbital distribution, as it is more symmetrical maybe?

S-orbitals lie closer to the nucleus. Since the lone pairs are regions of high electron density they may be stablised by keeping them in an s-orbital closer to the nucleus.


text:
Effect of adding pure solid/liquid in heterogenous equilibrium
QUESTION

Consider the following reaction as an example:
$$\ce{NaNO3(s) <=> NaNO2(s) + 1/2O2(g)}$$
Once this reaction has achieved equilibrium and then if we added pure $\ce{NaNO3}$ to this system, then is it correct to say that it will favour forward reaction?

My Thoughts
I have gone through a lot of questions related to this topic in Chemistry Stack Exchange and all of them have explained in detail that why it will not affect the value of equilibrium constant (as their "concentration" would not change and I fully agree with that), but no one answers the core question.
That if we add or remove a pure solid/liquid then why wouldn't "moles" of other compounds in the reaction change which are included in equilibrium expression?

I meant that let's say we add some moles of nano3 (s) to the above example then some moles of both nano2 and o2 would form by standard reaction stoichiometry.

Then- 1) Wouldn't change in moles of o2 affect reaction quotient? (Keeping Temperature constant, I know 'Kp' wouldn't change but Reaction Quotient 'Q' should change and hence forward reaction should be more favoured (have higher rate than backward reaction)

If what I said above is true, then wouldn't that mean that changing amount of solid also affects equilibrium but not Kp

It isn't concentration but activity

Where is the carbon on the right side of the chemical equation coming from?

Sorry, it should be o2, I will correct it.

If you add some new amount of pure $\ce{NaNO3}$ to such a system in equilibrium, you would produce more decomposition, Where is the problem ?

The implicit unit activities of pure solid compounds imply they remain or are created as pure. Things may get more complicated when nitrate anions are being replaced by nitrite gradually, so activity of neither of solids would be unit.

I will revive this topic because it also haunted me. The question is:

[OP] Once this reaction has achieved equilibrium and then if we added pure $\ce{NaNO3}$ to this system, then is it correct to say that it will favour forward reaction?

The answer is no, and nothing will happen. I will justify this from a classical thermodynamic framework.

1. Thermodynamic Equilibrium In heterogeneous reactions, like this one, the number of degrees of freedom is given by Gibbs rule for reacting systems
\begin{equation}
  F = 2 - \text{[Num. of phases]} + \text{[Num. of components]} -
  \text{[Num. of chemical reactions]} \tag{1}
\end{equation}
We have:

Three components: $\ce{NaNO3}$, $\ce{NaNO2}$, and $\ce{O2}$.
Three phases: gas from $\ce{O2}$, one solid phase of $\ce{NaNO3}$, and another solid phase of $\ce{NaNO2}$.
One chemical reaction.

Hence, Eq. (1) yields
$$ F = 2 - 3 + 3 - 1 \to \boxed{F = 1} \tag{2} $$
Thus, the equilibrium state is fixed with one intensive variable. Say say we are in the lab, and we have a beaker of a volume $V$ at a temperature $T$. We put initially some initial amount $n_\ce{NaNO3}$ of sodium nitrate.
The equilibrium constant at temperature $T$ is denoted by $K$, and its value is
\begin{equation}
  K(T) = \exp\left[-\frac{\Delta_\mathrm{r}G^\circ(T)}{RT}\right] \tag{3}
\end{equation}
Now we apply the law of mass action
\begin{align}
  K(T) &= \prod_j a_j^{\nu_j} \\
  &= \dfrac{a_\ce{NaNO2} \sqrt{a_\ce{O2}}}{a_\ce{NaNO3}} \\
  &= \sqrt{a_\ce{O2}} \\
  &= \sqrt{\frac{f_\ce{O2}}{f_\ce{O2}^\circ}} \\
  &= \sqrt{\dfrac{\phi_\ce{O2}P_\ce{O2}}{P^\circ}} \\
  &= \sqrt{\dfrac{P_\ce{O2}}{P^\circ}} \rightarrow
  \boxed{P_\ce{O2} = K^2(T)P^\circ} \tag{4}
\end{align}
since:

The activities of solid pure phases are one.
The gas phase behaves as an ideal gas, so the fugacity coefficient $\phi_\ce{O2} = 1$.
The standard pressure is denoted by $P^\circ$.

The gas phase is composed only of oxygen, thus, the amount of oxygen using Eq. (4) is
\begin{equation}
  n_\ce{O2} = \frac{P_\ce{O2}V}{RT} \rightarrow
  \boxed{n_\ce{O2}(T) =
  \left[\frac{K^2(T)V}{RT}\right]P^\circ} \tag{5}
\end{equation}
There is a little assumption in the calculation, regarding the volume. In this equilibrium state, the two solid phases will take some space of the beaker, so the volume $V$ in Eq. (5) won't be the volume of the beaker, but a little less.

2. Analysis Eq. (5) affirms that for a fixed temperature $T$ in a volume $V$, we can only have this amount of oxygen in an equilibrium state. Initially we had $n_\ce{NaNO3}$ in the beaker, so given an infinite time when the law of mass action will hold, some $n_\ce{NaNO3}$ will be consumed. How much? By stoichiometry, until we have $n_\ce{O2}$ in the gas phase. Once this value is reached, even though we keep adding $n_\ce{NaNO3}$, no more $\ce{O2}$ can be present in the gas phase.
If we take seriously the fact that adding $n_\ce{NaNO3}$ will take some volume of the beaker, then Eq. (5) will still hold, but $n_\ce{O2}$ will change. If $V$ decreases, then $n_\ce{O2}$ must decrease. We have the funny situation that we add reactant and the reaction will proceed in the reverse direction. For practical purposes, the volume of solids are negligible, and the system will not change its state.

If 0 deg C ice is added to ice+water in equilibrium, it will not suddenly start freezing nor melting. Adding more of pure solid (isobarically, it is not exactly true for isochoric scenario) does not change its chemical potential nor activity, as these are intensive, not extensive properties.

That nothing will happen is NOT CORRECT. Addition of more of a phase in a form that it can react will increase the rate of the forward reaction and provide more sites for the increased rate of the reverse reaction.  A living example: boil a small pot of water until the equilibrium VP is met.  Then boil some huge pots of water until drops are condensing and evaporating all over the kitchen. At the same T the VP is the same.  In solid state reactions the difference can be control or an explosion.

I could be also said that equilibrium depends on intensive properties, like activity, concentration, partial pressure, fugacity. Not on extensive properties like mass or volume.

@Jimchmst Hello Jimchmst. I am not sure that I have understood you case. Your example of VLE is not a case of chemical reaction, but phase transition, so I do not think it applies. Generally, adding a reactant increases the rate of reaction, which we agree. My conclusion applies to this particular question, where only a unique partial pressure can exist at equilbrium.

True. At equilibrium forward and reverse reactions ae equal that is Thermo. What those rates are is kinetics. Adding nitrate in an appropriate form increases the forward rate increased O2 and Nitrite increases the reverse rate to reestablish equilibrium. All this takes time ie kinetics. Phase changes are as much a chemical reaction as a physical reaction; the same rules apply [the First and Second Laws]. The thermo and Phase Rule approach is exquisite but is equilibrium and needs attention to kinetics.


text:
How to oxidise iron in spinach to iron(III)
I'm planning an experiment to test the iron content of spinach. I plan to ash spinach, grind it, then dilute it in a solution which would hopefully oxidise the iron ions in it to $\ce{Fe^3+}.$ This is in order to get a reaction with ammonium thiocyanate solution in the end, which will allow me to determine concentration by colorimetry.
What should I dilute my ashed spinach into to oxidise the iron in it? Preferably something safe with the thiocyanate.

Use ceric ammonium sulfate.[1,2]  Quoting from reference 2:

2 Reagents
(i) 0.25 per cent ceric ammonium sulphate in 8.8 N sulphuric acid.
(ii) 40 per cent potassium thiocyanate solution.
(iii) (a) Stock standard ferrous iron solution - 100 μg iron per ml.
(b) Working standard ferrous iron solution - 0.3 to 10 μg iron per ml
2.2 Procedure
To five ml of clear water sample in a test tube 0.5 ml of ceric ammonium sulphate solution is added and mixed thoroughly by shaking. Finally, 0.5 ml of potassium thiocyanate solution is added and again mixed properly. Exactly 5 minutes after addition of thiocyanate solution, the absorbance of ferric thiocyanate complex is measured at 480 nm in a spectrophotometer (Spekol, Carl-Zeiss, Jena).
A series of freshly prepared working standard ferrous iron solutions (upto 10 ppm iron) were treated similarly and the absorbances measured at 480 nm. From the calibration curve, plotted with concentrations of different standard iron solutions against absorbances, the iron content of the unknown sample is evaluated.

References

Willard, H. H.; Young, P. Ceric Sulfate as a Volumetric Oxidizing Agent. Ii. Determination of Iron. J. Am. Chem. Soc. 1928, 50 (5), 1334–1338. DOI: 10.1021/ja01392a013.

Goswami, D. C.; Kalita, H. Rapid Determination of Iron in Water by Modified Thiocyanate Method. DSJ 1988, 38 (2), 177–182. DOI: 10.14429/dsj.38.4835.
Also available via CiteSeerx. (The pdf icon has a cached version of the article.)

@OscarLanzi Thanks for the answer, unfortunately doesn't look like I have access to ceric ammonium sulfate.

That is too bad, this is a pretty standard choice.  To get a better answer I think you will need to provide more input.  I assume you have some chemical knowledge and thus can identify some oxidizing agents to which you do have access.  We can then pick the best one.

https://www.sigmaaldrich.com/catalog/substance/ammoniumceriumivsulfatedihydrate632551037847911?lang=en&region=US

@OscarLanzi I am thinking maybe sulphuric acid would be a good choice. However it would need to be quite concentrated to work from my understanding. Do you think 1 molar sulphuric acid would be sufficient? Obviously I would later dilute it with de-ionised water since thiocyanate and concentrated acids don't go well.

One molar sulfuric acid is far from sufficient.  It would be completely dissociated into aqueous hydrogen and bisulfate ions, and neither of the latter will do the job you want.  Usually when chemists want sulfuric acid as an oxidant, they don't mess around.  They go for the fully concentrated stuff.

I see. My problem is that concentrated acids and ammonium thiocyanate do not go well together.


text:
Determine if stuff is Potassium Alum
How can one determine (at home) if the product shown below is Potassium Alum? I held a piece of it in a gas flame (assuming that the flame would become violet if potassium were present), but there was no change in the color of the flame. [Got product for use as astringent and antiseptic after-shave -- and people say (do not know why) that for such use, it has to be potassium-(not something else)-alum.]

Update: As mentioned in the comments to the chosen answer (by @James Gaidis), the product is ammonia alum. Just to be double sure, I repeated the tests with McCormick's Alum powder whose ingredient is potassium alum. (1) The picture of the flame test is below.

(2) With the McCormick Alum, the test involving sodium bicarbonate resulted in sound and definitely no ammonia gas.  (3) The solution of Fatkari had a salty taste that became sour in a second or so; the K-alum was tasteless and became sour in a second or so.

Take some muddy water and see if it settles down faster. Alum is generally used for puriyfing water in this way.

Need to know if the 'X' in @gaidis  XAl(SO4)2⋅12H2O, is potassium or something else.

See if it aids in the coagulation of blood loss in minor cuts like Potassium Alum is supposed to, if you fancy an experiment

@Autrey it that a definite test -- meaning, is it known that other Alums -- sodium, ammonia, chromium, and ferric -- do not aid in coagulation?

From Wikipedia:

An alum is a type of chemical compound, usually a hydrated double sulfate salt of aluminum with the general formula $\ce{XAl(SO4)2 · 12 H2O},$ where $\ce{X}$ is a monovalent cation such as potassium or ammonium.

There are also chromium alums and ferric alums, but the color in your sample does not seem to suggest either of them.
The alums melt at relatively low temperatures, because they already have a lot of water in the crystal structure, so ammonium alum and potassium alum can not be differentiated on the basis of melting points. Solubilities are fairly close, too:
$$
\begin{array}{llc}
\hline
\text{Compound} & \text{mp}/\pu{°C} & \text{Solubility}/\% \\
\hline
\ce{(NH4)Al(SO4)2 · 12 H2O} & 93 & 15 \\
\ce{KAl(SO4)2 · 12 H2O} & 92.5 & 11 \\
\hline
\end{array}
$$
If there is no violet color to a flame, there may be no potassium, but the test can be sensitive and may be indeterminate. Eliminate the possibility that the alum is an ammonium salt by dissolving some in water and adding a strong alkali, then sniffing it gently. If you smell ammonia, you have an ammonia alum; it will still work as a flocculant for muddy water, but is different chemically. A strong alkali readily available at home, well, at the grocery store anyway, is sodium carbonate (washing soda).

In steel vessel, added water and put crystal in smaller steel cup inside vessel; heated water to under 95C while continuously moving smaller cup (with tongs). Melted -- then solidified before water got to 75C. MP for Na-alum is 61.  For the ammonia-odor test, does pure baking soda (100% Na-bicarbonate) constitute strong alkali?  Got product for use as astringent and antiseptic after-shave -- and people say (do not know why) that for such use, it has to be K-alum.

Put crystal in water and stirred vigorously for over a minute. Removed crystal. Water had no odor; taste was not as strong as I had expected. Dumped in sodium bicarbonate. Reaction made sound and did emit ammonia odor.

@gaidis please consider updating the post with the equation for the reaction between  the X-alums and sodium bi-carbonate.


text:
What would happen if water is to flow through a microchannel where its top is hydrophilic and bottom hydrophobic?
What would happen if water is to flow through a microchannel where its top is hydrophilic and bottom hydrophobic?
If an external pressure is applied, for example, a syringe pump is used, it is expected that water will flow through the microchannel, even if the channel is completely hydrophobic.
I wonder what is the situation when only the capillary force is the driving force, when the two surfaces of the channel are different in their wettability.


text:
Who came up with C1V1 = C2V2?
I was wondering if there is an original scientific publication showing why $C_1V_1 = C_2V_2$ works.
Or perhaps, who described/used it first?

Maybe this would be better asked on [hsm.se].

This is probably the extension of the definition of the word "concentration" which was probably used already in the Middle Age, with masses of course (instead of moles}.There is probably no original scientific publication of the definition of the concentration.

Very difficult to find out because this formula is a special case of general mass balance equations. We will have to find who came up with mass balance equation?

Even the ancient merchants realized 10 sacks with 50 coins each means 500 coins and that if you put 100 coins into a sack then 5 sacks is enough. Why to scientifically show that it works, when everyday life showed it almost 2000 years ago ?

I’m voting to close this question because this is a history question, not a chemistry one.

tl;dr It's so simple it was likely never published as a "research result", since it stems quite directly from definitions.

It's a small step from the conservation of matter and the definition of concentration.
By definition: $C = \frac{n}{V}$, thus $n = CV$. If you dilute with pure solvent, you don't change the amount of solute that you have, in other words: $n_1 = n_2$. Using the previous equation, you get $C_1V_1 = C_2V_2$.

I don't see how this answers the question. OP is asking about the origins of the expressions in terms of history, not why it's mathematically correct.

My point is that it's so simple it was likely never published as a "research result", since it stems quite directly from definitions.


text:
Why do batteries drain so quickly if they are forced to draw more current than they normally do?
When a battery is pushed to use twice the current it normally does, it lasts for less than half as long before dying...
In fact, batteries often come with a 'C' rating that gives you an idea of how long they might last when run at a high current level...
What is going on inside the battery when this occurs?  What is the chemistry, so to speak?

The chemical reaction the drives a battery does not depend on the currant that is produced. It is only a question of kinetics. If the contact between the poles is good, the currant is high, and the battery is soon discharged. If there is an important resistance between the poles, the currant is low and the battery will be operational a long time. But the chemical composition of the electrodes does not depend on the intensity of the currant.

High current leads to increased temperature, leading to increased parasitic internal discharge, which leads to further temperature increase.

Batteries store chemical energy. They have a finite amount of it. if you use that energy faster (all other things being equal that is what "higher current" means) then the capacity will be reduced faster. That's the simple explanation. If you are asking why does the energy sometimes decrease faster than expected given that basic observation, you need to clarify the question.

The capacity of a cell is measured toward the terminal cut off voltage. If e.g.a double current is drained from a cell, the external cell voltage is decrease by the doubled voltage drop on the cell internal resistance: $$U_\mathrm{e}=U_0 - R \cdot I$$
That caused reaching the terminal voltage sooner then in the half time.
It is additionally leading to the ohmic internal energy wasting $$P=R \cdot I^2$$
But this energy loss affects the useful energy output, not the capacity ( directly).
Additionally, higher than optimal current causes lowering of the immediate cell open voltage. Limited rate of diffusion of electroactive components and/or of electrode reactions are not able to maintain electrode equilibrium potentials.
In result, even the $U_0$ (non linearly) decreases with the higher current and the terminal $U_e$ is reached after providing smaller total charge.

Also, energy is wasted in the internal heating of a cell. At low currents, the cell is hardly warmed, but at high currents, the voltage dropped across the cell's internal resistance is wasted as heat. This is just a "common sense" explanation of the answer.

@DrMoishe Pippin This is involved in RI, as P=RI^2.

Yes, I was just making the point that the cell gets noticeably hot at high drain.


text:
Find the major products when these cyclohexenone bromides react with caustic potash in alcoholic medium:

Question:

Also find the order of the rates of the given reactions.

Answer: The answer of this question is given as option 'D' ie. all products are identical, however I am getting the answer as option 'C'.

My approach: I figured out that the under the presence of the given reagent the substrates will undergo elimination reaction via the E2 mechanism, in reactions (2) and (3), the same alkenes will be formed and the ketone will tautomerise to form a highly stable phenol, but in the first reaction I made Cyclohex-2,5-dien-one as the product, I am not able to understand how will this covert to phenol?

For the comparison of the rates I figured out that reaction (2) will have the fastest rate due to a highly acidic hydrogen present (the anion formed is in extended resonance with the double bond and the ketone), however I am not able to compare between the rates of the first and the third reactions.

Am I thinking in the right direction or is there some mistake? Please help.

@Karl Can you explain by a diagram? I did not get what you wrote.

Nope, your homework: Draw the detailed step-by-step diagram of the keto-enol step in (2) and (3), then try to do sth. very similar with (1). ;-)

@Karl thank you , I got the answer as phenol

Can you comment on the rates of the reactions?

No clue, honestly. Zilch. But there are others around here who are smarter in OC than me. ;-)

One of them is probably faster to eliminate than the other two.

That's obvious textbook mistake - they didn't consider tautomerisation. Still, that should be rather potassium phenolate.

@user55119 how to we compare between rates of reaction 1 and 3?

Actually I think this is simple: Tautomerisation comes second, you don't have a particularily strong base, that only works if you can close the aromatic ring at the same time. Bromide is a good leaving group, so I say this is E1, and (1) has the intermediate cation next to the carbonly, that's bad, (3) has the cation next to the double bond, which is OK, but mesomery is not much of an improvement, and (2) is best. ?


text:
Where do the three negative oxygens in a phosphate group get their missing electron from?
A video I watched showed the construction of a phosphate group using a Lewis structure.
The oxygen with the double bond fills its octet group.
The three other oxygens lack an electron.
The teacher merely pointed out this lack and added the missing electrons (shown in red).

Where do these missing electrons come from? Are they stolen from passing hydrogens?

A person with net worth above $1,000,000,000;$$ is called a billionaire. Where do these money come from? It does not matter. Same thing here.

Do you mean my question was not worth asking?

Any question is worth asking. That's how you learn, after all. But then again, some questions will be answered with a "yes", some with a "no", and some with "that's a wrong question to ask, because..."

And some questions will be answered clearly and helpfully  (See @Maurice, below).

Note that your teacher is using an outdated depiction of phosphate. It is better to rotate the bottom oxygen in the leftmost picture 90° clockwise before making a bond. The resulting bond is a single bond, not a double bond in keeping with calculation and experiment.

Phosphorous has 5 valence electrons - what happens to the fifth one (if there's not a double bond)?

Well ! A neutral substance like $\ce{PO4}$ does not exist. It only exists as an ion with three negative charges : $\ce{PO4^{3-}}$. These charges can come from three atoms of the first column like Hydrogen $\ce{H}$  or sodium $\ce{Na}$. These atoms become the cations  $\ce{H^+}$ and $\ce{Na^+}$ after this electron "gift". So the ion phosphate is always surrounded by three ions like $\ce{H+}$ or $\ce{Na+}$. This is why the phosphate ions may be obtained by dissolving neutral substances like $\ce{Na3PO4}$ or $\ce{H3PO4}$ in water. They will never exist alone.


text:
The Curious Case of the equivalent weight
Please read the reference points given at the end.
I understand the following about equivalent weight. Firstly it is given by,
$Equivalent\:Weight= \frac{Molar\:Weight}{Valence\:Factor}$
And secondly, the Valence Factor is equal to the acidity/basicity in case of a acid/base and the change in oxidation number of the compounds in the most cases.
However I encountered the following equation,
$$\ce{Cu + HNO3(aq, dil)  -> Cu(NO3)2 + NO(g) + H2O(l)}$$
Clearly, in this case we see two roles of nitric acid.
I tried breaking down the equation into half reactions as follows:

Oxidation of copper:
$$\ce{Cu -> Cu^{+2} + 2e-}.....(1)$$
Dissociation of Nitric acid
$$\ce{HNO3(aq) -> H+(aq) + NO3-(aq)}.....(2)$$
Reduction of Nitric acid to Nitrogen Monoxide
$$\ce{HNO3(aq) + 3H+(aq) + 3e- -> NO(g) + 2H2O(l)}.....(3)$$

Performing $(1)\times3,\,(2)\times6,\,(3)\times2$ we get the balanced equation which is:
$$\ce{3Cu + 8HNO3(aq,dil) -> 3Cu(NO3)2(aq) + 2NO(g) + 4H2O(l)}$$
Now I started getting confused, as to how do we calculate the valence factor and hence the equivalent weight for this reaction. I tried doing as follows:
$$Valence\,factor,\,n\:= n_{acid}\:+\:n_{reduction}$$
Where after multiplication of the equation we get $n_{acid}\:=\:n(H^+)\:=6$ and $\:n_{reduction}\:=\:n(e^-)\:=6\:$
Now the total number of nitric acid molecules used up is $8$. So using the above mentioned formula we can state $$Equivalent\:Weight\:=\:\frac{8\times63}{12}$$
But this doesn't seem to be right. Can someone please explain this scenario.
Okay, now here are a few reference points:

I know that the system of equivalents, normality,valence factor etc is now obsolete and discouraged by the IUPAC, but even then it is of importance theoretically at many junior levels of chemistry where theoretical knowledge is more focussed on.
A similar question that I referred: The equivalent weight of HNO3 in the following reaction, however this question firstly doesn't have a satisfactory answer and secondly does not address the methodology involved in the calculation of equivalent weight in such cases. Moreover the accepted answer says that no nitrate ion is formed on the RHS, which is completely wrong as discussed above.

If you want to solve this problem, you must use and stay with ions, and not present the equation with neutral molecules. The equation to be used will be :$$\ce{3 Cu + 2 NO3^- + 8 H+ -> 3 Cu^{2+} + 2 NO + 4 H2O}$$ And then the rest of the calculation will go without problem.

@Maurice, I don't understand how the rest of the calculation would go without problem. Are you saying that we just have to look at the conversion of NO3- to NO?

Just to add some more information for you with some clarification. I gather that you are asked to calculate the equivalent weight of nitric acid here.
$$\ce{Cu + HNO3(aq, dil)  -> Cu(NO3)2 + NO(g) + H2O(l)}$$
So first thing to keep in mind that the equivalent weight is determined with respect to a single role of the molecule of interest. An exception is a disproportionation reaction, where the normality calculation is slightly tricky. In your case, the only role nitric acid is playing is that of an oxidizing agent not like an acid. The nitrate anion of the copper nitrate has to be ignored.
The equivalent weight of nitric acid= Molecular weight of nitric acid/ 3
The equivalent weight of copper = Atomic weight/2
I have never heard of "valence factor" or so-called n-factor- all these obsolete terms are coined by the local textbook writers. Delete these terms once you pass the exams from your mind.

Thank you for the answer. But how do you determine which role to pick?

There is no dual role, nitric acid is only acting as an oxidizing agent here.

Please forget about equivalents and normal solutions ! It is an old theory, that was abandoned in the middle of the $20$th century. It has been replaced by moles, molarity, and similar parameters. A solution containing $n$ moles of a solute in one liter water has a concentration expressed in mole per liter, that can be printed on the label of the flask. The concentration can also be calculated in equivalents per liter, and the result is called $n'$ times normal. But this value of concentration depends on the particular choice of reaction in which the solute will be used. For example, if a solution of $\ce{KMnO4}$ $0.1$ M is used to titrate some sulfite solution, its concentration in equivalents units is $0.5$ N if used in acidic medium, and only $\ce{0.3}$ N in a basic solution. So please forget about equivalents and normal solutions. Prefer using moles and molar solutions !

As a student the OP doesn't have a choice. The teacher/book uses the concept and the student must strive to understand the concept.

How will the teacher react if the student comes to him saying that a $0.1$ M $\ce{KMnO4}$  solution is sometimes $0.5$ N or sometimes $0.3$ N ?

Maurice -- LOL... Isn't that due to quantum mechanics? // I truly understand your point. However if the student is in a class where the teacher is grading the student, then the student doesn't have a choice. So if I'm in a geography class and the teacher tells me that the earth is flat, then for that class the earth is flat. // Teaching using equivalent weights still seems to have a firm following in India.

@ MaxW. There are plenty of ways for proving that the Earth is not flat. If you want it, I can explain you how I proceed so that my own students in my own high school has to admit by their eyes that the Earth is round.  And the quantum mechanics has nothing to do with stoichiometry, Nothing.

As I have already mentioned and as @MaxW says, I don't really have a choice, given the choice, I'd naturally go for moles and molarity(yes, even without knowing that this is discouraged) but, the question is pretty explicitly aimed at using this concept only. So could you please suggest what methodology was used when this was in practice? Or was this the exception that led to the discarding of this topic.

Moreover, Rutherford's model, Plum-Pudding Model etc are examples of theories that were discarded, even bohr's model cannot explain everything and is vague, but we still study that and not go on to the schrödinger's model directly do we? So I suppose that it is taught just so we know how things proceeded in the past, and how they are what they are today. And best way to know the shortcomings of a method is to break your head over it and understand where it fails. But what to discard what to not, is not a student's choice.

@ Tesla's Coil. Bohr's model is not vague, it is wrong. But it has the advantage of being simple, similar to the familiar solar system, and giving results that are not ridiculous. Of course, Schrödinger equation should be taught first. But it cannot be done in schools where integral calculus is not taught. So it is better to teach a poor model, than no model at all.

As you noted, IUPAC does discourage using the equivalent weight concept because the value is fuzzy. Consider the following reactions:

Dissociation of Nitric acid
$$\ce{HNO3 ->[aq] H+ + NO3-}$$
Reduction of Nitrate acid to Nitrogen dioxide
$$\ce{NO3^- + 2H+ + e- ->[aq] NO2(g) + H2O}$$
Reduction of Nitrate acid to Nitrogen Monoxide
$$\ce{NO3^- + 4H+ + 3e- ->[aq] NO(g) + 2H2O}$$

For reaction 1, $\ce{HNO3}$ loses one proton, thus for the acid base reaction the acid/base equivalence is 1.
For reaction 2, $\ce{HNO3}$ loses one electron, thus for that reduction reaction the redox equivalence is 1.
For reaction 3, $\ce{HNO3}$ loses three electrons, thus for that reduction reaction the redox equivalence is 3.
So the problem here is what number, 1 or 3, do you pick for the equivalent factor?

So back when this concept was in practice, what did people do for such reactions? Did they leave it out or did they actually use some method of judging the equivalence factor? Or did this one remain forever answered and added to the disowning of Normality notion?


text:
Are chemical treated as ideal gases when trying to figure out heat of formation using calorimetry?
At 36:48 of this video lecture, The professor says that the reaction moving from reactants to products keeping temperature of both same and under standard condition is an isothermal constant volume process.
Some some four minutes later (41:57), he uses the sum of energy of whole cycle property and equates, the $U_{rx}$ to the other definition of other definition of heat change given by calorimeter using the law that sum of internal changes over a loop process is zero.
He gets:
$$ \Delta U_{rx} = - C_{v}^{calo} \Delta T$$
where $C_{v}^{calo}$ is energy change of calorimeter.
Now, what I find weird is that by the definition of the path-3 being isothermal, shouldn't $$ \Delta U_{rx}$$ be zero..?
My resolution for this was that maybe the gas not really ideal and that it may change energy loss by expanding but I'm not sure how I would justify this more authoritatively. So, my question is, when we talk about these chemistry reactions, do we treat gases as non-ideal?

When a chemical reaction occurs, internal energy is a function not only of temperature.  It is also a function of chemical composition.

I'm not quite sure what I'm searching for because the 'heat of formation' is regarding enthalpy not internal energy or is there some link between the two which makes what I said in correct?

Well, if $h^0$ is the molar heat of formation at the reference conditions of $T^0=298 K$ and $P^0=1\ bar$, then the molar internal energy at this reference state can be taken as $u^0=h^0-RT^0$, and the molar internal energy at temperature T can be expressed as $u=u^0+C_v(T-T^0)=h^0-RT^0+C_v(T-T^0)$

Hmm, that makes sense.. I think the internal energy change I found was the internal energy formation. I think if I divide that expression by moles of the quantity which was formed, then I'd get molar quantities. I'm not quite sure. If you can give me a reference on where I could find a discussion about this concept, then I'll try study that and eventually write an answer here.

See Moran et al, Fundamentals of Engineering Thermodynamics, Section 13.2.  p823 has a subsection Analyzing Dlosed Systems that considers internal energies.  Also see Smith and Van Ness, Introduction to Chemical Engineering Thermodynamics, Section 4.4 onward.

You wrote "I think the internal energy change I found was the internal energy formation."  But CT is not the heat of formation by any stretch of the imagination.  In fact the heat of formation has virtually nothing to do with the heat capacity.  It is the result of forming chemical bonds from the elements comprising the chemical compound being considered.  For the elemental substances themselves in their natural state at the reference state, the heat of formation is zero (which certainly has nothing to do with the heat capacity)..

I agree somewhat.. but in the thermodynamics course I did, they do the reaction in adiabatic conditions of the calorimeter, find the total internal energy change is zero, and then devise another path, a part of which is isothermal so as to find the internal energy change at a constant temperature.

Let us continue this discussion in chat.

Internal energy of any gas is given by
$$U=U_\text{trans} + U_\text{rotational} + U_\text{vibrational} + U_\text{intermolecular} + U_\text{electronic} + U_\text{relativistic} + U_\text{bonds}$$
Last three aren't affected by ordinary heating. And $U_\text{intermolecular}$ for ideal gas is zero. That's why for ideal gas $U$ is only function of temperature. But during a chemical reaction (even if it involves an ideal gas)
$U_\text{bonds}$ comes into play. The formation and destruction of bonds results in a decrease of potential energy.
Therefore, for a chemical reaction, even if the temperature remains constant the internal energy changes.


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formation of a compound due to photo-isomerisation
I was reading through a topic called photoisomerism, where I read that when a molecule is hit by a photon it changes its structure to a more stable one. Is this a one-way reaction, or is it reversible (as in is it an equilibrium)? If it was an equilibrium, what was stopping it from forming before the photon hit it?
update
since the question was a bit ambiguous, my question was after the light that was shone was removed, if the equilibrium still exists, then what was the hindrance that did not allow such equilibrium to exist before the light was shone. (i am not sure if such a reaction exists, my question is more of a hypothetical question, asking if such a reaction exists then what was hindrance from reaching equilibrium (that was achieved after the shone light was removed)  before the light was shone)(if no such reaction can exist please also explain why so)

What molecule? And you have heard about activation energy?

I'm surprised to see this question getting downvoted, as it has merit.  It is a deceptively difficult question, and probably too broad to be answered briefly.  I would like to see someone try to answer it, myself.

@Karl might be retinol, although OP would have to confirm.

@Bertram The question is (still) very sloppily and unclearly written. It could easily be improved, and then one could give a good answer.

@Karl I do know the concept of activation energy. hence I have asked in a condition of equilibrium and not a one-directional reaction. it is very evident that in a single direction reaction it is nothing but a photochemical reaction. I am asking in an equilibrium reaction (if it is possible)(with k less than 100 and greater than 0.1) what was the hindrance of the photoisomer not forming before light was shone on the molecule. and sorry this was the first time I am asking on stack exchange so the ambiguity in the answer

If the structure the molecules changes into is more stable, then the equillibrium must be very much on that side. And if a reaction does not reach its equillibrium, the hindrance always includes some activation energy. I'm sure you know that, so what is your question?

I'm asking if on shining the light if it reaches an equilibrium. what was not allowing it before shining this light from reaching this equilibrium. the point of equilibrium is that it should still be in equilibrium even after the light that is shone is removed. (I'm not sure if such reaction exists that a new equilibrium is formed after light which is shone is removed, I'm asking if such a reaction exists why was the equilibrium not reached after the light was shone)

(You need to work on your grammar and syntax a bit, it's very hard to follow.) Generally there is one equillibrium with the light on, and another one with the light off, AND it could be that the reaction reaches the equillibrium, or not. How different those two equillibria are also strongly depends on the wavelength and intensity. Not sure if I understand your question correctly, but I think there is no general answer.

‘Is this a one-way reaction, or is it reversible (as in is it an equilibrium)?’ – this can only be answered for the specific, individual reaction. Hence why needs detail.

Stilbene has been extensively studied with pump-probe absorption spectroscopy as well as time resolved fluorescence.

In the ground state (S0) there is a large activation energy to go from trans  to cis and vice versa so that both species are stable at room temperature. This is not the case in the first excited state (S1)and the trans converts to the cis in a few tens of picoseconds and it is even faster for the cis excited state to convert to the trans.
The potential energy scheme is shown below (Not to scale). On the left is the trans species with a large barrier to for the cis (on the right). This has a higher energy than the trans but also has a large barrier to form the trans.
Detailed analysis suggest that the reaction occurs by a lengthening of the double bond before twisting. Exciting the trans puts it into the first singlet excited state (S1) where there is a small barrier on the pathway to cis, but as it happens there is an intersection with the ground state (by 3ps label & twisted species) and from here either the ground state trans or ground state cis can be formed. If the cis ground state is excited then there is effectively no barrier to form the trans ground state and the excited cis molecule lasts for only a few ps.

For a review or organic photochemistry calculations see Bernardi et al. Chemical Society Reviews, p321, 1996, in particular fig 1. A lot of the original experimental work was done by the Hochstrasser group, see Annu. Rev. Phys. Chem. 2006. 57:37–64 for many references.


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How does current start to flow when galvanic cell is first connected?
Using the Daniell Cell as an example, where in the steady state we expect zinc to be oxidised and copper to be reduced spontaneously due to the overall reduction in Gibbs free energy, how does the system "know" to start moving electrons from the anode to the cathode? What is the underlying physical mechanism?
When the two half cells are first connected via the external circuit, it seems that they would not yet "know" anything about each other; each half cell would not already "know" it is anode or cathode (e.g. zinc could be the cathode in a different galvanic cell) and so how would the system immediately start driving the electrons in the right direction?
In other answers on this site (e.g. https://chemistry.stackexchange.com/a/28638/98580) I've seen statements like "the metal atoms of one half-cell are able to induce reduction of the metal cations of the other half-cell; conversely stated, the metal cations of one half-cell are able to oxidize the metal atoms of the other half-cell".
However, as a physicist this is somewhat unsatisfying, as the phrase "induce reduction" seems to bundle up a whole lot of physical processes in the half cells and the external circuit into a very high level concept. Fundamentally, it must come down to electrical forces and thermodynamics to explain the overall behaviour of the system.
My understanding is that prior to connecting the external circuit, each electrode is in equilibrium with its electrolyte, undergoing both oxidation and reduction at the same time; so each electrode could both push and pull electrons to/from a connected wire. The moment the external circuit is connected, do electrons start flowing from the zinc electrode to the copper electrode because zinc has a greater "tendency" (than copper) to push electrons onto a connected wire, and copper has a greater "tendency" (than zinc) to pull electrons from a connected wire? That would then sound like the galvanic cell gets started via a statistical process and is sustained by electrical charging of the electrodes thereafter.

If you connected a single cell to a very sensitive chargemeter, you would know wether the zero current equillibrium is reductive or oxidative. But who cares? Ions in an electrolyte are by many orders of magnitude less mobile than electrons in a wire.

@poutnik why do the redox reactions "try" to maintain a particular potential? Is there some underlying physics governing that? And could you elaborate on the origin of the current that disbalances the potentials and what that means from a microscopic perspective?

My comments are moved to an answer.

The electrode redox reactions of the Daniell cell:
$$\ce{Zn(s) <=> Zn^2+ + 2 e- }$$
$$\ce{Cu(s) <=> Cu^2+ + 2 e- }$$
maintain at the respective electrode a particular potential where both opposite reactions have the same rate, implying there are no external galvanic causes that affect this potential. Rates of oxidation electrode reactions exponentially grow with growing potential and vice versa for reduction.
This applies on potentials near the equilibrium one, when the reaction rate is not limited by the diffusion yet. Diffusion cuts off this exponential growth.
The faster of the opposite reactions either pushes electrons to an electrode either pulls electrons from it until the electrode potential reaches the equilibrium potential. At this potential, rates of both reactions are equal. Even before electrodes are ever connected, these electrode potentials are in place. Electrodes do not need any external galvanic circuit.
A current flowing between 2 galvanically connected places of different potentials is not part of electrochemistry, but a part of physics which you are already familiar with.
The connected external circuit is then just following the Ohm law $I=\frac UR$
The current is shifting electrode potentials by moving electrons, what is the same process as in electrostatics, where  potentials of capacitor electrodes are changing, when galvanically connected.
As the resulting  potentials are not the equilibrium ones any more, one of the electrode opposite redox reactions takes the upper hand, leading to oxidation at the anode and reduction at cathode.  Particularly, the potential of the $\ce{Zn}$ electrode with the lower equilibrium potential gets higher and starts oxidation, making the electrode the anode.
$$\ce{Zn(s) <=>> Zn^2+ + 2 e-  }$$
Similarly, the potential of the $\ce{Cu}$ electrode with the higher equilibrium potential gets lower and starts reduction, making the electrode the cathode.
$$\ce{Cu(s) <<=> Cu^2+ + 2 e- }$$
This leads to a kind of race on the zinc electrode in providing electrons on the chemistry side and taking them by the circuitry side. And vice versa for the copper electrode. The actual electrode potential under the load is a kind of the score of this race.
Electronic chips are known for the internal use of electronic charge pumps to get different voltage levels than externally provided. E.g. once famous 8bit CPU Z80 uses external voltage +5V and 0V, internally generating also +12V,-12V,-5V.  Electrodes in electrochemical cells can be considered in some sense as chemically powered charge pumps with different target potentials.
Great basic knowledge of chemistry ( and mainly of physics ) is Hyperphysics-Chemistry-electrochemistry, in form of linked cheat sheets. I would recommend electrochemistry chapters of textbooks of a physical chemistry. I have once studied a translation of Moore, EN natives would suggest much more.  More advanced sources are here: SE Chemistry resources-for-learning-chemistry

Thanks @Poutnik, I am slowly getting there. So would it be correct to say that the microscopic situation, when the external wire is first connected, is that the Zn electrode is truly at a lower electric potential than the Cu electrode, and hence there are literally electric forces acting on the electrons as per the usual electrodynamical theory?

As another question, can Cu ever be the anode? E.g. if one half cell had a F electrode, would Cu then be oxidised whilst F is reduced?

F is not good example, as fluorine would blow up near any electrolyte and solvent out of the container being extremely oxidative. But if Cu is used against Ag/Ag+, or Hg/Hg2^2+, or against a half-cell with an inert electrode and all-in-solution redox system, they yes, Cu/Cu^2+ half cell can be an anode. Note  that a half-cell is the whole redox system around 1 electrode, while an electrode in wider sense is the same as a half-cell, in narrower sense it is just that conductive stick, that may(Zn) or may not ( Pt, glassy C ) be involved in redox reactions.

Thanks again. My chemistry knowledge is sorely lacking; would you recommend any particular texts that have a good treatment of this subject? I'd be interested in both introductory and more advanced treatments, especially from a fundamental physical perspective.

See the A update

BTW, electrons don't actually actually "flow" very fast. The drift velocity of electrons is incredibly slow, on the order of a millimeter every 5 or ten seconds in a room-temperature conductor. The propagation of current is like the particles in Newton's Cradle, where the electric field bumps each electron down the line, a much faster process. Furthermore, electrons are swept out of the conductor by the electromagnetic force, which propagates on the surface of the conductor at a large fraction of the speed of light in vacuuo, C. The fraction is the velocity factor, perhaps 0.5 to 0.75 C.
So the answer to your question is not as clear-cut as it might seem. You could say the reaction propagates at the speed of the EMF, C times the velocity factor in the connecting wire. However, one then must take into account the buildup of slow-moving ions, creating a space charge slowing things down.


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Oxidation state of d block elements
Why d block elements do not show +1 oxidation state ? Only copper shows +1 , whereas oxidation number starts from +3 in scandium .

Are you sure about that? https://imgur.com/L5UZNx9

@nilay Ghosh yes, I'm pretty sure , you may also check that .

And I'm pretty sure you're wrong. d block elements show even negative oxidation states, also zero, +1 and pretty much whatever you'd wish.

Tl, dr:  There is more to getting iron(I) than meets the eye.  The aquo-nitrosyl complex doesn't do it.  To get there we need a nonaqueous complex formed under strongly reducing conditions.
As noted in a comment to another answer, this question is not as cut and dried as it may appear, especially in the case of iron.  As explained in this answer, it was long thought that the "brown ring" iron-nitrosyl complex was iron(I) bound to $\ce{NO^+}$, but in 2002 (three decades after the Cotton-Wilkinson text) spectroscopic studies revealed that it actually consists of iron(III) bound and antiferromagnetically coupled to $\ce{NO^-}$.  The reader may want to digress to an addendum at the bottom, in which iron is rendered so highly electropositive that this outcome is really no surprise.
Here the case of iron is explored in more detail, using the work of Keilwerth et al. [1].  Keilwerth et al. performed extensive spectroscopic and electrochemical studies of a series of iron complexes with nitrous oxide and tris[2-(3-mesitylimidazol-2-ylidene)ethyl]-
amine ($\text{TIMEN}^{\text{Mes}}$), which have the chemical formula $\ce{(TIMEN^{\text{Mes}})FeNO(CH3CN)^q}$ with varying charge states $q$.  Initially the complex is formed with $q=+3$.  When this tri-cation is reduced -- by zinc, with one electron, by magnesium, with two electrons, or by sodium amalgam, with three electrons -- the nitrisyl group in the resulting reduced species is identified as antiferromagnetically coupled $\ce{NO^-}$, with the iron oxidation state being decreased from +3 to +2 to +1 as the number of accepted electrons increases.  Thus the sodium-reduced species, with three added electrons, is found to indeed contain iron(I), bonded to $\ce{NO^-}$ rather than $\ce{NO^+}$.  As an aside, and consistent with the addendum, it takes a reducing agent as strong as magnesium even to reach iron(II).
Reference
1.
Martin Keilwerth, Johannes Hohenberger, Frank W. Heinemann, Jörg Sutter, Andreas Scheurer, Huayi Fang, Eckhard Bill, Frank Neese, Shengfa Ye, and Karsten Meyer, "A Series of Iron Nitrosyl Complexes {Fe−NO}6−9 and a Fleeting {Fe−
NO}10 Intermediate en Route to a Metalacyclic Iron Nitrosoalkane", J. Am. Chem. Soc. 2019, 141, 17217−17235.  pdf

Addendum:  The electropositivity of iron
We usually think of "highly electropositive elements" as alkali or alkaline earth metals, plus a few others such as aluminum and some rare earths.  But iron is actually not far behind.  It is actually a quite electropositive element for which even the +2 oxidation state is a low one in an electronegative environment.  What is often called $\ce{FeO}$ is actually not a pure iron(II) oxide, but a nonstoichiometric compound containing a significant amount of iron(III).  Moreover, when cooled to ambient conditions this oxide is only meta-stable, tending to decompose to form $\ce{Fe3O4}$.  Under certain naturally occurring conditions, iron(II) can even displace hydrogen from water, again forming $\ce{Fe3O4}$ instead of $\ce{FeO}$.  And of course, rusting produces a combination of $\ce{Fe3O4}$ and various iron(III) compounds.  Given these properties, we should realize that to stabilize iron into a low oxidation state requires low-electronegativity ligands and strongly reducing conditions.  We see this in the main body of the answer.

Cotton and Wilkinson state that

$\ce{Cr(I)}$ exists in the ion $\ce{[Cr(bipy)3]+}$
$\ce{Mn(I)}$ exist in $\ce{Mn(CO)5Cl}$ and $\ce{K5Mn(CN)6}$
$\ce{Fe(I)}$ exists in the ion $\ce{[Fe(H2O)5NO]^{2+}}$*
$\ce{Co(I)}$ exists in the ion $\ce{[Co(bipy)_{3}]^+}$ and in the molecule $\ce{CoBr(PR_3)_3}$

Ref.: F. A. Cotton, G. Wilkinson, Advanced Inorganic Chemistry, Interscience Publ., 3rd Ed. 1972
*This complex, however, is no longer considered a metal(I) compound, although iron(I) is still known.  Another answer explores that case in more detail.

There are indications that iron bound to a nitrosyl ligand actually imparts a negative charge to the latter, so that the quoted iron complex is Fe(III).

In general, transition metals do not have a common +1 oxidation state because they have a $\ce{ns^2}$ valence shell. It would be more stable for metals to lose 2 electrons instead of 1. There are exceptions to this rule. Group 11 elements, Copper, Silver and Gold, do have a common +1 oxidation state. This is because instead of a $\ce{nd^9ns^2}$ configuration, they have a $\ce{nd^{10}ns^1}$ configuration. As such, they readily lose only 1 electron to form a stable cation with a +1 charge.
I would not be looking at Group 12 elements since strictly speaking, they are not transition elements. However, even if they are part of the discussion, only mercury has a common +1 oxidation state in the form of $\ce{Hg^{2+}2}$.


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Why did ammonium chloride fumes turned pink in this synthesis?
I was watching the synthesis of tetrasulfur tetranitride performed by Tom on his Youtube channel, Explosion & fire. He followed the basic route: adding ammonia gas to sulfur chlorides ($\ce{SCl2}$ and $\ce{S2Cl2}$). At first, everything was going normal, purple-black precipitate on the flask with white $\ce{NH4Cl}$ fumes emitting as expected and white $\ce{HCl}$ fumes resulting from hydrolysis of sulfur chlorides. But suddenly a problem occurred. The fumes turned pink.  The following is a sequence in chronological order (snapshots from his videos; temperature of ppt. in brackets):

white $\ce{NH4Cl + HCl}$ fumes on top of precipitate ($\pu{T = 25 ^\circ C}$)

fumes turned pink ($\pu{T = 27-28 ^\circ C}$)

fumes entering the condenser from flask turning it red ($\pu{T = 35 ^\circ C}$)

fumes escaping the condenser ($\pu{T > 35 ^\circ C}$)

condenser changed, $\ce{NH4Cl}$ pellets observed above it, fuming doesn't stop ($\pu{T > 35 ^\circ C}$)

Eventually, after so much fuming and snowing, the end-product was achieved (when fumes turned golden brown, the original color of $\ce{S4N4}$). So, why did ammonium chloride fumes turned pink (when it was evolving)? Is it due to the compounds in the precipitate? If so, what species could have caused that color?
I did some research and found out that $\ce{S4N4}$ is thermochromic(changes color according to temperature). From britannica:

Tetrasulfur tetranitride forms thermochromic crystals, which are crystals that change colour with temperature. They are red at temperatures above 100 °C (210 °F), orange at 25 °C (80 °F), and colorless at −190 °C (−310 °F).

There is no mention of $\ce{S4N4}$ turning pink, so I proposed pink to be an intermediate color (orange -> pink -> red) and since it is a volatile compound, it was turning the fumes pink. But if the compound was pink, why did it later turn golden-brown, the original color of $\ce{S4N4}$? The precipitate was a purple-brown-black sludge, so it was definite not the end product. Thus my proposal contradicts.
prepchem says:

Tetrasulfur tetranitride is prepared by dissolving 1 volume of sulfur
dichloride in 8 or 10 volumes of carbon disulfide, cooling, and
passing in dry ammonia gas until the dark brown powdery
precipitate which forms at first has dissolved and an
orange-yellow solution results which contains light-colored flocks of ammonium chloride.

Here, there is no mention of pink. What species could have caused fumes to turn pink in color?
I recommend to watch the synthesis in details on his second channel, Extraction & Ire: Part 1 and Part 2

Take white paint and add a bit of red. What colour will you see?

@Alchimista Pink. So, what red compound would have caused the fumes to turn pink?

Doesn't your text mention that due to thermochromism of tetrasulphur tetranitride red can be a possibile color? I mean the part in the box

@Alchimista that red color is possible at temperatures above 100 C. The reaction flask was at 35 C.

$\ce{NH3}$ and $\ce{SCl2}$ form $\ce{NH4Cl}$ which evolves as white fumes. $\ce{NH4Cl}$, when passed through $\ce{SCl2}$ (cherry-red liquid), gives the appearance of dense-pink fumes
$$\ce{\underset{(cherry red)}{6SCl2} + 16 NH3 -> S4N4 +2S + \underset{(white)}{12NH4Cl}}$$
That is why the fumes of $\ce{NH4Cl}$ turned pink

For formatting, See here and here. For a more detailed MathJax guide, look here, minor other details

While it explains the color change, (1) why did it take some time for the color to change? Also, the change was sudden and not gradual. (2) what was the reason for the temperature spike? The color changed as the temperature was rising, suggesting thermochromosity(?) (3) the precipitate was a purple-black sludge indicating that it was a mixture of products, by products and side products not just red sulfur chloride.


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Negative s - orbital chatacter
I am learning about how to calculate s-orbital character. While doing so, I came across a formula
$$\cos\theta = \frac{s}{s-1}.$$
For cycloproprane, I used $\theta = 60^\circ,$ and got $s = -1.$

How can the s-orbital character be negative?
What does the negative sign mean?

Question which I want to solve
Find out the stability of the following carbanion:

I found Does negative sign of s-orbital exist?, but failed to understand it. If that answers my questions, please tell.

The s-orbital character itself is not much of a thing. It does not worth thinking about.

Actually I want to solve a question which involves stability of carbanion. For that I need to find out which is more electronegative. Since electronegativith is proportional to s orbital character. Therefore I want to calculate s orbital character

Post the actual question here, since what you do may not be the right method and you might be walking down the wrong road

@Safdar I have edited the question. Please see.

As much as I am aware, the Coulson's formula is only applicable for angles >90 *.

@Safdar https://chemistry.stackexchange.com/questions/138297/is-hybridization-of-the-cyclopropyl-anion-sp5 Can any stablity order be observed?

@Aditya Roychowdhury I am sorry. I am just knowing this formula ( I don't even know who discovered it). I tried to search a lot for its limitations on the internet but got nothing. Can you please share some resource with me if you are having.

https://chemistry.stackexchange.com/questions/15671/utility-of-bents-rule-what-can-bents-rule-explain-that-other-qualitative-con

The formula only works when the orbital is directed along the bond line between two atoms. Since no orbitals can give you a $60^o$ bond angle, the orbitals are directed outside the triangle, forming a “banana” bond and the formula is being used inappropriately. So the negative sign in your Question #1 is not acceptable. We must do something different.
Let’s look at the H-C-H bond angle. The H-C-H bond angle in cyclopropane is $115^o$ Ref 1. Using that angle, and assuming that the orbitals are directed exactly between the atoms, s = 29.7% for the C-H bond, which is broken to form the carbanion. The acidity of cyclopropane is also indicated by its pKa: 46.
Cyclopropane is interesting because you might think, starting with atomic orbitals, that such a structure would not form, since the smallest angle you could imagine for a carbon atom would be $90^o$; how could you possibly get $60^o$?
Well, that's where your imagination comes into play. Don't go to negative s-bonds (that's too imaginative). Consider how the simple (i.e., positive) addition of individual atomic orbitals could lead to something like the bonding that you do observe:

With simple $sp^3$ orbitals, you can point one pair from each carbon just outside the ring (there will still be considerable overlap: Figure b) and each carbon will be attached to 2 hydrogens above and below the ring with the other two $sp^3$ orbitals. These ring bonds are called banana bonds - not directly in line with the atoms, but bonding nevertheless. Figure a) shows a more complicated bond scheme where the carbon is hybridized $sp^2p$ with the $p$ obital in the plane of the carbons: you get two bonds and one antibond (the whites overlap, the reds overlap, and the red-white antibonds). But then you get three $sp^2$ orbitals poking right into the ring to form a 3-atom bond - again, the bonding is not in line with the atoms. (The hydrogens are bound with $sp^2$ bonds, above and below the ring. You might use the equation to determine the s and p character of these C-H bonds if you know the H-C-H angle (Ref 1).
A very interesting (11 slides) presentation about the bonding is made at Ref 2.
Ref 1: https://cyclopropanelasalle.weebly.com/structure.html
Ref 2: https://www.slideshare.net/DanielMorton26/using-molecular-orbital-theory-to-explain-bonding-in-cyclopropane
But the upshot is that all (most?) equations have limited ranges of usefulness. Whenever some experimenter makes something that goes beyond, a new equation is needed.
Cyclohexane, at the other end of your question, has a pKa of 52 (very non-acidic!), and a probable H-C-H bond angle of $109.5^o$, and s-character in the C-H bond of 25% (note that the molecule has a chair structure and is not planar).
Other pKa’s of interest are:
acetylene, pKa =15, C-H bond is $sp$, 50% s
ethylene, pKa = 44, C-H bond is $sp^2$, 33% s
methane, pKa = 48, C-H bond is $sp^3$, 25% s
(interesting that cyclohexane is similarly hybridized but even less acidic)
Perhaps the best answer to the stability question is that the smaller rings are more acidic, i.e., have more stable anions because the acidic proton has more s-character in the more strained ring.
Question #2 discussed sign in your link. “Sign” (+/-) in the equation for s-character is a measure of amount: how much s-character, in %, is in the bond. But when describing orbitals, we use + and - as indicators of the polarity of the wavefunction, not the amount of s or p character. Plusses combine in such a way that electrons flow from one orbital to another without hindrance, in a bond. A - polarity next to a + requires a node: no electrons there at the node, therefore no bonding. So s-orbitals are designated with one sign (either + or -, depending on whether it is bonding with an adjacent atom; higher levels have internal nodes, but that is no concern for us here). p-Orbitals have a node at the nucleus which is indicated by calling one lobe + and the other lobe -. Bonding molecular orbitals have the same sign of the wavefunction adjacent; in the picture, red-red and blue-blue (in (c)) are bonding; red-blue (in (a) and (b)) is antibonding because the wavefunction splits between the atoms.

Thank you. It has solved most of my confusion. So, about the sign, I conclude from your answer that negative sign of s - orbital character has no effect on its magnitude. Eg. -1 s-orbital character is more than 0.33 s-orbital character. So, please tell if this is correct or not.

Yes, correct. A negative sign on an s-orbital is still 1 (100%) in magnitude, just nonbonding relative to some other orbital which is designated with a + sign. An sp orbital has 50% s-character whether it is designated + or -; the + signifies bonding with an adjacent (+) orbital and a - sign signifies nonbonding (to an orbital designated +). Antibonding, bonding and nonbonding refer to the energy levels of the molecular orbitals. Just as you can have different % of s-character, you can have different degrees of overlap and stabilization energy, or none at all.


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Why does dimethyl sulfoxide have higher boiling point than acetone?
They have very similar structure except for a central atom, but the difference in boiling points is very large $(\pu{186 ^\circ C}$ for DMSO and $\pu{56 ^\circ C}$ for acetone).
How is central atom impacting the boiling point?

Checkout the geometry of both molecules and also identify the polar bonds. Now figure out how dipole moment and boiling points relate.

Firstly, sulfur has a larger atomic radius than carbon, so we would expect DMSO (Ddimethyl sulfoxide) to have a much larger and thus polarisable electron cloud than acetone. The London dispersion forces between DMSO are thus stronger, causing DMSO to have a higher boiling point.
Secondly, if we compare the dipole moments of the $\ce{S=O}$ and $\ce{C=O}$ bonds, the $\ce{S=O}$ bond is much more polarised. This is because the $\mathrm{3p-2p}$ overlap between $\ce{S}$ and $\ce{O}$ is much less favourable than the $\mathrm{2p-2p}$ overlap between $\ce{C}$ and $\ce{O}$. As such, the resonance structure on the right is actually the more significant resonance structure. This makes the $\ce{S=O}$ bond much more polarised as compared to the $\ce{C=O}$ bond, causing the permanent dipole-permanent dipole interactions between DMSO to be much stronger than the same interactions between acetone molecules.

Lastly, we would realise that DMSO has a trigonal pyramidal shape while acetone is planar. Due to this, the net dipole moment in DMSO is likely to be stronger due to it being a more asymmetric molecule.

Sulphur has higher EN than Carbon

@user85426: There is another point you forgot, the boiling point is also depends on the molar mass of the compound. Difference in MM: $78.13$ versus $\pu{58.08 gmol-1}$.

I have mentioned the point about it having a larger atomic radius i.e. an electronic cloud. A higher molar mass is often used to indirectly justify the larger electron cloud and hence, I feel that it is more apt to directly compare electron cloud rather than molar mass.

@MathewMahindaratne Sorry for off topic, due to the very poor pi overlap of 3p-2p as large different in size and energy, can I say for methyl phosphate, the predominant resonance structure has 4 formal charges, which 3 oxygens have -1 charge, and Phosphorus has +1 charge, explaining the why P is tetrahedral (sp3 hybridized)?

@Question: I think it's better you post your question as a new one. In that way, it draws the attention of others who are more qualified to answer based on molecular orbital theory.


text:
Jahn Teller Effect on Metal Complexes
I've been learning how the Jahn Teller Distortion effects the orbitals in Metal complexes and how the splitting of eg and t2g orbitals happens.
But the book mentions that the effect is strong for electron configurations where the eg orbital has 1 or 3 electrons. I can't figure out what that's the case. Why isn't the effect strong with 2 electrons in the eg orbital as well?

If the orbitals of the metal core is distorted. the amount of splitting is not high enough to promote the pairing of energy. therefore the 2 electrons will go to two separate eg orbitals. let the splitting energy due to this distortion be β1. the electron in the lowers orbital will have an energy loss of β1/2 and the electron in the higher orbital will have an energy gain of β1/2 and therefore the net gain in the energy is 0. which does not give any additional stability to the complex(as there is no release of energy).
hence the distortion does not have a strong effect.


text:
Determining the formula of sodium carbonate hydrate using acid-base titration
I found this question online at mathsmadeeasy.com:

Sodium carbonate exists in hydrated form, $\ce{Na2CO3 · x H2O},$ in the solid state. $\pu{3.5 g}$ of a sodium carbonate sample was dissolved in water and the volume made up to $\pu{250 cm3}.$ $\pu{25.0 cm3}$ of this solution was titrated against $\pu{0.1 mol dm-3}$ $\ce{HCl}$ and $\pu{24.5 cm3}$ of the acid were required. Calculate the value of $x$ given the equation:
$$\ce{Na2CO3 + 2 HCl -> 2 NaCl + CO2 + H2O}.$$

I have spent around 20 minutes trying to figure out the final steps to this question. I was able to workout the amount of sodium carbonate in $\pu{250 cm3}$ as $\pu{1.225E-2 mol},$ but I don't know how to use this to finish off the question, please help.

If you do not say what you have tried, we cannot say what you did wrong.

The problem is not too challenging. Think of mass balance. Given the fact that you have determined the moles of sodium carbonate. Determine the grams of sodium carbonate. It should be less than 3.5 g. Where is the rest of the mass coming from? I guess it is water. Add these steps in the question, and hopefully we will find a solution.

@ FP. You have done most of the job. You are nearly at the end. You have found that the amount of $\ce{Na2CO3}$ is $ n = 0.01225$ mol. The only thing you still have to calculate is the molar mass of the initial compound : $M = m/n = 3.5$ g/$0.01225$ mol = $286$ g/mol.
As anhydrous $\ce{Na2CO3}$ has a molar mass of $106$ g/mol, the difference $286$ g/mol - $106$ g/mol = $180$ g/mol is $x$ times the molar mass of $\ce{H2O}$, which is $18$ g/mol. So $x$ = $180/18 = 10$, and the formula of your initial compound is $\ce{Na2CO3·10H2O}$

This is a different way to calculate $x$. Suppose amount of $\ce{Na2CO3}$ in $\pu{3.5 g}$ of sample is $n$ $\pu{mol}$. Thus, the amount of water in the same sample is $n$ $\pu{mol}$. Since molar masses of $\ce{Na2CO3}$ and $\ce{H2O}$ are $105.99$ and $\pu{18.02 g mol-1}$, respectively,
$$105.99n + 18.02nx = 3.5 \tag1$$
Since most of the measurements are in two significant figures, it is okay to round these numbers. Thus, after simplification:
$$18.0x = \frac{3.5}{n} -106 \tag2$$
By your calculations, you found $n = 1.225 \times 10^{-2}$. Applied this values in the equation $(2)$:
$$18.0x = \frac{3.5}{1.225 \times 10^{-2}} -106 \approx 180 \ \Rightarrow \ \therefore \ x = \frac{180}{18} = 10$$
Thus, your original hydrate is $\ce{Na2CO3.10H2O}$.